• Power Plant Engineering

[A Textbook for Diploma, Degree and A.M.J.E.

Section B Students]

MR Nagpal
?.Sr Engineering (Mechanical)
rech. (Mech. Engg.)
F 7. Mechanical Engineering Department
i.A., Institute of Engineering
ARIDAfiAD (Haryana)

me Drawing
me ToilJ'i- . ing
we Design
igineering arA lesign
Forming Process

Khanna Pub1ishe
Published by:
Romesh Chander Khanna
for KHANNA PUBLISHERS
2-B, Nath Market, Nai Sarak
Delhi- 110 006 Undia)

© All Rights Reserved

This book or part thereof cannot be translated or reproduced in any
form (except for review or criticism) without the written permission
of the Author and the Publishers.

ISBN No.: 81.7409-155-6

I

Fifteenth Edition
Fourth Reprint: 2006

C 'nper Typeset by:

;ts Computers,
D-W17. .;ayal Pur, Delhi 110 094

Printed at:
3•.: . htP.-inters, Phari Bhoja,
Delhi 110006 .
 DEDICATED
TO
MY LATE RESPECTED MOTHER
 Preface
This book is quite suitable for degree and diploma engineering
students of many Universities and for A.M.I.E. (Sec. B). The book
has been revised and brought up-to-date to suit the requirements
of the students. More problems have been solved on steam power
plant and hydro-power plant. The overall objective is to present
the subject matter in more simple language. Objective type problems
have been added to make the subject matter more clear.
Although carefully prepared and reviewed the book may con-
thin some errors. The author will be thankful to the readers for
bringing them to his notice. Suggestions for improvement of the
book shall be gratefully received and appreciated.
The author is thankful to Sh. Romesh Chander Khanna for his
most co-operative and painstaking attitude for bringing out this
edition in a very short time.

—G.R. NAGPAL
 Contents
Chapter
Pages
1. Sources of Energy
1-52
1.1. Power
1.2. Sources of Energy
1.2.1. Conventional and Non-Conventional.
Sources of Energy
1.3. Fuels
.
1.3.1. Solid Fuels
1.4. Calorific Value of Fuels
1.5. To Calculate Approximate Flue Gas Loss
... 10
1.6. Types of Coal
1.7. Liquid Fuels
1.7.1. Oil Properties
1.8. Advantages of Liquid Fuels over Solid Fuels
1.9. Gaseous Fuels
1.10. Advantages of Gaseous Fuels. over Solid Fuels
1.10.1. Composition of Liquid Fuels .
1.11. Comparison of Sources of Power
1.12. Sources of Energy in India
1.12.1. Conservation of Energy
1.13. Combustion of Fuels
1.14. Products of Combustion
1.15. Combustion Chemistry
1.16. Combustion of Gaseous Fuels
1.17. Weight of Air Required for Complete Combustion
of Fuel
1.18. Coal Selection
1.18.1. Ranking and grading of coal
1.19. Composition of Solid Fuels . ... 34
1:19.1. Ash
1.19.2. Volatile Matter
1.20. Weight of Excess Air Supplied
1.21. Requirements of Fuel
1.22. Principal Stages of Combustion
1.23. Complete Combustion
1.24. Incomplete Combustion
...38
1.24.1. Weight of Carbon in Flue Gases
1.24.2. Weight of Fuel Gas per kg of Fuel Burnt
1.25. Conditions for Proper Burning (combustion) of Fuel ...39
1.26. Temperature of Fuel Combustion
Problems
.. .50
 (viii)
2. Power Plant Economics 53-139
2.1. Power Plant .53
2.2. Types of Power Plants
2.3. Requirement of Plant Design
2.4. Useful Life of a Power Plant
2.5. Comparison of Public Supply and Private
Generating Plant .0. 55
2.6. Prediction of Future Loads . . .56
2.7. Terms and Definitions .0. 57
2.8. Power Plant Capacity
2.8.1. Feasibility of Electric Power Plant
2.9. Principles of Power Plant Design . ..63
2.10. Economic Load Sharing between Base Load and
Peak Load Power Stations . ..64
2.11. Type of Loads ...65
2.12. Typical Load Curves
2.13. Cost of Electrical Energy .0. 66
2.14. Energy Rates (Tariffs)
2.14.1. Types of Tariffs ...69
2.15. Economics in Plant Selection ...72
2.16. Economic of Power Generation
2.17. Plant Performance and Operation Characteristics
2.18. Economic Load Sharing .0. 77
2.19. Condition for Maximum Efficiency .
2.20. Choice of Power Station ... 80
2.21. Effect of Variable Load on Power Plant Operation
and Design . . .82
Problems ...135
3. Steam Power Plant 140-315
3.0. Introduction ... 140
3.1. Essentials of Steam Power Pit. uipment ...140
3.1.1. Power Station Design ...142
3.1.2. Characteristics of Steam Power Plant ...142
3.2. Coal Handling 143
3.2.1. Dewatering of Coal
3.3. Fuel Burning Furnaces ...149
3.3.1. Types of Furnaces . ...149
3.4. Method of Fuel Firing ... 150
3.4.1. Hand Firing ...150
3.4.2. Mechanical Firing (Strokers) ... 153
3.4.3. Types of Stokers ...154
3.5. Automatic Boiler Control ...158
3.6. Pulverised Coal ... 159
3.6.1. Ball mill .0. 160
 (ix)
3.6.2. Ball and race mill
3.6.3. Shaft Mill
3.7. Pulverised Coal Firing
3.8. Pulverised Coal Burners
16(
3.8.1. Cyclone Fired Boilers
3.9. Water Walls
3.10. Ash Disposal
3.10.1. Ash Handling Equipment
3.11. Smoke and Dust Removal
3.12. Types of Dust Collectors
3.12.1. Fly Ash Scrubber
18
3.12.2. Fluidised Bed Combustion (FBC)
18
3.12.3.Types of FBC Systems
3.13. Draught 1
3.13.1. Comparisons of Forced and Induced
Draughts
18'
3.13.2. Draught Measurement
3.14. Chimney
3.15. Calculation of Chimney Height
3.16. Methods of Burning Fuel Oil
...19E
3.16. (a) Fuel Oil and Gas Handling
3.16. (b) Gas Burners 19
3.17. Slag Removal ..20C
• . .20C
3.18. Economiser
...201
3.18.1. Soot Blower
.203
3.18.2. Air Preheater
• ..203
3.18.3. Heat Transfer in Economiser and Air
Preheater
.204
3.19. Super Heater
...205
3.19.1. Sugden Superheater
...209
3.20. Advantages of Super-heated Steam
...209
3.21. Super-heat Control
.209
3.22. Feed Water Treatment
...210
3.23. Methods of Feed Water Treatment
...212
3.24. To Determine Blow Down
...217
3.25. pH Value of Water
..218
3.26. Analysis of Water
...218
3.27. Feed Water Heaters
...219
3.28. Steam Condensers
...220
&29. Types of Steam Condensers
.220
3.29.1. Surface Condensers
...220
3.29.2. Jet Condensers
..222
3.29.3. Types of Jet Condensers
...224
3.30. Condenser Cooling Water Supply
.225
3.30.1. River or Sea
..226
3.30.2. Cooling Ponds
.226
 (x)
3.30.3. Cooling Towers ...227
3.31. Maintenance of Cooling Towers • ..230
3.32. Condenser Efficiency .230
3.33. Vacuum Efficiency ...231
3.34. Condenser Pressure ... 231
3.35. To Calculate the Weight of Cooling Water ...232
3.36. Selection of a Condenser ..234
3.37. Sources of Air in a Condenser .234
3.37. (a) Effects of Air Leakage ...234
3.38. Air Extraction Pump ...235
3.39, Condenser Auxiliaries ...235
3.40. Condenser Performance ...236
3.41. Steam Separator ..236
3.42. Steam Trap ...238
3.43. Steam Turbines ...239
3.44. Advantages of Steam Turbine Over Steam Engine ...241
3.45. Steam Turbine Capacity .242
3.45.1. Nominal Rating ...242
3.45.2. Capability .242
1.46 Steam Turbine Governing ...243
.47 Steam Turbine Performance .244
3.47.1. Steam Turbine Testing ..245
3.47.2. Choice of Steam Turbine ...245
1.48. Steam Turbine Generators ...246
3.48.1; Steam Turbine Specifications ...246
1.49 Boilers ...246
1.50. ms of Boiler ...248
3.50. (a) Babcock and Wilcox Boiler .:.251
3.50. (b) Cochran Boiler ...251
3.51. Lancashire Boiler ...252
3.51. (a) Scotch marine bqiler ...253
3.51. (b) Cornish Boiler ..255
3.52. Boiler Mountings and Accessories ..255
3.53. Flue Gas and Water Flow .255
3.54. Causes of Heat Loss in Boilers .256
3.55. Thermal Efficiency of Boiler ...256
3.56. Boiler Performance .257
3.56.1. Selection of Fuel for Boiler ...257
3.56.2. Equivalent Evaporation ..258
3.57. Boiler Trial ...258
3.58. Boiler Maintenance ..259
3.59. Control and Measuring Instruments • ..260
3.59.1. Soot Blowers .260
3.60. Circulation of Water in Boilers ..260
3.61. Feed Water Regulators ...261
3.62. High Pressure Boiler .262
3.62.1 Unique Features of High Pressure Boilers ...262
 (Xi)
3.62.2 Advantage
3.63. Types of High Pressure Boilers ...263

3.63.1. (a) Advantages of . .. 263
High Pressure Boilers
3.63.2. Selection of Boiler ...265
3.64. Modern Trends in Generating Steam ...266
3.65. Gas Fired Boilers . ..266
3.65.1. Selection of boiler (steam generator) . ..267
3.66. Piping System ...267
3.67. Types of Piping System
3.68. Size and Strength of Pipe ...268
3.69. Insulation
3.70. Material for P , ...269
3.71. E xpansion Bends ...269
3.72. Pipe Fittings ...270
3.73. Pipe Joints .. .270
3.74 Valves ...271
3.75. Principles of Steam Power Plant Design .. .272
3.76. Factors Affecting Steam Plant Design ...273
3.77. Site Selection . ..273
3.77.1 Controls at Steam Power Plant ...274

3.77.2. Feed Water Control- ...274
3.78. Industrial Steam Turbines ...275
3.79. Overall Thermal Efficiency ...275
3.80. heat Flow ...277
3.81. Cost of Steam Power Plant ...278
3.82. H eat Balance Sheet for Boiler ...280
3.83. Useful Life of Components ...281
3.83.1. Power Plant Pumps ...284
3.84. Plant Layout... 285
3.85. Terms and Definitions ...285
3.86. Modern Steam Power Station ...286
3.87. Ways of Increasing Thermal Efficiency of a Steam ...287
Power Plant
3.88. Indian Boiler Act
3.89. Thermal Power Stations in India ...290
3.90. Super Thermal Power Stations ...293
3.90. (a) Singrau ...293
Super Thermal Power Plant
3.91. Korba Super Thermal Power Station ...294
3.92. Thermal Power Plants Environmental Control ...294
3.93. Com missioning of Plants ...294
Problems .. .296
...312
1. Diesel Engine Power Plant
4.0. I ntroduction 316_-355
4.1. Classification of ...316
Engines Internal Combustion (I.C.)
4.2. Four Stroke Diesel Engine ...318
...321
 (xii)
4.3. Two-stroke Diesel Engine ...322
4.4. Application of Internal Combustion Engines ...324
4.5. I.C. Engine Terminology .324
4.6. Engine Performance • .325
4.7. Heat Balance Sheet ...326
4.8. Diesel Engine Power Plant Auxiliaries ...331
4.9. Internal Combustion Engine Cooling Methods ...332
4.9.1. Cooling Methods ...333
4.10. Lubrication ...337
4.11. Engine Starting Methods ...338
4.12. Starting Procedure ...339
4.12.1. Stopping the Engine ...339
4.13 Starting Aids ...339
4.13.1. Warming up of Diesel Engine ...340
4.14. I.C. Engine Fuel ...340
4.15. Fuel Supply ...340
4.16. Diesel Engine Fuel Injection System ...341
4.17. Fuel Injection Nozzle • .343
4.18. Filter and Silencer Installation ...343
4.19. Advantages of Diesel Engine Poyer Plant ..344
4.20, Site Selection ...345
4.21. Layout •.345
4.22. Applications of Diesel Engine Plants • . .346
4.23. Cost of Diesel Power Plant ...346
4.24. Testing Diesel Power Plant Performance .346
4.25. Log Sheet .348
4.26. Advantages of I.C. Engine Over Steam Engine .348
4.27. Plant Maintenance ...348
4.28. Specific Fuel Consumption ...349
4.29. Comparison of a Diesel Engine and Petrol Engine ...349
4.30. Supercharging ...349
4.31. Advantages of Supercharging ...350
4.32. Factors Affecting Engine Performance ...351
4.33. Combustion Phenomenon in C.I. Engines ...351
4.34. Comparison of Gas Turbine with Reciprocating I.C.
Engine ... 351
Problems ... 354

S. Nuclear Power Plant 356-397

5.1. Nuclear Energy ...356
5.2. Chain Reaction .. 353
5.3. Fertile Material •. .359
5.4. Unit of Radioactivity (Curie) ... 359
5.5. Parts of a Nuclear Reactor . . 361
5.5.1. Nuclear Fuel ... 361
5.5.2. Moderator ... 363
 (Xiii)
5.5.3. Moderating Ratio ...363
5.5.4. Control Rods • .364
5.5.5. Reflector ... 365
5.5.6. Reactor Vessel ... 365
5.5.7. Biological Shielding ...365
5.5.1. (a) Coolant ...365
5.5.8. Coolant Cycles ...366
5.5.9. Reactor Core ...366
5.6, Conservation Ratio
... 367
5.7 Neutron Flux
367
5.8. Classification of Reactions
...367
5.8.1. Design of Nuclear Reactor ...368
5.9. Main Component of a Nuclear Power Plant
5.10. Boiling Water Reactor (B.W.R.) ...368
5.11. Pressurised Water Reactor (P.W.R.) ...369
5.12. Sodium Graphite Reactor (S.G.R.) ..369
5.13. Fast Breeder Reactor (F.B.R.) ...370
5.14. Coolants for Fast Breeder Reactors ...371
• ...372
5.15. Waste Disposal
5.16. Homogeneous Reactor ...372
...373
5.17. Heat Exchanger
...374
5.17.1. Candu Reactor ...374
5.18. Gas Cooled Reactor
...375
5.18.1. Objectives of R and D in Nuclear Energy ...375
5.19. Breeding ...376
5.19.1. Electron Volt (eV) ...376
5.20. Thermal Neutrons ...377
5.21. Fast Neutrons
...377
5.22. Burn up
5.23. Cost of Nuclear Power Plant ...377
... 377
5.24. Nuclear Power Station in India
...378
5.25. Light Water Reactors (LWR) and
Heavy Water Reactors (HWR) ...383
5.26. Importance of Heavy Water .383
5.27. Advantages of Nuclear Power Plant
.384
5.28. Site Selection .385
5.29. Comparison of Nuclear Power Plant and Steam
Power Plant .386
5.30. Multiplication Factor 86,
5.31. Uranium Enrichment ...387
5.32. Power of a Nuclear Reactor ...389
5.33. Reactor Power Control ...391
5.34. Nuclear Power Plant Economics .391
5.35. Safety Measures for Nuclear Power Plants ... 392
5.36. Site Selection and Commissioning Procedure .393
Problems .396
 (xiv)
6. H ydro-Electric Power Plant
398-498
6.1. Water Power
...398
6.2. Application of Hydro Power Plant . .400
6.3. Essential Feature or Elements of Hydro Electric
Power Plant
.. .400
6.3.1. Types of Dam ...402
6.4. Selection of Site for a Darn . ..406
6.5. To Calculate Penstock Thickness ...407
6.5.1. Number of Penstocks ...407
6.5.2. Anchor Blocks for Penstocks ...408
6.5.3. Penstock Joints ...408
6.6. Layout of Hydra Power Plant ...410
6.7. Classification of Hydro-electric Power Plants ...411
6.8. Advantages of Hydro-electric Power Plant ...415
6.9. Mini and Micro Hydro-Power Plants ...416
6.10. Draft Tube
...417
6.10.1. Types of Draft Tubes ...417
6.11. Surge Tanks
...418
6.12. Safety Measures in Hydro Power Station ...419
6.13. Hydraulic Turbines
...420
6.13.1. Classification ...420
6.13.1. (a) Based on action of water on moving
blades
6.13.2. Direction of flow of water
6.13.3. Position of shaft ...426
6.134. Head of Water ...426
6.13.5. Classification based on speed ...426
6.13.5. (a) Head on turbine.. ...427
6.13.6. Classification based on specific speed ...427
6.14. Choice of Turbines,
...427
6.15. Comparison of Pelton Wheel and Francis Turbine ...428
6.16. Turbine Governing
...429
6.17. Performance of Water Turbine
...430
6.18. Efficiency ...435
6.18.1. Volumetric efficiency ...435
6.18.2. Hydraulic efficiency . ...435
6.18.3. Mechanical efficiency ...435
6.18.4. Overall efficiency ( ii ) ...436
6.19. Coupling of Turbine and Generator ...436
6.19.1. Speed and Pressure Regulation ...437
6.19.2. Power Generated ...437
6.20. Synchronous Speed ...437
6.21. Cavitation ...437
6.22. Operating Characteristics 439
.0.
6.23. Efficiency Load Curves . ..439
 (XV)
6.24. Site Selection
6.25. Comparison of Hydro Electric Power Plant and .440
Steam Power Plant
6.25.1. Cost of Hydro Power ...441
6.26. Hydro Steam Inter -connected System ...442
6.27. Tunnels ...444
6.28. Flumes
6.29. Spillway and Gates in Darns .446
6.30. Gates .446
631. Types of Valves .448
6.32. Average Life of Various Components .449
6.33. Hydraulic Accumulator ...451
6.34. Cost of Hydr oelectric Power Plant ...451
6.35. Hydrology • ..452
6.36. Hydrologic Cycle ...452
6.37. Hydrograph .452
6.37.1. Flow Duration Curve • .•453
6.38. Mass Curve .453
6.38.1. Unit Hydrograph ... 454
6.38.2. Factors affecting Run-off .454
6.38.3. Run-off Estimation :..455
6.38.4. Flood Run-off ...456
6.39. Controls in Hy dro-Electric Plants ...457
6.40. Surveys Needed for Hy .458
dropower Plants
6.41. Control Room Functions ... 459
6.42: Mechanical Equipment .460
6.43. Switchgear
6.43. (a) Hydro Power Plant in India ...461
6.43.1. Hydro Power in India .466
6.43.2. Preventive Maintenance of Hydro Electric ...466
Power Plant
6.43.3. Electrical and Mechanical Equipment .468
6.43.4. Hydel-ThermalMix ...469
6.44. Economic Loading of Hydr o-Power Plants ... 470
6.44.1. Run-of-river plant in co ...471
mbination with steam
power plant
6.44.2. Pump storage plant in combination with .472
steam power plant
6.45. Power House Planning ...472
6.46. Surface Power Plants • . .485
6.47. Under Ground Power Plants • ..486
6.48. Components of Underground Power Plant • ..487
6.49. Types of Underground Power Plants ...487
6.50. Largest Underground Power Plant (Napth.)a .488
Jhakhrj Hydel Power Project)
6.51. Advantages and Dis-advantages of Underground ...488
 (xvi)
• Power Plant . . .48
Problems ... 495

7. Gas Turbine Power Plant 499-525

7.0. Introduction .499
7.0. (a) Classification of Gas Turbine Plants ...499
7.1. Elements of a Simple Gas Turbine Plants ...500
7.2. Terms and Definitions .502
7.3. Engine Efficiency or Turbine Efficiency ...503
7.4. Starting of Gas Turbine Power Plant .504
7.5. Fuels ..504
7.5.1. Qualities of Fuel .504
7.6. Comparison of Kerosene Oil and Gasoline .505
7.7. Air Fuel Ratio ...505
7.8. Gas Turbine Cycles .505
7.8.1. Open Cycle ...505
7.8.2. Refinement of Simple Open Cycle ...506
7.8.3. Closed Cycle ...508
7.9. Some Other Possible Arrangements ...508
7.10 Semi-closed Cycle Gas Turbine ...510
7.11. Compressors ...510
7.12. Air Rate ...511
7.12.1. Factors Affecting Air-rate ...511
7.13. Free Piston Gas Generators Turbine System ...512
7.14. Advantages of Closed Cycle Gas Turbines ...513
7.15. Advantages of Open Cycle ...514
7.16. Relative Thermal Efficiency at Part Load ...515
7.17. Jet Propulsion ...515
7.18. Specific Thrust ...516
7.19. Applications of Gas Turbine ...516
7.20. Advantages of Gas Turbine Power Plant ...516
7.21. Layout ...517
7.22. Advantages of Gas Turbine over Steam Turbine ...518
7.23. Gas Turbine Cycle Efficiency ...518
7.24. Effect of Blade Friction ..520
7.25. Improvement in Open Cycle ...521
7.26. Combined Working of Gas Turbine Plant and
Steam Power Plant ...523
7.27. Gas Turbine Power Plants in India ...523
Problems ...524

8. Instrumentation 526-538
8.0. Introduction ...526
8.1. Classification of Instruments ...526
8.2. Measurement of Pressure ...5N
.2. Temperature Measurements
 (Xi. • t i)

S.4. Flow Measurement

8.5. Fuel Measurcuient .. 5 29
8.6. Speed Measurement... 530
8.7. Level Indicators .530
8.8. Gas Anal sis ... 530
8.9. Calorjuet p rs Fuel and -Stearn .531
...531
8.10. Selection of Instruments
8.11. Electrical Instruments ...sa 1
. 532
8.12. lnstrwnentat ion and Controls in Steam Power
Stations
.532
8.13. lustrunients and Controls Arrangement
8.14 Orsat Apparatus .533
8.15. Ox y gen Meter . .534
.534
8.16. Impurity Measuring Instruments
...535
8.17. Measurements of Smoke and Dust
.536
8.17.1. Photo cell-type smoke meters
..,537
8.17.2. Reflected Light dust recorder ...537
... 537
Problems
... 537
9. Miscellaneous Problems
539-566
9.1. 'Magneto-hydro Dynamic (MHD) Generator ...539
9.2. Fuel Cell
. ...541
9.3. The Indian Electricity Act 1910
...542
9.4. Indian Electricity Rules, 1956
9.5. Energy Cycles .545
...547
9.5.1. Carnot Cycle
...548
9.5.2. Rankine Cycle
...548
9.5.3. Reheat Cycle
...549
9.5.4. Regenerative Cycle
9.5.5. Binary Vapour Cycle .550
. ...551
9.5.6. Reheat-regenerative Cycle ...552
9.6. Installation of PowerPlant
9.7. Blo-Gas ...555
9.7.1. Biogas Manure Plant ...558
...561
9.8. Combined Working of Different types of Power
Plants
9.9. Economy of operation ...562
9.10. Efficiency of Power Plants ...564
Problems . 565
...566
10. Major Electrical Equipment in Power Plant
567-595
10.1. Introduction
10.2. Generator ...567
10.3. Exciter ...567
10.4. Generator Constants ...568
10.5. Generator Cooling Methods ...569
...569
—2
 (xFLU)

10.5.1. Parallel Running of Alternators .570

10.6. Power Transformers .. .570
10.7. Reactors .572
10.S. Location of Reactors .572
10.9. Circuit-Breakers .573
10.10. Earthing of a Power System ... 575
10.11. Layout of Electrical Equipment .576
10.12. Protective Devices for a Power Plant ...578
10.13. Characteristics of Relays .578
10.14. Types of Relays .579
10.15. Voltages Regulation of Transmission Lines ... 580
10.16. Transmission of Electric Power .581
10.16.1. Transmission Lines .585
10.17. Systems of Electrical Energy Transmission ...587
10.18. AC Power Distribution ...588
10.19. Practical Working Voltage .589
10.20. Over-head and Under Ground Power Transmission
Systems .589
10.21. Conductors ... 590
10.21.1. Disposition of Conductors .590
10.22. Electric Power System Stability .591
10.23. Control Room .591
. . .592
10.24. La y out of Power System
10.25. Factors Affecting Power Generation and
Distribution .593
Problems .594

11. Non-Conventiona l Sources of Energy 596-631

- .596
no. Introduction
11-1. Solar Energy
11.2. Solar Radiation Measurement . .599
.599
11.3. Solar Constant
11.4. Solar Energy Collectors .599
11.5. Flat Plate Collectors .600
11.6. Concentrating Collector .601
11.7. Line Focusing Collectors .601
11.8. Point Focusing Collector .602
11.9. Advantages and Disadvantages of Concentrating
Collectors .. .602
11.10. Solar Thermal Power Generation ..603
11.11. Low Temperature Thermal Power Generation .603
11.12. Tower Concept Type Solar Power Plant .604
11.13. Photovoltaic (PV) Cells or Solar Cells .,.605
.606
11.14. Solar Pumping
.607
11.15. Wind-Energy.
.608
11.16. Wind Mills


'xix)
11.16.1. Basic Components of a Wind Energy
Conversion System (WECS)
11.17. Site Selection for Wind Mill Units ...611
...613
11.17.1. Performance of Wind Machines
11.18, Bioniass ...613
11.19. Biomass Gasification ...614
11.20. Tidal Power ...615
11.20.1. Tidal Power Plants ...615
11.21. Classification ...616
...616
11.22. Advantages and Disadvantages of Tidal Power ...617
11.22.1. Regulation of Tidal Power Supply
11.23. Biogas ...618
11.24. Classification of Bio-gas Plants ...618
...618
11.24.1. Factors affecting Bio-digestion or
Generation of Gas
11.24.2. Thermionic Generation ..'.619
...620
11-24.3. Thermiorjc Converter
11.25. Types of M111)Generators ...620
... 621
11.25.1. Open cycle generator
11.25.2 Closed cycle systems ...622
11.25.3. Advantages of MHD Sys ... 622
tem ...623
11.25.4. Combination of MIlD power plant and
steam power plant
11.26. Fuel Cell ...623
11.27. Therm o-Electric Power ... 624
11.28. Therm ... 624
Thermo-electric Power Generator
11.29, Therm o-electric Materials ... 625
11.30. M . 626
Methods for Maintaining Biogas Production
Problems ..629
..630
12. Environment Pollution and its Control
12.1. Introduction 632-648
12.2. Steam Power Plant Pollutants ...632
12.3. Control of Pollutants ...633
.635
12.4. Control of Particulate Matter
12.5. Control of SO, ...636
12.5.1. Wet Scrubber .637
12.5.2. Catalytic Oxidation ...638
12.5.3. Magnesiun- Oxide Scrubbing ...639
12.6. Control of NO, ...639
...640
12,7. Control of Waste Waters from Steam Power Plant ...641
12.8, - '.11utants from Nuclear Power Plants their Effects
Z1110 Control
12.9. Pollution and Noise Control
12.10 St a11c13rdj . itio15 ... 644
for
Environmental Pollution ..644
12.11. Thermal Pollution
..644
12.12. Cleaning of Ventilation Air at Atomic Power
Stations .645
1213. Fuel econom y in furnaces of boilers .6 16
Problems .618

13. Direct Energy Conversion Systems 649-670

13.1. Introduction .649
13,2. Thermo-electric Conversion S yst rn .619
13.3. Performance Analysis of Tlicrmo-ck'ct rc Power
Generator .651
13.4 Thermo electric Materials .652
13.5. Anal ysis of Thermionic Generator .653
13.6. Electro-gas D y namic Generator (EGD) • .654
13.7. Power Output of MUD Generator .655
13.8. Materials for MHD Generator .659
13.9. Electrode Materials
13.10. Materials for Channel .661
13.11. Batteries .66 1
13.12. Division of batteries .661
13.13. Battery Principle .661
13.14. Types of batteries .662
13.15. flattery Equivalent Circuit Models .662
13.16. Types of Battery Arrangenwnts • .66:3
13.17. Power of Battery .6134
13.17.1. Energy efficiency (ii ) of battery • .661
13.17.2. Cycle life of battery • .661
13.18. Principal and anal ysis of H2, 02 (Hydrogen-Oxygen)
fuel ccli .664
13.19. Types ' .666
13.20. Applications .666
13.21. Output .666
Problems .670

14. Objective Type Questions 671-700

APPENDIX A—CONVERSION TABLES .701-702

APPENDIX B—PROPERTIES OF DRY
SATURATED STEAM .703-707

I-...
 Introduction
The global energy crisis has attracted the attention of the de-
veloping and deelopecl countries to explore and find out new means
of energy sources to meet this ever-increasing demand of the man-
kind as the conventional main sources of energ y viz., coal and oil
would exhaust after certain period of time. Electrical energy is an
important index of country's economic and technological progress.
In India the installed power capacity has been increased by more
than 13 times during the last 30 years. i.e. from 2,300 to 31,000
MW in 1980. The power demand has also gone up steeply. If we
analyse this load growth, two factors will con out more promi-
nently apart from others.
• Growth of industries.
• Increase in domestic load due to nw and extensive urban
residential area and vertical growth in existing area with
increased and intensive use of electrical gadgets.
The acceleration of the process of industrialisation and urbani-
sat i on following the determined effort of the developing countries
like India to improve their economic well-being has inevitabl y lead
to larger and larger demands for energy. The central problem, there-
fore, is how a country like India can ensure adequate supplies of
energy of fuel, its economic growth into global environment of scar-
city of desired fuels and high costs.
In our country the main sources oferiergv are fossil fuels, hydel,
and nuclear power even though solar energy, wind power and tidal
power offer hopeful technological opportunities.
Coal will remain as main source of energy for severni decades
to some. While discovers' of more oil and gas is not ruled out, the
question is what production level can he reached and for how long
can it be sustained. A large untapped hydro-potential exists in our
country. It is found that total h ydro-potential could be equivalent
to 75,400 M\V at 6O'2 load factor of which about 10 1,' to 1 2'' has
been exploited so far. hydra power being a renewable sourL(-(f
energy nu.ist receive a high priority in our energ y development pro-
gramme.
The use ofIiio-gas for lighting and irrigation opens Uj) new pos-
sibilities for self-contained rural communities. In the long run over
energy economy would have to be built aroei !and-based hiojuass
fuels and the plentiful sunshine which we receive virtuall y tlirwgli-
out, the year. Uwe succeed in tapping solar energy oil n
 (.rxu)
scale for lift irrigation purposes we may indeed achieve a break
through in our rural economy by successful utilising our coal, oil,
gas uranium and hydro are commonly called conventional energy
sources.
World over the trend has been towards utilising the existing
resources like hydle and coal and at the same time developing nu-
clear power generation capabilities. Developed countries have taken
a long term view of their power needs and have gone in for a judi-
cious mix of these sources.
The various Lources of power generation should be used to match
particular circumstances and needs. Nuclear power is a clean source
of energy and hazards can be minimised. Dams have longer life
than thermal power plants. While the demand for electrical power
is on the rise, the over all known resources of petroleum are on the
decline. It is therefore imperative that the consumption of petro-
leiim products in the field of power generation, transmission and
distribution is conserved to the maximum extent.
In India, the present petroleum may not last long. Alternative
sources of energy must therefore be found out so far the known
sources of unconventional energy are nuclear fusion, solar energy,
geo-thermal energy, wind power, tidal wave energy and biogas tech-
nical. In a tropical country with abundant solar energy. If we suc-
ced in tapping solar energy on a massive scale for lift irrigation
purposes, we may indeed achieve a break-through in our rural
economy by successfully utilising our under-ground water resources.
All efforts should be made to conserve conventional sources and to
develop non-conventional sources of energy.
In our country industrial sector consumes the maximum dcc-
rical power about 52% of the total power. Next of industrial sector
is the transport sector.
In our country the installed capacity was nearly 2300 MW in
1950 and will be about 65000 MW by 1991. Power sector proposes
to induct an additional capacity of about 38000 MW during eighth
five year plan which will need an investment of about Rs. 128000
crores. This is a very huge amount and therefore all efforts should
be made to conserve energy. There is a need to make best use of
our present capacity. Plants load factor must be increased. Efforts
should be made to reduce transmission and distribution losses. We
must also look into alternative sources of energy like solar, bio-
mass, photovoltaic energy and micro-hydel projects.
In our country the per capita energy production is ver y low.
For example the per capita annual average production of electricity
 (Xxiii)
during 1986-88 in UK was 5300 KWH, France 6300 KWH, Ger-
maii•' 6800 KWH as against that India's generation was onl y 260
KWh. The peak power demand by the end of the ninth plan (2001-
02) will be 104000 MW. This would require an instafled capacity of
about 170000 M\V. As such about 100000 MW of additional gener-
ating capacity would need to be installed during the decade 1992-
2002.

Emphasis needs to be given to

• tapping of hydel power. H y dro-thermal mix has alread y
de-
clined from the desirable 40:60 ratio to 27 : 73. Therefore it
is desired that private investors should be offered hydel
projects in preference to coal, gas-based thermal power
plants and uneconomical diesel and naptha-based power
stations.
• expansion of power plant program already entrusted to the
central power corporations and state electricity boards.
• tap new and renewable sources of energ y . The energy from
the sun, wind and Biomass and alleviate the pressure on
traditional sources that have been considerably depleted.
Economical means to harvst such new energy sources need
to be developed.
 1

Sources of Energy

1.1 Power
Power plays a great role wherever man lives and works--in
industry, agriculture transportation etc. Power provides our homes
with light and heat. The living standard and prosperity of a nation
var y directly with increase in use of power.

As tchnology is advancing the consumption of power is steadily

This necessitates that ill to tire existing sources of
power such as coal, water, petroleum etc. other sources of energy
hotild be searched out and new and more eflicient ways of produc-
ing energy should be devised. Nuclear energy has enlarged the
world's power resources, The energ y released by burning one
kilogram ofuramum is equivalent to the energ y obtained by burning
4500 tonnes of high grade coal. In our country the total generating
a
cpacity at the beginning of first five year plan was 2.3 million k\V.
'Fhj capacity was raised to 3.4 million kW by the end of first five
Year plan 5.6 million kW b y end of second plant and 10.5 million
kW b y tire end of third plan. In 1969-70 it was 15.5 million kW.
During the fourth and fifth plan the generating capacity targets are
of tile order of 20 million kW and 10 million kW respectively In our
country the natural resources are found ill lignite, oil, hydro
sources and nuclear fuels. These rcSouFCLS should be exploited in
the most efficient form b y using improved technology so that power
at clre;tper rates becomes available, which will help toaccelerate the
growth of industry. It is fou that these resources are L eit ted in an
rifieven form in the country which requires interliikag the grid
Yt ( • I1) ofadjoi Il nig states so that power generation by hydro,
tlieriiiitl arid nuclear plants could be vell co-ordinateti This will
('iureacoi1s( I nt supply of power to all consumers throughout the
country, In northern part of tire country hvdro power is the main
source available w hereas iii Madhya Pradesh and Eastern
llr.rr.ltr'i coal and hvdi'o sources are available But in Gujarat
jild \Vtstcrji MaIr:ir:ishrtra 1r tict1lv in,
'I hei etr ' re, (wcr should be generated either i'c€- are available.
by nuclear power
2 POWER PLANT

plants or b y thermal power plants depending upon the relative

economics. Mysore and Kerala have abundant hydro resources.
Andhra Pradesh has got both hydro and coal resources and Tamil
Nadu depends upon the thermal power at Neyveli. West Bengal,
Southern Bihar and Orissa have abundant coal resources and,
therefore, power generation by thermal power stations is cheaper.
Solar energy in India has an ideal geographical situations. The
northern and central parts of the country receive bright sunshine
oil average for more than eight hours daily. This amounts to heat
equivalent to more than 200 MW per square kilometre. This shows
that there is all potential for developing solar thermal
power plants.
Thermal !)o'eI product i on in India costs more than nuclear
power after the recent rise in oil prices. So far as nuclear power is
concerned India is fairly well endowed organisationally as "vell as
resourcewise. The resources re available in large quantities in the
form of Uranium and Thorium. Uranium deposits have been located
in Bihar, Rajasthan and Tamil Nadu and thorium reserves have
been found in Monazite beach sands of Kerala and in Ranchi.
Assuming effective prevention of radioactive hazards it is a clean
source of power which does not contributed to air pollution. Thus it
is observed that in our country different power resources are avail-
able in different states and, therefore, in order to ensure constant
supply to all the consumers throughout the year interlinking of
various power plants in essential so that spate capacity olone power
station can be utilized by the other. With this objective in view
programme for integrated power generation on regional basis by
interlinking the power stations of adjoining states has started
working and successful functioning of regional grids may finall give
.wa to Nat ional Grid system. Five regions already have intercon-
nected power systems. The northern power grid s y stem covers UT.,
Punjab, Har y ana, Bhakra Management Board (BMB ). Jammu,
DESU, iaasthun and M.P. Systems with a capacit y of 550-I MW.
The construction of Biidarpur Jaipur 220kV line which will connect
the Badarpur. DESU, 13MB s ystera with the Rajasthan s y stem is
being completed. Completion of this line will further stabilise inter-
connected working in the northern region. The southern region
Orissa grid covets Andhra Pradesh, Kerala, Karnataka, Tamil
Nadu and Orissa systems with 5525 MV capacity. The Western grid
covers Maharashtra, Gujarat and 'larapur s y stems with 3865 MW
capacity. Eastern grid covers West Bengal, Bihar, and DVC systems
with 3200 MW capacity. Assam, Meghalaya and Tripura with an
installed capacity of 178 MW are inter connected. An adequate grid
system has the following advantages
 SOURCES OF ENERGY

1. It enables the base load to be supplied by the most

economical power stations and peak demand to be sup-
plied b y more expensive power stations.
2. It provides security against all normal operating hazards
with a smaller margin of spare capacity thereby saving on
overall capital expenditure.
Planning for power generation, capacity addition and power
systems should be based oil assumption that the power systems
in each region will operate in close integration and power will be
exchanged between systems confirming to well established stand-
ards which will lead to optual operation of the integrated systems,
the ultimate objective being fi.mation of national power grid.
Following factors can help in improving the generation of
electric power:
(i) To improve power plant load factor.
(ii) To lay stress on conservation of energy.
(iii) To give due importance to renewable sources of energy.
(iv) To improve coal supply to power plants.
(v) To establish more gas based power plants.
(vi) To promote the private agencies to produce electric power.
(vi) To reduce power transmission losses.
(uiii) To use an efficient control and operation of power plants.
1.2 Source of Energy
The various sour'es of energy are as follows:
1. Fuels. The fuels are broadly classified as follows
(i)Solid fuels. Various solid fuels used are wood, coal including
bituminous coal, anthracite, lignite, peat, etc.
(ii)Liquid fuels. Liquid fuels include petroleum and its deriva-
tives.
(iii)Gaseous fuels. Gaseous fuels consist ofnatural gas, producer
gas, blast furnace gas, coal gas etc.
2. Energy Stored in Water. The potential energy of water at
higher level is utilized for the generation of electrical energy.
Water power is quite cheap where water is available in abun-
dance. Although capital cost of hydroelectric power plants is higher
as compared to other types of power plants but their operating costs
are quite low.
3. Nuclear Energy. Controlled fission of heavier unstable
atoms such as U 23 , Th232 and artificial elements Pu 239 liberate large
amount of heat energy. This enormous release of energy from a
relatively small mass of nuclear fuels makes this source of energy
of great interest. The energy released by the complete fission of one
 POWER PLANT
4

kilogram of U 235 is equal to the heat energy obtained by burning

4500 tonnes of high grade coal. However, there are some difficulties
in the use of nuclear energy namely high capital cost of nuclear
power plants, limited availability of raw materials difficulties as-
sociated with disposal of radioactive wastes and dearth of well-
trained personnel to handle the nuclear power plants.
Nuclear power can cater to the future needs of energy. Three
stages of Indian nuclear power programme are as follows
(i) natural uranium fuelled pressurised heavy water reactors.
(ii) fast breeder reactors utilising plutonium based fuel.
advanced nuclear power systems for utilisation of
(iii)
thorium.
4. Wind Power. Wind power call made use of where wind at
suitable velocity is available. Wind power is capable of generating
small amounts of electrical energy. It is successfully employed for
pumping water from deep wells. Wind power has served many
countries as a source of power in early days and were called as wind
mills. 'l'he propulsive power of wind can be used to drive multi-
bladed turbine wheel. Wind turbines prove to be costly if designed
to run at all wind speeds. They usually'start running at wind speeds
just enough to overcoflW the system losses and develop full power
at the prevailing speed for the locality. Wind is the cheapest source
of pwxcr.
Wiiul energy is a renewable source of energy. The wind power
systems are fl on-polluting. However wind energy is noisy in opera-
tion and large area is required to install wind mills. Wind energy is
weak and fluctuating in nature.
In India, wind velocity along coast line has a range of 10-16 km
ph ad a survey of wind power has revealed that wind power call
used to pump water from deepwells or for generating electric energy
in smaller amount. Modern wind mills are capable of working on
velocities as low as 3-7 kinph velocity while maximum efficiency is
attained at a velocity ranging between 10-12 kmph.
5. Solar Energy. The heat energy contained in the rays of sun
is utilised to boil water and generate steam which call used to
drive prime movers to generate electrical energy.
The facts speak in favour of solar energy. The world's reserves
of coal. oil and gas will get exhausted with ill few decades. Atomic
energy involves considerable hazards and nuclear fusion has not yet
overcome all the problems of even fundamental research. Compared
with these technologies the feasibility of which is still uncertain and
iuntested, the technical utilisation of solar energy can prove very
useful. Utilisation of solar energy is of great importance to India
since it lies in a temperatUr( climate oftlie region of the world where
 SOURCES OF ENERGY 5

sunlight is abundant for a major part oft l i

e year. The basic research
in solar energy is being carried out in universities and educational
and research institutions. Public sector institutions like Bharat
Heavy Electricals Limited and Central Electronics Limjd are
carrying out a co-ordinated programme of research in solar energy.
Some of the fields in which solar energy can be used are as follows:
(i) Solar power plants used for electric power generation
(ii) Solar water pumps used for pumping water.
(iii) Solar water heaters used for
water heating.
(iv) Solar cabinet type driers for drying of food grains.
(v) Solar kilns for drying wood etc.
Solar energy is effective, only during day time and if power
supply is to be made during night also then some reservoir of energy
such as storage battery or heat accumulator should be used. Solar
energy cannot be used during cloudy weather and rainy season,
solar energy is used efficiently there would be enough power to meet
if
the increasing power demand for several years to come. At present
harnessing, storage and use of solar energy i much more expensive
than using fossil fuel (coal oil, gas etc.). It is believed that solar
power can become economically feasible with the following aims
achieved:
1. Availability of better heat Collectors.
2. Availability of improved materials and manufacturing
techniques.
3. Better techniques for storage and cheap distribution of
solar power.
While the energy crisis is mounting, any co-operation to find
alternative sources of energy is certainly laudable. India has the
potentialities of solar and wind energy. 'What it needs is the ad-
vanced technology to tap this potential.
Solar heat must cost less than conventional (oil base ( l) heat in
order to be economically feasible. When comparing both g
enerating
methods, it is important whether the solar plant operates in a
monovalent, i.e., as the only heat supplier, in a bivalent mode as a
fuel saver, connected to a conventional plant. In the first case, the
investment costs for the saved (oil) boiler should also be taken into
account. In both cases, the fossil fuel—oil, gas, coal, wood, etc. -
prices should be considered not only at their present level but also
with regard to future price increases (hue to inflation and other
factors.

Perhaps the great advantage of solar power is that the system

is ideally suited to the human environment being free from pollution
and noise. Besides, maintenance is cheap and convenient. Of all the
non-conventional energy sources, solar power is much cheaper for
cooking, street lighting and water heating.
6 POWER PLANT

6. Tidal Power. Ocean waves and tides contain large amount

of energy. Such tides rise and fall and water can be stored during
rise period and it can he discharged during fall. Due to low head of
water available low head hydroelectric plants can work successfully
Fig. 1.1 ShO\VS the schematic layout of a power plant using tidal
p.wer. These plants can utilise a head of just a few metres. During
high tide the height of tide is above that of tidal basin and the turbine
unit operates and generates power. During low tide the height of
tide is lower than that of the tidal basin. At this time water is allowed
to flow out to drive the turbine unit. The turbine unit does not
operate if the tide sea level and basin level are equal.
In India the possible sites, identified for tidal power plants are
as 1ujow

I::Jfle v
Generator

Low tide
_•:: .-TidoI bosn
-
= '
Turbinc - Gnerator
Fig. 1.1.

(i) Gulf of Cambay

(ii) Gulf of Kutch
(iii) Sunderban area in West Bengal.
The tidal range in the Gulf of Cambay is about 10.8 metre.
Whereas the maximum range in Gulf ofKutch is 7.5 metre. The tidal
range in Sunderban area is 4.3 metre.
Advantages. The various advantages of tidal power plants are
as follows
(i) The power generated does not depend on rain. Therefore
there is certainty of power supply as the tidal cycle is very
definite.
(ii) The tidal power plants are free from pollution.
(iii) Unhealthy wastes like ash, gases etc. are not produced.
(iv) These plants require lesser space.
(v) Such plants have a unique capacity to meet the peak
power demand effectively when they work in combination
with hydro power plants and steam power plants.


SOURCES OF ENERGY 7

Disadvantages. The various disadvantages of tidal power

plants are as follows
(i) The capital cost of tidal power plant (nearly Rs. 5000 per
kW) is .considerably large as compared to steam power
plant and hydro power plant.
(ii) The supply of power is not continuous as it depends upon
the timing of tides.
(iii) Tidal power plants are located away from load centres.
This increases power transportation cost.
7. Geothermal Energy. According to various theories the
earth has a molten core. The fact that volcanic action takes place in
many places on the surface of earth supports these theories. The
steam vents and hot springs come out of earth's surface. The steam
from such natural steam wells is used for the generation of power.
Fig. 1.2 shows schematic layout of power plant using steam from
steam wells. Steam drum separates moisture and solids from steam.
STEAM
DRUM COOLING
TURBINE TOWER
STEAM! GENERATOR _____
WELL I Fl 1

CONDENSER
'T' - TE
C, :DENSA
CIRCULATING
PUMP

r!. flfl

Fig. 1.2

TLJRBINEJ..L1 GEN.
CONDENSER

VAPOR1 I-lEA T
EXCHANGER
PUMP

GEOTHERMAL
REGION

Fig 1.2(A)
8 POWER PLANT

The hot water closed (Binary) s ystem shown in Fig. 1.2 (A) is
also used where temperature and pressure of water are not suffi-
cient to produce flash steam. In this system heat in water is used in
a closed c y cle. In this system Freon or Iso butane is used as working
fluid which is continuously circulated. This system has the ad-
vantage that lower hot water temperature can be used. Such system
is under development in USA and USSR.
8. Thermo-electric Plant. When the two junctions of loop of
two dissimilar metals are kept at different temperatures, an
electromotive force is developed and current starts flowing in the
loop. This is known as Seebeck effect. By using suitable materials
this method call be used for the generation of electrical energy in
small amounts.
1.2.1. Conventional and Non-conventional sources of
energy. The sources of energy used for mass generation of power
called conventional sources ofénergy are as follows
(i) Thermal (ii) Hydro-power
(iii) Nuclear power.
The non-conventional sources of energy used for generating
power in lesser magnitude are as follows
(i) Solar energy (ii) Wind power
(iii) Tidal power (iv) Rio-gas
(v) Magneto-Ilydro-dynamiC plant
(vi) Geo-thermal energy.
1.3 Fuels
Fuel is defined as any material which when burnt will produce
heat. Various fuels commonly used are as follows
1.3.1. Solid Fuels. Natural solid fuels include wood, peat,
lignite, bituminous coal and anthracite coal. The prepared solid
fuels are charcoal, coke and pulverised coal, Peat, Lignite,
Bituminous coal and Anthracite, coal are various varieties of coal.
Coal havingrelatively high percentage of volatile matter is called
soft coal and with lower percentage of volatile matter is called hard
coal. Wood can be burnt easily and gives maximum intensity of heat
very quickly, but is not suitable for boilers etc. because the calorific
value of wood is low (3000-4000 cal/kg).
Coal. The vegetable matter which accumulated under the earth
millions of years ago was subjected to the action of pressure and
heat. This changed the physical and chemical properties of matter
and it got converted into what we call as coal. In India coal is the
primary source of energy and Coal India is the controlling body for
the coal industry. India has reserves of both oil and coal. While coal
reserves in the country are sufficient to last for some hundreds of
years, oil represents only a fraction of total coal reserves. The power
 SOURCES OF ENERGY

sector COUSUWOS nearly 379r, of total coal produced aH

t) ri'ui:ji a-
ing being used in industries like steel, fertilizers, cc 'at, r, ilway.
etc. The coal bearing strata of India have been (L'-.s j uj
geologists into the following two main groups d 1v
(i) Golidwaria coal fields (ii) Tertiary coal fie'ds
Most ofGonclina coals are bituminous and sub-hit ominous in
quality. Gondwana coal fields are situated at Bengal, lhar, M.P.,
Orissa, Andhra Pradesh and Maharashtra Tertiary coa fi t lds are
situated at Madras, Assain, Kashmir and Rajasthan. Te ioiv coals
are mostl y lignite. Lignite is available in large quantil
it'- at Nevveli
in Madras. The coal reserves largely of bit umino rank of our
country are estimated to be about 112 x iO Mt in coil Seams of
thickness 0.5 ill above and up to a depth of 1200 mm
Indian coals are genera llycl1ai .act . j 5
(lifficult 1w high ash content and
The coals however h ive a vcrv low
sulphur contents. The Indian coal resources cor
0.8% of total coal (serves of the wrld L unIv about
basis. The occurri'ncc of coal in Our cc ,iiutrv Is .mng tenth oil
mainly c o nfined to
one quaclran On the eastern amid south ea -tern p,, t
sitating long transport haul distancc- in thus neces-
soutliem 0 parts of the country. 1 1rtift' - , er 11 an
td

1.4 Calorific Value of Fuel

The basic thermal characteristic of a fuel is its calurjfjc value
(heating value). The calorific value of fuel is defined as the amount
of heat produced when unit quantity one kilogram ofsoliul or liquid
fuel or one cubic metre of gaseous fuel) of fuel is completely burnt
under standard Condtlons, The calorjfj (
' value is expressed as kcal
per kg or kcal, per cubic metre of gaseous fuel. 'I'he calorific value of
a foal can be class ifle(l in two Ways
(i) Higher calorific value (II.C.\T.)
(it) Lower calorif i c value ([_CV)
The higher or gr oss of calorific value is the total amount of heat
produced when unit quantity of fu el is burnt
copletely
m -Ill(,
the
products of combustion have been cooled to room teiperature
generally 15 C. The lower calorific value is the net anhu,uru of heat
produced when unit quantity of fuel is completely burnt and the
products ofcomnbustjon are not cooled to room teinperat i n
i-at are
allowed to escape. There are several methods of findi.. tI- u ritic
value of fuel. According to Dulong formula the ci!
fuel is given by the following relation

H.C.V. =8080C + 34,500 H 2220 S

8 )

_-1
 POWER PLANT
10
where Cil.( ml represent the percentage by weight of Carbon,
h y drogen, ON -W11 iliI(l sulphur respectivelY.
The net )r I wer calorific value(L.C.V.) is obtained b y subtract-
ing from I IC V. the heat carried by the products of combustion
especially by stuani which call taken as 588.76 kcal/kg of water
vapours b.r:iwd due to burning of 1 kg of fuel.
L.C.V. - (11.(,'.V. - 588.76 x W) kcallkg.
where W is the amount of water vapours formed by the combustion
of 1 kg of hid.
The hiher calorific value at N.T.P. of various constituents of
fuek are a follows C = 8060 kcal/kg, S = 2220 kcal/kg,
ll . 3 1.500 kcallkg and CO 2430 kcal/kg.
1.5 To Calculate Approximate Flue Gas Loss
Percent age of net calorific value (Btu) lost in flue gas
(T1
=J1x

where Ki . constant 0.35 for Bituminous coal

0.37 fir Anthracite coal 0 39 for Coke
0.31 for Oil
T 1 - Temperature of exit gases F
T2 = Temperature of inlet air F
CO2 = percentage of CO2 in exit gases at point
where temperature T1 is measured.
The vtlue of constant K 1 becomes as follows if centigrade scale

= 0.63 for BituminoUs coal

0.68 for Anthracite coal .- 0.70 for Coke
= 0 56 for Oil.

1.6 Types of Coal

1. Peat. It consistS of decayed vegetable matter mainly decom-
pUS(5I
water plants amid inosse etc. It has high moisture content and
should lx dried before burning. Its approxiflmte composition is
C = 60, H - 58, 0 = 33. Ash
Its calorific value is 3500 kcalfkg.
SOURCES OF ENERGY 11

2, Lignite or Brown Coal. It is brown in colour, it burns with

a brightl y slightl y , smok y yellow flame. Its calorific value is 5000
kcal/kg and its approximate
proxiate composition is C = 67%, H = 5%, Q =
20% and Ash =
Neyveli in Tamil Nadu is the only major lignite mine in India
producing about 6 Mt. ofcoal. Developed as a complete project lignite
mined at. Neyveli is used primarily for power generation. Some
lignite is also used for production of fertilizers and briquettes.
Briquette fuel is used b y domestic and industrial consumers.
3. Bituminous Coal. It is soft, consists of large amount of
volatile matter and is widely used as fuel. It burns with a long yellow
and smoky flame. Its calorific value is 7800 kcal/kg and approximate
Composition is
C = 83.5%, H = 5%, 0 = 5%, Ash = 6.5%.
4. Sub-bituminous Coal. Sub-bituminous coal contains 12 to
25 1/c. moisture. It is of black-colour and the approximate calorific
value of this coal is 4600 kcal/kg.
5. Anthracite Coal. It is black in colour and burns with a short
bluish flame and the amount of ash p ......... ' ..... its burning is
very little. Its calorific value is 8500 kcaiikg and approximate
composition is as follows
C = 9017c, 0 = 2, H= 3(7 an-.: -\sli = 51/,.
Indian coals are known to he low in sulphur (0.3-0.9%) and high
in ash content (30-50%). The average value of calorific value varies
between 4000-5000 kcal/kg. Table 1.1 A) indicates the typical
analysis of Indian coal (average).

Table 1.1 (A)

Content Percentage (9J

Carbon 21-50
2.3-3.5 iI
Nitrogen 0.7- -1.4
Sulphur 0.3---0.9

Ash 30-50
Oxygen 17-11
_J-
Vo!atile 15-26
Moisture _ 7-20
Composition and properties of lignite found at Neyveli are
indicated in Table 11(B) on
I) air dried basis
(b) d - in. in- I (1) iv mineral matter
tter free) basis.


POWER PLANT
12
Table 1.1(B)
Proximate Analysis

Fixed curb- Volatile Fixed carb•

Moisture Ash Volatile
,natter on matter
1 30-35 L 52-60 40-48
3.5-7.5 40-45
10-30
Ultimate Analysis (% d.m.m.f.)
Oxygen g,

Calorific Value, kcaJkg

Ii A7rie

A typical composition of various types of coals is given in Table

1.11 (C).
Table 1.1 (C)

AFuelsp^xjmate Ana1YSS.% Witrrute Analysts,%

Ic calfkg

d Ash C Jl N2 S Ash Dn
olatile
atter rarb- I I repor busts
h-el
I
26 11 6 23 to 1.55 6
28 31 6 42 L i 6 4000

volatile

Medium
3516TiTh5
3 24 62 11 77

Sit urn rw:e Cool

5
J 4.5

5 1.5 05 11 7'00 7700

volatile _ J 1 7 6750 7100

36 49 ' 68.5 5.5 16.5 1.5
High 8
volatile 4_ J 5600 7000
4 59 6 295 1 05 4
19 31 46
bituminous
3 8.5 75 45 15 05 10 75007700

an thrac ite 1 7500 7700

7 865 25 3 05 0.5
Anthracite 2.5 3

1.7 Liquid Fuels

Liquid fuels include petroleum and its derivatives. Fractional
distillation of crude petroleum helps in separating it into its various
varieties such as gasoline, kerosene oil gas light diesel fuels and
residual oil. During distillation of petroleum gasoline is obtained
SOURCES OF ENERGY 13

between the range of 30 - 200°C whereas kerosene oil is obtained

between the range of 150 - 300°C and diesel oil is obtained between
200 - 320°C. The calorific value of the gasoline is 11,250 kcal/kg and
that of kerosene.oil is 11,100 kcal/kg. The calorific values of diesel
oil and fuel oil are 11,000 and 10,350 kcal/kg respectively.
The general index of classification of fuel oil is the specific
gravity of the oil. The specific gravity of fuel compared with water
at 60°F is expressed in the degrees API (American Petroleum In-
stitute) and can be found by the use of the following formula:
Degrees API I- 141.5 --
131.5
Sj gravity at 60 F
Specific gravity in API helps is finding the calorific value of fuel
using the following relation
BtuJlb of oil = 17687 + 57.7 (API gravity at 60°F).
The API gravity of commercial fuel oil varies from 10 to 40.
Liquid fuels are commercially classified as light, medium and heavy
domestic oil and as light, medium and heavy industrial oils. API
value of light and heavy domestic oils varies between 38-40 and
34-32 respectively whereas API value of heavy industrial oil is
between 14-16.
While burning fuel or oil it is desirable that it should be finely
atomised to ensure mixing of oil and air. To facilitate pumping and
for correct atomisation of fuel oil it is essential that viscosity of oil
should be lowered. This is achieved b y providing heating coils in the
oil storage tank to maintain temperature of about 40-50 C. The
oil leaving the storage tank is passed through another heater before
it enters the burner.
Commercial gasoline (petrol) and diesel oil are mixtures of
various hydrocarbons. The average properties of these fuels are as
follows
Gasoline

()A.Qgravity at 34
0.85

(iii) Viscosit y (CS) at IOC 0.72 4.5

at 40C 0.55
(&t) Cetane number

18 5 to

6s
h)Ojcfins
Aromatics 17
 A

POWER PLANT
14

1.7.1 Oil Properties. Petroleum or any of its derivatives is

usually analysed on the basis of the following properties
(i) Calorific value (ii) Viscosity
(iii) Specific heat (iv) Specific gravity
(v) Flash point.
It is the temperature at which oil vapour flashes or burns
steadily.
(vi) Pour point.
It is the lowest temperature at which oil flows.
(Lii) Coefficient of volumetric expansion.
(uiii) Carbon residue.
(ix) moisture and sediment.
(x) Sulphur content.
1.8 Advantages of Liquid Fuels over Solid Fuels
1. Handling of liquid fuels is easy and they require less
storage space.
2. Liquid fuels can be fired easily and the maximum
temperature is attained in lesser time as compared to solid
fuels. The solid fuels containing higher percentage of
moisture burn with great difficulty.
3. The solid fuels leave a large quantity of ash after burning
and the disposal of ash becomes a problem. Whereas the
liquid fuels leave very little ash after burning.
4. In case of liquid fuels the continuous firing the furnace is
achieved without any difficulty.
5. The combustion of liquid fuels is uniform.
6. The combustion of liquid fuels can be easily controlled.
Therefore, the change in load, can be easily met.
The disadvantages of liquid fuels are as follows
(i) They are costly as compared to solid fuels.
(ii) They require special type of burners for their burning.
(iii) Sometimes they give unpleasant odours.
(it; t There is danger of explosions.
In cold climates the oil stored iii tanks is to he heated ii
order to avoid stoppage of oil flow.

1.9 Gaseous Fuels

The gaseous fuels can be classified as follows
(i) Natural - Natural gas.
(ii) Prepared - coal gas, blast furnance gas, producer gas,
water gas etc.
(a) Natural Gas. Natural gas comes out of gas wells and
petroleum wells. It is mainly composed of methane (CH 4 ) = 85%,
ethane (C 2 116) 10% and other hydrocarbons = 5%. It is colourless
SOURCES OF ENERGY 15

and nun poisonous. The calorific value of nat aral gas is 5,2.-)
k cii L'cu 1)1 c met cc.

(/t) Mast Furnace Gas. This gas is ol)tairitc as a by-product

In-omn hl.t furnace used for producing pig irou its npproxmnitte
composition is C() = 30Y.. N 52. IL -: 3, (U 2. (1I
'lilt calirific value of this gas is 970 keal/cuhic ni,tw.

(c' Coal Gas. Coal gas is a b y -product I,tiiid (luring the

ilcstructivc- distillation ol coal. Its calorific value is 7600 kcal/cu
metre. Its approxiruitte composition is C11 1 - ii = -15 ('( ) =
6Q, N = G, (O 2 q and ((them' h ydrocarbons -:

(d) ProducerGas. Producer gas is produced tliirint incomplete

coiiihiistion of' coke in current of air. Its main cumtstitu&-iits art
nitrogen and carbon inotoxidi. Its approxiinatv coinpo.-.ition is N =
62'. Co = 2:3'.?. }I 61, CO 2 :/. Cl! 4 TV Its ciiloi-ihc value
1200 kral/cu mutt re.

(t' ,'atcr Gas. NY it is obtained by pissiugi Lili-t of teumn

through a nJcep bed of red hot co-e. Its ilium constituents ire
Co. CO and II.

10 Advantages of Gaseous Fuels over Solid Fuels

1. It is easy to control the length and n;iture of flun mid
hence temperature control is eusier.
2. Gansus fuels do not contain ash and thor Ircigri nut'±r,
and burn completel y . Their use is ec-.,itnuuuil torupared
Lo solid and Ii c iuicf fuels.. No ash removal is re1umred.
t. I lirohlilug of gaseous fuel is Uot 1e(fulrtl tlit-v can he
&asilv piped into the furnace.
1. Le.sc-r aniuuut 0tt-xces5 air is needed to burn tlitm coin-
plutely.
5. Greater cicanhiness m assured as the tot and smoke Is
practicall' nil. Gas fired burners operate ozi pressures
,
ranging from 0.15 to 1.5 kg/cm .

1.10.1 Composition of Liquid F'uels. Tin- comiipusition by

weight of lii1und fuels are givell in 514 1,2.


16 POWER PLANT

Table 1.2
- Specific Composttioo bvu'ei/zt
- ity - .--.- - -

ll€-.. fuel 0.95 86.1

I - '
11.8
1-.
j
-••
2]
od 0.87 J 8.3 - 128 0.9
Paratuie 0.79 8h.3 136 1 0.1
h'trt Wa sohlw 1 0.74 110

1.11 Comparison of Sources of Power

lit tin sources fir generating power in India are coal and
hvciro (water;. \ucloiir }R)WF is also bi'iig used successfull y . The
v j ri on SO.! rce- (f cow t r can he compared as follows
'lit . t.l cia hydro-electric power plant is highertlian
o st c,. . ) V.'cr plant, whereas the operating Cut of a steam power
-. t Ii a hvtlro pwer
o plant. The initial c ce-.t ofa nuclear
-. hni test and its operating cost call as low as that
• , , ari plant. The cost oferection of .i dte,'l plant is low
a. - Ii . ( I to cost eferectuta of steam plant
1' .' 'OI1) power station can be located at the load centre
i.\ii !'wct - plant has to he located where water is
vil.e1- . .' . :- quantities As the performance of a Ii dro )(iWCV
pLot ,,, the availabilit y of water which in turn mainly
not oral phenoineiton of rain, therefore, seasnaI
vturlatioi1 i ,aliilI great Iv affect the power output from a
livdropwer jt nt llie nuclear power plants are best suited for
areas %% 1 ,. ; It a rc''l..titiit from collieries and fuel costs are high The
amount 1 'aW I ateri,tl.- required for generating a given quantity
of ptlwe is the li'ti-t in a nuclear power plant.
ui India has the largest thorium reserves iii the world. It would
' 7 de ''lopnieiit of nuclear power. The other coLIltries having
significa:. t}ioriin reserve's are U.S.A. and Brazil. Although con-
VCi'siOiI of IcUlli into uranium 233 a fissionable form (if Uranium
Oil ceiliineicItu! sciil(' has not been achieved but it is hoped that it
would ii' 1 -.-.ihle to do so and thus ihirnitu would he siicci--sftjllv
used liii' pel. or geno-rotiun This will solve Indias JIOWP • prohlels
to great ext. ui 'I'he iuucc,'ar reactors using tlot'ium-ui-tinium cycle
are euoiiomue-.il tend have longer l;fc,'. The' utilization of' tliiii'jujii iii
fast breeder tors is qwte SuccOSslul. lle stetcin i e o'er stations
depend LI}IOIt el which is exli-iuistihl .Aceideng tee one'
the coa l i ''l It., in hiieli,e are e'.-t ituat ed tee 1)0 .il,c,te I .10()0() million
tomes exiltiling lignite', the lignite reserves being nearl y 2-1-12
million toiii • 5 The liveir, power plants depend upon water an
 SOURCES OF ENERGY
17

inexhaustible source. The water power potential of India is es-

timated to be the order of 40 x 10 3 MW.
(u) A hydro-electric power plant can be used for supplying peak
load. It can also be used as base load plant and in such case diesel
power plant or asteani power plant is used for supplying peak load.
Nuclear power plants are used as base load plants because they are
economical univ when used as such. Diesel power stations are of
limited generation capacity and are used for supplying peak loads.
The capacity of power plants using gas is also not large.
(v) In nuclear power plants the disposal of radio active wastes
is a costl y problem. In steam power stations there is nuisance of
smoke, ash etc. Hydro power plant is the cleanest way of producing
power.
(vi) A hydro-electric power plant can be easily started from cold
conditions as no warming up period is required.
(vii) Running cost of a hydro power plant is less than a steam
power plant. Running cost of nuclear power plant is perhaps least
of all.
(viii) The overall thermal efficiency of a diesel power plant is
higher than steam plant.
(ix) il y dro power plants are more reliable than steam power
plants and usually give some advance indication of loss of power,
hence their need be less spinning reserve.
(x) It takes about 4 to 8 years for planning and building of hydro
electro power plant whereas the time taken in case of nuclear power
station is 3 . 6 years and for steam power station 2-3 years. The hydro
power plant has long life whereas average life of nuclear power plant
is about 20 y ears and that of steam power plant about 20-25 years.
1.12 Sources of Energy in India
World wide rising prices of oil since 1973 have put considerable
strains oil economy and generated the urgent need to undertake
new efforts in research and technology to find new sources of energy.
On the other hand, the acceleration of the process of industrialisa-
tion and urbahisation following the determined effort ofthe develop-
ing countries like India to improve their economic well being is
inevitably leading to larger and larger demands ofelectrical energy.
The industriahIs((l Countries consume nearly 8O ofthe world's
total annual energy generation.
The per capita consumption of energy in India is quite low as
compared to U.S.A., Japan and other developed Countries.
 POWER PLANT
18

An American con s umes nearly as much energy in a single (lay

as two West Germans, or Australians, three Swedens or .Japanese.
six Yugoslay s, nine Mexicans or Cubans or 53 Indians.
Even this low demand of energy in our country has not been met
and power cuts have become common. Power shortage is mainly due
to the three separately identifiable factors as follows
(i) Due to inadequate installed capacity.
(ii) Arising from inability to get the best of installed capacity.
(iii) Inability to optimise the utilization of the available energy.
To improve electrical energy generation in our country cfThrt.s
should be made to increase the generation capacities instead of
increasing installed capacities. Our present generation ratio at
about 52% is one of the lowest in the world. In industrially advanced
countries the average generation ranges between 60% to 85% of
installed capacity while Austria, West. Germany and Switzerland
obtain 92%, 90% and 95% respectively.
The common sources of energy in our country are coal, water,
oil and gas. Nuclear power is recently being encouraged and has still
its teething troubles. Other non-coventiunal sources like wood,
hi-gas, solar, geothermal and tidal are still in experimental stage.
('oat is the most important commercial source ofenergy in India.
Thermal pow er generation call most reliable because coal reser-
ves i ll country are sufficient to last for some hundreds of years.
However the quality of coal produced must be improved. The Indian
il has on an average 18 to 22% ash content as compared to coal
available in Europe and U.S.A. which contains ash 8 to 10% whereas
coal available in Japan contains ash 6 to 8% Power generated
through super thermal power plants constructed near coal pitheads
cait be more economical.
The second important ource of power is water. But hydropower
has its shortcomings. Power generation from hydro-power plants
mainly depends upon the 'water reserve available' in the storage
reservoirs and face unavoidable constraints due to poor monsoon,
c y clic drought conditions and necessary withdrawal of water from
these i'eser'Oi' for irrigation purposes. A j udicious combination of
both hydro and thermal power is the optimum solution for proper
power generation. In India the development of pumped storage
plants is likely to assume importance in near future.
A vast hivdel potential exists in north-eastern region of our
country. The region's hvdel potential is estimated to hr about 21,000
MW which is nearl y 30% of the total hvdel potential (if the countr.
Arunachal alone pO5SC5SS 17,000 MW of hvdr''-potential cut of
''1000 MW otlivd'.'l potential of the re 101.
SOURCES OF ENERGY 19
Himachal Pradesh has vast hydel potential of about 20,000 riw
which is nearly one-fifth of our country's potential for generating
hydel power.
The reasonably assured uranium reserves in our country are
placed at about 34,000 tonnes of uranium oxide of which about
15,000 tonnes are considered to be economically exploitable. The
established uranium resources are capable of supporting natural
uranium reactors of about 8,000 MW of installed capacity. This
shows that the share of nuclear power generation in the total power
programme will remain modest. However when the fast breeder
reactor technology is fully developed and U-233 cycle used then
there will be a large increase in nuclear power as there exist an
estimated deposits of about 363,000 tonnes of thorium.
It is expected that in our country by the year 2000 A.D. the
installed capacity of nuclear power will be 10,000 MW. India has the
world's largest thorium reserves which will prove to be quite
economical fuel in fast breeder reactors.
By the end of century the share of nuclear power is expected io
increase from 2 to 10 per cent of total production. India at present
has five operation nuclear power stations two at Tarapur (near
Bombay) of 210 MWs capacity each, one unit of 220 MW of Kota in
Rajasthan and two at Kalpakkam (near Madras) of 235 MWs each.
In addition, two nuclear power stations of 235 MWs capacity
each at Narora (UP) and another two of 235 MW each at Kakrapar
(Gujarat) are in an advanced stage of construction. Work has com-
menced at Kaiga in Karnataka Rawabhatt in 'asthan on twin
units of 235 MWe. The second phase programme col... i instal-
lation of 500 MW capacity fast breeder reactor.
A nuclear power plant of 2000 MW capaciiy .0 units
each of 1000 MW will be installed at Koodankulan in il Nadu.
Nuclear power is playing a major role in meeting the increasing
electricity needs.
Bulk of power generation however would conic from the tradi-
tional sources of thermal and hydel.
Energy requirements are primarily met by coal, oil, hydropower,
gas and nuclear power. There has been a considerable increase in
production ol gas. The gas production was nearly 1.4 billion cubic
metre ill 1970-71 which rose to about 5 billion cubic metre in
1982-83. The main thrust of the nuclear power pr-ogramlne in the
sixth plan is to coutinue with the development of the natural
tirarliurn based ll\VR power plants and P1.11-SLIC of!? and activities
I at rig to t I W devel opine i it of F BR technology. Need for in ak i rig a ii
urgent :111(1 all out scientific efforts to promote the developriient of
 POWER PLANT
20
utilisation of solar and other forms of renewable energy to reduce
our dependence on fossil fuels (coal, oil, hydro) and to help safeguard
environment is essential.
The various aims of research and development (R and D) ac-
tivities in the energy sector are as follows
(i) To develop new techniques for the exploration of energy
resources.
(ii) To develop technologies to maximise energy production
and recovery.
(iii) Improvement and adaptation ofexisting technologies with
emphasis on conservation.
(iv) Development of technologies for harnessing alternative
sources of energy.
(u) To develop new and more efficient technologies for utilis-
ing indigenous onergy resources.
The commission for Additional sources of energy (CASE) under
the central Government is responsible for formulating policies and
programm es for the development of new and renewable sources of
energy.
The department of non-conventional Energy Sources is respon-
sible for activities initiated based oil recommendation of CASE
covering areas of bio-gas development, solar energy, wind energy,
energy from bio-mass etc.
Our country has embarked on an extremely ambitious
programme of oil exploration. Over 30% of the country's total
petroleum product consumption is in the from of diesel both in
stationary and vehicular engines. The total production of crude oil
in our country during 1984-85 i expected to be 29.7 million tonnes
against the requirement of about 45 million tonnes.
Natural gas is fast merging as all substitute for oil
and as an i mportant future energy source. India has vast reserves
of natural gas estimated at over one hundred billion cubic metres.
Natural gas is prOdUCe(l in Assam, Tripura, Gujarat and Western
offshore. Reserves of natural gas have also been established in the
K.G. Basin and Caveri Basin.
In view of the mounting problem of power shortages and the
economies available from using gas-based thermal power plants in
meeting power requirements the importance of gas-based power
generation acquires greater relevance.
Nat rid gas although a new-corner oil energy scene will play
a significant role in future power generation. Natural gas can easily
be used in place of liquid or solid energy source as fuel. Natural gas
is a premium sourCe of energy for domestic and commercial sector
SOURCES OF ENERGY 21

where the user is benefitted from assured uninterrupted supply and

the environmental compatibility.
Natural gas does not only provides the option of replacing liquid
petroleum in various sectors but offers great potential of conserva-
tion of energy in these applications due to its inherent properties
and also on account of up dated technology and equipment available
for utilisation of natural gas for achieving higher efficiency.
For India, solar and wind energy and use of bio-mass fuels are
generally agreed to be most relevant, as the potential of geothermal
energy and tidal energy etc. are rather limited. Development of wind
power offers a promoting prospect in meeting the energy require-
ments. India with its long coastline, blessed with steady winds
almost throughout the year has tremendous potentialities of wind
power which can be harnessed with immense benefits. Many in-
dustrialised countries like USA, UK, Japan and USSR have
gathered vast experience in designing and commercial operation of
wind power stations.
Utilization of solar energy is of great importance to India since
it lies in a temperature climate of region of world where sunlight is
abundant for a major part of the year. With the availability of
improved materials, manufacturing techniques and better heat
collectors in the near future solar power may become economically
feasible.
Generation of electricity using photo voltaic (PV) cells should b
exploited co in merci ally. The PV modules serve as an efficient source
of power supply to remove locations. These can also be used success-
fully by hospitals, schools, offices, railway stations, airports and
factories.
The thrust of the research efforts in our country in the solar
energy has bee,ji directed both towards solar thermal applications
and direct con ervation of solar electricity. Utilisation of solar
energy is of great importance to the country as it lies in tropical
climatic region where sunlight is abundant for a major part of the
year.
Under an action plan formulated by the Ministr y of Non-con-
ventional Sources of Energy, the targets for power generation from
non-conventional energy sources have been upgraded to 2,000 mega
watts from 600 mega watts for Eighth Plan.
The most promising and fast moving solar technology today is
that of solar cells, flat metallic blue chips made of highly pure silicon
that can convert sunlight into electricity. Though silicon is available
in abundance in every country of the world, the process of refining
silicon to more than 99.9% is extremely complex and the cost

POWER PLANT
22
extremely high. Researchers in leading laboratories and univer-
sities have achieved self-reliance in this vital technology area which
is essentially based the traditional and mature semiconductor
process industry.
These solar photovoltaic cells are being used in rural areas and
isolated locations for a variety of applications such as water pump-
ing for micro irrigation and drinking water supply, community and
street lighting, power sopplies for microwave repeater station,
communications equipment, radio and television receivers. A
programme is on for supply of one million solar lanterns and 50,000
deep well solar pumps in villages.
Table 1.1 (A) indicates expected power demand. To meet the
growing power demand emphasis will have to be laid on new and
Non-conventional methods of generation of electrical energy.

Table 1.2A
Electrical Energy Generation

Year _J__. Generation in Billion k

1970-71 61.2
1193

1987-83 217.2
1999-2000
Projected Dernand- 424/465

The use of bio-gas for lighting and irrigation opens up new

possibilities for self-contained rural communities. In the long run
our energy economy would have to be built around land based
biomass fuels and plentiful sun shine which we receive virtually
throughout the year.
In a bid to provi4e a simple and cheap method of energy
production, the scientists have developed a nuiber of designs of
biogas plants which make use of wide range of agricultural wastes
like animal dung, human excreta, vegetable waster, water hyacinth
and produce fuel in the form of gas, simultaneously with high quality
manure.
The installed capacity in power sector in India is indicated in
Table 1.3.
Table 1.3

7otal

I 1951 575 1.261 - 1.836

1,917 4,653

SOURCES OF ENERGY 23
0 ,3837,906 - ì 420 14.709
1978-79 13.000 I 19,5501020 33,570
L?34 20,000 31,000 1900 53,000 -

To improve power generation the generation capacity of power

plants must be increased. Following fictors help in improving fill'
generating capacit y of thermal power station.
(i) Adequate suppl y ofa reasonable good and uniform quality
of coal. The coal not onl y has ash but other extra neous
materials like stones, silica, alumina and shale which
should be removed before hand.
(ii) Improved maintenance of power equipment.
(iii) Use of better operation and maintenance techniques.
(iv) Better spare part management.
(v) Reduction of transmission and distribution losses,
(vi) Prevention of theft and pilferages of electrical energy.
B y the end of Sixth Five Year Plan (1880-1885) India would have
a total installed power capacity of about 50,700 MW with actual
power generation around 33 million kW on account of a very low
generating capacit y of about 5217(.
The installed capacity will be nearly 78,000 MW by 1990.
In view ofrapid increase in the cost ofenergy, the utilities should
make efforts to focus attention of the consumers towards energy
conservation. Particular attention should he paid to installation of'
LT capacitors which reduce losses along distribution lines.
A better balance between hydel and thermal power is required
This will reduce the cost of electrical power. The hydel thermal mix
was 40 : 60 at the end of fifth plant, was 33,7 66.3 at the end of'
sixth plan and is likely to be 30.7 69.3 at the end of seventh plan.
Without, adequate hydel back up the overall cost of' meet i rig the
power demand will be expensive. It is, therefore, necessar y to take
up corrective measures during the course of seventh and successive
plans towards a better hvdel thermal balance.
Small hvdel units should be installed because they can provide
economic power supply to rural and remote areas in a decentral ised
manner. The potential of these units is said to be about SOOt) MW
out of'whieh the installed capacity is only I GO M\V. -
.
B % generating electrical energ y from non-conventional suurces
like inch, solar energy , and bio-gas it would become nu,n'h more
('aS:e r to meet the energy requirement of people living in rural and
f)r flcmoj areas in 0 decentralised manner. India has vast potential
to harm'ss ener g y from these sources,
.'\ f).%, 'xp'rimemital solar power plants in the range ci iric' to 5
k\V have been in.t.th!ed in a number of villages mu Andhra l'radi'.',hi,
 24
POWER PLANT
(ion, Karnataka, Taniilnadu, Uttar Pradesh, Lakshadweep and
Tripura. Two larger plants of 20 to 25 kW are tinder installation in
Haryana and Orissa.
The total potential of wind resources is estin l atecl to be about
20,000 MW for power generation. At present six wind power
generating units with an aggregate capacity of about 6 MW have
been set up in Gujarat, Maharashtra, Tamil Nadu and Orissa.
Following measures can help in meeting the increasing power
demand
(1) Conventional sources of ener f,y are being consumed at
faster rate. These fuels must he saved for commercial
applications. Therefore harnessing of renewable sources
of energy like solar energy wind energy bio-gns, ocean
energy must be encouraged continuously oil long term
basis. They are non-polluting and well suited for
decentralised use.
(ii) Sustained efforts should be made throughystem s im-
provement to reduce losses and to improve transmission
efficiency.
(iii) Renovation and inodernistjorj schemes in existing power
stations should be thought ofas a continuing process.
(iv) On going power generation schemes should be comrnis-
sioned at a faster rate.
(v) To maintain the liydel thermal mix at the desired level of
40 60 more hydro projects should be taken up. Although
it takes more time to build hydro power plants and their
initial installation cost is more but the operational cost of
hydro-power plants is much less than thermal power
plants.
(vi) Energy conservation should be achieved by improving
plant load factor and by reducing power transmission and
distribution losses.
(vii) National grid should be set up. This will help in proper
distribution of load among various power plants.
(viii) The power plants should be run at as high toad factors is
possible.
In India, power demand has been assessed at about 48 OW arid
62 GW for the Eighth and Ninth Plan respectively. Eighth five year
plan is expected to provide additional 38000 MW power. Strengthen-
ing of transmission net work is therefore essential so that benefits
can be reaped optimally from targetted additional power generation
during 1990-2000 A.D. Effective transmission systems are required
to deliver power from regional projects to user states and to ex-
SOURCES OF ENERGY 25

change power between inter connected systems. Use of UHV and

EHV lines should be introduced.
Out country can not depend on using fossil fuels to large extent
in the century to come. Conservation of petroleum products is the
subject deserving the highest priority. The oil exploration effort,
both offshore and on shore—be intensified. Non-conventional sour-
ces of energy such as wind energy, solar energy, electricity genera-
tion from use of agricultural wastes etc. should be continuously
explored and exploited. This can help our country in becoming self
reliant in power generation.
Increasing energy demand, depleting fuel resources and grow-
ing environmental pollution have led to the development of alter-
nate energy sources. These include several renewable sources such
as solar, wind, hydro, etc. as well as depletable energy sources as
geothermal and synthetic fuels. The need of the hour is to exploit
the non-depleting sources of energy that are environmentally ac-
ceptable.
Electrical energy is a resource that has been in short supply. It
is a yard stick measuring Industrial development as also quality of
life. Currently the total energy spply picture has been dominated
by conventional 'fossil' fuels with contribution also from hydropower
and nuclear sources. Other sources, termed as 'non-conventional' or
renewable are not yet developed to any great degree and therefore
cannot be compared to conventional sources. One major hurdle in
achieving commercial status in these renewable sources is that they
are widely distributed and relatively diffused. Supplies are essen-
tially limitless. Renewable sources however Fold much promise in
an energy-starved world. Present trends have shown that capital
Costs for renewable energy projects are decreasing and its reliability
is increasing. With increasing debate on the balance of the environ-
ment and energy economics, it is clear now that the potential and
contribution of renewable energy sources will be it key factor in
future developmental issues.
In addition to the conventional hydro source, the renewable
energy sources include, inter alia, the following areas of interest
solar, wind, geothermal, biomass ocean and hydrogen. In our
country some of these such as solar, wind and biomass have seen
some development. But these efforts have been sporadic and have
not yet been integrated into the mainstream.
It is desirable
(a) that the conventional energy sources, particularly the
fossil sources are exhaustible and are to be consumed at
regulated level only;
—4
26 POWER PLANT

(b) that renewable energy sources have to be relied upon to

the maximum extent possible;
(c) that the subject of energy from non-conventional energy
sources should be given utmost priority on commercial
basis and for captive needs;
(d) that all energy generation in future shall be environment
pollution free.
In our country the peak load demand will be about 86000 MW
by the end of 8th plan period. To meet this demand matching
Ldditional generating capacity should be installed. Present peak
level generation of about 65% of the installed capacity should be
improved to about 75%.
Other areas which require attention are as follows
(i) reduction of transmission losses
(ii) power factor improvement;
(iii) use of equipment with higher mechanical and electrical
efficiency.
1.12.1 Conservation of Energy
Due to prohibitive cost of creating additional sources for power
generation countries like India have a particular need to conserve
energy. Some of the methods by which electrical energy can be
conserved are as follows:
(i) Minimisation of transmission and distribution losses.
(ii) Minimisation of coal burnt in thermal power plant.
(iii) Use of high efficiency motors.
(iv) Optical reactive power scheduling.
(v) Efficient energy system planning.
(vi) Improving power plant load factor by better operation and
maintenance procedures.
(vii) Optimum utilisation of installed capacity. Conservation of
electrical energy remains in the circumstances the most
viable option because it is cheaper in terms of investment
and also because the gestation period of conservation
measures is short and results quicker.
In our country the energy conservation has been identified as
priority area of activity in power generation for the eighth plan
period (1990-95) and beyond.
Conservation of electrical energy helps to maintain demand-
supply equilibrium. There is huge waste of electrical energy due to
low power factor and improper distribution net work. Proper ways
and means should be used to reduce transmission and distribution
losses. So far petroleum products are concerned out of total con-
sumption of petroleum products in our country about 30% is
produced indigenously and balance 70. is imported from other
SOURCES OF ENERGY 27

petroleum producing countries. The world is mostly depending on

petroleum fuel for its energy requirements. Fossil fuels are getting
depleted at fast rate. This requires that alternate, new reliable and
eco-friendly sources of energy should be found which can help in
melting the injreasing electrical energy demand.
Economic development is not possible without electric power
which used to he a luxury earlier but is now part and parcel of
common man's life.
In our country per capita energy consumption is only 300 units
against nearly 3000 units abroad. Although electric energy con-
sumption of common man is increasing but we are unable to meet
out due to resources crunch.
Our country has vast hydro energy potential yet we have not
been able to tap the same. Inexhaustible Hydroenergy if properly
utilised can save other sources of energy such as coal and petroleum
products.
Biomass—a major clean and convenient source of energy when
converted into modern energy carriers like fluid fuels and
electricity—has given new hopes to scientists all over the world to
sustain the ultimate, irreducible essence of the universe. Defined as
all organic matters except fossil fuels, biomass includes all crop and
forest products, animal matters, microbial cell mass, residues and
by products that are renewable. It serves as food, feed, fibre, bed-
ding, structural material, soil organic matter and fuel.
Biomass comes in as a potential source of energy to meet this
ever-increasing demand. Bio-energy plantations on rural degraded
land can make local population self-sufficient in their energy re-
quirements. Domestic, agricultural and industrial energy needs can
also be met through this decentraliséd power generating system.
In fact, the third world is already deriving 43% of its energy from
biomass and over two billion people are almost totally reliant on
biomass fuels for their energy needs. The dispersed rural 70% of the
world population are also sustaining on biomass. If all families in
developing countries that are now using biomass fuels were to
change to kerosene, the third world's demand for oil would rise by
about 20%.
At present biomass-energy production is associated with
agricultural and forestry activities. In industrial countries, a major
part of the bio-energy component is produced from residues from
papc nd timber industries. There is also a growing realisation that
those countries with large reserves of agricultural land can utilise
this resource to produce energy, thus cutting oil import require-
ments and also reducing crop surpluses if necessary. Large-scale

28 POWER PLANT

ethanol programmes using maize in the USA and sugarcane in

Brazil have deniostrated that substantial quantities of liquid fuel
can be generated from biomass.
Many technologies exist for converting biomass to heat energy
through direct combustion, or to liquid or gaseous fuels through
thermo-chemical, extraction or biological processes. The sugar and
starch, cellulose, lignin and other constituents of biomass may be
burnt directly to produce heat or mechanical work in an external
combustion engine.
To cope up the shortage in electric power generation, it is ob-
served that the MHD system is a promising feature for developing,
countries like India.
1.13 Combustion of Fuels
It deals with various reactions taking place concerning different
elements which constitute the fuel.The combustion of fuels may be
defined as a chemical combination of oxygen in the atmospheric air
and hydro-carbons. It is usually expressed both
(i) qualitatively and (ii) quantitatively
by equations known as chemical equations which indicate the na-
ture of chemical reactions taking place.
Adequate supply of oxygen is very essential for the complete
combustion of a fuel in order to obtain maximum amount of heat
from a fuel.
Combustion of fuels is accomplished by mixing fuel and air at
elevated temperature. The combustion process may be simply ex-
pressed as follows:
Fuel + Air = Products of combustion + Heat liberated
T he oxygen contained in the air unites chemically with carbon,
hydrogen and other elements in fuel to produce heat. The amount
of heat liberated during the conibustion process depends on the
amount of oxidation of the constituents of fuel and the nature of fuel
(chemical composition of fuel).
In order that the combustion of fuel may take place with high
efficiency, the following conditions must be fulfilled:
1. The amount of air supplied should be such that it is
sufficient to burn the fuel completely. Complete combus-
tion of fuel means complete oxidation of all the combus-
tible material in the fuel. A deficiency of air causes
incomplete combustion of fuel which results in consider-
able unburnt fuel being discharged from the furnace
whereas too much supply of air simply dilutes the gases
and cools the furnace.


SOURCES OF ENERGY 29
2. The air and fuel should be thoroughly mixed so that each
combustible particle comes in intimate contact with the
oxygen contained in the air.
3. The fuel should remain in the furnace for sufficient time
till it get burnt completely.
4. The temperature in the furnace should be high enough to
ignite the incoming air fuel mixture.
1.14 Products of Combustion
The complete combustion of fuel produces varicus gases such as
carbon dioxide (CO2), sulphur dioxide (SO 2 ), water vapour nitrogen
(Ni ) and oxygen (Oz). Nitrogen comes from air supply and oxygen
from excess air. Water vapour is produced from the following three
sources
(i) Moisture originally contained in the coal
(ii) Vapour produced by combustion of hydrogen
(iii) The water vapour of atmospheric humidity.
If all the carbon present in the fuel does not get burn completely
then carbon monoxide (C) •.mced. The flue gases will have
considerable amount of carbon monoxide in them if the oxygen
supply is less However, large excess of air would mean that a large
amount of sensible heat wu!d in flue gases. Analysis of flue
gases give a correct idea of how the fuel is burning.
1.15 Combustion Chemistry
The combustion process involves chemical reactions. The com-
bustible elements in fuels consists of carbon, hydrogen and sulphur.
The chemical equations represent the combustion of C, H2, S,
CH4 , etc., are described as follows:
(i) Combustion of Carbon
C + 02 4 CO2
Substituting the values of m&ecular weight in equation.
12 + 16 x 2 = 12 + 16 x 2
12 + 32 = 12 + 32
8 8
1 1+—
8 11
+
I=

This means that 1 kg of carbon requires 8/3 kg of oxygen for its

complete combustion and produces 1113 kg of carbon dioxide.

30 POWER PLANT

If the amount of oxygen supplied is not sufficient the combustion

of carbon is incomplete and the product of combustion will be carbon
monoxide.
2C + Oz - 2C0
2x 12+ 16x2=2(12+ 16)
24 + 32 = 56
4
1+= 7
which means that 1 kg of carbon needs 4/3 kg of oxygen to produce
7/3 kg of carbon monoxide. Further burning of CO produces CO2.
2C0 + 02 -* 2CO2
2(12+ 16)+ 16x2-2(12+32)
56+32 -+ 88
11
1---
+4
This means that 1 kg of CO needs 4/7 kg of oxygen and produces
1/7 kg of CO2.
i ii) Combustion of Hydrogen. Burning of hydrogen with
oxygen produces water vapours,
21-12 +02 --4 2H20
2(1+ 1)+16X2-2(2+16)
4+32-436
1 +8-9
This means that 1 kg of hydrogen combins with 8 kg of oxygen
k
to produce 9 of water.
(iii) Combustion of Sulphur. When sulphur burns with
oxven it produces sulphur dioxide.
S +02 SO2
32+16x2-432+ 16x2
32 ± 32 - 32 + 32 -+ 64
1 kg of sulphur + 1 kg of oxygen - 2 kg of sulphur dioxide
(iv) Combustion of Methane (Clii)
CH. + 202 -4 CO2 + 21120
(12 4) + 2x 32 - (12 + 32) *2 x 18


SOURCES OF ENERGY 31

16 + 64 - 44 + 36
11 9

1 kg methane.+ 4 kg oxygen -4 1114 kg.

Carbon dioxide + 9/4 kg of water.
The various values are summarised in Table 1.4.

Table 1.4.
Substance (Oxygen Products of Combustion (kg).
(I kg) reqd. kg.)
CO CO2 1120 SO2
C 8/3 - 11/3 -
CO 4/7 - 11/7 -.
H2 8 - -. 9..
S 1 -•. .- 2.
CH 4 4 . - 11/4 . 9/4 .

1.16 Combustion of Gaseous Fue!s.

Gaseous fuels are usually measured by volume (in cubic metrés)
The various chemical equations are described as follows:
U) Combustion of Hydrogen
2H2 +02 -+2H 20 •.
2 vol. + 1 vol. -* 2 vol.

1 cu-metre + cu-metre -41 cu-metre. rlIjS means that one

cu-metre of hydrogen requires cu-metre of oxygen to - produce one

cu-metre of water. . . . . . . ..
(ii) Combustion of Methane. When meLhanëMarsh Gas)
burns with oxygen it gives CO 2. . . ..
CH4+202-21-120 +CO2
1 vol. + 2 vol. - 2 vol. +1 vol. .. .

1 cu-metre + cu-metre - 2 cu-metre + 1 cubic-metre. .

Thus 1 cu-metre of methane needs 2 cu-metre of oxygen for

combustion and produces 2 cu-metre of water and .1 cu-metre of

32 POWER PLANT

(iii) Combustion of CO. When carbon monoxide bui as in

oxygen, it gives CO2.
2C0+02-*2CO2
2 vol. + 1 vol. -4 2 vol.
1 cu-metre + cu-metre - 1 cubic metre.

Thus one cu-metre of CO needs cu-metre of oxygen to produce

one cu-metre of CO2.
(iv) Combustion of Ethylene ( C 2114). When C211 4 burns in
oxygen it gives CO 2 and H20
C 2 H 4 + 302 -, 2CO2 + 21-120
1 vol. + 3 vol. —* + 2 vol. + 2 vol.
1 cu-metre + 3 cu-metre -4 2 cu-metre + 2 cu-metre.
• Thus 1 cu-metre of ethylene combines with 3 cu-metre of oxygen
to produce 2 cu-metre of CO 2 and 2 cu-metre of water vapours.
• The combustion products of various gaseous fuels are sum-
marised in Table 1.5.
Table 1.5

Gas (1 cu-metre) Oxygen reqd. (cu-

metre)
Products of combustion (cu-metre) 7
112 I 1 I — I

Cu 4 2 1 2

Co 1 -
2
C2IL. 3

1.17 Weight of Air Required for Complete Combustion of

Fuel
The weight of air required for the complete combustion of fuel
is calculated from the analysis of fuel. To calculate the amount of
air required for complete combustion of fuel, firstly the oxygen
required for burning each of the constituent fuel is calculated and
then the air-required is found out. The atmospheric air consists of
oxygen, nitrogen and small amount of carbon dioxide, and other
gases such as neon, argon, krypton etc. For calculating the air
 SOURCES OF ENERGY 33
required for burning a fuel, the following composition of air can be
taken:
By weight: Oxygen = 23%
Nitrogen = 77%
By volume: Oxygen = 21%
Nitrogen = 79%
If it is found that the fuel already contains some amount of
oxygen, then it should be deducted from the' calculated value of
oxygen.
1.18 Coal Selection
While selecting coal for steam power plant the following proper-
ties should be considered:
1. Size and Grade. The size and grade of coal will determine the
type of equipment to be used for burning the coal.
2. Heating Value. The coal selected should have high heating
value (calorific value).
3. Contents of moisture, volatile matter, fixed carbon ash and
sulphur. The slagging characteristics depend on ash temperature
and corrosion characteristics depend on sulphur contents.
4. Coking and caking tendency of-coal that is retention of
original shape during combustion VS softening.
5. Physical properties such as resistance to degradation and size
consistency.
6. Various constituents indicated by approximate the ultimate
analysis.
7.Grindability i.e. the ease with which a coal can be pulverised.
Grindability index is expressed by a number.
1.18.1 Ranking and grading of coal
According to ASME and ASTM,
(i) Higher ranking of coal is done on the basis of fixed carbon
percentage (dry basis).
(ii) Lower ranking is done on the heating value on the moist
basis.
For example a coal having 6% C and a calorific value of 5000
kcal/kg is ranked as (60-500) rank.
Grading of coal is done on the following basis:
(i) Size
(ii) Heating value
(iii) Ash content
(iv)Ash softening temperature


POWER PLANT
34

(t') Sulphur content.

For example a coal of grade written as 6-10 cm, 500-A8-F24 S
1.7 means
(a) coal has a size of 5-10 cm
(b) coal heating value is 5000 k-cal/kg
(c) coal has ash content 8 to 10%
(d) ash softening temp is 2400°F
(e) sulphur content ofcoal is 1.7%.
119 Composition of Solid Fuels
The various constituents of solid fuels are carbon, hydrogen,
oxygen, sulphur, nitrogen and mineral matter.
Following methods of analysis are used to determine the coin-
position of coal
1. Ultimate analysis. 2. Proximate analysis.
1. Ultimate Analysis. The, analysis is used to express in per-
centage by weight of carbon, hydrogen, nitrogen, sulphur, oxygen
and dsli and their sum is taken as 100%. Moisture is expressed
separately. This analysis enables to find the amount of air required
for the combustion of 1 kg of coal and to calculate the heating value
of coal.
The ultimate analysis of most of the coals indicates the following
ranges of various constituents.
Constituents: C H2 02 S N2 Ash
Percentage: 50-90% 2-5.5% 2-40% 0.5-3% 0.5-7% 2-30%
2. Proximate Analysis. This analysis is used to determine the
following components:
(i) Moisture
(ii) Volatile matter (carbon combined with hydrogen and
other gases that are driven off on heating).
(iii) Ash
(iv) Fixed carbon
They are expressed percentage by weight and their sum is taken
as 100%. Sulphur is expressed separately.
To find the volatile matter I gm of finely divided coal free from
moisture is heated in a crucible for 7 minutes to about 950 ± 20'C.
The crucible is then cooled and the difference in weight indicates
the amount of volatile matter. The sample is then burnt in an open
pan so that it gets burnt completely. The amount of residue left
behind is ash. Weight of original sample minus weight of moisture,
volatile matter and ash gives the weight of fixed carbon.
Ultimate analysis and proximate analysis are expressed in
terms of:
 SOURCES OF ENERGY 35
(i) Coal 'as received' or 'as fired': Coal 'as fired' is in the same
conditions as it comes out of the bunkers.
(ii) Coal 'moisture free' or 'dry'.
(iii) Coal 'moisture and ash free' or 'combustible'.
The ultimate.analysis of coal is a more precise test to Ad the
chemical composition of coal whereas proximate analysis of coal
gives good indication about heating and burning properties of coal.
The proximate analysis of most of the coal indicates the follow-
inganges of various constituents.

Constituents: I Fixed C M •'iture

1.19.1 Ash
• Ash is the combustion product of mineral matters presents in
the coal. It comprises mainly of silica (Si0 2 ), alumina (Al 203) and
ferric oxide with varying amounts of other oxides such as CaO, MgO,
NaO etc. High ash content in coai is undesirable in general.
A coal with high ash content has following properties:
(i) is harder and stronger
(ii) has lower calorific value
(iii) produces more slag (impurities) in the blast furnace when
coke made out of it is used therein.
Ash content of the coal is reduced by its washing.
1.19.2 Volatile Matter
Certain gases like CO, CO 2 , CH 4 H2, N2 , 02, hydrocarbons etc.
are present in the coal which conies out during its heating These
are called the volatile matter of the coal.
The coal with higher volatile matter content has following
properties
(i) ignites easily i.e. it has lower ignition temperature
(ii) burns with long smoky yellow flame
(iii) has lower calorific value
(iv) will give more quantity of coke oven gas when it is heated
in absence of air
Cu) will require larger furnace volume for its combustion
(ui) has a higher tendency ofcatching fire (due to low tempera-
ture exothermic oidation) when stored in open space.
1.20 Weight of Excess Air Supplied
The weight of excess air required during combustion of coal is
calculated from the weight ofunused oxygen in flue gases after CO
is present in flue gases is burnt to CO2.
 POWER PLANT
36

Let W 1 = Weight of flue gases per kg of fuel

W 2 = Percentage weight of oxygen present in flue gases

W3 = Percentage weight of CO present in flue gases

As 1 kg of CO need 4/7 kg of oxygen. to burn to CO2.
Oxygen required to burn W3 kg of CO to CO2.
• W3 4
= x =W 4 (say)
1 7
W2
Excess Oxygen (W5) =- W4
Weight of excess oxygen per kg of fuel (W6)
= Weight of excess oxygen per kg flue gas x Weight of flue gas
per kg of fuel
W6 = W5 X W1

Therefore, the weight of excess air supplied

x 100
= w6
Proper control of the right amount of excess air maintains
optimum combustion efficiency. CO 2 and 02 in combustion gases are
index of excess air. Air feed should be controlled so that optimum
amount of COz or 02 is produced. Smoke formation and slagging of
boiler surfaces also play an important part in determining the
optimum excess air. Practical conditions of fuel type furnace arran-
gements and heat transfer arrangement determine the total amount
of air needed for complete combustion.

COKE
ANTHRACITE
(02

I 08ITJMINOUS

NATURAL GAS -
0 50 100 150
- EXCESS AIR(/.

Fig. 1.3
The amount of CO2 in flue gases depends on type-of fuel and
excess air supplied to the furnace. Fig. 1.3 shows typical variation


SOURCES OF ENERGY
37
Of CO2 in flue gases (per cent by volume) and excess air (percent) for
complete combustion of various types of fuels.
The total amount of air needed for complete combustion of a fuel
depends on following factors:
(i) Type of fuel.
(ii) Furnace arrangements.
(iii) Heat transfer surface arrangement.
Typical values of excess air supplied, expressed as percentage
of the quantity theoretically required are as follows:
Hand fired boiler furnace : 100
Mechanically stokered furnace : 40
Petrol engine : 20
Oil engine : 20
1.21 Requirements of Fuel
A fuel should possess the following requirements:
(i) Calorific value. The fuel selected should have high calorific
value.
(ii) Price. It should be cheap. -
(iii) Operating efficiency. The fuel should burn n •ffectively.
It should produce minimum amount of dust, smoke, slagging and
clinkering. In case of coal a careful study should he de about
volatile matter, ash, sulphur, moisture, ash fusion te mperature ash
analysis and grinding and coking characteristics
(iv) Refuse disposal. The fuel should produce minimum ash on
burning. In general oil and gas, produce ash in very small quantities
and do not present any refuse disposal problem -whereas . coal
Produces sufficient amount of ash and, th.erefare, ash disposal
equipment is required where coal is used as fuel.
(u) Handling cost. The handling cost shoulcibe minimum. Han-
dling cost of coal at power station is maximum and gas requires
minimum handling cost whereas handling cost ofoil is intermediate.
(vi) Operating labour cost. The operating labour cost is maxi-
mum in coal fired plants whereas it is minimum where gas is used
as fuel.
1.22 Principal Stages of Combustion
The combustion of fuel is a complicated physical and chemical
process in which the combustible elements of the fuel combine with
the oxygen of air witH the evolution of heat attended by a sharp rise
in temperature and formation of flame. During the burning of any
fuel two stages are observed
(i) Ignition (ii) oxnbustion.
 POWER PLANT
38

Ignition is the period during which the fuel is gradually raised

in temperature. On attaining a definite temperature the fuel is
ignited and stable combustion sets in. When solid fuel is introduced
into the furnace, moisture is first removed and the volatiles begin
to be liberated. The resultant gaseous products of the fuel decom-
position are gradually heated to the ignition point and burn in a
flame over the solid part of the fuel. Combustion of' the gaseous
substances, heats the coke which begins to burn stably when the
ignition points is reached. At this stage maximum temperature is
generated. Burning down is the final stage in the combustion of solid
fuel. Gasification and the combustion of solid elements are com-
pleted is this stage and enough heat is generated to maintain
combustion at a sufficiently high temperature. Liquid fuel should
first atornised to increase its area of contact with the air. An
dditional phase in the firing of liquid fuel is evaporation. A drop of
fuel ii entering the spray of burning fuel is heated and gradually
reaches the temperature at which the fuel components begin to
evaporate. The gases formed burn in the oxygen of air and increase
the temperature of drop. At a definite temperature the molecules of
the drop begin to disintegrate. The drop begins to burn at its surface.
The ox y gen penetrates inside where partial combustion takes place
and the gases formed inside the drop firther intensity the combus-
tion.
During different stages of combustion Qf. fuel the requiI
tity of air should be supplied. Complete and incomplete corn
of fuel take place depending upon the quantity of air supplieL
1.23 Complete Combustion
It is process in which the combustible elements of fuel combine
chemically with the oxygen of air at a definite temperature.
The flue gases produced consist of CO 2, SO2 , water vapour
(1120), oxygen (02) and nitrogen (N2).
1.24 Incomplete Combustion
\ deficiency in air supplied causes in complete combustion of
luet which results in considerable unburnt fuel being discharged
from the furnace along with ash and slag. The presence of carbon
monoxide gas (CO) in the combustion products indicates in complete
combustion. When liquid or gaseous fuel is fired incomplete combus-
..n is accompanied by soot formation.
1.24.1 Weight of Carbon in flue gases
The weight of carbon contained in 1 kg of flue gases can be
calculated from the amounts of CO2 and CO present in it.
During the complete combustion of C to CO2
 SOURCES OF ENERGY
39
C + 0 2 = CO2
12+16x2=12+16x2
12 +32 = 44
3244
+ 12 12
8 -11
1+33

Thus 1 kg of carbon on combustion produces 11/3 kg of CO2.

Hence 1 kg of CO 2 will contain 3/11 kg of carbon.
Now when carbon burns in insufficient supply of oxygen then
the combustion of carbon is incomplete and products of combustion
will be carbon monoxide.
2CO2=2CO
2x12 +16X2=2(1 9 f 16)
24 + 32 = 56 or 1 +- =
24 24
47

Thus 1 kg of carbon produces 7/3 kg of CO. Hence 1 kg of CO

contains 3/7 kg of carbon.

W i weight of carbon in 1 kg of flue gases

= 11 CO2 + 7CO
1.24.2 Weight of flue gas per kg of fuel burnt. The actual
weight of dry flue gases can be obtained by comparing the weight of
carbon present in flue gases with the weight of carbon in the fuel.
Let, 1V2 = weight of carbon in 1 kg of fuel
W = Weight of flue gas per kg of fuel burt
= w21w1
where W1 = Weight of carbon in 1 kg of flue gases.

1.25 Conditions for Proper Burning (Combustion) of Fuel

The v ariousconditioris that should be established for proper
burning of fuel are as follows.:
(i) Corret mixing and ratio of fuel and air
(ii) Ehough time to burn the fuel compltelv
(mu) High temperature flame


POWER PLANT
40

(iv) Turbulent mixing of fuel and air

(v) Proper proportioning of furnace dimensions.
1.26 Temperature of Fuel Combustion
The combustion of fuel is always accompanied by heat losses.
Therefore, the real temperature or actual temperature of combus-
tion is always lower than the theoretical temperature of combustion
which is obtained in ideal cases without heat losses. Table 1.5 shows
the theoretical temperature of combustion for various fuels in de-
gree centigrade.
Table 1.6

offue Exces AIR coefficient

LU 1.3 1.5 2.0
.\nthracito 2270 1845 1665 1300
1.ign_ 1875 1590 1425 1150
1700 1510 1370 1110
LPeat
1265
I Fuel oil 2125 1740 1580
'"37
Jt

Example I.I. The percentage composition of a sample of coal

was found to be as follows:
C = 85%; H2 = 3%;
02 = 2%; Ash = 10%.
Determine the minimum weight of air required for the complete
combustion of one kg of coal.
Solution.
Substanc.' Weight per kg of Oxyg required Oxygen per kg of fuel
uel_ per kg substance
C 0.85 0.85 x = 2.266

0.03 8 0.03x8=0.24
H2
- 002

Total oxygen required = 2.266 + 0.24 - 0.02 = 2.486 kg.

Weight of air required for complete combustion of 1 kg of fuel
100 10.8 kg. Ans.
= 2.486 x
Example 1.2. The percentage composition by weight of ci sample
of coal; was found to be as follows:

SOURCES OF ENERGY 41
C=24%;H2=5%,
02= 8%; Ash = 63%.
It was also observed that the dry flue gas had the following
composition by volume:
CO2 = 10%;C0=2%;
02 = 13%;N2 = 75%.
Determine the following:
(a) Minimum weight of air required for complete combustion
oil kg of coal:
(b) Weight of excess air required per kg of coal.
Solution. (a)
Substance Weight per kg of coal j O.'.ygen required per I 02 reqd. per kg of
C
I 0.24 8 0.64

Ash i

Total oxygen required = 0.64 + 0.40 - 0.80 = 0.96 kg.

Minimum weight of air required per kg of coal
= 0.096 x 100 = 4.18 kg. Ans.
(b) The composition of dry flue ga given b y volume. It can
be converted into composition by weight as follows:

Gas Volume per m 2 of Molecular weight

I Proportional Weight per kg of
flue gas (V) (M,) weight (W) flue was (W)
W=VxM W

9.10
0.02
L 4.40 0.146
28 0.56
0.13 32 _ 4.16 0.140
28 21.00 0.696
I W=30.12

Weight of carbon per kg of coal = 0.24 kg

Weight of carbon per kg of the flue gas
= Weight of carbon in 0.146 kg of CO2
+ Weight of carbon in 0.018 kg of CO
—5


POWER PLANT
42

= x 0.146 + x 0.018
11 7
= 004 + 0.008 = 0.048 kg.
•. Weight of flue gas per kg of coal
0.24 -
= -- = a kr'
0.048
Weight of excess oxygen per kg of flue gas
= Amount of oxygen in flue gas -- Oxygen
require(I to burn CO

= 0.140 - x 0.018 = 0.13 kg.

Weight of excess oxygen per kg of coal

5 x 0.13 = 0.65 kg.
Weight of excess air per kg of coI
= 0.65 x 100 2.83 kg. Ans.

Example 1.3. Calculate the amount of air required to burn one

and product of combustion for a foil lb ptrecn loge
kg of fuel
composition of which is given as follows
C = 80'k; 112 20k•.
Solution.
Subsiance weig!ij0'et J" T'
fuel

C O.S 8/3 0.8 813-2,13 '1

---.-- ----...- .-- .

Total oxygen required = 2.13 + 1,6 3.73 kg.

Weight or air required = :3.73 x' = 16.2 kg. Ans.

Products of Combustion
C+O-4CO2 12=32 - 12+32
12=32 - 44


SOURCES OF ENERGY 43

1 kg of C + kg of 02 kg of CO2

kgof02 gives CO2=-kg

v
2
2.13 kgof0 gi es CO2 r xx2.13 = 2.93 kg

CO 2 = 2.93 kg/kg offuel. Ans.

Similarly, 2112 + 02 -, 21120
4 + 32--436
1+8---)9

1 kgofH2+8kgofO2=9kgofH2O
=
1120 X 16 1.8 kg/kg of fuel. Ans.

Example 1.4. A gas used as fuel has the following composition

by volume:
H 2 = 27%; CO2 = 18%
CO= 12.517r; Gilt =2.5%
N 2 = 40%.
Calculate the volume of air required for complete combustion of
one cubic-metre of the gas.
Solution. The various values are as follows:
Name of gas Vol per co. ,nelre
of/uci
Fo, rd. per ,, (f 1 02metre
reqd. per cu-
o/'fuei
co,zstLfI,'?zt
0.27 1 0.135

CO2 0.18 - -

L
CO

CU4
- 0.125
__ 1
2
0,062

N2 0.40 - -

Total oxygen required = 0.135 + 0.062 -- 0.05 = 0.247

Volume of air = 0.247 x -°- = 1.176 m 3 . Ans.

Example 1.5. The percentage composition of a sample of coal is

found to be as follows

44 POWER PLP NT

C=88%,H 2 =4.3% 02=3%

N2=0.7%,S=1% Ash =2%
(a) Calculate the minimum weight ofair required for complete
combustion of one kg of this coal.
(b) If 40% excess air is supplied, calculate the percentage
composition by volume of the dry flue gases.
Solution. (a)

Substance Weight Weight of Weight of Weight of products of Combustion

per kg of oxygen oxygen re-
coal (kg) per kg of quired
substance (kg)
(kg)
CO2 SO2 N3
C 0.88 8/3 •2.34 _3.23 - -
02 - 0.04 - 0.04 - - -'
H2 0.043 8 0.344 - - - -
N2 0.007 - - -- -' 0.007
S 0.001 1 0.01 0.02 -
Ash 0.02
Total oxygen required
2.34 + 0.04 + 0.344 + 0.01 = 2.734 kg.
Minimum amount of air required
= 2.734 x 100 = 11.88 kg. Axis.
(b) As 40% excess air is supplied
N2 in actual air supply
=1. j 88x-x1.4=12.8 kg.

Total nitrogen = 12.8 + 0.007 = 12.807 kg.

Excess oxygen = 11.88 x x 0.4 = 1.1 kg.

Substances Weight per kg Molecular Parts by volume Percentage

of coal Weight (a) volume
Li X 100
CO2 3.23 44 3-0.73 12.88%

SO2 0.02 64 -
= 0.0003 0.05%


SOURCES OF ENERGY 45

02 1.1 32 1.1
32
N2 12.8 28 12.8046 81.07%
28
Total 0.5673 100%
Example 1.6. The percentage composition by weight of a sample
of coal is given as below.
C=65.50%; 112=6.65%
02= 17.50%; S= 1.80%
Using Dulong formula, calculate the calorific value of coal.
Solution. According to Dulong's formula, the higher calorific
value (H.C.V.) is given by the following relation:
H.C.V. = ooc + 34,500 [H - + 2220 s}
1001
=x 65.50 + 34,500 (6.65- 17.50) + 2220 x 1.801
100 18080
100 J520,924 + 153,870 + .39961 = 6787.90 kcal/kg
=
Steam produced = 0.0665 x 9 = 0.5985 kg.
Lower calorific value (L.C.V.) = H.C.V. - 0.5989 x 588.76
= 6787.90 - 0.5985 x 588.76 = 6787.90 - 352.37
= 6435.53 kcal/kg. Ans.
Example 1.7. A boiler uses coal of the following composition
C=89%; 112=4%; 02=3.8%
If CO2 records read 10% calculate the percentage of excess air
supplied per kg of coal.

Substance per Mal. Weight Proportional vol. Oxygen reqd. Thy prod uct
kg of coal composition by vol. ofcombustion
by vol.
C=0.89 12 0.89 0.074 0.074
- -0O74
1
H2 = 0.04 2 0.01
-002
02 0.038 32 0.0380.012
i2- -OO12

Solution. Total oxygen required

= 0.074 + 0.01 - 0.012 = 0.0828

POWER PLANT
46

Minimum air required by volume = 0.0828 x = 0.394

Volume of N2 with minimum air = 0.394 - 0.0828 = 0.3112

Let V = Volume of excess air
Percentage CO2 in dry products
Co2
- CO2 + N2 + V
0.074 x 100
10 _
0.074 + 0.3112+ v
V= 0.2548
= 0.394
0.2548 x 100 = 64.7% Ans.
Excess air

Example I.S. An oil engine uses oil having gravirnet nc analysis

as follows
C=0.85;H2 =0.14 ;Ash =0.02
The ratio of air supplied to fuel burnt is 30: 1.
Determine the mass of various constituents of wet products and
the percentage composition of dry products.
Solution. Fuel supplied = 1 kg
Total air supplied = 30 kg
23 = 6.9 kg
Total oxygen supplied = 30 x yC0

Total nitrogen supplied = 30 x = 23.1 kg

Various products of combustion are as follows:

Con . Afass fnun uJ Products of t wnhuslwn (kg)
stituent kg/kg of 02 rcqd.
fuel kg -
iu2oTJQ:
C 0.85 Is x 0.S5 11
-x0.6 -
=2.27 =3.11
112 I 0.14 0.14 x 8 - F x 0.14 - -
1 : 12 1,26
Ash
6.9-
I __L=-- -_ L
I :u.0 I 3.39 _!L._L...1 ............? ..
 Wet products and dry products are calculated as follows:

WdprductsTh Dry Products

Co, i sttue'it kJIJ - Co,zstj1uen kujffuei Corpitzpn
C01 I I.II CG2 3.11 10.4
iio .126
N2 23.1 N2 23.1 777
02 3.51 0-2 3.51 11.9
Total 30.98 -29.72 - - iOOJ

Example 1.9. The volumetric analysis of a certain flue gas given

by Orsat apparatus is as follows:
CO2= 15%;C0=1%
02 = 5% Nitrogen = 79%.
Find the analysis of the flue gas by weight.
Solution. The analysis of flue gas by weight is determined as
follows

Name of Volzu;w per JMokcular Proportional Percentage ec,n-

gas ,n of flue gas zteigh1 u-eight position b y weight
I (It) C = A xB

J
(A)
IC
0.1.541 0.15 x 41

= 6.6 =21.56.
CO 0.01 28 0.01 X 28 0.28
-100
= 0 28
02 0.05 32 0.05 x 32
Li.-.----- -
100

--
N 2 079 28 0792S 2212
X 100

:i 12 =7229
.-'- L. 1=00

Example 1.10. A sample of coal has the following composition

by weight C = 707c, Hydrogen 8%, nitrogen 3%, oxygen 7%, sulphur
2% and ash 1017P.
Lkter,n inc higher corific value and lower calo&ifie value offuel.
Solution. C = weight of carbon per kg of coal = 0.7
112 = 0.08; N = 0.03: 02 0.07

48 POWER PLANT

S = 0.02; Ash 0. 1.

H.C.V. = 8080C + 34,500 (112 - + 2220S

= 8080 x 0.7 + 34,500 (0.08 - 12220 x 0.02
= 8770 kcal/kg
L.C.V. = HCV - (911 2 x 586)
= 8770 - (9 x 0.08 x 586) = 8348 kcal/kg.
Example 1.11. A boiler furnace using 50% excess air burns coal
with following composition:
C = 0.77
H 2 = 0.05
02=0.08
S = 0.02
N2 = 0.02
Ash 0.06
The flue gases enter the chimney at 324°C and atmospheric temp.
is 16°C.
C,, = 1.007 kJ/kg for 02, N2 and air
= 1.05 kJ/kg for CO2 and SO2 from flue gas
Heat carried away per kg of moisture from flue gas
= 2930 kJ/kg
Determine the heat carried away by the flue gases in kJ/kg of
coal.
Solution.
= Minimum amount of air required to burn one kg of coal
8
= 100 [( C + 8H2 + S] - 02]
23
100 [(8
x 0.77 + 8 x 0.05 + 0.02)_ 0.08]
= 23 3
= 10.48 kg
M = Mass of air required to burn one kg of coal
= + Excess air
= 10.48 + x 10.48 = 10.48 + 5.24

-;

SOURCES OF ENERGY
49
= 15.72 kg.
mi = Mass of CO 2 produced
= x C as 1 kg of carbon produced kg of CO2
3-
-
=
x 0.77

= 2.82 kg
P112 = Mass of 1-120 produced

= 9 x Fl2 as one kg of hydrogen produces 9 kg of water

= 9 x 0.05
= 0.45 kg
= mass of SO 2 produced
= 2 x S as one kg of sulphur produces 2 of SO2 k
=2x0.02
0.04 kg
1114 = Mass of excess 02 produced per kg of coal
23
x excess air supplied
=
23
x 5.24
=
= 1.2 kg
1115 = mass of N 2 produced
77
x Actual air supplied

77
x 15.72
=
12.1 kg
= Temp. of flue gases entering the chiiuiey = 324°C
T 2 = 16°C
II I = Heat carried away by CO2
= Mass x Specific heat x Rise in temperature
= mi x C, X (7' - 7'2)
= 2.82 x 1.05 x(324 - 16)
 POWER PLAN
50

= 912 kJ/kg
112 = Heat carried away by S02
= fl13 X C, x (Ti - '2)
= 0.04 x 1.05 x (324 -. 16)
12.94 kJ/kg
Heat carried away by excess 02
= m4 x C,, x (Ti - T2)
= 1.2 x 1.007 x (324 - 16)
= 372 kJ/kg
11 4 Heat carried away by N2
= M5 Cp (Ti - T2)
= 12.1 x' 1.007 x (324- 16)
3753 kJlkg
Ks Heat carried away by moisture
=
= 2930 kJx fl12
=2930x0.45
= 1318.5 kJ
H = Total heat carried away be flue gas
=H1 +112+1131-1141-115
= 912 + 12.94 + 372 +3753 + 1318.5
= 6368.5 kJ/kg of coal.

PROBLEMS
1.1. Name and explain the various sources of energy. Compare the
various sources of energy.
1.2. What are the various types of solid fuels? Describe Bituminous,
Lignite and Anthracite varieties of coal.
1.3. What do you understand by higher calorific value (H.C.V.) and
lower calorific value (L.C.V.) of a fuel? Explain Dulong's formula
to find H.C.V.
1.4. What is meant by composition of fuel? Give percentage composi-
t ion of some of liquid fuels. Explain ultimate analysis to find
various constituents of a solid fuel.
1.5. What are the various advantages of liquid fuels and gaseous
fuels over solid fuels?

SOURCES OF ENERGY 51
1.6. Write short notes on the following: -
(i) Degrees A.P.I. of liquid fuel
(ii) Firing qualities of coal
(iii) Combustion of fuels
(iv) Types of gaseous fuels
(u) Ultimate analysis
(vi) Proximate analysis
(vii) Products of combustion
(viii) Requirements of a fuel.
1.7. A boiler uses an oil with a calorific value of 9000 kcaVkg. The
analysis of the oil is 85 percent carbon and 15 per cent hydrogen.
The air supplied is ible the theoretical mass required for the
complete combustion u .he oil. Calculate the mass of exhaust
gases per kg of oil burnt.
1.8. The percentage composition by weight of a sample ofcoal is given
below:
C=70%, H2=6%
02 = 22% 8=2%
Using Dulong formula, determine the calorific value of coal.
1.9. Explain the method to find the weigh-. ,.,f excess air required for
the combustion of a fuel.
1.10. Write short notes on the following:
(a) Principal stages of fuel combustiors.
(b) Theoretical temperature of combwion of a fuel.
1.11.A boiler uses fuel oil. Gravimetric analysis : Carbon 0.86 and
hydrogen 0.14 at the rate of consumption is 500 kg/hour. The air
supplied is 25% in excess of theoretical minimum air required for
complete combustion. What is the total amount of air supplied
per hour? [Ans. 9250 kg/hI
1.12. The ultimate analysis of a sample of coal gives in percentage
composition by weight. C = 66%, H 2 6%. 0 = 19% and S =
Find the calorific value of coal using l)ulong's formula.
.13. Discuss the sources of energy in India.
1.14. Write short notes on the following:
(a) Conservation of energy
(b) Tidal power
(c) Solar energy
(d) Geothermal ..nergy.
1.15.A fuel contain- the following . rcentage of combustibles by
weight
Carbon 84%, and h ydrogen 4.1%.
If the air used for hirning of the coal in a boiler is 16.2 b y per kg
of fuel, fin- the tut heat carried awa y by dry flue gases and if
they escape at 300 C The specific heat of (.'O2_ 02, N2 are 0.213
and 0.219-050 respectivel y l'uul the minimum amount of air
required for the complete c()n'l 'tLstion of 1 kg of this fuel and the
excess oxygen supplied. 1A.M.1.E. I9711
1.16. Write short notes on the following:
(a) Liquid fuel properties
 POWER PLANT
52

(b) Conservation of energy

() Advantages of liquid fuels over solid fuels.
1.17. Discuss the conditions for proper burning of fuel.
1.18. Describe how to find the following:
(a) Weight of carbon in flue gases
(b) Weight of flue gas per kg of fuel burnt.
1.19. Define the following:
(a) Minimum air
(b) Excess air
(c) Products of combustion
(d) Conventional and non-conventional sources of energy.
1.20. A sample of producer gas has the following analysis by volume:
CO2 = 24%, H2 = 14%, CH4 = 5%
CO2 = 6%,02 = 2%, N2 = 49%
Calculate the air required for complete combustion of one cubic
metre of fuel. Also find the volume of dry flue gas.
1.21. Explain the advantages ofhaving a common grid for all the power
stations in a region.
 KI

Power Plant Economics

2.1 Power Plant

Power plant is an assembly of equipment that produces and
delivers mechanical and electrical energy. Electrical equipment of
a power station includes generators, transformers, switch gears and
control gears. Fig. 2.1 shows the main part of a power system.

O/5TAI9UT,N
$V8-SrAr,o,.

r1i ii
CONSUMERS

1125ERVICE
MAIN
fu(rfl,b,,rnn
S../O4y

/fA'V I711$2,'.v fEEDEP

SFORt4ER
-d
J2/JJ v

r 94 VSM/$ S/ON
MANSMISSiOlV 5C/8 -STA TION

Fig. 2.1
From the economic point of view it is desirable that when large
amount of electric potver is to be transmitted over long distance it
should be transmitted at a voltage higher than the distribution
voltage. The voltage for transmission should be so chosen that it
gives best efficiency, regulation and economy. Step up transformer
is used to step up the generation voltage to transmission voltage
which is usually 132 kV At the transmission sub station the voltage
is stepped down to ledi.imvoltage usually 33 or 3.3 kV. The feeders
carry the power to e dstrjbution sub-stations. Feeders should not
be tapped for direct supply. The function of transformers at
the distribution sub-station is to step down the voltage to low
 POWER PLANT
54
distribution voltage which is usually 400 to 230 V. Distributors are
uscd to supply power to the consumers.
Transmission of electric power over long distances can be done
most economically by using extra high voltage (E.H.V.) lines. In the
world today many A.C. extra high voltage lines are in operation.
These E.H.V. lines operate at voltages higher than the high voltage
lines i.e. 230 kV. The E.H.V. lines are now in operation in Europe,
USA and Canada and oplrate at 330 kV, 400 kV, 500 kV and 700
W. Still higher voltage 1000 kV level are in the experimental stage.
In India there is no E.H.V. line so far but it is hoped that soon such
lines in the form of Super Grid will be developed.
2.2 Types of Power Plants
Based upon the various factors the power plants are classified
as follows
1. On the basis of fuel used
(i) Steam Power Plant
(a) condensing power plant
(b) non-condensin g power plant
(ii) Diesel power plant
(iii) Nuclear power plant
(iv) Hydro electric power plant
(o) Gas-turbine power plant
2. On the basis of nature of load
(i) Base load plant
(ii) Peak load power plant
3. On the basis of location
(i) Central power station
(ii) jolted power station
4. On the lasis of service rendered.
(j) Stationary
(ii) Locomotive.

2.3 Requirement of Plant Design

The factors to be kept in view while designing a power station
are follows:
1. Economy of expenditure i.e. minimum
(i) Capital cost
(ii) Operating and maintenance cost.
.2. Safety of plant and personnel
3. Reliability
4, Efficiency
5. Ease of maintenance
6. Good working conditions
7. Minimum transmiss ion loss.


POWER PLANT ECONOMICS

55
2.4 Useful Life of a Power Plant
Every power plant wears as the time proceeds and it becomes
less fit for use. The deterioration of the equipment takes place
because of age of service , wear and tear and corrosion. By a thorough
programme of preventive maintenance and repairs it is possible to
keep the power station in good conditions to get proper service from
it. A power plant becomes obsolete when it can be replaced by one
of more modern design which operates at a reduction in total annual
costs. Therefore, useful life of a power plant is that after which
repairs become so frequent and expensive that it is found economical
to replace the power plant by new one. Useful life of a conventional
thermal power plant is 20 to 25 years? the useful life of nuclear power
plant is 15 to 20 years and useful life of diesel power plant is about
15 years. The useful life of some of the equipment of steam power
plant is indicated in Table 2.1.
Table 2.1.
qpzt
Steam turbine

Fire tubes 10-15

Boilers
L Water Lube 20
Coal and ash machinery 10-20
Purnus
Feed water heal
20
lOr--30
10-20

9ilIIiIT15--20 T
and in.,trun_J_ 10-
2.5 Comparison of Public Supply and Private
Generating Plant
Industrial concerns ma y generate their own power or may
purchase power from public supply company. The two are compared
as f011QWs
(a) Public Supply
(i) Reliability of power is assured and over-load power
demand can be available at short notices.
(ii) It is cheaper to purchase power from public supply com-
pany.
 POWER PLANT
56

(iii) The space required for the installation of power generat-

ing unit is considerably large. The same space can be
saved and utilised for some other purpose such as for the
expansion of industry.
(h) Private Generating Plant
(i) Industries where power d l:,s.. otis small and where power
is required continuously ILiCh as in hospitals, private
power generation is preferred. In such cases power
generation by diesel power plant is economical.
(ii) Industries where wastes produced can be used as fuel,
prefer to generate their lower power. For example in sugar
mills the left-over of sugarcane called begasso can be
burnt as fuel in boiler which can be used for steam genera-
tion.
(iii) In industries like sugar mills and textile mills steam is
rcquirtd for processing work. Therefore, such industries
generate their own power by steam turbines so that steam
leaving the turbines can be used for processing work.

.6 Prediction of Future Loads

When a power station is to he installedin a particular area it is
desirable that maximum power demand of that area should be
known. This help in deciding the capacity of the power station.
Although it is difficult to forecast exactly the future load require-
ments of the area but approximate estimate about power demand
should be made. Two methods are used to forecast the load require-
ments.
(i) Statistical method (ii) Field survey method.
In statistical method data of annual maximum demand pertain-
to the area is collected for past se'eral years and from this data
t'acted future load can be judged. In field survey method existing
requirements of the area for different loads such as industrial,
agricultural, municipal and residential are found out. Then the
future load requirements are decided taking into account the
various factors like population growth, standard of living of the
people, climate of the regions and industrial development.
The load prediction or forecasting may be done for
(i) Short term covering a period of 4 to 5 years.
(ii) Medium term, covering a period of about S to 10 years.
years or more.
(iii) Long term, covering a period of about 20
For the installation of a new power project or for the expansion
of the existing power plant, it is necessary to estimate the total
amount of load that would he required to he met for variOuS pur-
poses. The economics of tile installation or expansion of a power
plant calls for the correct prediction of load. The usual practice
 POWER PLANT ECONOMICS 57

followed in the hydro-power plant planning is that the full Potential

of the project is developed in stages. The power required for inune-
diate demand is developed in 1st stage and remaining potential
being des-eloped in subsequent stages.
There are number of formulae used for estimating Power
generation requirements

Scheer formula for estimating the power generation require_

ments are as follows
tog 11) G = K 0.15 log jtj M
where G = Annual growth in generation (per cent)
M = Per capita generation
K = Constant
= 0.02 (po
pulation growth rate) + 1.33
2.7 Terms and Definitions
u Connected Load. It is the sum of ratings in kilowatts (kW)
Of equipment installed in the consumer's premises. If a consumer
has connections for 4 lamps of 60 watts (W) each, and power point
of 500 W and a radio COnsuming 60W, then the total connected load
of the consumer -
=4x60+500. 60= 240+500
i-60=800W
(ii) Maximum Demand. It is the maximum load which a
COOS LI Ifler USCS
at any time. It can be less than or equal to connected
load Ifall the devices fitted in consumer's house run to their fullest
extent sinI ultaneously then the
maxil l'uni demand will be equal to
connected load. But generally the actual maximum demand is less
than the connected load because all tli€
'':i ces never run at full load
at the same time. Maximum deman(j' .1
a power station is the
maximum load on the power station in it period.

(hi) Demand Factor.

It is defined as the ratio of maximum
demand to connected loa(I.
(i c) Load Curve. It is graphical re p r esentation bet wt'en load
10
kilowatt(k\V) and time in hours. It shows variation of load on the
Power station. When it is plotted for 24 hours of it div it is called
daily load curve and if the time considered is of'one
y ear (8760 hours
then it is called annual load CUrVC. The areas under the load curve
represents the energy generated in the period considered. The area
under the curve divided by the total number of hours gives the
average load oil power station. The peak load indicated by the
load curve represents the maximum demand on the power station.

—6
 POWER PLANT
58

AiE RA
LOA u
..t

9AS LOAD

TM (SOURS

Fig. 2.2 Load Curve

Load curves give full information about the incoming load and
help to decide the installed capacity of' the power station and to
decide the economical sizes of various generating units. They also
help to estimate the generating cost and to decide the operating
schedule of the power station i.e. the sequence in which different
generating units should be run. Fig. 2.2 shows a load curve.
(v) Load Factor. It is defined as the ratio of average load to
maximum demand. Load factors and demand factors are always less
than unity. Load factors play all part oil cosi of
generation per unit. The higher the load factor the lesser will he the
cost of generation per unit for the same maximum deniaiul.
Load factors for different types of consumers are as follows
(i) Residential load 10- 15
(ii) Commercial load 25-30%
(iii) Municipal load 25
(iv) Industrial load
(a) Small scale industries 30-50
(b) Medium size industries 55-60%
(c) Heavy Industries70-807
Base load plants run oil high load factor whereas the load
factor of peak load plants is usually low.

(vi) Base Load and Peak Load Power Plants. The

pOWCI

plants work at different load factors. The power plants used to

supply the load of the base portion of load curve are called base load
power plants. Base load power plants run throughout the y ear, are
of large capacity and run at high load factors and are highly efficient.
The fixed and semi-fixed cost of these plants is tisuallv high. The
power plants which supply the load on the top portion of load curve
are called peak load plants. The y are of smaller capacity. run for a
 POWER PLANT ECONOMICS 59
short period in the year and work at load low factors. Peak load
plants should be capable of quick starting.
Hydro and nuclear power stations are usuall y classified as base
load power stations. Thermal power stations ma y he taken as
intermediate power stations whereas diesel power stations are
usually classified as peak load stations. Parallel operation of dif-
ferent power stations and the co-ordination of generation electricity
leads to considerable saving in comparison with the same load fid
by independent power stations.
To meet the fluctuating power demand the power should be
produced conforming to the demands. These da ys except a few
isolated projects hydel power is used along with conventional steam
power plant.. In hydel power, regulation can be easily achieved by
restricting the discharge through the water turbine and this can be
achieved without much trouble. Time taken to activate a hydro
power station is 5 to 15 minutes. The time of heating up a boiler of
steam power station varies from 2 to 10 hours depending upon size.
Therefore, it is now accepted practice that power s y stem should
consist of steam and hydel power each supplementing the other.
During combined working of hydro-power plant and steam
power plant the hydro plant. with ample water storage should be
used as base load plant and steam power plant should be used as
peak load plant. If the amount of water available at hydro power
plant site is not sufficient then steam power plant should supply the
base load and hydro power plant should supply peak load.
The major advantage of steam power plants is that they can bc
located near the load centre. This reduces the transmission losses
and cost of transmission lines. In hydro -power plants there is more
or less dependence on the availability of water which in turn
depends oil natural phenomenon ofrain. Although the operating
cost of a hydroelectric power Plant is very less but excessive, (us-
tance from the load centre ma y some time prohibit. the Use of such
a plant in favour ofa thermal power plant. Some hydro-powir plants
are supplemented by steam power plant or diesel engine power
plant.
The requirements cia base load power plant are as follows
(i) Its capital cost should be low.
(ii) It should be able to supply the load continuously.
(iii) Its operation cost should be low as it has to operate most
f the time.
(iv) its maintenance cost should be low.
The requirement of a peak load plant are as follows

60 POWER PLANT

(a) It should be capable of being started from cold conditions

within minimum time.
(b) Its operating cost should be low.
(c) Capital cost involved should be minimum.
(vii)Plant Capacity Factor. It is defined as the ratio of actual
energy produced in kilowatt hours (kWh) to the maximum possible
energy that could have been produced during the same period.
Plant capacity factor ---
Cxt
where E = Energy produced (kWh) in a given period.
C = Capacity of the plant in kW
t = total number of hours in the given period.
(viii) Plant Use Factor. 1t is defined as the ratio of energy
produced in given time to the maximum possible energy that could
have been produced during the actual number of hours the plant
was in operation.
Plant use factor = E
Cx ti
where ti = actual number of hours the plant has been in operation.
(ix) Diversity Factor. It is defined as the ratio of sum of
individual maximum demand to the simultaneous maximum
demand of a system. Usually the maximum demand of various
coisuiners do not occur at the same time and simultaneous maxi-
mum demand is less than their total maximum demand. Power
station should be capable of supplying.the simultaneous maximum
demand. Diversity factor is more than unity.
The high value of load factor, demand factor, diversity factor
and capacity factor are always desireable for economic operation of
the power plant and to produce energy at cheaper rate.
Some typical demand factors are mentioned in Table 2.2.
Table 2.2
Type of Consumer I Demand Factor

LightigRedence) - upto'J4kW 1.00

to 1/4 kW 0.60

over I kW 0.50
Lghtirig coinincrctl Schools, Hostels, 0.50
Small Industry, Theatres 0.60
Restaurants, Offices, Stores 0.70
 POWER PLANT ECONOMICS
61
General Power Service uptoionjJ
10 to 20 H.P. 0.65
20 to 100 H.P. 0.55
above 100 H.P. 0.50
Some typical diversity factors are mentioned in Table 2.3.
Table 2.3

jettnetors
Lighting !ightu G'ncral Poui'r
J

(x) Load Duration Curve. Load duration curve represents

re-arrangement of all the load elements of chronological load curve
in the order of descending magnitude. This curve is derived from the
chronological load curve. Consider typical daily load curve (Fig. 2.3)
for a power station. It is observed that the maximum load on power
station is 32 kW from 2 to 6P.M. This is plotted in Fig. 2.4. Similarly
other loads of the load curve are plotted in descending order in the
same figure. This is called load duration curve.

)I KW
2 I KW

xW

6A. SAM IPM 2PM 6 P 6AM

TINt (HOUPS)

Fig. 2.3

The area under the load duration curve and the corresponding
chronological load curve is equal and epresents total energy
delivered by the generation station. Load duration curve gives a
clear analysis of generating power economically. Proper selection of
base load power plants and peak load power plants becomes easier.
 POWER PLANT
62

rIME HOURS -b.

Fig. 2.4

2.8 Power Plant Capacity

The capacity of power plant must be equal to atleast the peak
load. In smaller power plants it is desirable to instal two generating
units each being capable of supplying maximum demand so that if
one unit is not Working clue to repair or breakdown the other is able
to maintain uninterrupted supply of energy. In case of large power
plants using several generating units the total installed capacity is
kept equal to the expected maximum demand plus the capacity of
the two largest generating units. The number of generating units
hr.u1d he kepts two or more than two so that in the event of
br.akdown or maintenance etc. of a unit the power can be supplied
by other unit without interruption. While designing the power
station provision should be made for the installation of more
generating units dependinj upon the expected rate of increase of'
maximum demand over the next few years. In a power plant with
several generating units most of the units needed to indicate the
variation of load so that the different generating units call placed
in operation at the desired time. Plant capacity can be decided by
studying the load duration curve and the anticipated future
demand.
Power plant capacity depends upon the following factors
(i ) Maximum demand of consumers at present.
(ii) Type of load : such as
(a) Private
(b) Public
(c) Industrial
(d) Commercial
(t') Domestic
(fl Railways.
(iii) Future load conditions : Expected future electric power
demand for at least next five years should be known.
(ii) Availability of fuel.
POWER PLANT ECONOMICS 63

(v) Total cost of power plant.

(vi) l'ossibititv of inter connecting the power plant to other
power plants.
The advantage of having big power-plants is that it can directly
generate high voltage required for long distance transportation. In
this case, the loss of energy is in two ways (i) long distance
transportation, and (ii) transformation from high voltage into low
voltage before use. In case of smaller plants, transportation loss is
.sscr because of short distance transportation but the transforina-
ion needs to be done twice, ti:., W from low voltage into high voltage
before transportation, and (ii) from high voltage into low voltage
before use. Additionally, construction of number of smaller power-
plants may, combined together, require more capital and infrastruc-
ture investments. Thus, a relative economic anal ysis along with
other considerations like regional resource availability, socio-
economic development, etc. would lead to an optimum capacity of a
power-plant.

2.8.1 Feasibility of Electric Power Plant

Following factors should be considered while installing a new
power 1)llflt
(i ) Estimate of p roba 1)1 I' load
(it) Future load conditions
(iii) Power plant capacity
(it) Total cost of power plant
v ) Annual ru mining cost
(vi) The rate at which power will be sold to the consumer
(vii) 'l'vpe of fuel to he used.

2.9 Principles of Power Plant Design

While designing a power plant the following factors should be
considered
(i) Low capital cost
(ii) Rd iahilitv of suppivimig power
(iii) Low maintenance cost
(it)) Low operating cost
(v) lEigh efticiency.
(ti) Low cost of energ y generated
Wit) Reserve capacit y to meet future power demand
(1-iii) Simplicit y of design.

The al)OVC factors depend upon power plant site, availabilit y of

raw material, availabilit y of water, type of load, maximum power
demand, generating equipment etc

64 POWER PLANT

2.10 Economic Load Sharing between Base Load and Peak

Load Power Stations
Consider a system having load duration curve shown in Fig. 2.5.
The load to be supplied by a base load power plant and peak load
power plant.

Let C 1 = Annual cost of base

load power station
j C 2 = Annual cost of peak
load power station
C04 C 1 = Ri kW + P2 kWh

C2=R 2 kW+ P 1 kWh

Assuming R 1 > R 2 and
TIME (HOU R S) . P i <P2

Fig. 2.5 Let M Maximum demand of

the system
E = Total number of unit (kWh) generated
Ali
= Peak load of base load station
= No of units (kWh) generated by base load station
M2 = Peak load on peak load station = M-
E2 = Units generated b y peak load station = E --
Ci = R 1M 1 + PiE1
C 2 = 112 Of - M 1 ) + P 2 (F] E1)

P = Total annual cost of operation of the system

C=C1+C2

C = R 1M1 + Pi E 1 + 112 (M - M 1 ) * i ( F] F]1)

Now for minimum cost
dC
dM1
dC dEj dE1
This gives = 0 = R 1 + P 1 - 112 - I'2

dE 1 R1-R2
= ------ hours.
dM 1 P2-P1
E 1 the number of units generated by the base load station is
represented by area under straight line AB
 POWER PLANT ECONOMICS
65

dE1= Area of strip ABCJJZ (jjf 1 xAJJ

AB=2J=-iTi
dM 1 Pz-P1
This indicate that for economic load sharing the peak load
power station should run fr hours in one year.

2.11 Type of Loads

The various t y pe of loads are as follows
Residential Load. It includes domestic lights, power needed for
domestic appliances suèh as radio, television, water heaters,
refrigerators, electric cookers and small motor for pumping water.
Commercial Load. It includes lighting for shop, advertisements
and electrical appliances used in shops and restaurants etc.
Industrial Loud. It consists of load demand of various in-
(IUStI1CS.
Municipal Load It consists of street lighting, power required
for water suppl y and draige purjoses.
Irrigation Load. Electrical Power need for pumps driven by
electric motors to supply water to fields is included in this type of
load.
Traction Load. It includes train trolley, buses and rail-
ways.

2.12 Typical Load Curves

Load curve indicates the power demand at different intervals of
time. With the help of load curve the peak load can be ascertained
and hence the capacity of power plant can be judged. Fig. 2.6 shows
typical load curve for industrial load. The load starts increasing
from 5 A.M. and it is maximum at about 8 A.M., when all the
industries are working. It is constant till noon ajid falls for some
time due to lunch break. After that it is again rising and then starts
falling at about 5 P.M.
The urban load is high at ibout 9 A.M. It is less at noon and
then starts rising and is maximum between 5.30 to 6 P.M. (Fig. 2.7)
The street lighting load (Fig. 28) is nearly constant.
Domestic load is maximum in the evening and is minimum
during late hours of night (Fig. 2.9).


66 POWER PLANT

tT.i I t[I H
$2 6 72 6 12
A.M. 11.
.
12 6 12 6 72
All. RI-f.
1'
IkL
12 6 12 6 72
4.11. R.
TIME -
Fig. 2.6 Fig. 2.7 Fig. 2.8

LIE TROPOUTAW AREA LOAD

rwi

12 6 $2 6 12 12 6 12 6 72 6 12 6 12 6 72 6 72
1W. A.M. An A.M. (7M. 4.11. Pd-I.

Fig. 2.9 2.10

This curve varies with cities and season. For metropolitan areas
a typical daily load curve is shown in Fig. 2.10.
2.13 Cost of Electrical Energy
The total cost of electrical energy generated b y a power station
can be sub-divided as follows
1. Fixed cost. It includes the following cost:
(a) Capital cost of power plant.
(i) Cost of land
(ii) Cost of building;
(iii) Cost of equipment
(iv) Cost of Installation;
(t') Cost of designing and planning the station.
(b) Capital cost of primary distribution system which in-
cludes cost of sub-stations, cost of transmission lines etc.
Fixed cost consists of the following:
(i) Interest, taxes and insurance on the capital cost.
(ii) General management cost.
(iii) Depreciation cost.

Depreciation cost is the amount to be set aside per year from

income to meet the depreciation caused by the age of service, wear
and tear of machinery. Depreciation amount collected per year helps
in replacing and repairing the equipment.
POWER PLANT ECONOMICS 67

There are several methods of calculating the annual deprecia-

tion cost. The most commonly used methods are as follows
1. Straight line method. 2. Sinking fund method.
(a) Straight Line Method. According to this method it is
assumed that depreciation occurs uniforml y according to a straight
line law as shown in Fig. 2.11 (a). The annual amount to be set aside
is calculated by the following expression

Annual depreciation reserve

where A = Principal suL.capital cost of plant)

S = Salvage value or Residual value of power plant at the
end of years.
,i = Probable duration of useful life of power plant in
years.
The amount collected every year as depreciation fund does not
depend on the interest it may draw. Any interest if earned by the
depreciation amount is considered as income.
(b) Sinking Fund Method. In this method the annual amount
to be set aside per year consist of annual instalments plus the
interest on all the instalments.
Annual depreciation reserve

= (P - S)
[(1 + r)' - 1
where r = rate of compound interest.
Fig. 2.11(b) shows the depreciation curve.

TOTAL
N.., EPREC/AT/ON rio....
I-1
e

VALVE
L/EIN YEAAS
STRAIGHT LFNE .4ETHoO J(flKIP1 b VP1Q M& 'MUO

Fig. 2.11

Let P = Principal sum (capital cost of the plant)

S = Salvage value
n = Useful life in yep.-s.

68 POWER PLANT

r = Rate of interest
A = Annual depreciation amount.
Then the amount A after earning interest for one year
=A +Ar =A (1 + r)
The amount A after earning interest for two years
=A(1+r)+Ax(1+r)xrA(1+r)2
The amount A set aside after one year will earn interest for
(n - 1) years.
Therefore, the sum of the amounts saved together with interest
earnings should be equal to (P - S)
(P-S)=A+A(1+r)+A(1+r)2A(1+r)n
(P-S)=AI1+(1+r)+(1+r)2 .... (1+r)'']
Multiplying both sides by(1 + r)
(1+r)(P-S)=Al(1+r)+(1+r)2+(1+r)3(1+r)'J ...(ii)
Substracting the two equations from each other, we get

[(1 + r)' - 1
2. Energy Cost. It consists of the following costs:
(i) Cost of fuel
(ii) Cost of operating labour;
(iii) Cost of maintenance labour and materials
(iv) Cost of supplies such as:
(a) water for feeding boilers, for condensers and for
generai use,
(b) lubricating oils;
(c) water treatment chemicals etc.
In power generation activities minimum annual costs are
achieved by a proper balance of the fixed and operating costs. The
fundamental way to compare alternate schemes of power generation
is to compare their total annual costs.
3. Customer Charges. The ccst utcluded in these charges
depend upon the number of customers. The various costs to be
considered are as follows:
(i) Capital cost of secondary distribution system and
depreciation cost, taxes and interest on this capital cost.
ii) Cost of inspection and maintenance of distribution lines
and the transformers.
 POWER PLANT ECONOMICS
69
(iii) Cost of labour required for meter reading and office work.
(iv) CosL of publicity.
4. Investor's Profit. The investor expects a satisfactory return
on the capital investment. The rate of profit varies according to the
business conditions prevailing in different localities.
Cost of power generation can be reduced by adopting the follow-
ing economical meosures:
(i) By reducing initial investment in Jie power plant.
(ii) B y selecting generating units of adequate capacity.
(iii) the power plant at maximum possible load
factor.
(iv) By increasing efficiency offuel burning devices so that cost
of fuel used is reduced.
(v) By simplifying the operation of the power plant so that
fewer power operating men are required.
(vi) By installing the power plant as near the load centre as
possible.
(vii) By reducing transmission and distribution losses.
2.14 Energy Rates (Tariffs)
Energy rates are the different methods of charging the con-
sumers for the consumption of electricity. It is desirable to charge
the consumer according to his maximum demand (kW) and the
energy consumed (kWh). The tariff chosen should recover the fixed
cost, operating cost and profit etc. incurred in generating the electri-
cal energy.
Requirements of a Tariff. Tariff should satisfy the following
requirements
() It should be easier to understand.
(ii) It should provide low rates for high consumption.
(iii) It should enco-age the consumers having high load fac-
tors.
(iv) It should take into account maximum demand charges
and energy charges.
(v) It should provide less charges for power connections than
for lighting.
(vi) It should avoid the complication of separate wiring and
metering connections.
2.14.1 Types of Tariffs
The various types of tariffs are as follows
(a) Flat demand rate
(b) Straight line meter rate
(c) Step meter rate


70 POWER PLANT

(d) Block rate tariff

(e) IWO part tariff
(f') Three part tariff.
'l'he various types of tariffs can be derived from the following
general equation
Y = DX + EZ + C
where Y = Total amount of bill for the period considered.
D = Rate per kW of maximum demand.
X = Maximum demand in kW.
E = Energy rate per kW.
Z = Energy consumed in kWh during the given period.
C = Constant amount to be charged from the consumer during
each billing period
Various t y pe of tariffs are as follows
(i) Flat demand rate. It. is based on the number of IafllJ)s
installed and a fixed number of hours of use per month or per year.
The rate is expressed as it certain price per liinip or per unit of
demand (k\V) of the consumer. This energ y rate eliminates the u.
of metering equipment. It is expressed b y the expression
Y = DX

"I
C
I.)
-I 2
4 X

0
4

'd1.LY CC O2 J-ENERGY CONSUMED (Z)

(a) (H
Fig. 2.12 Flat demand rate.

(ii) Straight line meter rate. According to this energy rate the
amount to be charged from the consumer (kp(llds upon the energy
consumed in kWh which is recorded b y a means of a kilowatt hour
met 'r. It is expressed in the firiti
EZ
This rate suffers from a drawback that conswla'r using no
it

energy will not pa' an y amount although he has incurred slim.

POWER PLANT ECONOMICS 71

expense to the power station due to its readiness to serve him.

Secondly since the rate per kWh is fixed, this tariff , (foes not en-
courage the consumer to use more power (Fig. 2.13.

I
t N

It. a
0
C
'I.
C,
4
I-.
0

ENERGY CONSJMEO(Z) - ENERGY CONSUMED(Z)

Fig, 2.13 Straight meter rate.

(iii) Step meter rate. According to this tariff the charge for
energy consumption goes down as the energy consumption becomes
more. This tariff is expressed as toltows (Fig. 2.14).

f
t

0 z
0
4

._..*_ [lV (OU'. N. Z I -.- EN ERY C ONSUM[OZ )

Fig 2 14 Step motor rate.

,.fl
0
5
'., C)

EN1 Rio co lll^

-.. (NRY CON.lJI(D (

Fig. 2 15
 IV

72 POWER PLANT

Y=EZ ifO5Z5A
Y=E 1 Z 1 ifAZ.!5B
Y=E2Z2 ifBZ2C
and so on. Where E, E 1 , E2 are ihe energy rate per kWh and A, B
and C are the limits of energy consumption.
(iv Block Rate Tariff. According to this tariff a certain price
per units (kWh) is charged for all or any part of block of each unit
succeeding blocks of energy the corresponding unit charges
decrease.
It is expressed by the expression
Y= EZ 1 + E 2Z2 E3Z3 + .......

where Ej, E2, E:3.... are unit energy charges for energy blocks of
magnitude Z i , Z2. Z3,.... respectively (Fig. 2.15).
(v) Two part Tariff (Hopkinson Drnand Rate). In this tariff
the total charges are based on the maximum demand and energy
consumed. It is expressed as
Y = D. X + EZ
A separate meter is required to record the maximum demand.
This tariff is used for industrial loads.
(vi) Three-part Tariff (Doherty Rate). According to this tariff
the customer pays some fixed amount in addition to the charges for
maximum demand and energy consumed. The fixed amount to be
charged depends upon the occasional increase in fuel price, rise in
wages of labour etc. It is expressed by the expression
Y = l)X + EZ + C.
2.15 Economics in Plant Selection
A power plant should be reliable. The capacit y ofa power plant
depends upon the power demand. The capacity of a power plant
should be more than predicted maximum demand. It is desirable
that the number ofgenerating units should be two or more than two.
The number of generating units should be so chosen that the plant
capacity is used efficiently. Generating cost for large size units
running at high load factor is substantially low. However, the unit
has to be operated near its point of maximum econom y for most of
the time through a proper load sharing programme. Too many stand
bys increase the capital investment and raise the overall cost of
generation.
The thermal efficienc y and operatingcost ofa steam power plant
depend upon the steam conditions such as throttle pressure and
temperature.
 POWER PLANT ECONOMICS
73

The efficiency of a boiler is maximum at rated capacity. Boiler

fitted with heat recovering devices like air preheater, economiser
etc. gives efficiency of the order of 90%. But the cost of additional
equipment (air preheater economiser) has to be balanced against
gain in operating cost.
Power can be produced at low cost from a hydropower plant
provided water is available in large quantities. The capital cost per
unit installed is higher if the quantity of water available is small.
While installing a hydropower plant cost of land, cost ofwater rights,
and civil engineering works cost should be properly considered as
they involve large capital expenditure.
The other factor which influences the choice of hydropower plant
is the cost of power transmission lines and the loss of energy in
transmission. The planning, design and construction of a hydro
plant is difficult and takes sufficient time.
The nuclear power plant should be installed in an area having
limited conventional power resources. Further a nuclear 'power
plant should he located in a remote or . unpopulated are to avoid
damage due to radioactive leakage during an accident and also the
disposal of radioactive waste should be easy and a large quantity of
water should be available at the site selected. Nuclear power be-
comes competitive with conventional coal fired steam power plant
above the unit size of 500 MW.
The capital cost of a nuclear power plant is more than a steam
power plant of comparable size. Nuclear power plants require less
-.pace as compared to any other plant of equivalent size. The cost of
maintenance of the plant is high.
The diesel power plant can be easily located at the load centre.
The choice of the diesel power plant depends upon thermodynamic
considerations. The engine efficiency improves with compression
ratio but higher pressure necessitates heavier construction ofequip-
merit with increased cost. Diesel power plants are quite suitable for
smaller outputs. The gas turbine power plant is also suitable for
smaller outputs. The cost of a gas turbine plant is relatively low.
The cost of gas turbine increases as the sample plant is modified by
the inclusion of equipment like regenerator, reheater, and inter-
cooler although there is an improvement in efficiency of the plant
by the above equipment. This plant is quite useful for regions where
gaseous fuel is available in large quantities.
In order to meet the variable load the prime movers and gener-
ators have to act fairly quickly to take up or shed load without
variation of the voltage or frequency of the system. This requires
that suppl y of fuel to the prime mover should he carried out by the
action of a governor. Diesel and hydro-power plants are quick to
74 POWER PLANT

respond to load variation as the control supply is only for the prime
mover. III steam power plant control is required for the boilers as
well as turbine. Boiler control ma y be manual or automatic for
feeding air, feed water fuel etc. Boiler control takes time to act and
therefore, steam powers plants cannot take up the variable load
quickly. Further to Cope with variable load operation it is necessary
for the power station to keep reserve plant ready to tnuntain
reliability and continuit.y of power supply at all times. To supply
variable load combined workingofpower stations is also economical.
For example to supply a load the base load may he supplied by
a steam pow?r plant and peak load ma y he supplied by a hydropower
plant or diesel power plant.
The size and number of generating units should be so chosen
that each will operate on about full load or the load at which it gives
maximum efficiency. The reserve required would onl y he one unit
of the largest size. In a power station neither there should be only
one generating unit nor should there he a large number of small sets
ofcliffercnt sizes. In steam power plant generating sets of80 to 500
MW are quite commonly used whereas the maximum size of (liesel
power plant generating sets is about 4000 kW. Hdro electric
generating sets up to a capacity of' 200 M\V are in use in U.S.A.
2.16 Economic of Power Generation
Economy is the main principle of design of'a power plo:it. Power
plant economics is important in controlling the total power costs to
the consumer. Power should be supplied to the consumer at the
lowest possible cost per kWh. The total cost of power generation is
made up of fixed cost and operating cost. Fixed cost consists of
interest on capital, taxes, insurance and management cost. Operat-
ing cost consists of cost of fuel labour, repairs, stores and super-
vision. The cost of power generation can he reduced by,
U) Selecting equipment of longer life and proper capacities.
(ii) Running the power station at high lad factor.
(iii) Increasing the efficiency of the power plant.
(ii Carrying out proper maintenance of power plant equip-
nient to avoid plant breakdowns.
(r) Keeping proper supervision as a good supervision is
reflected in lesser breakdowns and extended plant life.
(vi) Using a plant of simple design that (lees not need highly
skilled personnel.
Power plant selection depends upon the fixed cost and operating
cost. The fuel costs are relativel y low and fixed cost and operation
and maintenance charges are quite high in a case ofa nuclear power
plant. The fuel cost in quite high in a diesel power plant and for
hydro power plant the fixed charges are high of the order of 70 to
POWER PLANT ECONOMICS 75

80 of the cost of generation. Fuel is the heaviest items of operating

cost in a steam power station. A typical proportion ofgenerating cost
for a steam power station is as follows
Fuel cost = 30 to 4017p
Fixed charges for the plant = 50 to 60%
Operation and maintenance cost = 5 to 10%
T power generating units should be run at about full load or
the load at which they can give maximum efficiency. The way of
deciding the size and number of generating units in the power
station is to choose the number of sets to fit the load curve as closely
at possible. It is necessary fora power statibn to maintain reliability
and continuity of power supply at all times. In an electric power
plant the capital cost of the generating equipments increases with
an increase in efficiency. The benefit of such increase ill capital
investment, will be realised in lower fuel costs as the consumption
of fuel decreases with an increase in cycle efficiency.
Fig. 2. 1 (a) shows the variation of fixed cost and operation cost
wi'th investnent.

Annuol
cost

—investment (efficiency)
(a)

0
Li

- Capacity
(b)
Fig. 2.15
76 POWER PLANT

Fig. 2.15 (b) shows the variation of various costs of power plant
versus its capacity.
Graph A shows variation of engineering and labour cost where
as graph B indicates material cost and graph C indicates total cost.
Unit Cost = Total cost
Capacity
2.17 Plant Performance and Operation Characteristics
Boilers, turbines, generators etc. of a power stations should
work efficiently. Some curves are plotted to observe their perfor-
mance. The various curves used are as follows
I. Input output curve. Performance of a power station is most
precisely described by the input-output curve which is a graphical
representation between the net energy output (L) and input M. The
input is generally expressed as millions of BTU/hr or kcal/hr and
load or output is expressed in megawatts. The input to hydroplant
is measured in cusecs or cubic metre per second of water.
INPUT - OUTPUT ahQvE EFFICIENCY CURVE

iz

10

0 OUTPUT 0 OUTPUT L

(W (bJ

0 OUTPUT
(c)
Fig. 2.16
 POWER PLANT ECONOMICS 77

Fig. 2.16 (a) shows input-output curve. In order to keep the

apparatus functioning at zero load a certain input (Ia) is required to
meet frictional and heat losses.
2. Efficiency curve. The ratio of output of power station to input
is called efficiency. The efficiency curve is obtained by plotting
efficiency against output. It is shown in Fig. 2.16 (b).
3. Heat rate curve. The ratio of input to output is known as heat
rate (HR).
HR = IlL
Heat rate curve is obtained by plotting values of heat rate
against corresponding value of output. Fig. 2.16 (c) shows heat rate
curve.
4. Incremental rate curve. Incremental rate (IR) is defined as
dl
IR
dL
Incremental rate is obtained by plotting values of I.R. against
corresponding values of output. Fig. 2.16 (c) shows I.R. curve. This
curve expresses additional energy required to produce an added unit
of output at the given load.
2.18 Economic Load Sharing
The total load to be supplied by the power station should be
economically distributed among the various generating units in-
stalled at the power station. Consider a power station having two
generating units A and B. Fig. 2.14 (a) shows the input outjut-cur-
'es of power generating unit A and B which supply in parallel a
common load. Although unitA requires less input for a given output
it is not essential that unit A should be loaded first and then unit B.
For economical loading the combined input ofunitsA and B should
be plotted against load on unitA for a constant total load. Let a total
load of 4 MW is to be supplied by units A and B.
= 4 MW = L A + L8
where L = Combined output
LA = Output of unit A
L = Output of unit B.
Let the unit B supplies total load of MW and Unit A supplies
zero load.
Now corresponding to zero load on unit A and 4 MW on unit B
the values of input to unit A ('A) and input to unit B (Ij) can be
determined respectively from Fig. 2.17 (a) and thus value of
(l + l can be plotted against zero load on unit A. Again let 2 MW


78 POWER PLANT

is supplied by unit A and 2 M is supplied by unit B (so that total

load remains 4 MW) then values of 1A and 1 can be determined B
corresponding to 2 MW load on each unit and value Of ( I A + IB) can
be plotted against 2 M load on unit as shown in Fig. 2.14 (b). In
this way curve for a total load of 4 MW can be plotted corresponding
to different output of unit A.

tdOC
8001

/ 12OO OPTIM(/Mt.OAO
OCUSFOR 4.',.t6

400 800

q200 ca Jo:

2 4 6 8 10 %.11 4 6 8 FO
Uv/T LOAD (LA OR L8) £0.40 0/i? VN/ A(LA)-F.4W
(t4EGA WA

i) (h)
Fig. 2.17

Similarl y curves for total load of 8 MW, 12 MW etc. can be

plotted. In these curves there is at least one point where combined
input is minimum for a given total load. -Corresponding to this point
f minimum, the load of unit A can be found. Then the load on unit
Ii will be difference of total load and load on unit A.
This method is not easily applicable for power stations, where
more than two power generating units supply the load.
Consider two units A and B sharing a total load Lc.
Let, Ic = Combined input = IA + 'B
= Input to unit A + Input to unit B
and Lc = Combined output = LA + LB
= Output of unit A + Output of unit B.
DifThrentiating(t), we get
dlc dIA dIR0
= dL, (.lI-A r 4c. =0 1
L dLA j
As Ic is constant,

POWER PLAN r ECONOMICS 79
(11,1 - d.IB
(lL\ (1)
dL.1
dIR diR dL8
But
dL . ,j C/LB aL
Now L( = LA + LB
dL dLjj
(IL A (IL A dL%
But L ( ' is constant,
dL

CILA - dLB
dL,4
1dLB
dL.. 3
Substituting in (2), we get
(11 11 (/Ij
dLA - dL
F'ron(1) and (4)
(11,4 - di
(ILA dLB
Therefore, to achieve best economy in load sharing, the slopes
Of the input-output curves for each unit must he equal at L.4 and
LB (Fig. 2.18). In other words, the incremental heat rates are equal.

2.19 Condition for Maximum Efficiency

The load at which efficiency will be maximum the heat rate will
be minimum at that load as efficiency is inverse of heat rate.
Efficiency = LII
Heat rate (H.R.) = ilL
Therefore, for minimum value of heat rate

(HR) = 0
dL
• d(
dLL

Ldi - JilL
0

Ldl = IdL

80 POWER PLANT

Output(L)
Fig. 2.18

I dL
L 1L
This shows that (t (iciency will be naximum at a load where heat
rate is equal to incremental heat rate.
2.20 Choice of Power Station
In a power station energy is converted from one form or another
electrical energy. The power plant should be able to meet power
demand efficiently.
The various factors to be considered while choosing the type of
wer plant are as follows
(i) Type of fuel available. If the site where power plant is to be
installed if near the coal mines then steam power plant is preferred
whereas a hydro power plant is chosen if water is available in larger
quantity. Nuclear power plants are located near river or sea so that
nuclear waste can be disposed off easily. Diesel power plant is used
for supplying smaller loads.
(ii) Power plant site. The power plant site should satisfy the
following requirements
(a) Cost of land as well as taxes on land should he low.
(b) It should be nearer to load centre so that cost of transmit-
ting energy is low.
(c) It should be accessible by road, rail or sea so that transpor-
tation of fuel etc. is easier.

POWER PLANT ECONOMICS 81

l) Sufficient quantity of cooling water should be available

near the site.
(e) The site selected should be away from thickly populated
areas in order to avoid atmospheric pollution and to
reduco noise.
(f') Sufficient space for future expansion of power station
should be available near the site.
g) Site sub-soil conditions should be such that foundation can
be made at reasonable depth.
(iii) Type of load. It is essential for a power plant to maintain
reliability and continuity of power supply at all times.
If the load varies and is not large a diesel power plant should be
chosen. If a hydro power plant of small or suitable capacit y is made
available it can also be used for supplying varying load as it can be
started in shorter period and can take up load quickly. The large
base load is supplied by hydro power plant, steam power plant or
nuclear power plant. The nuclear power plants cannot be operated
efficiently at varying loads. Power plant capacit y must be equal to
at least peak load.
(it) Generating units. There should not be only one generating
unit nor there should be large number of small generating units. It
is economical to use a few generating units of larger size than using
number of small size generating units for the same total capacity.
The aim should be to have generating unit of different capacities
which will suitably fit in load curve so that most of the generating
sets when in use can be operated at nearly full load, There should
beat least two generating sets whatever the total capacity of power
plant maybe.
Steam generators of 80 to 200 MW are in use in big power plants
whereas hydro-electric generators sets up to 200 MW are in use.
(t) Cost. The initial cost of hydro power plant is high although
its operating cost is low. The capital cost of a nuclear power plant is
also high. But higher capacity of a nuclear power plant is an
advantage. A gas turbine power plant is less expensive in initial plus
operating cots than the equivalent steam power plant. The total
cost of diesel power plant per kW of the iristalleeacityi s about
20 to 30% less than that of a steam power plant u. uivalent size.
It is always desirable to run the power station at high load factor
so that the installed capacity of power plant is used to the best
possible extent and maximum possible amount ofenergy is produced
which will lower the total Cost both fixed and operating costs. A poor
load factor increases cost per kilowatt hour.
82 POWER PLANT

2.21 Effect of Variable Load on Power Plant

Operation and Design
The variable load problem affects power plant design and opera-
tion and the cost of generation. To Supply variable load a careful
study of load duration curve is needed. This will help to decide the
capacit y of base load plant and peak load plant. The base load plant
should he run at high load factor. The peak load plant should be of
as small capacit y as possible to reduce cost per kilowatt hour (kWh).
Economical load sharing between base load plant and peak load
plant is desirable. Steam power plants and nuclear power are
preferred as base load plants whereas diesel power plant and hydro
power plant. can be used as peak load plant. Il ydro power pint with
larger water storage can also be used as base load plant. If the whole
of load is to be supplied by the same power plant then the prime
movers and generators should act fairly quickly and take up or shed
load without variation of the voltage or frequency of the system. An
important requirement is threfore the control of supply of fuel to
the prime moverby the action of governor. The equipment to be used
for variable load conditions should be so designed that it operates
at lower loads with nearly same efficiency as at full load. The size
of the generators should be so chosen as to suit and fit into the
portions of the predicted load curve. If the load conditions differ too
much from time ideal the cost of energ y increases.

Example 2.1. A power station has a maximum demand of

80 x 10 k Wand daily load curve is defined as follows

06 I688J2Jj21iFlJTh26f2224
t
I
L±.L_jL_L°
50 ...........
(a) Determine the load factors of power station.
• (Ii) What is the load factor of standby equipment rated at 25
MW that takes up all load in excess of 60 MW ? Also
calculate its use factor.
Solution. (a) Load curve is shown iii Fig. 2.19. Energy
generated (area under the load curve)
=40x6+50x2+60x4+ 50x2 *70x4
•1 8 x 4 +40 x 2
= 1360 M\Vh = 1360 x 10 kWh.
1300 x 103
Average load = = 56,666 kW.

Maxinu.imn demand = 80 x 10 3 M.


POWER PLANT ECONOMICS 83

Averagjoad 56,666
Load Factor 0.71. Ans.
- Max. demand = 80,000 =

1
0C
0
4

TIME CHOURS
Fig. 2.19

(h) Now the standby equipment supplies

10 MW (70 - 60 = 10) for 4 hours
and 20 I1\V (80 - 60 = 20) for 4 hours.

Energy generated by standby equipment

=(10x4+20x4)x10 3 =120X 10kWh
Time for which standby equipment remains in operation (from
the load curve) = 8 hours

Average load = = 15 x 103 kW

5 X_103
Load factor = = 0.75. Ans.
20 x
E
Use factor =
C x ti
where E = Energy generated
C = Capacity of the standby equipment.
= Number of hours the plant has been in operation.
120x 10
Use factor = 0.6. Ans.
25 x 103 x 8

Example 2.2. A central power station has annual factors as

fOl/OLS
Load factor = 60c7c Capacity factor = 40n/
Use flictor = 45%,
t'ou'i'r station has a maximum demand of 1.5,000 hW.

84 POWER PLANT

Deter,nznL'
(a) Annual energy production
(b) Reserve capacity Over and above peak load
(c) flours per year not in service.
Solution.

(a) Load factor = Maximum demand

0.6 = Avergeo
15,000
Average load = 15,000 x 0.6 = 9000 kW.
Annual energy produced
E = 9000 x 8760 = 78.84 x 10 6 kWh.
E
b) Capacity factor =
here C = capacit y of the plant
t= time in hours in one year = 8760 hours
78.84 x 106
0.4- Cx8760

4xI = 22,500kW
0.4 x 8760
Reserved capacit y = 22,500 - 15,000 = 7500 k\V.
(c) Use factor =
Cx ti
where ti = Actual number of hours of the year fbr which the plant
remains in Operation.
0.45 - 7884 x 106
22,500 x
78.84 x
ti-
= = 7786 hours.
22,500 x 0.15
Hours per y ear not in service = 6760- 7786 = 971 hours.

Example 2.3. Find the diversity factor of power station ie/eh

Supplies the follou'ing loads
Load A . Motor load of 100 k V working between lOAM. 0/1(1 (;
P.M.
Load B .' Lighting load of GO kW between 6 P.M. and 10A.M.
Load C .' Pumping load of 40 kW between 4 P.M. and lOAM.


POWER PLANT ECONOMICS 85

Solution. In this case sum of maximum demands = 100 + 60 +

40 = 200 kW and the simultaneous maximum demand is found as
follows

Time (kW) Total load (kW)

10A.M. to 4 P.M. 10( in,tor) 100
4 P.i1. 106 1'. 100 mtor)
+ 'pumping) 140
6 P.M. t -10 - 1 , N 1. 6 0 dight
+ 40 iiurnjg 10_0.
10 P.M. to lOAM. 40 (pumping) 40

The above valueshow that simultaneous maximum demand on

the power station is 140 kW.

Diversity factor = = 1.43.

Example 2.4. The annual peak load on a 30 MW power station

is 25 MW. The power station supplies loads having maximum
demands of 10MW, 8.5 MW, 5 MW and 4.5 MW. The annual load
factor is 45%.
Find:
(a) Average load,
(b) Energy supplied treryear,
(c) Diversity firtor,
(d) Demnand /ictor.
Solution. Capacity of power station = 30 MW
Maximum demand on power station = 25 MW
(a) Load factor = - -Av2ra
Mctxiinuxn demand
Average load = i.45 x 25 = 11.25 MW.
(b) Energy supplied per year
= Average load x Number of hours in one year
= (11.25 x 1000) x 8760 = 98.55 x 106 kWh.
(c) Diversity factor
Sum of individuals maximum demands
Simultaneous maximum demand
10 + 8.5 + 5 + 4.5 28 112
- 25 25
(d) Demand factor of power station
Maximum demand 25
-- Connected load - 10 + 8.5 + 5 + 4.5- 28
25 - 0.89
 POWER PLANT
86

Example 2.5. For a power station the yearly load duration curve
is a straight line from 30,000 to 4,000 kW. To meet the load three
turbo-generator are installed. The capacity of two generators is
15,000 kW each and the third is rated at 5,000 kW. Determine the
following:
(a) Load factor,
(b) Capacity factor or plant factor,
(c) Maximum demand.
Solution. As shown in Fig. 2.20 the load curve is a straight line
from 30,000 to 4,000 kW
Average load
Load factor
- Maximum demand
From the given load duration curve.

30,000 kW

cl
0
-.

4000 kV
4 87601iours
Time__-
Fig. 2.20

Energy generated per year

= Area under the curve

= 4000.x 8760 + - x 8760 x 2600

= 8760 (4000 + 13,000) = 8760 x 17,000 kWh

= Energy eerated per year
8760
8760 x 17000
Average load 8760 - = 17,000 kW
=
Maximum demand = 30,000 kW.

Load factor == 0.57 = 57%

30, 000

Capacity factor = Capacity of plant x ñme

 POWER PLANT ECONOMICS
87

Capacity of plant = 15,000 + 15,000 + 5,000 = 35,000 kW

8760 x 17,000 - 17,000 -
Capacity factor = 35,000 8760 - . 35,000 - 0.49 49%.

Example 2.6. A power station has two 60 MW units each

running for 7000 hours a year and one 30 MW unit running for 1500
hours a year. The energy produced per year is 700 x 10 kWh.
Calculate the following
(a) P/unt load factor
(h) Plant use factor or utilisation factor.
Solution. (a) Total capacity of power plant,
C = 60 + 60 + 30 = 150 MW = (150 x 1000) kW.
x io
Average load = kW.
$760
(8760 being the number of hours in one year)

Load factor = iygjpad

Maximum demand
(Assuming power stations capacity.equal to maximum demand)
700x 106
= 8760x150x 1000 = 0.53 = 53%. Ans.
(b) Actual energy generated = 700 x 10 6 kWh.
Energy that could be generated by two 60 MW units and one 30
MW units
=2<60x7000+30 x 150084x10 4 +45x 103
= 88.5 x 104 MWh = 88.5 x 101 x 103 kWh
= 885 x 10 6 kWh

Plant use factor - 7oo x 106 = 0.79 = 79%. Ans.

- 885 x 106
Example 2.7. A base load power station and standby power
station share a common load as follows
Base load station annual output = 150 x 10 6kWh
Base load station capacity = 35,111V
Maxim urn demo ad on base load station = 30,1111'
Standby station capacity = 18 MW
Standby station annual output = 14 x 10' kWh
Maximum demand (peak load) on standby station = 15 MW
Determine the following for both power station
(a) Load factor
('apacit)' factor (Plant factor).


POWER PLANT
88

Solution. (a) Base load station

Average load = = 17,100 kW.

Aver eload
Load factor =
Maximum demand
17,100 170 0.57
30x10003OO
Energygjrated
- plant x 8760
Capacity factors = Capacity of

= 150 LJP = 0.49 = 49%

35.8760
(b) Standby power station
6
Annual average load - 8760 1600 kW.

Avcruage load
Annual load factor = --_
Maximum demand
1600 = x 100 = 10.7%
15,000 150
geratCd - 14x10
Capacity factor = Ener gycity x 8760- 18 x 1000 x 8760

= 0.09 = 9%. Ans.

Example 2.8. A power station is to supply three region of load

hose peak loads are 20 MW, 15 MW and 25 MW. The annual load
factor is 50% and the diversity factor of tile load at the station is 1.5.
Determine the follow jng:
(a) Maximum demand on the station,
(h) Installed capacity suggesting number of unit.
(c) Annual energy supplied.
Solution. (a) Diversity factor
Sum of individual maximum demand
Simultaneous maximum demand

Simultaneous maximum demand

Sum of individual maximum demand
- Diversity factor
20 t 1±J2. = 40 MW.
-- 15

POWER PLANT ECONOMICS
89
Hence maximum demand on station = 40 MW.
(b) Installed capacity of the plant can be taken equal to sum
of individual maximum demand.
Installed capacity 20 + 15 + 25 = 60 MW.
Two units each of capacity 30 MW can be installed.
(c)Load factor =Averageoa4
Maximum demand
Average load = Load factor x Maximum demand
= 0.5 x 40 x 1000 = 20 x 1000 kW.
Energy produced per year
= 20 x 1000 x 8760 = 175.2 x 106 kWh.
Example 2.9. A power station is said to have a use factor of47%
and capacity factor of 40%. How many hours did it operate during
the year.
Solution. Let E = Energy produced
C = Capacity of plant
ti = Number of hours the plant has been in
operation
t = 8760 hours (number of hours in an year)
E
C xt1
E

Dividing (ii) by (i), we get

0.40 E Cxt1 t1
0.47_Cx8760X E 8760
tj = 0.40 x8760 = 74t,5
-
hours.

Example 2.10. Determine the generating cost per unit of 8 MW
power station with the following data: -
Capital cost = Rs. 40 x 10'
Annual cost of fuel = Rs. 80,000
Annual wages and taxes = Rs. 90,000
Interest and depreciation = 10%
Annual load factor = 45%
Solution. Interest and depreciation

—8
 POWER PLANT
90

= -x4x 105 =Rs. 4 x io

0
Cost of fuel + Cost of wages and taxes = 80,000 + 90,000
= Rs. 170,000 = Rs. 1.7 x 105
- Average load
Load factor
- Maximum demand
Average load = Load factor x Maximum demand
= 0.45 x 8 = 3.6 MW = 3600 kW
Energy produced per year
=3600x8760 kWh =31X 10 6 kWh.
Total cost = (4 x - 1.7 x 10) = Rs. 5.7 x iø
Energy cost per kWh
= 5.7 x 10' x 100 1.84 paise.
31x 106
Example 2.11. Ina 60 MW steam power station working at 40%
load factor the energy cost is found to be 1.5 paiselkWh. Calculate
the cost of energy if the power station load factor is improved to 50%.
Due to increased energy generation the fuel cost increases the annual,
generation cost by 6%.
Solution. Load factor = Average load
Maximum demand
Average load = (0.4 x 60,000) kW
Annual energy generated
= 0.4 x 60,000 x 8760 kWh = 210 x 106 kWh

Total annual cost 210 x 106 x = Rs. 315 x io

Now when the load factor is 50%.

Annual energy generated
= 0.5 x 60,000 x 8760 = 262 x 106 kWh.
Annual cost at 50% load factor
= 315 x 104 x 1.06 = Rs. 333 x iO
333 x iO
Cost per kWh 262 x 106 = 1.23 paise. Ans.
 POWER PLANT ECONOMICS
91'
Example 2.12. A 150 H.P. motor is required to operate at full
load for 800 hours per year, at half load for 1800 hours per year and
to be shut down for the remainder of the time. The motors available
are
Motor M: Cost Rs. 11,000
Efficiency at full load = 88%
Efficiency at half load = 85%
Motor N
Efficiency at full load 90%
Efficiency at half load = 8791,
Determine the maximum price which could economically be paid
for motor N if the energy rate is 8 paise per kWh. Interest and
depreciation cost may be taken as 12% per year.
Solution. Motor M: Output at full load
= 150 x 0.735 x 0.88 = 96 kW.
(Assuming 1 H.P. = 735 watts)
Output at half load = 75 x 0.735 x 0.85 = 47 W.
Energy consumed per year
= 96X800+47X 180O. 161,400 kWh
Cost of energy = 161,400 x = Es. 12,912

Interest and depreciation = io x 11,000 = Rs. 1320.

Total cost = 12,912 + 1320 = Es. 14,232
Motor N: Output at full load
= 150 x 0.735 x 0.9 = 99 kw.
Output at half load = 75 x 0.735 x 0.87 = 48 W.
Energy consumed per year
= 99 x 800 + 48 x 1800 = 166,600 W.
Cost of Energy = 166,600 x Rs. 13,328.
Let the cost of motor = C
Interest and dnrecjatjon = 12
j-xC

Total cost = Rs. 113,328+100x C]

Assuming annual cost of motor Al and motor N to be equal

 POWER PLANT
92

14,232 = 13,328 + 12x C

C = Rs. 7533. Ans.

Example 2.13. The load of a residential consumer for a day was

found to be as follows:
From 12.00 midnight to 6A.M. No load
From 6A.M. to 8A.M. 120W
From 6A.M. to 9.30A.M. 540 W
From 930A.M. to 11.30A.M. No load
From 1130A.M. to2P.M. 480W
From 2 P.M. to 3.30 P.M. No load
From 3.30 P.M. to 5P.M. 120 W
From 5 P.M. to 6.30 P.M. 360 W
From 6.30 A.M. to 9A.M. 480 W
From 9P.M. to 12 mid night 120 W
Calculate
(a) Energy consumed
(b) Load factor.
Solution. Energy consumed during the day
= 120 x 3 + 540 x 1.5 + 480 x 2.5 + 120 x 1.5
+ 360 x 1.5 + 480 x 2.5 + 120 x 3
= 4470 Wh = 4.470 kWh.
= 4470
= 186 W.
Average load
Maximum demand = 540 W.
- Averjd
Load factor - Maxithum demand
Ans.

Example 2.14. In a steam power station having a ,naximum

demand of20 MW the boiler efficiency and turbine efficiency are 80%
and 90 01c. respectively.
If the coal consumption is 1 kg per unit of energy generated and
cost of coal Rs. 40 per tonne ; determine:
(a) The thermal efficiency of power station,
(b) Cost of coal per year.
Annual load factor is 50%.
Solution. (a) Thermal efficiency
= Boiler efficiency x Turbine efficiency
 POWER PLANT ECONOMICS
93

= 0.8 x 0.9 = 0.72 or 72%

(b) Load factor = Average load
Maximum demand
Average load = 0.5 x 20 = 10 MW
Energy generated per year
= (10 x 1000) x 8760 = 87.6 x 10 6 kWh.
106 x 1
Coal consumption = 87.6 x
1000
= (87.6 X 10) tonnes.
Annual cost of coal = (87.6 x 10) x 40 = Rs. 3,504,000.
Example 2.15. Determine the annual cost of a feed water sof-
tener fro,n the following data
Cost = Rs. 80,000
Salvage value = 5%
Life = 10 years.
Annual repair and maintenance cost = Rs. 2500
Annual cost of chemicals = Rs. 5,000
Labour cost per month = Rs. 300
Interest on sinking fund = 5%.
Solution. Capital cost,
P = Rs. 80,000

Salvage value, S Rs. x 80,000

n = Life 10 years.
Rate of interest on sinking fund, r = 5%.
Annual sinking fund payment
r
=(P-S)[--
L(' + r)' -1

= 0.95 x 80,000 x _0.05 1

[(I+ o.5)1°_ij

= Rs. 0.95 x8 x 104 0.05 ) Rs. 6040

(
Total cost per year
Annual sinking fund = Rs. 6040
Annual repair and
maintenance cost = Rs. 2500

94 POWER PLANT

Annual cost of chemicals = Rs. 5000

Annual labour cost = Rs. 3600
Total cost = Rs. 17,140.

Example 2.16. A 500 kW electric power station cost Rs. 800 per
kW installed. The plant supplies 150 kWfor 5000 hours of the year,
400 kW for 1000 hours and 25 kW for the remaining period. Deter-
mine the cost of production per unit of electric energy. The fixed
charges are 10% and operating charges 7paise per kWh.
Solution. Energy supplied
= 5000 x 150 + 1000 x 400 + 2760 x 25
= 1,219,000 kWh

Fixed cost = 10x (800 x 500) = Rs. 40,000

( 40,000 \
Cost per kWh
=1\ 1,219,000J 10.3 paise.

Example 2.17. (a) Compute the monthly bill and unit energy
cost for a total consumption of 1600 kWh and a maximum demand
of 10 kW using Hopkinson demand rate quoted as follows:
Demand Rates
First kilo-watts of maximum demand Rs. 75 per kW per month.
Next 5 kW of maximum demand at Rs. 12 per kWper month.
Excess over 6 kW of maximum demand at Rs. 10 per kW per
month.
Energy Rates
First 50 kWh at 15 paise per kWh.
Next 50 kWh at 12 paise per kWh.
Next 300 kWh at 8 paise per kWh.
Next 500 kWh at 6paise per kWh.
Excess over 900 kWh at 4 poise per kWh.
(b) Also find the lowest possible bill for a month for 3 days and
the unit energy cost on the given energy consumption.
Solution. (a) Demand charge = 15 + 5 x 12 + 4 x 10
= 15 + 60 + 40 = Rs. 115.
Energy charges = 50 x 0.15 + 50 x 0.12 + 300 x 0.08
+ 500 x 0.06 + 700 x 0.04
 POWER PLANT ECONOMICS 95
= Rs. 95.50
Monthly bill = 115 + 95.50 = 210.50
Unit energy cost = 210,50
1600
= 13.15 paise per kWh.
(b) The lowest bill occurs when the demand is maximum
which is possible at 100% load factor.
Maximum demand = Average load

= 30x24 = 2.2 kW.

Demand charge = 15 + 1.22 x 12 = 15 + 14.64 = Rs. 29.64
Energy charge = Rs. 95.50
Minimum monthly bill
= 29.64 + 95.50 = Rs. 125.14.
Unit energy cost -- 25.14 -
1600 - 7.8 paise per kWh.

Example 2.18. The annual cost of operating a 15,000 kW

thermal power station are as follows:
Cost of plant = Rs. 900 per kW
Interest, insurance taxes on plant = 5%
Depreciation =5%
Cost of primary distribution system = Rs. 500,000
Interest, insurance, taxes and depreciation on primary distribu-
tion system =5%
Cost of secondary distribution system = Rs. 900,000
Interest, taxes, insurance and depreciation on secondary dis-
tribution system = 5%
Maintenance of secondary distribution system = 180,000.
Plant Maintenance cost
(i) Fixed cost = Rs. 30,000
(ii) Variable cost = Rs. 40,000
Operating costs = Rs. 600,000
Cost of coal = Rs. 60 per tonnes
Consumption of coal = 30,000 tonnes
Dividend to stock-holders = Rs. 10,00,000
Energy loss in transmission = 10%
Maximum demand = 14,000 kW
Diversity factor = 1.5
 POWER PLANT
96

Load factor = 70%

Device a two part tariff
Solution. Maximum demand = 14,000 kW
Average load
Load factor = 0.7 Maximum demand

Average load = 0.7 x 14,000

Energy generated per year = 14,000 x 0.7 x 8760
= 85.8 x 106 kWh
Cost of plant = Capacity of plant x Cost per kW
= 15,000 x 900 = Rs. 1,35,00,000
Interest, insurances, taxes on plant
=100 x 13,500,000 = Rs. 675,000

Plant depreciation =x 13,500,000 = Rs. 67.5,000

Interest, insurance, taxes, depreciation on primary distribu t ion

system
=x 500,000 Rs. 25,000

Cost of secondary distribution system

= Rs. 900,000.
Interest, insurance, taxes depreciation on secondary dis-
tribution system
=x 900,000 Rs. 45,000

Cost of coal = 60 x 30,000 = Rs. 18,00,000

Fixed Costs
Interest, taxes and insurance on plant = Rs. 675,000
Plant depreciation = Rs. 675,000
Intciest, taxes, insurance and depreciation
on primary distribution system = Rs. 25.000
On secondary distribution system = Rs. 45,000
Fixed part of plant maintenance = Rs. 30,000
Dividend of stock-holder = Rs. 10,00,000

Total fixed cost = Rs. 24,50,000


POWER PLANT ECONOMICS 97

Sum of maximum demand of consumers

= Diversity factor x Maximum demand
= 1.5x 14,000 = 21,000 kW
Chargeper kW per year
24,50,000 = Rs 116.6 per kW per year.
21,000
Variable Costs
Cost of coal = Rs. 18,00,000
Plant maintenance = Rs. 40,000
Variable part of operating cost = Rs. 600,000
Total variable cost = Rs. 26,20,000
Energy loss in transmission = 10%
Net energy transmitted
=0.9x85.8x106=77.2xlO6kWh.

Rate per kWh = = 3.4 paise per kWh.
72.2 x 100

' Example 2.19. The annual costs expected by a utility system in

supplying certain residential suburb of 40,000 customers are as
follows:
Fixed charges = Rs. 2,40,00,000
Energy charges = Rs. 17,86,000
Customer charges = Rs. 210,000
Profit = Rs. 168,000
Maximum demand =5000 kW
Energy registered on
customer's metres = 17 x 106 kWh
Diversity factor =4
1i
Form a three charge rate putting 1 profit in fixed charges n
j of
energy charges and I in customer charges.
Solution. Sum of maximum demands of customers
=5000x4 =20,000kW

Fixed cost = 2,400,000 + 1 x 168,000 = 12.000.

Cost per kW per year

- 24,42,000 - Rs 122.10
- 20,000 -
98 POWER PLANT

Energy cost = 17,86,000 + x 1,68,000 = Rs. 18,70,000

Energy rate- -187,00,000

17 106 --11 paise per kWh.

Customer charge = 210,000 + x 168,000 = Rs. 252,000
Charges per customer per year
- 252,000 =
- 40,000 6.30.

Example 2.20. An input-output curve of a 10 MW station is

expressed as follows:
1= 106(I6+8L+0.4L2)
where I is in kcal per hour and L is in mega watts.
(a) Without plotting any curve find the load at which the
maximum efficiency occurs.
(b) Find the increase in input required to increase station
output from 3 to 5 M by means of the input-output curve
and also by incremental rate curve.
Solution. (a) I= 10 6 (10 + 8L + 0.4 L2)
=lO6[4+8+o.4L]

Efficiency Output
In -L
put -
1
11=
106(+8+0.4LJ
Now for maximum value efficiency
411 - 0
dL
Differentiating (1), we get
- 1o6(_+ 0.4 1
drl L2
dL

POWER PLANT ECONOMICS 99

_106[3+0.4]=0

or Lo
-i=O.4
L
10-0.4L2=0
L2==25

L =5 MW.
(b) When load L = 3 MW
Input 13=106110+8x3+0.4x321
= 106 [10 + 24 + 3.61 = 37.6 x 106 kcal/hour.
When load L=5MW
Input 15 = 106 [10 1+ 8 x 5 + 0.4 x 52]
= 106 110+40+101
= 60 x 106 kcal/hour:
Increase in input required = 1 - 13
= (60 - 37.6) x 10 6 = 22.4 x 106 kcal/hr.
From incremental rate curve
When load varies from 3 to 5 Mw, the incremental rate may be
considered to be straight line and the average height of area under
the curve between 3 MW and 5 MW would be a
= 4 MW

8L+0.4L2)

Increment rate, ir, L) x 106

JR = (8 + 0.ô 4) x 106, when load = 4 MW

= (8 + 3.2) x 106 (112) x 106
Hence total increase in input
=411.2 x 2) x 10 6 = 22.4 x 106 kcalihour.
Thus increase in input needed to increase stations output from
3 5 MW in both cases is same. This shows that the incremental
to
rate curve can be assumed to be straight line.

100 POWER PLANT

Example 2.21. The incremental fuel costs for two generating

units A and B of a power plant are given by the following questions:
dFA = 0.O6PA + 11.4
dPA
dFB
dPB =0.O7PB+10
where P is in mega-watts and F is in rupees per hour.
(a) Find the economic loading of the two units when the total
load to be supplied by the power station is 150 MW.
(b) Find the loss in fuel cost per hour if load is equally shared
by the two units.
Solution. (a) Let PA = Load supplied by unit A
Pa = Load supplied by unit B
PA+PB= 150
For economic loading of the two units
dFA - dFB
dPAdPR
0.06PA+11.14=0.07P8+10
Solving (1) and (2)
PA = 70 MW
P = 80 MW.
(b) If the load is equally shared by the two units that is if each
unit supplies 75 MW, then increase in cost for unit A

=s:.o.o6 PA + 11.4) dPA

75
= [0.03 PA + 11.4 PA]7 R.. 78.75 per hour.
Increase in cost for unit B will be

=f(0.07Pa+ 10)dP2

= [0.035 P1 + 10 Pa1
so = - Rs. 77.12 per hour.
This shows that in case of unit B there is decrease in cost.
Hence net increase in cost due departure from economic dis-
tribution = 78.75 - 77.12 = Rs. 1.63 per hour.
POWER PLANT ECONOMICS 101

Example2.22.AnY undertaking consumes 6 x 10 6 kWh peryear

and its maximum demand is 2000 kW. It is offered two tariffs.
(a) Rs. 80 per kW of maximum demand plus 3 paise per kWh.
(b) A flat rate of6paise per kWh.
Calculate the annual cost of energy.
Solution. (a) (According to first tariff the cost of energy)
=2000x80+-jx6X 106
= 160,000 + 180,000 = Es. 340,000
(b) Cost of energy according to flat rate
=x 6 x 106 = Es. 360,000

Example 2.23. Two lamps are to be compared:

(a) Cost of first lamp is Re. I and it takes 100 watts.
(b) Cost of second lamp is Rs. 4 and it takes 60 watts.
Both lamps are of equal candle power and each has a useful life
of 100 hours. Which lamp will prove economical if the energy is
charged at Rs. 70 per kW of maximum demand per year plus 5 paise
per KWh?
At what load factor both the lamps will be equally advantageous?
Solution. (a) First Lamp
Cost of lamp per hour
== 0.1 paise
Maximum demand per hour
=J=0.1kW.
Maximum demand charge per hour
=0.1x 70x100
7860 =0.O8paise
Energy consumed per hour
= 0.1 x 1 = 0.1 kWh.
Energy charge per hour
= 0.1 x 5 = 0.5 paise
Total cost per hour = 0.1 +0.08 + 0.5 = 0.68 paise.
 102 POWER PLANT

(b) Second Lamp

Cost of lamp per hour
4x100
= 1000 = 0.4 paise
Maximum demand per hour

== 0.06 kW
Maximum demand charge per hour
= 0.06.x 70 x 100 = 0.048 paise.
• Energy consumed per hour 0.06 x 1 = 0.06 kWh.
Energy charge per hour = 0.06 x 5 = 0.30 paise.
Total cost per hour = 0.4 + 0.048 + 0.3 = 0.748 paise.
Therefore, first lamp is economical.
Let x be the load factor at which both lamps become equally
advantageous. Only maximum demand charge changes with load
factor.
0.1+ 0' 08 0.5 = 0.4 + 0.048 +0 3
x x
x = 0.32 or 32%.
Example 2.24.A new factory having a minimum demand of 100
k Wanda load factor of25% is comparing two power supply agencies.
(a) Public supply tariff is Rs. 40 per kW of maximum demand
plus 2 paise per kWh.
Capital cost = Rs. 70,000
Interest and depreciation = 10%
(b) Private oil engine generating station.
Capital Cost = Rs. 250,000
Fuel consumption = 0.3 kg per kWh
Cost of fuel = Rs. 70 per tonne
Wages = 0.4 poise per kWh
Maintenance cost = 0.3 poise perk Wh
Interest and deprecjatjcn = 15%.
Solution.
Load factor = Average load
Maximum demand
Average load = 'oad factor x Maximum demand

POWER PLANT ECONOMICS 103

=0.25x700= 175 kW.

Energy consumed per year
= 175 x 8760 = 153.3 x io kWh.
(a) Public Supply
Maximum demand charges per year = 40 x 700 = Rs. 28,000.
Energy charge per year
x 153.3 x 104= 30,660
=
Interest and depreciation

= x 70,000 = Rs. 7,000.

Total cost = Rs. (28,000 + 30,660 + 7,000)
= Rs. 65,660
Energy cost per kWh
65,660
= 153.3 x io 100 = 429 paise
(b) Private oil engine generating station
Fuel consumption = 0.3 x 153.3 x 104 = 460 tonnes
1000
Cost of fuel = 460 x 70 = Rs. 32,000
Cost of wages and maintenance
100
= [0.4+0.3) x 153.3 x 10 = Rs. 10,731.

Interest and depreciation

= x 250,000 = Rs. 37,500

Total cost = Rs. [33,203 + 10,731 + 37,500]
Rs. 80,431
Energy cost per kWh
= 80,431
x 100 = 5.2 paise.
153.3 x 104
Example 2.25. A diesel engine power plant has one 700kW and
two 500 kW generating units. The fuel consumption is 0.28 kg. per
kWh and the calorific value of fuel oil is 10,200 heal per kg. Estimate

104 POWER PLANT

the fuel acquired fora month and overall efficiency ofthe plant. Plant
capacity factor = 40%.
Solution. Capacity factor = E
where c = capacity of the plant
=700+2x500= 1700 kW
t = time (hours) in the given period
= 30 x 24 = 720 hours.
E
- 2700 x 720
E = 0.4 x 1720 x 720 = 48,96,000 kWh.
Fuel oil consumption = 48,96,000 x 0.28 = 137,088 kg.
- Output
Overall efficiency
Input
Output = 48.96,000 kWh = (48,96,000 x 860) kcal
48,96,000 x 860
Overall efficiency = 0.3 or M. Ans.
137,088 10,200
Example 2.26. The incremental fuel costs of two generating
units A and B of a power station are given by the following expres-
sions:
dFA
^iPA =0.01 PA+2.25

dFB
=0.0J5Ps+1.5
dPB
where F is in rupees per hour and P it is MW. Determine incremental
fuel cost and loading schedule for minimum cost if the total load to
be supplied is to be 50 MW 175 MW and 200MW. Both units operate
at all times and maximum and minimum load on each unit is to be
100 MW and 15 MW respectively.
Solution. Let P = total load
PA = load supplied by unit A
RB = load supplied by unit B

P = PA I-PB
As the incremental fuel cost of unit A is higher than that of B so
for a total load of 50 MW the unit A should supply 15 MW.
 POWER PLANT ECONOMICS
105
PA=15MW
PB=35MW
dFA
= 0.01 x 15 + 2.25 = Rs. 2.40 per MWII
dPA
dFB
^PB = 0.015x 35 + 1.5 = Rs. 2.025 per MWh

when the load increases the unit B will continue to supply the
additional load till = 2.40.
dP

2.40 = 0.015 P8 + 1.5

PB9-6OMW
When P=75MW
PA = 15 MW and P8 60 MW
For total load P = 175 MW
PA+PB= 175
.(i)
and for economical loading
dFA - dF8
dPA - dP8
0.01 PA + 2.25 = 0.015 PB + 1.5
.(ii)
Solving (i) and (ii), we get
PA = 75 MW
P8 = 100 MW.
When the total load = 200 MW
PA = 100 MW
P8 = 100 MW.
dFA
= 0.01 x 100 = 2.25 Rs. 3.25 per IfWh
dFB
= 0.015 x 100 + 1.5 = Rs. 3 per MWh.

Example 2.27. A daily load

curve which exhibited a 20 minute
peak load of 6000kW is drawn to scale.
1 cm = 2 hour 1 cm =
400 kW. The area under the curve is found
to be 75 square centimetres. Find the load factor.
—9


POWER PLANT
106
4
Solution. 1 square cm = 400 x 2 = 800 kWh.

Average load = 8002-4-----=2500kW

x 75

Load factor = ygjoad = 2500 = 0.41. Ans.

Peak load

Example 2.28. In a power distribution system a certain feeder

supplies three distribution transformers each one supplying a group
of transformers whose connected loads are as follows:

Transformer I 1 Transformer 2 Transformer 3

Residence Lighting General Power seruice
Store lighting and Power

(j)IORW,5H.P. (005
kW (i)10H.P.,5kW
e;;1.1 bW (ii20 kW (it) 15 H.P.

Assuming motor efficiency 70% and suitable demand factor and

diversity factor from Tables 2.2 and 2.3. Calculate maximum feeder
load.
Solution. Demand factor
= Individual Maximum Demand
Connected Load
Individual Maximum Demand
= Demand factor x Connected load
Transformer 1
Sum of individual maximum demand
= 10 x 0.70 + xO.746 x 0.75 + 4 x 0.70

= 7 + 4 + 2.8 = 13.8 kW
Sum of individual maximum demand
Diversity factor = Simultaneous maximum demand

•. Simultaneous maximum demand on transformer 1

Sum of individual maximum demand
M' - - Diversity factor
== 9.2 kW.
Similarly simultaneous maximum demand on transformer 2
and transformer 3 is given by M2 and M3 as follows:


POWER PLANT ECONOMICS 107

= 0.5 x 0.6 + 20 x 0.5 -10.3

M2
3.5 3.5
- = 2.94 kW.
10 x 0.746 15 x 0.746 x 0.651
_0X0.75+5x0.60+
Ms=j{ 0.70

[8 + 3 + 10.4] = = 14.2 kW.
= 1.5 1.0
Assuming a diversity factor of 1.3 between transformers.
Maximum feeder load
- M 1 + M2 + M3
- 1.3
- 9.2 + 2.94 + 14.2 - 20.8 kW. Ans.
—
1.3 -
Example 2.29. A power station has to supply loud as follows:
Time (Hours) 0-6 6-12 12-14 14-18 18-24
Load (MW) 30 90 60 100 50
(a) Draw the load curve.
(b) Draw load-duration curve.
(c) Choose suitable generating units to supply the load.
(d) Calculate load factor.
(e) Calculate plant capacity.
Solution. The load curve is shown in Fig. 2.21 and the load
duration curve is shown in Fig. 2.22.

T IME (HOURS) -

Fig. 2.21

108 POWER PLANT

a
a

TIME (HOURS)

Fig. 2.22 Load duration curve.

Energygenerated =30x6+90x64-60x2+ 100x4+50x6

= 1540 MWh = 1540 x 103 kWh
1540 x
Average Load = iokW
24
Maximum demand 100 x 103 kW

Load factor 1540 x 10= 0.64 Ans.

24 x 100 x io
To supply the load three generating units each of 30 MW
capacity and one generating of 10 MW capacity should be selected.
One additional unit should be kept as standby. Its capacity should
be equal to the capacity of biggest set, i.e. 30 MW.
Load duration curve will indicate the operational schedule of
different generating units. The operational schedule will be as
follows:
(i) One generating unit of 30 MW will run for 24 hours.
(ii) Second generating unit of 30 MW will run for 18 hours.
(iii) Third generating unit of 30 MW will run for 10 hours.
(iv) Fourth generating unit of 10 MW will run for 4 hours.
E
Plant capacity factor =
CXt
where E = Energy generated (kWh)
C = Capacity of the plant (kW)
=30x4+lxlO=13OMW= 130x 10' kW.
t = Number of hours in the given period = 24 hours.


POWER PLANT ECONOMICS 109

Plant capacity factor

- 1540 x io
= 0.49. Ans.
- 130 x x 24
Example 2.30. It is proposed to supply a load with a maximum
demand of 100 MW and load factor of 40%. Choice is to be made from
Nuclear power plant, Hydropower plant and steam power plant.
Calculate the overall cost per kWh in case of each scheme.

Solution.
Nuclear Power Plant
Capital cost = 2500 x 103 x 100 = Rs. 25 x
Interest = Rs. x 24 x iø

Depreciation = Rs.x 25 x iO

Annual fixed cost (Interest + Depreciation)

= Rs. 2 x 5/100 x 25 x 10 = Rs. 25 x 106
Energy generated per year
= Average load x 8760
= Load factor x Maximum demand x 8760
= 0.4 x 100 x 103 x 8760 = 350.4 x 10 6 kWh.
Running cost/kWh = Operating cost/kWh + Transmission and
distribution cost/kWh
= 4 + 0.1 = 4.1 paise.
Overall cost/kWh = Running cost/kWh + Fixed cost/kWh
25 x 106 x 100
= 4.1+ --
350.4 x 106
= 4.1 + 7.1 = 11.2 paise.
 POWER PLANT
110

Hydropower Plant
Capital cost = 1800 x 100 x 10 3 = Rs. 18 x
Interest=_X18X107r9X106

Depreciation =x 18 x iø = Rs. 7.2 x 106

Annual fixed cost = 9 x 10 6 + 7.2 x 10 6 = 16.2 x 106

Running cost/kWh = 2 + 0.4 = 2.4 paise.
16,2 x 106 x 10
Overall cost/kWh = 2.4 +- = 2.4 + .1 R = 7 paise
350.4 x 106
Steam Power Plant
Capital cost = Rs. 900 x 100 x 10 = Rs. 9 x

Interest = Rs.x 9 x 107

Depreciation = Rs.x 9 x 107

Annual fixed co;t = 2 x x 9 x 107 = Rs. 10.8 x 106

Running cost/kWh = 7 -+ 0.1 = 7.1 paise

Overall costfkWh Running cost/kWh + fixed costJkWh
10.8x 106 x 100 = 10.2 paise
= 7.1 + -
30.4 x 106
Therefore, overall cost/kWh is minimum in case of hydropower
plant.
Example 2.31. Following data pertains to a power plant of 100
MW i nstalled capcicitv.
Capital cost = Rs. 9001kW installed
Interest and depreciation 12%
Annual load factor = 60';7t
Annual capacity factor = 50%
Annual running charges = Rs. 15 x 106
Energ y consumed by the power plant auxiliaries = 60..
Calculate the following
(a) Reserve capacity

POWER PLANT ECONOMICS 111

(b) Cost per kWh.

Average load
Solution. (a) Load factor= Maximum demand
E
Capacity factor =

where E = Energy produced = Average load x Time

C = Capacity
t = Time
Average load
Capacity factor = Capacity
Load factor = Average load < Capacity
Capacity factor Maximum demand Average load
06 - Capacity
0.5 - Maximum demand
Maximum demand
- 0.5 x Capacity 0.5 x 150 = 125 MW
0.6 0.6
Reserve Capacity = 150 - 125 = 25 MW. Ans.
(b) Load factor Maximum demand
Average load = Load factor x Maximum demand
= 0.6 x 125 = 75 MW = 75 x 10 kW.
Energy produced per year
= 75 x 103 x 8760 = 657 x 106 kWh
Energy used by auxiliaries
=x 657x 106

Net energy sent out = 657 x 10 6 - x 657 x 106

= 617.6 x 106 kWh.

Annual Interest and depreciation (Fixed Cost)
-12 x 150 x 103 x 700 = Rs. 12.6 x 105

Total annual cost = Fixed cost + Running cost

= 126 x + 15 x10 6= Rs. 27.6 x 106

112 POWER PLANT
.

Cost of generation = 27.6 x 106 x 100 = 4.47 paise. Ans.

617 x 106
Example 2.32. It is proposed to supply a load either by a hydro
power plant or by a steam power plant. The following data may be
assumed as shown below:
Reserve capacity for hydropower plant is 30% and for steam
power station is 20%. Determine the load factor at which the overall
cost per kWh would be same for both the power plants.

Capital cost per kW I Es. 1800 I Rs. 1000

Solution. Let, Maximum demand

= M kW
Load factor =F
Average load =MxF
Energy generated per year Average load x 8760
Energy generated per year = M x F x 8760 kWh.
Hydropower Plant
Reserve capacity = 30%
Installed capacity 1.3 M kW
Capital cost = Rs. 1800 . = Rs. 2340 M.
Fixed cost per year (Interest and depreciation)
=Rs.-j1 x2340M=Rs. 234M
Overall cost/kWh
= Running cost per kWh + Fixed cost per kWh
( 234M
=3+MF876öX 100)Paise.

Steam Power Plant

Reserve capacity = 20%
Installed capacity = 1.2 M kW
Capital cost = Rs. 1000 x 1.2 M = Rs. 1200 M
 POWER PLANT ECONOMICS
113
Fixed cost per year = (Interest + Depreciation)
Rs. - L4- 1200M=Rs. 168M
1
Overall cost per kWh = Running cost per kWh + Fixed cost per
kWh
= 6+ 168 M x iooj paise.
MFx 8760

As the overall cost per kWh is same for both the power stations.
234M 168Mx100
3+MFxl00=6+ MFx8
66Mx 1000
MFx8760
600
Fx87603
F=0.25
% Load factor = 25%. Ans.
Example 2.33. The input output curve of 100 MWpower station
is expressed as follows:
1= 10 (100+2L +0.00041!)
where I is in kcal/hr and L is in MW.
Determine (a) Input, heat rate and efficiency when load is 40
jlI%V
(b) Load at which efficiency is maximum.
Solution.
(a) 1= 106(100+2L+00004L3)
When L= Load =40MW
i = 106 (100 + 2 x40 + 0.0004 x 403)
= 106 x 205.6 kcal/hr.
Heat rate (I) = 106(100 +
2 + 0.0004 x L2)

= 10' (100 ± 2 x.0004 x 40)

kcal/MWh.
= 102.64 :< 10

FJficiv O -utput 40 x 10 x 860

=-------•-----

Input = 10 x
205.6
 - POWER PLANT
114

(As 1 kWh = 860 kcal)

(b) Efficiency will be maximum when heat rate is equal to
incremental rate.
i = 106 (100 + 2L + 0.0004 x L3)
H.R. = f = 10 6 + 2 + 0.0004 L 2 ](L

Incremental rate, dL = 106 ( + 2 + 0.00012 L2)

106 100 + 2 + 0.0004 L 2] 106 (2 + 0.0012 L2)

+ 0.0004 L 2 = 0.00012 L2

= 0.0008 L2

L2 = 0.0008= 125,000
L = 50 MW. Ans.

Example 2.34. Calculate the cost of generation per kWh for a

power station having the following data:
Installed capacity of the plant 120 MW.
Capital Cost Rs. 96 x 106
Rate of intere.t and depreciation = 14%
Annual cost of fuel oil, salaries and taxation = Rs. 12 x iO
Local factor = 40% *
Also find the saving in cost per kWh if the ann 'load factor is
raised to 50%.
Averagjoad
Solution. Load factor = Maximum dernan.
Assuming maximum demand equal to c. )acity of the power
plant
0.4 = Average load
120
Average load = 120 x 0.4 = 48 MW.
Energy generated per annum
= 48 x 103 x 8760 = 420.38 x 106 kWh
POWER PLANT ECONOMICS 115
Interest and depreciation (Fixed cost)

=.jx96x106=Rs.1344x104

Running cost = Cost of fuel oil, salaries and taxation.

Total annual cost = Fixed cost + Running cost
= 1344 x 104+ 12 x 10 6 = Rs. 25.44 x 106
Cost per kWh = 25.44 x 106x 100
420.48 x 106 = 6 paise. Ans.

When the load factor is r. d to 50%

0.5 = Average load
Maximum demand
Average load = 0.5 x Maximum demand
= 0.5 x 120 = 60 Mw.
Energy produced per year
60 x 10 3 x 8760 =525,6 x 106 kWh.
Total annual cost will remain same.
Total annual cost = Rs. 25.44 x 106
106 x 100
Cost, per kWh = 25.44x = 482 paise.
5 25.6 x 106
Saving in cost per kWh = 6 -4.82 = 1.18 paise. Ans.

Example 2.35. The input output curve of a 10MW power station

is expressed as follows:
i=106(8+L+0.4L2)
where I is in kcal per hour and L is in MW.
Determine the average heat rate of the power station for a day if
it operates at its full capacity for 12 hours and is kept running at zero
load for remaining 12 hours. Also calculate the saving per kWh of
energy produced if the energy is generated at a constant 24 hours
load (10O7(. load factor).
Solution. (a) Energy generated,
E= 10 12+0x 12= 120MWh.
Input I = 110 + it)
where iio = Input when load is 10 MW
Jo = Input when load is zero MW
4

116 POWER PLANT

110= 106 (8+8x10+0.4x 10 10)x 12

=lO6xl28xl2kcal
I=110+10=l36xl2xlO6kcal.
. Average heat rate
- Input _.L.136x12x 106
- Output - E - 120
= 136 x 106 kcal/MWh.
(b) Total energy generated during 24 hours
=E=12OMWh.
120
Average load --- = 5 MW

Input l5 = 106 (8+ 8x5 +0.4x52)

= 106 (58) kcal per hour.

Heat rate = x 10 6 = 11.6 x 106 kcal/MWh

Saving = (13.6- 11.6) 106

= 2 x 10 kcalThIWh. Ans.

Example 2.36. A 20 H.P. condensate pump motor has been

burnt beyond repair. Two alternatives have been proposed to replace
it. Manufacturer A offers to replace the motor for Rs. 5000. The
efficiency of this motor is 90% at full load and 80% at half load.
Manufacturer B offers a motor for Rs. 4200. Its efficiencies are 80%
and 84% at full load and half load respectively.
The life of each motor is 20 years, salvage value 5%, interest on
capital cost 5% and the motor operates 20% of time at full load and
75% at half load.
The annual maintenance cost of motor A is Rs. 400 and that of
motor B is Rs. 500. If the energy rate 10 paise per kWh, find which
motor is more economical to buy ?Also determine at what energy rate
the two alternatives become equal.
Solution. Motor A
Capital cost, (P) = Rs. 5000
Salvage value, (S) = 5 x P

Life, (n) = 20 years.

POWER PLANT ECONOMICS 117
Depreciation per year
_?!
- n
0.95 x500
=Rs.237
= 20
5
Annual interest x 5000 = Rs. 250
=
The motor operates at full load for x 8760) hours and a half
load for x hours in one year.
8760j
Energy cost =[25xO.735xx_!_x_]

+25x O.735xx8760_J._1P. ] -
4 0.85 100
= Rs. (4471 + 14,203) = Rs. 18,674.
Annual maintenance cost = Rs. 400
Total cost = Rs. [237 + 250 4- 1,674 + 4001 = Rs. 19,561.
Motor B
Depreciation per year = 0.95 x 4200 Rs. 199
20
Interest -j-j x 4200 = Rs. 210

Energy cost = [ 25 x 0.735 x 1- x 8760 x 101

+[25 x 0.735 x 3 x 8760 x io l

—
4 682 100
= Rs. [4734 + 14,7221 = Rs. 19,456.
Annual maintenance cost = Rs. 500
Total cost = Its. [199 + 210 + 19,456 + 5001
= Rs. 20,365.
Hence motor A is economical.
(b) Let the energy rate = r paise/kWh
At this rate the two alternatives becomes equal.
Annual cost of motor A
CA=Rs.[237+25O+25XO735X_0X_L


POWER PLANT
118

x + 25 x 0.735 x 8760 x x +400]

100 4 085
Annual cost of motor B
CB = Rs.[199+210x25X0.735XX8760XxiI5o+

3 1 x r 5001
25 x 0.735 x 8760 x 4 x
0.82 100
Equating CA = GB, we get
r= 28 paise. Ans.

• Example 2.37. A diesel power station has ful consumption 0.2

kg per kWh. If the calorific value of the oil is 11,000 kcal per kg
determine the overall efficiency of the power station.
Solution. For 1 kWh output
Heat input = 11,000 x 0.2 = 2200 kcal.
Now 1 kWh = 862 kcal.
= 862 - 39.2%. Ans.
Overall efficiency = Output
Input 2200 -

Example 2.38. A steam power station has an installed ca

of 120 MW and a maximum demand of 100 MW. The coal consu,
tion is 0.4 kg per kWh and cost of coal is Rs. 80 per tonne. The annw
expenses on salary bill of staff and other overhead charges excluding
cost of coal is Rs. 50 x iü. The power station works at a load factor
of 0.5 and the capital cost o/the power station is Rs. 4 x 10 5. If the
rate of interest and depreciation is 10% determine the cost ofgenerat-
ing per kWh.
Solution. Maximum demand = 100 MW
Load factor = 0.5
Average load = 100 x 0.5 = 50 MW.
= 50 x 1000 = 50,000 kW.
Energy produced per year
= 50,000 x 8760 = 438 x 106 kWh.

Coal consumption = 438 x 10 6 x 0.4 = 1752 x 106 tonnes.

Annual Cost
102 x 80 = Rs. 14,016 x 102
(i) Cost of coal = 1752 x


POWER PLANT ECONOMICS 119

(ii) Salaries = Rs. 50 x 105

(iii) Interest and depreciation
=x4x10= Rs. 4x10

Total cost = Rs. 14,016 x 103 + Rs. 50 x 105

+ Rs. 4 x io = Rs. 19,056 x
Cost of generation per kWh
19,056 x 103
x 100 = 4.35 paise. Ans.
- 438x106

Example 2.39. A 200 MW thermal power station is to supply

power to a system having maximum and minimum demand 140 MW
and 40 MW respectively during the year. Assuming load duration
curve to be a straight line, determine the following:
(i) Load factor
(ii) Capacity factor (Plant factor).
Solution. C = Capacity of power plant 200 MW
Maximum demand = 140
140 MW
Minimum demand = 40 MW
t = Time per year
= 8760 hours
Load duration curve is
Time (own in Fig. 2.23.
Fig. 2.23 Energy supplied per year (E)
8760x40+ 8760 x 100
=
2
= 8760 x 90 MWh
8760 x 90
Average load = 8760 = 90 MW
Average load = -. = 0.64
Load factor = Maximum demand 140
E 8760x90 =0.45.
Capacity factor =
= 200 8760
Example 2.40. Calculate the cost of electrical energy generated
per kWh at 100% load factor, 75% load factor, 50% load factor and
at 25% load factor for a steam power plant. The fixed cost is Rs. 438
per kWof installed capacity per year and the fuel and operating costs


120 POWER PLANT

are 5 paise per kWh geerated. Plot the curve between cost of energy
per kWh and load factor.
Solution.
Load factor = 100%
C1 = Fixed cost per hour = 43,800 - 5 paise.
8760 -
1 kW plant is available at a cost of 5 paise per hour.
E Energy generated per hour at 100% load
factor
= 1 x 1 = 1 kWh.
C2 = Fuel and operating cost per hour
= E x 5 1 x 5 = 5 paise
C = Total cost per kWh produced per hour.
= C 1 + C2 = 5 + 5 = 10 paise.
Similarly the total costs per kWh at other load factors are as
follows:
Load fac. Energy Fixed Fuel and Total cost Cost per kWh
tor % Produced cost operating per hour produced per hour
per hour (Paise) cost per (Paise) (Paise)
(kWh) hour
100 1 5 5 10 10

75 075 5 3.75 875

8.75 x = 11.66
50 0.5 52.5 7.5
x .! = 15
25 0.25 5 1.25 6.25
6.25 x = 25
1 25

'.-

CL
Cj

'1'
0
C-)

Load factor ?.=

Fig. 2.24
POWER PLANT ECONOMICS 121
Fig. 2.24 shows the variation of cost of energy per kWh
generated with respect to load factor.
This shows that cost per kWh increases with decrease in load
factor.
Example 2.41. In a steam power plant the capital cost ofpower
generation equipment is Rs. 25 x io. The useful life of the plant is
30 years and salvage value of the plant to Rs. .1 x io. Determine by
sinking fund method the amount to be saved annually for replace-
ment if the rate of annual compound interest is 6%.
Solution. P = Capital cost = Rs. 20 x iO
S = Salvage value = Rs. 1 x
n = Useful life =30 years
r = Compound interest
A = Amount to be saved per year for replacement

A=(=(29x105_1x105)x0.06
(1+r)'-1 (1+0.06)°-1
= Rs. 24,000. Ans.
Example 2.42. A hydra power plant is to be used as peak load
plant at an annual load factor of 30%. The electrical energy obtained
during the year is 750 x 105 kWh. Determine the maximum demand.
lithe plant capacity factor is 24% find reserve capacity of the
plant.
Solution. E = Energy generated = 750 x 10 5 kWh

Average .load - 750 x i05 = 8560 kW

- 8760
where 8760 is the number of hours in year.
Load factor = 30%
M = Maximum demand
Load factor- Average load
- Maximum demand
M=---28.53okW
0.3
C = Capacity of plant
Capacity factor
= C x8760

12Z POWER PLANT

0.24 - 750 x 105

C 8760
C = 35,667 kW
Reserve capacity = C - M = 35,667 - 28,530 = 7137 kW.
Example 2.43. A power supply system has a hydro-power plant
of 12 MW capacity and a diesel power plant 0130 MW capacity. The
hydro power plant has a pondage provision and can store water to
generate 20 MWh. For the coming week the estimated power in river
flow is 3 MW and expected load to be supplied is as follows:

42 20
30 . 60

8 18
Total = 168 hours
(a) Calculate total energy that can be generated by hydrc
power plant.
(b) How the load shall be shared by the two power plants?
Solution.
(a) E 1 = Energy which can be generated in coming week.
= 3 x 168 = 504 MWh
As the hydro power plant has a provision to store 120 MWh.
E2 = Total energy that the hydro power plant can generate
=E+120=504+12Q=624MWh.
(b) The hydro power plant will be started when the load
exceeds 30 MW.
E = Energy generated by diesel engine

=8x 18+15x20+25x50+30x60
= 144 + 300 + 1250 + 1800 = 3494 MWh.
The hydro power plant will run at its full capacity of 12 MW for
20 hours. Energy generated by hydro power plant
= 12 x 20 = 240 MWh.
Example 2.44. A steam power plant is to supply load for 24
hours as follows:
Time SAM. to 10 10AM. to 5 5P.M. to 11 II P.M. to 5
- A.M. P.M. P.M.__- AM.
Load(AM) 1 50 100 35 15
POWER PLANT ECONOMICS 123

(a) The energy rate is 14paise per kWh and cost of input is 15
N.E. per 6000 kcal. The thermal efficiency of the plant is
35% at 100 MW, 30% at 59 MW, 28% at 35 MW and 20%
at 15 MW.
Determine the net revenue earned.
(b) lithe above load is supplied by a combination of steam
power plant and pump storage plant and steam power
plant runs at constant load with 35% efficiency. Calculate
the capacity of steam power plant and percentage increase
in revenue earned.
The overall efficiency of pump storage plant is 79% and cost of
energy input and cost of selling power is same as mentioned above.
Solution.
(a) E = Energy generated by steam power plant in 24 hours.
= 40 x 5 + 100 x 7 + 35 x 6 + 15 x 6
= 250 + 700 + 210 + 90 = 1250 MWh
= 1250 x 103 kWh
r = Energy rate = 14 paise per kWh
C 1 = Cost of selling power = E x r

=1250x10xj=175x10rup.s
C2 = Input to steam plant in 24 liouis
100x7 50x5 35x6 Yx
0.35 + 0.3 + 0.28 + 0.2
= 2000 + 833 + 750 + 450 = 4033 MWh.
= 4033 x 103 kWh = 4033 x 103 x 860 kcal.
= 3468 x 106 kcal.
(as 1 kWh = 860 kcal.)
S = Input rate = 15 paise per 6000 kcal.
= Cost of input energy
3116 x
= 86.7 x 103
= 6000 100
New revenue earned from steam power plant = C 1 - C2
= 175 x iø - 86.7 x 10
= 88.3 x 10 3 rupees per day.

Igo POWER PLAN1

(b) Let = Capacity of steam power plant when it works in

combination with pumps storage plant.
= Energy used from thermal plant to pump water
of pump storage plant during off-peak period.
=(100-x)x 7
E2 = Energy supplied by pump storage plant during
off-speak period:
=[(x-50)X5+(x-35)X6+(X-15)X6]x07
But E, =E2
(100-x) x 7 = [(x - 50)x5 +(x -35) x6 + (x- 15)x6Jx0.7
x = 57 MW
Thus the base load of 57 MW should be taken by steam power
plant
C1 = Cost of selling power
= 175 x 103 rupees (as already found)
E 2 = Energy input to steam plant in 24 hours
- x x 24 - 57 x 24
- 0.35 - 0.35
=3902MWh=3908x 103 kWh
= 3908 x 103 x 860 kcal = 3360 x 10 6 kcal.
C2 = Cost of input energy
- 3360 x 10 6 15
-. 6000 x-:j_84x10rupees
Net revenue earned = C1 - C2 = 175 x iø - 84 x
= 91 x 102 rupees per day
% increase in revenue earned
91 x 103 - 88.3 x iO
x 100
= 88.3
= 3.05%.
Example 2.45. The following data is supplied for a power plant:
Annual load factor = 65%
Capacity factor = 55%
Installed capacity of the plant = 200 MW
POWER PLANT ECONOMICS 125

Capital cost of the plant = Rs. 140 x 106

Annual cost of coal, oil, taxes and salaries = Rs. 21 x 106
Rate of Interest =5% of the capital
Rate of depreciation = 601c of the capital
Units of energy used in running the plant auxiliaries = 4% of
total units generated.
Determine:
(a) Reserve capacity
(b) Generating cost.
Solution. C = Capital of the plant
L = Average load
M = Maximum demand
t = Hours in one year = 8760
Load factor = Average load
Maximum demand

O.65=h
L=0.65xM
E = Energy generated in one year
x 8760 = 0.65 xMx8760

Capacity factor =
CXt
0.65xMx8760
0.5..= 200x8760
M= 169 MW
Reserve capacity= C - M = 200 - 169
= 31 MW
0.69 x 169 x 8760
= 962 x 103 kWh = 962 x 10 6 kWh.
Now 4% of E is required to run the plant auxiliaries.

E 1 = Total energy generated

100 )

16 POWEP PLANT

=11+ I__)x962x1o6
-... .,
=lOOlxlO6kWh
C i - Interest per year
x 130 x 10 = Rs. 7 x 106.
=
C.Depreciation...per year
1Xi40x!o6=RS.S.4x 106

C3 = Running cost per year

= Rs. 21 x106
C = Total cost = C 1 + C2 + C3
(7 + 8.4 + 21) 10 6 = Rs. 36.4 x 106
C 364 x1O6
Cost of generation x 100
- -
= 3.64 paise per kWh.

Example 2.46. The estimated costs of two power stations A and

B running parallel are
Rs. (130 x kW + 0.028 kWh) and
Rs. (125 x kW + 0.032 kWh) respectively. The power stations
sipply power to a system whose maximum load is 120 MW and
minimum load is 20 MWduryzg the year. The load varies as straight
line. Determine for minimum cost of generation.
(a) installed capacity of each station.
(b) Annual load factor and capacity factor of each station.
(c) Cost of generation.
Solution.
(a) C.4 = Cost of power station A
= 130 x kW + 0.028 kWh
Let Ai = 130
B 1 = 0.028
Cu Cost of power station B
= 125 kW + 0.032 kWh.
Let A4=125
B2 = 0.032

POWER PLANT ECONOMICS 127

H—Time in hours for which base load power plant is to be

operated for minimum overall cost
Ai-Az
- B2 -B1
130 - 125
= 1250 hours
= 0.032 - 0.028
Now as shown in Fig. 2.25

12,

.S0
2

Fig. 2.25

Let P1 = Peak load on base load plant

P2 = Peak load on peak load plant
From triangles JRT and JKN.
NK JN
TRJT
1250 120-P1
8760 - 120 -20
P i = 106 MW
P2 = 120 - 106 = 14 MW
(b) For base load plant
Load factor =- E1
P 1 x 8760
128, POWER PLANT
where Ei Actual units generated
= Area NKRSON
= Area NKRT + Area TRSO

= (NK+R7) xNT+RTxRS

= (1250 + 8760) x (106- 20) + 8760 x 20

= 605630 MWh.
Load factor = 605630
= 0.65
x 8760
Capacity factor Ei
= Pi x 8760
- 605630 = 0.65
106 x8760
For Peak Load Plant
Load factor E2
P2x8760 =Area sJKN

=x(120- 160)x 1250

= 8750 MWh
Load factor = 8750
14 x 8760
=0.07=7%
E2
Capacity factor
= ,2 x 8760
= 0.07 = 7%
Since there is no reserve capacity.
(c) For base load plant
E1 = Total energy generated
= 605630 MWh 605.630 x 10 6 kWh
C1 = Cost
= 130 x 103 x 106 + 0.028 x 605.630 x 106
= Rs. 30 x 106
For peak load plant


POWER PLANT ECONOMICS

129
E2 = Total energy generated
=8750MWh =8.750x 106 kWh
C = Cost
= 125 x 14 x 103 + 0.032 x 8.750 x 106
= Rs. 2 x 106
E = Total energy generated
E = E1 + E2 = ( 605.630 + 8.750) x 106
= 614.38 x 106
C = Total Cost
=Ci+C2 =30 x 106 +2x 106
= Ks. 32 X.
Cost per kWh =

32 x 106 x 100
= 5.2 paise.
6 14.38 x 106
Example 2.47. A common load is to be shared by two power
plants. One power plant is a base load plant with 30 MW installed
capacity and other power plant is a standby plant with 20 MW
capacity. The yearly output of the base load plant is 130 x 10 kWh
and that of standby plant is 9 x 10 6 kWh. The peak load taken by the
standb y plant is 15 MW and this plant works for 2800 hours during
the year. The base load plant takes a peak 0125 MW. Determine the
following for both plants:
(a) Annual load factor
(h) Plant use factor
(c) Capacity factor.
Solution.
Standby Power Plant
C = Capacity of plant
= 20 MW
E = Energy generated per year
9 x 10 6 kWh
I = Hours of working per year
= 2800 hours
1 lours in y ear 8760

4130 POWER PLANT

M = Maximum demand
= 15MW
(a) Annual load factor
E 9x106
MxT 15x1000x8760
= 0.068 or 6.8%.
M
(b) Plant use factor = - = 15
= 0.75 or 75%.
(c) Capacity factor = Cxt
9 x 106
16 or 16%.
20000 x 2800 = 0.
Base Load Power Plane
C = Capacity of plant
= 30 MW
E = Energy generated per year
= 130 x 106 kWh
M = Maximum demand
= 25 MW.
(a) Annual load factor
E 130x106
M x T 25 x 1000 x 8760
= 0.593 or 59.3%.
M 25
(b) Plant use factor =-
= 0.83 or 831/(,.
E
(c) Capacity factor =

= - --- - = 0 49 49 44
30000 x 8760
Example 2.48. An t'leetricsnpp!y company has the /oIlouing
annual expenses
Generation Rs. 80 x to'
Transmission Rs. 25 x to'
Distribution Rs. 20 x 10
Fuel Rs. 30 x 10'
Repair etc. Rs. 3.5 x
 POWER PLANT ECONOMICS 131

The number of units generated per year are 430 x 106 kWh. The
consumers have an aggregate ,naxi,num demand of 80 MW. The
fixed charges for generation. transmission, distribution, fuel, repair
etc. are 80%, 90%, 9%, 15% and 50% respectively. Losses in trans-
mission and distribution are 10%.
Determine a two part tariff to be charged from the consumers.
Solution. M = Maximum demand
=80 MW = 80 x 10'kW
E = Energy generated per year
= 430 x 106 kWh.
The fixed and running charges are calculated as follows:
Rem Total Fixed charges Ru
charges Amount 'Ye Amount
(Rs.) I (Rsi

-Generation 80 64 x 20 16X IO
Transmission -25 x 104 90 22.5 x iO 10 2.5'< io
Distribution 20 x 10 5 95 19 x le 5 1. 10 5
Fuel 30 x 105 15 4.5 x iø 85 Z5.5x
Repairs !f__ 3.5 50 1.75 x 10" 50 1.75 x
Total 339x 10 301 x
C1 = Fixed charges
= Rs. 339 x io
Fixed charges per kV of maximum demand
- C1 - 339 x iO
M 80x103
= Rs. 42.37
C2 = Running charges = Rs. 301 x 10
Transmission losses = 10%
Net energy supplied (E)
= 430 x 1 6 x 0.9
Running charges per unit
C2 301 x 10 x 100
- E - 430 x 101; x 0.9
= 0.77 paise.

I

J32 POWER PLANT

Therefore the two part tariff is Rs. 42.37 per kW of maximum

demand plus 0.77 paise per kWh.
Example 2.49. Determine the thermal efficiency of a steam
power plant and its coal bill per annum using the following data.
Maximum demand = 24000 kW
Load factor = 40%
Boiler efficiency = 90%
Turbine efficiency = 92%
Coal consumption = 0.87 kg /Unit
Price of coal = Rs. 280 per tonne.
Solution. i = Thermal efficiency
= Boiler efficiency x Turbing efficiency
=0.9x0.92
-0.83
= Average Load
Load factor Maximum Demand
Average Load 0.4 x 24000
= 9600 kW
E = Energy generated in a year
= 9600 x 8760
=84lxlO5kWh
E x 0.87 x 280
Cost of coal per year = 1000

841 x 105 x 0.87 x 280

- 1000
= Rs. 205 x 105.
Example 2.50. The following data relates to a steam power
plant.
Capacity of the plant = 100 MW
Capital Cost Rs. 1800 per kWinstalled
Maximum demand = 80 MW
Interest and depreciation = 10% on capital
Fuel cost = Rs. 80 per tonne
Fuel consumption = 1.3 kg/kwh
Salaries, wages and maintenance = Rs. 8 x 10 per year
POWER PLANT ECONOMICS
133
Load factor = 50%
Determine the cost of generation per kwh.
Solution.
C = Capacity of the power plant
= 100 MW= 100 1000 kW
Capital investment
= 100x1000X18OO Rs. 18x iø
Ci = Interest and depreciation

18 x 106
Average load = Maximum Demand x Load factor
=SOXO.S=4OMW4OX1000kW
E = Energy generated per year = Average loac
x Time in hours
= 40 x 1000 x 8760
= 3504 x 10 5 kWh
W = Weight of fuel consumed
= 1.3xE
= 1.3 X 3504 x
= 4555x 10 5 kg
= 4555 x 102 tonne
C2 = Cost of fuel
=Wx80
= 4555 x 10 2 x 80 = Rs. 3644 x io
C3 = Salaries, wages and maintenan
= Rs. 8x105
S = Total investment
=C1+C2+C3 •;
=18x106+3644x104+8x10
. S.
5'
= Rs. 55.24 x 106
0 •5 •
Cost of generation = S -
•
 POWER PLANT
134

55.24 x 106
- 3504 x 105
= 15.76 paise per kWh.

Example 2.51. Two electrical units are used for same purpose.
Cost of first unit = Rs. 7000 and it takes 90 kW.
Cost of second unit = Rs. 18000 and it takes 50 kW.
Useful life of each unit =38 x 10 hours.
Energy rate = Rs. 100 per kWof maximum demand per year and
8 poise per kWh.
Find which unit will be economical if both units run at full load.
Solution. First Unit.
C 1 = Capital cost per hour
I
==7000Rs. 0.184
38 x io
Maximum demand = 90 kW
Rate for maximum demand = Rs. 100 per kW
C2 = Charge for maximum demand per hour
90 x 100
= 8760 = Rs. 1.028

E l = Energy produced per hour = 90 x 1 = 90 kWh.

C3 = Energy charges per hour
= E 1 x Rate

= 90 x = Rs. 7.20

C = Total charge per hour for the operation

of the unit
= 0.184 + 1.028 + 7.20 = Rs. 8.412
Second unit
C 1 = Capital cost per hour = 18000 = 0.47
38 x iO
C2 = Charge for maximum demand per hour
- - Rs. 0.57
- 8760 --
 POWER PLANT ECONOMICS
135
C3 Energy charge per hour

= (50 x 1) ><

= Rs. 4.0
C = Total charges per hour
= C1 + C2 + C3
= 0.47 + 0.57 + 4.0 = Rs. 5.04
Hence second unit is economical.
Example 2.52. The maximum (peak) load on a thermal power
plant of 60 MW capacity is 50 MW at an annual load factor of 50%.
The loads having maximum demands of25 MW, 20 MW, 8 MW and
5 M are connected to the power station.
Determine : (a) Average load on power station (b) Energy
generated per year (c) Demand factor (d) Diversity factor.
Solution.
(a) Load factor = _ A v era g e loa4
Maximum demand
Average load = 0.5 x 50 = 25 MW
(b) E = Energy generated per year
= Average load x 8760
= 25 x 1000 x 8760
= 219 x 106 kWh.
(c) Demand factor = Maximum demand
Connected load
=-
25 + 20 + 8 + 5
= = 0.86

(d) Diversity factor =M

1
where MI = Sum of individual maximum demands
=25+20+8+558M\V
M2 = Simultaneous maximum demand = 50 MW

Diversit y factor = 1.16

PROBLEMS
2.1. Epla,n the following terms as applied to power
 POWER PLANT

(i) Demand factor,

(ii) Connected load,
(iii) Maximum Demand,
(iv) Use factor,
(v) Capacity factor.
2.2. (a) What is meant by load curve ? What is its significance in
power generation?
(b) Explain the difference between load curve and load duration
curve.
2.3. What is meant by load factor and diversity factor? Prove that
an increase in diversity of load improves the load factor of a
power system.
2.4. What do you understand by power plant economics? Explain the
fixed costs and operating cost of a power station.
2.5. What is meant by depreciation of a power station? Explain the
straight line method and sinking fund method of calculating
depreciation?
2.6. What is meant by tariff? What are the various types of tariffs in
common use? Explain the two part tariff.
2.7. Explain the terms (i) heat rate. (ii) incremental rate of a power
plant. What is the significance'of incremental rate curve?
2.8. Explain how economical load division is obtained between the
two alternators of a power stations?
Prove that for economical load sharing the incremental rates of
the two are equal.
2.9. Power stations A and B are of equal capacities. State which
power station is running economical if the two power stations
are working at load factor of 70% and 60% respectively.
tAns. Power Station Al
2.10. Write short notes on the following
(a) Types of loads
(b) Types of power plants
(c) Base load power lant
(d) Peak load power plant.
2.11. The input-output curve of a 150 MW station is expressed by the
formula,
i = 106 (250 + 5L + 0.03L3) + L4
where I is in kcal per hour and L is in mega-watts. Find the load
at which minimum heat rate occurs and check with the plot.
[Ans. 73.8 MWI

2.12. The input-output curves of three generating units are given by

the following formulae:
Units Input-Output Curve
A I=106(10+0.04L2+0.02L3)
B I = 106 (8 + 8 L + 0.4 L2)
C I=106(1O-6L)


POWER PLANT ECONOMICS 137

where I is in kcal per hour and L in the MW.
The capacity of unit A, B, C and 10, 10 and 6 MW respect ivelv.
Plot the individual and combined incremental rate curves on a
common graph and devise a combined loading schedule for the
three units when total loads to be supplied are 6 MW, IS M\V,
12 M\V, 24 MW and 26 MW.

2.13. Calculate the unit Cost of production of electric energy for a

power station for which data are supplied as follows
Capacity = 50 MW
Cost per kW = Rs. 600
Load factor 40%
Interest and depreciation = 10%
Cost of fuel, taxation and salaries = Rs. 36 x 105.
[Ans. 3.71 patsej
2.14. Estimate the generating cost per unit supplied from a power
plant having the following data:
Plant capacity = 120 MW
Capital cost = Rs. 600 x 106
Annual load factor = 40%
Annual cost of fuel, taxation, oil and salaries
= Rs. 600,000
Interest ind depreciation = 10%
(Ans. 1.33 paiset
2,15. Prove that for maximum efficiency of a power plant the in-
cremental rate is equal to heat rate.
2.16. Estimate the generating cost per unit supplied from :i power
plant having data
Output per year = 4 x 10'kWh
Load factor = 50%
Annual fixed charges = Rs. 40 per kW
Annual running charges = 4 paise per kWh.
2.17. A 50 MW generating station has the following data
Capital cost = Rs. 15 x 100
Annual taxation = Rs, 0.4 x 105
Annual salaries and wages = Rs. 1.2 x 106
Cost of coal = Rs. 65 per tonne
Calorific value of coal = 5500 kcallkg.
Rate of interest and
depreciation = 12%
Plant heat rate = 33,000 kcal/kWh
at 100% capacit y and 40000 kcallkWh at 60%.
Calculate the generating cost/kWh at 100% and 60" capacity
factor.
(A. At. LE. 1979- -
2.18. Write short notes on the following:
(o) Economics in plant selection
(b) Power plant capacity
(c) E.H.V. lines.
—41,

138 POWER PLANT

219. A svsteni with a maximum demand of 100,000 kW and load

factor of 30% is to be supplied by either (a) a steam station alone
or (b) a steam station in cOfljuflCtiOfl with a water storage
scheme, the latter supplying 100 million units with a maximum
output of 40,000 kW. The capital cost of the steam and storage
station are Rs. 1200 per kW and Rs. 1050 per kW respectively.
The corresponding operating cost are 15 paise and 3 paise per
kWh respectively. The interest on capital cost is 15 per annum.
Calculate the overall generating cost per kWh and state which
of the two projects will he economical. (A.M.tb]. 1976)
2.20. Discuss the factors which affect choice of power station.
2.21. Find the cost of generation per kWh from the following data
Capacit y of a plant = 120 MW
Capital cost = Rs. 1.200 per kW installed
Interest and depreciation = 10% on capital
Fuel consumption = 1.2 kg/kWh
Fuel cost = R—s. 40 per tonne.
Salaries, wages, repair and maintenance = Rs. 600,000 per year.
The maximum demand is 80 MW and load factor is •l0
(A.M.I,E. 1981)
2.22. A power plant has the following annual factor
Load factor = 70%
Capacit y factor = 50%
Use factor = 60%
Maximum demand = 20 MW
Find out
(a) Annual energy production
(h) Reserve capacity over and above peak load
(c) hours during which the ilnt is not, in service per year.
(A. Al. 198 1)
2.23 The following data is supplied for a power plant
Capacity = 150 MW
Capital cost = Rs. 1800 Per kW
Interest and depreciation = 8% of the capital cost
Annual running charges = Rs. 18 l0'
Profit = 70% of the capital
Energy consumed to run
power plant auxiliaries = 5 of energy generated
Annual capacity factor = 0.5
Annual load factor = 0.55
Determine the following;
(a) Reserve capacity
(b) Cost of generation per kWh.
2.24. Why is it necessary to predict the future load demand. What are
the methods of load forecasting ?
2.25. Explain the effect of variable load on power plant operation and
design.
2.26. (a) An Electric Supply undertaking has the following data.
Power generated = 500 x 10 kWh
Maximum demand = 150 x 10 > kW
 POWER PLANT ECONOMICS 139

Cost of generation = Rs. 32 x 10

Cost of transmission line Rs 650 x 10
Cost Of line of distribution = Rs. 280 x 10
Cost of fuel Its. = 550 x 10 4
Out of these 10'. and 6', 6 and 907c are running charges and
remaining is a fixed charge. The transmission and distribution
loss is 109-. Calculate two part tariff.
(b) If the load factor of the plant is raised to 55 for the same
maximum demand, calculate the percentage saving in overall
Cost perkWh.
2.27. The cost of an electric motor is Its 5000. It is to be overhauled
two tunes during life of 10 years. On each overhauling a sum of
Its 1000 is spent Calculate depreciation cost perycar if residual
cost of motor after 10 years is Rs. 800. [Ans. Rs. 6201
2.28. A power plant supplies power to four regions of loads of 10 MW,
20 MW, 15 MW and 18 MW Find maximum demand ifdiversitv
factor is 1 3.

A ., ... ..
'.,4 . .•..)_.
: -, . . I' •

C .

"I

f t'.' ,,

•i•'•'
C..
 S

Steam Power Plant

3.0 Introduction
Steam is an important medium of producing mechanical energy.
Steam has the advantage that it can be raised from water which is
available in abundance it does not react much with the materials of
the equipment of power plant and is stable at the temperature
required in the plant. Steam is used to drive steam engines, steam
turbines etc. Steam power station is mbst suitable where coal is
available in abundance. Thermal electrical power generation is one
of the major method. Out of total power developed in India about
60% is thermal. For a thermal power plant the range of pressure
may vary from 10 kg/cm 2 to super critical pressures and the range
of temperature may be from 250' C to 650'C.
The average all India Plant load factor (P.L.F.) of thermal power
plants in 1987-88 has been worked out to be 56.4% which is the
highest P.L.F. recorded by thermal sector so far.
3.1. Essentials of Steam Power Plant Equipment
A steam power plant must'have following equipments

(VAt
RA6( L,.J

A".k
AR Wf -

$'4 Ifq

H P £ P CO1OfN5A If
14TfR kE41ER IXIRACIICN PW.1P

Fig. 3.1


STEAM POWER PLANT

141
(i) A furnace to burn the fuel.
(ii) Steam generator or boiler containing water. heat
generated in the furnace is utilized to convert water in
steam.
(Liz) Main power unit such as an engine
or turbine to use the
heat energy of steam and perform work.
(ic) Piping system to conve y steam and water.
In addition to the above equipment the plant requires various
auxiliaries and accessories depending upon the availability afwater,
fuel and the service for which the plant is intended.
The flow sheet of thermal power plant consists of the following
four main circuits
(i) Feed water and steam flow circuit
(ii) Coal and ash circuit
(iii) Air and gas ci i •cu it
(it) Cooling water circuit.
A steam power plant using steam as working substance works
basically on Rankine cycle.
Steam is generated in it boiler, expanded i niine mover
and condensed in the condenser and ftd into t1 n ,ain.
The difk'rent t y pes of s y stems and Componi '('d in steam
power plant are as Ibllows
U) 11gb pressure boiler
(u) I'rinn mover
(ill) ( ii(hasurs and cooling towers
(ii) ('nd haiicllin svs(enl
i') Ash and dust handling svsteni
(ti) Draught svsteni
(iii Feed water purification plant
(iii i ) Pu in ping syste mu
(ix) Air preheater, cc. I 01111 se r; super liea te r, feed licaturs
Fig. 11 shows aschemiatic arcmqernum o r,quipmeaofi stean,
power station. Coal received in coal storage yard of power stat Ion i
transferred in thi furnace by coal handling unit. Ilc'at produced thu
to burning of coal is utilised in Converting water contaimii'd in homIer
drum into steam at suitable pressure and temperature The stealli
generated is passed through the superheater. Superheateci steam
then flows through the turbine. After doing work in the turbine the
press ore of St Ca iii is red cm ced Stea rn leaving the tu rbm e passes
through the condenser which Ii11imi1ain the low pressure olsteam at
the exhaust of turbine Stea m prescire in the condenser depends
upon now rate and temperature of cooling water and on efl'ctive
ness of air removal equipment. Water circulating through the con-
denser ma y he taken from the various sources such as river, lake or
sea. If sufficient quantit y of water is not available the lint Water
14' POWER PLANT

coming out of the condenser may be cooled in cooling towers and

circulated again through the condenser.
Bled steam taken from the turbine at suitable extraction points
is sent to low pressure and high pressure water heaters.
Air taken froii the atmosphere is first passed through the air
pre-heater, where it is heated by flue gases. The hot air then passes
through the furnace. The flue gases after passing over boiler and
superheater tubes, flow throughthe dust collector and then through
economiser, air pre-heater and finally they are exhausted to the
atmosphere through the chimney.
Steam condensing system consists of the following
(1) Condenser (ii) Cooling water
(iii) Cooling tower (iv) Hot well
(v) Condenser cooling water pump
(vi) Condensate air extraction pump
(vii) Air extraction pump (viii) Boiler feed pump
(ix) Make up water pump.
3.1.1. Power Station Design
Power station design requires wide experience. A satisfactory
design consists of the following Steps
(i) Selection of site.
(ii) E( i mat ion of capacit y of power station.
cfectiiin of turbines and their auxiliaries.
i ; Selection of boilers, and their auxiliaries.
(t') Design of fuel handling system.
(vi) Selection of condensers.
(vii) Design of cooling system.
(viii) Design of piping system to carry steam and water.
(ix) Selection of electrical-generator.
(x) Design and control of instruments.
(xi) Design of layout of power station.
Quality of coal used in steam power station plays an
important role in the design of power plant. The various
factors to be considered while designing the boilers and
coal handling units are as follosvs
(i) Slagging and erosion properties of ash.
(ii) Ninisture in the coal. Excessive moisture creates addition-
al prOl)leflIS particularly in case of pu I yen sed fuel power
plants.
(iii) Ruriiing characteristic of coal-
(iv) Corrosi ve nature of ash.
3.1.2. Characteristics of Steam Power Plant
The desirable characteristic for a steam power plant are as
follows
STEAM POWER PLANT 143
(i) Higher efficiency. (ii) Lower cost.
(iii) Abilit y to burn cil especially of high ash content, and
inferior coals.
(ii') Reduced enviror ental impact in terms of air pollution.
(t') Reduced water i tuirernent.
(vi) I ligher reliabihtv and availability.
3.2 Coal Handling
Coal delivery equipment is one ofthe major components of plant
cost. The various steps involved in coal handling are as follu.s
(See Fig. 3.2)
(i) Coal delivery (ii) Unloading
(iii) Preparation (it') Transfer
(v) Outdoor storage (vi) Covered storage
(vii) In plant handling (['iii) Weighing and n1eauriiig
(ix) Feeding the coal into furnace.
COAL ThL1VhRY

UNLOADING

/'/?M'A IFA TlOV 1

TRA.V.cFKII

OIJTI)O011 STORAGE
(!)EAI) S7OIMGI.;

COVb.RKI) .TTORA G/.

(1.1 Vb. ST(IMC:)

IN PL.4NT FIA.VI)I.I.VG

EIG ti/VU
AN!)
A1/.1SU/?f,V(;

b'tJI?VA(:E
Fig 32.


POWER PLANT

Coal D.'livery. The coal Ironi suppl y fniiltS is doliveri'd l

ships Or boats to powerstations situated ri,ar tosea orriver :hereas
oaI is supplied b y rail or trucks to the power stations which ore
situated away from sea or river. The transportation of coal b y trucks
is used if the railwa y facilities are not available,
it) Unloading. The type of equipment to be used for unloading
the coal received at the power station depends on how coal is
received at the power station. If coal is delivered by trucks, there is
no need of unloading device as the trucks laity dump the coal to the
outdoor storage. Coal is easil y handled it' the lift trucks with SCOOp
are used. In case the coal is brought b y railway wagons, ships or
boats, the ii nload ing ma y he done b y car slia kes, iota ly ear d urn pci's
cranes, grab buckets and coal lcrclei-ators. Rotar y car dumpers
althougii costl y
are quite efficient for unloading closed wagons.
(iii Preparation. When the coal delivered is in the form of big
lumps and it is not of pro per size, the preparation (sizing) ofcoal can
be achieved hv crushers, breakers, sizers driers and magnetic
separat )r.
(j) Transfer. After preparation coal is transferred to the dead
storage he itetn.S c' fthe filliaving system's
1. flelt conVecol's. 2. Screw conveyor's
1. lu('k"t elevators. 4. Grab bucket elevators.
. Skip hoisLs. 6. Flight conveyor.
1. Belt conveyor.
CM Fig. :1.3 shows a belt coo-
GE COAL veyor. It Consist,,; of' an
RO! 'CR endless belt moving over
)
- (rollers). itI
- tance it supporting roller
r - is provided at tl centre.
........---. The belt is made un of'
rubber or canvas.
ii H Conveyor is suitable for
the t ra itsf_'r' of coal over
Fig. 33.
long distances. It is used
in in en i urn and large
power plants. The initial cost of the s y stem is not high and power
consumption is also low. The inclination at which cod can hc
successfull y ekvated b y belt COIIVe\'Oi' is about 21) Average speed
of belt cun'vurs varies between 200--300 m'.p.rn, This cuovevir is
pre ferred haii other types.
 STEAM POWER PLANT
145
Advantages of belt conveyor
1. Its operation is smooth and clean.
2. It requires less Power as compared to other types of
systems.
3. Large quantities of coal can be discharged quickly and
continuously.
4. Material can be transported on moderates inclines.
2. Screw conveyor. It Consists of an endless helicoid screw
fitted to a shaft (Fig. 3.4). The screw while rotating in a trough
transfers the coal from feeding end to the discharge end.
This system is suitable, where coal is to be transferred over
shorter distance and space limitations exist. The initial cost of the
system is low. It suffers from the drawbacks that the power con-
sumption is high and there is considerable wear of screw. Rotation
of screw varies between 75-125 r.p.m.
3. Bucket elevator. It consists of buckets fixed to a chain (Fig.
3.6). The chain moves over two wheels. The coal is carried by the
buckets from bottom and discharged at the top.
COAL INLET

EVSC/IARGE END
Fig. 3.4 Screw conveyor.

4. Grab bucket elevator. It lifts and transfers coal on a single

rail oi track from one point to the other. The coal lifted by grab
buckets is transferred to overhead bunker or storage. This system
requires less power for operation and requires minimum main-
tenance.
The grab bucket conveyor can be used with crane or tower as
shown in Fig. 3.4 (a. Although the initial cost of this system is high
but operating cost is less.

GRAB
CRA N E

Fig 3.4 (a).

146 POWER PLANT

5. Skip hoist. It consists of a vertical or inclined hoistwav a

bucket or a car guided by a frame and a cable for hoisting the bucket.
The bucket is held in up right position. It is simple and compact
method of elevating coal or ash. Fig. 3.5 shows a skip hoist.
6. Flight conveyor. It consists of one or two strands of chain
to which steel scraper or flights are attached which scrap the coal
through a trough having identical shape. This coal is discharged in
the bottom of trough. It is low in first cost but has large energy
consumption. There is considerable wear.
Skip hoist and bucket elevators lift the coal verticall y while
Belts and flight conveyors move the coal horizontally or on inclines.

Fig. 3.5

Fig. 3.5 (a) shows a flight conveyor. Flight conveyors possess the
following advantages
(i) They can be used to transfer coal as well as ash.
(ii) The speed of conveyor can be regulated easily.
(iii) They have a rugged construction.
(iv) They need little operational care.
Disadvantages. Various disadvantages of flight couvevors are
as follows
(i) There is more wear clue to dragging action.
(ii) Power consumption is more.
 STEAM POWER PLANT 147

Scrapper

RcPer -Chain

Coal
Fig. 3.5 (a).

(iii) Maintenance cost is high.

(lv) Due to abrasive nature of material handled the speed of
conveyors is low (10 to 30 mlmin).
(v) Storage of coal. It is
desirable that sufficient quantity
of coal should be stored. Storage
of coal gives protection against
the interruption of coal supplies
when there is delay in transpor-

ti
tation of coal or due to strikes in
coal mines. Also when the Prices
are low, the coal can be purchased
and stored for future use. The COAL/WZET
amount of coal to be stored
depends on the availabilit y of
space for storage, transportation
fci!iies, the amount of coal that
will whether away and nearness Fig. 3.6
to coal niines of the power station. -
Usuall y coal required for one month operation of power plant is
stored in case of power stations situated at longer distance from the
collieries whereas coal need for about 15 days is stored in case of
power station situated near to collieries. Storage of coal for longer
periods is not advantageous because it blocks the capital and results
in deterioration of the qualit y of coal.
The coal received at the power station is stored in dead storage
in the form of piles laid directl y on the ground.
The coat stored has the tendenc y to whether (to combine with
oxygen of air) and during this process coal toss some of its heating
value and- ignition qualit y . J)ue to low oxidation the coal ma y ignite
s>ttivuiisly. Thi s is avot(hI•cl by storing coal in the form of pilvs

148 POWER PLANT

which consist of thick and compact layers of coal

so that air cannot pass through the coal piles. This
will minimise the reaction between coal and
AR oxygen. The other alternative is to allow the air to
pass through layers of coal so that air may remove
the heat of reaction and avoid burning. In case the
coal is to be stored for longer periods the outer
surface of piles may be sealed with asphalt or fine
coal.
The coal is stored by the following methods
(i) Stocking the coal in heats. The coal is piled on
the ground upto 10-12 m height. The pile top
should be given a slope in the direction in which
F 3
Fig. 7 the rain may be drained off. The sealing of stored
pile is desirable in order to avoid the oxidation of coal after packing
n air tight layer of coal.
Asphalt, fine coal dust and bituminous coating are the materials
commonly used for this purpose.
(ii) Under water storage. The possibility of slow oxidation and
spontaneous combustion can be completely eliminated by storing
the coat under water.
Coal should be stored at a site located on solid ground, well
drained, free of standing water preferably on high ground not
subjected to flooding.
(vi) In Plant Handling. From the dead storage the coal is
brought to covered storage (Live storage) (bins or bunkers). A
cylindrical bunker shown in Fig. 3.7. In plant handling may include
the equipment such as belt conveyors, screw conveyors, bucket
elevators etc. to transfer the coal. Weigh lorries hoppers and auto-
matic scales are used to recoid the quantity of coal delivered to the
furnace.
(vii) Coal weighing methods. Weigh lorries, hoppers and
automatic scales are used to weigh the quantity coal. The commonly
used methods to weigh the coal are as follows
(j) Mechanical (ii) Pneumatic (iii) Electronic.
The Mechanical method works on a suitable lever system
mounted on knife edges and bearings connected to a resistance in
the form of a spring of pendulum. The pneumatic weighters use a
pneumatic transmitter weight head and the corresponding air pres-
sure determined by the load applied. The electronic weighing
machines make use of load cells that produce voltage signals propor-
tional to the load applied.
 STEAM POWER PLANT
149
The important factor considered in selecting fuel handling sys-
tems are as follows
(i) Plant flue rate, (ii) Plant location in respect to fuel shipping,
(iii) Storage area available.

3.2.1 Dewatering of Coal

Excessive surface moisture of coal reduces and heating value of
coal and creates handling problems. The coal should therefore be
dewatered to produce clean coal. Cleaning of coal has the following
advantages:
(i) Improved heating value.
(ii) Easier crushing and pulverising.
(iii) Improved boiler performance.
(iv) Less ash to handle.
(u) Easier handling.
(vi) Reduced transportation cost.

3.3 Fuel Burning Furnaces

Fuel is burnt in a confined space called furnace. The furnace
provides supports and enclosure for burning equipment. Solid fuels
such as coal, coke, wood etc. are burnt by means of stokers where as
burners are used to burn powdered (Pulverised) c m liquid
fuels. Solid fuels require a grate in the furnace to .old the bed of
fuel.
3.3.1 Types of Furnaces. According to the methou ox firing fuel
furnaces are classified into two categories:
(i) Grate fired furnaces (ii) Chamber fired furnaces.
Grate fired furnaces. They are used to bn solid fuels. They
may have a stationary or a movable bed of fuel.
These furnaces are classified as under depending upon the
method used to fire the fuel and remove ash and slag.
(i) Hand fired (ii) Semi-mechanized
(iii) Stocker fired.
Hand fired and semi-mechanised furnaces are designed with
stationary fire grates and stoker furnaces with travelling grates or
stokers.
Chamber fired furnaces. They are used to burn pulverised
fuel, liquid and gaseous fuels.
Furnace shape and size depends upon the following factors:
(i) Type of fuel to be burnt.
(ii) Type of firing to be used.
(iii) Amount of heat to be recovered.
(iv) Amount of steam to be, produced and its conditions.
(v) Pressure and temperature desired.
(vi) Grate area required.
 POWER PLANT
150

(vii) Ash fusion temperature.

(viii) Flame length.
(ix) Amount of excess air to be used.
Simply furnace walls consists of an interior face of refractory
material such as fireclay, silica, alumina, kaolin and diaspore, an
intermediate layer of insulating materials such as magnesia with
the exterior casing made up of steel sheet. Insulating materials
reduce the heat loss from furnace but raise the refractory tempera-
ture. Smaller boilers used solid refractory walls but they are air
cooled. In larger units, bigger boilers use water cooled furnaces.
To burn fuels completely, the burning equipment should fulfil
the following conditions
1. The flame temperature in the furnace should be high
enough to ignite the incoming fuel and air. Continuous
and reliable ignition of fuel is desirable.
2. For complete combustion the fuel and air should be
thoroughly mixed by it.
3. The fuel burning equipment should be capable to regulate
the rate of fuel feed.
4. To complete the burning process the fuel should remain
in the furnace for sufficient time.
5. The fuel and air supply should be regulated to ach.
optimum air fuel ratios.
6. Coal firing equipment should have means to hold
discharge the ash.
Following factors should be considered while selecting a suitable
combustion equipment for a particular type of fuel:
(i) Grate area required over which the fuel burns.
(ii) Mixing 1rrangement for air and fuel.
(iii) Amount of primary and secondary air required.
(iv) Arrangement to counter the effects of caping in fuel or of
low ash fusion temperature.
(v) Dependability and easier operation.
(vi) Operating and maintenance cost.
3.4 Method of Fuel Firing
The solid fuels are fired into the furnace by the following
methods:
1. Hand firing. 2. Mechanical firing.
3.4.1 Hand Firing
This is a simple method of firing coal into the furnace. It requires
no capital investment. It is used for smaller plants. This method of
fuel firing is discontinuous process, and there is a limit to the size
of furnace which can be efficiently fired by this method. Adjustments
 STEAM POWER PLANT
151
arc to be made every time for the supply of air when fresh coal is fed
into furnace.
Hand Fired c;r(ztE5. A hand fired grate is used to support the
fuel bed and admit air for combustion. While burning coal the total
area of air openings varies from 30 to 50% of the total grate area.
The grate area required for an installation depends upon various
factors such as its heating surface, the rating at which it is to be
operated and the t ype of fuel burnt by it. The width of air openings
varies from 3 to 12 mm.
The construction of the grate should be i that it is kept
uniformly cool by incoming air. It should allow asii c freely.
Hand fired grates are made up of cast iron. The various types of
hand fired grates are shown in Fig. 3.8. In large turnaces vertical
shaking grates of circular type are used.

tuna C -,

00 0

SAWDUST. GRATE
Fig. 3.8

The main characteristic of a grate fired furnaces are the heat

liberation per unit of grate area and per unit of volume. The heat
liberation per unit area of fire grate area is calculated as follows
WxC
A

where H = Heat liberation per unit of fire grate area

IV = Rate of fuel consumption (kg/sec)


152 POWER PLANT

C = Lower heating value of fuel (kcallkg)

A = Fire grate area (m2)
The heat liberation per unit of furnace volume is given by the
following expression
H - w C

where H = Heat liberation per unit volume

W = Rate of fuel consumption (kg/sec)
C = Lower heating value of fuel (kcal/kg)
V = Volume of furnace (m3).
These two characteristics depend on the following factors:
(i) Grade of fuel (ii) Design of furnace
(iii) Method of combustion.
Fig. 3.9 shows a hand fire grate fur-
nace with a stationary fuel bed. The grate
divides it into the furnace space in which
the fuel is fired and an ash pit through
which the necessary air required for com-
bustion is supplied. The grate is arranged
horizontally and supports a stationary bed
ofburnin;fuel. The fuel is charged by hand
through the fire door. The total space in the
grate used for the passage of air is called Fig. 3.9
its useful section.
In a hand fired furnace the fuel is periodically shovelled on to
the fuel bed burning on the grate, and is heated up by the burning
fuel and hot masonry of the furnace. The fuel dries, and then evolves
gaseous matter (volatiles combustibles) which rise into the furnace
space and mix with air and burn forming a flame. The fuel left on
the grate gradually transforms into coke and burns-up. Ash remains
on the grate which drops through it into ash pit from which it is
removed at regular intervals. Hand fired furnaces are simple in
design and can burn the fuel successfully but they have some
disadvantages also mentioned below:
(i) The efficiency of a hand fired furnace is low.
(ii) Attending to furnace requires hard manual labour.
(iii) Study process of fuel feed is not maintained.
Cleaning of hand fired furnaces may be mechanised by use of
rocking grate bars as shown in Fig. 3.9 (a). The grate bars loosen
the slag and cause some of it to drop together with the ash into the
bunker without disturbing the process of combustion.
 STEAM POWER PLANT
153

-
Lever
'. •%'•:..
'• Grate bars

. .

Fig. 3.9 (a)

3.4.2 Mechanical Firing (Stokers)

Mechanical stokers are commonly used to feed solid fuels into
the furnace in medium and large size power plants.
The various advantages of stoker firing are as follows:
(i) Large quantities of fuel can be fed into the furnace. Thus
greater combustion capacity is achieved.
(ii) Poorer grades of fuel can be burnt easily.
(iii) Stoker save labour of handling ash and are self-cleaning.
(iv) By using stokers better furnace conditions can be main-
tamed by feeding coal at a Uniform rate.
(u) Stokers save coal and increase the efficiency of coal firing.
The main disadvantages of stokers are their more costs of
operation and repairing resulting from high furnace temperatures.
Principles of Stokers. The working of various types of stokers
is based on the following two principles:
FLAM(
4Q

GRAT( ______

I 1 11
RIMARYAR
Fig. 3. 10

1. Overfeed Principle. According to this principle (Fig. 3.10) the

primary air enters the grate from the bottom. The air while moving
through the grate openings gets heated up and air while moving
through the grate openings gets heated up and the grate is cooled.
The hot air that moves through a layer ofash and picks up additional
energy. The air then passes through a layer of incandescent coke
where oxygen reacts with coke to form-CO 2 and water vapours
accompanying the air react with incandescent coke to form
CO2 , Co and free H 2 . The gases leaving the surface of fuel bed
contain volatile matter of raw fuel and gases like CO 2 , CO, H2, N2
and H2 0. Then additional air known as secondary air is supplied to
burn the combustible gases. The combustion gases entering the
—12
154 * POWER PLANT

boiler consist of N2, CO2 , 02 and 1120 and also CO if the combustion
is not complete.
T1AME5 QYAIR
AS

(E(N COAL -

PRIMARY AIR

Fg. 3.11

2. Underfeed Principle. Fig. 3.11 shows underfeed principle. In

underfeed principle air entering through the holes in the grate
comes in contact with the raw coal (green coal). Then it passes
through the incandescent coke where reactions similar to overfeed
system take place. The gases produced then passes through a layer
of ash. The secondary air is supplied to burn the combustible gases.
Underfeed principle is suitable for burning the semi-bituminous and
bituminous coals.
3.4.3 Types of Stokers
The various types of stokers are as follows
Stokers

Overfeed Underfeed

-
Conveyor Spreader Single Retort Multi-Retort
Stoker Stoker Stoker Stoker

Chain Grate Travelling Grate

Stoker Stoker

Charging of fuel into the furnace is mechanised by means of

stokers of various types. They are installed above the fire doors
underneath the bunkers which supply the fuel. The bunkers receive
the fuel from a conveyor.
(i) Chain Grate Stoker. Chain grate stoker and travelling
grate stoker differ only in grate construction. A chain grate stoker
(Fig. 3.12) consists of an endless chain which forms a support for the
fuel bed.
STEAM POWER PLANT 155

AIR FOR
I /

• Ai
I'VI E TS
I!
Fig. 3.12.

The chain travels over two sprocket wheels, one at the front and
one at the rear of furnace. The travelling chain receives coal at its
front end through a hopper and carries it into the furnace. The ash
is tipped from the rear end of chain. The speed of grate (chain) can
be adjusted to suit the firing condition. The air required for combus-
tion enters through the air inlets situated below the grate. Stokers
are used for burning non-coking free burning high volatile high ash
coals. Although initial cost of this stoker is high but operation and
maintenance cost is low.
The travelling grate stoker also uses an endless chain but differs
in that it carries small grate bars which actuall rt the fuel
fed. It is used to burn lignite, very small sizes ul an rac'. coke
breeze etc.
The stokers are suitable for low ratings .. must
be burnt before it reaches the rear of the furnace. With forced
draught, rate of combustion is nearly 30 to 50 lb of coal per square
foot of grate area per hour, for bituminous 20 to 35 pounds per
square foot per hour for anthracite.

CA1
HOPPER

OVER
cc c

AIR
r

Fig. 3.13.

I5 POWER PLANT

(ii) Spreader Stoker. A spreader stoker is shown in Fig. 3.13.

In this stoker the coal from the hopper is fed on to a feeder which
measures the coal in accordance to the requirements. Feeder is a
rotating drum fitted with blades. Feeders can be reciprocating rams,
endless belts, spiral worms etc. From the feeder the coal drops on to
spreader distributor which spread the coal over the furnace. The
spreader system should distribute the coal evenly over the entire
grate area. The spreader speed depends on the size of coal.
Advantages
The various advantages of spreader stoker are as follows:
1. Its operation cost is low.
2. A wide variety of coal can be burnt easily by this stoker.
3. A thin fuel bed on the grate is helpful in meeting the
fluctuating loads.
4. Ash under the fire is cooled by the incoming air and this
minimises clinkering.
5. The fuel burns rapidly and there is little coking with
coking fuels.
Disadvantages
1. The spreader does not work satisfactorily with varying
size of coal.
2. In this stoker the coal burns in suspension and due to this
fly ash is discharged with flue gases which requires an
efficient dust collecting equipment.

COAL

-kr-
Trbø II
AIR DUCT

Fig. 3.14.

(iii) Multi-retort Stoker. A multi-retort stoker is shown in Fig.

3.14. The coal falling from the hopper is pushed forward during the
inward stroke of stoker ram. The distributing rams (pushers) then


STEAM POWER PLANT

157
slowly move the entire coal bed down the length ofstoker. The length
of stroke of pushers can be varied as desired. The slope of stroke
helps in moving the fuel bed and this fuel bed movement keeps it
slightly agitated to break up clinker formation. The primary air
enters the fuel bed from main wind box situated below the stoker.
Partly burnt coal moves on to the extension grate. A thinner fuel
bed on the extension grate requires lower air pressure under it. The
air entering from the main wind box into the extension grate wind
box is regulated by an air damper.
As sufficient amount of coal always remains on the grate, this
stoker can be used under large boilers (upto 500,000 lb per hr
capacity) to obtain high rates of combustion. Due to thick fuel bed
the air supplied from the main wind box should be at higher
pressure.
Example 3.1. A chain grate stoker is used to burn bituminous
cool having 10% moisture and 10% ash. The higher calorific value
of coal is 7000 kcal 1kg. The steam generator produces steam at the
rate of 10,000 kg per hour. It uses 600 kcal to evaporate 1 kg of feed
water entering the boiler and super heat it to the final temperature,
calculate the following:
(a) Hourly coal supply
(b) Grate area
(c) Grate length if grate width is 5 metres.
Assume overall steam generator efficienc', %. This - ker can
burn the coal at the rate of15 x 10 15 kcalp'r sq. ,,.. r.
Solution. Heat absorbed by water and steam
= 100,000 x 600 = 60 x 10 6 kea er hr.
As the steam generator efficiency = 80%.
Heat to be produced
- 60 x 106
- 0.8 = 75 x 106 kcal per hour.
Amount of coal required
- 75x106
- 7000 x 1000 = 10.71 tonnes.
75 x 106
Grate area = 15 x 10= 50 sq. metres.
Grate width = 5 metres.
Grate length = = 10 metres.
 POWER PLANT
158

15 Automatic Boiler Control

By moans of automatic combustion control it becomes easy to
maintain a constant steam pressu re and uniform furnace draught
and supply of air or fuel call regulated to meet the changes in
steam demand. The boiler operation becomes more flexible and
better efficiency of combustion is achieved. This saves manual
labour also.

5TAMp,Ec5IJFE'1
,4V ANc c.uE S ..
/ TO FAN HAND
I
I' P15 Tc'R

BOfER I

r SECOND4RY
T/0N5 I AIR FAN
N rOMSLJS T/OAI.
-f Cl/4/I8ER

- - ITOIZI
MOTOR
I/AND RF6LJLATOR
_____---%-, StIPPLY

;!R 5U"PY
FAN
V# Nt
I

Fig. 3.15.

Hagan system of automatic combustion control is shown in Fig.

:3.15. Master relay R 1 , is se'nsitive to small vanations in steam
pressure and is connected to steam pressure gauge. A fall in pres-
sure operates the master m-elayl?t which in turn operates the servo-
motor coupled to the vanes of the induced draught U. D.) fall open
them slightly and simultaneously the secondary air fall gets
opened proportionately. By this readjustment of induced draught
takes place and stabilised conditions in the combustion chamber get
changed. These changes operate relay R 2 to alter the position of
forced draught fall to adjust the position of forced
'Iraught fall so that stable conditions in combustion chamber
are maintained. This change causes more air to flow through pas-
sage which in turn opratCs relay Rj. This causes stoker motor to
itpply extra fuel into the furnace. In case ofaii increase of pressure
STEAM POWER PLANT 159

of steam the above process is reversed. I land regulators are provided

to servo motors and master relay for manual control of system.
3.6 Pulverised Coal
Coal is pulverised (powdered), to increase its surface exposure
thus permitting rapid combustion. Efficient use of coal depends
greatly on the combustion process employed.
For large scale gneration of energy the efficient method of
burning coal is confined still to pulverised coal combustion. The
pulverised coal is obtained by grinding the raw coal in pulverising
mills. The various pulverising mills used are as follows:
(i) Ball mill (ii) Hammer mill
(iii) Ball and race mill (iv) Bowl mill.
The essential functions of pulverising mills are as follows
(i) Drying of the coal (ii) Grinding
(iii) Separation of particles of the desired size.
Proper drying of raw coal which may contain moisture is neces-
sary for effective grinding.
The coal pulverising mills reduce coal to powder form by three
actions as follows
(i) Impact (ii) Attrition (abrasion)
(iii) Crushing.
Most of the mills use all the above mentioned all the three
actions in varying degrees. In impact type mills 1' break the
coal into smaller pieces whereas in attrition t pe the coat pieces
which rub against each other or metal surfaces to disintegrate. In
crushing type mills coal caught between metal r ......aces gets
broken into pieces. The crushing mills use steel balls in a container.
Those balls act as crushing elements.

Fig. 3.16.

160 POWER PLANT

43.6.1 Ball mill

A line diagram of ball mill using two classifiers is shown in Fig.
3.16. It consists of a slowly rotating drum which is partly filled with
steel balls. Raw coal from feeders is supplied to the classifiers from
where it moves to the drum by means of a screw conveyor. As the
drum rotates the coal gets pulverised due to the combined impact
between coal and steel balls. Hot air is introduced into the drum.
The powdered coal is picked up by the air and the coal air mixture
enters the classifiers, where sharp changes in the direction of the
mixture throw out the oversized coal particles. The over-sized par-
ticles are returned to the drum. The coal air mixture from the
classifier moves to the exhauster fan and then it is supplied to the
burners.

PULVERISED COAL
TO BURNERS

RAWCOAL FEED

ROTATING •
I • •
CLASSIFIER
• •I

SPRING

UPPER
RACE
GRINDING
ELEMENTS
4.

-lOT PRIMARy I
AIR SUPPLY BALL 1101 PRIMARY
AIR SUPPLY
LOWER-1
RACE I rGLAR

WORM

Fig. 3.16(a)


STEAM POWER PLANT 161

3.6.2 Ball and race mill

Fig. 3.16 (a) shows a ball and race mill. In this mill the coal
passes between the rotating elements again and again until it has
been pulverised to desired degree of fineness. The coal is crushed
between two moving surfaces namely balls and races. The upper
stationary race and lower rotating race driven by a worm and gear
hold the balls between them. The raw coal supplied falls on the inner
side of the races. The moving balls and races catch coal between
them to crush it to a powder. The necessary force needed for
crushing is applied with the help of springs. The hot air supplied
picks up the coal dust as it flows between the balls and races, and
then enters the classifier. Where oversized coal particles are
returned for further grinding, where as the coal particles of required
size are discharged from the top of classifier.
In this mill coal is pulverised by a combination of crushing,
impact and attrition between the grinding surfaces. The advantages
of this mill are as follows
(i) Lower capital cost (ii) Lower power consumption
(iii) Lower space required (it') Lower weight.
However in this mill there is greater wear as compared to other
pulverisers.
The use of pulverised coal has now become the standard method
of firing in the large boilers. The pulverised coal burns with some
advantages that result in economic and flexible operation of steam
boilers.
CYCLONE
INTERMEDIATE BUNKER

BUNKER

AUTOMATIC
BALANCE
FEEDER

BALL MILL AIR HEAT FR

Fig. 3.17 (a).

Preparation of pulveried fuel with an intermediate-hunker is

shown in Fig. 3.17 (a). The fuel moves to the automatic balance and
then to the feeder and [)all mill through w1' ich hot air is blown It
dries the pu}vcriscdcoal and carries it from the mill ti the separator.

1 182 POWER PLANT

The air fed to the ball mill is heated in the air heater. In the
separator dust (fine pulverised coal) is separated from large coal
particles which are returned to the ball mill for regrinding. The dust
moves to the cyclone. Most of the dust (about 90%) from cyclone
moves to bunker. The remaining dust is mixed with air and fed to
the burner.
Coal is generlIy ground in low speed ball tube mill. It is filled
to 20-35% of its volume. With steel balls having diameter varying
from 30-60 mm. The steel balls crush and ground the lumps of coal.
The average speed ofrotation of tube or drum is about 18-2.0 R.P.M.
[Fig. 3.17 (b)].

FUEL IN
.I1IIj1uIi
HOTAIR
1

ARMOUR

Fig. 3.17(b) Ball tube mill.

Advantages
The advantages of using pulverised coal are as follows
1. It becomes easy to burn wide variety of coal. Low grade
coal can be burnt easily.
2. Powdered coal has more heating surface area. They per-
mits rapids and high rates of combustion.
3. Pulverised coal firing requires low percentage of excess air.
4. By using pulverised coat, rate of combustion can be ad-
justed easily to meet the varying toad.
5. The system is free from clinker troubles.
6. It can utilise highly preheated air (of the order of 700F)
successfully which promotes rapid flame propagation.
7. As the fuel pulverising equipment is located outside the
furnace, therefore it can be repaired without cooling the
unit down.
8. High temperature can be produced in furnace.
Disadvantages
1. It requires additional equipment to pulverise the coal. The
initial and maintenance cost of the equipment is high.
STEAM POWER PLANT 163

2. Pulverised coal firing produces fly ash (fine dust) which

requires a separate fly ash removal equipment-
3. The furnace for this type of firing has to be carefully
designed to withstand for burning the pulverised fuel
because combustion takes place while the fuel is in
suspension.
4. The flame temperatures are high and conventional types
of refractory lined furnaces are inadequate. It is desirable
to provide water cooled walls for the safety of the furnaces.
5. There are more chances of explosion as coal burns like a
gas.
6. Pulverised fuel fired furnaces designed to burn a par-
ticular type of coal can not be used to any other type of coal
with same efficiency.
7. The size of coal is limited. The particle size of coal used in
pulverised coal furnace is limited to 70 to 100 microns.
3.6.3 Shaft Mill
Fig. 3.17 (a) shows fuel pulverisation with a shaft mill. The fuel
from bunker is moved to feeder via automatic balance. Then from
duct fuel goes to mill where it is crushed by beaters secured on the
spindle of the mill rotor.

Auk
bQ

Fe


164 POWER PLANT

The pulverised fuel is dried up and then blown into shaft by hot
r.ir. Secondary air is delivered into the furnace through holes to burn
the fuel completely.
3.7 Pulverised Coal Firing
Pulverised coal firing is done by two system:
(i) Unit System or Direct System.
(ii) Bin or Central System.
Unit System. In this system (Fig. 3.18) the raw coal from the
coal bunker drops on to the feeder.

SECONDARY AIR FURNACE

PRIMARY AIR
BUNKER COAL ii
RAW COAL
T BURNET
IPULVERISING
MILL
FEEDER -
FAN
HOT AIR

Fig. 3.18

Hot air is passed through coal in the feeder to dry the coal. The
coal is then transferred to the pulverising mill where it is pulverised.
Primary air is supplied to the mill, by the fan. The mixture of
pulverised coal and primary air then flows to burner where secon-
dary air is added. The unit system is so called from the fact that
each burner or a burner group and pulveriser constitute a unit.

R(TuRl AID

, ran—,
WENT I ,.i I
PL' VER/SED (O.4 8I.'Q

PR,MARr AIR

SE(QO4Rv.N

Fig. 3.19.
 STEAM POWER PLANT
165
Advantages
I. The system is simple and cheaper than the central system.
2. There is direct control of combustion from the pulverising
mill.
3. Coal transportation system is simple.

Bin or Central System. It is shown in Fig. 3.19. Crushed coal

from the raw coal bunker is fed by gravity to a dryer where hot air
is passed through the coal to dry it. The dryer may use waste flue
gases, preheated air or bleeder steam as drying agent. The dry coal
is then transferred to the pulverising mill. The pulvérised coal
obtained is transferred to the pulverised coal bunker (bin). The
transporting air is separated from the coal in the cyclone separator.
The primary air is mixed with the coal at the feeder and the mixture
is supplied to the burner.

Advantages
I. The pulverising mill grinds the coal at a steady rate
irrespective of boiler feed.
2. There is always some coal in reserve. Thus any occasional
breakdown in the coal supply will not effect the coal feed
to the burner.
3. For a given boiler capacity pulverising mill " mall
capacity will be required as compared to unit sy ;tem.

Disadvantages
I. The initial cost of the system is high.
2. Coal transportation system is quite complicated.
3. The system requires more space.
To a large extent the performance of pulverised fuel system
depends upon the mill performance. The pulverised mill should
satisfy the following requirements:
1. It should deliver the rated tonnage of coal.
2. Pulverised coal produced by it should be of satisfactory
fineness over a wide range of capacities.
3. It should be quiet in operation.
4. Its power consumption should be low.
5. Maintenance cost of the mill should be low.
Fig. 3.19 (a) shows the equipments for unit and central sS'stem
of pulverised coal handling plant.
 POWER PLANT
16

RAW COAL

PRIMARY CRUSHER

MAGNETIC SEPARATOR

COAL DRIER

COAL BUNKERS

UNIT SYSTEM CENTRAL SYSTEM

SCALE
SCALE

PULVERISER
PULVE RISER

CENTRAL BIN
7 Is
BURNERS FEEDER

FURNACE

FURNACE ,
BURNERS

Fig. 3.19 (a)

3.8 Pulverised Coal Burners

Burners are used to burn the pulverised coal. The main dif-
ference between the various burners lies in the rapidity of air-coal
mixing i.e., turbulence. For bituminous coals the turbulent type of
burner is used whereas for low volatile coals the burners with long
flame should be used. A pulverised coal burner should satisfy the
following requirements
(i) It should mix the coal and primary air thoroughly and
should bringthis mixture before it enters the furnace in
contact with additional air known as secondary air to
create sufficient turbulence.
(ii) It should deliver and air to the furnace in right proportions
and should maintain stable ignition of coal air mixture
STEAM POWER PLANT 167

and control flame shape and travel in the furnace. The flame shape
is controlled by the secondary air vanes and other control adjust-
ments incorporated into the burner. Secondary air if supplied in too
much quantity may cool the mixture and prevent its heating to
ignition temperature.

COLD (TEMPERING) AIR

FROM FORCED DRAFT FAN HOT AIR FROM
BOILER AIR HEATER
TEMPERING AJR,
DAMPER
FURNACE

RAW COAL
BUNKER
_7 PULVERIZED FUEL
BURNERS
BURNER WINDBOX

EEDER

\ PULVERIZED FUEL
CONTROL AND AIR PIPING
DAMPER
PULVERIZER
- PRIMARY
AIR FAN

Fig. 3.19(b) Shows typical firing system in pulverise-

(iii) Coal air mixture should move away from 4' ,rner at a
rate equal to flame front travel in order to avL.I, iash back
into the burner.
The various types of burners are as follows
1. Long Flame Burner (U-Flame Burner). In this burner air
and coal mixture travels a considerable distance thus providing
sufficient time for complete combustion [Fig. 3.20 (a)].
2. Short Flame Burner (Turbulent Burner). It is shown in
Fig. 3.20 W. The burner is fitted iii the furnace will and the flame
enters the furnace horizontally.
3. Tangential Burner. A tangential burner is shown in Fig.
3.20 (c). In this s ystem one burner is fitted attach corner of the
furnace. The inclination of the burner is so made that the flame
produced are tangential to an imaginary circle at the centre.
4. Cyclone Burner. It is shown in Fig. 3.20 (d). This burner
uses crushed coal intend of pulverised coal. Its advantages are as
follows
 S

168 POWER PLANT

I Coal and primary

air

Pr

Z. Secondary
air
huh

Secondary
'I
air
(a) (b)

Coal and
Primary air Cool and
primary
Sec air air
sec.—"
air

ec
Furnace
-.1 SECTION AT A-A
A
(c)
Fig. 3.20

BOILER DRUM
CONVECTION
SUPER HEATER

ECONOMIZER._
-
•o.
I I'll
AIR
PREHEATER 7WATER WALL

:7
BOILER
TUBES
Ill P ULVERISED COAL
BURNER
TO STACK ULJ


STEAM POWER PLANT 169

(i) It saves the cost of pulverisation because of a crusher

needs less power than a pulveriser.
(ii) Problem of fly ash is reduced. Ash produced is in the
molten form and due to inclination of furnace it flows to
an appropriate disposal system.
Fig. 3.21 (a) shows a pulverised coal-fired boiler.

RAW COAL
BUNKER

FEEDER
CONVEYOR
PRIMARY
AIR
Rota

IICRUSHER J I—'-1
BREAKER VALVE

______ d It C
T CYCLONE
FURNACE
f .
COAL PIPE
41,121
CONVEYOR
Fig. 3.21 (b) Shows a typical firing system fo r ., ,urnace.

3.8.1 Cyclone Fired Boilers. In cyclon 'red } ! ors the fur-

nace is arranged as a horizontal cylinder. The L-rtsed fuel is bed
along the periphery of the cylinder. The hot ga.e travel axiall y into
the water, tube section having a tight helix path. The temperature
generated in the combustion zone is quite high and because of this
the tubes are coated with fused ash which goes on collecting the ash
particles doing in the flue gases. The out going gases contain
particles less than 20 microns.
The c y clone furnaces can successfully burn coals having low ash
fusion temperature. The cyclone furnace is operated under combus-
tion air pressure of 700 to 1000 mm of water gauge.
C y ci no tired boilers have the following advantages () Quick
load variations can be easily handled. (U) Nearly 85 17c of ash in co.il
is burnt in the form of liquid slag. The ash can be removed in the
molten form. (iii) The slag can be used as a building material. (iv)
Fl y -ash problem is reduced to mach lower limits

3.9 Water Walls

Larger central station type boilers have water cooled furnaces.
The combustion space of furnace is shielded wholl y or partiall y by
small diameter tubes placed side by side. Water from the boiler , Is
—13

170 POWER PLANT

made to circulate through these tubes which connect lower and

upper headers of boiler.
The provision of water walls is advantageous due to following
reasons : (1) These walls provide a protection to the furnace against
high temperatures. (2) They avoid the erosion of the refractory
material and insulation. (3) The evaporation capacity of the boiler
is increased.

IA'SULATING CONCRITE
rMANOED METAL LATH
NIGH TEMPPLASTIC/#15LATIOn
MAGNESIA BLOCK
CASING

Touching tubes.

CAST
L. .r...
BRICK
NSi'L A nON
BLANKET
4_^^ INSUtA''
CASING
Half radiant cast ief.ctoiy.

HIGH TEMPERA TLac'( 'LA7

li/SLit A 7/H6 BR/CK
MAGNESIA OtOCK
C.45/4

Tangent tubes-flat files.

Fig. 3.22
The tubes are attached with the refractory materials on the
inside or partially embedded into it. Fig. 3.22 shows the various
water walls arrangement.
3.10 Ash Disposal
A large quantity of ash is, produced in steam power plants using
coal. Ash produced in about 10 to 20% of the total coal burnt in the
furnace. Handling of ash is a problem because ash coming out of the
furnace is too hot, it is dusty and irritating to handle and is
accompanied by some poisonous gases. It is desirable to quench the
ash before handling due to following reasons: (1) Quenching reduces
the temperature of ash. (2) It reduces the corrosive action of ash. (3)
Ash forms clinkers by fusing in large lumps and by quenching
STEAM POWER PLANT 171

clinkers will disintegrate. (4) Quenching reduces the dust accom-

panying the ash.
Handling of ash includes its removal from the furnace, loading
on the conveyors and delivered to the fill from where it can be
disposed oft
3.10.1 Ash Handling equipment
Mechanical means are required for the disposal of ash. The
handling equipment should perform the following functions : (1)
Capital investment, operating and maintenance charges of the
equipment should be low. (2) It should be able to handle large
quantities of ash. (3) Clinkers, soot, dust etc. create troubles, the
equipment should be able to handle them smoothly. (4) The equip-
ment used should remove the ash from the furnace, load it to the
conveying system to deliver the ash to a dumping site or storage and
finally it should have means to dispose of the stored ash. (5) The
equipment should be corrosion and wear resistant.
Fig. 3.23 shows a general layout of ash handling and dust
collection system. The commonly used ash handling systems are as
follows:

CHI-tEY

ASH DrSCHARGE
EQUIPMENT

Fig. 3.23

(I) Hydraulic system (ii) Pneumatic system

(w' Mechanical sem.
The commonly used ash discharge equipment is as follows:
(i) Rail road cars (ii) Motor truck
(iii) Barge.
The various methods used for the disposal of ash are as
follows:
 1724 POWER PLANT

(i) Hydraulic System. In this system, ash from the furnace

grate falls into a system of water possessing high velocity and is
carried to the sumps. It is generally used in large power plants.
Hydraulic system is of two types namely low pressure hydraulic
system used for continuous removal of ash and high pressure system
which is used for intermittent ash disposal. Fig. 3.24 shows
hydraulic system.

BOILERS

WATER_

SUMPS

BOILERS
Fig. 3.24

In this method water at sufficient prssure is used to take away

the ash to sump. Where water and ash are separated. The ash is
then transferred to the dump site in wagons, rail cars or trucks. The
loading of ash may be through a belt conveyor, grab buckets. If there
is an ash basement with ash hopper the ash can fall, directly in ash
car or conveying system.
(ii) Water Jetting. Water jetting of ash is shown in Fig. 3.25.
In this method a low pressure jet of water coming out of the
quenching nozzle is used to cool the ash. The ash falls into a trough
and is then removed.
(iii) Ash Sluice Ways add Ash Sump System. This system
shown diagrammatically in Fig. 3.26 used high pressure (H.P.)
pump to supply high pressure (H.P.) water jets which carr y ash from
the furnace bottom through ash sluices (channels) constructed in
basement floor to ash sump fitted with screen. The screen divides
the ash sump into compartments for coarse and fine ash. The fine
ash passes through the screen and moves into the dust SUrnp (D.S.).
Dust slurry pump (l).S. pump) carries the dust through dust pump
(D.P.), suction pipe and dust delivery (D.D.) pipe to the disposal site.
Overhead crane having grab bucket is used to remove coarse ash.
A.F.N. represents ash feeding nozzle and S.B.N. represents sub way
booster nozzle and D.A. means draining apron.
(iv) Pneumatic system. In this system (Fig. 3.27) ash from the
boiler furnace outlet falls into a crusher where larger ash particles
are crushed to small sizes. The ash is then carried b y a high velocity
air or steam to the point of delivery. Air leaving the ash separator


STEAM POWER PLANT 173

is passed through filter to remove dust etc. so that the exhauster

handles clean air which will protect the blades of the exhauster.

FURNACE

NOZZLE

STOKER

T.ROUGH___1
WATER JETTING
Fig. 3 25

H. P. WATER PIPING ?HEAOC .E

rF
D. PIPE
HOPPER

H.P.P UMP
__ I ID
BT D S PUIP

JLR S.8.N.
A.F.N. ,LiI

-.
SCREE

ASH SLUICE ASH SUMP

N
-OUST FROM
PRECIPITATOR
OVERFLOW
WAY
.P. SUCTION
PIPE

Fig. 3.26

(t') Mechanical ash handling system. Fig. 3.27 ((1) shows a

mechanical ash handling s ystem. In this s y stem ash cooled by water
seal falls on the belt conve y or and is carried out continuousl y to the
bunker. The ash is then removed to the dumping site from the ash
bunker with the help of trucks.
174 POWER PLANT

cc

ir
uJ
x
'U

cr

Fig. 3.27

3OILER
FU RNA CE S

ASH
U 6H

BELT CON V EYOR BUNKER

cII1 TRUCK

Fig. 3.27 (a)

STEAM POWER PLANT 175

Efficient Combustion of Coal

The factors which affect the efficient combustion of coal are as
follows:
1. Type of coal. The important factors which are considered for
the selection of coal are as follows:
(i) Sizing (ii) Caking (iii) Swelling properties (iv) Ash fusion
temperature.
The characteristics which control the selection of coal for a
particular combustion equipment are as follows:
(i) Size of coal
(ii) Ultimate and proximate analysis
(iii) Resistance of degradation
(iv) Grindability
(u) Caking characteristics
(vi) Slagging characteristics
(vii) Deterioration during storage
(viii) Corrosive characteristics
(ix) Ash Content.
The average ash content in Indian coal is about 20%. It is
therefore desirable to design the furnace in such a way as to burn
the coal of high ash content. The high ash content in coal has the
following:
(i) It reduces thermal efficiency of the boiler as loss of heat
through unburnt carbon, excessive clinker formation and
heat in ashes in considerably high.
(ii) There is difficulty of hot ash disposal.
(iii) It increases sue of plant.
(iv) It increases transportation cost of fuel per unit of heat
produced.
(v) It makes the control difficult due to irregular combustion.
High as content fuels can be used more economically in
pulverised form. Pulverised fuel burning increases the
thermal efficiency as high as 90% and controls can be
simplified by just adjusting the position of burners in
pulverised fuel boilers. The recent steam power plants in
India are generally designed to use the pulverised coal.
2. Type of Combustion equipment. It includes the following:
(i) Type of furnace
(ii) Method of coal firing such as : (a) Hand tiring (b) Stoker
firing (c) Pulverised fuel firing.
(iii) Method of air supply to the furnace. It is necessary to
provide adequate quantity of secondary air with sufficient
turbulence.
(it') Type of burners used.
 176 POWER PLANT
S
(v) Mixing arrangement of fuel and air.
The flames over the bed are duel to the burning of volatile gases,
lower the volatile content in the coal, shorter will be the flame. If
the volatiles burn up intensely high temperature is generated over
the furnace bed and helps to burn the carbon completely and vice
versa.
For complete burning of volatiles and prevent unburnt carbon
going with ash adequate quantity of secondary air with sufficient
turbulence should be provided.
3.11 Smoke and Dust Removal
Ill fed furnaces the products ofcombust.ion contain particles
of solid matter floating in suspension. This may be smoke or dust.
The production of smoke indicates that combustion conditions are
faulty and amount of smoke produced call reduced by improving
the furnace design.
In spreader stokers and pulverised coal fired furnaces the coal
is burnt in suspension and due to this dust in the form of fly ash is
produced. The size of dust particles is designated in microns
(1 M = 0.001 mm). Dust particles are mainly ash particles called fly
ash intermixed with some quantity of carbon ash material called
cinders. Gas borne particles larger than 1 t in diameter are called
dust and when such particles become greater in size than 100t they
are called cinders.
Smoke is produced due to the incomplete combustion of fuels,
smoke particks are less than 10p in size.
The disposal smoke to the atmosphere is not desirable due to
the. following reasons
1. A smoky atmosphere is "ealthful than smoke free air.
2. Smoke is produced due to incomplete combustion of coal.
This will create a big economic loss due to loss of heating
value of coal.
3. In a smoky atmosphere lower standards of clean, - are
prevalent. Buildings, clothings, furniture etc. becomes
dirt y due to smoke. Smoke corrodes the metals and
darkens the paints.
To avoid smoke nuisance the coal should be completely burnt in
the furnace.
The presence of dense smoke indicates poor furnace conditions
and a loss in el'ticiencv and capitcityofa boiler plant A small amount
of smoke leaving chimne Ny. shows good furnace conditions whereas
smokeless chimne y does necessarily mean a better efficiency in
the boiler room.
 STEAM POWER PLANT
177
To avoid the atmospheric pollution the fly ash must be removed
from the gaseous products of combustion before they leaves the
chimney.
The removal p f dust and cinders from the flue gas is usually
effected by commercial dust collectors which are installed between
the boiler outlet and chimney usually in the chimney side of air
preheater.
3.12 Types of Dust Collectors
The various types of dust collectors are as follows:
1. Mechanical dust collectors.
2. Electrical dust collectors.
Mechanical dust collectors. Mechanical dust collectors are
sub-divided into wet and dry types. In wet type collectors also known
as scrubbers water sprays are used to wash dust from the air. The
basic principles of mechanical dust collectors are shown in Fig. 3.28.
As shown in Fig. 3.28 (a) by increasing the cross-sectional area of
duct through which dust laden gases are passing, the velocity of
gases is reduced and causes heavier dust particles to fall down.
Changing the direction of flow [Fig. 3.28 (b)] of flue gases causes the
heavier particles of settle out. Sometime baffles are provided as
shown in Fig. 3.28 (c) to separate the heavier particles.
Mechanical dust collectors may be wet type or dry type. Wet type
dust collectors called scrubbers make use of water sprays to wash
the dust from flue gases.

(a) (b) (c)

Fag. 3.28
Dry type dust collectors include gravitational, cyc ' one, louvred
and baffle dust collectors.
A cyclone dust collector is shown in Fig. 3.29. This Collector uses
a downward flowing vortex for dust laden gases along the inner
walls. The clean gas leaves from an inner upward flowing vortex.
The dust particles fall to the bottom due to centrifuging action.
Electrostatic Precipitators. It has two sets of electrodes,
insulated from each other, that maintain an electrostatic field
 POWER PLANT

between them at high voltage (Fig. 3.30). The flue gases are made
to pass between these two sets ofelectrodes. The electric field ionises
the dust particles that pass through it attracting them to the
electrode of opposite charge. The other electrode is maintained at a
negative potential of 30,000 to 60,000 volts. The dust particles are
removed from the collecting electrode by rapping the electrode
periodically. The electrostatic precipitator is costly but has low
maintenance cost and is frequently employed with pulverised coal
fired power stations for its effectiveness on very fine ash particles
and is superior to that of any other type.
The principal characteristics of an ash collector is the degree of
collection.
il = Degree of collection
G1—G2
C'
- Cl - C2
Cl
where Gi = Quantity of ash entering an ash collector per unit
time (kgls)
G2 = Quantity of uncollected ash passing through the
collector per unit time (kg/s)
C, Concentration of ash in the gases at the inlet to the
=
ash collector (kg/m3)
C2 = Ash concentration at the exist (kg/rn3).
Depending on the type ot fuel and the power of boiler the ash
collection in industrial boilers and thermal power stations can be
effected by mechanical ash collectors, fly ash scrubbers and electros-
tatic precipitators.
For fly ash scrubbers of large importance is the content of free
lime (CaO) in the ash. With a high concentration of CaO the ash can
be cemented and impair the operation of a scrubber.
The efficiency of operation of gas cleaning devices depends
largely on the physico-chemical properties of the collected ash and
of the entering waste gases.
Following are the principal characteristics of the fly ash
(i) Density
(ii) Dispersity (Particle size)
(iii) Electric resistance (For electrostatic precipitators)
(iv) Coalescence of ash particles.
STEAM POWER PLANT 179
Due to increasing boiler size and low sulphur high ash content
coal the problem of collecting fly ash is becoming increasingly
complex. Fly ash can range from very fine to very coarse size
depending on the source. Particles colour varies from light tan to
grey to black. Tan colour indicates presence of ion oxide while dark
shades indicate presence of unburnt carbon. Fly ash particles size
varies between 1 micron (iji) to 300.t. Fly ash concentration in flue
gases depends upon mainly the following factors:
(i) Coal composition.
(ii) Boiler design and capacity.
Percentage of ash in coat directly contributes to fly ash emission
while boiler design and nr'ration determine the percentage
retained in the furnace as botiin ash and fly ash carried away by
flue gas. Fly ash concentration widely varies around 20-90 g/mm3
depending on coal and boiler design. Fly ash particle size distribu-
tion depends primarily on the type of boiler such as pulverised coal
fired boiler typically produces coarser particles then cyclone type
boilers. Electrostatic precipitator (ESP) is quite commonly used for
removal of fly ash from flue gases.

Purified gas

Houi

Inlet

ii
180 POWER PLANT

3.12.1 Fly Ash Scrubber

Fig. 3.28 (a) shows a fly wash centrifugal scrubber. It is similar
to a mechanical ash collector but has a flowing water film on its
inner walls. Due to this film, the collected ash is removed more
rapidly from the apparatus to the bin and there is less possibility
for secondary. Capture of collected dust particles by the gas flow.
The degree of ash collection in scrubbers varies from 0.82 to 0.90.
The dust laden gas enters through the inlet pipe.
Cinder Catcher. Cinder catcher [Fig. 3.28 (b)J is used to
remove dust and cinders from the gas. In this catcher the dust laden
gas is made to strike a series of vertical baffles that change its
direction and reduce its velocity. The separated dust and cinders fall
to the hopper for removal. Cinder catchers are ordinarily used with
stoker firing.
cBQftI2S

Dust
Ladden
IJ I
GasCIeari
li
1111111
U Gas

.1

Dust Hopper

Dust -
Fig. 3.28 (b).
3.12.2 Fluidised Bed Cbmbustion (FBC)
Burning of pulverised coal has some problems such as particle
size of coal used in pulverised firing is limited to 70-100 microns,
the pulverised fuel fired furnances designed to burn a particular can
not be used other type of coal with same efficiency, the generation
of high temp. about (1650 ' C) in the furnace creates number of
problems like slag formation on super heater, evaporation of alkali
metals in ash and its deposition on heat transfer surfaces, formation
Of SO2 and NO in large amount.
Fluidised Bed combustion system can burn any fuel including
low grade coals (even containing 70% ash), oil, gas or municipal
waste. Improved desuiphurisation and low NO emission are its
main characteristics. Fig. 3.28 (c) shows basic principle of Fluidised


STEAM POWER PLANT

181
bed combustion (FBC) system. The fuel and inert material dolomite
are fed on a distribution plate and air is supplied from the bottom
of distribution plate. The air is supplied at high velocity so that solid
feed material remains in suspension condition during burning. The
heat produced is used to heat water flowing through the tube and
convert water into steam. During burning SO 2 formed is absorbed
by the dolomite and thus prevents its escape with the exhaust gases.
The molten slag is tapped from the top surface of the bed. The bed
temperature is nearly 800-900C which is ideal for sulphur reten-
tion addition of limestone or dolomite to the bed brings down SO2
emission level to about 15% of that in conventional firing methods.
FLUE GASES

STEAM —.0 oO'I NASH

FUEL AND I: . -. . -
DOLOMITE j,o ?' -
bQ..o.'
—TUBES
WATER
DISTRIBUTOR
PLATE
N
'r
AIR
Fig. 3.28 (c
The athount of NO is produced is also reduced because of low
temperature of bed and low excess air as compared to pulverised
fuel firing.
The inert material should be resistant to heat and disintegra-
tion and should have similar density as that of coal. Limestone, or
dolomite, fused alumina, sintered ash are commonly used as inert
materials.
Various advantages of FBC system are as follows:
(i) FBC system can use any type of low grade fuel including
municipal wastes and therefore is a cheaper method of
power generation.
(ii) It is easier to control the amount of SO 2
and NO formed
during burning. Low emission of SO 2 and NO will help in
controlling the undesirable effects of SO 7 and NO during
combustion. SO 2 emission is nearly 15% of that in conven-
tional firing methods.
(iii) There is a saving of about 10% in operating cost and 15%
in the capital cost of the power plant.


S
182 POWER PLANT

The size of coal used has pronounced effect on the opera-

(iv)
tion and performance of FBC system. The particle size
preferred is 6 to 13 mm but even 50 mm size coal can also
be used in this system.
3.12.3 Types of FBC systems
FBC systems are of following types : (i) Atmospheric FBC
system : (a) Over feed system (b) Under feed system.
In this system the pressure inside the bed is atmospheric.
Fig. 3.28 (d) shows commercial circulation FBC system. The
solid fuel is made to enter the furnace from the side of walls. .The
low velocity (LV), medium velocity (MV) and high velocity (HV) air
is supplied at different points along the sloping surface of the
distribution ash is collected from the ash port. The burning is
efficient because of high lateral turbulence.
(ii) Pressurised FBC system. In this system pressurised air is
used for fluidisation and combustion. This system has the following
advantages: (a) High burning rates. (b),Improved desulphurisation
and low NO emission. (c) Considerable reduction in cost.
TO BOILER

SOLID I I I . - TART UP
FUELN H SBURNER

DEFLECTOR
WALL
Lid IT c
ii

MATTER —•

DISTRIBUTE - LIQUID

________ FUEL
L4 M4 AIR
AIR AIR
H4
ASH
AIRrORT

ASH

Fig. 3.28 (c

3.13 Draught
The purpose of draught is as follows:
(i) To supply required amount of air to the furnace for the
combustion of fuel. The amount of fuel that can be burnt
per square foot of grate area depends upon the quantity of
air circulated through fuel bed.
(ii) To remove the gaseous products of combustion.
 STEAM POWER PLANT
183
Draught is defined as the difference between absolute gas pres-
sure at any point in a gas flow passage and the ambient (same
elevation) atmospheric pressure. Draught is plus if <Pgos and
it is minus Pat . > Pgas. Draught is achieved by small pressure
difference which causes the flow of air or gas to take place. It is
measured in millimetre (mm) of water.
kflpjr

"IS , B0 40-80 kV dc

Cleaned gas
Rectifie

Gas teal
Inlet H.4 Cu^rrent[ gas
Tr
Oust laden
gas - electr

jE,n,tt ing

Ilecting
electrode
Duet

Fig. 3.29.
Fig. 3.30.

If only a chimney is used to create the necessary draught, the

system is called natural draught system and if in addition to
chimney a forced draught (F.D.) fan or an induced draught (I.D.),
fan or both are used the system is called mechanical draught system.
Fans or chimneys produce positive pressure and is called available
draught whereas fuel bed resistance, turbulence and friction in air
ducts, gas breechings, chimney, etc. create negative pressure and is
called the required draught.
The various types of draught systems are as follows
(i) Natural draught (ii) Mechanical draught
(iii) Steam Jet draught.
Natural Draught. Natural draught system is used in boilers
of smaller capacities. Natural draught is created by the difference
in weight of column ofcold external air and that ofa similar column
of hot gases in the chimney. This system is dependent upon the
height of chimney and average temperature of the gases in the
chimney. Fig. 3.31 shows natural draught system.
 POWER PLANT
I B4

cc
COAL CIWAINEY

MOURNACC

Fig. 3.31.

Now-a-days the chimney is not used for creating draught in

steam power-plants as it has no flexibility, the total draught
produced is insufficient for high generating capacity. By using
chimney draught can be increased by allowing the flue gases to leave
the combustion chamber at higher temperature and this reduces the
overall efficiency of the power plant. The chimney is, therefore, used
only to discharge the flue gases.
Mechanical Draught
In boilers of larger capacities, fans are employed to create the
necessary draught in order to reduce the height of chimney, to obtain
draught that is independent of weather conditions and to control the
draught easily.
Mechanical draft may be induced, forced or balanced draft.
Induced draught system shown in Fig. 3.32 (a) is created by
chimney and fan located in the gas passage on the chimney side of
the boiler. In this system gas movement is achieved as result of a
vaccum.
The various pressures indicated are as follows
P 1 = Inlet pressure of forced draft fan.
P2 = Outlet pressure of forced draft fan.
P = Pressure below grate.
P4 = Pressure above the grate.

P5 = Ihlet pressure of induced draft fan.

P6 = Outlet pressure of induced draft fan.
Induced draught is not as simple and direct as forced because
fans used in induced draft system operate in gases of much higher
temperature (nearly 500 - 904 F). This becomes more expensive.

I,'
 STEAM POWER PLANT
185
The fan sucks in gas from the boiler side and discharges it to the
chimney (stack).
The draught produced is independent of the temperature of the
hot gases an d, therefore, the gases may be discharged as cold as
possible after recovering as much heat possible in air preheater and
economiser.
In forced drauLc vstem [Fig. 3.32 (b)] the fan installed near
the boiler base supplies the air at a pressure above that of atmos-
phere and deIivers it through air duct to the furnace. Most high
rating combustion equipment employs forced draught fans for sup-
plying air to the furnace. Forced draught is used in under fed stokers
carrying a thick fuel bed.

t
Li
-,i
DRUM
LO. r.4,v

()

a
r1 cHflNey

(b)
GR4T
F 0 F,1N

/o\ - lOrAN

(c)

84L4PVCED 0Au6,,7 5Y57j

Fig. 3.32
—14
 POWER PLANT
186

Balanced draught system is a combination of induced and farced

air tli ruugh the
draught s y stems. The forced draught fan forces the
fuel bed oil to the top of grate and the induced draught fan sucks in
gases from the boiler side and discharges them to the chiiiiney. This
system is used where pressure above fire is
slightly below atmos-
pheric, Fig. 3.32 (c) shows this system.
Fig 3.32 shows the pressure distribution for the balanced sys-
tem. t
Boiler
Chimney

Forced
fan
I nduced
f.

u _!
Grote
conomiser

Air
Furnace pre-heciter
e
Atmosph ric
r pressure (mr

P,
4=-^
P5
Fig. 3.32 Balanced draft.

Construction such as shielding or water cooling or water protect

the bearings of fans. Secondary, the fans handle gases laden with
dust which causes were of blades. In forced draught system the fans
handle coot and clean air and the fan can be located where con-
venient. Balanced draft system is more efficient. In balanced draft
system about 0.1 inch water vaccum is maintained over the fuel bed.
Multivane centrifugal fans are generally used for moving largi
volume of air and gases. The performance of a fail on th€
shapes of blades which are, in general, of three types
1. Backward curved blades.
2. Forward curved blades.
3. Straight radial blades.

(0 S q*
[Various types of blade forms are shown in Fig. 3.331

Backward
cu,vzd
Forward
curved
Fig. 3.33.
Rod, al
 STEAM POWER PLANT
187
The air leaving the tips of ackward curved blades possesses low
velocity. This makes them suitable for high rotor speed. Fans with
backward curved blades are used in forced Iraught system. Forward
and radial blades are used in induced draught fans.
3.13.1 Comparisons of forced and induced draughts
The forced draught system has tie following advantages as
compared to induced draught
i) The induced draught handles more volume of gases and
at elevated temperature. Therefore, the size and power
required for induced draught fan is more than forced fan.
iii) Improved route of burning of fuel is achieved by using
forced draught because there is more uniform flow of air
through the grate and furnace and also the air penetrates
better into the fire bed.
(iii) In case of induced draught when doors are opened for
firing there will be rush of cold air into the furnace and
this reduces the heat transmission
(iv) The induced dr:uight fan handles flue gases at high
temperature and us water cooled bearing are needed for
induced draught Ian.
Steam Jet Draught. Steam jet draught may be induced or
forced draught depending upon the location of steam jet producing
the draught.
Induced dra ight produced by steam jet is shov n Fig. 334 (a).
This system is used in locomotive boilers. Exhau i steam from the
engine enters the smoke box through a nozzle to ci te draught. The
air is induced through the flues, the grate and as. t to thy smoke
box.

STE.4M

A/R
SUCKCC, 111

5TE NCZZ(E
F 05FR
ST,. .,OLJcEQ OR4L6HT SYSTEM Jt T-FORCED 2RAt6H7 5Y5 TEl"
(a)
(b)
Fig. 3.34.
188 POWER PLANT

Fig. 3.34 (b) shows a forced draught developed by steam jet.

Steam from the boiler is passed through a throttle valve, throttle
pressure being 1.5 to 2 kg/cm 2 gauge. Then the steam passes
through a nozzle projecting in diffuser pipe. The steam comes out of
nozzle with great velocity and drags a column of air along with it
thus allowing the fresh air to enter. The mixture of steam and air
possesses high kinetic energy and passes through the diffuser pipe.
The kinetic energy gets converted into pressure energy and thus air
is forced through the coal bed, furnace and flows to the chimney.
Steam jet is system simple, requires less space and is economical.
But it can be used only if steam at high pressure is available.
3.13.2 Draught Measurement
A U-tube manometer shown in Fig. 3.35 is employed to measure
the draught. One end of manometer is open to atmosphere and the
other end is connected to the boiler furnace, chimney or any other
point in the gas flow passage where draught is to be measured. Due
to pressure difference on the two sides of the tube the level of water
in the two sides is altered to balance the pressure. The difference in
level a in the two siders measures the draught.
7' C111fl/V)' Draft-loss. It is the
pressure loss caused by
friction between two
open to points in the gas flow
atmosphere path.
Draft-system flow
resistance. The follow-
ing relation may be used
to evaluate the flow
pressure pattern in a
draft system.
Fig. 3.35

Pf + Ps = PA + P0 + PV
where PF = Total fan effect pressure
Ps = Net stack effect
(Chimney ± vertical passages)
PA = Draft pressure loss on air side.
= Sum of friction losses in following:
(i) Airducts, bends (ii) Air heaters
(iii) Stoker be
 STEAM POWER PLANT 189
(iv) Secondary air pressure at fuel burner.
Po = Draft pressure loss on gas side.
This is sum of friction losses in the following:
(i) Gas ducts, bends (ii) Economiser
(iii) Air heater (iv) Super heater
(v) Boiler setting (vi) Chimney
Pv = Gas exit velocity pressure.

3.14 Chimney
Chimneys are made up or steel of bricks and concrete. Concrete
chimneys are more popular. The average life of concrete chimneys
is 50 years and that of steel chimneys is ah''. v':rs depending
upon the care taken to prevent corrosion. The net . .a of chimney
depends on the following factors:
(i) Volume of gases to be discharged when
at maximum rating.
tr
boilers operate
(ii) Draught to be produced.
Chimney generally denotes brick concrete construction whereas
stack means steel construction.
Fig. 3.36 (a) shows a brick chimney and Fig. 3.36 (b) shows
reinforced chimney. Chimney is provided with a lighting conductor,
aircraft warning lights, and various means of access and inspection.
The various advantages of steel chimneys over masonry chim-
neys are as follows
(i) Lower overall cost (ii) Slightly high efficiency
(iii) Easier construction (iv) Requires lesser space
(u) Lighter in weight.
R.C.C. chimneys are also becoming popular.
3.15 Calculation of Chimney Height
Air is required to burn coal or fuel in the furnace. We know
C + 02 = CO2
This shows that volume of chimney gases produced by the
complete combustion of 1 kg of coal or oil is same as that of air
required to support combustion if the temperatune of flue gases and
air is same. -
Let H = Height of chimney above grate level in metres.
W = Weight of air in kg required per kg of fuel.
T = Average absolute temperature of chimney gases in K.
T1 = Absolute temperature of air outside the chimney in K.
190 POWER PLANT

(IW4

(a) (b)
Fig. 3.36.
Weight of chimney gases produced
= (W x 1) kg of fuel burnt
rno Of chimney gases at 231K
= volume of 1 kg air at 273K

= RT=. 29.27 x 273 x 0.7734

P 1.03 x iO4
Therefore, volume of one kg of air at TiK

= 0.7734 x

Thus volume of W kg of air at Ti K

- 0.7734 'x T 1 x W
273
Mass
...
Densit y of tirat Ti K = Vomc
u
-273 273
-IV'x
W - 0,7734x
Therefo cssure at the gate due toa column of cool air of 11
nietres height
'73 273
Ii 1293 x x Ii kgliw (i)
= U;7734i x
\'olutneof chiniiey gases at 'I' K p er kg of air


STEAM POWER PLANT 191
T
0.7734 X
27
Volume of( IV' 1) kg of chimne y gases at T K

07731 x X W

Density of chimne y gases at T K

Mass 273
(W + 1)
xTxW
1 273
= 1.293 IX
I F
-, Ii 1
-Ia.

Iheretore, pressure at the grate y a clun1n of hot gases of

height 11 met ri. = flnsit x ii kg/m
• 1.293.1IX IIkg'm

Therefore. pressure I' c:ttimit the draught

iI -73 fj
H
i' -- 1.293 x

1 '73
1.293 . xlIL
7'
1 Hi it
3:,.>!! . T km.

If this drauzht is ii mm of water :05 no i'ured h y a U-tube water

limlininettli,
-
; 3 f/i
H wi, i X nini of water

We know I k/,u t -= I iuumii of water ('oJiumun

Let /o - licight of column of loot gases
W 1 .173
1.293 p
. '
I I
°' H
W 'I


192 POWER PLANT

* H [[i (W+1 1
353
h' =
W+1 1
1.293 x 273 x1,— —x-

h'=H[1,i)x_
Example 3.2. A chimney is 28 metres high and the temperature
of hot gases inside the chimney is 320C. The ternperaiuie of outside
air is 27C and furnace as supplied with 15 kg of air per kg of coal
burnt. Calculate
(a) Draught in mm of water.
(b) Draught height in metres of hot gases.
Solution. We have

(a) h = 353 xf .[±. _f:_

=353x28
1(27+273)- ...x
116 1
-353 x28 --- -
-1300 15 593
= 353 x 28 (3.33 x 10 3 - 1.8 x 10 lj
= 353 x 28 x (1.50 x 10 3 = 15.1 mm of water.
(b) Ii
[ W T 1 r15 593 1
I Li x_ l ]= 28 [Tx
= 24.88 metres.
Example 3.3. Determine the height of chimney to produce a
static draught of 22 mm of water if the mean fl oe gas temperature
in a chimney is 290C and ambient temperature in boiler house is
20 C. The gas constant (R) for air is 29.26 kgfm 1kg K and /)r
chimney flue gas is 26.2 kgfin 1kg K. Assume barometer reading as
760 mm of,nercury.
Solution. Let
P = Absolute pressure of gas (kg/nY)
V= Volume of gas (m2)
R= Gas constant
T = Absolute temperature of gas ( K)
Now PV=RT


STEAM POWER PLANT 193

11=
P
Difference of pressure (1P)
= Height of chimney (H)
- Pi1u) kg/mm2.
A P = 22 mm of water = 22 kg/in'.
(As 1 kg/rn 2 = 1 mm of water)

= P
=I

1.033x104 3
= 29.26x(273+205 = 1.2 kg/rn
(P = Pressure = 760 mm of Hg = 1.033 kg/cm2)
1 P = 1.033 x 101
p, = = = 0.69 kg/rn

Height of chimne y (TI) = = 22

(p - pp,) 1.2-0.69
= 43.1 metres. Ans.
Example 3.4. Determine the height of chimney to gel net
draught of 12 mm if the total draft losses are 4mm. The temperature
of air is 25C and the temperature of chimney gases is 300C. The
moss of air used per kg of fuel used is 18 kg. One kg of
air occupies
a vol ii me of 0. 7734 1713 at N.T.P.
Solution. Let
H = Height of chimney in metre
A = Gross sectional area of chimney in rn2.
Density of air = 1 273 3
27 3+25 1.24 kg/rn
X

Density of chimney gases at 0C

18+1 - 19
- 18 x 0.7734 - 18 x 0.7734
Density of chimne y gases at 300CC
- 18
X
- 18 0734 273+300 273-06k/rn
-.5g
Mass of chimne y gases = A x H x 0.65 kg.


194 POWER PLANT

11 x 1.2 . 1 kg.
Mass of equal column of external air A x
Difference in masses = 1.24 All 0.65 All 0.59 A x 1 kg.

P = Draft in kin2 = = 0.59 11 k g/ i,12

= 0.59 11 mm of water.
(As 1 kg/rn 2 = 1 mm of water)

0.5911=16
11 = 27.1 metres. Ans.
Example 3.4. (a) A chimney proc/aces a c/ru ugh t of 1.8 (-111 of
It, cite I, when temperature of /7ue gases is 280°C and anl/)tent tempera.
tare is 21°C. The flue gases formed per kg of fuel burnt are 21 kg.
Taking diameter of chirn nev as 1.77n c/etermi,u' the mass of/be gases
flowing through the c/urn ,iey.
Solution.
It Draught
1.8 cm -
IS mm of water
T - Temperature of flue gases
280° f 273
= 553°K
= Ambient Temperature
= 21 -i 273
= 294"K,
W = Mass of air reqd. per kg of fuel
=24-1
= 23 kg

h=353H[-(.-jx

18 = 353 H[- - x
23 553
^23-)
H = 34.1 in of air
ii' = Height of column of hot gases
r w ' T 1
=HLix 1J
 I

STEAM POWER PLANT 195

[23 553
x
=
= 28.9 in of air
C = Velocity of gases
=

=
= 23.5 m/s
Wg = mass of flue gases
=AXCX
where A = Area of chimney
Tt
=-x(1.7) 2 fly)--

1
x
=
=33[__ x 1

= 0.67 kg/m 3
(17)2 x 23,5 x 0.67

= 35.7 kg/s.
3.16 Methods of Burning Fuel Oil
Fuel oil is burnt by means of burners. The function of a burner
is to atomise the fuel and mix it with proper amount of air for
combustion. The various types of burners used are as follows

CCIT
-

Fig. 3.37
. 196 ... POWER PLANT

(a) Vaporising burners (b) Atomising burners.

(A) Vaporising burners. In vaporising burners, the fuel is
vaporised by heating. Such burners are used in domestic and in-
dustrial applications. Vaporising burners may be classified as fol-
lows:
(i) Rotating cup type burner.
(ii) Atmospheric pressure atomising burner
(iii) Recirculation burner
(iv) Wick type burner.
Fig. 3.38 (ci) shows a rotary cup oil burner. This burner sprays
oil on fast turning cup to break up oil film at rim by centrifugal force.
It mixes with secondary air spinning in opposite direction.
R

AIR NOZZLE

SPARY OIL.
CONE

.A1ISNG1I

cup

AIR
Hg. 3.38 (a)

Vaporising oil burners

(i) Vaporise the fuel before ignition
(ii) Mix the vaporised fuel thoroughly with air
liii) Minimise soot formation.
They (i) produce more heat by burning more amount of fuel per
hour.
(ii) can give efficient combustion at part load.
(B )Atomisingburners. These burners atomise the fuel oil to split
it into very fine particles before combustion. Fuel oil is atomised to
expose the maximum possible surface to combustion reactions.
Mechanical spraying can be done by the following ways
(i) Forcing oil under pressure through small orifices.
(ii) Using pressurised steam or air to break up the oil.
(iii) Breaking oil film into tiny drops centrifugal force.
Atomising fuel burners can be classified as follows:
 STEAM POWER PLANT
197

(a) Mechanical or oil pressure atomising burners.

(b) Steam or high pressure air atomising burners.
(c) Low pressure air atomising burners.
These burners (i) atomise the fuel into fine particles to equal
size.
(ii) supply air in required quantity.
(iii) minimise soot formation.
They (i) give high combustion intensity
(ii) give high thermal efficiency
(iii) operate efficiently at varying load.
Mechanical Atomising Burners
Mechanical atomising burners have the following four principal
parts:
(i) Atomiser
(ii) Air register
(iii) Diffuser
(iv) Burner throat opening
Atomiser breaks up oil mechanically into a fine uniform spray
that will burn with minimum of excess air when projected into the
furnace. Air register supplies air needed for combustion. Diffuser is
a hollow metal cone mounted near the furnace end of the atomiser
assembly. It stablilises the flame to prevent it from being blown
away from the atomiser tip. Burner throat opening is made of
refractory and is circular and concentric with burner outlet. The
atomiser and diffuser assembly should be so positioned. That the
flame clears the throat opening sufficiently to avoid striking.
Steam atomising burners are of two types:
(i) Outside mix type
(ii) Inside mix type
In outside mix type burners oil is ejected through outside of
holes and is blasted by a high velocity jet of steam from other hole.
Mixing takes place outside the burner as shown in Fig. 3.38.1.
In case of inside mixing type burners steam and oil are mixed
inside the burner before mixture is projected in the furnace as shown
in Fig. 3.38.2. These burners provide high efficiency at high firing
rates.

oil Out

Slewn

Fig. 3.38.1 Fig. 3.38.2

198 POWER PLANT

Low pressure burners use air pressure between 0.015 bar to 0.15
bar. They are simple in shape and are quite commonly used.
(i) Blast -atorn.jsation. In this method the oil is atoinised by
compressed air or steam. Steam atomising burners may he inside
burners or outside burners.

Fig. 3.38 (b).

 STEAM POWER PLANT 199

Ia the inside burners the steam and (0l oue in contact inside
the burner and atomisation of mixture tik place while passing
through the orifice of the burner. Fig. 3.37 (a) shows an injector
inside mixing burner. In outside burner, the oil is pumped to the oil
orifice and is picked up and atomised by jet of steam outside the
burner.
(ii) Mechanical atom isation. I n this method the fuel oil is
atoinised b y subjecting it to a high pressure and passing it through
an orifice. Fig. 3.38 (h) shows Babcock and Wilcox mechanical
atomising burner. The fuel oil flowing through the centre tube is
atomised and discharged through tangential slots in the sprayer
plate. The oil then passes a conical chamber with an orifice at its
apex and due to whirling action oil leaves the orifice in the form of
a hollw tone of minute particles.

Fig. 3.38 (C).

Fig. 3.38 (c.) shows spring loaded piston type burner also called
oil pressure t y pe burner. The oil comes out of nozzle in the form of
spray
3.16 (a) Fuel oil and gas handling
Fuel handling system is designed depending on the type and
nature of ueI. I'i. 3.38d) shows liquid fuel handling arrangement-
Fuel storage tanks of concrete or steel are located near the power
plant. Underground tanks are usuall y preferred. A vent pipe open
to the atmosphere allows the storage tank to breathe. The level
indicator is used to record the level of oil in the tank. Oil pumped
from the tank is first passed through strainers, and then through
heaters to bring the liqi.iid fuel to the conditions necessary for the
burners. Man-hole is provided for cleaning the storage tank.
Pipes are used to handle the gas in power plants using gas as
fuel. The pipe line may divide into two parallel lines each fitted with
the following
(. 1) meters (ii) regulators
(lit) stop valves.
Plants call justify storing gas on their premises. However
underground storage proves feasible in some areas. ManY plants use


200 POWER PLANT

natural gas as fuel. For best econom y the pipe lines should be kept
flowing at full capacity.

Control
valve

I Burners

Excess Discharge
Boiler pressure
Strainers
relief line
(return)
Level
indicators ØPressure

j
Vent pipe

21 rr_ID. lanhole
Strainer
Oil
pumps
Heaters

Storage tank

Heating
cod

Fig. 3.38 (

3.16(b) Gas Burners

Gas burners are used to mix air and fuel gas in the desired
proportions before ignition. Gas as a fuel has the following ad-
vantages:
(i) It can be handled easily.
(ii) Furnace temperature Gas
can be easily dontrolled.
(iii) There is no pollution. .

Two types of burners are

shown in Fig. 3.39 (a)Fig. 3.39 h).
A long flame results by using
burners shown at Fig. 3.39 (a).
When h.LL the gas and air are
under pct , ,re burner shown in as
Fig. 3.39 ti.' is ased.
Fig. 3.39 (a) Fig. 3.39 (b)
3.17 Slag Removal
Ash is removed iii dry state in case offurnaces burning coals of
high fusion temperature and in the liquid form if the coal used is of
low fusion temperature. Ash when in the form of liquid slag is
 STEAM POWER PLANT
201
collected on the furnace by the various methods shown. Slag col-
lected on the furnace floor is tapped off through a slag spout in the
side wall Fig. 3391 or it may be removed at the rear of furnace. (Fig.
3.40). Two-stage furnace arrangement (Fig. 3.41) and (Fig. 3.42) is
also used to remove the slag. In primary furnace nearly complete
combustion of fuel takes place and temperature is quite high and
cooling takes place in secondary furnace.

Stag
-,
ii
Fig. 3.39
Fig. 3.40

Secondary Primary
furnace

Slog

Fig. 3.41 Fig. 3.42

.18 Economiser
In order to utilise the heat accompanying combustion gases
?aving the furnace, the gases are passed through the heat recovery
I1 ip111e11t such as economiser and air preheater.
.15
202
POWER PLANT I
b
Economiser. Economiser is a device intended for heating the
feed water by means of flue gases from boiler. There are steaming
economisers in which the water is raised to the boiling point and
partially (10-20%) evaporates and non-steaming economisers in
which the temperature of water is below the boiling point by
20 - 30°C. The advantages of an economiser are as follows:

TUBES

-WATER

WATER

Fig. 3.43

(i) It reduces the losses of heat with the flue gases.

(ii) It reduces ths consumption of fuel.
(iii) It improves the efficiency of the boiler installation.
Economiser may have iron or steel tubes. Smooth or ribbed. Fig.
3.43 shows a general view of an iron economiser.
Flue gases flow over the tubes.
FLUE GAS

WATER
QUTLEr

WATER
/AIL ET

FLUE 645

Fig. 3.44

Fig. 3.44 shows an economiser. It consists of series of steel tubes

through which the feed water flow. The combustion gases pass over
the tubes and transfer some of their heat to the feed water. The
boiler efficiency rises by about 1% for each 10°F rise in feed water
temperature. Economisers may be parallel-flow or counter-flow,
when the gas flow and water-flow are in the same direction
 STEAM POWER PLANT 203

economisers are called parallel-flow whereas in counter-flow

economiser the gas flow and water-flow are in the opposite direction.
Installation of an economiser depends on its initial cost, type of
boiler and nature of feed water used.
Fig. 3.44 (a) shows by-pass arrangement which enables to iso-
late or include the economiser in the path of the flue gases.
Ec on am iser

00000000 0000000
0 0000000 0000000
000000000000000
00000000 0000000
00000000 0000000.\
0000000
• 00000000 0000000
00000000 0000000
0001,0000

S gases
,__
,7; -7l I O77 I.-7,

- Lancashin
boiler
Fig. 3.44 (a)

3.18.1 Soot Blower

Soot blower is a device which blows off the soot from the
economiser tubes. Soot which gets deposited on the economiser
tubes due to the flow of flue-gases over the tubes. The soo.t obstructs
the transfer of heat from flue gases of the feed water.
A soot blower consists of tube having a number ofnozzles. High
pressure steam is blown through the nozzles which strikes against
the tubes and removes the soot.
3.18.2 Air Preheater
It consists of plates or tubes with hot gases on one side and air
on the other. It preheats the air to be supplied to the furnace.
Preheated air accelerates the combustion and facilitates the burn-
ing of coal. Degree of preheating depends on the type of fuel, type
of fuel burnirg equipment and the rating at which the boiler and
furnace operated. The principal benefits of preheating the air are
increased thermal efficiency and increased steam capacity per
square metre of boiler surface. There are two types of au
preheater:
1. Tubular type 2. Plate type
 POWER PLAN r
264
A tubular type air preheater as shown in Fig. 3.45. After leaving
the boiler or economiser the gaseous products of combustion travel
through the inside of the tubes of air preheater in a direction
opposite to that of air travel and transfer some of their heat to the
air to be supplied to the furnace. Thus the air gets initially heated
before being supplied to the furnace.
The gases reverse their direction near the bottom of the air
heater, and a soot hopper is fitted to the bottom of air heater casing
to collect soot.
In plate type air preheater the air absorbs heat from the hot
gases being swept through the heater at high velocity on the opposite
side of a plate.
Finally the products of com-
d'WE 64
IlVI El
bustion leave the stack (chimney)
to make their passage to the at-
mosphere. It is desirable that the
4TED
temperature of the gases leaving
4 Ife the stack should be kept as low as
posible to keep the heat loss to
T the stack at minimum.
iy installing economiser and
AIR air preheater less fuel is required
INL E
per unit mass of steam raised and
boiler efficiency is increased. The
TO justifiable cost of economiser and
51/INEY air preheater depends upon the

He giri in boiler efficiency.

Fig. 3.45. Air preheater should be used
where a study of costs indicates
that some money can be saved or
some beneficial action on combustion can be obtained by its use.
Some factors that need to be taken into account in examining a case
for justification of air preheat system are as follows
(i) Improvement in combustion efficiency.
(ii) Cost of the equipment and estimated maintenance cost.
(iii) Cost of extra draft.
(iv) The extent to which air can be preheated.
(v) Saving in-heat discharged to the chimney.
3.18.3 Heat transfer in economiser and air preheaters
In economiser water is passed through the tubes and hot flue
gases pass over the tubes whereas in air preheater air passes
through the tubes and flue gases pass over the tubes and transfer
the heat to the air. The most effective use of given heat transfer
surface is obtained when two fluids (water or air and flue gases)
STEAM POWER PLANT 205

travel in the opposite direction (counter flow). Fig. 3.46 (a) shows
the variation of temperature along the flow paths of the fluids.

oh0
OCj

Fig.3 16(a).

The heat transfer from the hot flue gases to water in case of
economiser and to air in case of air preheater is found as follows
Q = UA.0,,,
where Q = Quantity of heat transferred
U = Overall heat transfer coefficient
A = Area of heat transfer surface
= Log Mean temperature diticconce
- 0, -
log 0,
61,
Or,, = Outlet temperature of cold fluid
0,., = Inlet temperature of cold fluid
= Outlet temperature of hot fluid
Oh, = Inlet temperature of hot fluid
0, = Temperature difference at inlet = Oh, - 0,
0,, = Temperature difference at outlet = Oh, -
3.19 Super Heater
Thesteam produced in the boiler is nearly saturated. This steam
as such should not be used in the turbine because the dryness
fraction of the steam leaving boiler will be low. This results in the
presence of moisture which causes corrosion of turbine blades etc.
To raise the temperature of steam super-heater is used. It consists
Of several tube circuits in parallel with one or more return bends
connected between headers. Super-heater tubes range from 1 to 2
inch of diameter. Super-heater supplies steam at constant tempera-

206 POWER PLANT

ture at different loads. The use of super-heated steam increases

turbine efficiency.
There are three types of super-heaters.:
1. Convective super-heater 2. Radiant super-heater
3. Combination of the two.

Fig. 3.46. Fig. 3.47

Convective super-heater makes use of heat in the gases entirely

by convectiL. whereas a radiant super-heater is placed in the
furnace and wall tubes and receive heat from the burning fuel
through radiation process.

Fig. 3.48
The final temperature of steam depends upon the gas flow rate,
quantity of gas flow and the temperature of the gases leaving the


STEAM POWER PLANT 207

super-heater section. The flue gas temperature should be nearly

175C higher than the temperature of super-heated steam. Material
used for super-heater tubes should have high temperature strength
and high resistance to oxidation. Special steel alloys such a
chromium molybdenum alloy is used for tubes of super-heater for
modern high pressure boilers.
According to location the super-heaters are classified as follows:
(i) Over deck (ii) Inter deck
(iii) Inter tube (iv) Inter bank.
Fig. 3.46 shows over deck location of a super-heater and inter
deck location is shown in Fig. 3.47. Inter tubes and interbank
location of super-heater is shown in Fig. 3.48 and Fig. 3.49 respec-
tively.

Fig. 3.49

Inter deck super-heaters are essentiai!v a... ;ective super-

heater. The heat transfer conditions in a super-heater vary with
load. When load is decreased the gas mass flow decreases propor-
tionately and in a convective superheater fewer degrees of super-
heat are obtained whereas in a radiant superheat steam receives
more heat than at higher toads. A
radiant superheater has a falling
characteristic with increased steam
output of boiler. Modern boilers using
high pressures use combination of
convective and radiant superheater.
Fig. 3.50 shows the locating of both
convective and radiant super-heaters
in a bent tube boiler.

Fig. 3.50

b 208 POWER PLANT

1..

\\
E.2
\\'
()
E ma

I1__
3. -o
0. 0.
E-
E
0 ci
2 -i
(.0 -
C

C
(Al
I? 3
E
0
0 0
0

(0
E —0
0 u_ Q(


STEAM POWER PLANT 209

3.19.1 Sugden Superheater
Fig. 3.50 (a) shows sugdon's super heater used in a Lancashire
boiler. This super heater uses two steel headers to which are
attached solid drawn 'U' tubes of steel. These tubes are arranged in
groups of fourand one pair of the headers generally carries ten of
these groups or total of forty tubes. The steam from the boiler enters
and leaves the headers as shown by the arrows. It shows how the
steam pipes may be arranged so as to pass the steam through the
superheater or direct to the main steam pipe.
3.20 Advantages of Super-heated Steam
Super-heated steam is vapour whose temperature has been
increased above that of its boiling point at that pressure.
The various advantages of using super-heated steam are as
follows
(i) Super-heated steam has an increased capacity to work
duo to a higher heat contact. Therefore, an economy in
steam consumption in steam turbines and steam engine
achieved.
(ii) Super-heating raises the over all efficiency of the plant.
The temperature of the super-heated steam t g higher
it gives a higher thermal efficiency when . or working
a prime mover.
(iii) Super-heating of steam avoids the ci sinn 0: turbine
blades in the last stages of expansion of sam. In order to
avoid blade erosion it is desirable to limit the moisture
content 10 to 12% in the exhaust of the steam turbines.
3.21 Super-heat Control
It is desirable that there should be a close control over the final
temperature of steam over a reasonably wide range of load.
The various methods employed to achieve this are as follows
1. Use of Desuperheater. To control the temperature of steam
a desuperheater (attemperator) is used. In the clesuperheater (Fig.
3.51) some quantity of cold water is injected into around the pipe
carrying the-steam. This causes the evapora' 'ii of water so injected
and thus the temperature of steam is lowered
2. Use of Tilting Burners. Tilting burners furnace are
used to regulate the temperature of gases loavin rnace Fig.
3.52).
3. Use of Dampers. Dampers are provided to control the
direction and flow of hot combustion gases in order to vary the
quantity of gas passing through the super-heater.

210 POWER PLANT

Desu per
heater
Steam

Drum

Super heater
sections

Fig. 3.51 Fig. 3.52

4. Use of Auxiliary Burners. Auxiliary burners (Fig. 3.53) can

be used to control degree of super-heat.
5. Twin Furnace. Arrangenient [Fig. 3.541 mav be used for the
control of super-heating temperature.
3.22 Feed Water Treatment
Natural water supplies contain solid, liquid and gaseous im-
purities and as such this water cannot be used for the generation of
steam in the boilers. This requires that the impurities present in
the raw water should be removed before it can he used in the boilers.
Although in steam power plants the main condensate returns
to the boiler as feed water but make up water is needed to replace
the losses due to blow down, leakages etc. in the cycle.
The various impurcties present in the natural water (raw water)
may be in the following forms
(i) Dissolve salts such as carbonate, sulphates chlorides of
calcium, sodium and magnesium. Sometimes some iron,
aluminium or silica salts are also present.
(ii) Dissolved gases such as carbon dioxide, oxygen and SO2.
(iii) Mineral acids.
(iv) Suspended matter such as alumina and silica may he
present as mud and salt.
In the water intended for steam boilers all the impurities are
harmful especially if present in considerable quantities. The salts
of calcium and magnesium are extremely harmful and when water
containing them is heated and steam generated they precipitate as
solid residue and form a hard scale oil surface. The scale so
formed is not desirable as it hampers the processof heat transfer.
STEAM POWER PLANT 211
Presence of mineral acids in water is always undesirable as it may
result in chemical reaction with the boiler materials.
These impurities may cause the following troubles

Fig. 3.53 Fig. 3.54.

(i) Scale formation (ii) Corrosion

(iii) Foaming and Priming (iv) EmbritLlement.
1. Scale Formation. Impurities present in the water may cause
scale formation in the boiler drums of header tubes and feed water
piping system. This will reduce the heat transfer rate a id will cause
over-heating of tubes which may result in blistering ' .i rupturing.
The scale formation in the feed water pipes chokes the flow which
requires higher pressure to maintain the water flow .cale is due
mainly to salts of calcium and magnesium. When scale has formed,
the tubes should cleaned with water or electric powered rotary
brushes and cutters -' nushed through the tubes during
boiler overhauls.
2. Corrosion. Corn ir in the boiler shell, tubes,
plates due to acidity pres. vater. This reduces the life of
construction materials. Corrosion is the destructive conversion of
metal into oxides or salts. Corrosion takes place due to the presence
of oxygen, carbon dioxide or chlorides dissolved in water. Corrosion
duo to oxygen produces small pits.
3. Foaming and Priming. A layer of foam is caused in the
boiler drum by soluble and insoluble salts and other organic im-
purities which are carried iq suspension. Foaming prevents the free
escape of steam bubbles as they rise to the surface of water. Oil aud
other impurities which may be present in boiler water may cause
foaming.
212 POWER PLANT

Priming or carrying-over is the passing of small water particles

with steam as it leaves the boiler. These water particles make the
steam unfit for use in engines. This is caused by the impurities in
water, a high water level and the method of operation of the boiler.
To prevent foaming and priming the following precautions
should be observed:
(i) Oil, soap and other suspended impurities should not be
present in boiler water.
(ii) Various valves should not be opened suddenly to maxi-
mum.
(iii) Water in the boiler should be at its minimum possible
level.
4. Embrittlement. Caustic einbrittlement is caused due to
caustic impurities present in water. Presence of certain concentra-
tion of sodium hydroxide causes embrittlement. Due to this the
boiler metal becomes brittle and inner cracks appear along the
seams below the water level.
3.23 . Methods of feed Water Treatment
It is desirable that the water to be used in the boiler should be
free from various impurities.. The impure water is chemically
treated in different ways depending upon the nature and concentra-
tion of impurities. The different treatments adopted to remove the
various impurities are as follows
1. Mechanical Treatment. It includes sedimentation,
coagulation and filtration. Suspended matter can be removed easily
by these processes. Sedimentation involves allowing the water to
stand quietly for some time. In this way the solid matter settles
down and is removed periodically. In case of coagulation some
coagulants like aluminium sulphate, sodium aluminate or ferrous
sulphate are added to the impure water. This removes the minute
colloidal suspensions.' Filtration consists in passing the water
through filters. The suspended matter adheres to the filter material.
The filters may be either gravity filter or pressure filter.
Fig. 3.54 (a) shows a pressure filter. The raw water is passed
through alum pot and then through a tank containing fine sand,
graded layer of gravel. These filters can be easily installed on the
pipe line as much lesser space is required.
2. Thermal Treatment. It includes distillation and deaerative
heating of water. By these processes dissolved gases of water are
removed.
Deaerating Heater. A tray type deaerating heater is shown in
Fig. 3.55. In this heater feed water after passing through the vent
condenser is sprayed upwards in the spray pipe. Water falls in the


STEAM POWER PLANT 213

F(owmeter

Inlet
valve
Venturi Fine sand
nozzle

Backward Gravel
water to
Alum waste
pot Strainer

Raw water

Filtered
water
Backward outlet
supply

Filtered water
to waste
Fig. 354 (a).
form of uniform showers over the heating trays and air separating
trays and finally gets collected in the storage space. Steam enters
the heater through a nozzle fitted in the side of heater shell. The
entire space between the shell and tray compartment gets filled with
steam. The steam makes its way downwards through the perfora-
tions in the top plate of tray compartment. While flowing downward
the steam comes in contact with the falling water. Most of the steam
condenses in between the spray and heating trays. From the bottom
of heating trays, the remaining steam and separated gases such as
oxygen etc. flow to the vent condenser. The steam used for heating
may be the main turbine bled steam or may be from other sources.
Storage tank with controls helps to add make up water when needed
to maintain the feed water flow.
Closed Feed water heater
Fig. 3.55 (A) shows a closed feed water heater in which steam
bled from steam turbine is used for heating the feed water. This
process is also called regenerative process and increases the efficien-
cy of the generation cycle.
214 POWER PLANT

3. ChenicaI Treatment. It includes addition of some chemi-

cals to cause precipitation or impurities.

VENT WA7ER COA/TROI
VALVE

VENT __
CONDENSER

5PR4 y PIPE
uLJuijuuI
LJLJLJLJLJ I
HE.4TIWG TRAYS UUuIJL, UI .STEAAI
.4/P 5EPARAT/N6— 4-J L3L_JL.I ,
TRAYS LJLJLLJ

OVLR FLOW _______

L OAT CAGE

TO 5ORAGE
• TA WI(
T9 FEED
WATER P1114P

Fig. 3.55

Turbine

ICondenser
Boiler

Heaters
Pump

"Condensate
Hot weii_- extraction pump

Fig. 3.55 (a) FOR

(a) Internal Treatment. In this process the reagents are added

to water present in boiier. Various reagents used are sodium car-
bonate, sodium phosphate, sodium hydroxide etc. Sodium phos-
phate when added removes carbonate impurities.
CaCO3 + 2Na 3 PO4 - Ca 3(PO) 2 .L 3Na2CO3

STEAM POWER PLANT 215

Internal treatment is suitable for low pressure (about 300 psi.

boilers.
(b) External Treatment. In this treatment raw water is
received in a tank where the reagent are mixed. The reaction
becomes rapid if the water is heated before the addition of reagents.
The various reagents used in external treatment are as
follows
I Lime Soda Treatment. In this method lime Ca(OH) 2 is added
to remove the carbonate hardness of water and soda ash Na 2 CO 3 is
added to remove the sulphate hardness.
Ca(HCO 3 )9 + Ca(OH) —3 2CaCO 3 I + 2H20
MgSO i -- Na2CO 3 —4 MgCO 3 .L + Na2SO4
(ii) Ion Exchange Process. In this process sodium zeolite
N 2 Z (Na 2 Al 2 Si 2 02) is added. It reacts with calcium and mag-
nesium salts to form zeolites. Zeolite is reproduced on the addition
of brine (sodium chloride) solution.
CaSO 4 t Na 2 Z - Na 2 SO 4 + CaZ
CaZ + 2NaCI - N 2 Z + CaC12
A zeolite softener Fig. 3.56 consists of a shell holding a bed of
Ective sodium zeolite supported by layers ofgraded gravel lying over
, water distribution and collection s y stem. The raw water enters
the zeolite softner at the top. It flows downwards through the bed
ilthough the system can be by-passed while back wbing, ifneces-
sarv. The figure also shows arrangements for th periodic back
washing with brine needed by this system.

Row AirFloat
e f
water re control

storage

, Teolite -Gravel
Graded
gravel Brine tank

Back wash ----Orifice plates Soft water

flow 61 L outlet
Controller..
Rinse water
I
i0rajn

flow controller

F. 3.56 Ion exchange system.



216 POWER PLANT

4. Demineralisation. It is used to remove mineral contents of

water. In this s ystem (Fig. 3.56 (01 raw water enters the hydrogen
zeolite exchanger (cation removal) at the top and then flows to the
anion exchanger and degasified and finally passes through silicon
absorber.
In spite of the best treatment some quantity of dissolved and
suspended impurities enter the boiler. Table 3.0 indicates the
recommended boiler water concentration of impurities at different
operating pressures (p).
Table 3.0

- Pkg/crn 2 Silica PPM Suspended 14/kalinity PPM Total Solids

solids PPM i PPM
0-25 100 65OT 3000
25-50 60 150 500 2500
50-100 8 40 ______ 20C
Above 100 1 4 70

The above values show that safe lijnit of concentration decreases

with increase in boiler working plessure Concentration of these
solids can be reduced to a great extent by blowing down some
quantity of water from the botLoin of boiler drum. Periodical blow
down helps in maintaining the concentration of impurities within
the allowable limits.
5. Blow down. Water entering the boiler may contain some
dissolved solids. The concentration of these solids goes on increasing
as the water is vaporised. Beyond a certain limit of concentration,
these solids may cause foaming and priming. The concentration of
these solids can be reduced by drawing off some of the quantity of
the boiler water from thQ bottom of boiler drum. This is called
blowing down and discharged water is known as blow down.
nc-t1C,C::Q

1xc

Fig. 3.56 (a)

 STEAM POWER PLANT
217
The blow down mainly contains the undesirable impurities
which concentrate at the bottom of drum. As a result of blow down
the concentration of these impurities inside the boiler drum can be
temporarily reduced. Therefore a boiler may have periodical blow
down so that the concentration of impurities can be kept within
permissible limits.
3.24 To determine blow down
Draining of some of the boiler water carrying excessive con-
centration of solids and replacing it with fresh feed water keep the
solid concentration within safe limits: This is called blowing down
the boiler and the discharged water is called blow down.
The blow down is calculated as follows
K = Blow down % = WI Wf
where IV = weight of water blow down from boiler in
kg per hour
IVf= weight of feed water supplied to boiler for
steam and blow down in kg per hour.
Blow down ma y be continuous or*
intermittent. Continuous blow
down is about 1 to 10% of the incoming feed water.
The impurities in blow down are found by the following rela-
tion
M = M1 + M 2 + M3
where M = Impurities in blow down (weight of
blow down) x (Allowable salt concentration) ppm.
Mi = Feed water impurities = (weight of feed
water) x (Impurity concentration) ppm.
M2 = Internal treatment chemicals
M3 = Steam impurities = (weight of steam) x
(Impurity concentration) ppm.

BOILER

—16
Fig. 3.56 (b)

POWER PLANT
218
Fig.
3.56 (b) shows mass balance of impurities and treatment
chemical s, in boiler.
3.25 pH Value of Water
It is the logarithm of the recipuocal of hydrogen ion conceittra-
hon in water. in water either hydroxyl ions (011 - or h ydrogen ions
1)

(H) predominate causing either an alkaline or acidic condition.

Aciditv or alkalinity is measured in pH values ranging from Ito 14.
pH value I is strongly acidic. pH 14 is strongly basic and p11 7
indicates a neutral solution. pH value of a sample of water can be
Measured b y a pH meter.
3.26 Analys i s of Water
Analysis of water is carried out to determine the quantity of
ill)purities and other chemical substances in a sample of water. The
data contained in a t y pical water analysis report is as follows

Water Analysis Report

Serial No Date....
Source of sample Quantity of Sample .....
Date of collection Date of analysis.....
I
ii-
1. Colour water
L2 upenkd hI

4., Organic matter --

5. - Chloroform, oil etc
alue_
flIcaiLI1I(.y t.n.t.

8. Hardness as CaCO
9. Silica as SiO2
10. Iron as Fe-203
11. Sodium -
i2iii ie1ii1ii—It
13 . Calcium
................. -
14. Carbonates as CO3 -
15. Bicarbonates as llCOj
Ilvdroxide as
L_
17. Sulphate as SO4
Chloride as Cl
19. Nitride as NO:3
20. Carbon dioxide S --
STEAM POWER PLANT 219

3.27 Feed Water Heaters

Feed water heaters are used to heat the feed water before it is
supplied to the boiler.
Heating of boiler feed water serves thefollowing purposes
1. It causes scale forming dissolved salts to precipitate out-
side that boiler.
2. It removes dissolved gases such as oxygen and carbon
dioxide which corrode boiler metal.
3. B y using pre-heated feed water, the steaming capacity of
boiler is increased.
4. It avoids the thermal stresses which can be induced in the
boiler surface by cold water entering a hot drum.
There are two types of feed water heaters:
1. Contact o r open heaters. In these heaters feed water get
mixed with the heating steam. These heaters include tray type
heater (Fig. 352) and jet type heater.
2. Surface or closed heaters. Surface heaters hve shell and
tube construction, and may be steam tube type or water tube type.
Generall y water tube type heaters are used. They ordinaril y use
bled steam for heating. Heating the feed water with steam at a lower
pressure than boiler pressur' raises the over-all efficiency of the
plant. Fig. 3.56 (c) shows a closed feed water in which water flow
through tubes and steam flows all az'ound the tubes. Steam while
following transfers its heat to the feed water. Surface heater has
construction similar to a surface consider except that surface heater
is designed for higher pressure and temperatures.

Condensote Water
inlet
Fig. 3.56 (c).
Chemical tests are performed at frequent intervals on samples
of water to have information of some improtant items such as
variation in mineral characteristics of raw water suppl y , condition
of boiler water with respect to treatment adjustment and blow down
adjustment percentage ofcoridensate and make up in the feed water

220 POWER PLANT
b
steam purity and correction of feed water and boiler water for
corrosion control etc.
3.28 Steam Condensers
A steam condenser is meant to receive the exhaust steam from
the turbine or engine, condense it and maintain a pressure at the
exhaust lower than atmospheric. Some extra work is obtained due
to exhaust at a pressure lower than the atmospheric. This improves
the efficiency at a pressure lower than the atmospheric. This im-
proves the efficiency of the plant. Air inside the condenser should
be pumped out continuously in order to maintain the vacuum. The
condensation of steam occurs in the range of 25'C to 38'C.
Steam pressure in a condenser depends mainly on the flow rate
and temperature of the cooling water and on the effectiveness of air
removal equipment. Some of the advantages of a steam condenser
are as follows:
(i) It increases power output. Power plant cycle improves in
efficiency as the turbine exhaust pressure drops.
(ii) It recovers most of the feed water (in case of surface
condenser) which is available at 45'C to 50'C. This save
the amount of fuel to be burnt in boiler.
(iii) The use-of condenser, decreases the size of boiler installa-
tion.
The main disadvantage of condenser is that it adds to the initial
cost of power plant as the condenser requires additional equipment
such as cooling tower or cooling pond, vacuum pump, water circulat-
ing pump etc.
The vacuum obtainable in a condenser is governed by the outlet
water temperature which in turn varies with the amount ofcondens-
ing water used per kg of steam and its initial temperature. Air
entertainment in the condenser has its effect upon the vacuum. The
addition of air lowers the vacuum.
3.29 Types of Steam Condensers
Steam condensers are of two types:
(i) Surface condensers (ii) Jet condensers.
3.29.1 Surface Condensers
In surface condensers there is no direct contact between the
steam and cooling water and the condensate can be re-used in the
boiler. In such condenser even impure water can be used for cooling
purpose whereas the cooling water must be pure in jet condensers.
Although the capital cost and the space needed is more in surface
condensers but it is justified by the saving in running cost and
increase in efficiency of plant achieved by using this condenser.
Depending upon the position of condensate extraction pump, flow of
 STEAM POWER PLANT
221
condensate and arrangement of tubes the surface condensers may
be classified as follows
(i) Down flow type. Fig. 3.56(d) shows a sectional
view of down
flow condenser. Steam enters at the top and flows downward. The
water flowing through the tubes in one direction lower half comes
out in the opposite direction in the upper half. Fig. 3.57 shows a
longitudinal section of a two pass down-flow condenser.

Steam and
Air

Air and
Steam

Fig. 3.56 (ca.

Exh(rst

Cover
Plate
Baffle
Wa t e, Plate
Box

Condensate Air Cooling Water

Fig. 3.57.
 POWER PLANT
222
Fig. 3.58 shows i central flow
(ii) Central flow condenser.
condenser. In this condenser the steam passages are all around the
b
periphery of the shell. Air is pumped away from the centre of the
condenser. The condensate moves radially towards the centre of
tube nest. Some of the exhaust steam while moving towards the
centre meets the undercooled condensate and pre-heats it thus
reducing undercooling.
(iii) Evaporation condenser. In this condenser (Fig. 3.59)
steam to be condensed is passed through a series of tubes and the
cooling water falls over these tube in the form of spray. A steam of
air flows over the tubes to increase evaporation of cooling water
which further increases the condensation of steam.
Steam and
Air

onden.qa ;e

Fig. 3.58.

Steam

Air

(ati2'1 ate -

Fig. 3.59.
STEAM POWER PLANT 223

Advantages and disadvantages of a surface condenser

The various advantages of a surface condenser are as follows
ii The condensate can be used as boiler feed water.
(ii) Cooling water of even poor quality can be used because the
cooling water does not come in direct contact with steam.
(iii) High vacuum (about 73.5 cm of Hg) can be obtained in the
surface condenser. This increases the thermal efticiencv
of the plant.
The various disadvantages oi the surface condenser are as
follows
(i The capital cost is more.
(ii) The maintenance cost and running cost of this condenser
is high.
(iii) It is bulky and requires more space.
Requirements of a modern surface condenser. The re-
quirements of ideal surface condenser used for power plants are as
follows
(1) The steam entering the condenser should he evenly dis-
tributed over the whole cooling surface of the u .uc1en'er
vessel with minimum pressure loss.
(Li) The amount of cooling water being circulated in t he Con-
denser should be so regulated that the temperature of
cooling water leaving the condenser is equivalent to
saturation temperature of steam corresponding to steam
pressure in the condenser.
This will help in preventing under cooling of' condensate.
(iii) The deposition of dirt on the outer surface of tubes should
be prevented.
This is achieved by passing the cooling water through the
tubes and allowing the steam to flow over the tubes.
(iv) There should be no air leakage into the condenser because
presence of air destroys the vacuum in the condenser and
thus reduces the work obtained per kg of steam. If there
is leakage of air into the condenser air extraction pump
should he used to remove air as rapidly as possible.
3.29.2. Jet Condensers. In jet condensers the exhaust steam
and cooling water come in direct contact with each other. The
temperature of cooling water and the condensate is same when
leaving the condensers.
Elements of the jet condenser are as follows
(i) Nozzles or distributors for the condensing water.
(ii) Steam inlet.
(iii) Mixing chambers : They may be (a) parallel flow tvpt ih
counter flow type depending on whether the steam and
 224 POWER PLANT

water move in the same direction before condensation or whether

the flows are opposite.
(it') Hot well.
In jet condensers the condensing water is called injection water.
3.29.3 Types of jet condensers
(i) Low level jet condensers (Parallel now type). In this
condenser (Fig. 3.60) water is sprayed through jets and it mixes with
steam. The air is removed at the top by an air pump. In counter flow
type of condenser the cooling water flows in the downward direction
and the steam to be condensed moves upward.
Exhaust
ctccUt?
Air
Extraction
-oolinq Water
Inlet

Condensate
- r— Outlet
Fig. 3.60.
(ii) High level or Barometric condenser. Fig. 3.61 shows a
high level jet condenser. The condenser shell is placed at a height
of 10.33 ni (barometric height) above the hot well. As compared to
low level jet condenser this condenser does not flood the engine if
the water extraction pump fails. A separate air pump is used to
remove the air.
(iii) Ejector Condenser. Fig. 3.62 shows an ejector condenser.
n this condenser cold water is discharged under a head of about 5
.0 6 m through a series of convergent nozzles. The steam and air
enter the condenser through a non-return valve. Steam gets con-
densed by mixing with water. Pressure energy is partly converted
into kinetic energy at the converging cones. In the diverging cone
the kinetic energy is partly converted into pressure energy and a
pressure higher than atmospheric pressure is achieved so as to
discharge the condensate to the hot well.
 STEAM POWER PLANT
225

Air Pump
Suction

Tail
Pipe

Th.jcct ion
Pump

Coo linq Pond

Fig. 3.61

t er
In 1 et

Non Return
onvera1ng Va lye
Cones
Exhaust
Steam
Dierifl!1
Cone

vnioclyzrac to
hot Well
Fig. 3.62

3.30 Condenser Cooling Water Supply

A condenser is a device in which exhaust steam from steam
engines or steam turbines is condensed and the heat energy given
up by the steam during condensation is taken up by the cooling
 ROWER PLANT
226

water. The water coming out of the condenser is hot and is cooled in
order that it may be recirculated through the condenser.
There are two types of condensers namely surface condenser
and jet condenser. The surface condensers are preferred because the
cooling water and exhaust stem do not mix with each other. The
cooling water passes through the tubes and steam passes over the
outer surface of tubes. The steam leaves the condenser in the form
of condensate which is re-used as boiler feed water. The amount of
cooling water required is usually quite large such as in large steam
power plants millions of gallons of cooling water per hour is required
for the condenser. Therefore, the source of cooling water chosen
should be able to supply the required quantity of cooling water. The
coi:ling water supply is made by the following sources:
Ci) River or sea (ii) Cooling ponds (iii) Cooling towers.
3.30.1 River or sea
Large power stations require enough quantity of cooling water
per hour. Such plants are usually located near a river or sea. The
water is constantly drawn from the river by the pump, filtered and
circulated through the condenser: Hot coolant is discharged back
into the river (Fig. 3.63).

F/t TeR

PUMA

Fig. 363

3.30.2 Cooling ponds

In this system (Fig. 3.64) warm condensing water from the
condenser is sprayed through nozzles over a pond of large area and
cooling effect is mainly due to evaporation from the surface of water.
In this system sufficient amount of water is lost by evaporation and
windage.
Some of the factors which influence the rate of heat dissipation
from a cooling pond are as follows:
(i) Area and depth of pond
(ii) Temperature of water entering thçond
 227
STEAM POWER PLANT

(iii) Atmospheric temperature

(iv) Wind velocity
(v) Relative humidity
(vi) Shape and size of water spray nozzles.

PLAN

SECTIONAL ELEVATION
COOL fMI- PONO

Fig. 364

3.30.3 Cooling Towers

The different types of cooling towers are as follows:
1. Atmospheric cooling tower.
2. Natural draught cooling towers.
3. Forced or induced draught cooling tower.
Atmospheric cooling tower. In this cooling tower hot water
is allowed to fall over louvers. The air flowing across in transverse
direction cools the falling water. These towers are used for small
capacity power plants such as diesel power plants. Fig. 3.64 (a)
shows an atmospheric cooling tower.
Natural draught cooling towers. In natural draught cooling
tower, the hot water from the condenser is pumped to the troughs
and nozzles situated near the bottom. Troughs spray the water
which falls in the form of droplets into a pond situated at bottom of
the tower. The air enters the cooling tower from air openings
provided near the base, rises upward and take up heat of falling
water.

228 POWER PLANT

Air

rffLcuvers
Water—=='J_______
(n
water
out
Fig. 3.64 (a)

URBINE
&IRIH J) lUNJCI.ING UNIT

WATER
PUMP

kK
NO EN 5€ P
CIRCULATING
PUMP

Fig. 3.65

INICT
HURDLESS

Fig. 3.66
Cooling towers may be made up of timber, .concrete or steel. A
concrete hyperbolic cooling tower is shown in Fig. 3.66. Fig. 3.67
shows the water circulation from the cooling tower to the condenser.
Fig. 3.65 shows the cooling tower in which the position of turbine
has also been shown. The system consists of turbine, condenser.
Circulating pump, tank, additional water pump and sprinkling unit.

STEAM POWER PLANT
229
The air is delivered through the holes in the side walls of the tower.
The circulating water is delivered to the upper part of the watering
unit where it flows down and gives its heat to the surrounding air.
The cooled water flows into the tank and is circulated through the
condenser. Towers made up of the concrete are preferred because
they are stable against larger air pressure, their maintenance cost
is low and they have larger capacities.

WATER

Fig. 3.67

Forced draught cooling towers. In this tower draught fan is

installed at the bottom of tower. The hot water from the condenser
enters the nozzles. The water is sprayFd over the tower filling slats
and the rising air cools the waters. A forced draught cooling tower
is shown in Fig. 3.68.

,PAY(LJA1FNATOR HdTA/R

SPRAY
- hT WAT!R
OW(Rft P/6 _______ :• ____
SLArS

Cool.o I k

CA rci,' 8AI
Fig. 3.68

The various factors that affect cooling of water .i a cc:

are as follows
(i) Size and height oi cooling tower
(ii) Veloity of air entering the tower.
(iii) Temperature of hot water coming out of condenser.
(iv) Temperature of air.
(u) Humidity of air.

POWER PLANT
230

(vi) Accessibility of air to various parts of cooling tower.

Mechanical draft cooling towers may be forced draft cooling
towers or induced draft towers. Fig. 3.68(o) shows an induced draft
cooling tower. The hot water is allowed to pass through. The draft
fan installed at top of tower draw air through the tower. The air
moving the upward direction cools the water. Induced draft towers
produces less noise.
In the cooling tower water cools by
(i) evaporation. (ii) heat transfer to the air.

Hot Air
on
Motor
Eliminator
I Hot
Nozzle
Water

Cold -
Water
Fig. 3.68 (a)

Most of the cooling (about 75(fl takes place by evaporation.

Make up water should be continuously added to the tower collecting
basin to replace the water lost by evaporation and spray carry over.
3.31 Maintenance of Cooling Towers
In order to achieve , the desired cooling and to reduce the
depreciation costs the regular maintenance of cooling towers is
essential. The fins, motors housing etc. should be inspected from
time to time. Motor bearings should be greased and gear boxes oiled.
Any unusual noise or vibrations in them should be corrected imme-
diately. At least once in a year motor's gear boxes should be checked
for structuralweakness. The circulating water should be tested for
hardness and should kept free from impurities to avoid scale forma-
tions and to avoid corrosive action of water. The water spraying
nozzles should be inspected regularly for clogging.
3.32 Condenser Efficiency
Condenser efficiency is defined as follows
Condenser efficiency (Ti)
-
Rise in teninerature of Cooling......
Temperature corresponding to vacuum in condenser

STEAM POWER PLANT 231

- Inlet temperature of cooling water

T2 -
= T3 - Ti
where Ti Inlet temperature of cooling water
= Outlet temperature of cooling water
T3 = Temperature corresponding to vacuum in the
condenser.
3.33 Vacuum Efficiency
Vacuum efficiency is defined as ratio of actual vacuum to
theoretical vacuum. Vacuum efficiency (i) is defined as follows:

= P1
'2
where PI = Actual vacuum in condenser
P2 = Theoretical vacuum in condenser.
Theoretical vacuum in the condenser is the vacuum if no air is
present in it.
3.34 Condenser Pressure
The total condenser pressure (P) is given by the relation
P =P +P
where P. = Steam saturation pressure corresponding to steam
temperature
Pa = Air pressure.
Air must be removed constantly to keep P0 low.
Example 3.5. The vacizurn in a condenser is 68 cm. of Hg with
barometer reading 76 cm of Hg. lithe inlet and outlet temperatures
of cooling water to a condenser are 28'C and 42 C respectively,
calculate the condenser efficiency.
Solution. Ti = 28C
T2 =42'C
Rise in temperature = 42 - 28 = 14C
Absolute pressure in the condenser
= 76-68 8 c of Hg = 0.108 kg/cm2,
Saturation temperature corresponding to 0.108 kg/cm 2 = 47 C.

0

232 POWER PLANT

Condenser efficiency =

= x 100 = 73.6%.

3.35. To Calculate the Weight of Cooling Water

The weight of cooling water required for condenser to condense
a given amount of steam is calculated as follows:
Let W = Weight of cooling water required (kg) to condense
W, (kg) of steam
Ti = Inlet temperature of cooling water in C
= Outlet temperature of cooling water in C
T3 = Temperature of condensate in C
q = Dryness fraction of steam entering the
condenser
L = Latent heat of steam entering the condenser
H, =Heat of water at pressure entering the condenser
T= Temperati-- of water at the pressure of steam
entering the condenser in C
H = Total heat of 1 kg of steam entering the condenser
in kcal = H + qL
H 1 = Total heat of condensate leaving the condenser
(kcal)
Heat gained by cooling water = W (T2 - Ti ) kcal.
Heat lost by steam = Wj (H - H i) kcal.

7'2 - Tj

7'2 - T1
Now in place ofH, and d 1 the corresponding temperature Tand
T3 respectively can be substituted with sufficient accuracy.
• _________
-
STEAM POWER PLANT 233

Example 3.6. Ca/cu/cite the quantity of cooling water required

in leglinin for a surface condenser to condense 18 hg of steam per
minute. The dryness fraction of steam is 0.9 and the temperature of
steam entering the condenser is 37 C. The inlet and outlet tempera-
ture of cooling water are 18C and 32'C respective/v. The condensate
temperature is 37 C. The latent heat of steam should be taken (IS 576
kcai / kg.
Solution. W1 = yeight of steam to be condensed/minute
= 18 kg
W = weight of cooling water required
Ti = 18C ; T2 = 32'C
T3 = 37CC T = 37C
q = 0.9 ; L = 576 kcallkg
W(T2 Ti) = %V i (T+qL + T3
W-
-. T2-T1
18 ( 37 ± 0.9 , x 5 76
= 666 kg per mm.
32-18
Example 3.7. Calculate the quantity of cooling water required
for a jet condenser to condense 40 kg of steam ncr mi,iite. The
vacuum in condenser is 710 mm of mercury (Barometer 760 ,nmn of
mercury). The inlet temperature of cooling water is 14 C. The latent
heat of steam 576 is kcal/kg.
Solution. W1 = weight of steam = 40 kg
W weight (kg) of cooling water required/minute
Ti = 14'C
Absolute pressure in the condenser
= 760- 710 = 50 mm of mercury

=x 0.01359 = 0.068 kg/cm2

Now T3 = Temperature of condensate

= 38C (corresponding to 0.068 kg/cm 2 from
steam tables)
= Outlet temperature of cooling water
=T3=38C
—17

234 POWER PLANT

%'' (T2 - TO = W I (T + qL - T3)

- 1 (T + qLT3)
-
IV W
-

= 40(38 + 1
= 960 kg.
g.

3.36 Selection of a Condenser

The selection of a condenser depends upon various factors. The
floor space required by a jet condenser is less than a surface
condenser. The first cost and maintenance cost of jet condenser is
less than an equivalent surface condenser. The main advantage of
a surface condenser is that it recovers the distilled condensate for
boiler feed water whereas in a jet condensr the condensate gets
mixed with cooling water anti, therefore, cannot be used for boiler
feed water. Surface condensers are most commonly used in power
plants.
3.37 Sources of Air in a Condenser
Air may enter the condenser through different sources and it
should be removed continuously in order to maintain vacuum inside
the condenser. As a condenser is required to maintain low pressure
of exhaust steam which is possible only if a partial vacuum exists
inside the condenser.
The various sources of air in the condenser are as follows
(i) Air may enter the condenser through various joints of
different parts where internal pressure is less than atmos-
pheric pressure.
(ii) The feed water in boiler may contain some air in it.
Therefore, the e,çhaust steam may carry some amount of
air along with it.
(iii) In case ofjet condensers, the air dissolved in cooling water
enters the condenser.
3.37 (a) Effects of Air Leakage
Various effects of air leakage in the condenser are as follows
(i It increases the pressure in the condenser and reduces the
work done per kg of steam.
(ii) The heat transfer rates are greatly reduced because air
offers high resistance to heat flow. This requires more
quantity of cooling water to maintain heat transfer rates.
iii The pressure of air lowers the partial pressure of steam
and its corresponding temperature. The latent heat of
steam increases at low pressure. Therefore more quantity


STEAM POWER PLANT 235

of water is required to condense one kg of steam as the quantity of

latent heat removed is more.
Therefore air should be removed from the condenser. The air
from the condenser is removed with the help of air pumps.
3.38 Air Extraction Pump
Air pump is used to remove the
air from the condenser. Edward's
..... .,- air pump (Fig. 3.69) is generally
I used for this purpose. It consists of
a piston (D) which has a conical
- . head (E). The bottom of the pump
-- ..... casing is also of conical shape in
A [ order that piston head can be easily
C seated. The piston slides inside the
i barrel (C) having a cover (B) which
hasa number of delivery valves (A).
U. Pa- ge(G)is connected to the con
...-.. denser. On the down stroke of the
piston a partial acuum is
Fig.3.69
produced above it since delivery
valves are closed and sealed by
water. When the piston uncovers
the ports (E) air and condensate from the condenser rush into the
space above the piston. As the piston further move own the conical
part (E) displaces the condensate which has colic .ed in the bottom
portion of the pump and forces it to flow into he upper portion
through the ports (F). Now when the piston move upward it raises
the pressure slightly over that of the atmosphere and delivery valve
(A) gets opened and allows air and condensate to move out and flow
over weir H) and finally to hot well. The relief valve (K) provided
at the base of cylinder is used to release the pressure if due to some
reasons the pressure below the piston exceed the atmospheric pres-
sure.
3.39 Condenser Auxiliaries
The auxiliaries are required for the condenser to function
properly. They are as follows
(i) Cooling water suppl y pump to maintain the required flow
of cooling water in the condenser.
(ii) Condensate pump to remove the condensate from the
condenser.
(iii) Feed water pump to supply the feed water to boiler.
(iL) Air removal pump to remove the air from the condenser.
 POWER PLANT
b 236

3.40 Condenser Performance

The condenser performance depends greatly on the cleanliness
of the tube heat transfer surface. Cooling water carries in debris and
impurities that foul up the tube interiors and reduce heat transfer
rate. Tubes should be cleaned periodically.
If the cooling water flows should be accidentally interrupted a
relief valve must release exhaust steam to atmosphere to prevent
building up a destructive pressure in condenser shell.
Air must be removed constantly to keep air pressure low.
A condensing installation is shown in Fig. 3.69 (a). In this
system the cooling water is forced through condenser by circulating
pump. The condensate moves from the bottom of the surface coolers
of steam ejectors and then to the system of regenerative heaters of
the turbine. The function ofejector is to suck air out of the condenser.
The ejector coolers use the condensate as a cooling medium to
condense the live steam in the ejector. This condensate is then
returned to the condenser. An air valve is used to prevent an
excessive increase in pressure in the condenser.

To atmosphere steam
from turbine
....._Livt steam
t

Co nd enser
Steam jet Water
ejector gouge glass

Condensate Pump
CircutotnQ pump
Fig. 3.69 (a)

3.41 Steam Separator

To avoid the turbine blades from the harmful effects of moisture
it is desirable that steam entering the turbine should not contain
drops of water. To achieve this steam separator is located in the
supply line close to the turbine. The steam separator removes
suspended drops or slug of water from the steam.
STEAM POWER PLANT
237

The steam separators may be classified as follows

(i) Reverse Current Separator. A reverse current system
separator is shown in Fig. 3.70. In this separator the direction of
steam is suddenly changed. This causes the heavier water particles
to fall down into the chamber of separator while steam being lighter
flows out.

STEAM -- STEAM
INL E 7 CLI TLE 7

Fig. 3.70.

(ii) Separator with baffle plates. Iii this separator a few

baffle plates are placed in the. Path of steam. The water particles
adhere to these plates.

S team •\\ ( Stearn

m

Water
gouge aBaffle

Water

r Q fl

Fig. 3.70 (a)

C
238 POWER PLANT

Fig. 3.70 (a) shows baffle plate steam separator. Steam entering
the separator moves down and strikes the baffle plates and gets
deflected water particles fall to the bottom of separator and are
drained off. Steam leaving the separator is free from moisture.
(iii) Separator with screens. In this separator, the water
particles are separated by mechanical filtration.
3.42 Steam Trap
The water formed due to partial condensation of steam is pipe
line should be removed to avoid damage. It is done by locating a
steam trap. Traps are used on steam mains, headers, separator etc.
where they remove water formed due to condensation. Trap allows
automatic removal of water but permits now flow of steam. The
various types of steam traps in common use are as follows
(i) Ball float steam trap.
(ii) Inverted bucket type steam trap.
(iii) Thermostatic traps.
(iv) Expansion of orifice traps.

• CONDANSATE
I •i— INSET
8,1L L
FLOAT

W.4 T1/?
::::;Z?72')mrA CJTLET

Vt.
Fig. 3.71

A ball float type steam trap is shown in Fig. 3.71. It is quite

simple in principle and operation. The rising level of water in the
trap lifts the ball float. This causes the opening of valve and the
water gets discharged. After the water has been discharged the float
drops thus closing the valve.
Fig. 3.72 shows steam trap piping arrangement. The gate valve
can be used to shut off the steam trap when it is to he repaired and
removed. During this time the globe by pass valve is utilised for
regulating the flow of water.

TEAM POWER PLANT 239

STE.4M lE4rEO
(dU/PMENT

; f 'dGA7E VAcVE
57Q/4'EP

I.'

0 SC EL/NE CATE

Fig 3.72

3.43. Steam Turbines

Steam turbine is a heat engine which uses the heat energy
stored in steam and performs work. The main parts of a steam
turbine are as follows
(i) A rotor on the circumference of which a series of blades or
buckets are attached. To a great extent the performance of the
turbine depends upon the design a:d construction of blades. The
blades should be so designed that tiey are able to withstand the
action of steam and the centrifugal . .rce caused by high speed. As
the steam pressure drops the length and size of blades should be
increased in order to accommodate the increase in volume. The
variousmaterials used for the construction of blades depend upon
the conditions under which they operate. Steel or alloys are the
materials generall y used.
(ii) Bearing to support the shaft.
(iii) Metallic casing which surrounds blades, nozzles, rotor etc.
(it) Governor to contro.l the speed.
U') Lubricating oil system.
Steam from nozzles is directed against blades thus causing the
rotation. The steam attains high velocity during its expansion in
nozzles and this velocit y energy of the steam is converted into
mechanical energy b y the turbine. As a thermal prime mover, the
240 POWER PLANT

thermal efficiency of turbine is the usual work energy appearing as

shaft power presented as a percentage of the heat energy available.
High pressure steam is sent in through the throttle valve of the
turbine. From it comes torque energy at the shaft, exhaust steam,
extracted steam, mechanical friction and radiation.
Depending upon the methods of using steam, arrangement and
construction of blades, nozzle and steam passages, the steam tur-
bines can be classified as follows
1. According to the action of steam
(i) Impulse (ii) Reaction
(iii) Impulse and reaction.
In impulse turbine (Fig. 3.73) the steam expands in the station-
ary nozzles and attains high velocity. The resulting high velocity
steam impinges against the blades which alter the direction ofstcam
jet thus changing the momentum ofjet and causing impulsive force
on the blades. In reaction turbine steam enters the fast moving
blades on the rotor from stationary nozzles. Further expansion of
steam through nozzles shaped blades changes the momentum of
steam and causes a reaction force dn the blades. Commercial tur-
bines make use of combination of impulse and reaction forces
because steam can be used more efficientl y by using the impulse and
reaction blading on the same shaft. Fig. 3.74 shows an impulsed
reaction turbine.

Fig. 3.73
2. According to the direction of steam flow
(i) Axial (ii) Radial
(iii) Mixed.
3. According to pressure of exhaust
(t) Condensing (ii) Non-condensing
(iii) Bleeder.
4. According to pressure of entering steam
(i) Low pressure (ii) High pressure
(iii) Mixed pressure.

STEAM POWER PLANT 241

"VVIV6
& ADES CL E4Qt.'c

S :'E4/.1 .. ii u ii II I I
ENTRy L,1LJI II II I I EXHAUST

'OTORI U-. -5IN6

Fig. 3.74

5. According to step reductions

(i) Single stage (ii) Multi-stage
6. According to method of drive such as
(i) direct connected (ii) geared.
3.44 Advantages of Steam Turbine Over Steam Engine
The various advantages of steam turbine are as follows
(i It requires less space.
(ii) Absence of various links such as piston, piston rod, cross
head etc. make the mechanism simple. It is quiet and
smooth in operation.
(iii) Its over-load capacity is large.
(iv) It can be designed for much greater capacities as com-
pared to steam engine. Steam turbines can be built iii sizes
ranging from a few horse power to over 200,000 horse
power in single units.
(v) The internal lubrication is not required in steam turbine.
This reduces to the cost of lubrication.
(vi) ia steam turbine the steam consumption does not increase
with increase in years of service.
(vii) In steam turbine power is generated at uniform rate,
therefore, flywheel is not needed.
(viii) It can be designed for much higher speed a greater range
of speed.
(ix) The thermod ynamic efficiency of steam turbine is higher.


242 POWER PLANT

3.45. Steam Turbine Capacity

'rte capacities of small turbines and coupled generators vary
from 5Q0 to 7500 kW whereas large turbo alternators have capacity
varying from 10 to 90 MW. Some of preferred sizes of such units are
indicated in Table 3.1. Very large size units have capacities up to
500 MW.
Table 3.1

Tu rbo alternator size

(MW)
Throttle Ternperatur' I Throttle I'.c.-ur,
CJ - -- (kg/c
40 -
- __482 60
- ----;-
482 60
40 510 87
510 87
__560___ 100

Generating units of 200 MW capacity are becoming quite corn-

mon. The steam consumption by steam turbines depends upon
steam pressure, and temperature at the inlet, exhaust pressure
number of bleeding stages etc. The steam consumption of large
turbines is about 3.5 to 5 kg per kWh.
Generator kW
Turbine kW =
Generator efficiency
Generators of larger size should be used because of the following
reasons:
(i) Higher efficiency.
(ii) Lower cost per unit capacity.
(iii) Lower space requirement per unit capacity.
3.45.1 Nominal rating
It ishc declared power capacit of turbine expected to be
maximum load.
3.45.2 Capability
The capability of steam turbine is the maximum continuous out
put for a clean turbine operating under specified throttle and ex-
haust conditions with full extraction at any openings if provided.
The difference between capability and rating is considered to he
overload capacity. A common practice is to design a turbine for
capability of 1257, nominal rating and to provide a generator that
will absorb rated power at 0.8 power factor. By raising power factor
to unity the generator will absorb the full turbine capability.

STEAM POWER PLANT 243

3.46 Steam Turbine Governing

Governing of steam turbine means to regulate the supply of
steam to the turbine in order to maintain speed of rotation sensibly
constant under varying load conditions. Some of the methods
employed are as follows
(i) Bypass governing. (ii) Nozzle control governing.
(iii) Throttle governing.
Fig. 3.75 shows by pass governing arrangement. In this system
the steam enters the turbine chest (C) through a valve (V) controlled
by governor. In case of loads of greater than economic load a bypass
valve (V1 ) opens and allows steam to pass from the first stage nozzle
box into the steam belt (S).

Nozz le
Box

Throttle
Valve(V)
Stecor
lelt
(f;)
Fig. 3.75

Steam
Fig. 3.76
244 POWER PLANT

Fig. 3.76 shows nozzle governing arrangement. In this method

of governing the supply of steam of various nozzle groups Ni, N2,
and N3 is regulated by means of valves V1 , V2 and V3 respectively.
Fig. 3.77 shows nozzle governing control sysieni. Fig. 3.78 shows
throttle governing. In this method ofgoverning the double beat valve
is used to regulate the flow of steam into the turbine. When the load
on the turbine decreases, its speed will try to increase. This will
cause the fly bar to move outward which will in return operate the
lever arm and thus the double beat valve will get moved to control
the supply of steam to turbine. In this case the valve will get so
adjusted that less amount of steam flows to turbine.

Bar Lift
Valve St
Valve Li ft Bar

!\ozle chest_ 4l _J4—Nozzle Group

Balance Rod
Fig. 3.77

3.47 Steam Turbine Performance

Turbine performance can be expressed by the following fac-
tors
(i) The steam flow process through the unit—expansion line
or condition curve.
(ii) The steam flow rate through the unit.
(iii) Thermal efficiency.
(iv) Losses such as exhaust, mechanical, generator, radiation
etc.
Mechanical losses include bearing losses, oil pump losses and
generator bearing losses. Generator losses include will electrical
and mechanical losses. Exhaust losses include the kinetic energy of
the steam as it leaves the last stage and the pressure drop from the
exit of last stage to the condenser stage.
For successful operation of a steam turbine it is desirable to
supply steam at constant pressure and temperature. Steam pres.
STEAM POWER PLANT 245

sure can be easily regulated by means of safety valve fitted on the

boiler. The steam temperature may try to fluctuate because of the
following reasons:
(i) Variation in heat produced due to varying amounts of fuel
burnt according to changing loads.
(ii) Fluctuation in quantity of excess air.
(iii) Variation in moisture content and temperature of air
entering the furnace.
(iu) Variation in temperature of feed water.
(u) The varying condition of cleanliness of heat absorbing
surface.
The efficiency of steam turbines can be increased:
(i) By using super heated steam.
(ii) Use of bled steam reduces the heat rejected to the con-
denser and this increases the turbine efficiency.
3.47.1 Steam Turbine Testing
Steam turbine tests are made for the following
(i) Power (ii) Valve setting
(iii) Speed regulation (eu) Over speed trip setting
(u) Running balance.
Steam condition is determined by pressure gauge, and ther-
mometer where steam is super heated. The accep nce test as
ordinarily performed is a check on (a) Output. (b) S in rate or heat
consumption, (c) Speed regulation, (d) Over speec trip setting.
Periodic checks for thermal efficiency and ba arrying ability
are made. Steam used should be clean. Unclean steam represented
by dust carry over from super heater may cause a slow loss of load
carrying ability.
Thermal efficiency of steam turbine depends on the following
factors
(i) Steam pressure and temperature at throttle valve of tur-
bine.
(ii) Exhaust steam pressure and temperature.
(iii) Number of bleedings.
Lubricating oil should be changed or cleaned after 4 to 6 months.
3.47.2 Choice of Steam Turbine
The choice of steam turbine depends on the following factors
(i) Capacity of plant
(ii) Plant load factor and capacity factor
(iii) Thermal efficiency
(iv) Reliability
 POWER PLANT
246
1
(v) Location of plant with reference to availability of water for
condensate.
3.48 Steam Turbine Generators
A generator converts the mechanical shaft energy it receive from
the turbine into electrical energy. Steam turbine driven ac.
synchronous generators (alternators) are of two or fouK pole designs.
These are three phase measuring machines offering economic
advantages in generation and transmission. Generator losses ap-
pearing as heat must be constantly removed to avoid damaging the
windings. Large generators have cylindrical rotors with minimum
of heat dissipation surface and so they have forced ventilation to
remove the heat. Large generators generally use an enclosed system
with air or hydrogen coolant. The gas picks up the heat from the
generator any gives it up to. the circulating water in the heat
exchanger.
3.48.1 Steam Turbine Specifications
Steam turbine specifications consist of the following
(i) Turbine rating. It includes
(a) Turbine kilowatts
(b) Generator kilovolt amperes
(c) Generator Voltage
(d) Phases
(e) Frequency
(1) Power factor
(g) Excitor characteristics.
(ii) Steam conditions. It includes the following
(a) Initial steam pressure, and Temperature
(b) Reheat pressure and temperature
(c) Exhaust pressure.
(iii) Steam extraction arrangement such as automatic or non-
automatic extraction.
(iv) Accessories such as stop and throttle valve, tachometer
etc.
U') Governing arrangement.

3.49 Boilers
Boiler is an apparatus used to produce steam. Thermal energy
released by combustion of fuel is transferred to water which
vaporises and gets converted into steam at the desired pressure and
temperature. The steam produced is used for:
(i producing mechanical work by expanding it in steam
engine or steam turbine.


STEAM POWER PLANT 247

(ii) heating the residential and industrial buildings

(iii) performing certain processes in the sugar mills, chemical
and textile industries.
Steam to
t me Control surface

Boiler

Water

Fuel
Fig. 3.78 (a)

Stearn Control

err
Turbine

Insulation V
steam

Fig. 3.78 (b) Shows flow of steam through the steam turbine.

Boiler is a closed vessel in which water is converted into steam

by the application of heat. Usually boilers are coal or oil fired. A
boiler should fulfil the following requirements:
(i) Safety. The boiler should be safe under operating conditions.
(ii) Accessibility. The various parts of the boiler should be
accessible for repair and maintenance.
(iii) Capacity. The boiler should be capable of supplying steam
according to the requirements.
'tv) Efficiency. To permit efficient operation, the boiler should
be able to absorb a maximum amount of heat produced due to
burning of fuel in the furnace.


248 POWER PLANT

Turbine shatt

Pivot Fly bar

._..j_...To turbine

arm
4 / J 11- I1
DOUblQ__H

t
Steam
Fig. 3.78.

G0 It should be simple in construction and its maintenance cost

should be low.
(vi) Its initial cost should be low.
(vii) The boiler should have no joints exposed to flames.
(viii) The boiler should be capable of quick starting and loading.
The performance of a boiler may be measured in terms of its
evaporative capacity also called power of a boiler. It is defined as
the amount of water evaporated or steam produced in kg per hour.
It may also be expressed in kg per kg of fue l burnt or kg1hr/rn 2 of
heating surface.
3.50 Types of boilers
The boilers can be classified according to the following criteria.
1. According to flow of 'ater and hot gases
(i) Water tube. (ii) Fire tube.
In water tube boilers, water circulates through the tubes and
hot products of combustion flow over these tubes. In fire tube boiler
the hot products of combustion pass through the tubes which are
surrounded by water. Fire tube boilers have low initial cost, and are
more compacts. But they are more likely to explosion, water volume
is large and due to poor circulation they cannot meet quickly the
change in steam demand. For the same output the outer shell of fire
tube boilers is much larger than the shell of water-tube boiler. Water
tube boilers require less weight of metal for a given size, are less
liable to explosion, produce higher pressure, are accessible and can
response quickly to change in steam demand. Tubes and drums of
water-tube boilers are smaller than that of firetube boilers and due
to smaller size of drum higher pressure can be used easily. Water-
STEAM POWER PLANT 249
tube boilers require lesser floor space. The efficiency of water-tube
boilers is more.

Fig. 3.79

The basic principle ofwater-tube boiler is shown in Fig. 3.79 and

that of fire-tube boiler is shown in Fig. 3.80. ExaniRs ofvater tube
boilers are Babcock and Wilcox and La \l. ut Boilers whereas
Cornish, Locomotive and Cochran Boilers are fire tubes boilers.
Water tube boilers are classified as I ows:
1. Horizontal straight tube boile
(a) Longitudinal drum (b) Cross-drum.
2. Bent tube boilers
(a) Two drum (b) Three drun
(c) Low head three drum (d) Four drum.
3. Cyclone fired boilers
Various advantages of water tube boilers are as follows
(i) High pressure of the order of 140 kg/cm2 Call obtained.
(ii) Heating surface is large. Therefore steam Carl
generated easily.
(iii) Large beating surface can be obtained b y use of large
number of tubes.
(iv) Because of high movement of water in the tubes the rate
of heat transfer becomes large resulting into a greater
efficiency.
Fire tube boilers are classified as follows:
1. External furnace
(i) Horizontal return tubular
(ii) Short fire box
(iii) Compact.
EIM

250 POWER PLANT

2. Internal furnace
(1) Horizontal tubular
(ii) Short fire box (b) Locomotive (c) Compact (d) Scotch.
(ii) Vertical tubular
(a) Straight vertical shell, vertical tube
(b) Cochran (vertical shell) horizontal tube.
Various advantages of fire tube boilers are as follows
(i) Low cost
(ii) Fluctuations of steam demand can be met easily
(iii) It is compact in size.
5TgAM

FURNACE 8E0

AIR

Fig. 3.80

Stop valvr- , 7Super h,Ot€d

steam
Boiler
drum
Water level
nd cot or,.

FeedwtCr Down take

inlet header
Uptake - Battles
header Header
Tube

Furnacee'BIOwott
valve
h pit

Mud box
Cleaning
door

Fig. 3.8
 STEAM POWER PLANT
251

The disadvantages of fire tube boilers are as follows

(i) The steam obtained is generally wet.
(ii) Larger steam pressures cannot be obtained. These boilers
can produce a maximum pressure of about 17 kg/cm' .
(iii) More time is needed to generate steam. The output of the
boiler is also limited.
3.50 (a) Babcock and Wilcox Boiler
Fig. 3.81 shows a Babcock and Wilcox boiler. It is a water tube
boiler. It consists of a drum, super heater, headers, tubes etc. The
headers are inter-connected by means of tubes which form the real
heating element of the boiler. The fuel is burnt on the grate and hot
gases ofcombustion pass over the tubes and heat the water circulat-
ing through the tubes. The baffles provide a zig-zag path for the flue
gases. The boiler is fitted with various mounting and access,jes.
The hot water and steam rise up through the uptake header into the
boiler shell where steam separates from water and collects in the
steam space. The cold water flows down into the tubes through the
downtake header. Superheater is used to superheat the steam. The
superheated steam can be taken out to the steam stop valve. Any
sediment or mud in water gets collected in mud box l is blown off
from time to time through the blow off valve.

(C/NEC 770t1 FOR -STEAM 5 TQP VALVLr

5.4ErY V4L.r / 1 ç

CCW6U5 TON
CH4/IBER
SMOKE

",P.A' BOX

T .-._

Fig. 3.82
3.50 (b) Cochran Boiler
A brief outline of this boiler is shown in Fig. 3.82. The coal burns
on the grate inside the fire box. The flue gases after passing through
 POWER PLANT
252

the combustion chamber make their way through the fire tubes. The
gases heat the water and convert it into steam hich collects in
steam space, and can be taken out through steam stop valve. The
flue gases escape to the atmosphere after passing through smoke
box and stack. Man-hole is provided to clean the boiler when neces-
sary. A fire brick lining is provided to prevent the shell from being
damaged. It is a vertical type boiler, requires lesser space and is
quite compact in design. In such boilers the ratio of heating surface
to great area ranges between 10 to about 25 and steam can be
produced upto a pressure of 10 kg/cm 2 . This boiler is made in size
upto about 2.743 metres diameter and 5.791 metres height and the
maximum evaporative capacity from and at 100C is 4730 kg per
hour.
2. According to position of furnace
(i) Internally fired (ii) Externally fired
In internally fired boilers the grate combustion chamber are
enclosed within the boiler shell whereas in case of extremely fired
boilers and furnace and grate are separated from the boiler shell.
3. According to the position of principle axis
(i) Vertical (ii) Horizontal
(iii) Inclined.
4. According to application
(i) Stationary (ii) Mobile, (Marine, Locomotive).
5. According to the circulating water
(i) Natural circulation (ii) Forced circulation.
6. According to steam pressure
(i) Low pressure (ii) Medium pressure
(iii) Higher pressure.
3.51 Lancashire Boiler
Fig. 3.83 shows a brief outline of this boiler. The coal is fed
through the fire hole on the gate (G) where its combustion takes
place. F.B. represents Fire-box. The flue gases move through the
entire length of the furnace tubes (T) upto the back end then move
downward and flow along the bottom flue and finally letive to the
atmosphere through a chimney. The surface tubes are nearly one
metre in diameter, the diameter being larger at the front end and
smaller at the rear end.
The two furnace tubes are surrounded by water on all sides. The
boiler shell is filled with substantial quantity and is supported over
the brickwork. The steam produced collects in the space above the
water level. Blow cock (B.C.) is provided to empty the boiler so that
repairing, inspection or cleaning can be carried out when desired.
Such boilers are gr'nerally constructed for working pressure upto
17.5 kg/cm 2 and evaporation capacity of 8300 kg hour.

STEAM POWER PLANT 253

JTT

L1 1tt

ii •
l'
1i:1 i1
ii

Fig. 383.

3.51 (a) Scotch marine boiler

It is a fire tube boiler. It is compact, more efficient and has the
ability to use any type of water. These boilers ma y be
(i) single ended
and (ii) double ended.
Single ended boiler as shown in Fig. 3.83 (a) ma y have one to
four furnaces. A single ended boiler has a length up to 3.5 metres
whereas double ended boiler man y up to 6.5 metre long.

254 POWER PLANT
a

mny rSar stay

Shell

Smoke I--CombustIOn
box .. - chamber

_
--1

\ I--
L-

Furnace

Fig. 3.83 (a) Scotch marine boiler.

The furnace tubes, smoke tubes and the combustion chamber,

all being surrounded by water, give a very large heating surface area
in proportion to the cubical size of boiler. The water circulates
around the smoke tubes. Te level of water is maintained a little
above the combustion chamber. The flue gases, from the combustion
chamber, are forwarded by draught through the smoke tubes, and
finally up the chimney. Fig. 3.83 (a) shows this boiler.

Fire tube

Bottom
flu

Fig 3 83 (b) Corntsh boiler.

STEAM POWER PLANT 255

3.51 (b) Cornish Boiler

l'i. :3.3 ih I shows this boiler. Its construction is similar to
1 cashtre boiler. It is a fire tube boiler. It has onl y one flue tuhi
compared to Laucasitire boiler which has two flue tubes. The
ii iieter of flue tube ma y he about 0.6 times that of' shell The
diameter of boiler ma y be 1 to 2 in and length 5 to 7.5 metre. As
compared to Lancashire boiler the capacity of co rnish boiler is less.
3.52 Boiler Mountings and Accessories
Mountings are the components used for the safety of boi icr. The
boiler requires the lolloving mountings a feed check valve to prevent
the return of water from boiler in base the feed pump is not
operating, a steun stop valve to regulate the flow of steam from the
boiler, safet y valve (at least two) to protect the boiler from pressures
higher than the designed value, a blow off valve to empty the boiler
when needed and to discharge the mud and sediments that collect
in the boiler, water level inclicdtors to show the water level inside
the boiler, a pressure gauge to indicate the pressure ofsteamn in the
boiler, fusible plugs to protect the boiler against low water level
OLER ilor WATEF? ECOt.1MIcER FEED WATER
--------------------- ------ -- -----
I HorFLuE GASES

I CASES AIR
TO PRE HEATER
COWNEY

tIorAIR
DRAFT
EOU,PiEN r

SLJPE1 HLA TL

H OrFLUE
GASES
STLAM
Fig. 384

loileI' accessories are used to operate it efficiently. The various

accessories used are as follows feed pumps to suppl y water in th.c
boilct ecoilumnisers to heat the feed water before it enters the boiler,
air pre-licaters, to heat the air before it enters the furnace, super-
heaters to heat the steam above saturation temperature, and draft
cqui niElil to supply air to furnace.
3.53 Flue Gas and Water Flow
Fig 3.1 shows the flue gas flow aril feed waterfiow. Steam flow
in boiler unit. The heat of flue gases is utilised in heating the feed
water in i-conomliiser and also in heating the air to be supplied to
boiler fur n ace.
256 POWER PLANT

.54 Causes of Heat Loss in Boilers

The various causes of heat loss in a boiler are as follows:
(t) Loss duo to heat carried away in chimney gases.
(ii) Loss due to incomplete combustion of fuel. This can be due
to insufficient air supply or fuel bed being in poor condi-
tions.
(iii) Loss due to moisture formed by combustion of hydrogen
in fuel.
(iv) Loss due to moisture in coal.
(u) Loss due to radiation and convection from boiler.
(vi) Heat loss due to unburnt fuel which is passed on to the
ash pit.
3.55 Thermal Efficiency of Boiler
Thermal efficiency of a boiler is defined as the ratio of heat
actually utilised by water in generation of steam to the heat supplied
during combustion of fuel.
- W(H2—.H1)
flThh
w x CV

where W = Weight of steam produced, kg

Hi = Enthalpy of feed water kcal/kg
112 = Total heat of steam kcallkg at the
generation pressure
W 1 = Weight of fuel burnt, kg
C.V. = Calorific value of fuel kcallkg.
Although the major portion of
heat produced during combustion
of fuel is utilised to heat th feed
water for conversion into steam Oil ized
PuLver
00 Jcoa1
but some portion of heat produced
is lost in flue gases leaving the
chimney and some amount of heat LL TNat
gas
is lost due to radiation from fur-
nace to the outside. The boiler ef-
ficiency is nearly 75% without heat
recovery equipment and it varies
from 85 to 90% with heat recovery
equipment. —C
Boiler efficiency can be im-
proved by taking into account the
following factors Fig. 3.85.
STEAM POWER PLANT 257

(i) Proper supervision should be carried out.

(ii) Toomuch excessive air should not be supplied.
(iii) Boiler, superheater, economiser and air heater surfaces
should be cleaned at proper intervals.
(it') Type of fuel used also controls boiler efficiency. Fig. 3.85
compares the performance in burning bituminous coal, oil
or natural gas in the boiler.
Efficiency () indicates steam generating unit efficienc y and C
represents per cent of rated capacity.
(L) Proper supply of excess air.
(vi) Minimising combustible in refuse.
(vii) Minimising combustible in flue gas.
(viii) Maintaining required cleanliness of heat transfer surfaces.
Efficiency of a boiler depends on the following 1ictors
(i) Steam pressure
(ii) Calorific value of fuel
(iii) Efictiveness in burning of fuel.
Indication of the effective burning of fuel is given by the per-
centage of carbon dioxide (CO2) in the flue gases. The percent-
age of C Q to give maximum efficienc y and satisfactory
operation will depend on
(a) type of fuel such as oil, gas, coal.
(h) method of firing.
(iv) Capacity of boilers
(U) Use of heat recovery apparatus such s air preheater,
economiser etc.
3.56 Boiler Performance
The performance of a steam boiler ma y be expressed in terms of
the following
(i) If release per cubic metre of furnace volume.
(ii) Efficiency.
(iii) heat transferred per square metre of the heating surface
area per hQur.
(it-) of combustion in kcal per square metre of the grate
area per hour for solid fuel.
U) Amount (kg) of steam produced per hour.
3.56.1 Selection of Fuel for Boiler
In choosing the fuel for boiler the following economic considera-
tions are important
(i) Plant efficiency
(ii) Cost of fuel per billion killocalories
(iii) Cost of fuel burning equipment.
258 POWER PLANT

di Operation and maintenance cost of fuel hurtiing equip-

ment aI1(1 heat absorbing apparatus.
r) Cost of fuel preparation and storage equipment.
ii) Fuel transportation cost.
3.56L2 Equivalent. Evaporation
To compare evaporative capacity or perforilnince of different
boilers working under clifferint conditions it is desirable to provide
a commonii I)k1Si' so that water be supposed to he evaporated under
standard conditions.
The standard conditions adopted are temperature of feed water
100°C and converted into dr y and saturated steam at 100°C. As per
these standard conditions 1 kg of water at 100°C requires 2257 kJ
to get converted to steam at 100°C.
Therefore equivalent evaporation is defined as the amount of
wate r eva ( ) rated from water at 100°C to dry and saturated steam
at lOUC. However j ) the evaporation rate of' boiler may iliSi) he
expressed in terms of kg of steam per kg of fuel.
(ii) The capacity of a boiler may also be expressed in terms of
total heat added per hour.
Factor of evaporation (Fe)
It is defined as follows
1)'i
Fe =
112
where H = Heat received by 1 kg of water under working comb-
t ions.
112 = Heat received by one kg of water evaporated from and at
100°C.
Heat losses in a boiler 'plant
The heat losses which ma y take place in a boiler plant are as
follows
(i heat loss clue to incomplete combustion of' fuel
(ii) heat loss clue to unburnt fuel
(iii) convection and radiation heat loss
(iv heat lost in flue gases.

3.57 Boiler Trial

Steam is generated in boilers under certain conditions of inlet
water and exist steam while a certain rate of' fuel is being burnt. In
order to study the performance of boiler the experiments are con-
ducted by operating the boiler for a certain length of' time and
recording the data. This procedure is known as boiler trial. A heat
STEAM POWER PLANT 259

balance sheet is prepared which indicates the heat supplied heat

utilised through various sources and heat wasted.
The main objects of boiler trial are as follows
(i) To determine the thermal efficiency of the boiler when
working at a definite pressure.
(ii) To draw up heat balance sheet for the boiler to check the
performance of the boiler.
Example 3.8. Determine the boiler efficienc y if the boiler
gene/cite 20,000 kg of steam at 10.2 kglc,n 7 dry and saturated The
amount of coal used is 2100 kg and tile calorific value of coal is 8600
kcctl / kg. The feed water temperature is 105C.
Solution. Total heat of steam at 10.2 kg/cm2
= 664 kcal/kg
23,000ç664_-_105 = 0.71
- 2100 x 8600
-71%. Ans.
3.58 Boiler Maintenance
Regular maintenance and careful supervision of working of
Various components of boiler result in the efficient operation of the
boiler. The various factors which should be carefully observed for
the proper maintenance of the boiler are as follows
1. The combustion equipment should be so adj .ted that the
temperature in the furnace does not exceed, the designed
value. The air supply to the furnace shouhi be in correct
proportion.
2. The water level in the boiler should not be allowed to fall
beyond the minimum level.
3. The temperature should change slowly and uniformly in
the various parts of the boiler. Rapid changes in tempera-
ture cause unequal expansion.
4. The water used for steam generation should he free from
scale forming impurities because the scale if deposited
oil and boiler shell does not allow the heat transfer
to take place effectively and sometimes causes over heat-
ing of tubes and boiler shell which may result in tubes
failure and boiler explosion.
5. Soot and ash deposited in tubes on gas side should be
removed regularly.
6. Bearings of pumps, stokers, pulverisers and fans etc.
should be lubricated regularly.

260 POWER PLANT

7. At least once in a year the internal inspection of boiler is

necessary. The boiler should be inspected for corrosion.
cracks, leaks and other irregularities.
8. Piping s ystem, joints and valves should be checked for
leakage.
3.59 Control and Measuring Instruments
The various control and measuring instruments provided in a
boiler are as follows
(i) Steam flow metres.
(ii) Thermometers showing temperature of gas, steam and
water.
(iii) Draught meters.
(iv) Devices to analyse flue gases
3.59.1 Soot blowers
In a system power plant the heating surfaces of boilers especial-
ly coal fired water tube boiler have a tendency to become coated with
a layer of soot, cinder and fl yash. This loosely adhering la yer does
not allow the heat transfer to take place properly and should be
removed after it has built up sufficiently to result in a significant
increase in flue gas temperature. Soot blower is used to remove this
layer of soot, cinder and flyash.
Soot blowers should be operated frequently enough to keep the
tubes clean.
3.60 Circulation of Water in Boilers
Reliable circulation of water in the boiler is essential for proper
boiler design. Circulation means continuous flow ofwater in a closed
circuit designed to create continuous motion of the steam and water
in the boiler and intensively heat from the heating surface. Heat
transfer rate is improved and boiler walls are protected from chemi-
cal destruction by continuous motion of water as it washes of the
bubbles of steam from the heating surface. Circulation of water in
boiler is carried out in two ways
(i) Natured circulation. (ii) Forced circulation.
Natural circulation of water is on the different densities of water
and steam and water mixture. Fig. 3.85 (a) shows the simplest
natural circulation arrangement consisting of drum, a header and
two pipes. As heat is supplied the steam starts forming in left hand
pipe. The steam and water mixture rises and separates into steam
and water, steam accumulating in the drum above water. The
amount of steam produced in the right hand pipe is small also and
formation of steam is less intensive. Due to this reason the density
of steam and water mixture in right hand pipe will he greater than
in left hand pipe and therefore a continuous flow of water from right
hand pipe to left hand pipe, will take place.
STEAM POWER PLANT 261

Heat
Supply

Fig, 3.85 (a)

In forced circulation of water pumps are used to maintain the

continuous flow of water in the boiler. The relation between the
amount of water passing through a circulation circuit in a definite
time and the amount of steam generated in the circuit during the
same time is called circulation ratio. The value of circulation ratio
varies from 8 to 50 in natural circulation water tube boilers. Boilers
with natural circulation arrangement require larger diameter tubes
and also natural circulation system does not ensure safe and reliable
operation. Boilers with forced circulation require smaller diameter
tubes thereby reducing the weight of boiler. The system is helpful
in giving the most rational shape to the boiler, providing the re-
quired rate of circulation and in such boilers circulation ratio is
small.
3.61 Feed Water Regulators
Water, fuel and air are the three variables entering into the
production of steam in boiler. It is necessary that feed water should
flow into the boiler almost as rapidly as steam flows out. Maintain-
ing a fixed water level is not advisable. Every change in load on
boiler must be met with a change in feed water input. Feed water
regulation should be automatic as it cannot be clone successfully by
hand.
The commonly used feed-water regulators are of following
types
(i) Float type (ii) Vapour pressure type
(iii) Thermostatic expansion type.
Fig. 3.85 (h) shows a float type feed water regulator. This
regulator has a float chamber piped to boiler drum. It is installed
opposite the normal water level of the boiler so that level can be
duplicated in the float chamber. With the change in level the motion
of float mechanically opens or closes a balanced regulating valve in
feed water line.

262 POWER PLANT

17 't
Biier D:"a'

FCCd Water

1eci1 Water
Va
Fig, 3.85(b)

3.62 High Pressure Boiler

The present tendency is towards the use ofhigh pressure boilers.
The boiler pressures have reached as high as 20 kg/cm 2 and steam
temperature 550 C. By using high pressure boilers, low grade fuels
can be burnt easil y , a saving in cotly construction material can be
achieved and a higher efficiency s possible. high pressure boilers
are water-tube boilers, use pulverised coal firing and in such boilers
furnace walls are water cooled and water tubes from the walls of the
furnace. They use slag type furnace.
3.62.1 Unique features of high pressure boilers
The high pressure boilers possess special features which make
them more advantageous. The unique features are as follows.
1. Method of water circulation.-In these boilers water cir-
culationis maintained with the help of pump which forces the water
through the boiler plant. However sub-critical boilers may use
natural circulation.
Circulation b y a pump ensures
(i) ensures positive circulation
(ii) increases evaporative capacity of the boiler
and (iii) less number of drums will be required.
2. Type of tubing. These boilers generally use several set of
tuhings through which water passes. This arrangement helps
(i) to reduce the pressure loss
and (ii) to achieve better control over the quality of the steam.
3. Improved method of heating. In high pressure boilers
following improved methods of heating may be used to increase the
heat transfer rate.
(I) To save heat by evaporation of water above critical pres-
sure of the steam.
 STEAM POWER PLANT 263

(ii) To heat water b y mixing the superheated steam. It gives

high heat transfer coefficient.
(iii) To use high water velocity inside the tubes and to increase
gas velocity above sonic velocity. This will increase over
all heat transfer coefficient.
3.62.2 Advantages
The various features (advantages) of such boilers which make
them useful are as follows
(i) In high pressure boilers pump are used to maintain forced
circulation of water through the tubes of the boiler. This
ensures positive circulation of water and increases
evaporative capacity of the boiler and less number of
steam drums will be required.
(it) The heat of combustion is utilised more efficiently by the
use of small diameter tube in large nuiii: in multi-
ple CiFCUit.
II0 Piessurised combustion is used which increases rate of
firing of fuel thus increasing the rate of heat release.
Due to compactness less floor. spate is required.
(i) It. is economical to use high pressure and high tempera-
ture steam produced b y such boilers,
(ii) All the parts are uniformly heated, therefore, the danger
of over heating is reduced.
i) The efficienc y of tile plant is increased to about •lO
3.63 Types of High Pressure Boilers
The various t y pes of high pressure boilers c-':'111101113- used are
as follow
CCNV(C T/N
SL'P(R I1A TER

&A1 1/ STEAM

- I tiri7vER

I. 1jT .r.Eo
'o
')1' SE(T"2V

Fçj. 3.85 (C)



24 POWER PLANT

(i) La Mont Boiler (ii) Loeffler Boiler

(iii) Benson Boiler (iv) Velox Boiler.
La Mont Boiler. Fig. 3.85 (c) shows the schematic arrangement
of the boiler. Feed water from hot well is supplied to the separatin g-
drum (boiler) through the economiser. Water circulating punip
maintains the forced circulation of water through the tubes in
evaporation section. The steam produced flows to the drum. Dry
steam from the top of drum is then passed through a convection
superheater and the superheated steam flows to the turbine. This
type of boiler has a working pressure of 170 kg!cin2.

5 TEA . To
pRfMr.AIO vR

-1
ECO)V OM/5ER 1

I
CON V5C TON i
5L/PERHATR

-4 RADIANT I

.SUPERHEATER

STEAM
IRCuAT/NG
PUMP [HI
ORLJPI
Fig. 3.86

• Loeffler Boiler. In this boiler (Fig. 3.86) the drum is placed

away from the furnace. Feed water passing through the economiser
moves to the drum through which superheated steam is also pass-
ing. The superheated steiiii converts the water into saturated
steam. A steam circulating 1,unip is used to circulate steam through
radiant and convective superheaters, where steam is heated to the
required temperature. A 1,.irt of this steam flows to the prime mover
and the remaining steam flows to the drum. Such boilers have a
working pressure of 135 kg/cm2.
Benson Boiler. The schematic arrangement of the boiler is
shown in Fig. 3.87. This boiler does not use any drum. The feed water
after circulating through the economiser tubes flows through the
radiant parallel tube section (T) to evaporate partly. The steam
water mixture produced then moves to the transit section (TS)
where this mixture is converted into steam. The steam is now passed
through the convection superheater(CS). Such boilers have a work-
ing pressure of 230 kg/cur.
 STEAM POWER PLANT 265

ECONOMISER
5rg4/.1
To
PR'#4EMOVER r

II

AEED POMp
1._

Fig. 3.87

STEAM To
PRIME MOYER
CONVECTION
SUPER YEA TER-

STEAM
54R4 TIMS .-
SEC TIC/V

Areo
PUMP
IWC

L±J Fig. 3.88

Ecodya4wsER

Velox Boiler. In this boiler (Fig. 3.88) pressurised combustion

is used. The axial flow compressor (AC) driven by gas turbine (GT)
raises the incoming air from atmospheric pressure to furnace pres-
sure. The combustion gases after heating the water and steam flow
through the gas turbine to the atmosphere. The feed water after
passing through the economiser is pumped by a water circulating
pump (WC) to the tube evaporating section (T. Steam separated in
steam separating section flows to the superheater.
This boiler can generate a pressure of about 84 kg/cm2.
3.63.1 (a) Advantages of High Pressure Boilers
The various advantages of high pressure boilers are as
follows
266 POWER PLANT

(i) The steam can be raised quickly to meet the variable load.
(ii) By using high pressure boilers the efficiency of steam
power plant is increased upto about 40%.
(iii) By using forced circulation, there is more freedom in the
arrangement of furnace, tubes and boiler components.
(iv) The space required is less.
(v) Light weight tubes with better heating surface arrange-
ment can be used.
(vi) The tendency of scale formation is eliminated clue to high
velocity of water through the tubes.
(vii) Various parts are uniformly heated. Therefore, the danger
of overheating is redqced and thermal stress problem is
simplified.
3.63.2 Selection of boiler
Selection of boiler depends upon the following factors
(1) Type of fuel
(ii) Space available
(iii) Cost of fuel
(it) Controls necessary to enable the load to be picked up
quickly
(t) Desireability of using heat recovery equipment such as air
preheater, economiser
(ti) Type of load to be supplied by the boiler.
3,64 Modern Trends in Generating Steam
Some of the special features used in generating steam using
high pressure boilers are as follows
(i) Use of forced circulation of water inside the boiler.
ii) Use of small tubes in large number in order to increase
the surface area of heat receiving surface.
(iii) Treating the feed water thoroughi .N before use.
(it') Use of multiple tubes circuits.
(u) Use of pressurised combustion.
(ti) Use of superheaters.
The process ofsteam generation can be divided into three stages
namely
(i) heating of water upto boiling point
(ii) evaporation of boilitg water and its conversion into dry
saturated steam,
(iii) transformation of dr y saturated steam into superheated
steam.
These stages are shown on temperature (T) and heat input
diagram (see Fig. 3.88 (a)). At point. I the ice is heated. As point 2,
ire starts melting. At point 3 entire ice has incited. Further heatiiit.


STEAM POWER PLANT 267

causes water to boil at point 4. At point 5 the saturated steam is

obtained. The steam is thin super heated as shown at point 6. Use
of super heatedsteam in steam engines and steam turbines raises
the efficiency. $

Temp.

Heat input—..
Fig. 3.88 (a)

3.65 Gas Fired Boilers

Industrial gases like coal gas, producer gas, blast furnace gas
and sewage gas are also used in firing steam boi l Where such
gases are available in abundance, their use proves economical.
Gas fired boilers have the following advantages
(L) Gases can be burnt with low excess air r "ing in higher
efficiency.
(ii) Storage of fuel (gases) is not required.
(iii) As no ash is produced, therefore, problern of ash disposal
is eliminated.
(iv) It is a clean system of firing as soot and smoke produced
as almost negligible:
(v) The system does not require higher chimney.
3.65.1 Selection of boiler (steam generator)
Boiler is used to supply steam to the turbine. While selecting a
the following points should be considered.
. Type of load.
Amount and type of fuel to be burnt.
pace available for boiler installation.
st of boiler.
The . of boiler depends upon the following factors
Operating pressure.
'n erating temperature.
I firing.
e of boiler—Generally water tube boilers are used in
central power plants.
(e) Efficienc y of boiler.
tf Amount of steam to be generated.

268 POWER PLANT
a
3.66 Piping System
Piping system is an essential part of steam power plant. It is
employed to transit water, steam, air, oil and vapour from one-place
equipment to the other. The piping system should fulfil the following
requirements:
(i) Piping system should be of necessary size to carry the
required flow of fluids.
(ii) Pipes carrying the fluids at high temperature should be
able to withstand the temperature and expansion caused
due to the temperature changes.
(iii) Piping system must withstand the pressure to which it is
subjected.
(iv) For smooth and safe operation of plant and for easy
inspection and maintenance it is desirable to use mini-
mum length of pipes consistent with the requirements of
the power station. Also a pipe should be run as direct and
straight as possible. The number of fitting and bends
required to make the necessary connections should be
kept at a minimum to avoid pressure drop.
(v) The piping system used for steam should be installed in
such a way that horizontal runs must slope in the direction
of steam flow with steam traps at all points where water
may accumulate to make it easy to drain the same and to
prevent air accumulation of water during steam flow.
3.67 Types of Piping System
A power plant uses many fluids like water, steam, air, oil, gas
etc. during its operation. This requires a variety of integrated piping
'systems mentioned as follows
(i) Steam' piping—raw water, feed water, condensate and
condenser cooling water.
(ii) Steam piping—Main, reheat, bleed exhaust steam.
(iii) Blow of piping—Boiler, evaporator, feed treatment.
(iv) Miscellaneous piping—Water treatment, service water,
lubricating oil drains compressed air etc.
Brackets attached to the building columns and hangers
suspended from beams usually support piping in a power plant.
3.68 Size and Strength of Pipe
The size of the pipe depends upon the velocity of fluid in the pipe
and the internal pressure of fluid and thermal expansion limita-
tions.
Wrought iron and mild steel pipes used for different pressures
are graded as follows
STEAM POWER PLANT 269

1. Standard.
2. Extra heavy or extra strong.
3. Double extra heavy or double extra strong.
Pipes up to-12 inch (300 mm) diameter are designated by their
nominal inside diameter and pipes above 300 mm diameter are
specified by their outside diameter.
Various codes and standards specify minimum pipe diamen-
sions and materials to meet certain requirements. The codes should
be consulted for allowable stress values. They depend on tempera-
ture and vary for different types of steel. Pipe diameter affects the
fluid speed for a given mass flow rate. High temperature steam in
piping has the following effects:
(i) It decreases allowable stress.
(ii) It accelerates oxidation and corrosion.
(iii) It produces expansion.
. (iv) It makes the pipe material creep.
3.69 Insulation
The pipes carrying the fluids at high temperature should be
properly insulated to avoid heat loss to the sun ings. The
insulating material should have low co-efficient of t.iermal conduc-
tivity. It should not damage the pipe material and s;:rnild he able to
resist the temperature to which it is subjected. It s .i be easily
moulded and applied and have the requisite mechanical strength.
High temperature line can be effectively insulated by 85%
magnesia block in varying thickness depending upon the tempera-
ture. The various materials in common use 85% magnesia, asbestos
fibre, cork, hair felt, mineral wool, glass wool, expanded, mica etc.
85% magnesia contains 85% carbonate of magnesium and 15%
binder and is used for temperature upto 325C. Asbestos blankets
can be used upto 500'C. Mineral wool can be used up to tempera-
tures of 875°C. Mineral wool is made by blowing steam through
fused clay lime stone of furnace slag to fibrise it. Expanded mica,
hair felt and cork can be used up to 125C. Glass wool can be used
upto temperatures of 350C. It is made by blowing steam through
streams of molten glass.
3.70 Material for Pipes
The various materials used for the pipes are as follows:
(i) Cast Iron. Cast iron pipes are used underground for water
and drainage systems and in other places where problem of cor-
rosion is excessive. Cast iron pipes are used for water services upto
a pressure of 15 kg/cm2.
(ii) Wrought Iron. Pipes made up of wrought iron are used for
condensate, feed water and blow off lines. Such pipes are used . for
270 POWER PLANT

low and medium pressure range and should not be used when
pressure is more than 250 p.s.i.
(iii) Wrought Steel. Most of the pipes used in power station are
made up of wrought steel. Wrought steel pipes are cheaper.
(iv) Alloys Steel. For high temperature flows pipes are made
up of alloy steel.
Chromium steel pipes are used for temperature higher than
525CC. For temperatures between 400-525C carbon molybdenum
steel may be used.
(v) Copper and Brass. Pipes made up of copper and brass are
costly and are mostly used for oil lines. Brass pipes are used upto
pressure 20 kg/cm2.
3.71 Expansion Bends
In order to allow for the expansion in pipes due to temperature
changes, expansion bends are used in pipe lines. The various types
of expansion bends used are shown in Fig. 3.89.

90.BENO U-BEND DOuBLE OFFSET

LI-SEND

I •'.

EXPANSION U-BEND
DeL/SuE o'F5E7 EXPANSION LOOP
BEND

Fig. 3.89

3.72 Pipe Fittings

Joint fittings are necessary to assemble piping system and make
connections. Fittings are made in variety of forms such as screwed
or welded fittings which are generally used in sizes upto 90 mm
whereas flanged or welded fittings are used for large sizes of pipes.
The various fittings include elbows, bend, tees, crosses, plugs bush-
ings and reducers. Fig. 3.90 shows various fittings.
Elbow is used to change the direction of two pipes which it
connects. Tee is used for joining two pipes running in the same
 STEAM POWER PLANT 271

direction and having an outlet for a branch pipe. Plugs and caps are
used to close the ends of fittings and pipes. Reducers are used tojoin
pipes of different sizes.
Fittings made up of cast iron, malleable iron, steel, steel alloys
and brass are in common use. The materials used for fittings depend
on service as regards pressure and temperature.

2 3 4 5

MED11>T^_1 Fig. 3.9&

In Fig. 3.90 various fittings shown are as follows:

1. Elbow (90) 2. Tee
3. Elbow (45 ' ) 4. Double t'-'inch elbow
5. Side outlet elbow 6. Side ' t Tee
7. Cross 8. Lateral
9. Reducer.
3.73 Pipe Joints
Various types ofpipejoints are screwed joint ( 91), welded
joint (Fig. 3.92), socket and spigot joint (Fig ), expansion joint
(Fig. 3.94) and flanged joint. Flanged fittings are generil1v used for
larger size of pipe and for high temperature and hig} essure work.

Fig. 391 Fig. 3.92

The selcct(on of type of pipe joint depends upon the following
factors
(1) Size of pipe.

272 POWER PLANT
a
(ii) Pressure and temperature of fluid inside the pipe.
(iii) Initial cost.
(iv) Ease of removal of sections between joints.
(v) Future maintenance.

ket End
End
Fig. 3.93

Fig. 3.94

3.74 Valves
Various types of valves used in the pipe line are gate valve, globe
valve, angle valve automatic stop valve, reducing valve, check valve
and back pressure valve. Out of these first four are called stop
valves. They are used to stop the flow of fluids. The function of
reducing valve is to change pressure in a steam lute. Check valve
permits flow in one direction only. Back pressure valve is used in
STEAM POWER PLANT 273

connection with the exhaust piping of an engine to permit undue

rise in pressure.
Small size valve upto 75 mm are usually made up of brass.
Material used, in valve of larger size is either cast iron or cast steel
or forged steel. Brass is used for valve seats, discs and spindles of
valve used for steam or water.
3.75 Principles of Steam Power Plant Design
The essential principles of steam power plant design are as
follows
(i) Low capital cost
(ii) Reliability
(iii) Low operating and maintenance cost
(iv) High thermal efficiency
(v) Accessibility
(vi) A simple design.
The power plant should be designed such that it can be expanded
if required. It should be simple in design. Use of automatic equip-
ment is desired to reduce the labour cost. Heat recovery devices
should be used wherever possible. Total capacity of the plant should
be subdivided into four or five generating units so that during
reduced load periods some of the units may be stopped.
3.76 Factors Affecting Steam Plant Design
The various factors which affect the design of a steam power
plant are as follows
(i) Steam pressure and temperature
(ii) Capacity of power plant
(iii) Ratings of generating units
(iv) Thermodynamic cycle
(v) Voltage generation.
The trend is towards using higher pressure and temperatures
of steam. This improves the thermal efficiency. The capacity of the
plant can be determined by studying the load duration curve and
anticipated future load demand. The size of turbo-generator
depends on the following:
(1) Rate of growth of load.
(ii) Availability of condensing water.
(iii) Space available.
Larger size turbo generator sets should be used.
The voltage of generation is usually 11 kV; though 22 kV and 33
kV are also used. In central power plants water tube boilers are
commonly used. The type of fuel used in a boiler will influence the
design and efficiency of the boiler plant considerably.
The steam power plant needs a lot of space for (a) storing the
fuel (b) elaborated arrangements for fuel handling (c) ash handling

- 274 POWER PLANT

(d) lay out of high-pressure boilers (e) steam turbines generators

(g) control switch boards (ii) cooling water arrangements.
In steam power plant the A.C. generators are driven b y turbines
with condensing arrangement at a high vacuum of about 73.6 cm:
Hg. The size of the generating set may vary between 10 MW to 500
MW or even higher. Modern generators are 2 pole with 3000 R.P.M.
as speed and 50 cycles per second as frequency.

3.77 Site Selection

The various factors to be considered while selecting the site for
steam power plant are as follows
(i) Distance from coal mines. Steam power plant should be
situated near the coal mine so that cost of transportation of coal is
low. If the plant cannot be located
* near coal fields the plant in such
case should be connected by rail or road to the coal mines so that
transportation of coal is easy.
(ii) Distance from load centre. The power plant should be
located near the load centre so that the cost of transmission lines
and losses occurring in them are less.
wo Availability of water. Water should be available in large
quantities Ifwatcr available is pure it will be useful because impure
water needs purification. The water required is about 560 x 103 kg
of every one tonne of coal burnt.
(iL) Ash disposal. Near to powerstation site, large space should
be available fi)r the ash disposal.
(e) Distance from popular area. Smoke and other gases
produced due to combustion of coal pollute the atmosphere. There-
fore, the plant should be situated, as far as possible, away from the
densel y populated area.
(vi) The site selected should be capable of Supporting a large
building and heavy machinery.
3.77.1 Controls at steam power plant
Generally the electric load on power plant varies in all
manner. The control provided at a power plant help in meeting the
variable load successfully. Controls for the following are provided at
a steam power plant to run the plant smoothly.
(i ) Fuel (ii) Air
(iii) Feed water (iv) Steam
(t.') Ash (ii) Flue gases
(vii) Furnace temperature (viii) Condenser cooling water.
 STEAM POWER PLANT 275

3.77.2 Feed water control

The supply of feed water depends upon the plant load. Some
of the automatic method used to control feed water flow are as
follows
(i) Single element pilot operated system
(ii) Single element self operated system
(iii) Two element pilot-operated system
(iv) Three element pilot-operated system.
Fig. 3.94 (a) shows a single element pilot operated feed water
control system. This allows manual correction of water level with
flow variation. In this system the water level recorder (R) operates
an air pilot valve which through a relay and indicator (P) controls
the feed water regulating valve.
stecim
PegdMzter
Rauati-n•:i

Dnum

Feed

Pe lay

SU Z,/
—44 Fig. 3.94 (a).

The pilot valve (C) can be automatically reset and this makes it
possible to hold boiler drum level constant at all rates ofsteamn flow.
Pilot valve's manual adjustment allows lowering the water level at
low load and vice versa.
3.78 Industrial Steam Turbines
These turbines in addition to power generation supply some
steam for manufacturing processes that (steam) would otherwise be
wasted. Some such turbines are as follows:
1. Pass Out or Extraction Turbines. In these turbines (Fig.
3.95) the high pressure steam from the boiler enters the High
 276 POWER PLANT

_Pressure (H.P.) stage where it expands. A part of the steam leaving

the high pressure stage is utilised for some processing work such as
feed water heating, paper making, dyeing etc. Remaining steam
enters low pressure stage and expands there and finally goes to
condenser. This turbine is capable to meet variations in steam
demands.

H I - -

$ TO
To PR0CES

FEED
WATER TO CONLNSER

Fig. 3.95

2. Back Pressure Turbines. In this turbine (Fig. 3.96) the

steam after expansion is used for processing work and then from
processing plant it flows to condenser from which it is pumped back
to the boiler: This turbine is used where the steam demand is
constant.

Fig. 3.96

3. Exhaust Turbine. In this turbine the steam leaving the

steam engine enters the turbine where it expands thus further
utilizing the steam leaving the engine which (steam) would other-
wise be wasted. Fig. 3.97 shows such air arrangement.
4. Mixed Pressure Turbines. Sometimes in rolling mills and
colliery cngines the waste steam is intermittent and the pressure of
exhaust steam in intermediate between turbine inlet steam pres-
sure and that of turbine exhaust steam. The exhaust steam and
 STEAM POWER PLANT 217

main steam enter the turbine where steam expands. Fig. 3.98 shows
such turbine arrangement.

(XI?AUST
STEAM

- l 1
STEAM
ENGINE

70 CONDENSER rOCONdWSER
Fig. 3.97 Fig. 3.98

Fig. 3.98 (a) shows the use of industrial steam turbines to apply
process steam and generate by product power. C1 and C2 are the
condensers and P1 and P2 are the condenser pumps.
Condensing
Non-condensing sin extraction
turbine turbine Condensing double
Steam extroctpon turbine
High
Boiler pressure

-
drive Intermediate
pressure
Process
______ steom
uses
L
Heater pressure
it r

J
- Water
Feed pumD

Fig. 3.98 (a)

3.79 Overall Thermal Efficiency

The overall efficiency of steam power plant depends upon the
efficiency of boiler, turbine and alternator. The heat produced due
to the burning of coal is not fully utilised for generating electrical
energy because there are heat losses in the boiler turbine and
mechanical and electrical losses in the turbine. The overall thermal
efficiency of steam power station is given by the following relation
H
• •Wxt.V.
il = Overall efficiency
where H = Heat equivalent per kWh.
W = Amount of coal consumed per kWh.
278 POWER PLANT

C.V. = Calorific value of coal.

3.80 Heat Flow
A power plant consists mainly of equipment for energy flow and
transformation. The basic modes of heat transfer are as follows
(a) Conduction (b) Convention
(c) Radiation.
All the three forms of heat transfer are found in power plant
engineering usually in combination with each other. The principal
heat transfer calculations are associated with the following
(i) Radiation from fuel bed and luminous gases to boiler tubes
and water walls.
(ii) Radiation and conduction in heat insulators like refrac-
tories and pipe coverings.
(iii) Conduction of heat through boiler, economiser and air
preheater surfaces.
(iv) Conduction of heat through condenser or heater tubes.
(v) Heat convection from combustion region to more remote
boiler tubes, economiser, tubes and air preheater surfaces.
Example 3.9. The overall thermal efficiency of a 40 MW turbo
alternator is 30%. If the load factor of the power station is 50% n
coal burnt has a calorific value of 6800 kcal, determine the lln

(i) Coal consumption per kWh.

(ii) Coal consumption per day.
Solution.
Average load
Load factor
iaximurn demand
- Average load
0.50
- 40x1000
assuming maximum demand equal to capacity of the power station.
Average load = 0.5 x 40 x 1000 = 20,000 kW
Energy produced per day = Average load x 24
= 20,000 x 24 = 48 x 10 4 kWh.
Overall thermal efficiency
860
- W x 8600
W = Coal consuThption per kWh.
(l kWh =S6Okcal)
STEAM POWER PLANT 279
860
0.3- iv-
x 8600

%V=--8680600 =O42kg Axis

0.3 x
Coal required per day •
= Energy produced x
= 48 x 104x 0.42 kg.
i:..
48 x 104 x 0.42
= tonnes

= 201 tonnes. Ans.

Example 3.10. In a steam power station the coal consumption
is 0.4 kg per kWh output. If the calorific value of coal is 6800 kcal
per kg, boiler efficiency is 70% and mechanical and electrical efficien-
cy of alternator is 90%. Prepare roughly the heat balance sheet for
the power station.
Solution.
Let output = 1 kWh 4
Heat input = 0.4 x 6100 = 2720 kcal.

Electrical energy input = 3.16 kWh

Losses. (i) Boiler house loss

= 3.16 x 0.3 = 0.948
(as boiler efficiency is 70%).
Heat output of steam = 3.16 - 0.948 = 2.212 kWh.

Input to alternator = 111 kWh.

=
(ii) Loss in alternator
= 1.11 - 1 = 0.11 kWh.
(iii) Loss in turbine = 2.212 - 1.11 = 1.102 kWh.
This loss may be considered as rejected to cooling water.
Hence, output = 1 kWh.
Loss in boiler 0.948 kWh.
Loss in alternator = 0.11 kWh.
Loss in turbine = 1.102 kWh.
Total = 1 + 0948 + 0.11 + 1.102 = 3.61 kWh.
280 POWER PLANT

Example 3.11. The daily output of a steam power station is

18 x 10' kWh. lithe coal consumption is 700 tonnes per day, calculate
the thermal efficiency of power station.
Assume calorific value of coal 8500 kcal 1kg.
Solution. Coal used = 700 x 1000 kg.
Energy input per day = 700 x 100Q x 8500 x 595 = 107 kcal
Energy output per day
= 18 x 10 5 kWh
= 18 x 10 5 x 860 kcal.
(as 1 kWh 860 kcal).
- Output - 18 x 10 5 x 860
Efficiency - Input - 595x107 =26%.

3.81 Cost of Steam Power Plant

In recent days vast improvements have been made in generating
electric power from steam. About 0.45 kg of coal is needed to
produce 1 kWh of energy. It is observed that larger capacity power
plants can utilise the thermal energy more efficiently than the
smaller capacity plant. In the design of a thermal power station
future availability of coal and its price has to be taken into account.
A steam power station may cost about Rs. 1600 per kW of
capacity. A typical sub-division of investment cost of a steam power
station is as follows
(i) Turbo-generators and condensers —25%
(ii) Load building and foundations —25%
(iii) Boiler plant —18%
(iv) Fuel handling-S
(v) Piping —5%
(vi) Switch yard, switching and wiring —16%
(vii) Miscellaneous —5%
The investment cost includes the following costs
(i) Cost of land (ii) Cost of building (iii) Cost of mechanical
and electrical equipment and their installation.
Conditions affecting the investment depend on a number of
factors
(a) Characteristics of the site.
(b) Size and number of power generating units.
(c) Fuel storing and fuel handling methods.
 STEAM POWER PLANT . 281

3.82 Heat Balance Sheet for Boiler

It indicates the heat supplied, heat utilised through various
sources and he-it wasted.
Prepare the heat balance sheet for a boiler having the following
data:
Steam pressure = 13 kg/cm2
Water evaporated = 300,000 kg
Coal fired = 30,000 kg
Superheat temperature = 300'C
Air temperature
(a) To air preheater = 45C
(b) From air preheater = 90'C
Feed water temperature
(a) To economiser = 50CC
(b) Form econc.miser =
Flue gas exit temperature = 115C
Boiler house temperature = 25C
Mean specific heat of
the flue gases - 0.25
Coal analysis (as fired)
C = 64%; = 4.5%
02 = 5.5% ; H2 = 1.5%
S = 0.7%;
Ash = 9.8%
Total moisture = 14%
Calorific value of coal = 6800 kcallkg
Duration of trial = 8 hours
Flue gas (dry) analysis
C09 = 12.6%; CO = 0.4%
02 = 6%; N2 = 81.1%.
Solution. (0 To find weight of water evaporated.
Let W = Actual weight of dry flue gas per kg of coal
11CO2+802+7 (CO +N 2 ) (
xC^ — s
3 C0)
11 x 12.5 + 8 x 6 + 7 (0.4 + 81.1) x (064 0.007
.
3(125 + 0.4) + 1.83
= 12.6 kg

Wi = Total weight of flue gas per kg of coal

= Actual weight of dry flue gas
+ Moisture evaporated from coal
—20
 POWER PLANT

+ Water produced from H

= 12.6 + 0.14 + 9(0.045) = 13.14 kg.
Now ash content 9.8%
W2 = Weight of fuel burnt

= 1 - 0.098 = 0.902
W3 = Weight of air supplied per kg of coal fired
W3 = IV I - W2 = 13.14 - 0.902 = 12.238 kg
= Weight of water evaporated per kg of fuel fired
300,000
k
30,000 - 10 g.
(ii) To Calculate Heat Absorbed:
H1 = Heat absorbed by air preheater per kg of
coal fired
= Weight of air x Temperature rise x Specific heat
= w9 x (90 - 45) x 0.24
= 12.238 x 45 x 0.24 = 132 kg cal.
H2 = Heat absorbed by boiler
= Weight of water evaporated per kg of coal
x (Total heat in 1 kg of steam
- Heat in feed water - Heat in air)
= 10 [666 - (120 - 0) - 1321
= 4140 kg cal.
H3 = Heat absorbed by economiser
= Weight of water evaporated
x Temperature rise
= 10 1120 - 501 = 700 kg cal.
H4 = Heat absorbed by superheater
= Weight of water evaporated x [Super heat
temperature - saturated temperature]
x Specific heat
= 10 [300 - 1901 x 0.5
= 10x 110 x 0.5 = 550 kcal.
 283
STEAM POWER PLANT

(iii) To Calculate Heat Lost:

H0 = Heat lost in dry flue gases
= Weight of dry flue gases per kg of coal
x Temperature rise x Specific heat
W x (115 - 25) x 0.25
= 12.6 x 90 x 0.25 = 283.6 kg cal
H5 = Heat loss due to moisture in fuel
= w (100 - t) + L + k (T - 100)
where w is the weight of moisture evaporated from coal
= 0.14 [(100 x 25) + 539 + (115 + 100) x 0.471
= 16 kg cal
(Assuming specific heat as 0.47)
H7 = Heat lost due to water formed by
combustion of hydrogen
= 9H [(100 - t) + L.+ k (T - 10l
where H is weight of hydrogen per kg of fuel, the atmospheric
temperature. T is exist gas temperature and k is the specific heat of
steam and L is latent heat of evaporation
= 9 x 0.045 ((100 -25) + 539 + (115- 102) x 0.471
= 0.4 x 621 = 248.4 kg cal
H8 = Heat lost due to incomplete combustion of carbon

= C X C.V.
(
C() CO
CO
where C is the weight of carbon per kg of fuel, and CO and CO2
represent the per cent of volume of carbon monoxide and carbon
dioxide in flue gases respectively and C.V. is the calorific value of
coal
0.4 ] = 130 kg cal
= 0.64 x 6800[

H9 = Heat unaccounted for losses

=68OO_jHi+H2+H3+H4+H5+H6+17+h'8l
= 6800 - 6170 = 630 kcal.
Example 3.12. A boiler with maximum continuous steam rating
of 1.2 x 10 kg per hour is fired with bituminous coal having I0'4
moisture and 10% ash and a higher calorific value of 7500 kcal per
J84 POWER PLANT

kg. If it takes 1000 kcal to evaporate 1 kg of feed water entering the

boiler and superheat it to final temperature and if overall efficiency
of boiler is 80% determine the following
(i) Hourly coal supply.
(ii) Grate length of building dimensions limit grate width to 6
metre. Given that
that the maximum heat produced is 3 x 106
kcal per square metre per hour.
Solution. H = Heat required to convert water into steam
= 1.2 x 10 3 x 1000 = 1.2 x 10 s kcal per hour
Efficiency (q) or boiler = 0.8
Heat to be produced by the fuel -
- 0.8
H 1 x 108

H 1 = 1.5 x 10 8 kcal per hour

108
Coal required kg = tonnes
- 7500 7500 x 10
-= 20 tonnes per hour
Grate area H
J.5 x108 - 50 square metres
= 3 10 6 - 3
Grate length = Lo = 8.33 metres. Ans.

3.83 Useful Life of Components

Approximate useful life- of some of the components of a steam
power plant are mentioned in Table 3.2
Table 3.2
 STEAM POWER PLANT
285
3.83.1 Power Plant Pumps
In a steam power plant pumps are used for the following ser-
vices
(i) Boiler feed (ii) Circulating water
(iii) Evaporator feed (iv) Condensate
(vi Well water (vi) Ash sluicing,
(viz) Fuel oil.

Power plant pumps are classified as follows:

C1s
() Reciprocating (a Direct acting
(b) Power
(a) Vane
(ii) Rotary (b) Screw
(c) Gear
(d) Irobe
(iu ) Contrfuga1 (a) Volute
b Difiser
(c) Axial flow
(d) Mixed flow
3.84 Plant Layout
Fig. 3.99 shows the layout of power plant. The turbine room
should be sufficiently wide to accommodate various auxiliari In
boiler room enough space should he left for repairs and
tenance. Various equipment should be so placed that acces-
the parts is easy.

7194q0
I J_& 4 17 Mv. MRS
Coat
S/T( BA.QD
ROO.,'

*RQANGCM T

Fig. 3.99

Basic arrangements of various equipments used in power sta-

tion differ from station to station. The following factors should be
considered while installing various components
1. All similar items such as turbines, boiler, transformers
bunker bays and other mechanical and electrical coin-
- -
 POWER PLANT
286
ponents are arranged in parallel lines and at right angles rows
individual boiler, turbo-generator, transformers etc.
2. Chimneys should he erected independently of the station
building with chimney serving two or more boilers.
3. Main flue, draft fans and outdoor precipitators should be
located behind the boiler house.
4. Cirulating water supply, coal supply to bunkers and
lifting equipment should be properly placed.
5. Outdoor generator and unit transformers should be in
front of the turbine house.
Fig. 3.100 shows the cross-section of a typical station housing of
about 60 MW capacity with unit boilers.
3.85 Terms and Definitions
Ii Load Shedding.
Power load on the system in shed if insuf-
ient generating power plant is available to meet the demand for
current load.

BOIL
BUNKER
FURNACE
L) A RA TOP
/ ECONOMISER
CONTROL ROOM

WE
PiI V
/'/RT
PRECIPITATOR

/ A.S. J \. p .F MILL \ ID FAN

GENERATO R F.D FAN
TRANSFORMER

Fig. 3.100

(ii) Furnace rating. It is defined its the heat input to the

furnace expressed usually in kcal/hour per rn 1 Off'-1 1-11 ace volume.
(iii) Lagging. It iS prOCC5S of applying insulating material to the
outer parts of equipment (in which some hot fluid is flowing) to
prevent heat loss and distortion due to unequal temperatures.
(ju) Inhibitor. These are the chemical substances as added to
the turbine oil which check or retard the occurrence of undesirable
properties.
..
STEAM POWER PLANT 287

v) Grit Arrestor. It is the device used for removing grit from

boiler gases before they flow to chimney. The removal of grit and
dust from flue gases is called flue gases treatment,
(vi) Electrostatic Precipitator. The equipment is used for
cleaning boiler flue gases. In this the dust particles of flue gases are
electrically attracted to the metal tubes placed in the path of flue
gases.
(vii) Feed water. Water suitable for feeding to a boiler for steam
generation.
tiii) Make up Water. It is the purified water (distilled or
chemically treated) for replacing losses the f! system.
(ix) Raw Water. It is the unpurified water obtaiiiJ directly
from a natural source. This call used as boiler water after proper
purification.
3.86 Modern Steam Power Station
Central steam power station of larger capacities use higher
boiler pressure with super-heated steam as the super heated steam
contains more heat energy than saturated steam at the same pres-
sure and provides more energy to the turbine for conversion into
electric energy. They use large capacity boilers, and improved
methods of coal firing such as stoker filing or pu. : coal firing.
ilic coal used is of good quality as l.' coal such as those
having high ash contents (about 18—:.0 and above and higher
moisttite contents iahout 30 or more ire not pr- , -. , , reil. High ash
contents in coal reduce the heating val . .al and more labour is
required for the removal of ash from the furnace. Ash with low
fusing, or melting temperature forms clinkers when subjected to
hnti temperatures of fuel bed. The clinkers so formed interfere with
l stokers. The coal to be burnt oil
he movement of fuel o i
fired grates and o il stokers should have a minimum of about 4
to G if ash. Moisture also reduces the heat value of coal. Coals with
7 to 12' moisture are generally burnt oil grate .i:d ivelling
grate stokers. Coal analysis is done to know its CO!:uSitiOn.
Small size of coal have greater tendency to h,,ld the moisture.
Etlicient combustion control equipment is used. ne shape, size and
ttmperature of the furnace used depends on the t y pe of coal to be
burnt, type of burner and its location and t y pe of boiler used.
Furnace mayb e air cooled or water cooled. Induced draft fans and
toned draft fans are used to regulate the flow of air to the furnace
and for exhaust of flue gases. In larger power stations heat recovery
equipment such as economiser and air preheater are used. The flue
gases are made to pass through these devices so that some of their
heat energy ma y be recovered by heating feed water in the
economiser and air supply to the furnace in the air preheater. For
 288 POWER PLANT

eflcient operation of the plant it is essential that the water to be

used in boiler should be free from impurities as improperly treated
water causes corrosion and scale formation etc. which may lead to
failure of the plant. Circulating and feed pump may be reciprocating
or centrifugal type. Multi-stage centrifugal pumps are commonly
used for high pressure. Regenerative feed water heating increases
the thermal efficiency of the plant. In regenerative feed heating
steam is bled from the turbine at different pressure and is used to
heat feed water in the feed water heaters. Condensers are used to
increase the horse power and efficiency of the plant. Air removal
from the condenser is very essential as presence of air in the
condenser reduces the heat transfer action in surface condenser. In
steam turbine plant using surface condenser the air leakage should
not be more than 5 kg for 10000 kg of steam condensed. The main
sources of air in the condenser are:
1. Air dissolved in boiler, feed water is carried by the steam
to the turbine and from there it enters the condenser with
exhaust steam.
2. Injection water ofjet condensers ma y bring some amount
of air dissolved into it. -
3. Air may leak in through the joint; turbine packing gland
or exhaust nozzle connections.
4. Condensing water leakage through tubes is another
source.
l'eecl regu!.itor is employed to maintain the level of water in the
boiler. In such plants back pressure or extraction turbines are used.
3.87 Ways of Increasing Thermal Efficiency of a Steam
Power Plant
The thermal efficienc y of steam power plant can be increased
by the following ways
(i, An increase in the initial pressure of steam raises the
thermal efficiency.
(ii) The thermal efficiency can be increased by raising the
initial temperature of the steam without changing the
pressure.
(iii) Intermediate reheating of steam improves the thermal
efficiency of'the plant. As already mentioned an increase
in the initial pressure of steam improves the thermal
efficienc y of steam power plant but the wetness fraction
of such steam also grows at the end of expansion. The
wetness is so high that it causes wear ofthe blades of the
last stages of a steam turbine. This shortcoming
eliminated b y intermediate reheating of the steam. The
steam from boiler after partial expansion in the first stage

STEAM POWER PLANT 289

turbine is fed into super heater, where it is reheated at a constant

pressure either to the temperature of live steam or to a temperature
slightly below that one. The reheated steam then flows to the last
stages of the turbine where the process of expansion is completed
(Fig. 3.101).

Turbine
Boiler

'ondenser

FEED PUMP

Fig. 3.101
(iv) Thermal efficiency of steam power plant can be improved
by carrying out regenerative heating of the feed water.
Such heating of the water is carried out by using the heat
of steam partl y tapped from the turbine. In this method
(Fig. 3.102) the steam from boiler flows intt am tur-
bine. After partial expansion e of sonic
the ste'nis tapped
from the first stage of the turbine and dire, d to feed
water heater and then to feed tank. The remaini. .g steam
enters the second stage of the turbine where it continues
to expand. At the outlet from the second stage some of the
steam is directed into water heater and then to feed tank.
The other part of steam goes to the third stage of the
turbine and expands there to the final pressure and enters
the condenser. The condensate is delivered by pump to
feed tank.

So E 1 e

Condenser

Fig. 3.102
 POWER PLANT
290

3.78 Indian Boiler Act

The Indian Boiler Act was passed in 1923 and later amended in
1953. Some ut its clauses are as follows
1. Definitions. Following are some of the important definitions
mentioned in the act
(1 )Arcthf itt. Accident means explosion of the boiler or steam pipe
or any damage to the above which reduce the strength and makes
them liable to explode.
(ii) Boiler. Boiler is a closed vessel having capacity more than
22.75 litres and used for generating steam under pressure. It in-
cludes mountings and other fittings attached to such vessel.

(iii) Economiser. It means any part of a feed pipe that is wholly

or partially exposed to the action of flue gases to recover waste heat.

(ti) Feed pipe. It means any pipe under pressure through which
'ed water passes directly to the boiler and that this pipe is not
.ntegral part of the boiler.
(v) Steam pipe. It is any pipe through which steam passes from
the boiler to the prime mover and the steam pressure exceed
,
3.5 kg/cm.
2. BoilerRcgistratiofl. The boiler cannot be fixed unless it has
heen registered. The owner of the boiler has to apply for the registra-
ii, a of boiler to the chief inspector of the boiler. The inspector will
then exanune the boiler and submit the report to the chief inspector.
If the boiler is approved for registration acertificate is issued to time
owner for time use of boiler for a period of 12 months at a given
maximum pressure. The boiler registration number is mentioned in
the certificate.
3. Restriction on Use of Boiler. Restriction on the use of boiler
areas follows
(i) No owimerof a boiler shall use unless it has been registered.
(iio lfthe boiler hasbcen transferred from one state to another.
state it should not be used until the transfer has been
reported in the prescribed manner.
(ju) The boiler should not be used at a pressure than the
maximum pressure recorded in the boiler certificate.
Time boiler should be in the charge of a person holding
competenc y certificate.
-1 Renewal of Certificate. A certificate authorising the use of
boiler shall he renewed under the following condition,,;
()It expir y of the period for which it was granted.
(it) When an y accident occurs to tin- boilers.
m jj \Vhieim the boiler has been moved to another state.

STEAM POWER PLANT 291

(iv) When any structural alteration, addition or renewal is

made in or to the boiler.
(t) When the steam pipe or pipes of the boiler are found to be
in dangerous conditions.

5. Transfer of a Boiler. When a boiler is transferred from one

state to another state, the fact is noted in the register. The new
owner of the boiler shall apply to the chief inspector of the state in
which the boiler is to be installed for the registration of transfer.
The chief inspector then obtains the necessary records from the state
from which the boiler was transferred. On receipts of the record,
entry is made in the registers of inspector and chief inspector.
6. Report of Accident. If some accident occurs to the boiler
owner or person incharge thereof shall, within 24 hours of accident,
inform to the inspector in writing giving full details about it and
injuries, if any caused to the boiler or to the steam pipe or to any
person. The inspector will then investigate the matter and decide
whether the boiler can be reused at the original or reduced pressure
without repairs or pending the completion of any repair or altera-
s
tion. At the time of investigation the in pector can ask questions
from any body and every person shall be bound to answer trudy to
the best of his knowledge and ability.
7. Repairs of Boiler. Sanction of the chief inspector should be
obtained before taking a repair in hand. However a few water tubes
or smoke tubes may in emergency condition be renewed pending the
sanction of the chief inspector. Extensive repairs such as renewal of
furnace, plate and fire boxes etc. should be carried out under the
supervision of the inspector.
8. Alterations and Renewals to Boilers. No structural al-
teration, addition or renewal be made in or to any boiler unless such
alterations, additions or renewal have been permitted in writing by
the chief inspector.
9. Penalties
a) Minor Penalties. The Act provides penalties up to Rs. 100
for the following offences:
(i) To use a boiler without obtaining certificate or a
provisional order.
(ii) To refuse to surrender a certificate.
(iii) To use a boiler which has been transferred from one state
to another state without such transfer having been
reported.
(b) Penalties for illegal use ofboiler. If the owner ofthe boiler
runs the boiler without certificate or runs the boiler at a higher
presire than the allowed shall be liable to the fine up to Rs. 500
 292 POWER PLANT

and in the case of continuing the offence he may be fined further at

the rate of Rs. 100 for each day after the first day on which he is
convicted of the offence.
(c) Other Penalties. A person can be fined up to Rs. 500 for the
following offences:
(i) To use boiler after alterations without informing about
such alteration.
(ii) To fail to mark the registration number of the boiler.
(iii) To make alterations in a boiler or its steam pipe without
obtaining prior permissions.
(iv) To fail to report any accident to a boiler or steam pipe that
might have occurred.
(v) To tamper with a safety valve of the boiler having a
malafide intention..
(d) Penalty for tempering with registration number
(i) The owner can be fined up to Rs. 500 if he removes,
changes, defaces, or renders invisible the registration
number of the boiler.
(ii) The owner of the boiler S can be fined up to two years
imprisonment or fined or both jibe marks a fraudulent
igistration number on the boiler.
10. Boiler Mountings. The following mountings shall be fitted
on the boiler for the safety of the boiler
(i) Safet y valve—Two
(ii) Water level indicators—Two
(iii) Steam pressure gauges—Two
(iv) Steam stop valve—one
(u) Feed check valve—one
(vi) Blow off cock—one,
(vii) Feed pump—one
(viii) Fusible plug—one
(ix) Valve after super heater—one
(x) For cleaning of boiler manholes, hand holes and sight
holes shall be provided as necessary.
11. Hydraulic Test of Boiler. Each boiler is tested hydrauli-
cally in the presence of the inspector. The hydraulic test pressure is
twice the approved working pressure or one and a half' times
working pressure or one and a half time working pressure plugs
4 kg/cm 2 whichever is less. Each piece before it is fitted in its position
is subjected to a hydraulic test at a pressure of the boiler. The boiler
drum after welding is subjected to a h ydraulic test pressure of one
and a half times the working pressure for a time (not less than half
an hour) sufficient to inspect various seams and connections. If there
 STEAM POWER PLANT
293
is no leakage the inspector allows to use the boiler up to the working
pressure.
3.89 Thermal Power Stations in India
Some of the thermal power stations installed in the country or
under the process of installation are as follows:

3.90 Super Thermal Power Stations

In view of the persisting power crisis in the country and the
resulting acute power shortage the Government of India has placed
emphasis on constructing large super thermal power stations
preferably near mine heads, with ultimate installed capacity of
2000-3000 MW each. The major super thermal power stations
planned in the initial stage are Singrauli, Korba, Ramagudam,
Neyveli, Taicher and Farakka.
 POWER PLANT
294

* 3.90 (a) Singrauli Super Thermal Power Plant

The first of these super thermal power station is Singrauli in
Mirzapur district in the South East Uttar Pradesh. This station
conveniently located near the coal belt thus reducing its dependency
on coal transport, one of the biggest bottle-necks in the power
generation The first stage of Sirigrauli thermal power station
foresees installation of 3 x 200 MW and 2 x 200 MW turbo alter-
nator sets which will he supplied by BHEL.
There are two units each of 500 MW capacity. The plant has a
total capacity of 2000 MW. It will supply power to UP., Rajasthan,
Punjab, Hary ana, HP, and J & K. This project has been executed
by National Thermal Power Corporation (NTPC).
3.91 Korba Super Thermal Power Station
This power station will have a total capacity of 2100 WM. The
first 200 MW unit ofthis power station has started generating power
from December 1982. The next two units each of 200 MW capacity
• has been commissioned by the end of 1983, Korba super thermal
power station is being constructed by National Thermal Power
Corporation (NTPC) near coalpit heads in the central sector. It is
located two kilometres down steam of Darri Barrage to the right of
right bank canal of the Hasdep river in Katghora Tebsil of Bilaspur
district. This power station will supply power to Madhya Pradesh,
Maharashtra, West Gujarat and Goa. For transmitting power 1960
km long 400 KV AC transmission lines will be established. In the
first phase four transmission lines Korba, Bhillai-I, 13 kin
Korba-Korba (West), 190 km long Korha Bhillai-II and 262 km long
Bhillai-Koradi will be constructed. About 8 kin from the power
station site are the Kusumunda n ines which will supply coal fo r the
power station. When the first three 100 MW units of this power
station come up they win use up 8000 tonnes of coal per day and
when the next three 500 MW units start power generation by
December 1988 it will require 2800 tonnes of coal everyday. For
transportation of this huge quantity of coal, a completely automatic
system called merry-go round rail system will be used.
3.92 Thermal Power Plants Environmental
Control
Fuel burnt at thermal power plants contain harmful impurities
which are ejected into the environment as gaseous and solid com-
ponents of combustible products and can adversely affect the atmos-
phere and water. Toxic substances contained in the flue gases
discharge from chimneys of thermal power plants can produce
harmful effects on the whole of complex of living nature. The flue
gases may contain the following
(i) oxides of carbon and hydrogen.
STEAM POWER PLANT 295

(ii) fly ash.

(iii) solid particles of unburnt. fuel.
(iv) oxides of sulphur and nitrogen.
Thermal power plants consume more than of all the fuels
produced and thus can significantly affect the local environment and
the whole of bio-sphere comprising the atmospheric layer near the
earth's surface and upper layers of soil and water basins. The
influence of thermal power plants on the surroundings is deter-
mined by following factors
(i) Ejection of the flue gases, heat and contaminated water.
(ii) Type of fuel used.
(iii) Method of combustion.
(iv) Type of furnace used.
The fraction of solid particles carried off from the furnace with
the flue gases (K) depends on the type of furnace as indicated below

Type of Furwce K)
Horizontal cyclone furnace -0.15
Furnace with vertical
primary chambers . ).7 --0.4
Two chamber furnaces -
Open furnaces with hydraulic
ash disposal 0.7-0.85
Chamber furnaces with
dry ash disposal 0.94
U) Devices used for ejection of flue gases into the atmosphere.
(vi) Efficiency of dust collecting and gas cleaning plants.
The basic characteristic for calculation of environmental effects
of effluents from power generating power plar tue emission of
a particular pollutant per unit time. The toxic substances present
in the flue gases may have harmful effects on vegetation, animals,
people, buildings and structures. For example vegetables are most
sensitive to the content ofSO2 gas in the atmosphere. The toxic effect
of SO 2 gas is associated with deterioration of the surfaces of leaves.
People living in NO2 contaminated areas suffer from reduced
respiratory function, have a higher incidence of respirator y diseases
and exhibit certain changes in the peripheric blood.
The environmental control of the atmosphere at thermal power
plants is mainly aimed at minimising the discharge of toxic substan-
ces into the atmosphere. This will preserve the purity ofatmop}iere
and water basins.
This can be achieved as follows
(i) By decreasing the discharge of solid ash particles Ash
contents of various fuels is different. Modern ash cIIee-

296 POWER PLANT

tors used in steam power plants have a high degree of ash

collection and can thus reduce to a great extent the ash
particles ejected into the atmosphere.
(ii)Contamination of the atmosphere with sulphur oxide can
be prevented both by removing sulphur from fuel and
applying means to clean the flue gases from sulphurous
compounds.
(iii) By properly burning the fuel in the furnace so that com-
plete combustion takes place.
(iv) The impurities should be diluted to concentrations which
can do virtually no harm to both nature and man.
(v) By using better quality of fuel.
(vi) By selection proper equipment like a 1i collectors, chim-
neys etc. and to ensure proper operati of the eqbiipmcnt
so that discharge of effluents to the surroundings is min-
Mum.
The control of the atmosphere at thermal power plants is mainly
aimed at minimising the discharge of toxic substances into the
atmosphere.
3.93 Commissioning of Plants
The present trend is towards the commissioning of entire
mechanical section as well as electrical section simultaneously.
Commissioning of plant is done in the presence of the repre-
sentatives of manufacturers. Procedure for commissioning of steam
power plants is as follows.
(i) Boiler. Before putting a new boiler on load it is desirable as
per Indian Boilers act to conduct a hydraulic test of the boiler.
Hydraulic test for low pressure boilers is carried out at one and half
times the rated pressure of the boiler plus 3.3 kg/cm 2 . A pressure
gauge supplied by the Boiler Inspector is mounted on the boiler.
Checks for proper functioning of outlet of steam blow off, water
treatment plant, valves etc. are made. A heat balance sheet is
prepared to check the performance of boiler.
(ii) Turbine. Steam is admitted to the turbine to turn it slowly.
Turbine speed is then gradually increased with a close watch being
kept on bearing temperature until full speed is reached. Adjust-
ments in lubrication system and governing system are made, checks
for reading of turbine stage pressures are carried out.
(iii) Condensers. Exhaust outlet of turbine is connected to the
condenser. Various auxiliaries of condenser are checked for proper
functioning.
STEAM POWER PLANT 297

(iv) Coal conveying system. The coal conveying system should

be checked for smooth running and for its capacity to deliver the
required quantity of coal.
(v) Circulating water system. Water distribution system for
cooling towers is checked and adjusted for even flow. The service
feed water may be put in operation and performance of pumps may
be checked for pressures and output.
(vi) Alternators. The drying out test is carried out until the
insulation resistance is steady at constant temperature as specified
by the manufacturer. Other tests such as test for insulation of rotor,
over voltage test, phase rotation test, testing of exciters, ventilating
system etc. are carried out.
(vii) Water level floats, alarms, automatic controls and other
auxiliary equipment should be checked for proper operation.
(viii) Foundations for boiler, turbine condenser alternator etc.
should be of proper materials with suitable arrangements for
preventing the vibrations.
In our country National Thermal Power Corporation (NTPC) is
the largest producer of electric powei with a commissioned capacity
of about 10335 MW and a transmission net work of more than 16000
circuit krns. till 1991.
NTPC has set up five super thermal power plants namely
(i) Singrauli (ii) Korba
(iii) Ramagundam (iv) Vindhachal
(v) Rihand
and two gas based combined cycle power plants at
(i) Ahta (ii) Auraiya
At Farakha stage I of three 200 MW units is completed and work
for setting up two 500 MW units is in progress. First unit of2lO MW
of National Capital Thermal Power project at Dadri (Ghaziabad,
U.P.) has been completed and this power plant when fully commis-
sioned will have a total coal based capacity of 840 MW. The 720 MW
Badarpur thermal power plant at New Delhi and 270 MW BALCO
captive power plant at Korba in Madhya Pradesh are also being
managed by NTPC.
Power generation must be in tune with the times and NTPC is
planning to enter Solar Thermal power era. It is presently working
on a solar power plant to be located in Rajasthan.
NTPC has adopted numerous new technologies such as
(i) High voltage direct current (HVDC) transmission system.
(ii) Use of fly ash for construction of dykes.
(iii) Distribution digital control system.
(iv) Micro-processor based system for boilers and turbines.
—21

28 POWER PLANT

(t') Combined cycle for gas based power plants.

Environmental planning and preservation of ecological balance
continues to be a matter of priority for NTPC. Effluents from power
plants are neutralised to ensure that the unacceptable levels of
effluents are not discharge into the surrounding eco-system.
Some of the power plants commissioned and managed by NTPC
are as follows:

Power VV

Some of the other thermal power plants in construction stage

being looked after by NTPC are as follows:
(i) Kahalgaon (Bhagalpur, Bihar)
It is a coal based plant.
(ii) Talcher(Dhenkanal, Orissa)
It is also a coal based plant.
(iii) Kawas (Surat, Gujrat)
It is a gas based plant of 630 MW capacity.
(iv) Dadri (Ghaziabad, U.P.)
It is a gas based plant.
Gas based power plants in operation and managed by NTPC are
s follows
(i) Dadri (U.P.) of 131 MW capacity. Its total capacity will be
817 MW on completion.
(ii) Anta (Rajasthan) of 413 MW capacity.
(iii) Auraiya (U.P.) of 652 MW capacity.
Example 3.13. The following data is supplied for a boiler
plant:
(i) Boiler
Mass of coal supplied = 220 kg /hr.
Calorific value of coal = 7200k cal /kg
Mass of feed water = 200 kg/hr
Enthalpy of steam produced = 670k cal/kg
(ii) Economiser
Inlet temperature of feed water = 18'C
Outlet temperature of feed water = 88C
Atmospheric temperature = 19 C
Temperature of flue gas entering 350'C
Mass of flue gases = 4150 kg/hr.


STEAM POWER PLANT 299

Determine the following

(a) Efficiency of boiler
(b) Efficiency of economiser
(c) Efficiency of whole boiler plant.
Solution. (a) W = Mass of coal supplied = 220 kg/hr
C = Calorific value of coal
= 7200 kcal/kg.
H 1 = Heat obtained by combustion of coal
= %V x C = 220 x 7200
= 1584 x 10 3 k. cal/hr.
= Mass of feed water = 2000 kg/hr.
H2 = Heat utilised in evaporation of steam
= Wi (Ii - hi)
where h = Enthalpy steam = 670 k. cal/kg.
h 1 = Enthalpy of feed water = 88 k. cal/kg.
112 = 2000 (670 88)
= 2000 x 582 = 1164 x 103 k:
= 1164 x 10 k cal.
112 1164x10
----x iO3 x 100
Boiler Efficienc y = - x 100 = -584
H 1 1
= 73.5%. Ans.
(a) Ti = Inlet temperature of feed water = 18C
T2 = Outlet temperature of feed water = 88C
H3 = Heat utilised by economiser
= W1 (?'2 - Ti ) = 2000 (88 - 18)
= 14 x 10 1 k. cal/hr.
W2 = Mass of flue gases = 4150 kg/hr.
T3 = Temperature of flue gases = 350'C
T4 = Atmospheric temperature 19C
K = Specific heat of flue gases = 0.24
H 4 = Heat supplied to economiser
= W2 (T3 - T4 ) x K = 4150 (350- 19) x 0.24
= 4150 x 331 x 0.24 = 329 x 103 k cal/hr.

Mi POWER PLANT

Economiser efficiency
113 14x 104
(b) 100= X 10
329x103
= 42.5%. Axis.
(c) H = Total heat utilised
= 112+113 = 1164 x iO + 14 x io
= 1304 x iO' kcal/hr
= Over all efficiency

= x 100

= 1304 x 103
X 100 = 82.3%. Ans.
1584 x
This shows that by installing economiser the efficiency of the
boiler plant is increased.
Example 3.14. Determine the quantity of air required per kg of
coal burnt in a steam power plant furnace fitted with a 62 ni high
stack. The draft jiroduced is 38 mm of water and temperature of flue
gases is 419'C. Boiler house temperature is 29CC.
Solution. h = Draught = 38 mm of water
W = Weight of air required per kg of coal
burnt
Ti = Absolute temperature of air outside the
chimneys
= 29 + 273 = 302K
T = Average absolute teniperature of
chimney (stack) gases.
=410+273=683K
62 m.
H = Height of chimney =
1 (w+i ii
h=353H[_

i (w^1
jj
38=353x62 _ j xj
1302 w
16 kg.
STEAM POWER PLANT 301

Example 3.15. The following observations refer to a surface

condenser:
Weight of condensate = 1200 kg/hr.
Weight of cooling water = 48,000 kg hr
Mean temperature of condensation = 35C
Condenser vacuum = 700 mm
Barometer reading = 760 mm
Inlet temperature of cooling water = 20CC
Outlet temperature of cooling water = 30CC
Temperature of hot well = 29C
Calculate the following
(a) Weight of air per rn 3 of condenser volume
(b) State of steam entering condenser
(c) Vacuum efficiency.
Solution. P = Absolute pressure in condenser
= 760 - 700 = 60 mm
= 0.8 kg/cm2
Tj = Mean temperature of condensation
= 35C
P 1 = Partial pressure of steam at 35C
= 0.057 kg/cm2
P2 = Partial pressure of air
P - P 1 = 0.08 - 0.057 = 0.023 kg/cm2
Let V = Volume of condenser
rn = Weight of air present in condenser
Using gas equation
P2 V mRT1
'7' P2 0.023 x 10
V - KT 29.27 x (273 + 35)
= 0.025 kg/m'. Ans.
q = Dryness fraction of steam entering the
condenser
= Amount of cooling water
H, = Heat gained by cooling water
= W1 (T2 - T3 ) 48,000 (30 - 20)
= 48 x 10' kcal
302 POWER PLANT

W2 = Amount of steam entering the condenser

= 1200 kg/hr
L = Latent heat of steam at 0.08 kg/cm2
= 574 kcaL'kg (From steam tables)
T = Temperature of steam at 0.08 kg/cm2
= 41.16C
T4 = Temperature of hot well = 29°C
H2 = Heat lost by steam
= W2 (qL -f- T- T41
= 1200 lq x 574 + 41.16- 291
Hi=H2
48 x 10' = 1200 E q x 574 + (41.16-29)]
q=0.67
700
Vacuum efficiency = Ans.
760 - O.057x760 = 0.976 97.6%.
Example 3.16. The data given below refers to a simple steam
power plant (Pig. 3.103).

IM

Boiler Turbine
-3
Condenoer

Th condenrat
PumP
Fig. 3.103
Power of/)unh/) = 100 11. P.
Rate O/ste(iFfl flow = 8 x 10' kg/hr.
The entlzalpr and velocity fr the fluids at different points ofeycle
are tfl(liCute(l in table 3.2 (A)
Table 3.2 (A)
- I.e,tto,i r:fltha!J). (/.(/kg.%4.joeIi.y (_f7!Ie) I
Steam leaving theh.iI,r(l--IP YQ - 80_
,Ste.tiii t. flterIlIg tho , tilrhii'2 2) - . 776 81)
STEAM POWER PLANT 303

Steam leaving the turbine and entering 550

the condenser :)-3) - ------------------1--- - - -----__L --
Water leaving condenser and entering 70
- -
The entliulpy and velocity of hot water leaving the pump and
entering the boiler (5-5) are negligible. Heat loss from the turbine to
the atmosphere is 40000 heal / hr.
Calculate the following
(ci) f/eat trails/er per hour in pipe line joining boiler and
turbine.
(b) 1/eat transfer per hour in pipe line joining boiler and
condenser.
(c) Power output ofturbine.
Assume all the items to he working at the same level.
Solution. (a) Qi = Heat transfer in pipe line between boiler
and turbine
-
= (H, - 112) x m
where H, = 780 kcal/kg
776 kcal/kg
m = Rate of steam flow = 8 x 10 kg/hr

Qi -. (780 776) x 8 x 101

32 x io kcallhr
(b) W - work done by pomp
400 11.1'. = 400 x 75 x 3600 kgm/hr
- 400 x 75 x 3600 kcaVhr.
427
- 25 x 10' kcatfhr
11, - Enthalpy of water entering the boiler
+ Enthalpyof water leaving the pump
=25x 10 1 +70x8x 10
= 585 x 10 1 kcal/hr
11, - Enthalpy ofsteani leaving the boiler
- 780 x 8 x iO n. 624 x 10 kcal/hr
11 = Heat SUj)plied by tF.e boiler
= - lI-i = 624 x 10 585 x 10
=3655 x 10 1 kcal/hr


304
a POWER PLANT
(c) Q = Heat loss from turbine to atmosphere
40,000
8 x 10' 0.5 kcal/kg of steam flow.
Applying general energy equation
H vi
Vi
As all the items work at the same level
Z2 =
= 80 rn/sec
1)3 =160 rn/sec
H2 = 776 kcal/kg
H3 = 550 kcallkg
W = work done

= (776 550) + - 0.5

2x9.81x427 2x9.8x427
- 226 + (8060j -. 0 5
- 2x9.81x427
= 223 kcaVkg
= 223 x 427 = 95,221 -kgm/kg
= 95,221 x 8 104 kgm/hour
H.P.= 95.221x8x iO
= 28,213.
75x 3600
Example 3.17. In a steam power plant an endless rope haulage
raises 300 tonnes of coal through a vertical distance of 80 metres in
one hour. lithe efficiency ofthe haulage system is 40% and the System
runs for 8 hours per day for 5 days, calculate the following:
(a) H.P. output of system
(b) Power input
(c) Weekly cost of coal raising. The charge for electricity is 25
poise per kWh.
Solution. t Time = 1 hour = 3600 sec.
P = Power output
= nigh = 300 x 1000 x 9.8 x 8
 STEAM POWER PLANT
305
= 2352)( 10']
2352x 10'
= 65 x 103 w
3600
(As 1 W = I J per sec)

Power output (H. P.)= 88

735.5
'1 = Efficiency = 0.4
Power input = = 220 H.P.
Hours per week =8x5=40
Power o ut - 65x103
Power input = -
1 0.4
=162x1O3W
E = Energy consumed
162x io
- j — x4O=648o kWh.
Rate per kWh = 25 paise

Cost 6480 x 25= Rs. 1620.

Example 3.18.A steam power plant uses coal 5000 kg/hr. The
heat conversion efficiency is 30% and calorific value of coal is 7000
k. cal/kg, calculate the electric energy produced per day.
Solution. W1 = Amount of coal used = 5000 kg/h
C = Calorific value of cual
7000 kcal/kg
and Conversion efficiency = 0.3
H 1 = Net heat utilised
= W.C. 1 = 5000 x 7000 x 0.3
= 10.5 x 10 6 kcal/hr
E= Energy produced per hour
kWh
860
(As 1 kWh = 860 kcal)
 POWER PLANT
306
a
E = .1: 0. 103 = 12,210 kWh.

Energy produced per day

=Ex24 12210x24
= 293 x 103 kWh = 293 MWh.
Example 3.19. Find the rate of flow of cooling water and the
cooling ratio for a surface condenser with the following data
Total amount of condensing steam = 17kg1sec.
Temperature of condensate = 25C
Inlet temperature of cooling water = 12C
Outlet temperature of cooling water = 20C
Enthalpy of steam at inlet to condenser = 2400 KJ/kg
Solution. W1 = Amount of steam = 17 kg/sec
W = Weight of cooling water
T = Temperature of condensate
T1 = Inlet temperature of cooling water
= 12'C
Outlet temperature of cooling water
= 20C
if = Enthalpy of steam at inlet to condenser
H1 = Enthalpy of condensate
C Heat capcity of water
= 4.19 KJ/kg deg.
Heat lost by steam = Heat gained by water
W (H -- H 1 ) = W (T2 - T i ) C
17 (2400 - 4.19 x 25) = W (20 - 12) x 4.19
W = 1164 kg
W = 1164
Cooling ratio = = 68.4 kg/kg.

Example 3.20. A steam power plant of 150kW capacity uses gas

of calorific value 1200 kccil/m3.
Determine the volume of gas required per hour when the plant is
running at full load conditions.
 ri

STEAM POWER PLANT 307

Take the overall efficiency of the plant as 30%.

Solution. E = Energy generated per hour
= 150 x 1 = 150 kWh
= Heat required for E
=Ex 860= 150 x860 kcal
(As 1 kWh = 860 kcal)
Let V = volume of gas required per hour
C.V. = Calorific value of gas
= 1200 kcal/m3
Tj = Efficiency of plant = 0.3

112 = Heat produced by burning the gas

112 = V C.V. xii = V 1200 xO.3
II I =112
150 x860 = V 1200x0.3
V = 358 m3.
Example 3.21. Two boilers one with a super heaterand the other
without a super heater are supplying equal quantities of steam into
a common main. The temperature of steam from the boiler with
super heater is 330C and that of steam in the main is 2600. lithe
pressure in the boiler and the main is 15 kg/cm 2 and specific heat of
superheated steam is 0.54 determine the quality of steam supplied
by boiler without super heater.
Solution.
Let q = Dryness fraction of steam supplied by
boiler without super heater
At 15 kg/cm2 (From steam tables)
t = Saturation temperature of steam
= 197.4C
L = Latent heat
= 466.7 kcal
= 200.7 kcal
Let us assume that each boiler supplies one kg of steam into the
common main-
11' :7 Heat of steam

- 308 POWER PLANT

= If + qL
= 200.7 + q x 466.7 kcal.
Now t = Temperature of super-heated steam
= 330 C
054
II = Heat of super heated steam
= 11 + Ci,, (t. - t)
= 667.4 + 0.54 (330 - 197.4) k.cal
= 739 k.cal
H1 =H' +H
= 200.7 + q x 466.7 + 739 k.cal.
Since the temperature of steam in the common main is 260'C
and saturation temperature is 197.4C it is obvious that steam is
still in super heated condition.
H2 1=Heat of 2 kg of steam in the main
= (H + C (t - t)1 x 2
= (667.4 + 0.54 (260 - 197.4) x 2
= 1402.4 k.cal.
Now H1 = H2
200.7 + q x 466.7 + 739 = 1402.4
q=0.91.
Example 3.22. Calculate the efficiency of boiler in which coal
consumption is 65 kg per hour and which generates 370 kg of steam
per hour at 0.93 dryness fraction and at a pressure of 8 bar ablso
lute. Coal used has the following composition per kg.
Carbon = 0.71
Hydrogen = 0.05
Oxygen = 0.11
Sulphur = 0.02
Ash =0.71
Feed water temperature = 24°C.
Solution.
P = Pressure of steam
= 8 bar
STEAM POWER PLANT
309
From steam tables
hf = sensible heat
= 721 kJ/kg
hfg Latent heat of steam
= 2046.5 kJ/kg
x = Dryness fraction
H = Heat of 1 kg of steam
= hf+ x. hfg
= 7211-0.93 x-2046.5
= 2624.3 kJ/kg
HL = Heat supplied to steam per kg
=H- heat contained in 1 kg of feed water
=2624.3- 1x4.18x(24_O)
= 2524kJ/kg
W = weight of steam produced per hour = 370 kg
112 = Heat of steam
= WxH 1 =370x2524
= 933880
Cv = Calorific value of coal

= 3380°C + 144000 + 9270S

-8)
(H
where C Amount of carbon per kg of coal
= 0.71
H = Amount of carbon per kg of coal
=0.05
0 = Amount of oxygen per kg of coal
= 0.11
S = Amount of sulphur per kg of coal
= 0.02
Cv = 33800 x 0.71 + 144000 10.05 --- + 9270 x 0.02
29367.4 kJ/kg
J
W i = Amount of coal burnt per hour
 POWER PLANT
-310

= 65 kg
113 = Total heat input
= c x Wi
= 29367.4 x 65
= 1908855 kJ
Yj = Boiler efficiency

= 112 x 100
H3
933880
= 1908855 x 100
= 49%.
Example 3.23. A boiler working at a pressure of 10 bar generates
2100kg of dry and saturated steam per hour. The feed water is heated
by an economiser to a temperature 91105°C. Coal consumed is 208
kg and calorific value of coal is 30200 kJ/ kg. If 12% of coal remains
unburnt determine:
(a) thermal efficiency of boiler
(b) thermal efficiency of boiler and grate combined.
Solution. x = Dryness fraction = 1
m Rate of production of steam
= 2100 kg/h
AT = Feed water temp. rise
= 105°C
P = Pressure
= 10 Bar
h = Heat of steam at 10 bar pressure
= hg
= 2776 kJ/kg
hfj = Heat contained in feed water
=1x4.18xAT
= lx 418 x 105
= 439 kJ/kg
h i = Heat used to produce 1 kg of steam in boiler.
= h - hfTI
= 2776 - 439
 STEAM POWER PLANT
311
= 2337 kJ/kg
Unburnt coal = 10%
W = Coal consumed = 208 kg
mi = Mass of coal actually burnt

= 208 x 90

= 187.2 kg
rn2 = mass of steam produced per kg of coal actually burnt.

mi
- 2100
- 187.2
= 11.2 kg
H1 = Total heat of steam
= 1712 X h1

=11.2x2337
= 26174.4 kJ
C = Calorific value of coal
= 30200
TIb = Boiler efficiency
H1
= - x 100

26174A
x 100
=
= 86.6%
ilg = Efficiency of boiler and grate
171

= w k-- = 2100x 2337 = 0.78 = 78% . Ans.

t, 208 30200
Example 3.24. Sketch and describe a steam power plant in-
dicating various parts of the plant.
Solution. In a steam power plant, fuel and air enter the power
plant and products of combustion leave the unit. There is a transfer
of heat to the cooling water, and work is done in the form of the
electrical energy leaving the power plant. The overall objective of a

POWER PLANT
312

power plant is to convert the availability (to do work) of the fuel into
work (in the form of electrical energy ) in the most efficient manner,
taking into consideration cost, space, safety, and environmental
concerns.
A schematic diagram of a steam power plant is shown in Fig.
3.104. High-pressure superheated steam leaves the boiler, which is
also referred to as a steam generator, and enters the turbine. The
steam expands in the turbine and, in doing so, does work, which
enables the turbine to drive the electric generator. The low-pressure
steam leaves the turbine and enters the condenser, where heat is
transferred from the steam (causing it to condense) to the cooling
water.
Stock
QSS OUt Air in
4irprehectgr
High pressure
super heated Turbine
steam

-- Economrser
I-jot
[w t e r
Hot
aIr- Low.press4ire
steam
.
- Super-heater

Fuel—
-JA
- - Condenser
Cooling
water
Out
__.1_
Steam boiler
Water

Pumps Cooling
water

Cooling water
from river or
take or cooling
tower

Fig. 3.104

PROBLEMS

3.1. (a) What are the different types of coal conveyors? Describe the
construction and operation of belt-conveyor and screw
conveyor.
(b) Describe a grab bucket elevator.

STEAM POWER PLANT 313

3.2. (a) What is meant by 'over feed' and 'under feed' principles of
firing coal?
(b) What are the different methods of firing coal ? Discuss the
advantages of mechanical methods of firing coal.
(c) Make neat sketch and explain the working of:
(i) Chain gratestoker. (ii) Spreader stoker.
(iii) Multi retort stoker.
((I) What is Fluidised Bed Combustion system. Sketch and
describe a Fluidised Bed Combustion (FBC) system. State
the advantages of FBC system.
3.3. Describe the various types of grates used with hand fired fur-
naces.
3.4. Name the various methods of ash handling. Describe the
pneumatic system of ash handling. Why it is essential to quench
the ash before handling?
3.5. (a) Describe the various methods used to fire pulverised coal.
(b) Make a neat sketch of ball and Race mill and explain its
working.
(c) State the advantages of pulverised fuel firing.
3.6. Name the different types of coal-pulverising mills. Describe
Ball-Mill.
3.7. Describe the various types of burners used to burn pulverised
coal.
3.8. Name various draught systems. Describe the operation of a
balanced draught system.
3.. What is the cause of smoke ? State the factors .cessarv for its
prevention.
3.10. Name the different types ofchinineys used. Star ic advantages
of steel chimney. Derive an expression for the height ofchirnney.
3.11.(a) What are the harmful effects caused by using impure water
in boilers? Describe the various methods of purif ying feed water.
(b) What is meant by make up water of boiler and how is this
water fed into a boiler?
3.12. Describe the various methods used to control the degree of
superheat. Name the advantages gained by using super-heat
Steam.
3.13. What is condenser ? Name the different types of condenser.
Describe the operation of(i) Surface condenser (ii) Jet condenser.
3.14. What are the different types of cooling towers used in . a steam
power plant. Discuss their specific advantages.
3.15. What is asteam trap? Where it is located? Describe Ball Float
steam trap.
3.16. What are the requirement ofa well designed pipe line in a steam
power plant. Name and describe the various expansion bends
used in piping steam.
3.17. What are the advantages of using large capacity boilers ?
Describe the operation of:
(i) Velox Boiler (u) Benson Boiler (iii) Loeffler Boiler.
—22
314 POWER PLANT

3.18. State the advantages and disadvantages of a steam power sta-

tion as compared to hydro-electric power station and nuclear
power station.
3.19. Describe the various factors which determine the location of a
steam power station.
3.20. Write short notes on the following:
(a) Cyclone and collector
(b) Industrial steam turbines
(c) Hydraulic test of boiler
(d) Draught fans
(e) Steam separator
(f) Economiser
(g) Cyclone fired boilers
(h) Pressure Filter.
(i) Air preheater
(j) Pipe fittings
(k) Heat flow in steam plant.
3.21. What is the difference between water-tube and fire tube boilers
? Describe the working principle of Cochran Boiler or Lancashire
Boiler.
3.22. (a) how will you classify various types of boilers?
(b) Wr short notes on the following:
(i) Efficiency of boiler
(ii) MaThtenance of boiler
(iii) Accessories of a boiler
(ii') Overall efficiency of steam power plant
(u) Steam turbine specifications
(vi) Feed water control
(vii) Causes of heat loss in boiler.
3.23. What is a superheater?
(a) Describe three types for superheaters?
(b) State the advantages of superheated steam.
3.24. (a) What is a steam turbine?
(b) How are steam turbines classified?
(c) Explain three methods of steam turbine governing.
3.25. Explain the methods used to increase thermal efficiency of a
steam power plant.
3.26. Write short notes on the following:
(a) pH value of water.
(b) Power plant pumps.
(c) Steam turbine capacity.
(d) Comparison of forced and induced draft system for boiler.
(e) Principles of steam power plant design.
(f) Korba super thermal power station.
(g) Singrauli super thermal power plant.
3.27. Determine the quantity of air per kg of coal burnt in a furnace
if the stack height is 58 m and draught produced is 35 mm of
water. The temperature of flue gases is 380 C.
STEAM POWER PLANT 315

Also calculate the draught produced in terms of height of a

column of gases if the boiler house temperature is 27CC.
3.28. Discuss piping system of a power plant.
3.29. What is blowing down of a boiler? How will you determine blow
down?
3.30. Describe a water tube boiler and a fire tube boiler.
3.31. Describe a feed water regulator.
3.32. In a boiler, give the flow diagram for
(a) Flue gas flow.
(b) Water steam flow.
3.33. State the advantages of high pressure boilers.
3.34. Write short notes on the following:
(a) Steam turbine performance
(b) Steam turbine generators.
3.35. Describe environmental control of steam power plants.
3.36. Write short notes on the following:
(i) Selection of boiler
(ii) Gas fired boilers
(iii) Modern trends in generating steam.
3.37. State the requirements of a fuel burning equipment.
3.38. State the factors to be considered while selecting a suitable
combustion equipment for a fuel.
3.39. Describe the handling of liquid fuels and gaseous fuels.
3.40. Sketch and describe two types of gas b' ners.
3.41. Sketch and describe a pressure filter f feed water treatment.
3.42. State the effects of air leakage in cond riser.
3.43. Write short notes on the following:
(i) Selection of a condenser
(ii) Sources of air in a condenser
(iii) Condenser auxiliaries.
3.44. Sketch and describe Edward's air extraction pump for a con-
denser.
3.45. Write short notes on the following:
(i) Boiler mountings.
(ii) Hydraulic test of boiler.
3.46. Sketch and describe a schematic arrangement of equipments of
a steam power plant.
3.47. What are the principal requirements of ash handling plant.
3.48. Discuss commissioning of steam power plant.
3.49. Sketch and describe the following:
(i) Cyclone dust collector
(u) Electrostatic precipitator
(iii) Cinder catcher.
(iv) Fly ash scrubber.
3.50. Write short notes on the following:
(i) Steam turbine testing
(ii) Choice of steam turbine.
 Diesel Engine Power Plant

4.0 Introduction
Diesel engine power plant is suitable for small and medium
outputs. It is used as central power station for smaller power
supplies and as a standby plants to hydro-electric power plants and
steam power plants.
The diesel power plants are commonly used where fuel prices or
reliability of supply favour oil over coal, where water supply is
limited, where loads are relatively small, and where electric line
service is unavailable or is available at too high rates. Diesel power
plants in common use have capacities up to about 5 MW.
Fig. 4.1 (a) shows various parts of an I.C. engine. The cylinder
is the main body of the engine where in direct combustion of fuel
takes place. The cylinder is stationary and the piston reciprocates
inside it. The connecting rod transmits the force given by the piston
to the crank, causing it to turn and'thus convert the reciprocating
notion of the piston into rotary motion of the crankshaft.
The valves may be provided
(i) at the top
or (ii) on the side of the engine cylinder.
Fig. 4.1 (b) shows a typical overhead valve assembly.
The cam lifts the push rod through cam follower and the push
rod actuates the rocker arm lever at one end. The other end of the
rocker arm then gets depressed and that opens the valve. The valve
returns to its seating by the spring after the cam has rotated. The
valve stem moves in a valve guide acts as a bearing.
On a four stroke engine, the inlet and exhaust valves operate
once percycle, i.e., in two revolutions of the crankshaft. Consequent-
ly , the cam shaft is driven by the crankshaft at exactly half its
rotational speed.

DIESEL ENGINE POWER PLANT 317

.tn9

Inlet Exhaust

Inlet v bauM
,alve
I
ribustuon
pace

Stan
I flys
is
P
in der

QQOfl
)ifl

oflflecting
rod

:rank pin

rank

Crank co
rank Shaft

Fig. 4.1 (a)

 POWER PLANT
318

h roq

uide

—Corn

Fig. 4.1 (b) Overhead valve mechanism.

4.1 Classification of Internal Combustion (l.C.) Engines

Internal combustion engines can be classified according to the
following criteria
1. Method of Ignition.
According to method of ignition : I.C. engines are of two types
(a) Spark ignition engines
(b) Compression ignition engines.
In spark ignition engines such as in petrol engines the air fuel
mixture is compressed and ignited at the end of compression stroke
by an electric spark. The compression ratio in such engines varies
between 5 to 8. In compression ignition engines or diesel engines as
they are often called air admitted into the cylinder is compressed.
The compression ratio being nearly 12 to 20. The temperature of air
becomes very high due to compression. At or near to the end of
DIESEL ENGINE POWER PLANT 319

compression stroke fuel is injected through an injection nozzle into

the hot air in the engine cylinder. Due to high temperature of air
the fuel oil burns. The burning gases expand do work on the piston
and hence on the load coupled to the engine. The gases are then
exhausted from the cylinder and this cycle is repeated. In I.C.
engines the charge of fuel and air in correct proportions should be
supplied and combustion products should be exhausted from the
cylinder when air expansion is complete in order that fresh charge
may enter the cylinder.
Usually well designed compression ignition engines shows
greater efficiency than spark ignition engines because of their
higher compression ratios. Part load efficiency of compression igni-
tion engines is higher.
2. Cycle of Operation. According to cycle of operation I.C. engines
are of two types:
(a) Two-stroke cycle engine.
(b) Four-stroke cycle engine.
The relative advantages and disadvantages of these engines are
as follows
(i) The working or power stroke is completed in two revolu-
tions of the crank shaft in four stroke cycle engine whereas
in two-stroke cycle engine the working stroke is completed
in one revolution. Thus the power obtained from a two-
stroke engine should be twice that of power obtained from
four-stroke engine but due to charge loss and power
needed to drive scavenge compressor the actual power
obtained from a two-stroke engine is 50 to 60% more than
four-stroke engine. As one working stroke is completed for
every revolution of crankshaft the turning moment on
crankshaft is more uniform in case of two stroke engine
and, therefore, a lighter flywheel serves the purpose.
(ii) Two-stroke engine is lighter is weight and requires less
space than a four-stroke engine of the same power. This
makes it suitable fir marine engines.
(iii) In two stroke engine the power needed to overcome fric-
tional resistance during suction and exhaust stroke is
saved.
(iv) In a two-stroke engine there is more noise andwear.
(v) The consumption of lubricating oil is greater in a two
stroke engine due to large amount of heat generated.
(vi) Two stroke engine is simple and its maintenance cost is
low.
..320 POWER PLANT

(vii) Scavenging is better in four-stroke engine.

(viii) In two-stroke engine the exhaust port remains open for a
very short time which results in incomplete scavenging and
thus dilution of fresh change.
(ix) Construction of combustion chamber is better and simple
in two-stroke engine.
3. Number of Cylinders. According to number of cylinders, they
are classified as single cylinder and mutli-cylinder engines.
Internal combustion engines may have more than one cylinders
such as 4, 6, 8 etc.
For any given engine the number of cylinders are fixed by the
output desired, space available and balancing and torque considera-
tions. With increase in number of cylinders the weight, cost, space
occupied and number of working parts of the engine increase. The
size of an engine is designated by the cylinder diameter (bore) stated
first followed by the length of stroke.
4. Arrangement of Cy linders. According to the arrangement of
cylinders the I.C. engines may be classified as Inline engines,
V-engines, radial engines, horizontal engines etc. (Fig. 4.1).
5. Speed. According to speed I.C. engines may be classified as
follows:
(i) Low speed (upto 350 R.P.M.)
(ii) Medium speed (From 350 to 1000 R.P.M.)
(iii) High speed (Above 1000 R.P.M.)

VERTICAL
tN-LI WE
V- TYPE
1IiI10 11 ojIl
,HORIZONTAL TYPE
(C)
(a) (b)

Fig. 4.1

6. Method of Cooling the Cylinder. According to the method of

cooling the cylinder IC engines are of two types
(i) Air cooled
(ii) Water cooled.


DIESEL ENGINE POWER PLANT

321
7. Purpose. According to the purpose for which to be used they
are classified as stationary, mobile and special.
4.2 Four-stroke Diesel Engine
If four-stroke diesel engine the four operations are completed in
two revolutions of crank shaft. The various operations are as follows:
(1) Suction Stroke. In this stroke in let valve (I.V.) remains open
[Fig. 4.2 (a)] and exhaust valve (E.V.) remains closed. The descent
ing piston draws in a fresh charge of air to fill the cylinder with it.
The air taken in during suction stroke is nearly at atmospheric
pressure. Line ab in the indicator diagram (Fig. 4.3) represents this
stroke.
(ii) Compression Stroke. In this stroke I.V. and E.V. remain
closed. Piston moves up and the air sucked in during suction stroke
is compressed to high pressure and temperature (nearly 3.5
kg/cm 2 and 600C). This stroke is represented by the line bc in
indicator diagram.
(iii) Expansion Stroke. During the stroke Fig. 4.2 (c), IV. and
E.V. remain closed. Injection of fuel through the fuel valve starts
just before the beginning of this stroke. Due to compression the
temperature of air inside the cylinder become. igh enough to ignite
the fuel as soon as it is injected. The fuel is admitted into the cylinder
gradually in such a way that fuel burns at constant pressure. In Fig.
4.3, cd represents the fuel burning operation. The ignited mixture
of air and fuel expands and forces the piston downward. Expansion
stroke is represented by de in Fig. 4.3.

1.V E.V I.V g,E.V 1.V.&EV rv LV.

OPEN CLO.SED •aosEv ClOSED CLOSE OPEN

AIR IN COMPRESSION WORKING EXHAUST

TAKE sTRO,c
(a) (b) (C) (d)
Fig. 4.2
 POWER PLANT
322

- (iv) Exhaust Stroke. This stroke is represented b

y vu in Fig. 4.3.
In this stroke E.V. remains open, Fig. 4.2 (d) and the rising piston
forces the burnt gases out of cylinder.
The exhaust of gases takes place at a pre.;sure little above the
atmospheric pressure because ofrestrictecl area ofexhaust passages
which do not allow the gases to move out of cylinder quickly. Fig.
4.4 shows the valve Liming diagram for a four-stroke diesel engine.
The approximate crank positions are shown when IV., EN., and
fuel valves open and close. I.D.C. represents (inner dead centre) and
O.D.C. (outer dead centre), I,V.O. represents (Inlet valve opens) and
I.V.C. represents (Inlet valve closes). Similarly E.V.O...icans ex-
haust valve open and E.V.C. means exhaust valve closes F.V.O.
represents fuel valve opens and F.V.C. represents fuel closes and
F.V.O. represents fuel valve opens.

EXPANSIO.V
LU çiJO
Cx t'kl 61,
bc. F111
;) .o
C
CY

Oo
COMPRESS ON
EXHAUST
VOLUME
Fig. 4.3 Fig. 4.4

4.3 Two-stroke Diesel Engine

The various operations of a two-stroke diesel engine are shown
iiil Fig. 4_5 During the downward movement of piston (down stroke)
the exhaust port is uncovered and tile removal of burnt gases takes
place Fig. 4.5 (a). Furthe r movement of the piston uncovers tile
transfer port Fig. 4.5 ibi. At this stage the crank case and cylinder
space arc- in direct coifliflUlliCfltiOfl. Tile slightly compressed air ill
the crank case is transferred to the c y linder (at a pressure of about
0.05 kg/ cn gauge) through the transfer port. While the transfer of
change from the crank case to the cvlincic-r is taking place the
removal of products ofcoinbustion is also taking place simultaneous-
till' rejection of burnt gases.
lv, i.e. the incoming charge ill hclpng Ill
this is known as scavenging. As the piston moves upward (up stroke)
the compression of air starts, Fig. .1.5 (c. Near the end of coilipres-
sion stroke [Fig. 4.5 (iIi the fuel is injected and ignition of fuel takes
place due to heat of compressed air. '['lien due to expansion of
products of combustion the piston inovc-s downvard. As the mit
port is uncovered ii fresh change grts (1(t-red ill the crank
case.


DIESEL ENGINE POWER PLANT 323

FUEL VALVE

-CYLINDER
1
EXHAUST PORT
TR4A'S PER
PORT
INLET PORT

Fig. 4.5 (a)

U ELECTOR
a

(b) (c)

Fig. 4.5

" TDC

UJI

e
b 91
US 0
VOLUME
8DC

Fig 4.6 Fig. 4.7



324 POWER PLANT

Fig. 4.6 shows the indicator diagram for two stroke diesel
engine. In this diagram bc represents the compression of air, cd
represents constant pressure combustion line, de represents expan-
sion and exhaust and scavenging are indicate1 by eab.
Fig. 4.7 shows valve timing diagram for two-stroke diesel en-
gine. TDC and BDC represents top dead centre and bottom dead
centre respectively. 1PO means inlet port opens and IPC means inlet
port closes, EPO represent exhaust port opens and EPS represents
exhaust port closes. FAS means fuel admission starts and FAE
means foci admission ends.
4.4 Application of Internal Combustion Engines
Internal combustion engines are used in stationary plants,
marine power plants, in various vehicles and aircrafts, their use in
mobile units being predominant, because of their low size and
weight and low fuel consumption.
4.5 I.C. Engine Terminology
The important terms used in an I.C. engine are shown in Fig.
4.8. The inside diameter of engine cylinder is known as bore Top
dead centre (TDC)
Valves in vertical engine
cover and inner dead
Cleoron.. - ____________,- Cylinder centre (IDC) in
volume-.....
Extreme position
horizontal engine
of position at top
is the extreme
position of the pis-
Cylinder ton on head side of
strrlq
the engine.
Extreme position Whereas bottom
of piston at bottom dead centre (BDC)
Piston rod i n vertical engine
and outer dead
Fig. 4.8 centre (ODC) in
horizontal engines
indicate the extreme position at the bottom of the cylinder. Stroke
is the distance between the two extreme positions of the piston.
Stroke is the distance between the two extreme positions of piston.
Let D = Bore
L = Stroke
V1 = Swept volume = (it/4) D2 x L.
Clearance volume is defincd as the space above the piston at top
dead centre.
V= Volume of cylinder = Vi + V
where Vc is the clearance volume.
 DIESEL ENGINE POWER PLANT
325
4.6 Engine Performance
(i) IMEP. In order to determine the power developed by the
engine, the indicator diagram of engine should be available. From
the area of indicator diagram it is possible to find an average gas
pressure which while acting on piston throughout one stroke would
account for the network done. This pressure is called indicated moan
effective pressure (I.M.E.P.).
(ii) IHP. The indicated horse power (I.H.P.) of the engine can be
calculated as follows:

I.H.P. Pm L.A.N. n
4500xk
where Pm = I.M.E.P. in kg/cm2
L Length of stroke in metres
A = Piston areas in
N = Speed in R. P.M.
n = Number of cylinders
k = 1 for two stroke engine
= 2 for four stroke engine.
(iii) Brake Horse Power (B.H.P.). Brake horse power is
defined as the net power available at the crankshaft. It is found by
measuring the output torque with a dynamometer.
B.H.P. = 2n NT
4500
where T = Torque in kgm.
N = Speed in R. P.M.
(iv) Frictional Horse Power (F.H.P.). The difference of I.H.P.
and B.H.P. is called F.H.P. It is utilised in overcoming frictional
resistance of rotating and sliding parts of the engine.
F.H.P. = IHP - BHP.
(v) Indicated Thermal Efficiency (1j) . It is defined as the
ratio of indicated work to thermal input.
- I.H.P. x 4500
wxC xJ
where W = Weight of fuel supplied in kg per minute.
Cu = Calorific value of fuel oil in kcal/kg.
J = Joules equivalent = 427.
 POWER PLANT
326

(vi) Brake Thermal Efficiency (Overall Efficiency). It is

defined as the ratio of brake output to thermal input.
B.H.P. x 4500
11b =
IVX c x J
(vii) Mechanical Efficiency (11,,). It is defined as the ratio of
B.H.P to I.H.P. Therefore, r = B.H.P./I.H.P.
4.7 Heat Balance Sheet
Heat balance sheet is a useful method to watch the performance
of the plant. Of all the heat supplied to an engine only part of it is
converted into useful work, the remaining goes as waste. The
distribution of the heat imparted to an engine is called as its heat
balance. The heat balance of an engine depends on a number of
factors among which load is primary importance. The heat balance
of an internal combustion erigine.shows that the cooling water and
exhaust gases carry away about 60-70% of heat produced during
combustion of fuel. In order to draw the heat balance sheet of
internal combustion engine, the engine is run at constant load and
constant speed and the indicator diagram is drawn with the help of
indicator. The following quantities are noted
1. The quantity of fuel consumed during a given period.
2. Quantity of cooling water and its outlet and inlet tempera-
tures.
3. Weight of exhaust gases.
4. Temperature of exhaust gases.
5. Temperature of flue gases supplied.
To calculate the heat in various items proceed as follows.
Heat is Fuel Supplied
Let %V = Weight of fuel consumed per minute in kg.
C = Lower calorific value of fuel, kcal per kg.
Then heat in fuel supplied per minute = %VC kcal.
The energy supplied to I.C. engine in the form of fuel input is
usually broken into following items:
(a) Heat energy absorbed in I.H.P.
The heat energy absorbed in indicated horse power, I.H.P. is
found by the following expression
Heat in I.HP per minute
=I.H.P. x 4500
--.--. - kcal.

DIESEL ENGINE POWER PLANT 327

(b) Heat rejected to cooling in water

Let W i = Weight of cooling water supplied per minute (kg)
Ti = Inlet temperature of cooling water in C
T2 = Output temperature of cooling water in C
Then heat rejected to cooling water = W 1 (T2 - T1)
(c) Heat carried away by exhaust gases
Let W = Weight of exhaust gases leaving per minute in kg.
(sum of weight of air and fuel supplied)
T3 = Temperature of flue gases supplied per minute C.
T4 = Temperature C of exhaust gases.
Kp = Mean specific heat at constant pressure of exhaust
gases
The heat carried away by exhaust gases
= W2 x Kp x (T 4 - 7'3) kg cal:
(d) heat unaccounted for (Heat lost due to friction, radiation etc.)
The heat balance sheet is drawnas follows
Item Head units A-ca!
I-

I feat abscrbei b y I I! P
!(b)Heectedtocoohngwater.
4
(c)tleat carried awa y bv exhaust g ases.
d)I [eat unaccounted for (b y difference)________
Total
A typical heat balance sheet at full load for Diesel cycle (com-
pression ignition) is as follows

(i) Useful work = 307c

(ii) Heat rejected to cooling water = 30%
(iii) Heat carried awa y by exhaust gases = 26%
(iv) Heat unaccounted (I-lent lost due to friction, radiation etc.)
= ioc.

Example 4.1.1,1 (2 gas engine the mean effective pressure (m.e.p.)

is 48 kg1cm 2 and the ratio of diameter of piston to stroke is
Calculate the size of four stroke cycle gas engine if it runs at 250
PPM. and its B.H.P. is 16. The mechanical efficiency of tile engine
is 80%.
Solution. Mechanical efficiency,

328 POWER PLANT
=
B.H.P.
flm
= I.H.P.
08-16
I.H.P.
I.H.P. 08 = 20.
Let D = Diameter of piston
L = Stroke
,,LAN.
I . H . P.
4500xk
20- 4.8 x LA. x 250
- 45Q0x2
LA. = 150
where L is in metres and A (area) is in cm2.
D 2
Now
If D is in centimetres.
L =
3 =3 D 1 metres

Now LA. =150

150=xDx--x.D2

3 D3
150= 800

D2 = 150 x 180 =
12,727
3xit
D = 22.7 cm. Ans.

and L=-xD=-x22.7

= 3 x 11.35 = 34 cm. Ans.

Example 4.2. The following observations refer to trial on a

fur-stroke cycle gas engine:
Mean effective pressure = 7 kg1cm2
Fuel gas supplied = 0.24 ,n3/rninute
Calorific value of the gas = 4400 kcal/,n3


DIESEL ENGINE POWER PLANT 329

St,i)k&.' =,50 cm.
Bore = 20 cm.
Speed = 300 R. P.M.
Brake load = 70 kg.
Radius of brake drum 0. metres
Determ inc the folluuin -.
(a)
(b) B.H.P.
(c) Mcchcinual (ft ictcncv.
(d) Thermal (//111 '1ev.
Solution. L -- Stroke length 0.5 rn
D = Bore = 20 cm.

Area (A) - . (20)2 311 c1112.

= mean effective pressure At 7 kg/cn

P. LA V
-. where h = 2
400 x k
7x0.5x3.14 300
. — 3 6,8.
=0 2
2rtNT 2rtx 300
BlIP = = 234
4500 4500
where T = lorque in kgm = 70 x 0.8 = 56 kgm.
Mechanical efficienc y (%)

x 100 1°0 -63 S';

= =36.6
Heat supplied in fueL/minute
= 0.24 x 4500 - 1080 kcal.
Thermal efficiency oil H.P. basis
B.H.P. x 4500 23.4 x 4500 x 100
1080xJ 1080 x427
= 22.7%.

Example 4.3. An internal combustion engine consumes 6 kg of

u'l per hour and!. El. P. of engines is 27. It uses 12kg of cooling water
er minute and th? inlet and outlet temperatures of water being 18 C
nd 48'C respectively. The exhaust gases raise the temperature of
.40 kg of water through 32C. The calorific valuoffw'l used is
0566 kcal per kg. Calculate the indicated thermal efi f cien cy and
I raw heat balance sheet.
23

- 330 POWER PLANT

Solui:ion. Heat in 10(1 Sit1)])li l °( l per minute -

- x 10.560 = 1056 kg cal.

Indicated thermal efficiency

x 4500
11 = --- -- x 100
x 11
27 x,1500 x 100
27' . ; Ans.
1056 x427
I Iet energ y absorbed in indicated horse power

- 1.11.1-1 . - x 450 27 x 4500

J 427
= 284 ken!
11 eat rr cc ted to cooling water
= 12 (18 1$)- 12 X 30 360 ken].
I Ilat -.otid avav b y t- xliau-,t ga-es
. :2 :368.8 kuitl.

Heat Balance Sheet

I111 i' - 1WEn

H-t scjIai it! tat-i 1O'.
Ih-,,t•:a: ,sead III'. 284 26.)O
•Ii.rlrji....tig'attr
Il-at, earl ' -.1 " : Lv ,'xhau.t 3 25.33

liiat i.,ated L,r !3 143 2 13.36

- Ili Ier,iice -
1056 10011.

Example 4.-I. A (I1L'l power lUtI()1l is to sopplv power (/t'lIE(Lfld

I,;') 1.'l I/f/i.- ,,i,rizll c//iet'ncv (ift/epviE.('rgen,-,-at1n f. ii Fi j i t.S
(,i,til(itij f/i,- fi)11011 iig
Wit Of diesel oil rlquiJ'E'd /)(? /1001.
'1/;-' 'etrw energy generated per too tie of (lie foil oil.
I/n.' i-u,,. . . (i/Ui.' o/'/iel oil used is 12,000 kcu/ 1kg.
uijtitiun
• Output
1111)111
:t. h\V
 DIESEL LNGINL POW&.H PLANT 331
Efficienc y = 40%
= Output =
0. 4
Input Input

= = 75 kW

Input per hour = 75 x 1 = 75 kWh.

Now 1 kWh = 860 kcal.
Input per hour = 75 x 860 = 64.500 kcal.
64,500
Fuel oil required =000 = kg.
(b) Input per tonne of fuel oil
= 1 x 1000 x 12,000 kcal = 12 x 10 6 kcal.
12 x 106
= 860 = 13,954 kWh

- . = Output
Efhc ency
Iiput
Output = Efficiency x Input
= 0.4 x 13,954 = 5581 kWh.
4.8 Diesel Engine Power Plant Auxiliaries
Auxiliary equipment consists of the following systems
1. Fuel supply s ystem. It consists of fuel tank for the storage of
fuel, fuel filters and pumps to transfer and inject the fuel. The fuel
oil may be supplied at the plant site by trucks, rail, road, tank, cars
etc.
2. Air intake and exhaust system. It consists of pipes for the
supply of air and exhaust of the gases. Filters are provided to remove
dust etc. from the incoming air. In the exhaust system silencer is
provided to reduce the noise.
Filters may be of dry type (made up of cloth, felt, glass, wool etc.)
or oil bath type. In oil bath type of filters the air is swept over or
through a bath of oil in order that the particles of dust get coated.
The duties of the air intake systems are as follows
(i) To clean the air intake supply.
(ii) To silence the intake air.
(i1) To supply air for super charging.
Th .ntake system must cause a minimum pressure loss to avoid
reducing engine capacit y Sod raising the specific fuel consumption.
Filters must be cleaned periodically to prevent Ieure loss from
 332 POWER PLANT

- clogging. Silencers must be used on some systems to reduce high

velocity air noises.
3. Cooling system. This system provides a proper amount of
Water circulation all around the engines to keep the temperature at
reasonable level. Pumps are used to discharge the water inside and
the hot water leaving the jacket is cooled in cooling ponds or other
devices and is recirculated again.
4. Lubricating system. Lubrication is essential to reduce friction
and wear of the rubbing parts. It includes lubricating oil tank,
pumps, filters and lubricating oil cooler.

AR n i t SURGE TANI(
41A1111
SiIEN111116W
0^ (1 JACKET
Lrç'M
TM
LT'
JA

11 1 1.5 TARTINO
OVER
FLOVj t I.lR TANK
I—i
AIR LO
I
COMPRESSOR LU8PICAThI
L TANK I
FILTER

COOLING
TOWER

FUEL TANK

RAW WATER PUMP

Fig. 4.9 (a)

5. Starting system. For the initial starting of engine the various

devices used are compressed air, battery, electric motor or self
starter. Fig. 4.9 (a) shows the auxiliary equipment of diesel engine
power plant.
4.9 Internal Combustion Engine Cooling Methods
Due to combustion of fuel in the engine cylinder the temperature
of burning gases is too high (nearly 15000 to 2000C). This tempera-
ture may cause the distortion of some of the engine parts such as
cylinder head and walls, piston and exhaust valves and may burn
the lubricating oil. Thus a cooling arrangement is essential to carry
 DIESEL ENGINE POWER PLANT 333
away some of the heat from the cylinder to avoid the over heating.
A well designed , cooling system should provide adequate cooling but
not excessive cooling. A cooling system should
(i) Absorb and dissipate the excess heat from the engine in
order at prevent damage to the engine.
(ii) Maintain sufficient high operating temperature so that
smooth and efficient operation of the engine take place.
It is observed that about 25 to 30% of the heat supplied is
absorbed by the cooling medium. Fig. 4.9 (b) indicates a typical heat
distribution for a reciprocating internal combustion engine.

Heating supplied (100%)

4. .1. 4.
Useful Cooling Radiation Friction
work (30%) . exhaust Loss
(28%) (32%) (10%)
Fig. 4.9(b)
The following points should be noted to achic.e good cooling of
diesel engine.
(i) Adequate quantity of water sF. Ad cortinuously flow
throughout the operation of the c igine.
(ii) The cooling water should not be :rosive to metals.
(iii) The cooling water used for cylinder jackets should be free
from scale forming impurities.
(iv) The temperature rise of cooling water should not be more
than 1 1C and the temperature of water leaving the engine
should be limited to 60CC.
4.9.1 Cooling methods
There are two methods of cooling the I.C. Engines.
(a) Air cooling
(b) Water cooling.
Air Cooling. It is a direct method of cooling. In air cooled
engines fins are cast on the cylinder head and cylinder barrel to
increase its exposed surface of contact with air. Air passes over fins
and carries away heat with it. Air for cooling the tins may be
obtained from blower or fan driven b y the engine. Air moment
relative to engine may be used to cool the engine as in case of motor
cycle engine. About 13 to 15' of heat is lost b y this method. Fig. 4.10
shows air cooling system. Simplicit y and lightness are the ad-
vantages of air cooling. But this s y stem is not as c[i'ctive as water

334 POWER PLANT

cooling. The rate olcoohng depends upon the velocity, quantity and
temperature of cooling air and size of surface being cooled.
Fig. 4.10 (a) shows position of valves, Fins and head in air
cooling system. This system is used in motor cycles, scooters and
aeroplans.

ton

nicer

Fig. 4.10 (a) Air cooling system

Water Cooling. It is the indirect method of cooling the engine.

The various cooling systems used are shown in Figs. 4.11 and 4.12.
Water alter circulating in water jackets (passages around the
cylinder. combustion chamber valves etc.) goes as waste (Fig, 4.111
or in recirculating method of cooling Wig. 4.121 water is continuously
circulated through water.jackets. Water takes up the heat and
leaves fr radiator where it is cooled for recirculation.
4 Water
out

Cylinder

Pi ston
Fins
WQ'i( L.{___
in -
\\ I- J jk

Fig. 4.11
Fq, 4.10

DIESEL ENGINE POWER PLANT 335

.i 'FINS
..h /ET
W4EfKrlsm - --

;
PUMP
DRAIN CCCI'

Fig 4.12

In statwtIarv dical cilguic plants . the water cooling systems

used are as follows
(i) Open or Single Circuit Svt ni. [if sYstem, lFig. 4.131
pump (Iraws tlti' water From (011mg pond and utit the main
enginejackets. \Vat i titer circulat lug t hrottgh the engine return to
the cooling pond
EP/6I'IE
ETh —'--
LO_ .

Fig 4 13

(II) ('/iis&'il or /),nhle ('i,ililt Svsteni 1 . 1 the'. ,vstcin {Ftg i I 1

raw water is made to flow through (li( . hl -at exnu: vheii it t.ikc.
.lj) the heat otjaukt.t water and return-, hack to the cuuliiii. pin1

(NI(/E

i 7i. - P,.
PU....p
Fig 4 14

I feat lot IW N ., : lter cooling N about 25 to 35 ihi imitimit of

heat lost is called jacket li,s The mate ot flow of waler '.hi,tilil h•
adju.tcd that tlit outlet ttritpiratirc if cooling V- , 1 d nit
('Xc('Ud GO ( it!I(I ike in t imiperatulmi of coolin g watci i lieiittd to

11 (' Tin water it-oil fir coolin g i ir j iis Almild Im . tree trout
i m put it its Water Ii it miut h. id, pit mm it it niti ' riir it I alit


- 336 POWER PLANT

uoi!irIg creati s troubles inrv cold weather. Cooling efficiency is

reduced (lilt.' to scaling in the pipes, jackets and radiator. Engine
efiicierv affected by power needcd to drive the water pump and
i'udator fan.
'losed S stein of cuolini is mostl y Used in power stations. A
closed ci iol I uig S y st etil conipri st'S tIli' following equipment.
A sare tank.
ii SofI water circulating pump.
Soft \V,ittr circulation pipe.
i ii') Soft Wjtt • I hi-at exchanger or cooler.
i t I Raw watir -dtenin plant.
(Li)}?lW Watcr circulation pUmp.

lit Raw water circulation pipe.

' it 1 Raw water cooling arrangement such as cooling tower.
ix Tlirmiin,ter for measuring inlet and outlet tempera-
t I I Fl

ii1ijurittirp regulator to control fil e excessive jacket

are.
xi ! Iv (livUe to control the excessive jacket temperature
'lijis v>tc;it lIiwn in Fig. -1.1-1 ut use soft water for jacket
ciilia. 'lia lt jc' kt v. itcr from the engine is passed thrugh the
ciiultr ill-it l.,Liiarecr Whc re it is could with the help of raw water.
The raw \vat I 10 turn is cooled liv coiilin i towers,

C0011119 tower
Soft .. ,j

-; Lcc
Wtr L iflicotr
--- Li ( Efl ( I i Pe
Z
rCW .- / LJ
J

lump

co!i h2
L' Raw %vcter
:m<e u rc waTer bud in

Fig. 4 14 ta
 DIESEL ENGINE POWER PLANT
337
4.10 Lubrication
Frictional forces causes wear and tear of rubbing parts of the
engine and thereby the life of the engine is reduced. This requires
that some substance should be introduced between the rubbing
surfaces in order to decrease the frictional force between them. Such
s ubstance is called lubricant The lubricant forms a thin film be-
tween the rubbing surfaces and prevents metal to metal contact.
The various parts ofan I.C. engine requiring lubrication are cylinder
walls and pistons, higend bearing and crank pins small end bearing
and gudgeon pins, main bearing cams and bearing valve tappet and
guides and timing gears etc. The functions of a lubricant are as
follows :
1. It reduces wear and tear of various moving parts by
minimising the force of friction and ensures smooth run-
lung of parts.
2. It helps the piston ring to seal the gases in the cylinder.
3. It removes the heat generated due to friction and keeps
the parts cool.
The various lubricants used in engines are of three types
(1) Liquid Lubricants.
(ii) Solid Lubricants
(It:) Semi-solid Lubricants
Liquid oils lubricants are most commonly used. Liquid
lubricants are of two t ypes: (o) Mineral Oils (h) Fatty oils. Graphite,
white lead and mica are the solid lubricants Semi solid lubricants
or greases as they are often called are made from mineral oils and
fatt v-oils.
A good lubricant should possess tile following properties
(t) It should not change its state with change in temperature.
(ii) It should maintain a Continuous films between the rub-
bing surfaces.
(iii) It should have high specific heat so that it can remove
maximum amount of heat.
(iv) It should be free from corrosive acids.
(v) The lubricant should be purified befoi cit enter the engine.
It should be free from dust, moisture, metallic chips, etc.
The lubricating oil consumed is nearl y 1%of Hlconsump-
tion.
The lubricating oil gets heated because of friction of moving
:;arts and should be cooled before recirculation The cooling water
used in (lie engine may be used for cooling the lubricant. Nearly
2 of heat of fuel is dissipated as heat which is removed by the
lubricating oil.
 POWER PLANT
338

Lubricating oil is purified bV following four methods

i ) Settling,
(ii) Centrifuging,
(iii) Filtering,
(iv) Chemical reclaiming. The centrifuging widel used gives
excellent purification when properly (lone.
Fig. 4.15 shows the lubricating oil external circuit.

HEATER

DIESEL
OIL COOLER J—iJ ICONTRIFU6At

I 11 II R

FLURIC11TN&
OIL TA5

Fig. 415

4.11 Engine Starting Methods

Spark ignition engines (Petrol engines) are ti-ecl iiiainlv in
smaller size where compression ratio to he our coini ill cranking is
only 5 to 7. hand and electric motor (6 -- 12 \', ci - ) cranking are
practical. Diesel engines are difficult to he started b y hand cranking
because of high compression required and therefore mechanical
cranking sy s tem is used.
The various methods used for the starting of diesel engine are
as follows
1. Compressed Air S y stem. Comp i'sst'd air ,X,40111 is used to
start large diesel engines. In this system compressed air at a
pressure of about 20 kg per sq. cm is supplied from all bottle to
the engine an inlet valve through the distributor or through inlet
manifold. In a multi-cylinder engine compressed air enters 0111'
cylinder and forces down the piston to turn the engine shaft.
Meanwhile the suction stroke of some other cylinder takes place and
the compressed air again pushes the piston of this c y linder and
causes the engine crank shaft assembly to rotate. Gradually the
engine gains momentum and by supplying fuel the engine will start
running.
2. Electric Starting. Electric starting arrangement con-osts of
an electric motor which -,I drives pillion which engages a toothed rim
oil flywheel. Electric power suppl y for the motor is made
yen FrI w the engine. In
all electric generator dri ven
available by a small
 DIESEL ENGINE POWER PLANT 339

case of small plants a storage battery of 12 to 36 volts is used to

supply power to the electric motor.
The electric motor disengages automatically after the engine
has started. The advantages of electric starting are its simplicity
and effectiveness.
3. Starting by an Auxiliary Engine. In this method a small petrol
engine is connected to the main engine through clutch and gear
arrangements. Firstly, the clutch is disengaged and petrol engine is
started by hand. Then clutch is gradually engaged and the main
engine is cranked for starting. Automatic disengagement of clutch
takes place after the main engine has started.

4.12 Starting Procedure

Actual process of starting the engine differs from engine to
engine. Some common steps for starting the engine are as follows:
1 Before starting the engine it is desirable to check fuel
s y stem, lubricating system and cooling water supply.
2. Depending upon the method ..( starting a check for the
same is essential. [fair.-art ing is t. I pressure ufair
should be checked arni also the air system should he
checked for possible le.kage. The 5 : rIge battery should
be checked if electric rnc::d
Ir is used for starting.
3. There should be no load oil engine.
4. Crank the engine and run it at slow speed for a few
minutes and again check the working of various systems
such as fuel, lubricating oil s ystem etc.
The speed of the engine should he grdullv increased till it
svnchronises with the bus bars. Then connec: . 'i'nerator to the
bus bars and finally increase the engine speed so t;.at it takes up
the desired load.
4.12.1 Stupping the Engine
The engine should not he stopped abruptly. Top stop the engine
the speed should be decreased gradually until no power is delivered
b y the alternator. Then the engine is disconnected fromthe bus bars
and is allowed to run idle for some time.

4.13 Starting Aids

Starting aids may be used during cold weather to obtain quicker
starting of the engine. Ethyl ether is mostly used as such aid. Glow
plugs are another starting aid. Glow plug forms a local hot spot thus
initiating the combustion of fuel even if the compression tempera-
ture of air is insufficient.
340 POWER PLANT

4.13.1 Warming up of Diesel Engine

The diesel engine should be allowed to warm up for four to five
minutes after the engine has started. During this time the following
points should he checked.
(i) To check whether the firing is correct in all cylinders
ii) To check the operation of fuel pump
(iii To check the cooling water s ystem, circulating water
pump etc.
(iv) To check the lubrication system
(t') To check the colour of exhaust gases etc. to know whether
the combustion is proper.
After these checks the engine should be put on load. Then the
speed of engine should be gradually increased in order to
syiihronise the incoming generator with the station bus bars.
4. 4 s.C. Engine Fuel
The internal combustion engines use both oil and gas fuels, the
former being predominant. The oils used are specified according to
the following properties
(i) Cetane number '(ii) Viscosity
(iii) Volatility (it') Pour point
(v) Flash and fire point (vi) Heating value
(vii) Distillation test (viii) Aniline point
(ix) Conradson Carbon (x) Ash
Cetane number indicates the ease with which fuel ignites when
injected whereas viscosity is significant in oil-handling, volatilit y is
an index of case with which the combustible fuel air mixture can be
prepared and pour point indicates the temperature at which oil
flows. Fire hazards of an oil depend on flash and the fire point.
Heating value meases the thermal energy in fuel and the distil-
lation test is carried out jo indicate whether oil contains any heavy
ends of the refining process which generally burn poorly. Aniline
point of oil fuel is used in calculating the diesel index which
measures ignition quality. Conradson carbon indicates fire extent
of components in oil with a tendency to form carbon deposits and
ash indicates the components of oil believed to cause cylinder wear.
4.15 Fuel Supply
The fuels used in I.C. engines are in liquid form. The y are
preferred because of their high calorific value and case of storage
and handling. The storage of oil fuel is simpler than the solid fuel.
The amount of fuel to be stored depends upon the service hours and
varies for different installations. Bulk storage and engine day tanks
hold the engine fuel. The fuel delivered tot he power plant is received
in storage tanks. Pumps draw the oil from storage tanks and supply
 DIESEL ENGINE POWER PLANT
341
it to the smaller day tanks from where the oil is supplied to the
engine as shown in Fig. 4.16.
Fr/EL
(JNLOAOwgG
FUEL
STORAGE
TANKS

TO ENGINE
Fig. 4.16
The fuel oil used should be free from impurities. Efforts should
be made to prevent contaminator of. [, feel. An important Step is
to reduce the number of times the fuel is handled Greater amount
of impurities settle down in the storage tank and remaining im-
purities are removed by passing ti oil through filters. Storage tank
may be located above the ground or underground. But underground
storage tanks are preferred. Fig. 4.17 shows an underground storage
tank. It is provided with coils, heated by steam or hot water to reduce
the viscosity and to lower the pumping cost. Main hole is provided
for internal access and repair. Vent pipe is provided to allow the
tank to breathe as it is filled or emptied. Level indicator measures
the quantity of oil in the tank, and an overflow line is provided to
control the quantity of oil.

4.16 Diesel Engine Fuel Injection System

OIL LEVEL
INOICA TOP
VENT PIPE
CONCRETE
ENGINE ROOM FOOP
Putt FROM OVER Ft OW
UNLOMLING
Pu'lp STEAM FOR
110(. HEATING
S TOPMGE
TANK

Fig. 4.17

- 342 POWER PLANT

The fuel injection system should be such that adequate quantity

of fuel oil is measured by it, atomised, injected and mixed with the
fuel oil because even the smallest particles of dirt can completely
damage the fuel injection system.
The various system used for injection of fuel are as follows
(i) Air injection
(ii) Solid (airless) injection
(i) Air Injection. In this system a multistage compressor
delivers the air at a pressure of 70 kg/cm' into the fuel nozzle. The
fuel supplied by the fuel pump into the fuel nozzle is thus discharged
into the engine cylinder.
(ii) Solid Injection. In this system the fuel is sprayed into the
engine cylinder at a pressure of about 100 to 140 kg/cm'. Solid
injection systems are available in three types
(i) Unit injector.
(ii) Pump injection.
(iii) Distributor injection.
ROCKER ARMS (i) Unit Injector. In this sys-
PUSH CONTROL
tem a pump plunger is actuated by
RACK a cam through a push rod and rock-
ROD er arm mechanism. The plung"r
moving in a barrel raises the p
PLUNGER sure of fuel oil meters the q
of fuel and controls the inj
timing. There is a spring lot.
CAM delivery value in the nozzle. Th
SHAFT
valve is actuated by the change in
Fig. 4.18 fuel oil pressure (Fig. 4.18).

INJECT/ON
NOZZLES

HIGH PRE.SUPE
FUEL L'NE

CONTROL
RACK
PUMP
CAMSHAFT
PUMP WITH AN INDIVIDUAL
CflINDER FOR EACH NOZZLE
Fig. 4.19
 DIESEL ENGINE POWER PLANT 343
(ii) Pump Injection. In this system individual pump is
provided for each nozzle. The pump measures the fuel charge and
controls the iiljection timing (Fig. 4.19).
(iii) Distributor Injection. In this s ystem (Fig. 420) a pump
measures and pressurises the fuel and supplies to it the various
nozzles through a distributor block.

METMIN
OISTRIBUT
AN PR. 15SU,?
&OCK
PUMP
-.

PPIM.4R y ptp
Fig. 420

4.17 Fuel Injection Nozzle

Fuel injection takes place through very fine holes in the nozzle
body. There are several types 0Ie1 injection nozzles. Two common
types are multihole nozzle [Fig. 4.21 (a)] and piutie nozzle [Fig. 4.21
(b)]. In multihole nozzle each spray orifice produce a dense and
compact. spray. In pintle nozzle, fuel conies out in the form of conical
spray.

MULTI HOLE NOZZLE P/NTLENOZZj(

(a) (b)
Fag. 4.21

4.18 Filter and Silencer Installation

Fig. 1.22,tiows a t y pical filter and silvnc-r installation for diesel
engine .'J'lu air system begins with an intake located outside the
building provided with a filter which ma y he oil impingement, oil
path of div tVpe filter. The function of the filter is to catch dirt hV

344 POWER PLANT

causing it to cling to the surface of filter material. A sileiicr

e is
provided between the engine and the intake.

Silencer
Exhaust
manifold 4ir duct
Filter

Louvres

Diese' engine-
Fig. 4.22

4.19 Advantages of Diesel Engine Power Plant

The various advantages of the diesel engine power plants are as
follows -
1. Plant la yout is simple.
2. In thisplant haiidliugoffuel is easier. Small storage space
for fuel is required, there is no refuse to be disposed off
and oil needed can be easily transported.
3. It can be located near load centre.
4. A diesel engine extracts more useful work from each heat
unit than other types or I.C. engines. Therefore, it be-
comes an attractive prime mover wherever first cost is
written off and operating cost is important.
5. The plant can lie quickly started and call up load in
very short time.
6. There are no standby losses.
7. It does not require large amount of water for cooling.
8. The plant is smaller in size than steam power plant for
the same capacity.
9. The operation of the plant is easy and less labour is needed
to operate the plant.
10. Compared to steam power plant using steam turbine, the
life of diesel power plant is longer.
11. Diesel engines operate at higher thermal efficiency as
compared to steam power plants.
Disadvantages
1. Diesel oil is costly
 DIESEL ENGINE POWER PLANT
345
2. The plant does not work satisfactorily under overload
conditions for longer times.
3. Lubrication cost is high.
4. The capacity of plant is limited.
4.20 Site Selection
While selection the site for diesel engine power plant the follow-
ing factors should be considered
1. Distance from load centre. The plant should be located near
the load centre. This will minimize the cost of transmission lines,
the maintenance and power losses through them.
2. AL'ailabiizty of water. Water should be avaiiahlo in sufficient
quantity at the site selected.
3. Foundation condition. Sub-soil conditions should be such that
a foundation at a reasonable depth should be capable of providing a
strong support to the engine.
4. Fuel transportation. The site selected should he near to the
source o f fuel supply so that transportation charges are low.
5. Access to site. The site selected should have road and rail
transportation facilities.
The site selected should be away from the town so that the
smoke and other gases coming out of the chimneys do not effect the
inhabitants,
4.21 Layout
La y out of diesel engine power plant is shown in Fig. 4.23.
Generally the various units are installed with parallel centre lines.
r
Storage & p ace for
Shop expansion

Wash room Unit no 3

Switch
board
[
Unitno?
1 0
Office

HU
I
L Unit no 1
1 0Air
compressor
Oil storage
Tanks

Front Entrance
Fig, 4.23
—24


_346 POWER PLANT

Some space is left for future expansion. Sufficient space should be

provided around the various units for dismantling and repairing the
engine. The engine room should he provided with adequate ventila-
tion. Fuel oil storage tanks may he located outside the main build-
ings.
4.22 Applications of Diesel Engine Plants
1. The y are quite suitable for mobile power generation and
are widely used in transportation systems consisting of
rail roads, ships, automobiles and aeroplanes.
2. They can be used for electrical power generation in
capacities from 100 to 5000 H.P.
3. They can be used as standby power plants.
4. They can be used a peak load plants for some other types
of power plants.
5. Industrial concerns where power requirement are small
sa y of the order of 500 kW, diesel power plants become
more economical due to their higher overall efficiency.
Diesel power plant is quite suitable at places where
(i) Fuel prices or reliability of fuel supply favour oil over coal.
(ii) Water upply is limited.
(iii) Loads are relatively small.
(iv) Power from other power plants such as steam, hydro
power plants etc. is not available or is available at too high
rates.
4.23 Cost of Diesel Power Plant
Cost of any power plant changes rapidly when there are infla-
tionary tends in the natiom's currency and cost becomes out-of-date
far more rapidly than technical information. A diesel engine power
plant may cost about Rs. 1500 to Rs. 2000/kW of capacity. The major
prt of the cost in diesel engine power plant is that of engine
generator set. Approximate sub-division of investment cost for
various items may be as follows
Jkrn - A pprutrnat.oct") I
U) Engine Generator Set 90
Jjj) Cooling system, fuel systein and other 10
auxiliaries
Land, building and foundation
Switchin g and wiring -

4.24 Testing Diesel Power Plant Performance

The performance of the engine is dependent on engine speed,
compression ratio, weight of inducted air and friction losses. The
110AVIv purchased equipment is tested for various standards set up
DIESEL ENGINE POWER PLANT 347

by the Indian Standards Institution and other such institutions.

Tests such as checking of preliminary caliberations, accuracy, of
tolerances methods, specific thermal performance, accuracy of
speed control, governor characteristics and cyclic irregularity are
conducted to know whether the equipment supplied is up to the
standards specified. Careful supervision of the equipment used for
recording temperature, pressure and electrical data are essential.
The temperature inside the engine should not be allowed to exceed
safe limits as diesel engine is an all metal machine and there is no
refractory protection. Incorrect working of pressure gauges, ther-
mometers and automatic warning signals is very harmful. For
testing the cycle of the engine mechanical indicators are used for
low speed and for higher speed electronic inclijtors are used.
Electronic indicators given pressure time data which can be con-
verted intu pressure volume (p.u.) data by graphical devices and
then .iean effective pressure, power, valve action etc., can be deter-
mined.
The important items can be measured for predicting the perfor-
mance and making energy balance. They are as follows
(i) Rate of fuel consumption
(ii) I.H.P. (iii) B.H.P.
(iv) Quantity of cooling water and its rise in temperature
(v) Quantity of air (vi) Atmospheric temperature
(vii) Temperature of exhaust gas
(viii) Orsat analysis.
To calculate air consumed by the engine the volumetric efficien-
cy is calculated. To watch the performance of the plant heat balance
sheet is drawn and for this flow of fuel, coolant, exhaust gases,
teiipurature of these flows, quantity of fuel and air recorded. B.H.P.
of the engine connected to the generator is calculated by finding the
output of the generator (measurable by electrical instruments) and
efficiency of the generator. The heat lost due to friction, radiation
etc. can be found from the heat balance sheet.

'S

/
B.H.P. B.H.P

(a.) (b) (C)
Fig. 4.24
 348 POWER PLANT

The typical performance ofa diesel engine is shown in Fig. 4.24.

The variation of mechanical efficiency(nm), brake thermal efficiency
(i) and specific fuel consumption (S) with B.H.P. is indicated in the
figure.
4.25 Log Sheet
It is the official record of instrument reading and operating
details entered up by the plant operator.
4.26 Advantages of I.C. Engine over Steam Engine
Both I.C. engine and steam engine are basically heat engines
but in I.C. engine the combustion of fuel takes place inside the
engine cylinder whereas the combustion of fuel in steam engine
takes place outside the cylinder. In I.C. engine the pressure and
temperature inside the cylinder is very high and therefore, construc-
tion material with better resistance are required. The various ad-
vantages of I.C. engine over steam engine are as folTows
1. I.C. engine has higher efficiency ranging from 35 to 40%
whereas the efficiency of steam engine lies between 15 to
20%.
2. I.C. engines has low weight to power ratio due to its
compact design.
3. I.C. engines are usually single acting and hence there is
no necessity of stuffing box glands for piston rod.
4. To start a steam engine firstly the boiler is to he fired and
steam to be raised whereas I.C. engine can be quickly
started.
'4.27 Plant Maintenance
Diesel engine power plant maintenance depends on various
factors. Careful supervision of the equipment used for recording
temperature pressure and electrical data are essential. The
temperature inside the engine should not be allowed toexceed the
safe limits as diesel engine is an all metal machine and there is no
refractory protection. The temperature, flow and quality of fuel oil
should be checked from time to time. The fuel oil must be cleaned
from dirt and other impurities by means of filters. Filters may have
fibre element, or cloth or fibre or a combination of cloth and fibre.
When filter element becomes choke it should be replaced by a new
one. Dirt in fuel oil ruins the fine lap of fuel injection pumps and
plugs the injection nozzle orifice. Occasionally, all the fuel should
be drained and the fuel tank cleaned thoroughly. The temperature
and flow of coolant, lubricating oil and exhaust gases should be
checked at regular interval.


DIESEL ENGINE POWER PLANT

349
4.28 Specific Fuel Consumption
One of the most important parameters used for the comparison
of engines and one which is based on power produced or delivered
is the specific fuel consumption. It is defined as the ratio of amount
of fuel (kg) used by the engine per hour to the horse power produced
or delivered b y the engine, when specific fuel consumption is based
on J.H.P. produced it is called indicated specific fuel consumption
and if specific fuel consumption is based in B.H.P. delivered it is
called brake specific fuel consumption.
4.29 Comparison of a Diesel Engine and Petrol Engine
Diesel engines are quite efficient at part loads as compared to
petrol engines and work efficiently in greater range.
A diesel engine is superior than a petrol engine because of the
following factors
(i) The compression ratio is higher than
a petrol engine and this increases the
efficiency of the engine. The variation
of air standard efficiency (,1) with ratio
of compression(,-) is shown in Fig. 4.25.
(ii) In diesel engine, the combustion take )7
place approximately at constant pres-
sure rather than at constant volume as
in a petrol engine. Fig. 4.25
(iii) No electric spark is required.
(iv) As diesel is cheaper than petrol, therefore, the power cost
in diesel engine is low.
4.30 Supercharging
The 1.11,1'. produced by an I.C. engine is almost dirctiv propor-
tional to the air consumed by the engine. Increasing the aircon-
sumption permits the greater quantities of fuel to he added and
results in greater power produced b y the engine. It is, therefore,
desirable that the engine should take in the greatest possible mass
of air. The supply of air is pumped into the cylinder at a pressure
greater than the atmospheric pressure and is called supercharging.
When greater quantity of air is supplied to an internal combustion
engine it would be able to develop more power for the same size and
conversely a small size engine fed with extra air would produce the
same power as a larger engine supplied with its normal air feed.
Supercharging is used to increase rated power output capacit y of a
given engine or to make the rating equal at high altitudes cor-
responding to the unsupercharged sea level rating.
Supercharging is done by installing a super charger between
engine intake and air inlet through air cleaner super charger is
merely a compressor which provides a denser charge to the engine
350 POWER PLANT

thereby enabling the consumption of a greater mass of charge with

the same total piston displacement. Power required to drive the
super charger is taken from the engine and thereby removes from
over all engine output some of the gain in power obtained through
supercharging.
There are two types of compressors that may be used as super
chargers. They are as follows
(i) Positive displacement type super chargers.
(ii) Centrifugal type super chargers.
Positive displacement type super chargers are further of three
types as follows
(a) Rotary type. (b) Screw type
(c) Piston and cylinder type.
In rotary type superchargers the air is compressed by a meshing
gear arrangement called Roots blower as shown in Fig. 4.26 or b y a

Air in Air out Air Air Out

Fig. 4.26 Fig. 4.27

rotating vane element as shown in Fig. 4.27. The air is taken from
intake and discharged at outlet end. It screw type supercharger the
air is trapped between inter meshing helical shaped gears and
forced out axially. In piston and cylinder type super-charger the
piston compresses the air in a cylinder whereas a centrifugal type
super-charger has an impeler running in a housing at a high speed,
centrifugal supercharger is commonly by used in reciprocating
power plants for aircraft.
4.31 Advantages of Supercharging
Due to a number of advantages of supercharging the modern
diesel engines used in diesel plants are generally supercharged. The
various advantages of supercharging are as follows
(i) For given output engine size is reduced.
(ii) Engine output can he increased by about 30 to 5017(.
(iii) The specific fuel consumption of a super charged engine
is less than natural aspirated engine. This is due to the
fact that combustion is supercharged engine is better duo
to better mixing of fuel and air.
DIESEL ENGINE POWER PLANT 351

(ju) Supercharged engine has higher mechanical efficiency.

(t') Supercharging reduces the possibility of knocking in
diesel engine.
4.32 Factors Affecting Engine Performance
Diesel engine call more work out of each heat with unit
than other engines.
Various factors which affect performance of a diesel engine are
as follows
(i) Amount of fuel burnt per minute.
(ii) Brake mean effective pressure.
(iii) Fuel injection system An efficient fuel injection svsteni is
needed. The required quantity of fuel should he measured out,
injected, atomised and mixed with combustion-air.
(iv) Combustion process.
(t) Fuel-air ratio.
(ti T y pe of engine such as two stroke or four stroke engine. Two
stroke en g ines are generally used in diesel power plants.
(ti ) Cooling method.
wiii) Size of cylinder.
4.33 Combustion Phenomenon in C.I. Engine
In C.I. engines the intake is air alone and the Ii. l is injected at
high preL ii' in the form of fine droplets near the I of compres-
sion. The normal compression ratios are in the range of 11 to 17.
The air fuel ratio used in C.I. engines lie between IS and 25 as
against 1-I in S.! (spark ignition) engines. Therefore C.I. engines are
bigger and heavier for the same power out put than S.I. engine.
In C.I. engine combustion occurs b y the high temperature
produced b y the compression of air i.e. it is an auto ignition. Each
minute droplet of fuel as it enters the highl y heated air of engine
c y linder is quickl y surrounded by all of its own vapour and
this in turn and all internal is inflamed at the surfuce
of envelope.
4.34 Comparison of Gas Turbine with Reciprocating I.C.
Engine
Gas turbines and reciprocating I.C. engines are used for former
griiiratwn.
Advantages of gas turbines over I.0 engines are as follows
Gas turbine has lesser number of parts.
 352
POWER PLANT
(it) Mechanical losses in gas turbines are less because in gas
turbine the single rotating unit Consists of a compressor
and a turbine together with a few main bearings com-
pared to complicated reciprocating mechanism with its
valve gear arrangement which are the prime sources of
losses due to ii rction. Further oil and fuel supply pumps
are not used thus reducing mechanical losses.
The life of gas tuihitie is longer than I.C. engine.
(ii) It is easier to carr y out heat transfer process.
(L') Gas turbine has largt• Power to weight. ratio.
This reduces cost of gas turbine. This makes it more suitable
prime mover in mobile power units particularly in air crafts.
(vi) Gas turbine is simple in Construction.
vii) The gas turbine is reliable in operation because balancing
of rotating masses bot.h static and dynamic, call very
accurately done and unlike reciprocating engines the tor-
sional vibration effects due to combustion load changes
and inertia effects are absent due to the steady flow nature
that renders continuous effect on the rotor blades of com-
pressor and turbine. -
Further the absence of valve and valve gears is another reason
for quiet running of gas turbines.
Wil l j In gas t u rbii e the pi rts that are to be lubricated are few
L

in numbers.
(lx) Maintenance is easier.
Example 4.6. A diesel engine has a brake thermal el/u ie,u-v of
3O. If the calorific ca/ac of fuel used in 10000 kcal 1kg, ca/cu/cjfe
the brake specific fuel consumption..
ion-.
Solution. = Brake thermal efficiency 0.3
I.H.P. hr = 632.5 kcal
= F P hr equivalent
Tm
x C.N.
it

where w = Specific fuel consumption per 11.1'. hr.

CV. = Calorific value of fuel 10,000 kcal/kg
632.5 --
0.3
w x 10.000
632.5
0.3 x 10,000 = 0.21 kg/11.1) hr. Ans.

Example 4.7. A s ix c y /i ,ith'r tue stFn/.?c crc/c man itt.' diesel

en'mn- u-,t/t 100 11201 here a - l' 1.2 0 ,nn strain' (/t'/ite'rs 200 13,11.1'. at


DIESEL ENGINE POWER PLANT

353
2000 R.P.M.and uses 100 kg of fuel per hour. If fli p
determine the following , is 240,
(a) Torque, (b) Mechanical efficiency, (c) indicated specific fuel
consumption.

Solution. (a) B.H.P . =

4- 560
where T = Torque
N = R.P.M.
200 =
4500
200 x4500
7 = - -- 71.7 kg.m
(h = Mechanical efficiency

240 - 0.83.
(e) Indicated specific fuel consumption =
where W = Amount of fuel used per hour.
Indicated specific fuel consumption

= = 2.41 kgII.fl.P. hour.

Example 4.8. A diesel engine develops 200 H.P. to over eo,ne

friction and deluers 1000 131fF. Air consumption is 90kg per in mute,
The air fuel ratio is 15 to 1. Find the following:
(a) 1111', (h) Mechanjecil efficiency , (c) Specific fuel consumption.
Solution. (a) B.H.P. = 1000
Fli p . = 200
IHP = BlIP + FHP = 1000 + 200 = 1200
(1,) II,,, Mechanical
= efficiency
= BlIP 1000
= = 0.83 = 83%
K = Air fuel ratio 15
W = Air consumed per hour
=90x60
= 5400 kg per hour
S = Amount of fuel consumed
W 5400
K 15
 POWER PLANT
354

= 360 kg per hour

Specific fuel consumption = S = 1200

= 0.3 kgITHP hr.

PROBLEMS

4.1. How will you classif y I.C. engines? Describe the working of two
stroking of two stroke and four stroke cycle diesel engines.
Discuss their relative merits and demerits.
4.2. What are the different methods of cooling diesel engine? Com-
pare air cooling and water cooling.
4.3. Describe the various methods used for starting diesel engine.
Describe in correct sequence the steps for starting and stopped
procedure.
4.4. Describe the auxiliary-equipment of diesel engine power plant.
4.5. Give the layout of a diesel engine plant.
4.6. What are the various methods of fuel injection? What precau-
tions should be observed to ensure that fuel injection is satisfac-
tory?
4.7. What are the various factors to be considered while selecting the
site for diesel engine power plant? Discuss the advantages and
disadvantages of the diesel power plant.
4.8. Compare I.C. engine with steam engine and state the advantage
of I.C. engine over steam engine.
4.9. What is the importance of heat balance sheet ? What are the
various items considered while drawing the heat balance sheet
ofl,C. engine? Give a typical heat balance at full load for an I.C.
engine.
4.10. Describe the procedure of testing diesel power plant perfor-
mance. How is plant maintenance carried out?
4.11. Write short notes on-the following:
(a) Lubrication of diesel power plant.
(b) Indicated Thermal Efficiency, I.H.P. and B.H.P.
(c) Applications diesel power plant.
(d) Warming up of diesel engine.
4.12. Describe a typical filter and silencer installation for a diesel
engine.
4.13. Define specific fuel consumption. Explain indicated specific fuel
consumption and brake specific fuel consumption.
4.14. A four stroke diesel engine gave the following test results at a
speed of 450 R.l'.M.
Mean effective pressure = 8.50 kg/cm2
Cylinder bore = 22 cm.
Stroke = 26 cm.
Specific fuel consumption = 0.32 kg(BHP/hr
Calorific value of fuel = 11800 kcal per kg.

DIESEL ENGINE POWER PLANT 355
Mechanical Efficiency = 38%
Determine the following:
(a) B.H.P.
(b) l.H.P.
(c) Indicated thermal efficiency.
(d) Brake thermal efficiency.
4.15. Compare a diesel and petrol engine.
4.16. Write short notes on the following:
(a) I.C. engine fuels.
(b) Cost of diesel power plant.
4.17. (a) What is supercharging ? What methods are used for super-
charging diesel engines?
(b) Discuss the advantages of supercharging.
4.18. (a) Under what conditions diesel generating plants are
preferred?
(b) On what factors is the size of the generating plant selected?
(c) Draw a net diagram of a cooling system used for diesel power
plants showing all the essential components. What are the
advantages ofdouble circuit over single circuit system? What
precautions should be taken to ensure that cooling is satisfac-
tory?
4.19. Name the methods used to purify lubricating oil.

• • : . : •
1J i'J., f
T O . t•jA
.)R' 11k. / 4.• IØ ;-i.
01 '
• .•...-. •
Jl • : •• k
• ,• . .. .-. ..... ,.
 5
Nuclear Power Plant

5.1 Nuclear Energy

As large amounts of coal and petroleum are being used to
produce energy, time maycome when their reserves may not be able
to meet the energy requirements. Thus there is tendency to seek
alternative sources of energy. The discovery that energy can be
liberated by the nuclear fission of materials like uranium (U),
plutonium (I'u), has opened up a new sources of power of great
importance. The heat produced due to fission of U and Pu is used to
heat water to generate steam which is used for running turbo-gen-
erator.
It has been found that one kilogram of U can produce as much
energy as can be produced by burning 4500 tonnes of high grade
coal. This shows that nuclear energy can be successfully employed
for producing low cost energy in abundance as required by the
expanding and industrialising population of future.
Wisely used nuclear energy can be of great benefit for mankind.
It can bridge the gap caused by inadequate coal and oil supplies. It
should be used to as muh extent as possible to solve power problem.
Sonic of the factors which go in favour of nuclear energy are as
follows
1. H y dro-electric power is of storage type and is largely
dependent on monsoons. The systems getting power from
such plants have to shed load during the period of low
rainfall.
2. Oil is mainly needed for transport, fertilizers and
petrochemicals and thus cannot be used in large quan-
tities for power generation.
3. Coal is available only in some parts of the country and
transportation of coal requires big investments.
4. Nuclear power is partially independent of geographical
factors, the only requirement being that there should be
reasonably good supply of water. Fuel transportation net-
 NUCLEAR POWER PLANT
357
works and larger storage facilities are not needed and
nuclear power plant is a clean source of power which does
not pollute the air if radio active hazards are effectively
prevented.
5. Large quantity of energy is released with consumption of
only a small amount of fuel.
World's first nuclear power plant was commissioned in 1954 in
U.S.S.R. Since then efforts are being made to make use of nuclear
power.
In India, it was Dr. H.J. Bhabha who put India on the road to
nuclear research, more than two decades ago. He had in his mind
not the destructive power of atom but using this new source of
immense energy for peaceful purposes like power production. India
at present has four nuclear power plants. First nuclear power plant
is at Tarapur. It has two boiling water reactors (B.W.R.) each of 200
MeW capacity and each uses enriched U as fuel. These two reactors
have been built with the help of U.S.A. The other two nuclear power
plants are at Rana Pratap Sagar in Rajasthan and at Kalpakkam
in Tamil Nadu. The fourth nuclear power plant has been built at
Narora in U.P.
Nuclear energy is the most useful power available to mankind
today. In large parts of the world it is becoming a predominant
source of electrical power and a versatile tool for use in many areas
of human endeavour. In India too atomic energy is being used to
generate electricity and to bring 0. .1 . 1 .1 1
:nent in industry,
­

agriculture, medicine and in other tields Lnrouh its varied applica-

tions.
The major centre for research and vdopnit . nt work in atomic
energy in our country is the Bhabha Atomic Research Centre
(BARC) at Trombay. The centre is the largest single scientific
establishment in India. Besides BARC three other national institu-
tions associated with some important aspects of atomic energy
programme are as follows:
(i) Tata Institute of Fundamental Research, Bombay.
(ii) Tata Memorial Centre Bombay.
(iii) Saha Institute of Nuclear Physics, Calcutta.
Nuclear power plants resemble, convential thermal power
plants insofar as they produce steam to drive a turbine whose
rotational energy is converted into electricity by means of a gener-
ator. In contrast to power plants fired by coal, oil, or gas, nuclear
power plant use the energy released from splitting atoms to convert
water into steam. The "fuel" that leads itself to this splitting proce-
dure is the uranium atoms, and its splitting, or fission, is engineered
within a reactor.

,.
 POWER PLANT
358

India went nuclear in 1956 when its first research reactor went
critical at Trombay. Six units are now under various phases of
operation construction or design two each at Kota, Kalpakkam and
Narora. They are all of the CANDU (Canadian-Deuterium-
Uranium) type, most suited to Indian conditions. India has limited
deposits of uranium and would not be dependent on foreign enrich-
ment facilities or a foreign supply to enriched fuel, (used in the
Tarapur power plant). CANDU reactors, which use fuel available
within the country, do not require large capital and operating
outlays for fuel enrichment. The nuclear power generation will be
about 2270 MW by 1991-92 in our country.
5.2 Chain Reaction
Uranium exist as isotopes of U 238, U234 and U235 . Out of these
isotopes U235 is most unstable. When a neutron is captured by a
nucleus of an atom of U 5 , it splits up roughly into two equal
fragments and about 2.5 neutrons are released and a large amount
of energy (nearly 200 million electron volts MeV) is produced. This
is called fission process. The neutrons so produced are very fast
moving neutrons and can be made t .o'fission other nuclei of U 235 thus
enabling a chain reaction to take place. When a large number of
fissions occurs, enormous amount of heat is produced.
The neutrons released have a very high velocity of the o
1.5 x 107 metres per second. The energy liberated in the
reaction is according to Einstein law
E = mc2
where E = Energy produced
rn = mass in grams
c = speed of light ih cm/sec equivalent to 3 x 1010 cm,/sec-
Out of 2.5 neutrons released in fission of each nuclei of U 235 , one
neutron is used to sustain the chain reaction, about 0.9 neutron is
captured by U238 which gets converted into fissionable material,
Pu239 and about 0.6 neutrons is partly absorbed by control rod
material, coolant moderator and partly escape from the reactor.
Production of the fissionable material Pit 239 during chain reaction
compensates the burn up of primary fuel U 235 U238 + neutron =
Pu239 . If thorium is used in the reactor core it produces fissionable
material U233.
Th 232 + Neutron .- U33
Pu239 and U233 so produced are fissionable material and can be used
as nuclear fuel and are known as secondary fuel. U 235 is called
primary fuel.


NUCLEAR POWER PLANT 359

The chain reaction producing a constant rate of heat energy can
Continue only ifthe neutron liberated by fission, balance the disposal
of neutrons by different ways listed below
I. Escape of neutrons from the fissionable materials.
2. Fission capture by U 2 , and Pu 239 and U233.
3. Non-fission capture by moderator, control rods, fission
fragments and by impurities etc.
If the neutrons produced in the chain reaction are less than the
neutrons disposed off in different ways, the chain reaction will stop.
Fig. 5.1 shows the chain reaction.

FISSION
FRAGMENT
ESCAPE

*23FA

NEUTRON
'YE
__
NEU -I.1z FAST
______NEUTRON

MOA

Fig. 5.1
5.3 Fertile Material
It is defined as the material which absorbs neutrons and under-
goes spontaneous changes which lead to the formation of fissionable
material. U 235 and Th 22 are fertile materials. They absorb neutrons
and produce fissionable materials Pu 239 and U 233 respectively.

t Ii1fiIIIJ
5.4 Unit of Radioactivity (Curie)

COOLANT
III IJIIUl•
oil
PEFLECTOP

121
MODERATO
PRE55URE
VESSEL
CONCRETE.
SN/EL DG'JG

Fig. 5.2
-COOLANT

360 POWER PLANT

- The basic unit of radioactivity is named as Curie. It is-the

activity (Rate of decay) of one gram ofradionctivity element radium.
It has been estimated that rate of decay of one gram of radium is
equal to 3.7 x 1010 disintegrations per second.
1 Curie = 3.7 x 1010 disintegrations per second. It describes the
intensity of radioactivity in a sample of material.

Co

c 'C

-'t:m' Tznl<

y '"n zZ

FueL )d8
P

Fig 53
 NUCLEAR POWER PLANT
33

5.5 Parts of a Nuclear Reactor

A nuclear reactor is an apparatus in which heat is produced due
to nuclear fission chain reaction. Fig. 5.2 shows the various parts .)f'
reactor, which are as follows
(i) Nuclear Fuel (ii) Moderator
(iii) Control Rods (iv) Reflector
(v) Reactors Vessel (vi) Biological Shielding
(vii) Coolant.
Fig. 5.3 shows a schematic diagram of nuclear reactor.
5.5.1 Nuclear Fuel
Fuel of a nuclear reactor should be fissionable material which
can be defined as an element or isotope whose nuclei can be caused
to undergo nuclear fission by nuclear bombardment and to produce
a fission chain reaction. It can be one or all of' the followin g U
U235 and Pu239.
Natural uranium found in earth crust contains three isotopes
namely U 2 , U 235 , U 238 and their average percentage is as follows
99.3%; U 2 - 0.79c; U 233 - Trace.
Out of these U215
is most unstable and is capable fsust.ajnjn
chain reaction and has been given the name as prim tr y fuel. U2
and Pu 2 '39 are artificially produced from Tb 232 and I respectively
and are called secondary fuel.
Pu 239 and U 233 so produced can be fissioned by thermal
neutrons. Nuclear fuel should not he expensive to fabricate. It
should be able to operate at high temperatures and should he
resistant to radiation damage.
Uranium deposits are found in various countries such as Congo,
Canada, U.S.A., U.S.S.R., U.K., Australia, Czechoslovakia and Por-
tugal etc.
The fuel should be protected from corrosion and erosion of the
coolant and for this it is encased in metal cladding generally stain-
less steel or aluminium.
Adequate arrangements should be made for fuel suppl y , charg-
ing or discharging and storing of the fuel.
For economical operation ofa nuclear power plant special atten-
ti hould be paid to reprocess the spent up (burnt) fuel elements
t, i.cover the unconsurned fuel. The spent up fuel elements are
intensively radioactive and emits some neutrons and gamma ra's
u1 ( i should be liandlet! carefully.

—25


POWER PLANT

In order preVeilt the COOt :uininatiOfl of the coolant by fission

products, . ' •kutivt coating or cladding must separate the fuel
from the c- trc.ain. Fuel elenient cladding should pOSSeSS the

It •.; dcl be able to wit litand high temperature within

H it , 1-actor,
(it It shuld have high r)rrsion resistance.
Ht It huid ZlctV0 high thermal conductivity.
(ii) It ShOuld not have a t&ndncv to absorb neutrons.
It should have ufficient strength to withstand the effect
of radiations to winch it is uhjcctvd.
I)ensity of various nuclear fuels is indicaled in Table 5.1.

'l'al)Ie 5.1
.ictij . •. . J

----------------------------------------------. isrs__
1S.G
2(j -

L . t239 - - 196 - -

Uric:iiiin exide( t ; 0 ^) is another important fuel element.

Uraniuii exl luc, th e following advantages O\ cr natural uranium:
it is ire tahIc' than natural uranium
(it The:. is no problem or phase change in cac' of uranium
:nl thercfre it can be used for higher temperatures.
(ill) It tIe' - ,t corrode as easily as natural uranium. -
(it ) it mere conipatill& with most of the coolants and is not
at. eLy II.:,.N2
(r) Tiler, - t.- eTeater cliniensional stability during use.
Uranium ix, possesses following disarivantaizes
(to It hi- low thermal conductivity
it
(H) It is more brittle than natural urarncini and therefore
c 0 break due to thermal stresses.
is euntIal
IL: er irlctc
Uranium oxide is it l,riitie ceramic produced as a powder :uid
then snter(d to birm fuel pellets.
Anoti'r jjel used in the nuclear reactor is uranictin carbide
UC). It j a lack ceramic used I ll the form of pe1Iet.
ociCl
Table 5.1 o indicatessuiiie of the p1Ysical prolc'iti'


NUCLEAR POWER PLANT 363

Table 5.1 (a)

I Fu,i Th
(lUCtltttY K. keal/kgC (C)
- coi/,n.hr (4 4
Natural uraniumI 263 0.037 19000 1130
Uranium oxide 18 JO078 I PL_ 2750
1L_1Jr3!iiujn -carbide T---- 20-6 13600 2350J
5.5.2 Moderator
In the chain reaction the neutrons produced are fast moving
neutrons. These fast moving neutrons are far less effective in
causing the fis-ion of U 2 ' and try to escape from the reactor. To
improve the utilization of these neutrons their speed is reduced. It
is done by colliding them with the nuclei of other material which is
lighter, does not capture the neutrons but scatters them. Each such
collision causes loss of energy, and the speed of the fast moving
neutrons is reduced. Such material is called Moderator. The slow
neutrons (Thermal Neutrons) so produced are easily captured by
the nuclear fuel and the chain reaction proceeds smoothly. Graphite,
heavy water and ber y llium arc generally used as moderator.
Reactors using enriched uranium do not require mod''rator. But
enriched uranium is costl y due to processing needed.
A moderator should process the following properties
W It should have high thermal conductivity.
(it) It should be available in large quantities in pure form.
(iii) It should have high melting point in case of solid
moderators and low melting point in case of liquid
moderators. Solid moderators should also possess good
strength and machinability.
(iv) It should provide good resistance to corrosion.
(v) It should be stable under heat and radiation.
nil It should be able to slow down neutrons.

5.5.3 Moderating Ratio

To characterise a moderator it is best to use so chled moderating
ratio which is the ratio of moderating power to the macroscopic
neuron apture coefficient. A high value of moderating ratio indi-
cates that the given substance is more suitable for slowing do ri the
neutrons in a reactor. Table 5.2 indicates the moderating ratio for
.-wnc of the material used as moderator.

364 POWER PLANT

Table 5.2

This shows that heavy water, carbon and, beryllium are the best
moderators.
Table 5.3 indicates density of various moderators.
Table 5.3

P
Moderator Density (gmicm3)
1120 1
1)20 1.1
C 1.65
Be - 1.85
Table 5.4 shows some of the physical constants of heavy water
(D20) and ordinary water (1120).
Table 5.4
constant
0.9982 nVc
Freezing temperature 276.82 273
Boiling temperature I374.5 373}(
Dissociatior Cünstant 0.3x10 -1x10'
Dielectric Constant at293}( 80.5 82
Snecific heat at 293K 1.018 1

5.5.4 Control Rods

The control and operation of a nuclear reactor is quite different
from a fossil and fuelled (coal or oil fired) furnace. The furnace is fed
continuously and the heat energy in the furnace is controlled by
regulating the fuel feed and the combustion air whereas a nuclear
reactor contains as much fuel as is sufficient to operate a large power
plant for some months. The consumption of this fuel and the power
level of the reactor depends upon its neutron flux in the reactor core.
The energy produced in the reactor due to fission of nuclear fuel
during chain reaction is so much that if it is not controlled properly
the entire core and surrounding structure may melt and radioactive
fission products may come out of the reactor thus making it unin-
habitable. This implies that we should have some means to control
the power of the reactor. This is done by means of control rods.
NUCLEAR POWER PLANT 365

Control rods in the cylindrical or sheet form are made of boron or

cadmium. These rods can be moved in and out of the holes in the
reactor core assembly. Their insertion absorbs more neutrons and
damps down the reacation and their withdrawal absorbs less
neutrons. Thus power of reaction is controlled by shifting control
rods which may be done manually or automatically.
Control rods should possess the following properties:
(i) They should have adequate heat transfer properties.
(ii) They should be stable under heat and radiation.
(iii) They should be corrosion resistant.
(iv) They should be sufficient strong and should be able to shut
down the reactor almost instantly under all conditions.
(v) They should have sufficient cross-sectional area for the
absorption.
5.5.5 Reflector
The neutrons produced during the fission process will be partly
absorbed by the fuel rods, moderator, coolant or structural material
etc. Neutrons left unabsorbed will try to leave the reactor core never
to return to it and will be lost. Such losses should be minimised. It
is done by surrounding the reactor core by a material called reflector
which will send the neutrons back into the core. The returned
neutrons can then cause more fission and improve the neutrons
economy of the reactor. Generally the reflector is made up of
graphite and beryllium.
5.5.6 Reactor Vessel
It is a strong walled container housing the core of the power
reactor. It contains moderator, reflector, thermal shielding and
control rods.
5.5.7 Biological Shielding
Shielding the radioactive zones in the reactor from possible
radiation hazard is essential to protect, the operating men from the
harmful effects. During fission of nuclear fuel, alpha particles, beta
pr.rticles, deadly gamma rays and neutrons are produced. Out of
these neutrons and gamma rays are of main significance. A protec-
tion must be provided against them. Thick layers of lead or concrete
are provided all round the reactor for stopping the gamma rays.
Thick layers of metals or plastics are sufficient to stop the alpha and
beta particles.
5.5.7 (a) Coolant
Coolant flows through and around the reactor core. It is used to
transfer the large amount of heat produced in the reactor due to
fission of the nuclear fuel during chain reaction. The coolant either
transfers its heat to another medium or if the coolant used is water
366 POWER PLANT

takes. up the heat and gets converted into steam in the reactor
which is directly sent to the turbine.
Coolant used should be stable under thermal condition. It
should have a low melting point and high boiling point. It should not
corrode the material with which it conies in contact. The coolant
should have high beat transfercoefficient. The radioactivity induced
in coolant by the neutrons bombardment should be nil. The various
fluids used as coolant,are water (light water or heavy water), gas
(Air, ('02, hydrogen, helium) and liquid metals such as sodium or
mixture of sodium and potassium and inorganic and organic fluids.
Power required to pump the coolant should be minimum. A
coolant of greater density and higher specific heat demands less
pumping power and water satisfies this condition to a great extent.
Water is it good coolant as it is available in large qualities can be
eailv handled, provides some lubrication also and offers no unusual
corrosion problems. But due to its low boiling point (2 12F at
atmospheric pressure) it. is to he kept under high pressure to keep
it in the liquid state to achwve a high that transfer efficiency. Water
when used as cooluit 11101.11d be free from impurities otherwise the
impurities ma y become radioactive and handling of water will be
difficult
5.5.8 Coolant Cycles
coolant while circulating through thc reactor passages take
up heat. produ..., lu..' to chain reaction and trnnskr this heat to the
f?cd water in three w..vs as follows
(a) Direct Cycle. III s y stem [Fig, 5.1 (a )l coolant 'wluch is
water leaves the reactor in the form of steam. Boiling water reactor
ues this system.
(1..) Single Circuit S ystc,n. In this s ystem ll'ig. 5.4 (b)l the coolant
transfers the heat to the feed water in the steam generator. This
system is used in pressurised reactor.
(c) Double Cir cu it S y stem. In this s y stem [Fig. 5.4 (c)] two
coolant are used. Primary coolant after circulating through the
reactor flows through the intermediate heat exchanger (lIIX) and
passes on its heat to the secondary coolant which transfers its heat
in the feed water in the steam generator. This system is used in
sodium graphite reactor and fast breeder reactor.
5.5.9. Reactor Core
Reactor core consists of fuel rods, moderator and space through
which the coolant flows.
NUCLEAR POWER PLANT 307

5.6 Conservation Ratio

It is defined as the ratio of number of secondar y fri itrllIs to
the number of consumed primar y fuel atoms .Areitr %kith a
conversion ratio above unity is known as a brecdrr reactor. Rrcedrr
reactor produces more fissionable material than it coilsunirs. lfth

STEAM HOT CO LANT TSEATEMAM

fissionable material produced is equal to or less thui th ("USU
the reactor is called converter reactor. -

PEAC7ORREAcToR
I
YGENERAIOR
FEED
EED
WATER Pump WATER
(ti) (1)

C; 0:0 LNAT : SECOLANT

ECONDARY _[- -5TEAMSTEAM
F. 5 1

pRIMARY

REA CrOP WX GENER TORFEED

- WATER
PUMP PUMP
Fig. 54 (c

5.7 Neutron Flux

It is a measure of the intensit y of neutron radiation and it is the
number of neutrons passing through I C111 2 of t given target in one
second. It is expressed as or, where u is numbür of neutrons br
cubic centimetre and r is velocity o f neutrons in cm/sec.

5.8 Classification of Reactors

The nuclear reactors can he classified as follows:
1. Neutron Energy. r)epi-ndi ng upon the energ y of t h
neutrons at the time the y are captured hvthe fuel to induce fission -.
the reactors can be named as follows
(a I bast Reactors. In such reactors fission is brought about by
fast (non moderated) neutrons.
Thermal Reactors or .S/ow Rem-furs. In these reactors the fast
moving neutrons are slowed down b y passing them through the
moderator. These slow moving neutrons are then captured b y the
fuel material to bring about the fission.
368 POWEfl PLANT

(,) Intcrrnedwte Reactors. In such reactors most of the fission

events are caused b y neutrons in the course of slowing down.
2. Type of Fuel Used. Nuclear reactor may use U 23 , U13' and
Th 232 as their fuels Tb 232 and U 238 get converted in fissionable
materials like U233 and Pu 239 respectively.
3. Type of Coolant Used. On the basis of coolant. used the
reactors may be classified as follows
a) Gas cooled reactor.
(b) Water (ordinary or heav y water) cooled reactors.
(c) Liquid metal cooled reactors.
4. Type of Moderator Used. On this basis the reactors may
he classified as follows
(a) Graphite reactors.
(b) Beryllium reactors.
(c( Water (ordinary or heav y water) reactors.
5. Type of Core. According to the t y pe of core used the reactor
may be classified as follows
((I) Ho,naeneoii.s , -eactor. In this reactor fuel and moderator
represent a uniform mixture such as an aqu3ous solution of a
uranium salt.
(h) .H&:ero'fleous recict,. In such reactor fuel rods areinserted
in moderator The fuel (lenhlnt.s are generally arranged in some
regular order forming a lattice.
5.8.1 Design of Nuclear Reactor
The basic factors considered during the design of it nuclear
P .Vet' reactor are as follows
U Tvv of reactor.
ii lvpe of fuel to be Used
(iii) Piwer rating of' reactor 111 MW.
Ut Coolant system.
(i) Control system.
(vi) Rates of neutron production and absorption.
(1-ii) Sill i'tv of react or.

5.9 Main Component of a Nuclear Power Plant

The main compo:enf sofa nuclear power plant are shown in Fig.
5.5. 'I'liese include nuclear reactor, heat exchanger (steam gener-
ator), turbine, electric generator and condenser. Reactor ofa nuclear
power plant is similar to the furnace of steam power plant. The heat
liberation in the reactor due to the nuclear fission of the fuel is taken
up by the coolant circulting through the reactor core. hot coolant
loaves the reactor at top and then flows through the tubes of steam
generator(boiler and passes on its heatto the feed water. The steam
produced is passed through the turbine and after work has been
NUCLEAR POWER PLANT 369

done by the expansion of steam in the turbine steam leaves the

turbine and flows to the condenser. Pumps are provided to maintain
the flow of coolant, condensate and feed water.
5.10 Boiling Water Reactor (B.W.R.)

"Or COOLANT TURBINE

CORE
STEAM I
GENERATO1 c0NOENWI
I

Lll.11ilII
II II
jt WATER

I: I L
...j COOLA!.IT FEED PUMP
PUMP
CONCRETE SHILDING
Fig. 5.5
Fig. 5.6 shows nuclear power plant using B.W.R. In this reactor
enriched uranium (enriched uranium contains more fissionable
isotope U235 then the naturally occurring percentage 0.7%) is used
as nuclear fuel and water is used as coolant. Water enters the
reactor at the bottom. It takes up the heat generated due to the
fission of fuel and gets converted into steam. Stearn leaves the
reactor at the top and flows into the turbine. Water also serves as
moderator. India's first nuclear power plant at Tarapur has two
reactors (each of 200 MW capacity) of boiling water reactor type.

CONTROL
RODS -.

GENERATOR

REACTOR
CORE -
CONDENSER

FEED PUMP

Fig. 5.6

5.11 Pressurised Water Reactor (P.W.R.)

A P.W.R. nuclear plant is shown in Fig. 5.7. It uses enriched U
as fuel. Water is used as coolant and moderator. Water
370 POWER PLANT

through the reactor core and takes up the heat liberated due to
nuclear fission of the fuel. In order that water may not boil (due to
its low boiling point 212 F at atmospheric con(litions) and remain in
liquid state it is kept under a pressure of about 1200 p.si.g. by the
pressu riser. This enables water to take up more heat from the
reactor. From the pressuriser water flows to the steam generator
where it passes or its heat to the feed water which in turn gets
converted into steam.

CONTROL
RODS
PRESSURISER

TURBINE GENERATOR
S U'
EACTOR
CORE
CONDENSER

COOLANT • FEED
PUMP PUMP

Fig. 5.7
5.12 Sodium Graphite Reactor (SGR)
The reactor shown in Fig. 5.8 uses two liquid metal coolants.
Liquid sodium (Na) serves as theprimary coolant and an alloy of
sodium potassium (NaK) as the secondary coolant.
Sodium melts at 208C and boils at 1625F. This enable to
achieve high outlet coolant temperature in the reactor at moderate
pressure nearly atmospheric which can he utilized in producing
steam of high temperature, thereby increasing the efficiency of the
plant. Sttani at temperature as high as 1000 has been obtained
b y this s ystem. This shows that by using liquid sodium as coolant
more electrical power carl he generated for a given quantity of the
fuel burn up. Secondl y low pressure in the primary and secondary
coolant circuits, permit the use of less expensive pressure vessel and
pipes etc. Further sodium can transfer its heat very easil y . The only
disadvantage in this system is that sodium becomes radioactive
while passing through the core and reacts chemically with water.
So it is not used directly to transfer its heat to the feed water, but a
secondary coolant is used. Primary coolant while passing through
the tubes of intermediate heat exchanges (I. fIX.) translrs its heat
to the secondary coolant. The secondary coolant then flows through
the tubes of steam generator and passes on its heat to the feedwater.
Graphite is used as moderator in this reactor. For heat exchanger
refer Fig. 5.10 (b) Liquid metals used as heat transfer media have
NUCLEAR POWER PLANT 371

certain advantages over other common liquids used for heat transfer
purposes. The various advantages of using liquid metals as heat
t.ranfer media are that theyhave relatively low melting points and
combine high densities with low vapour pressure at high tempera-
tures as well as with large thermal conductivities.
5.13 Fast Breeder Reactor (FBR)
CONTROL
RODS

L! Ic STEAM T URBINE

I STEAM
CORE -IH I I'll I I 1HXI 6ENER^17

UOLAN7 COOLANT FEED
PUMP PUMP PUMP
Fig. 5.8
Fig. 5.9 shows a fast breeder reactor system. In this reactor the
core containing U23*5 is surrounded by a blanket (a la yer of fertile
material placed outside the core) or fertile material In this
reactor no moderator is used. The fast moving neutrons liberated
due to fission of U23 are absorbed by U 235 which gets converted into
fissionable material Pu2,39 which is capable of sustaining chain
reaction. Thus this reactor is important because it breeds fissionable
material from fertile material U 238 available in large quantities.
Like sodium graphite nuclear reactor this reactor also uses two
liquid metal coolant circuits. Liquid sodium is used as primary
coolant when circulated through the tubes of intermediate heat
exchange transfers its heat to secondary coolant sodium potassium
allo y . The secondary coolant while flowing through the tubes of
steam generator transfer its heat to fied water.
Fast breeder reactors
are better than conven-
tional reactor both from E_ NarO
the point of view of safety -"Mx
and thermal efficiency. U238—.
For India which already is BLANKET
fast advancing towards COPE ________

self reliance in the field of (U2 IS) Na

nuclear power technology, ;= —FROM PUMP

the fast breeder reactor be-
comes inescapable in view Fig. 5.9
372 POWER PLANT

of the massive reserves of thorium and the finite limits of its

uranium resources. The research and development efforts in the fast
breeder reactor technology will have to be stepped up considerably
if nuclear power generation is to make any impact on the country's
total energy needs in the not too distant future.

5.14 Coolants for Fast Breeder Reactors

The commonly used coolants for fast breeder reactors are as
follows:
(i) Liquid metal (Na or NaK)
(ii) Helium (He)
(iii) Carbon dioxide.
Sodium has the following advantages:
(i) It has very low absorption cross-sectional area.
(ii) It possesses goQd heat transfer properties at high
temperature and low pressure.
(iii) It does not react on any of the structural materials used
in primary circuits.
5.15 Waste Disposal
Waste disposal problem is common in every industry. Wastes
from atomic energy installations are radioactive, create radioactive
hazard and require strong control to ensure that radioactivity is not
released into the atmosphere to avoid atmospheric pollution.
The wastes produced in a nuclear power plant may be in the
form of liquid, gas or solid and each is treated in a different manner:
Liquid Wastes. The disposal of liquid wastes is done in two
ways:
(i) Dilution. The liquid wastes are diluted with large quantities
of water and then released into the ground. This method suffers from
the drawback that there is a chance of contamination of under-
ground water if the dilution factor is not adequate.
(ii) Concentration to small volumes and storage. When the
dilution of radioactive liquid wastes is not desirable due to amount
or nature of isotopes, the liquid wastes are concentrated to small
volumes and stored in underground tanks. The tanks should be of
assured long term strength and leakage of liquid from the tanks
should not take place otherwise leakage or contents, from the tanks
may lead to significant underground water contamination.
Gaseous Wastes. Gaseous wastes can most easil y result in
atmospheric pollution. Gaseous wastes are generally diluted with
air, passed through filters and then released to atmosphere through
large stacks (chimneys).
NUCLEAR POWER PLANT 373
Solid Wastes. Solid wastes consist of scrap material or dis-
carded objects contaminated with radioactive matter. Those wastes
if combustible are burnt and the radioactive matter is mixed with
concrete, drummed and shipped for burial. N on-combstjble solid
wastes, are always buried deep in the ground.
5.16 Homogeneous Reactor
Fig. 5.10 shows a Homogeneous Aqueous Reactor (H.A.R.). In
this reactor heavy water is used as coolant and moderator. It makes
use of both fertile and fissionable material which circulate with
coolant. Fissionable U 238 solution is contained in one zone and slurry
of Thoriuni oxide and Deuterium oxide is contained in the other
zone. Main advantage of this reactor is that problems associated
with solid fuels elements are avoided. As heavy water is used as
moderator coolant there is good neutron economy. The disad-
vantages or this system are:
(i) Large amount of fuel is required.
(ii) Circulative fuel system causes the external components to
become radioactive.
(iii) High vapour pressure of water.

Fig. 5.10

In the figure, H.E. represents Heat Exchanger, B.U.E. repre-

sents Blanket Heat Exchanger, G.S. represents Gas Separator.


371r POWER PLANT

5.17 Heat Exchanger

'The heat exchanger used in sodium graphite reactor is Shown
iii Fig. 5.11. The two coolants eparat ed b y a thin wall move in the
opposite dirctions. The narrow space between the partition walls
is filhd with mercur y to ensure a high efficiency of heat transfer.
The mercnry circulates in a siall separate circuit. If some radioac-
ivity is observed in mercury this indicates leakage in the primary
circuit and serves as a warning.
5.17.1 Candu Reactor
Candu (Canadian-Deuterium uranium) reactor shown in Fig.
5.10 (a) uses heavy water (99.8 per cent Deuterium OXi(IO D.()) as
iiioderator and Natural uranium containing 0.7% U 3 is
used i S fueL 'I he reactor vessel a steel cylindei containing number
Of' tubes which are subjected to high internal pressure. The tubes
also called c}1aniiol, Contaill fuel elements and the pressurised
('flOhlflt flw-, along ihe channels mind around the fuel elements to
remove the heat genrated b y h5i(uI. The coolint flow is In opposite
directions in ad u at (l1aI1iWl
In tifls rcmtct'r refuelhiig rvn)o\ of spent fuel mind replaccir
by fresh fnh is carried out while the reactor is operating. NO -
( of' neutron absorber rods of cmulmiiiin a re provided for con
protection of' remtct ' r l he high teiliperature coolant leavi
reactor flows to steam generator to heat the feed water so tm
gets converted into stemlin rif l e coolant then returns to the reacto

to stUrfl
- carr
_..JL
r ut C _f q d w.-mtcr
R ac (or

Lf
- ---"I,-,',.
..L-JL)
Ftc)i;

Fuel L.jrJ)es
Moderate hot
l4iaVy water exchangei
moderator
Fig. 5 10(a)

NUCLEAR POWER PLANT 375

flt Urn
INLET
Fig 5.11

5.18 Gas Cooled Reactor

In this nac'or Fig. 5.12 carbon dioxide gas is used to carry
awa y the heat produced in the reactor. The g .. is circulated at a
pressure of about 7kg/i1 2 . The gas flowing up through each of the
channels round the elements leaves the reactor at the top and flows
to heat exchanger where it transfer its heat to water which gets
eom-k-l-ted into steam The gas is recirculated with the help of gas
hIu\vurs The ,ti':iin drives the turl:iiies which ill (lrive alter-
nator to generate elect ni cit V
5.18.1 Objectives of It and I) ill Energy
The Research and 1)&velepment (R&D) activities ill field of
nuclear energy ill country are directed towards the following:
(I) Development of nuclear fuels
(z) Development of reactor technologies for example fast
breeder reactor (FBR,j system
l
(i l) I )cvelopmt'xtt of heavy water production technology
fu ) Development of nuclear safet y devices

(u) Nuclear waste cusp sal

Nuclear fuel rc-pnosing
LH Safe operation of iiucl''ar pu\ver
III country the Atomic Energ y Commission AF' is respon-
sible for toi- inulation of pohcws and prograiu1rne- in the field of
nuclear encrg\ . The l3hahlia Atomic lieseiicIi ('mitre (lARC) at
'l'ruiohnv is tin- national ri-scan h centre icr research and develop
ment work III c'nergv and related disciplines-
Other institutes which provide research support in iiulcai
i'iiergv ine i Saha lntituti- of nuclear ph ysics tt Tam I u-tituti-
376 POWER PLANT

of fundamental research (iii) Nuclear Reactor research centre Kal-

pakkam.
4PE FLOOQ MCI GAS
HAGE )BE

IwarE

" TEAM

8I0.ER BIOLOGICAL CCCi GS

SHIELD _7 DUCT

STEAM

TUREINE ALTER.4t3
7

DE MS S P

OL ATEP

Fig. 5.12

5.19 Breeding
It is the process of producing fissionable material (fissionable
material) from a fertile material such as uranium 238 (U 238 ) and
thorium 232 (Th 232 ) by neutron absorption.
(U238) neutron Pu239
Th 232 4 neutron - U233

Pu 239 and U233 are fissionable materials and can be used in chain
reaction.
5.19.1 Electron Volt (eV)
The electron volt is the amount of energy required to raise the
potential of an electron by one volt.
One electron volt (eV) = 1.60203 x 10-12 erg.
= 1.60203 x 10 -19 Joules.
 NUCLEAR POWER PLANT
377
5.20 Thermal Neutrons
Such neutrons are in thermal equilibrium with the material in
which they are moving; for example in the moderator. They poss&'ss
a mean energv.of about 0.025 eV, at normal temperature (15C)
5.21 Fast Neutrons
Fast neutrons are those neutrons which have lost relatively
little energy since being produced in the fission process. The lower
limit of their energy is taken as 1.0 MeV (million electron volts.
The general neutrons energies are as follows
(a) Thermal neutrons –0.025 eV
(l) Intermediate neutrons - 1.0 to 0.1 MeV
(c) Fast neutrons- - 0.1 MeV or more,
Thermal neutrons are the most effective in causing fission and,
therefore it is desirable to slow down or moderate the fast neutrons
which normall y have an energy of about 1 MeV.
5.22 Burn Up
It is the amount of fissible material in a reactor that gets
destroyed clue to fission or neutron capture expressed as a percent-
age of the original quantity of fissionable materials.
5.23 Cost of Nuclear Power Plant
Nuclear power plant is economical if used as base load power
plant and run at higher load factors. The cost of nuclear power plant
is more at low load factors. The overall running cost of a nuclear
power plant of large capacity may be about 5 paisa per kWh but it
may be as high 15 paisa per kWh if the plant is of smaller capacity.
The capital cost of a nuclear power plant of larger capacity (say 250
MW) is nearly Rs. 2500 per kW installed. A typical sub-division of
cost is as follows
Item
(a) Capital cost of land, building and
4_ Approximate Cost %
62%
eujnt etc.
ib) Fuel cost 22%
Maintenance cost
(d) Interest on cap cost ___
The capital investment items include the following:
(i) Reactor plant: (a) Reactor vessel (b) Fuel and fuel handling
system, (c) Shielding. (ii) Coolant system. (iii) Steam turbines,
generators and the associated equipment. (iv) Cost of land and
construction costs.
The initial investment and capii.ii cost ofa nuclear power plant
is higher as compared to a thermal power plant. But the cost if
transport and handling of coal for a theri 1 power plant is much
—26
378 POWER PLANT

higher than the cost of nuclear fuel. Keeping into view the depletion
of fuel (coal, oil, gas) reserves and transportation of such fuels over
long distances, nuclear power plants can take an important place in
the development of power potentials.
5.24 Nuclear Power Station in India
The various nuclear power stations in India are as follows
(i) 'rFpUF Nuclear Power Station. It is India's first nuclear
power plant.. It has been built at Tarapur 60 miles north of Bombay
with American collaboration. It has two boiling water reactors each
of 200 MW capacity and uses enriched uranium as its fuel. It
supplies power to Gujarat and Maharashtra
Tarapur power plant is moving towards the stage of using mixed
oxide fuels as an alternative 'to uranium. This process involves
rec y cling of the plutonium contained in the spent fuel. In the last
couple of ears it has become necessary to limit the output of
reactors to save the fuel c y cle in view of the uncertainty ofenriclied
uraniulil supplies from the United States.
(it) Italia I'ratap Sagar (Rajasthan) Nuclear Station. It has
been built itt 12 miles south vest of Kota in Rajasthan with
Canadian collhurution. It has two reactors each of200 MW capacity
and uses natural uranium iii the form of oxide as fuel and heavy
water as moderator.
(iii) Kalpzrkkam Nuclear l'ower Station. It. is (lie third
nuclear pOwet' :iatiun ill and is being built at about 40 miles
from MO4UOS ( ov. It will he wholl y designed and constructid by
aliaii -i''H i.- aid engineers. It has two fast reactors each oI'235
M\V capach v uid will use natural uranium as its fuel.
The first uiu f 235 MW capacity has started generating power
from .1983 aiu Lt- 235 MW unit is conimissionud iii 1985
The pressnrised buo y water reactors will use natural uritniulli
available in plety in India. The two turbines and steam generators
at the Kalpakkain atomic power project. are the largest (opacity
eneratiI1g sets inalled in our countr y . In this l)owci station about
88SF local machi u p rv and equipment liave been used
(it') Narora Nuclear Power Station. It is India's fourth
nuclear jIO et station and IsbelilgiWilt at. Naroiii in hullindshitliir
District of' Uttar Pr'adesh. This plant will initiall y have two units of
235 MV 'ahi and proviiefl hias,beein made to expand its caj)icity
of 500 MWI It IS ex 1 ucted to be corupleteti by 1991.
This plant 'viii have t\yo reactors of the CAN1)U--- 'II \V
(Canadian De ti trium_Ura! in—l'russuristd heav y \Vatt.r
tern and will ue natural uiniiuni as its fuel. 'ibis plait will ho
and cunttui'tt.'(l by the Indian scientists iIi(I (fl'
NUCLEAR POWER PLANT 379

gineers. The two units are expected to be completed by 1989 and

1990 respectively.
This plant will use heavy water as moderator and coolant. This
plant will provide electricity at 90 paise per unit. Compared to the
previous designs of Rajasthan and Madras nuclear power plants the
design of this plant incorporates several improvements. This is said
to be a major effort towards evolving a standardised design of 235
MW reactors and a stepping stone towards the design of 500 MW
reactors. When fully commissioned plant's both units will provide
0 MW to Delhi, 30 MW to Haryana, 15 MW to Himachal Pradesh,
35 MW to Jammu and Kashmir, 55 MW to Punjab, 45 MW to
Rajasthan, 165 MW to Uttar Pradesh and 5 MW to Chandigarh. The
distribution of remaining power will depend on the consumer's
demands. In this plant one exclusion zone of 1.6 km radius has been
provided where no public habitation is permitted. Moderate seis-
inicity alluvial soil conditions in the region of Narora have been fully
taken into account in the design of the structure systems and
equipment in Narora power plant.
Narora stands as an example Of a well coordinated work with
important contributions from Bhabha Atomic Research Centre,
Heav y Water board, Nuclear Fuel Complex, Electronics Corpora-
tion of India Limited (ECIL) and other units of Department of
Atomic Energy and several private and public sector industries
Instrumentation and control systems are supplied by ECIL. Bharat
Heavy Electricals Limited (BHEL) is actively associated with
Nuclear Power Corporation of India., It has supplied steam gener-
ators, reactor headers and heat exchangers for Narora Atomic
Power Plant (NAPP) 1 and 2 (2 x 235 MW).
NAPP is the fore-runner of a whole new generation of nuclear
power plants that will come into operation in the next decade. The
design of this reactor incorporates several new safety features
ushering in the state of the art in reactor technology. The design
also incorporates two fast acting and independent reactor shut down
systems conceptually different from those of RAPP and MAPl.
Soue of the new systems introduced are as follows
() Emergency Core Cooling S y stem (ECCS).
(ii) Double Containment System.
(iii) Primary Shut off rod System (PSS).
(it) Secondary Shut off rod System (SSS).
m,v) kutomnatic Liquid Poison Addition S y stem (ALPAS).
(vi) i'OSt accident clean up system.
According to Department of Atomic Energy (DAE) the Narera
Atomic Power Plant (NAPP) has the following features.
380 POWER PLANT

(i) It does not pose safety and environmental problems for

the people living in its vicinity.
The safety measures are constantly reviewed to ensure that at
all times radiation exposure is well within limits not only to the
plant personnel but also to the public at large.
(ii) NAPP design meets all the requirement laid down in the
revised safety standards. The design of power plant in
corporates two independent fast acting shutdown systems
high pressure, intermediate pressure and low pressure
emergency core cooling systems to meet short and long
term requirements and double containment of the reactor
building.
Narora Atomic Power Plant (NAPP) is pressurised heavy water
reactor (PHWR) that has been provided with double containment.
The inner containment is of prestressed concrete designed to
withstand the full pressure of 1.25 kg/cm 2 that is likely to be
experienced in the event of an accident. The outer containment is of
reinforced cement concrete capable of withstanding the pressure of
0.07 kg1cm 2 . The angular space between the two containments is
normally maintained at a pressure below atmosphere to ensure that
any activity that might leak past primary containment is vented out
through the stock and not allowed to come out to the environment
in the immediate vicinity of the reactor building. The primary and
the secondary containments are provided with highly efficient flu-
teration systems which filter out the active fission Products before
any venting is done.
The moment containment gets pressurised it gets totally sealed
from the,..environmnent. Subsequently the pressure in the primary
containment is brought down with the help of the following
provisions.
(i) Pressure suppression pool at the basement of the reactor
building.
(ii) Special cooling fan units which are operated on electrical
power obtainable from emergency diesel generators.
The containment provisions are proof tested to establish that
they are capable of withstanding the pressures that are expected in
the case of an accident. Fig. 5.12 (a) shows primary and secondary
containment arrangement.
(iii) The cooling water to all the heavy water heat exchangers
is maintained in a closed loop so that failure in these do
not lead to escape of radioactivity very little water from
River Ganga would be drawn for cooling purposes and
most of water would be re-cycled. -
NUCLEAR POWER PLANT 381

Fu e l
condary
C 1.0 d ding nfainment
Primary at transport
con fain me system

Fig. 5.12 (a)

(iv) The power plant has a waste management plant and

waste burial facility within the plant area.
(ii) NAPP is the first pressurised heavy water reactor
(PHWR) in the world to have been provided with double
containment.
(vi) No radioactive effluent, treated or otherwise will be dis-
charged into Ganga River. Therefore there will be no
danger of pollution of the Ganga water.
(vii) An exclusion zone of 1.6 krnradius around the plant has
been provided where no habitation is permitted.
(viii)A comprehensive fire fighting system on par with any
modern power station has been provided at NAPP.
(ix) NAPP has safe foundations. It is located on the banks of
river Ganges on alluvial soil. The foundations of the plant
reach upto a depth where high relative densities and
bearing capacities are met. The foundations design can
cater to all requirements envisaged during life of plant.
(x) It is safe against earthquakes.
(xi) In the event of danger over heated core of the reactor
would be diffused with in a few seconds by two features
namely shut down through control rods followed by injec-
tion of boron rich water which will absorb the neutrons
and stop their reaction in the core. This is in addition to
other feature like double containment system povided in
the reactor.
Above features assure total radiation safety of the plant person-
nel, general public and the environment during the operation of
power plant. With the completion of NAPP it would make a useful
contribution to the Northern grid thereb y accelerating the pace of
development in this region.
Narora Atomic Power Plant is the fburth atomic power project
to be commissioned in India, the others in commercial operation
 Qai
- being TAPS 1 and 2 (Tarapur, Maharashtra 1969), RAPS 1 and 2
(Kota, Rajasthan 1973 and 1981) and MAPS 1 and 2 (Kalpakkam
Madras, Tamil Nadu 1983 and 1986). This power plant is meant to
generate electricity and supply the same to the distribution system
(grid) in Uttar Pradesh and other states in the northern region. It
has two units each with a capacity of 235 MW of which about seven
per cent will be used to run the in house equipment and the rest will
be fed into the grid; The net output from the power plant will be
about 435 MW. At this power plant all due precautions have been
taken in the design, construction, commissioning and operation of
the unit with safety as the over-riding consideration. Therefore
there appears to he no danger to the public from the operation of
this power plant.
(u) Kakarpar Nuclear Power Plant. This fifth nuclear power
plant of India is to be located at Kakarpar near Surat in Gujarat.
This power station will have four reactors each of 235 MW capacity.
The reactors proposed to be constructed at Kakarpar would be of the
Candu type natural uranium fuelled and heavy water moderated
reactors-incorporating the standardised basic design features of the
Narora reactors suitably adapted to local conditions. The fuel for
the power plant will be fabricated at the Nuclear Fuel complex,
Hyderabad. ThQpower plant is expected to be completed by 1991.
The Kakarpur unit has two fast shut down systems. The
prirnry one works by cadmium shut off rods at 14 locations which
drop down incase ofheat build up and render the reactor sub-critical
in two seconds. There are 12 liquid shut off rods as a back up, further
backed by slow acting automatic liquid poison addition system
which absorbs neutrons completely and stop the fissile reaction.
In case of sudden loss of coolant, heavy water inside the reactor,
there is an emergency core.cooling system which also stops the tissue
reaction, Lastly, the pressure suppression system in which cool
water under the reactor rises automatically to reduce pressure in
case it increases and a double containment wall ensures that no
radioactivity would be released at ground leycl even in case of an
unlikely accident.
The Department of' Atomic Energy (DAE) has also evolved
emergency preparedness plans for meeting any accident even after
all these safety measures. It ensures a high level of preparedness tc
face an accident including protecting the plant personnel and sur.
rounding population. There s no human settlement for five km bell
around a nuclear power installation as a mandatory provision.
(vi) Kaiga Atomic Power Plant. The sixth atomic power plant
will be located at Kaiga in Karnatka. Kaiga is located away fron
human habitation and is a well suited site for an atomic power plant
It will have two units of 235 MW each. It is expected to hc uiiiiiiius-
sioned by 1995.
This nuclear power plant will have CANDU t y pe i ut r-.. These
reactors have modern systems to preveut accidents. ' i Lint would
have two solid containment walls- -inner and outer to guard
against any leakage. The inner contat iiineiit wat I could w thst and a
pressure of 1.7 kg/cri 2 and could prevent the plant troni bursting.
The outer containment walls of the reinforced ceili nt concrete has
been design to withstand pressure ot0.07 kg/ui-n 2 . The annular space
between the two containment walls would he maintained at a lower
pressure below that of the atmosphere to ensure that no radioac-
tivit y leaked past the primary containments.
5.25 Light Water Reactors (LWR) and Heavy Water
Reactors (HWR)
Light water reactors use ordiiiarv water technicatl known as
light water) as coolant and moderator. The y are simpler and
cheaper. But they require enriched tirar .uuin a their fuel. Natural
uranium contains ().6 of flssionahle isotope and 99.: of
fertile jj238 and to use natural uranium in such reactors it is to he
enriched to about 3' U' and for this urani.iiu enrichment plant is
needed which requires huge investment and high operational ex-
penditure. fleav y water reactors use h'avv water as their coolant
and moderator. 'l'hev have the advantage of using natural uranium
as their fuel. such reactors have seine operattoti nrrhlem to. Iavv
water preparation plants require utticient investment and leakage
of heav y water must he vonled as heavy water is vei\ cost lv I (eavv
water required lit primary circuit- must I.e 90 pure and this
requires purification plants l'uavvw;tftrsliuld rut absorb moisture
as b absorbing moisture it getstlgiaded In order to have suflicierit
quantity of heavy water required fOr nuclear power plants, the work
is fast progressing in our countr y - 1 four heav y water plant. These
plants are situated at Kotah 101) torlrtes per y ear). Bitrodit (67 2
tomes), luticori i (71.3 torines) awl Talcher 67 tme' pr e ya'.
er
These plants will give our countr y an installed hivv water pruluc-
hon capacitY of about 300 tonnes per year

5.26 Importance of Heavy Water

The nuclear power plants of Kota in Rajasthin. KaleikL.tru in
Tamil Nadu and Nirora in U.P. use heavy.water as coolant and
moderator. All these projects have CANDU reactors using nat ii rd
uranium as fuel and heav y water as moderator. After this enriched
uranium natural water at larapur. the CANDU reactors are
reactor.
the second generation ut reactor,, in India's nuclear puver
prograulme. The CAN DU reactor will produce plutonium which will
384 POWER PLANT

Ile the core fuel for fast breeder reactor. In fact in breeder reactor
heavy water is used as moderator.
A CANDU reactor of 200 MW capacity requires about 220
tonnes of heavy water in the initial stages and about 18 to 24 tonnes
each year subsequently. Therefore, about one thousand tonnes of
heav y water will be required to start the different nuclear power
stations using heav y water. The total capacity of different heavy
water plants wilt he about 300 tonnes per year ifafl the heavy water
plant under construction start production. It is expected that heavy
water from domestic production will be available from Madras and
Narora atomic power plants. The management of the heavy water
system is a highl y complicated affair and requires utmost caution.
Heavy water is present in ordinary water in the ratio I 6000. One
of the methods of obtaining heavy water is electrolysis of ordinary
water.
5.27 Advantages of Nuclear Power Plant
The various advantages ofa nuclear power plant are as follows:
1. Space requirement of a nuclear power plant is less as
compared to other conventional power plants are of equal
size.
2. A nuclear power plant consumes very small quantity of
fuel Thus fuel transportation cost is less and large fuel
storage facilities are not needed. Further the nuclear
power plants will conserve the fossil fuels (coal, oil, gas
etc. , for other energ y need.
3. There is increased reliabilit y of operation.
4. N ueleir power plants are-not effected by adverse weather
conditions.
5 Nuclear power phints are well suited to meet large power
demands. TI cv give better performance at higher load
factors 80 to 90U.
6. Materials expenditure on metal structures, piping.
storage inecliani.ins are much lower for a nuclear power
plant 01,111 ;i coal burning power plant. For example for a
100 N1 \\ n niclear power plant the weight of machines and
inclLtm ins, weight of metal structures, weight of pipes
and fittings and weight of Illasonrv and bricking up re-
ire nearl y 700 tonnes, 901) tonnes. 200 tonnes and
500 tonnes respectivel y whereas for a 100 MW coal burn-
ng power plant the corresponding value are 2700 tonines,
1250 tinne. 300 tonnos and 1500 tonnes respectively.
Further area of construction site required for 101) MW
nuclear power plant is a h e ctares whereas for a 100 MW


NUCLEAR POWER PLANT

385
coal burning power plant the area of construction site is nearly 15
hectares.
7. It does not require large quantity of water.
Disadvantages
1. Initial cost of nuclear power plant is higher as compared
to hydro or steam power plant.
2. Nuclear power plants are not well suited for varying load
conditions.
3. Radioactive wastes if not disposed carefully may have bad
effect on the health of workers and other population
In a nuclear power plant the major problem faced is the disposal
of highl y radioactive waste in form of liquid, solid and gas without
any injury to the atmosphere. The preservation of waste for a long
time creates lot of difficulties and requires huge capital.
4. Maintenance cost of the plant is high.
5. It requires trained personnel to handle nuclear power
plants.
5.28 Site Selection
The various itt: t. considered while selecting the site for
nuclear plant are as follows
I Availability of water. At the power plant site an ample
quantity of water should be available for condenser cooling and
made up water required for steam generation. Therefore the site
should he nearer to a river, reservoir or sea.
2. Distance from load centre. The plant should be located near
the load centre. This will minimise the power losses in transmission
lines.
3. Distance from populated area. The power plant should be
located far away from populated area to avoid the radioactive
hazard.
4.Accessibility to site. The power plant should have rail androad
transportation facilities.
5. Waste disposal. The wastes of a nuclear pow'r plant are
radioactive and there should be sufficient .,pace near th plant site
For the disposal of wastes.
6. Safeguard against earthquakes. The site is classified into its
respective seismic zone 1, 2, 3, 4, or 5. The zone 5 being the most
;eismic and unsuitable for nuclear power plants. About 300 km of
-adius area around the proposed site is studied for its past history
)f tremors, and earth-quakes to assess the severest earth-quake
hat could occur for which the foundation building and equipment
;upports are designed accordingly. This ensures that the plant will
awd
-retain integrity of structure, piping and equipments should an
earthquake occur. The site selected should also take into account
the external natural events such as floods, including those 1w
up-stream dam failures and tropical cyclones.
The most important consideration in selecting a site for a
nuclear power plant is to ensure that the site-plant combination
does not pose radio logical or any hazards to either the puhbc, plant
personnel on the environment during normal operation of plant or
in the unlikel y event of an accident
The Atomic Energy Regulatory Board (AERU) has stipulated a
code of practice on safety in Nuclear Power l'lant site and several
safety guide lines for implementation.
5.29 Comparison of Nuclear Power Plant and
Steam Power Plant
The cost of electricity generation is nearly equal iii both these
power plants. The other advantages and disadvantages are as
follows
(i) The number of workman required for the operation of
nuclear power plant is nuich less than ii steam power
plant. This reduces the cost of operation.
(ii) The capitul cost of nuclear power plant falls harplv lithe
size of plant is increased. The capital cost as structural
materials, piping, storage mechanism etc. is iiiuch less in
nuclear power plant than similar expenditure of steam
power plant. I lowever, the expenditure of nuclear reactor
and building complex is iiiucli higher.
(iii) The cost of power generation by nuclear power plant
becomes co!nlnlitive with cost of steam power plant above
the unit size of about 500 MW.
5.30 Multiplication Factor
Multiplication factor is used to determi ne whether the chain
reaction will continue at a steady rate, increase or decrease. It is
given by the relation,
P
K..E

where K = Effective multiplication factor.

1' = Rate of production of neutrons.
A :.: Combined rate of absorption of neutrons.
E = Rate of leakage of neutrons.
K = 1 indicates that the chain reaction will continue at steady rate
(critical) K > 1 indicates that the chain reaction will be building up


(super critical) whereas < I shows that reaction will he dying down
(subcritical).
5.31 Uranium Enrichment
In some cases the reaction does not take place with naturaL
uraniu : t i ning only 0.71% of U23.

F In such cases it becomes essential to use uranium containing

higher content of U 335 . This is called U 2 concentration of uranium
enrichment.. The various methods of uranium enrichment are as
follows : ENRICHED
PRODUCT

0 0 00
0 0 0 0 0 00 0 0
GAS
0-
000 0 00 0 00 0 00 0 0 0 000 0 0
J
INPUT 00%:O%0oO %oO%o0O%%o
00
- _ 0
00
000 0 - -,
DEPLETED
PRODUCT
I
00 0 •

0 0 001
1
• HEA VY MOLECULES
• LIGHT MOLECULES ENRIcI
Fig. 513
PRODUCT I
1. T/i.' gaseous diffusion ,iithvd. This meth based oil
principle that the diffusion or penetration molt ilar ota gas with a
given molecular weight through a porous barri r is quicker than the
molecules of a heavier gas. Fig. 5.13 shows a p:iriting stage for
gaseous diffusion. Non-saturated uranium he:a-t1ouride (UF 6 ) is
used for gaseous diffusion. The diffusing molecules have small
difference in mass. The molecular weight of U215
F6 = 235 + 6 x 19 = :349 and that of U 235 F6 = 352. The initial mixture
is fed into the gap between the porous harrier. That part of the
material which passes through the barrier is enriched product,
enriched in U 23 F6 molecules and the remainder is depleted
product.
2. Thermal difñision method. In this method (Fig. 5. 14) a column
consisting of two concentric pipes is used. Liquid UF 6 is filled in the
space between the two pipes. Temperature ofone of the pipes is kept
high and that of other is kept low. Due to difference in temperature
the circulation of the liquid starts, the liquid rising along the hot
wall and falling along the cold wall. Thermal diffusion takes place
in the column. The light U 235 F6 molecules are concentrated at the
hot wall and high concentration of U 236 F6 is obtained in the upper
part of the column.
 POWER PLANT

Enriched
product

Magnetic field
rgion-
72 -71
Liquid
UF6
i IRA

=.. Enriched
- product -.zI
_J

Fig. 5.14 . Fig. 5.15

3. Electromagnetic Method. This method is based on the fact that

when ions moving at equal velocities along a straight line in the
same direction are passed through a magnetic field, they are acted
upon by forces perpendicular to the direction of ion movement and
the fiold
Let P = force acting on ion
e = charge on ion
u = velocity of ion
H = magnetic field strength
m = Lrn mass
R = radius of ion path
P=evH
As this force is centripetal
mt'2
R
l.a
R = eL'H

mv
R
eH
This shows that ions moving at equal velocities but different
masses mbve along circumferences of different radii (Fig. 5.15). Fig.
5.16 shows an electromagnetic separation unit for uranium isotopes.
A gaseous uranium compound is fed into the ion source, where
neutral atoms are ionised with the help of ion bombardment. The


NUCLEAR POWER PLANT 389

ions produced come out in the form of narrow beam after passing
through a number of silos. This beam enters the acceleration cham-
ber. These ions then enter a separation chamber where a magnetic
field is applied. Due to this magnetic field the ions of different
masses move-along different circumference.

U2).

Ion
source II
: 1/ I

Magnetic
field
region

Accelerating
electrodes
H.V. Power
supply

Fig. 5.16

(J,
0 0 0
. el o Light rncule
• 0 0 •I 0 0
i. • • • . .1
a *Heavy molecule
• I.I a.
I. •a 010.0111
0 010 .1
.•. 01.00
L.° 0°OjOO .•

Fig. 5.17

4. Centrifugation Method. This method is based on the fact that

when a mixture oftwo gases with different molecular weight is made
to move at a high speed in a centrifuge, the heavier gas is obtained
near the periphery (Fig. 5.17). UF 6 vapour may be filled in the
centrifuge and rotated to separate uranium isotopes.
5.32 Power of a Nuclear Reactor
In a nuclear reactor a large ,number of neutrons are incident on
nuclear fuel atoms, causing fission and producing energy. A control-


390 POWER PLANT

led chain reaction takes place so that the heat energy released can
be controlled.
Let V = Volume of energy
N = Fuel atoms/m 3
n = Average neutron density, i.e. number per
m3
cx = Fission cross section
= Neutron flux
v = Average speed of neutrons rn/sec.
F ission cross-section represents the probability of fission per
incident neutron. For example ify is the number of incident neutron
then those causing fission = a y Neutron flux is the, number of
neutrons crossing a plane of area one metre square held at right
angle to velocity v.
= fl x v
S = total fuel atoms in reactor = N.V.
It = number of incident neutrons per second
fuel atoms
= S x = nv. N.V.
x = Number of neutrons causing fission per
second
= It 3< (1 = nv. N.V.a.
Now 3.1 x 1010 fission per second produce a power of one watt
(See example 5.3)
P = Power of nuclear reactor
X n.v.fv.V.a.
= -- = watts. -
:3.1 x 1010 3.1 x 1010
Let the fuel used in the reaction be U2
Mass per atom of U235
At. weight of U235 235
------ k
Avogadro Number 6.02 x 1021
Mass of NV atoms = N.V. x Mass per atom
= x - ---- . = M kg (say)
- 2(
6 . 02x10 -


NUCLEAR POWER PLANT 391

6.02 x 1026 x ,,
NV = ------ -----
23

n.p,N.V.a.
J) = -------------
Now -- watts
- 3.1 x 1010
1()26 M x 582 x 10
ç x 6.02 x
- 3.1x10'0x235
12
= 4.8 x 10 Mo watts.

5.33 Reactor Power Control

The power released in a nuclear reactor is proportional to the
number of mole fissioned per unit time this number being in turn
proportional to density of the neutron flux in the reactor. The power
of a nuclear reactor can he controlled by shifting control rods which
may he either actuated manually or automatically.
Power control of a nuclear reactor is simpler than that of
conventional thermal power plant because power of a nuclear reac-
tor is a function of only one variable whereas power of a thermal
power plant depends on number of factors such as amount of fuel,
its moisture content, 2ff supply etc. This shows that power control
of thermal plant requires measuring and regulating several quan-
tities which is of course considerably more complicated.

5.34 Nuclear Power Plant Economics

• Major factors governing the role of nuclear power are its
economic development and availability of sufficient amount of
nuclear fuel.
It is important to extroct as much energy from a given amount
offuel as possible. The electrical energy extracted per unit ofarnount
of fuel or expensive moderator might be called the "material
efficiency". In a chain reactor the high material efficienc y as well as
high thermal efficienc y leads to low over all energy cost.
Since the most attractive aspect of nuclear energ y is the pos-
ibilitv of achieving fuel costs cuniderahlv below that for coal, all
llu(-l1'ar power svstem heingconsidered fur large scale power produc-
tion involve breel!:ig or systenis. This program in -
cludes Inc h'velopnent (lila technolog y of low neutron absorbing
stiictural ii atei'iak such aS zirconium, the use of special moder:it-
ing materials such al)() and the consideration of special piOl)leilIs
associated with tust reactors. In so fur as i.'cunoniic' factors are
uncenied ii is i1ecc.sarv to ( - nsi(l,'r neutron ('CIA1011) V in a general
wa such a that measured b y the converiou ratio oi the system.
The conversion ratio is (14'hned as the atoms of new fuel produced
in fertile materialperatoni of fuel burnt. Thccunversiui ratio Varies

392 POWER PLANT

• with the reactor design. Its values for different reactors are inch
cated in Table 5.5.
Table 5.5
Type of r'actor Conter,um ratio -
BWRI'WRandSGR I .-. L - ..
Aqueous thorium breeder - -. 1.2
Fast breeder reictor

5.35 Safety Measures for Nuclear Power Plants

Nuclear power plants should be located far away from the
populated area to avoid the radioactive hazard. A nuclear reactor
produces a and 3 particles, neutrons and y-quanta which can disturb
the normal functioning of living organisms. Nuclear power plants
involve radiation leaks, health hazard to workers and community,
and negative effect on surrounding frests.
At nuclear power plants there are three main sources ofradioac-
tive contamination of air.
(1) Fission, of nuclei of nuclear fuels.
(ii) The second source is due to the effect of neutron fluxes on
the heat carrier in the primary cooling system and on the
ambient air.
(iii) Third source of air contamination is damage of shells of
fuel elements.
This calls for special safety measures for a nuclear power plant.
Some of the safety measures are as follows.
(i) Nuclear power plant should be located away from human
habitation.
(ii) Quality of construction should be of required standards.
(iii) Wastewater from nuclei r power plant should be purified.
The water purification plants must have a high efficiency of
water purification and satisfy rigid requirements as regards the
volume of radioactive wastes disposed to burial.
(iv) An atomic power plant should have an extensive ventila-
tion system. The main purpose of this ventilation system
is to maintain the concentration of all radioactive mi- -
purities in the air below the permissible concentrations.
(v) An exclusion zone of 1.6 km radius around the plant
should be provided where no public habitation is per-
mitted.
(vi) The safety system of the plant should be such as to enable
safe shut down of the reactor whenever required. En-
gineered safety features are built into the station so that
during normal operation as well as during a severe design
 NUCLEAR POWER PLANT
393
basis accident the radiation dose at the exclusion zone boundary will
be within permissible limits as per internationally accepted values.
Adoption of a integral reactor vessel and end shield assemblies,
two independent shut down systems, a high pressure emergency
core cooling injection system and total double containment with
suppression pool are some of the significant design improvements
made in Narora Atomic Power Project (NAPP) design. With double
containment NAPP will be able to withstand seismic shocks.
In our country right from the beginning of nuclear power
programme envisaged by our great pioneer Homi Bhabha in peace-
ful uses of nuclear energy have adopted safety measures of using
double containment and moderation by heavy water one of the safest
moderators of the nuclear reactors.
(vii) Periodical checks be carried out to check that there is no
increase in radioactivity than permissible in the environ-
ment.
(viii) Wastes from nuclear power plant should be carefully dis-
posed off. There should be no danger of pollution of water
of river or sea where the wastes are disposed.
In nuclear power plant design, construction, commissioning and
operation are carried out as power international and national codes
of protection with an over-riding place given to regulatory processes
and safety of plant operating personnel, public and environment.
5.36 Site Selection and Commissioning Procedure
In order to study prospective sites for a nuclear power plant the
Department of Atomic Energy (DAE) of our country appoints a site
;election committee with experts from the following:
(i) Central Electricity Authority (CEA).
(ii) Atomic Minerals Division (AMD).
(iii) Health and safety group and the Reactor Safety Review
group of the Bhabha Atomic Research Centre (BARC).
(iv) Nuclear Power Corporation (NPC).
The committee carries out the study of sites proposed. The sites
re then visited, assessed and ranked. The recommendations of the
ommittee are then forwarded to DAE and the Atomic Energy
ommission (AEC) for final selection.
The trend is to locate a number of units in a cluster at a selected
ite. The highest rated units in India are presently of 500 MW. The
idiation dose at any site should not exceed 100 milligram per
iember of the public at 1.6 km boundary.
The commissioning process involves testing and making opera-
onal individually as well as in an integrated manner the various
stems such as electrical service water, heavy water, reactor
.27
34 POWER PLANT

regulating and protection, steam turbine and generator. To meet the

performance criteria including safe radiation levels in the plant area
and radioactive effluents during operation the stage-wise clearance
from Atomic Energy Regulatory Board (AERB) is mandatory before
filling heavy water, loading fuel making the reactor critical, raising
steam, synchronising and reaching levels of 25%, 50%, 75% and
100% of full power. The commissioning period , lasts for about two
years.
Example 5.1. Calculate the number of fission in uranium per
second required to produce 2 kW of power if energy released per
fission is 200 meV.
Solution. P = Power =2 kW
E = Energy released per fission = 200 MeV
=200x 10 6 eV=200x 10 6 x 1.6x 10-12
3.2 x 10 ergs
P = 2 kW = 2000 Watts = 2000 joules/sec.
= 2000 x 10 7 ergs/sec = 2 x 10 10 ergs/sec
N = Number of fissions per Sec.
P 2x10'0 = 6.25 x 10' 3 . Ans.
E 32x 10-4

Example 5.2. A nuclear reactor uses as fuel. If the mass of

fuel is 1.2 kg and neutron flux is 10 16 per sec. Calculate the power of
reactor.
Solution. M = Mass of fuel= 1.2 kg
• = Neutron flux = 1016/sec
P = Power of reactor
= 4.8 x 10-12 m 40 watts
= 4.8 x 10 12 x 1.2 x 1016
= 57.6 x 103 watts = 57.6 kW.

Example 5.3. Calculate the fission rate of E935 for producing

power of one watt if 200 MeV if energy is released per fission of
u.
Solution. P = Power = 1 watt
E = Energy released per fission of U 235 nucleus
 NUCLEAR POWER PLANT
395
= 200 MeV = 200 x 1.6 x 10-13 j
= 3.2 x 10 hl watt sec.
as 1MeV=1.6x1013J
Fission rate of producing one watt of power
1
E3.2X10-11 = 3.1 x 1010 fissions/sec.

Example 5.4. A railway engine is driven by atomic power at an

efficiency of 40% and develops an average power of 1600 kW during
8 hour run from one station to another. Determine how much U35
vould be consumed on the run if each atom on fission releases 20 0
Me V.
Solution. Output = 1600 kW
Efficiency = 0.4
- Output
04 -
Input
Input __i_p_
- 0.4 - 4000 kW = 4 x 106watts.
E Energy released per fission = 200 MeV
=200x1.6x1012J=3.2x10-11J
t = Time = 8 hours = 8 x 3600 seconds.
Input nuclear energy required = Input x t
rr4x106x8x3600J=1152x1Q9J
Number of U 235 atoms required for 8 hour run
- 115.2x io_ 115.2x io
- E _ 3 2 10... ji=36x102°
We know that 235 gm of U 5 contains 6.02 x 10 23 atoms
(Avogaciro's hypothesis).
Mass of U 235 consumed
- 36 x 1020 x 235 -
6.02 x 1028 - 1.4 gin.

Exa pIe 5.5. Determine the energy released by the fission of 1.5
gm of (J in kWh assuming that energy released per fission is 200
MeV.

396 POWER PLANT

Avogadro number = 6.025 x 1023

Mass of Uranium = 235 a.m.u.
Solution. n = Number of atoms in 1.5 gm of U235
- 1.5 x 6.025 x 1023
235
E 1 = Energy released per fission = 200 MeV
= 200 x 106 x 1.6 x 10 19
= 3.2 x 10-11 Joules
E = Energy released by 1.5 gm of U235
= n xEj Joules
- 1.5 x 6.025 x 1023 3.2 x 10
x606003kWh
- 235
=3.42xlO4 kWh. Ans.

Example 5.6. Determine the fission energy released when a

U-235 nucleus is fissioned by a thermal neutron and two fission
fragments and two neutrons are produced. The average binding
energy per nucleon is 8 MeV in the fissioned U-235 nucleus and 8.8
MeV in the fission fragments.
Solution. E = fission energy released
= 234 x 8.8 - 236 x 8
= 158.4 MeV.

PROBLEMS
5.1. What is a chain reaction? How it is controlled?
5.2. What is a nuclear reactor ? Describe the various parts of a
nuclear reactor.
5.3. What are different types of reactors commonly used in nuclear
power stations? Describe the fast breeder reactor? What are its
advantages over sodium graphite reactor?
5.4. How waste is disposed off in a nuclear power station ? What are
main difficulties in handling radioactive waste?
5.5. Discuss the various factors to be considered while selecting the
site for nuclear power station. Discuss its advantages and disad-
vantages.
5.6. Write short notes on the following:
(a) Boiling water reactor (B.W.R.)
(b) Pressurised water reactor (P.W.R.)
(c) Multiplication factor.
(d) Fertile and fissionable material.
 NUCLEAR POWER PLANT 397

5.7. What are the different components of a nuclear power plant ?

Explain the working of a nuclear power plant. What are the
different fuels used in such a power plant?
5.8. What is a Homogeneous Reactor ? Describe a Homogeneous
Aqueou Reactor (MAR.).
5.9. What is meant by Uranium enrichment? Describe some methods
of Uranium enrichment.
5.10. Compare the economic (cost) of nuclear power plant with steam
power plant.
5.11. Explain the terms 'Breeding' and 'Burn up'.
5.12. Make a neat sketch and explain the working of a gas cooled
reactor.
5.13. State the properties of control rods.
5.14. Explain the properties of moderator used in a nuclear reactor.
5.15. Explain the principle of operation of a sodium graphite reactor.
5.16. Discuss the factors which go in favour of nuclear power plant as
compared to other types of power plants.
5.17. Write short notes on various nuclear power plants in India.
518. Write short notes on the following:
(i) Multiplication factor.
(ii) Moderating ratio.
(iii) Conversion ratio.
5.19. Compare the control of nuclear reactor and steam power plant.
5.20. Discuss the economics of a nuclear power plant.
5.21. Discuss the safety measures in nuclear power plant.
5.22. Describe various types of nuclear fuels.
5.23. Describe the objectives of R and Din nuclear energy in India.
5.24. Discuss the safety measures provided at Narora Atomic Power
Plant (NAPP).
5.25. Describe the site selection and commissioning procedure of
Nuclear Power plants in our country.

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. ...
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J f1J) m0 - •1 - 't ,
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 Hydro-Electric Power PIan.

6.1. Water Power

Water is the cheapest source of power. It served as the source
of power to our civilization in its earlier days in the form of water
wheels. Faraday's discovery of electricity has proved to be very
useful to use water for producing electric power. A hydro-electric
power plant is aimed at harnessing ppwer from water flowing under
pressure.
Hydro or water power is important only next to thermal power.
varly 30 11 of the total power of the world is met by hydro-olectri:
power. This was initiated in India in 1897 with a run of river schem
near Darjeeling. The first major hydro-electric development of 4i
MW capacit named as Sivasamudram Scheme in Mysore wa;
commissioned in 1902. In 1914 a hydro power plant named Khopol.
project of 50 MW capacity was commissioned in Maharashtra. Up
to 1947 the hydro power capacity was about 500 MW.
Water power has some inherent advantages as follows
(i) Runningcost of hydro power plant is low as compared to
steam power plant or nuclear power plant of same
capacity.
(ii) The hydro plant system reliability is greater than that of
other power plants.
(iii) The hydraulic turbines can be put off and on in matter of
minutes. Nuclear power plants and steam power plants
lack this facility.
(iv) Modern hydropower plant equipment has a greater life
expectancy which is about 50 years or more whereas as
nuclear power plant has an effective life about 30 years.
(v) Steam power plants have problem of ash disposal where
as hydro power plants have no comparable problem.
 HYDRO-ELECTRIC POWER PLANT 399

(vi) Modern hydro generators give very high efficiency over a

considerable range of load.
Although the capital investment of a hydro-electric power plant
is more but the operating cost of this plant is minimum as compared
to other power plants and power produced by this plant is cheaper
than the power generated by other power plants using coal, oil etc.
Besides power generation this plant is quite useful for irrigation and
flood control. This plant can be used both as base load plant and
peak load plant.
A schematic diagram of hydro-electric power plant is shown in
Fig. 6.1. Water surface in the storage reservoir is known as head
race level or simply headrace. Penstocks or canals are used to bring
water frQm the dam of the turbines fitted in the power house which
is built at some lower level. Penstocks are made up of steel, wood or
reinforced concrete. Water centers the turbine through the inlet
valve. Hydraulic turbines convert the potential energy of water into
mechanical energy. The mechanical energy developed by the turbine
is used in running the electric generator which is directly coupled
to shaft of the turbine.

CA 7CHMEN1
AREA

NE

Fig. 6.1

Water power can be divided into two types as follows:

(i) Primary or firm power
(ii) Secondary or surplus power.
Primary power is the power corresponding to minimum stream
flow with due consideration to the effects ofpondage and load factor.
It is the power always available to supply the load. The secondary
power is available only when quantity of water and storage are
cient.
sufficient.

400 POWER PLANT
*
Hydropower is a conventional renewable source of energy which
is clean, free from pollution and generally having good environmen-
tal effect. The pace of utilization of hydro potential during the last
decades has been slow compared to total energy development. Large
investments, long gestation period and increased cost of power
transmission are major obstacles in the utilisation of hydro power
resources.
Hydro power is important next to thermal power. About 30% of
total power of the world is met by hydro electric power plants. The
total hydro potential of the world is about 5000 GW. There are some
countries in the world where almost entire power generation is
hydrobased. For example in Norway the hydropower forms 99% of
total installed capacity.
Power output from a hydro power plant depends on the following
three factors:
(i) Head (ii) Efficiency
(iii) Discharge.
Power generation mainly depends upon the quantity of water
available.
6.2 Application of Hydro Power Plant
Hydro electric power plant can be used as independent power
plant. But this will require large amount of water to be stored and
at the times of low water flows the hydro power plant will not be
able to meet the maximum load as otherwise the maximum capacity
of the station has to be based on the maximum flow of water and
this will not prove to be economical. Therefore, the present trend it
to use hydro electrical power with a steam power station in an
,interconnected system. This will result in reduction in capital cost
of hydro electric power plant as the size of reservoir may be reduced.
In the interconnected system at times of low water flows the hydro-
power plant can be used as peak k_ -!ant and base load should be
supplied by steam power plant whereas iiuring periods of high water
flows the steam plants and hydro plants reverse their roles, the
hydro power plant taking the base load and steam power plant
supplying the peak load.
6.3 Essential Feature or Elements of Hydro-electric Power
Plant
The essential elements of hydro-electric power plant are as
follows
(i) Reservoir (ii) Dam
(iii) Forebay (iv) Trash rack
(v) Water way (vi) Draft tube
(vii) Surge tank (viii) Spill way
(ix) Power house and equipment.
HYDRO-ELECTRIC POWER PLANT 401

1. Water Reservoir. It is the basic requirement of a hydro-

electric power plant. Water reservoir is used to store water which
may be utilized to run the turbines to produce electric power.
Reservoir may be natural such as lake or artificial reservoir can be
built by erecting a dam across the river. Water held in upstream
reservoir is called storage whereas water behind the dam at the
plant is called pondage.
2. Dam. A dam is structure of masonry or some other material
built at a suitable location across a river. The primary function of
the dam is to provide a head of water. It also creates pondage or
storage. Economy and safety are the basic requirement of a dam.
Dam should be capable of reseting pressure of water and should be
stable under all conditions.
Dams are classified based on the following factors
(i) Function (ii) Shape
(iii) Construction materials (iv) Design.
Based on functions the dams may be called as storage dams,
diversion dams or detention dams. Storage dams are mainly for
storing water whereas diversion dams are constructed to raise the
water level and to divert the river flow in another direction. Deten-
tion dams are primarily used to store flood waters.
Based on shape the dams may be of trapezoidal section, and arch
type. The commonly used materials for constructing dams are earth,
rock pieces, stone masonry, concrete and R.C.C. concrete dams may
be plain as well as steel reinforced, earthen and rockfill dams are
the three most popular categories of dams based on the material
classification.
According to structural design the dams may be classified as
(a) Gravity dam (b) Arch dam
(c) Buttress dam (d) Earthen dams
(e) Rock fill dam.
A gravity dam is one in which the retained water thrust is
resisted by gravity action whereas in arch darn the thrust is resisted
by the arch action. A buttress dam resists the water thrust with the
help of buttresses which support the water through an inclined
structural member such as a buttress. Based on hydraulic design
dams may be classified as:
(i) Non over-flow dams (ii) Over-flow da
In non over-flow type dam water is not allowed iii uv over top
of dam whereas in overflow type water is allowed to flow over top of
dam.
The various structural material used for dams may be concrete
or stone masonry, earth, rock fill or timber. The type selected
depends upon the topography of the site, foundation condition and
402 POWER PLANT

eonomics. The foundation should be sufficiently impervious to

prevent seepage of water under the dam. Timber dams are rarely
used and need constant inspection to keep them in good condition.
Masonry dams are quite popular and are of three major types:
(i) The solid gravity dam (ii) The buttress dam
(iii) The arched dam.
6.3.1 Types of Dam. The different types of dams are as
follows.
(i)Masonry Dam. Masonry dams may be sub-classified as
follows
(a) Gravity dam. This type of dam resists the pressure of water
by its weight. The construction material used for this dam, is solid
masonry or concrete. Fig. 6.2 shows a gravity dam.

HEAD
RACE

_f\ ' L RACE

Fig. 6.2 Fig. 6.3

A gravity dam is subjected to the following forces:

(i) Water pressure (ii) Weight of dam
(iii) Uplift pressure (iv) Earth pressure
(v) Reaction of the foundation
(vi) Earth-quake pressure (vii) Wind pressure.
These forces should be taken into account while analysing the
structural stability of dam.
(b) Arch dam. It resists the pressures of water partly due to its
weight and partly due to arch action. This type of dam is located in
relatively narrow valley with steep slopes suitable for arch abut-
ment (Fig. 6.3). The main advantage using an arch dam is that the
HYDROELECTRIC POWER PLANT 403

amount of masonry or concrete required is much less than a gravity

darn of coinpared height, consequently the material cost is much
less. This type of dam cannot use the over fall type spillway as in
gravity darn. The spillway has to be separately provided which
increases the cost of outlet works and overall cost of the project.
(c) Buttress Darn. A buttress dam consists of buttresses support-
ing a flat slab or reinforced concrete. This type of darn is selected
when the cost of reinforced concrete is high (Fig. 6.4).
Masonry darns have the aJ intages of maximum height, longest
life most economical in water conservation and lowest maintenance
cost as compared to earth dam.

Fig. 6.4

( ii) Earth Dams. Earth dams are used for smaller power plants.
They can be built safety and economically on all types of foundations
of earth and rock. They can be further sub-classified as follows
(a) Earth Dam (b) Rock Fill Dam.
Earthed dams are used when effective height ofdam is not large,
river banks are not steep and the site is unable to take the weight
of gravity dam. Fig. 6.5 shows earth dam.

REINFORCED CONCRETE
CORE WALL

STONE

IMPERVIOUS

X-1
.:'
LL
MATERIAL
WXL.
NEARRV

Rock

Fig. 6.5


404 POWER PLANT

(iii) Rock-fill dam. This time of dam is preferred where

adequate quantity of good rock is available near the dam site. It is
used extensively in remote locations. Various parts of dam are as
follows
(a) Loose rock fill.
(b) An up stream dry rubble cushion of laid up stone bonding
into the dumped rock.
(c) An up stream impervious membrane resting on the dry
rubble cushion.
Fig. 6.5 (a) shows cross section of a rockfill dam.

Water level rLoose Rockfjll

- -
Natural
M embrcire_. slope

ry rubble
Fig. 6.5 (a)

Various advantages of rock fill dam are as follows

i) 1-ugh resistance to earth quake.
(ii) Foundation rock need not be as strong as for gravity dams.

canal

5pifl wQy

Fore bQY jygr

pØer_
house

Fig. 6.6

3. Forehay. It acts as a sort of regulating reservoir temporarily.

Water is temporaril y stored in the forebay in the event of rejection
of load by the turbine and there is withdrawal of water from it when
 HYDRO-ELECTRIC POWER PLANT 405

load is increased. In diversion canal plants water of the river is

diverted away from the main channel through a diversion canal. The
end of canal is enlarged in the form of forebay as shown in Fig. 6.6.
The forebay is provided with some type of outlet structure to direct
water to pen stock.
Following are the parts of a typical forebay : (i) Entrarree bay or
basin ; (ii) Spillway ; (iii) Flushing sluice ; (iv) Screens ; (v) Valve
chamber or gate chamber; (vi) Penstock inlet.
4. Trash rack. It is provided for preventing the debris from
getting entry into the intakes from dam or from the forebay. Manual
cleaning or mechanical cleaning is used to remove the debris from
trash rack.
Trash rack is made up of steel bars and it is placed across the
intake to prevent the debris from going into the intake.
The spacing of bars depends upon the following factors
(i) Type of turbine; (ii) Size of floating material.
If floating material is large and height of trash rack structure
is more, mechanical cleaning is economical.
(iii) Velocity of flow through trash rack.
The velocity of flow (V) through. trash rack should be kept
with-in limits so that it does not cause great loss of head. It is given
by
V= 0.12 Vji
where h = difference in head.
Velocity greater than 90 cm/sec. may cause the trash rack
structure to vibrate if bars are not rigidly stiffened and supported.
5. Waterway. A waterway is used to carry water from the dam
to the power house. It includes canal, penstock (closed pipe) or
tunnel. Tunnel is made by cutting the mountains where topography
prevents the use of a canal or pipeline. Tunnel is made to save the
distance. Various appliances used to control the flow of water are
called head works or control works. Head works include gates,
valves, and trash rack etc. Gates in the dam are very useful in
discharging the excess water during flood period. The various types
of gates used are sluice gates, tainter gates and rolling gates.
Penstocks are made up of steel, reinforced concrete and wood.
For smaller developments cast iron pipes may be used. The intake
of penstock at the dam or forebay of canal shcmld be at a level low
enough to provide an adequate water seal under all conditions
particularly at low water. It is desirable that the penstock should
be sloping towards the power house and its grade may be varied as
desired to suit the topography. Sharp bends in the penstock should
be avoided because they cause loss of head and require special
406 POWER PLANT

anchorages. Generally penstocks are not covered because exposed

pipes are cheaper and easily accessible for repairs and maintenance.
Covered penstocks should be used where topography is such that
there is danger from slides of snow, rock, earth etc. The velocity of
water in penstock is about 2 rn/sec in low head power plants and
about 4 rn/sec for medium head power plants and 7 mlsec for high
head power plants. The diameter and cross-section area of a
penstock pipe would be smaller if it were possible for the velocity of
water to be high.
Penstocks may be of two types
(i) Exposed pen stocks (ii) Buried pen stocks
Advantages of exposed penstocks are as follows:
(i) It is less expensive to install
(ii) Maintenance is easier
(iii) Inspection is easier
(iv) Life is more.
In case of buried pen stock are preferred where land slide, snow
slide or falling rocks may take place. In very cold climates where
freezing is expected buried pen stocks are used. However corrosion
is more and life is less in case of.biried penstocks.
6.4 Selection of Site for a Dam
The selection of site for a dam depends upon the followiTz
factors
(i) Function of dam. (ii) Type of dam.
(iii) Cost of dam : The cost of construction and cost of
tenance of dani should be low. To reduce the cost of dam the le
of the dam should be small. This requires that the site should be oi
where river valley has a neck formation as shown in Fig. 6:6 (a) or
the dam should be located after the confluence of two rivers as shown
in Fig. 6.6 (b) taking the advantage of both the valleys to provide
larger storage capacity.

River
River River
(

I
(a) (b)
9Z r
HYDRO-ELECTRIC POWER PLANT 407

(iv) Geological features : Sub soil investigations should be car-

ried out to check that soil or rock strata would provide entirely
satisfactory fo.undation.
(o) Accessibility from the view point of transportation of
materials.
(vi)The area which would get submerged after the construction
of dam.
(vii) Safety : The dam should be safe against floods and
catastrophes like earth-quakes.
6.5 To Calculate Penstock Thickness
The thickness of steel penstock depends on water head and hoop
stress allowed in the material. It is calculated using the following
relation:

2.f t
where t = Penstock thickness
d = Diameter of penstock
1= Permissible stress
P = Pressure due to water including water
hammer
= wH
w being the specific weight of water and H head of water.
il = Joint efficiency.
The joint efficiency is about 80% for riveted joints and 90% for
welded joints.
6.5.1 Number of penstocks
A hydro power plant uses a number of water turbines which are
to be supplied water through penstocks. Following alternative
choices may be considered
(i) To use a single penstock for the whole of plant. Such a
penstock will have a manifold at its end with as many
branches as the number of turbines.
(ii) To use one penstock for each turbine separately. In this
case each turbine draws water independently directly
from the reservoir.
(iii) To provide multiple penstocks but each penstock supply-
irg water to atleast two turbines.
408 POWER PLANT

Following factors should be considered while selecting the num-

ber of penstocks to be used to supply water to the turbines
(i)Economy. From economic point of view if length of penstock
is short one penstock should be provided for each turbine. For longer
penstocks a single penstock or as few penstocks as possible may be
used.
(ii) Operational safety. Use of a single penstock is not desirable
because any damage to the penstock would necessitate a complete
shut down at all the turbines.
(iii) Transportation facilities. The size of penstock selected
should be such that transportation of penstock should be easier.
6.5.2 Anchor blocks for penstocks
Anchor blocks act as supports for penstocks. They are massive
concrete blocks encasing the penstocks at intervals to anchor down
the pipe to the ground securely. The blocks are provided at ill
horizontal and vertical bends of the penstock pipe. The anchor
blocks may completely encase the penstocks as shown in Fig. 6.6 (c)
or the anchor block may be construc .tcd up to centre of penstock as
shown in Fig. 6.6(d) anchor blocks provide necessary reaction to the
dynamic forces at the bends. In general the anchor blocks by trans-
mitting loads to the ground provide the necessary degree of stability
to the penstock.

Penstock

Penstock

IMP

Anchor
Block

(c) E.
Fig. 6.6

6.5.3 Penstock Joints

The penstocks may be joined together by
(a) riveting (b) welding
Welding has the following advantages
(i) It is quick
(ii) Joint is strong
(iii) Joint is hydraulically more efficient.
The welded penstocks are subjected to the following tests.
(i) Ultrasonic testing


HYDROELECTRIC POWER PLANT

409
(ii) Radiographic testing
(iii) Testing using Dye-penetrants.
Example 6.1. Determine the thickness of a steel penstock 2.5
metre in diameter if the water head is 200 metres. The allowable
stress in steel is 1100 k9/cm 2 and the e
fficienc y of the joint is 80%.
Solution. d = Diameter of penstock = 2.5 metres
P = u'H= 1000x200= 2x 105kg/M2
1= Allowable stress iioo kg/cm2
= 1100 x 10 4 kg/M2 = 11106 kg/M2
(n) Efficiency 0.8

2./ii
2x105x25
= 1
2X11X106 x08 3-.2 metre
100=2.8 cm. Ans.
6. The Power House and Eq
uipment. Power house consists
of the main building of hydro electric development where the con-
version of energy of water to electrical energy takes place. Some
important
follows items of equipment provided in the power house are as
(i) Turbines (ii) Generators
(iii) Governors
(iv) Relief valve for penstock fittings
(u) Gate valves
(ii) Flow measurement equipment
(vii) Air duct (viii) Water circulating pumps
(ix)Transformers
(x) Switch board equipment and instruments
(xi)Reactors
(xii) Low tension and high tension bus bar
(xiii) Oil circuit breakers
(xi) Outgoing Connections (xiv)Storage batteries
(xvii) Shops and offices. (xvi) Cranes
Pover house should be Structurally stable. The layout of the
ower house should be such that sufficient space is provided around
he equipment like turbines generators, governors, valves, pumps
tc. in order to facilitate the dismantling and repairing.
A power house has three distinct sub divisions in most cases.
(i) The sub-structure (ii) Intermediate structure
(iii) The super structure
 POWER PLANT
410

The sub structure of a power house is that part which extends

from top of generator to the soil or rock. It houses the most of the
generating equipments. The sub-structure has to accommodate the
draft tube in addition to other equipments in case of Francis and
Kaplan turbines.
The intermediate structure of a power house is that part of the
power house which extends from the top of the draft tube to top of
generator foundation.
The super structure is that part which is above the generator
level housing mostly the cranes for handling the heavy equipments
in the sub-structures.
Depending upon stibgrade a power house may be classified as
follows
(i) Resting on rock. (ii) Resting on soil.
The sub grade design will be based on the soil or rock condition
below the foundation.
Depending on the type of super structure and covering provided
and housing of generating units a power house may be classified as
follows
(i) Out-door type (ii) Semi out-door type
(iii) In-door type.
In out door type power houses the generators etc. are provided
with local steel casing for weather protection but are full y exposed.
In semi out door type a low roof or deck is provided immediately
over the generators, the other area being open. The indoor type
power station has the conventional -structure and all generating
system are covered under a roof. This is the most commonly used
type of power house.
7. The Tail Race. Tail-race is a waterway to lead the water
discharged from the turbine to the river. The water after doing work
on turbine runner passes through the draft tube to tail race. The
water held in thii tail race is called as tail-race water level.
6.6. Layout of Hydro Power Ilant
The layout of .i power plant depends on the following
fac.ors
(i) 1 e services to L performed by the power plant
(ii) burrouncling topography
(iii) Controlling economics.
These factors in turn govern the type and arrangement of dams,
spillways and conduit systems.
The power plant building encloses generators, f, 'cr
crane, control room, offices, ancillary rooms etc. The cranc 1rovidecl
in the power house building should be high enough to clear the
 HYDR O-ELECTRIC POWER PLANT
411
g
enerator and should be able to pull up turbine runner out ofspji
casing and place it on op eration floor for maintenance .
and switch gears are located outside the building. Tra nsfonners

ROOF

GANTRY CRANE

iRa P1510 RM€ 14

VA1V7
L

InE
w,
ELIEF VALVE

TURBINE
U BE
uNWAT(RIP.Q PIPE A

Fig. 6.7

Fig. 6.7 shows a typical layout of a hydro power plant using a

<aplan turbine. The height of building is decided by Considering the
equired clearance for the crane when handling largest part of the
quipifient. The spacing between generators units is determined
fter finding the overall dimensions of the generators and the
urbines.
.7 Classification of hy dra-electric Power Plants
The classification of hydro electric power plants depends on the
llowing factors
1. Quantity of Water. (ri I
'fRiverP/t,t This type ofpower
ant has no control over thl
flow and uses the water as it
f

comes. During the rainy season

r*AVELI. I '3__.,fr
CftflWL
high water flow is available and
GgP1cR,p 'I * .j0 cNtry if the power plant is not able to
use this large flow of water some
... quantity of water is allowed to
TRASH SCRCgx
1SCROLCA5L
flow over darn spillways as wast .
ORAl? YQ$L Whereas during dry season, due
to low rates of water, the power
Fig. 6,8
produced by such plants will be
low. It is shown in Fig. 6.8.
- (b) Storage Plant. This type of power plants has facilities for
storing water. During rainy seasons the excess water is stored in
the reservoir and it is released to supplement low rates of flow
during run off (dry) periods. The advantage of this plant that the
power generated by the plant during dry season will not be affected.
(b) Pumped Storage Plant. Pumped storage plant (Fig. 6.9) in
combination with hydo electric power plant is used for supplying
the sudden peak loa d of short duration - a few hours in year. The
water leaving the turbines of hydro electric power plant is stored in
tail-race pond. This water is pumped back to the head race reservoir
by means of reversible pump turbine sets and is used for power
generation at the peak load time. Pump storage plants are generally
inter connected with other plants such as steam power plants. The
off peak capacity of steam power station can h used for pumping
water in the head reservoir.

AM

PENSTOCK P^Ep ML
POWER HOUE
DAM

Fig. 6.9

A 400 MW pumped storage plant is at design stage at Kadam-

parai in Tamil Nadu. This will be built with the existing upper
liyar Reservoir at the lower pool, tFic higher pool being constructed
at Kadamparai river. There will be four 100 MW reversible units in
an underground power houe.
Advantages of pumped storage plants are as follows:
(i) The capital cost of pumped storage plant is low as com-
pared to other peaking units.
(ii) There is a great deal of flexibility in the operational
schedules of the system.
(iii) A pumped storage plant can pick up load rapidly and is
dependable.
(iv) They operate at higher load factors and improve the
overall efficiency of the system.
They are used as peak load plants.
Fig. 6.9 (a) shows the pumped storage power plant for peak load
in conjunction with steam power plant as base load plant.

•; • -

;. I.

-
I-

-U
 HYDRO-ELECTRIC POWER PLANT
413

u pp er basin
Steam power plant

Pressure U
pipe tI Motor or
Z— generator
Va(ve

Vove
Purnp(4JfJ Turbine

LI

Fig. 6.9 (a)

2. Availability of Head of Water.
According to head of water
available the hy dro-electric power plants
lows: can be classified as fol-
(a) Low Head Plant.
metres the plant is namedWhen the operating head is less than 15
as low head plant. This type of plant uses
vertical shaft Francis turbine or Kaplan turbine. A small dam is
built to provide the necessary head. A low head plant (canal water
power plant) is shown in Fig. 6.10.

CANAl.
DAM PENSTQCIçS

Fig. 6.10
(b) Medium Head Plant. When the operating head of water is
from 15 to 50 metres the power plant is called medium head power
plant. This type of plant uses Francis turbines. The forebay provided
at the beginning of penstock serves as water reservoir. The forebay
draws water from main reservoir through a canal or tunnel. Forebay
also stores the rejected water when the load on the turbine decreases
(Fig. 6.11).
(c) High Head Power Plant.
When the hed of water exceeds 50
metres the plant is known as high head power plant. A surge tank
is attached to the penstock to reduce the water hammer effect on
the penstock. Pelton turbines are used in such power plants. Fig.
6.12 shows a high head-power plant.


POWER PLANT
414

- -1
1

Fig. 6:11

DAM SURGE TANK

TOCK
cwEgE

Fig. 6.12

Following are the main component parts of a high head

hydroelectric power plant.
(i) Reservoir (ii) Darn
(iii) Intake stfucture such as (a) Control gates (b) Screen
(iv) Penstock (v) Surge tank
(vi) Power house (vii) Tail race channel.

Head LA A" Vent

Wonal Head

);9w fiad
(7a t a i/o io t

Filter Gate

Trash Packs

Head (ate Penstock

'ter

Fig. 6.13
 HY DRO-ELECTRIC POWER PLANT
415
Fig. 6-1:3 shows a typical high pressure intake head works
Trash racks remove floating or submerged debris dire
ctly at the
intake entrance. Filter gate balances water pressure fbr opening the
gate whereas air vent prevents penstock vacuum
3. Classification according to topography
(i Low land (ii) Hill y area
(iii) Mountatieous region.
4. Cl assification according to load supplied.
) Base load plant
(b) Peak load plant.
5. Cla ssification according to capacit y of plant
(i Micro hydel plants - upto 5 MW
(j) Medium capacity plants —5 to 100 MW
(iii) I ligh capacity plants - 101 to 1000 MW
(iv) S tiper plants - above 1000 MW.
6. Classification according to turbine characteristic i.e.
specific speed.
(i) High specific speed
(ii) Medium specific speed.
6.8 Advantages of H ydro-electric Power Plant
The major tactorwhich goin fav ourofhvclroclectrjc power plant
are as follows
1. Water is the cheapest an: fellable source ofgeneration of
electric power heca it exists as a free gift of nature
2. There are no ash . . poslI problems. Also the atmosphere
is not polluted because no smoke is produced in this plant
3. No fuel transport fion problem.
1. It can take up the loads quickl y and it is capable of meeting
the variable loads without any loss in efliciericy.
5. Its maintenance cost is low.
6. It requires less supervising staff
7. Auxiliaries needed in the plant are less as compared to
steam plant of equal size.
8. Running cost of the plant is low.
9. In addition to the power generation such plants are used
for irrigation and flood control purposes also.
10. Hvdro electric power plants have become economics coin-
petitors with steam power plants and will acquire more
economic advantage with the rise in price of coal and oil.
11. The life of the plant, is more and the effect of age is
COmparativel- siiiall oil overall efficiency of the plant.
Disadvantages. The various disadvantages are as follows
1. The power produced by tilepla nt
depends upon quantity
of water which in turn is dependent upon the rainfall, so
416 POWER PLANT
if the rainfall is in time and proper and the required
amount of water can be collected, the plant will function
satisfactorily otherwise not.
2. Hydro-electric plants are generally situated away from
the load centres. They require long transmission lines to
deliver power. Therefore, the cost of transmission lines
and losses in them will be more.
3. Initial cost of the plant is high.
4. It takes fairly long time for the erection of such plants.
6.9 Mini and Micro Hydro Power Plants
Micro hydro power plants (up to 100 kW capacity), mini hydro
power plants (up to 1 MW) and small plants (up to 10 to 15 MW) can
be successfully used for electrification of isolated pockets of villages
where the transmission system cannot easily reach. This includes
exploitation of mountain streams rivulets and canal drops. The
power from mini-hydro power plants and micro-hydro power plants
will be attractive as the energy generated can be consumed near the
source itself without a big network of sub transmission systems.
Micro/mini hydro development is an extremely valuable,
economic, renewable energy source in its endeavour to meet
decentralised rural need of the country.
The potential in small hydro-electric projects in various
countries ranges from 5 to 15% of the total hydro-potential of that
country. In India the estimated hydro energy potential ofmicro/mini
hydro power plants leads to an installed capacity of 10000 MW
spread over 5000 to 6000 schemes. Small hydro power plants (SHP)
which generate electricity in small scale are now being developed.
These power plants are an attractive renewable source of energy in
remote and hilly regions isolated from main grids.
Various advantages of small i. '-o-power plants (SHP) are as
follows
(i) Readily accessible source of renewable energy.
(ii) Can be installed making use of water head as low as 2 m
and above.
(iii) Does not involve setting up of large dams.
(iv) Least polluting.
(v) Limited initial investments and short gestation pèridds.
(vi) Reduced transmission losses.
The Indian Renewable Energy Development Agency (IREDA) is
encouraging both public and private organisations to set up
smalVininiJmicro hydel power plants.
In our country nearly 120 micro/mini small hydel schemes (up
to 3 MW capacity) with a total capacity of about 86 MW are in
operation. Nearly 117 small power plants of total 125 MW capacity
HYDROELECTRIC POWER PLANT 417

are under con .truction. Eighth five year plan proposals envisage a
capacity addition through such schemes of about 600 MW.
The potential of small hydro power plants (up to 5 MW capacity)
is estimated to be around 5000 MW. This remains largely untapped.
6.10 Draft Tube
It is an integral part of reaction turbine. Draft tube connects the
runner exit to tail race. The area of the top of the draft tube is same
as that of the runner to avoid shock and is of circular cross-section.
The water after doing work on the turbine runner passes through
the draft tube to the tail-race. Draft tube is a metallic pipe or
cQncrete tunnel having gradually increasing cross-sectional area
towards outlet to ensure that as little energy possibly is left in water
as it discharges into the tail-race. Draft tube provides a negative
section head at the runner outlet by which it becomes possible to
install the turbine above the tail-race level without an y loss of head.
Secondly the velocity of water leaving the runner is quite high. Thus
the kinetic energy of water due to this velocity will he lost if the
water is allowed to be discharged freely. So by passing water
through the draft tube, the outlet velocity of water is reduced
considerably and gain in useful pressure head is achieved, that is
the net working head on the turbine inr aes and thus output of
the turbine also increases
6.10.1 Types of Draft Tubes
1. Straight Divergent Tube. it is used by the low specific speed
vertical shaft Francis turbine. This draft tube improves the speed
regulation (luring falling load [Fig. 6.14 (a]. Its maximum cone
angle is S
SECTION AT
(I) (2)

(1
(a) J "^
' '
(2)

0?pw 0
(C) (d)
Fig. 614


418
POWER PLANT
2. Moody Sprathng Draft iii/'.
This draft tube [Fig. 6.14 (ô )j
has all that its conical portion at the Centre reduces the
whirl action of water moving with high velocity centre reduces.
3. Simple Elbow draft tube [Fig. 6.14 ((.)] and draft tube with
circular inlet and rectangular outlet lFig, 6,14 (d[ require lesser
excavation for their instaIIatiin arid thus cost of excavation is less
for such draft tubes.
Efficiency of draft tube is given Lw the following formula
- .
n Efficiency
where H = Actual CO UVCI - S j OI1 of kinetic head into
pressure head

2g 2g

Ill
2g
where V1 = Velocity ofwatr at inlet, of draft tube
V2 = Velocity of water at outlet of draft tube

= Loss of head in draft tube.

6.11 Surge Tanks
Surge tank is a storage reservoir fitted to the penstock at a point
near to the turbine. The functions of the storage tank call
summarized as follows
When the load oil turbine decreases, the gates ofthe turbine
are closed partly by the governor to adjust the rate of flow of water
in order to maintain the constant speed of the runner. When the
gates are closed the water is moving to the turbine has to move
backward. This backward moving water is stored in the surge tank.
Similarly, when the load oil turbine increases, the turbine gates
are opened by the governor and increased demand of water is partly
met by the water stored in the surge tank. Thus the surge tank
controls the water when the load on the turbine decreases and
supplies water when the load on the turbine increases. In this wa
surge tank controls the pressure variations resulting from the rapid
changes in water flow in penstock and thereby prevents water
hammer effects. The functions of a surge tank are similar to a
fl y wheel fitted to all engine.
The surge tank may be opened or closed at the top. If it is open
at the tope it should be lower than the level ofwater in the reservoir.


HYDRO-ELECTRIC POWER PLANT 419

The various types of surge tanks are shown in Fig. 6.15. The
surge tank shown in Fig. 6.15 (a) has a cylindrical storage reservoir
connected by a vertical branch of pipe to the penstock. The sur-
getank in Fig..6.15 (b) has a conical reservoir whereas surge tank
in Fig. 6.15 (r) has a bell mouth spillway to discharge the excess
water. Fig. 6.15 (d) shows a differcntial surge tank. It has a central
riser with a small hole at its lower end. The water enters the surge
tank through this hole and the function of the surge tank depends
upon the area of hole. The surge tank shown in Fig. 6.15 (e) consists
of separate galleries ; the upper one is o,ed to store water when the
load on the turbine dec., ses and the lower one to suppl y water
when the load on the turhin increases.

0VEP FLOW

PENSTOCK (b) (C)

RISER j'71
(d) (e)
Fig. 6.15

6.12 Safety Measures in Hydro Power Station

Safety measures are provided for the safe operation. The various
safety measures provided in a hydro power plant are as follows
(i) Surge tanks (ii) Pressure regulators
(iii) Screens (iv) Fish passes
(v) Sand traps (vi) Spillways.
Surge tanks are placed as near as possible to the water turbine.
The surge tanks are provided to overcome the inertia forces in the
supply pipe during fall of load and to act as reservoir of water during
increase of load. A pressure regulator is provided near the turbine
and its function is that when turbine gates are suddenly closed,
pressure surges so produced are kept within the safe limits of the
pipeline. Screens are used to prevent logs, fishes ice blocks and other
obstructive elements from entering the pipelines and turbines
whereas fish passes are provided along the dams so as to allow the
fishes to pass to upstream without obstructions. Sand traps are
provided to prevent the sand flowing with water in pipes.
A spillway is also considered as safety valve for dam. It dischar-
ges major flood water without damage to dam.
 420
POWER PLANT
6.13 Hydraulic Turbines
In a hydro power station water turbine are designed to work
well under the condition that the head does not fall below a certain
minimum level. The minimum head on the turbines determines the
dead storage behind the dam.
Hydraulic turbines have the following advantages.
(i) Simple in construction
(ii) Easily controllable
(iii) Efficient
(iv) Ability to work at peak load or stand by plant
(v They can work on load variations and can run for weeks
without much attention
(vi) They can start from cold conditions and pick up load in a
short time.
Main t ypes of turbines aie
(i) Impulse (ii) Reaction.
In impulse turbines the wheel passages are not completely
filled, water acting on wheel buckets is at atmospheric pressure and
is supplied at few points at the periphry ofwheel and kinetic energy
is supplied to the wheel.
In reaction turbines water passages are completely filled with
water, water acting on wheel vanes is under pressure greater than
atmospheric, water enters all round the periphery of wheel, and
energy in the form of both pressure and kinetic energy is utilised by
the wheel.
The essential parts of a reaction turbine are as follows:
(i) Runner .(ii) Guides and gates
(iii) Speed ring (iv) Casing or approach flume
(c') Draft tube.
Hydraulic turbine consists of the following parts
1.Runner. Runner or wheel is fitted with vanes on its periphery.
It rotates due to action of water gliding on the vanes.
2. Guiding mechanism
line (penstock) to the runner.It guides the water coming from the pipe
3. Tail race.
The water after passing over the moving vanes of
the runner flows to the tail-race.
6.13.1 Classification
Depending upon the various factors the water turbines are of
two types:
6.13.1 (a) Based on action of water on moving blades
(a) Impulse turbinr. (b) Reaction turbines
 421

1. Impulse turbine. In impulse turbines water coming out of the

nozzle at the end of penstock is made to strike a series of buckets
fitted on the periphery of a wheel or runner. Pressure energy of
water is converted entirely into kinetic energy. The pressure at the
inlet to the runner and discharge is same and is atmospheric
throughout. The majority of impulse turbines are horizontal shaft
turbines. Impulse turbines are suitable for water moving with high
velocity and for high heads. The efficiency of the impulse turbine is
not dependent upon the volume of water flow which may be supplied
to the entire wheel. Pelton turbine (Fig. 6.16) is an impulse turbine.
The component parts of a Pelton wheel are as follows

RUNNER
BUCKET

(V.

........

Fig. 6.16
(a) Buckets (b) Nozzle and needle valve
(c) Shaft (d) Penstock
(e) Tail race channel.
Buckets are the hemispherical cups bolted to a shaft. They are
the main component parts which absorb and transmit the energy of
water. Nozzle imparts a very high velocity to water. Needle valve is
used to control the water discharge to the turbine. Penstóck carries
water from the reservoir to the turbine.
Water coming out of the turbine .is discharged into the tail race.
Fig. 6.16 (a) shows multijet Pelton wheel arrangement.
Fig. 6.16(b) shows super structure of pelton wheel power house.
A power house is provided with a travelling crane which can span
the width of power house and also travel itsentire length. Crane
should be able to lift generator rotor, shaft and runner. The roof and
upper part of the super structure may be either steel truss or rigid
frame.
 422 POWER PLANT

Nflr
Flow

Ru

Fig. 6.16(a)

Crone
Water

ol Machine
roan, halt

YOIVCJ F. IL
H
chamber 1L -
Fig. 6.16(b)

Fig. 6.17 shows an impi1se turbine installation. HRL. means

headrace level, whereas TRL indicates tail race level.
HA

TPL

Fig. 6.17


HYDRO-ELECTRIC POWER PLANT 423

2. Reaction turbine. In a reaction turbine water enters all round

the peripher y of runner and the runner remains full of water every
time and water acting on wheel is under pressure which is greater
than atmospheric. The turbine may be submerged below the tail
race but the accessibilit y to the turbine becomes difficult. The
turbine, therefore, is installed above the tail-race and water after
doing work is discharged into the tail-race through a draft tube
which remains submerged deep in the tail-race. Before entering he
turbine water has both pressure energy and kinetic energy. The
forces on the rotating parts are due to change of both potential and
kinetic energy of water. The casing in a reaction turbine is essential
as pressure at the inlet is more than at the outlet. This necessitates
that the water should flow in a closed conduit to avoid splashing of
water and, therefore, casing is necessary in this turbine.

WICKET GATE
5HAF T

•
1_44 WATER LEVEL
.':

DRAFT TUBE

Fig. 6.18 (a)

Guide wheel Runner shaft

/ Spiral \ ( Runner
casing

Draft tube
Tail race

Fig. 6.18

Fig. 6.18 shows the position of various parts such as

(i) Spiral casing (ii) Guide wheel
hit Runner shaft (i) Runner
Lb Draft tube.
 424
POWER PLANT
Reaction turbines may be horizontal shaft type or vertical shaft
type turbine but large reaction turbines are usuall y of the vertical
types. These turbines are used for low and medium heads, Reaction
turbines include Francis turbine (Fig. 6.18 (a)), Kaplan and
Propeller turbines.
r Link
Fked êCentral link
vanes

Wheel
Guide
blade
de

a l-% h'

Guide - Threaded
blade ring Spindle

Fig. 6.19

Water
Water
inlet

Outlet

'uide
(odes

Fig. 6.19 (a)

Francis turbine. It is used for medium heads. It is an inward
radial flow reaction turbine. It consists of runner provided with
vanes, wicket gates (guide vanes), guide mechanism, draft tube and
spiral casing. The runner is made up ofC.I. for small output whereas
for large output the runner is made up of cast steel. Guide vanes
regulate the supply of water according to the load and also direct
water to enter into the runner vanes at a suitable angle. The spiral


HYDRO-ELECTRIC POWER PLANT

425
casing surrounds the guide wheel. In this turbine the water enters
the turbine through penstock connected to spiral casing. The water
is then direct to enter the guide vanes and from guide vanes it enters
the runner vanes. Finally the water is discharged through the draft
tube. The oérall efficiency of this turbine lies between 85 to 90ç.
Fig. 6.19 shows a Francis turbine.
As shown in Fig.. 6.19 (a) the water flows radially inward and
leaves at the centre axially. The pressure at outlet is quite low and
therefore a draft tube is used for discharge at outlet.
WICKET GATE RUNNER SHAFT

WA TER
1i 171j ii-\ CASING

TUBE

RUNNER
BLADE TO TAIL
RACE

Fig. 6.19(b)
Fig. 6.19 (b) shows sectional view of turbine. The blades are
attached to the hub ofrunner. The runner is horizontal and it rotates
about a vertical shaft. Water enters the turbine in a radial (inward)
direction through an arrangement of movable wicket gates that
control the water supply and act as stationary nozzles. The water is
discharged in downward axial direction.
6.13.2 Direction of flow of water
According to the direction of flow of water the turbines may be
classified as follows
(i) Tangential flow turbine.
(ii) Radial flow turbine
(a) Radial Outward (b) Radial Inward
(iii) Axial Flow Turbine
(iv) Mixed Flow Turbine (Radial and Axial)
In tangential flow turbines as in case of Pelton turbines the
water strikes the runner in a direction which is tangential to the
path of rotation.
In a radial flow turbine the water moves in a plane perpen-
dicular to the axis of the rotation. In inward flow turbine the water
enters at the outer periphery and discharges at the inner periphery.
In axial flow turbine the path of water during its flow is parallel to
the axis of turbine shaft whereas in case of mixed flow turbine the
water enters radially inward and discharge axially.
—29


426 POWER PLANT

- Table 6.1 summarises the flow directions of commonly used

turbines.
Table 6.1
Type of turbine Flow Direction
fler and Caplan turbines, Axial flow
Francis turbine Radial inward or mixed flow
Pelton turbines Tangential flow

6.13.3 Position of shaft

According to the position of shaft the turbines can be classifie
as follows
1. Vertical shaft turbine 2. Horizontal shaft turbines.
Vertical shaft turbines required lesser space. Horizontal shaf
turbine is easily accessible. There is not much difference in th(
efficienc y of these two types of turbines.
6.13.4 Head of Water
According to the head of water the hydraulic turbines are
classified as follows
(a) High head turbines (b) Medium head turbines
(c) Low head turbines.
Impulse turbines are essential high head turbines whereas the
reaction turbines are used for low and medium heads. The Peltor
turbine is an impulse turbine and is used for high heads and low
flows. Francis turbine is used for low to medium heads and flows
For low heads and high flows Kaplan turbine also called propeller
turbine is used. Table 6.2 shows the water heads for different type!
,of turbines.
Table 6.2
Turbine
Pelton turbine 150 to 300 rn

Francis turbine 60 to 150 m
Kaplan turbine
to 60 m

6.13.5 Classification based on speed.

• As the turbine and generator are directly coupled, the rated
speed of the turbine is same as the synchronous speed of the
generator given by

_P
where N = Speed in R.P.M.
f= Frequency of generation
HYDRO-ELECTRIC POWER PLANT 427

= 50 hertz (50 c.p.$) usually

p = No. of pairs of poles of the generator.
Therefore, the turbine speed is determined by the generator
speed.
6.13.5 (a) Head on turbine
The total head or gross head is the difference between head race
level and tail race level. Net head or effective head is the head
available at the inlet of the turbine. As shown in Fig. 6.19 (c).
Head Dam -

Penstock

To fl race
Fig. 6.19 (c)
H = GrossEiead
= /1 + Hf
where h = Net head
H1 = Frictional head

H.G.L. indicates hydraulic gradient line.

6.13.6 Classification based on specific speed
Turbines are never classified by their actual speed but always
on the basis of the specific speed.
Thf specific speed (Na) for a turbine is given by the formula

H415

TaUe 6.3 indicates the specific speeds of different turbines.

6.14 Choice of turbines
The various factors to be considered while selecting a turbine
re as follows
1. Working Head 2. Nature of Load
3. Output 4. Specific Speed
428 POWER PLANT
a
For low heads propeller of Kaplan turbine are used. Kaplan
turbine may be used under variable head and load conditions,
whereas propeller turbines work satisfactorily when the head
remains constant. For medium head Francis turbine is used and for
high heads Pelton turbine is suitable. Pelton turbine has the ad-
vantages that it works quite satisfactorily even under variable load
conditions. For low heads of 60m or less Kaplan turbines are used
for heads of 150 to 350 in turbines are used and for very
high heads 350 in above Pelton turbines are used. Pelton turbines
are simple and robust, their control is easy and their maintenance
charges are lcw. The various factors influencing the choice between
horizontal and vertical type of turbines are as follows
1. Relative cost of the plant.
2. Site conditions, i.e. space available for foundations and
buildings.
3. Layout of the plant.
Vertical shaft turbines require smaller foundations and fewer
machines are required for a given -output. Further in case of vertical
shaft turbine weight of rotating part acts in the same direction as
the axial hydraulic thrust. This requires a thrust bearing capable
of carrying a relative heavy load and capable of working at maxi-
mum runway speed. For the same power a higher speed unit may
be chosen as it is smaller in size occupies less space and cost of
turbine, generator and their installation is low. But higher speed is
not always profitable because higher speed would require a higher
specific speed turbine which is usually more liable to cavitation and
create other mechanical difficulties.
6.15 Comparison of Pelton Wheel and Francis Turbine
(i) It is easier to regulate Pelton wheel than Francis turbine.
(ii) The operating efficiencies specially between half and full
load are better in Francis turbine.
(iii) The parts of Pelton wheel are more easily accessible and,
therefore, it is easier to repair them.
(iv) For the same head the running speed of Francis turbine
is more than Pelton wheel. This reduces cost of generators
and dimensions of units.
(v)•The dimensions of Francis turbine are much less than
those of Pelton wheel under similar operating head.
Therefore, the power house of smaller size is required to
accommodate the same number of Francis turbines.
(vi) Francis turbine utilises full available head by using a
draft tube whereas the Pelton wheel has to be installed
above the maximum tail race which causes loss of head.


HYDROELECTRIC POWER PLANT
429

6.16 Turbine Governing

Governing of turbine means speed regulation. In the normal
condition the turbine should be run at constant speed irrespective
of changes in load. This is achieved by means of a governor called
oil pressure governor.
The various cotnponents of the oil pressure governor are as
follows
1. Servomotor or Relay Cylinder. In the servomotor a piston
moves due to oil pressure and the movement of piston rod is
transmitted to the controlling device of turbine such as spear rod in
case of impulse turbine and guide vanes in case of reaction turbine.
2. Pendulum and Actuator. It is a flyball mechanism and it is
operated by power taken from turbine main shaft.
3. Distribution value or control valve. It controls the supply of
oil to cylinders.
4. Oil Pump and Gear Pump. Oil pump pressurises the oil frori
the sump and sends it to the cylinder of distributing valve.
5. Pipes connecting oil pump with control valve and control valve
with relay cylinder.
When the turbine is running at its normal speed, the position of
lever arms the piston in relay cylinder and distributing valve will
be in their normal place (Fig. 6.20). The ports P and Q remain closed.
CENTRAL
SPINDLE
PENDULUM OR

SLEEVE

ç RIGID FULCRUM H
FROM TURBINE 01.3 TRIBUTIN&
MAIN SHAFT - - VALVE

GEAR PUMP

CIL SUMP
-r- =--qftl

I r=

,LJONPID
SERVOMOTOR OP To TURBINE GUIDE
RELAY CYLINDER t4ECHAM$UM
Fig. 6.20
430 POWER DLANT

When the load on the turbine increases, the speed of turbine

falls and due to this the central spindle of the actuator will rotate
at less speed bringing the fly balls down.
Since these balls are connected with the sleeve, the sleeve will
also come down. This causes the main lever to rise pulling upward
the pistons of distributing valve. The slight upward movement will
open the port P and the oil under pressure will rush from distribut-
ing valve to the relay cylinder and to the right side of the piston.
This causes the piston to move to the left. This movement is
transmitted to the controlling device or guide mechanism of turbine
casusing more water to enter the turbine which, therefore, will run
faster and the speed will begin to increase. As the speed becomes
normal again, the main lover, the pistons of distributing valve and
relay cylinder will occupy this normal positions.
Similarly when the load on the turbine decreases, the speed will
in..rease. To bring the speed back to normal the oil pressure gover-
nor acts quickly and operates the guide mechanism. Fig. 6.21 shows
the guide mechanism for an impulse turbine. The governor doses
the nozzle opening partially by pushing the needle in the nozzle thus
allowing less water to enter the turbine which, therefore, will run
Slow.

ROD OF NEEDLE TO BUCKET OP

PELTONWIIEEL

I.

FROM RELAY
CYLINDER f1T_,1 r-
J' NEEDLE

1 I NOZZLE

SUPPLY PIPE t
Fig. 6.21

6.17 Performance of Water Turbine

Performance of a turbine means working of the turbine under
different loading conditions. Although turbines are required to run
at constant speed but this is difficult to achieve because of fluctua-
tions in load. To accommodate varying loads the discharge is regu-
lated.
The important parameters for any particular turbine are
(i) Discharge (ii) Head
 IY, 0-ELECTRIC POWER PLANT 431

Efficiency (iv) Speed

(v) Power.
It is desirable to study the behaviour of these quantities with
respect to u4e another.
Such studies can be carried out under unit quantities like unit
\ver, unit speed and unit discharge.
H turbine characteristic like unit power, unit speed and unit
discharge help in studying the performance of turbines.
() Unit Power (Pa ). It is defined as the power produced by the
turbine when running under unit head (1 metre).

• We know
75
whore P Power produced by a turbine
Q = Discharge in cumecs
H = Net Head is metres
co = Unit weight of water.
Now Q=AxV=k1.D2.V
where D = diameter of the runner
V= Velocity
Voc 'i=k2J

Q=k 1 .k 2 .D2 " ':.D2\h

(OH
P=--xQ

wkb2N ii1=k3D2H

For unit power (P).

H =1
P
P =

(b) Unit speed (N a ). It is the speed developed by a turbine when

operating under a unit head (1 metre).

60
where N = speed of turbine in R.P.M.
Now V -ff
 POWER PLANT

V=k 3 'i.ii
iEDN

N=k2ff

For unit speed (Na)

H 1
N
N1 =-

(c) Unit discharge ((Pa). It is defined as the discharge of a turbin

working under a unit head (1 metre).
= A. V. = k i D 2 V
= k 1 D2 'J.= k 2 D2 'LU
For unit discharge (Pu).
H=1
Q=k2xD2
(P 1

(Pu =

• (d) Specific Speed (Ns). It is the' peed of a geometrically similar-

turbine running under a unit head and producing unit power.
Specific speed is quite useL 'llows:
(i) It helps in selecting type of turbine to be used for a
particular power station.
(ii) It permits to visualise the performance o t4ne.
(iii) Specific speed being known the normal runn&., d can
be determined.
P7opH
As already proved (p—D2 CIf ...(
pD2H:v2

Also V= and V 'iii

ThereforQ, D
 HYDRO-ELECTRIC POWER PLANT
433
Substituting in equation (i),
PoH312

\ H

N N5

where Ns specific speed

-
Table 6.3 shows the specific speeds for different turbines.
Table 6.3
Turbine

Fra
_L_ Reaction— L =iI
Example 6.2. A turbine develops 18,000 H.P. working under a
head of 520 in when running at 400 R.P.M. Ca/cu/cite the specific
speed of the, turbine. Specify the type of turbine to be used.
Solution. H = Head = 520 R.P.M.
N = R.P.M. = 400 R.P.M.
P = Horse Power = 18,000.

Specific speed =N5= N.

400\'18,000
N 21. 4
5205'4-
For this specific speed Pelton turbine is used.
Example 6.3. Calculate the specific speed of (I turbine and
suggest the type of turbine required for a river having a discharge of
250 litre/sec with an available head of 5 0 ,,zctre. Assume effieiencv
of turbine as 80 and speed 450 R.P.M
Solution. 0 = 420 litre/sec-
11
= 50 inetr


434 POWER PLANT

= efficiency = 80%
Tj

N= 450 R. P.M.
P = Horse power developed
- no)' 0.8 x 1 x 250 x 50 133.3
- 75 75
(Taking w = 1 kg/litre)
Now, N3 = specific speed

N .TP - 450
H514 505'
For this specific speed Pelton turbine is suitable.
Example 6.4. It is desired to build a hydro-electric power station
across a river having a discharge of 30,000 litre/second at a head of
10 rn. Assuming turbine efficiency 80% and speed ratio (Ks) as 0.83,
determine the following:
(a) Is it possible to use two turbines with a speed not less than
120 R.P.M. and specific speed not more than 350 R.P.M.?
(b) Specify the type of runner that can be used. Also calculate
the diameter of runner.
- ru4H
Solution. - 75
where p = Total power available
= Specific weight of water
= Discharge
H = Head (metre)
0.8 x 1 x 30,000 x 10 =3200
P= .
Using two turbine each of capacity 1600 H.P. the specific speed
(N8) of the turbine iscalculated as follows:
N#120 I1600
N8=--=

= 267 R.P.M.
For this speed Francis turbine is suitable
U = Tangential velocity of the runner
nDN
=K. 4WH—

HYDRO-ELECTRIC POWER PLANT 435
where K. speed ratio and D = diameter of runner
60xK I1 60x0.83 Vjj
D—
itxN - itxl2O
= 1.85 metres.
6.18 Efficiency
Different types of efficiencies used in connection with hydraulic
turbine are as follows
(i) Volumetric efficiency (ii) Hydraulic efficiency
(iii) Mechanical efiT cv.
6.18.1 Volumetric efit cy
Some of the water flowing in the turbine may leak through the
joints or some water may not be effective in imparting its energy to
the turbine. The leakage of water reduces the efficiency of the
turbine.
- -

where VoN, metric efficiency

Discharge doing useful work
OL = Leakage from the turbine.
6.18.2 Hydraulic efficiency
Hydraulic efficiency takes into account the loss of head that
takes place in the turbine due to incomplete conversion of head over
the blades into useful work. It is defined as follows
H—h
N
where = Hydraulic efficiency
11H

H = Net head utilised

h = Head not effectively utilised.
6.18.3 Mechanical efficiency
Mechanical efficiency takes into account the power loss due to
friction.
1M P1—P2

where P = Power produced by the turbine

P1 = Shaft power
P2 = Power loss.
436 POWER PLANT

6.18.4 Overall efficiency (TI)

Overall efficiency of the hydraulic turbine is about 90% and is
given by
= fl. X flH X flM
6.19 Coupling of Turbine and Generator
A hydraulic turbine is coupled with electrical A.C. generator in
the following two ways:

"cc
- . Fig. 6.22
(i) Direct Coupling (ii) Gear Drive.
Direct coupling system is used for low speed turbines. In this
system both the turbine and generator shaft are directly coupled
and run at same speed. Any fluctuation in turbine shaft speed is
directly reflected in generator. In gear drive system spur gears or
bevel gears are used to connect turbine shaft with generator shaft.
This arrangement is quite commonly used.

,t9 [Genercr

Hood race

Fig. 6.22 (a)

Fig. 6.22 shows a low head turbine installation.
 HYD ROELECTRIC POWER PLANT
437
Fig. 6.22 (a) shows a turbine ofsmaller out put installed directly
in the head race. The water enters the wheel to the runner and
discharges into tail race (TRL) through a vertical draft tube. At the
end of the turbine shaft a bevel gear is mounted which enables a
generator to be connected for power generation.
619.1 Speed and pressure regulation
When the load on the generator coupled to the turbine changes,
the governor comes into action to regulate the inflow of water into
the turbine. With the sudden closing of gates there is a pressure rise
in the pen stock. This pressure rise is counter balanced by
(a) installing surge tanks at the end of pen stock near the
power station.
(b) providing pressure release openings or valves on the pipe
line at suitable places.
The normal maximum pressure rise permissible is 10 to 20% for
Pelton turb. ines and up to about 30% for Francis turbines
6.19.2 Power generated

The generator coupled to the turbines produces electric power

corresponding to the power developed 1v the turbines less losses in
the sot. The speed of the power generating set depends on:
Ci) head (ii) specific speed of the turbine
(iii) power of the unit.
Reaction turbines are vertical and at the power plant sufficient
substructure and super structure is needed to accommodate the
necessary arrangement of draft tube etc. The impulse turbines are
either horizontal or vertical and do not require special substructure.
6.20 Synchronous Speed
It is defined as the speed of turbine which corresponds to the
generator speed or some multiple of the speed of the generator. It
depends on the way the turbine is coupled with the generator.
6.21 Cavitation
In water turbine the water is conveyed to it through penstocks.
When the water pressure at an y
point in the penstock reaches the
vapour pressure a large number of small bubbles of vapours are
formed. These bubbles are carried by the stream to higher pressure
zones where the vapours get condensed to liquid. Due to this a cavity
is formed and the surrounding liquid rushed to fill it. Such an action
causes a violent collapse of cavity and produces loud noise and very
high pressure. The pressure produced may be so high that it may
Cause pitting or tearing off the surface of the material. This
phenomenon which causes pitting of the metallic surface is called
cavitation because of the formation of cavities. Pitting may also
occur at following points in a turbine:
 M

POWER PLANT
438

(i) On the upper part of the walls of the draft tube.

(ii) Near the tip of the valves in impulse turbine.
(iii) On the back of the bucket of the Pelton turbine where
water leaves the runner.
Thoma has suggested a cavitation parameter for water turbines
and pumps.
a = Cavitation factor

- H
where H = Total effective head
Hb = Barometric head.
H5 = Suction head or height of pump about tail water level.
For cavitation free iunning a has to be greater than critical
value acm.
< 11b - us
(Ocrit)
H
H5 5 Fib - acn: H
is a function of specific speed.
When H is large H5 comes out to be negative and therei
becomes essential to provide the pump with negative suction h
i.e. the power house location has to be so fixed that pump operak-
under submerged conditions.

PI

(a) (b)
 HYDRO-ELECTRIC POWER PLANT
439

1?

F ig. 6.23 Constant speed curves.

6.22 Operating Characteristics
Operating characteristics or constant speed curves of turbine
exhibit the behaviour of water turbine when the turbine is working
for the generation of power under constant speed. Such curves show
the relationship between Power (P)
(TI) and discharge (Q) and efficiency and discharge (Q), efficiency
and power (P). These curves
are shown in Fig. 6.23.
6.23 Efficiency Load Curves
The efficiency load curves of various types of water turbines are
shown in Fig. 6.24. The efficiency load curve of a Pelton turbine
(impulse turbine) is shown in Fig. 6.25. The efficiency curve of a
Pelton turbine remains slightly lower than that of a Francis turbine
but is less effected by variations of load.
TQA&,-

I h.o
'a

Fig. 6.24
Fig. 6.25

Francis turbines are suitable for operation of the plant from 75%
load to full load. Propellor turbines with fixed blade construction are
used at fairly constant load between 759 and 100% of their capacity.
 POWER PLANT
440

Kaplan turbine is suitable for frequent part-load operation. The

efficiency of a Pelton wheel unit is about 88%.

6.24 Site Selection

The ideal site will be one where the dam will have the largest
catchment area to store water at high head and will be economical
in construction.
The various factors to be considered while selecting the site for
hydro-electric power plant are as follows:
1. Water available. Geological, geographical and meteorological
conditions of the site should be studied thoroughly. Daily, weekly
and monthly flow water over a period of years should be recorded.
Estimate should be made about the average quantity of water
available throughout the .year and also about maximum and mini-
mum quantity of water available during the year. These factors are
necessary to decide the capacity of the hydro-electric plant, setting
up of peak load plant such as steam, diesel or gas turbine plant and
to provide adequate spillways or gate-relief during the flood period.
2. Storage of water. Water used in the hydro electric power plant
is mainly dependent on the rain and since the rainfall is not regular
throughout the year, therefore, it is essential to store water to afford
a uniform output. Thus a storage reservoir is constructed at the site.
During rainy season the excess water is stored in the reservoir and
it is released to supplement low rates of flow during run-off periods
to maintain the output. The reservoir should have large catchnient
area so that water in it should never fall below the minimum level.
3. Head of water. Water isjarge quantities and at a sufficient
head should be available. For a given power output an increase in
effective head reduces , the quantity of water required to be passed
through the turbines.
4. Distance from Load Centre. It is desirable that the .power
plant should be setup near the load centres, so that costs of erection
of the transmission lines and their maintenance are low. However,
in hydro-electric power plants it may not be possible because this
plant will be located where sufficient quantity and head of water is
available.
5. Accessibility of the site. The site selected should have rail and
road transportation facilities.
6. The land of site should be cheap and rocky. The ideal site will
be one where the darn will have the largest catchment area to store
water at a high head and will be economical in construction. After
the location has been chosen the exact position of the various
stcuctures are fixed by considering the following factors
 HYDRO-ELECTRIC POWER PLANT
441

(a) Details of foundatjbn conditions

(h) Requirements of head, flow demands of storage capacity
(c) Arrangement and type of following:
(j) ham. (ii) Intake system
(iii) Conduits' (iv) Surge tanks
(v) Power house
(d) Cost of dam and project
(e) Transport facilities and accessibility of site.

6.25 Comparison of Hydro Electric Power Plant and Steam

Power Plant
1. The initial (cost of land, dam, diversion works, water rights,
rail, road and generating plant) of a hydro electric power plant is
nearly twice (or more) as that of steam power plant of equal size
but the labour, maintenance and repair costs of hydroplant are
much less than the stream power plant. In hydro-plant the costs of
dam and waterways constitute the major portion of plant cost..
Depending upon the conditions the cost of a hydroplant var y from
Rs. 700 to Rs. 1800 per kW.
2. For preliminary estimates the fixed charges for the hvdro-
power station usually vary from 9 to 12%.
Interest 6%
Depreciation 2%
Taxes and insurance 1%
Total . 9%
The fixed charges for a steam power plant are higher and may
he assumed from 12 to 14%. This is because of greater proportion of
equipment and machinery costs and the depreciation charges are
relatively high.
3. It is easy to locate steam power plant near the load centre and
this eliminates the need for long transmission lines and thus trans-
mission cost is less whereas power generated by a hydro plant is to
he transmitted over longer distance and hence at greater cost.
4. The time to put the most steam power stations into operation
is nearl y 30 minutes whereas a hydro electric power plant can be
put into operation in very short time ranging from few seconds to 3
or 5 minutes. Thus where power is to be supplied both by
hydroelectric power plant and steam power plant, the steam power
plant should be used as base load plant and hydro-power plant
should be used as peak load plant.
5. The efficienc y of steam power plant drops as it gets older.
6. Stearn station can be operated as desired to suit the
demand of the load s y stem whereas power to he generated from
442 POWER PLANT

hydro-plant is dependent on the quantity of water available in the

storage.
7. The cost of transportation of fuel is quite high in case of steam
power plant.
8. As compared to steam power plant the hydro-electric genera-
tion is the cleanest and most simple way.
6.25.1 Cost of Hydro Power
The power produced by a hydro electric power plant is the
cheapest but the initial cost of any hydro electric project is very high.
The cost of a hydro electric development includes the following
costs
(i) Cost of land and water rights.
(ii) Cost of railways and high way required for the construc-
tion work.
(iii) Cost of constructions and engineering supervision of
project.
(iL) Cost of equipment and building etc.
v) Cost of power transmission. equipment.
Example 6.4. The various costs fora hydro power project are as
follows
Darn 50 crores
Power house building and
others = 10 crores
Turbines and generators = 4 crores
Transmission line = 6 crores
Penstock =5crores.
Load acquisition = 0.7crore
Taxes and insurajzce = 1%
Interest rate = 5%
Maintenance cost per year 1.2 crores.
Power is generated from 5 units each of 30 MW capacity. The
firm power is 60 MW and is sold at 12 paise per kWh and secondary
power is supplied for 210 days at 60% load factr.
The secondary power is 80 MW and is sold at 6 paise per kWh.
Assuming useful life fir dam, generator etc. Find whether the
project is under loss or pro/)t.
Solution.


HYDRO-ELECTRIC POWER PLANT

443
(i) To calculate total depreciation cost

Budding

Rem
_____-
I Cost (crores
50 50

Generatorand tu'rbinc: ^^40

line 6 25
LPstock.
- 25
r = Rate of interest = 0.05
C1 = Depreciation cost for dam and building

0 ±J 0i r = 60x0.05
(l+r)-1 (1+O.5)°_J
60 x 0.05
=----j = 0.287 crore
where n = Life in years = 50
Cost of dam and building
50+1060crOres
C2 = Depreciation cost for generaL...
4x0.05 3x0.05
- (1 +0.05)20 - 1 0.65-1
-
=O.2Crore
= Depreciation cost for transmission line and
penstock
11x0.05 11x0.05
(1 + 0.25)2_ 1 - 3.39-1
= 0.23 crore
C - Total annual depreciation cost

0.287 + 0.12 +0 23 = rore

To (lcuIatc 1iitcrtst
• Jit•rt on capital c,.,t
Capital 5u 1() t 4 5-

l;Ilntein CO-;j i' • :tr

I.lX(S and In'tri::-.,...t
 POWER PLANT
444
1
x 75 = 0.75 crore
=
S = Total cost per year = C - S i + S2 + S3

= 0.637 + 3.75 + 1.2 + 0.75 = 6.337 crores.

(w) To find Income
Firm power = 60 x 103 kW
= 60 x 10 3 x 365 x 24 kWh
= 52.56 x 10 7 kWh
Ri = cost of firm power
7 1
=52.5x10

= Is. 630.72 x 105 = 6,31 crores.

Secondary power = 80 x iO kW at 60% load factor is supplied
for 200 days
104 kWh
=80x 103 xO.6x 200 x24 =23 x
R2 = Cost of secondary power

= 23 X 104 x -j = 138 x iø

= 1.38 crores.
R = Total income = R 1 + R2
= 5.31 + 1.38 =7.69 crores.
Net profit = R -• = 7.69 - 6.337 = 1.353 crores.
6.26 Hydro-steam Inter-connected System
A hydro power plant maybe used as a base load plant or as peak
load plant but the trend is towards use of hydro power plant with
steam power plant in an inter-connected system. Power from hydro
electric plants usually costs loss and, therefore, when flow of water
is adequate hydro power plant should be used as base load plant and
steam power plant should be used as peak load plant whereas during
low water in reservoir the functions of two plants should be reversed.
By interconnecting hydro power plant with steam power plant
a great deal of saving in cost is achieved and standby requirements
are reduced in an interconnected system. The interconnected svs-
tern should have a suitable load allocation plant set prehand for each
day. Hydro turbines can be started from cold in a few minutes while
steam units need about 5 to 8 hours.
HYDRO-ELECTRIC POWER PLANT
445

(a) Reinforced concrete tunnel. '(b) Pressure tunnel unreinforced.

Fig. 6.26

Fig. 6.27 Horse shoe tunnel.

TIE

T/1&QE
PLUME

Th'I8RE.
TRE5TLE

Fig. 6.28 Timber flume. Fig. 6 99 Steel lurn

 446 POWER PLANT

€27 Tunnels
The tunnels are generally of circular section (Fig. 6.26) or horse
shoe section (Fig. 6.27). The hand excavated tunnels are usually at
least 6 feet in diameter whereas the minimum economic diameter
for machihe excavated tunnels is nearly 8 feet. Tunnels act as open
channels when flowing partially full and act as pressure conduits
when flowing fully.
6.28 Flumes
Where it becomes difficult or expensive to construct canals, the
water is conve yed through flumes. The flume channel is so designed
that it is able to carry its own weight and that of water as beam
between supports. Flumes may be made up of wood, steel or rein-
forced concrete. Wooden flumes are generally, of rectangular cross-
section width being nearly twice the water depth. Fig. 6.28 shows
the cross-section of a wooden flume. Fig. 6.29 shows the cross-sec-
tion of a semicircular steel flume. Fig. 6.30 shows the ôross-section
of a concrete flume on concrete piers. Reinforced concrete flumes
have longer life and their maintenance cost is low whereas wooden
flume are of shorter life and depreciate more rapidly.
6.29 Spillway and Gates in Dams

Fig. 6.30 Concrete Flume.

Spillway. A spillway acts as safety valve for a dam. Spillways

and gates help in the passage of flood water without any damage to
the dam. They keep the reservoir level below the predetermined
maximum level. An additional storage can be made available above
the spillway crest with the help of gates. The various types of
spillways are as follows:
(i) Ogee Spillway. Fig. 6.31 shows an Ogee Spillway at design
capacity.
 HYDRO-ELECTRIC POWER PLANT
447

CURVAT RE

•.t... U.' Ih .'V1 .iWv. . tflL ir . TJEP?AI.SS

'G.cWE P

Fig. 6.31 Ogee spillway. Fig. 6.32 Chute spillway.

(ii) Chute Spillway. In this spillway (Fig. 6.32) the discharging

water flows into a steep sloped open channel called chute. The
channel is made of reinforced concrete slabs. This type of spillway
is suitable for earth dams.
(iii) Shaft Spillway. In this spillway (Fig. 6.33) water flowing
through a vertical shaft enters a horizontal conduit from where it
flows out of the dam.

ROUNOV LIP 47
ENrRANCE

VERT/AL
STi'RI5Ei)

CONCRETE
/NEO W4%L

Fig. 6.33 Shift spillway.

3ROLLtR

GATE
•j1 9RCLLEO
STEEL GUIDE

q ,86Eg E41FR

dM -
SEAT

Fig. 6.34 Siphon spillway. Fig. 6.35 Stoney Gate.

448 POWER PLANT

The drawback of this spillway is the hazard . 'of clogging with.

debris. Therefore, trash racks, floating booms and other protection
devices should be used to prevent the debris from entering the
spillway inlet.
(iv) Siphon Spillway. Fig. 6.34 shows the cross-section of a
siphon spillway (submerged outlet). This spillway is used for smaller
capacity and where space is limited. They have the advantage of
automaticall y maintaining water surface elevation within very close
limits.
6.30 Gates
Water is allowed into the penstock, canal or direct to the turbine
through the intake.
The following structures are provided with intake:
(i) Gate
(ii) Operating and hoisting mechanism
(iii) Trash screen
(iv) Cleaning mechanism for trash screen.
The various t ypes of gates commonly used are as follows
(a) Vertical lifi, gates (b) Radial gates
(c) Rolling gates (d) Drum gates.
(i) Vertical lift gates. Vertical sliding gates are used for smaller
installation. Fig. 6.35 shows stoney gate. The gate is lifted by mens
of hoisting cables.

CONCRETE I'/DGf
,r pg
SETI'E&W
C4&E n
PIE,?

GA IF
\\

Fig. 636 TatrOer gate.

(ii) Radial gates. Fig. 6.36 shows the cross-section of a radial or
tainter gate. A steel framework supports thefce to the gate, the
filice being a c y lindrical segment. The steel framework is pivoted on
triiiions set in the downstream portion of the piers on the spillway
HYDRO-ELECTRIC POWER PLANT 449

crest. In such gates friction is concentrated at a pin and is much less

than in sliding gates.
(iii) Rolling gates. Fig. 6.37 shows a rolling gate. It consists of
cylindrical drum made of steel. The lower portion of the gate is a
cylindrical segment and touches the spillway crest. The gate rolls
on the rack when force is applied by means of hoisting cables.
(iv) Drum gates. This gate (Fig. .38) which is a segment of a
cylinder can fit in the recess provided in the top of spillway. When
water enters the recess the gate is raised by water pressure upto
the closed position.
HOF5T
ROOM

CONCR6TE.
TOWER

INCLINO
P40<

11 \\O P4
I'c&..J SEAL

:
:G.

Fig. 6.37 Rolling gate. Fig. 6.38 Drum gate.

6.31 Types of Valves

Various types of valves used in hydrcl-power plants are as
follows
(i) Regulating valves (a) Needle valves; (b) Tube valves; (c).
Hollow get valves.
(ii) Open-shut valves (a) Gate valves ; (b) Fish tail valves, (c)
Butterfly valves ; (d) Rotary valves.
(iii) Energy dissipators
The gate valves are used to regulate the flow. For moderate
heads butterfly valves are preferred. For high head installations\
needle valves and tube valves are used. These valves are generally
placed at the down stream end of sluices, discharge taking place.
directly into the atmosphere.. Fig. 6.39 shows a needle valve it
consists of three water filled chambers M, N and P in which the

450 POWER PLANT

hydraulic pressure can be varied. The valve is opened and closed by

varying pressure in these chambers. The valve is opened by forcing
the needle to the left and this is done by increasing pressure in
chamber P and releasing in chambers M and N which are intercon-
nected. To close the valve pressure is increased in chambers M and
N while chamber P is exhausted to atmosphere.

Fig. 6.39. Needle valve.

Fig. 6.40
Fig. 6.40 shows tube valve. This valve is open or closed by
mechanical means. To open or close the valve cylinder (tube) is
moved towards or away from the valve seat with the help of screw
stem activated by a bevel gear. As compared to needle valve this
valve is lighter in weight and is shorter in length.
Frame Operation rod
Closed

Disc

Disc
Fig. 6.40 (a)
HYDRO-ELECTRIC POWER PLANT 451

Butterfly valve. This valve is used not only as intake gate for
penstock but also is used before the turbine to facilitate the inspec-
tion without dewatering the whole of the penstock line. Fig. 6.40 (a)
shows a butterfly valve. It consists of a lens shaped disc which can
be moved in the frame of the valve.
6.32 Average Life of Various Cómpönents
The approximate average life of various components of a
hydropower plant is indicated in Table 6.3.
Table 6.3

6.33 Hydraulic Accumulator

rL.. t ai L,.,.......I;... A..,.,..I.,...


_452 POWER PLANT

Hydraulic accumulator system (Fig. 6.41) pumps water into

reservoir with off-peak power. In this system the centrifugal pump
Hydraulic turbine and motor. generator are mounted on the same
shaft.
6.34 Cost of Hydroelectric Power Plant
Hydroelectric power plants cost nearly Rs. 700 to 1800 per kW
of capacity. The cost of dam, intake tail-race and resel-voir is more
as compared to other items of the power plant. A typical sub-division
of investment cost is as follows
(i) Dam, tail-race, reservoir, intake —36%
(ii) Turbine and. generators —20%
(iii) Land, building and foundations —29%
(iv) Switching and wiring
(u) Switchyard —4%
(vi) Miscellaneous —5%
6.35 Hydrology
Hydrology is the natural science that deals with the distribution
of water on land beneath the surfaceof earth. It deals with the solid,
liquid and vapour forms of water. The various conditions which
bring about the transformation of one form into another are dealt
under this science.
6.36 Hydrologic Cycle
The various processes involved in the transfer of moisture from
the sea to the land and back to the sea again constitute which is
called hydrologic cycle (Fig. 6.42). Hydrologic equation is expressed
as follows
P=R+E

JAN
HiI
LLL,:/
!ll / .
NW RAIN SJñA)f
COURSE GAUGE GAUGE
Sd.VEY STATION 5T.4 rKJN

Fig. 6.42
HYDRO-ELECTRIC POWER PLANT 453

where P = Precipitation
R = Run off
E Evaporation.
Precipitation. It includes all the water that falls from atmos-
phere to the earth surface. Mainly precipitation is of two types.
(i) Liquid precipitation (Rainfall)
(ii) Solid precipitation (Snow, Hail storm).
Run-off. Run-off is that portion of precipitation which makes
its way towards streams, lakes or oceans. Run off occurs only if the
rate of precipitation exceeds the rate at which water infiltrates into
the soil and after depressions small and large on the soil surface get
filled with water. Rainfall duration, its intensity and a real distribu-
tion influence the rate and volume of run-off.
Evaporation. Transfer of water from liquid to vapour state is
called evaporation.
Transpiration. It is a process by which water is.released to the
atmosphere by the plants.
6.37 Hydrograph
It is.a graphical representation between discharge (cubic metres
per second) through a river and time.
A hydrograph may be
plotted for several weeks or
even months. Hydrograph mdi-
SURFACE PECK cates the power available from
"'SIN
RUNOFF cessos the stream at different times of
OrSCHARGE UMB the day, week or year. Extreme
4 LIMB conditions of flow can also be
TANGEN studied from a hydrograph.
BASE F I.Ow Hydrograph of stream of river
- TIME - will depend on the charac-
teristics of the catchment and
Fig. 6.43 precipitation over the catch-
ment. Hydrograph will assess
the flood flow of rivers hence it is essential that anticipated
hydrograph could be drawn for river for a given storm. A typical
hydrograph is shown in Fig. 6.43. A hydrograph has a rising limb,
peak and receding limb or recession curve.
6.37.1 Flow Duration Curve
It is a method to represent the run off graphically. This curve is
plotted between flow available during a period versus the fraction
of time. The total power available at the site may be known by this
curve which can be drawn with the help of hydrograph from the
 POWER PLANT

available run off data. Fig. 6.43 (a) shows a typical flow duration
curve.
This curve is nearly useful in the analysis of the development of
water power.

C
U
a

% of Time
Fig. 6.43 (a).
6.38 Mass Curve
A reservoir is used tQ conserve water which can be used during
the period of deficiency. A mass curve is a convenient device to
determine storage requirement that is needed to produce a certain
dependable flow from fluctuating discharge of river by a reservoir.
Mass curve is a plot of cumulative
volume of water that can be stored
from stream flow versus time in days,
weeks or months. The slope of the mass
curve at any point represents the F U-
hange of volume per change of time in ________
)ther words the rate of Row at the Time
moment. Hence the mass curve issteep
t4as Curve
when the river flow is large and flat
when the river flow is small. Fig. 6.44 Fig. 6 4 4
shows a mass curve.
6.38.1 Unit Hydrograph
This type of hydrograph represents a volume of one inch of run
off resulting from a rainfall of some unit duration and specified a
real distribution.
Unit h,idrographs of very large floods differ some what from
those of small rainfalls. A unit hydrograph can be contructe frnjH
a hydrograph of the actual run off when there is uniform r.
intensity and uniform a real distribution. The nu!b4r unit
hydrograohs for a given basin is theoreticall y infinin' as t.l.u, i1IV
e one unit hydrograph for every possible duration nt raitttiH aul
'iery possible distribution pattern of rainfall iii the basin . l
 HYDRO-ELECTRIC POWER PLANT
.jI5
tice only a limited member of unit hydrographs are used for a given
basin.
A unit hydrograph is constructed as follows
Ci) Measure the total volume of run off from actual
- hydrograph.
(ii) Determine the ordinates of unit hydrograph by dividing
the ordinates of actual ran off by the total volume of run
off in cms. over the drainage area.. This gives unit
hvdrograph for the basin.
(iii) Determine effective duration of run off producing rain for
which the unit hydrograph (WI) is applicable by a study
of the rainfall records.
Unit hydrograph is a very useful tool in estimating the run off
from a basin for a storm of given duration. It helps to predict the
expected flood flow from a catchment if rainfall intensity in the,
catchment area is known.
Fig. 6.44.1 shows a typical unit hydrograph obtained from actual
hydrograph of run off.

Run off
hydrogroh

Unit -
ydrograph

Time (hours)
Fig 6.44.1
6.38.2 Factors affecting run off
Run-off from an area depends upon the following factors
Ci) Precipitation cha racteristics. They include type of storm, its
intensity, extent and duration.
(ii) .\ I'frokçn( characteristics. They include temperature
humidit y , wind, prc'.sure variation etc. Greater temperature and
'ind velocity help evaporation.
(iii) Catchment characteristics. Geographical features of catch-
ment such as size, shape and location of catchment produce sig-
nificant effects on the run-oft
4 5fi POWER PLANT

Two types of catchment shapes with different arrangement of

drainage channel in the area shown in Fig. 6.45. The fan shaped
catchment produces greater flood intensity than a fern shaped
catchment.
When rainfall takes place over the catchment water from the
tributaries reach the point P simultaneously creating a sudden rise
of discharge at Pin case of fan shaped catchment.
On the other hand, in case of fern shaped catchment when
rainfall takes place the tributaries are so distributed that water
from different points takes different time to reach the point P. Hence
rate of discharge at P is slow and takes place over a longer period.

Main Stream

Tributaries
Iater'Shed Boundary

(t )

(s) Fan shaped catchment.

(ii) Form shaped catchment
Fig. 6.45
The surface of the catchment also has substantial effect on
run-off. A base surface gives more run-off than a grassy surface.
(iv) Storage characteristics. Out of the total precipitation that
falls on the catchment area a part is stored within the area and does
not appear as run off. There may be a number of depressions, pools
and lakes which store part of the precipitation and detain it either
temporarily or permanently.
6.38.3 Run-off Estimation
The following three kinds of informations are generally required
for runoff statistics
(i) Annual, monthly or seasonal run-off
(ii) Extreme low flows
(iii) Flood run-off.
The following methods are used to determine the runoff from a
catchment:
(i) Erhpirical method (ii) Rational method
(iii) From hydrograph.

HYDRO-ELECTRIC POWER PLANT 457

(i) Empirical Method. The following few formulae are used to

determine run-off. These formulae have been derived from limited
regional use.
U)) Lacey formula R = p2
3048F
Y-'
(ii) Inglis De Souza formul,.
R = 0.85P - 304.8 .or That area
P(P - 177.8)
2540 For 'ion Ghat area
(iii) Khosla formula R = P -'4.811T
where R = Run off in mm
P = Rainfall in mm
T = Mean temperature in C
F = Monsoon duration factor
= 0.5 to 1.5 depending upon type of catchment
S = Catchment factor
= 0.25 for flat lands to 3.5 for hilly areas.
(ii) Rational Method. The following rational formula is used
to give a rainfall run-off relationship
R=KA.P.
where R = Run off in hectare cm
A = Area of catchment in hectares
P = Precipitation in cm
K = Coefficient taking losses into account.
(iii) Hydrograph method. It is a graphical representation
showing discharge (run-off) of flowing water respect to time for a
specified time. The time period for discharge may be hour, day, week
or month. The rate of flow at any instant can be read from
hydrograph.
6.38.4 Flood Run-off
It involves estimating the maximum peak value of the flow. Few
empirical formulae listed below are used for finding maximum
discharge.
(i) Dicken's formula Q =
(u) Inglis formula Q = 123A
+
where Q = Discharge in iii3'sec.
—31
 POWER PLANT
458

A = Catchment area in km2.

Calculation of H.P. and kW Power
Let H = net head of water in metres
= discharge m3/sec
il = overall efficiency of hydro power plant
w = specific weight of water in kg/m3
H.P. = horse power developed.

HP ±Ln 75
(Taking 1 metric H.P. = 75 kg m/sec)
kW power = 0.746 x H.P.
6.39 Controls in Hydro-electric Plants
The various controls provided in a hydraulic power pLnt are as
follows
(i) Hydraulic controls
(ii) Machine starting and stopping controls
(iii) Machine loading and frequency controls
(iv) Generator and system voltage control
(v) Machine protection.
Hydraulic controls. All water levels are indicated on the
control panels. Water passing through the turbine is recorded in the
control room through water flow indicator and recorder for in-
dividual units. The various hydraulic controls are as follows
(a) Primary and secondary storage level indicators.
(b) Flood control. . (c) Intake gate control.
(d) River flow control.
Machine starting and stopping controls. The water flowing
to the turbine is controlled by means of gates and valves provided
in the supplying conduit and at the turbine inlet. The water supply
to the turbine is regulated in accordance with generator load
through a governor system. During starting of turbines filling of
casing should be done gradually and by pass valves are provided to
limit the rate of water flow.
Machine loading and stopping controls. The load on the
machine is controlled by adjusting the governor speeder control or
by controlling the system frequency.
Generator and system voltage control. The electric power
should be supplied at proper voltage. The voltage control is exercised
through voltage regulators.
HYDRO-ELECTRIC POWER PLANT 459

Machine protection. In a power plant the protective devices

should be provided to guard against breakdown of turbo generator
and auxiliary services, like transformers, switchgear, overhead
lines etc. Protection measures are also needed to guard against
incorrect operation and failure of control system.
Automatic controls are more reliable, safe and efficient. The
control room should be designed for convenience of operation and
the equipment should be spaced to permit easy access.
6.40 Surveys needed for hydro-power plants
The following surveys are carried out for a hydro plant installa-
tion.
(A) General surveys— They constitute the following:
1. Topographical surveys such as
(i) The dam site topographical survey plans including the
area of accommodate dam, spillway, outlet works, diver-
sion works etc.
(ii) The reservoir submergence plans
(iii) Contour plan of
(a) surge tanks (5) tunnels
(c) penstock (d) tail race channels.
(iv) River surveys
(v) For a barrage structure detailed survey maps to cover the
area under the barrage.
2. Geological and foundation investigations. A study
should be carried out for the following:
(i) Secismic condition of the region
(ii) Water flow in reservoir
(iii) Sub-surface conditions
(iv) Foundation investigations for dams.
3. Metrological and hydrological studies
The following data should be collected
U) Precipitation, evaporation.
(ii) Water flows such as maximum and minimum water flows.
(iii) Rainfall—run off relation data.
(iv) Mean latitude, and longitude, mean elevation, mean
monthly temperature.
(v) Location of raingauge station near power plant site.
4. Construction material investigations
Data should be collected for the following:
(i) Materials to be used for
(a) Concrete and masonary dams
(b) Earth and rock fill dams.
(ii) Tests to be carried out for materials.

460 POWER PLANT
a
5. Transportation and communication. Data should be collected
for the following:
(a) Road, water and rail routes
(b) Telephone and telegraph lines.
6. Environmental considerations
(i) Navigation (ii) Fish culture.
(B) Special surveys
Special surveys include the following:
1. Load surveys
They are made of
(i) Power-station capacity
(ii) Power to be generated
(iii) Details tor major lods to be served
(iv) Peak load
(v) Load factor
(vi) Future energy demand
(vii) Inter connection with other power systems
2. Lay-out studies. This includes the following
(i) Type of power plant such as storage type, run of river plant
pumped storage type etc.
(ii) Structural components like
(a) Dams (b) Canals
(c) Tunnels (d) Penstock
(e) Draft tube (I) Surge tanks
(g) Power house (h) Switch yard.
6.41 Control room functions
• The control room is used to perform the following functions:
(i) Machine starting and stopping
(ii) Machine loading
(iii) Frequency control
(iv) Generator and system voltage control
(v) Machine running supervision
(vi) Hydraulic control.
The control room is generally situated in the power house.
Automatic and semiautomatic together with push—button control
systems are preferred.
6.42 Mechanical Equipment
The mechanical equipment in a hydro power plant consists of
the following:
(i) Pumps for different uses
(ii) Bearings of turbine and generator
(iii) Ventilation and cooling equipment with
(a) fans (b) blowers (c) compressors
(iv) Brake circuits for the generator


HYDRO-ELECTRIC POWER PLANT 461

(v) Lifting cranes

(vi) Governors for turbines
6.43 Switch gear
It is used for making and breaking the circuits. It may consist
of the following:
(i) Switches (ii) Isolators
(iii) Surge-arrestors (iv) Circuit breakers.
If the switch gear is at the generated voltage it is normally
located indoors whereas if it is at the transmission voltage it is usual
to locate it outdoors in switch yard. Switch gear and transformers
located outside should be provided with adequate lightning protec-
tion.
Example 6.5. A reaction turbine is supplied with 100 cu m of
water per second and works uzder a maximum head of 120 m at 350
R.P.M. Assuming overall efi fciency of the plant 80% and specific
weight of water 1000kg/rn 3 ; calculate the horsepower developed and
power in kW.
Solution. H = 150 in
o = 100 m3/sec
to = iø kg/M3
ii .= 0.8
Horsepower developed
04H r - 1000 100 120x0.8
75 - 75
= 128,000
Power in kW = 128,000 x 0.746 = 95,488 kW. Ans.
Example 6.6. (a) Calculate the total energy in kWh which can
be generated from a hydro power station having following data
Reservoir area = 2.5 sq. km
Capacity =5x106m3
Net head of water at the turbines = 80 in
Turbine efficiency = 80%
Generator efficiency 90%
(b) Also find by how many metre the level of reservoir will fall if
a load of 20,000 kW is supplied for 5 hours?
Solution. (a) Work done/sec.
= 1000 x 5 x 10 6 x 80 kg rn/sec.
= 40 x 10 10 kg rn/sec.


POWER PLANT
462

- 40x 1010x0.8x0.9 = 0.38 x 1010

H.P. Developed = ____
75
Energy produce kW sec 0.38 x 10 10 x 0.746 = 0.28 x iO'
Energy produced (kWh)
- 0.28 x 10'
—0 = 7.78 x 10 5 kWh.
- 3600
(b) Let the fall in level of reservoir = h metre.
Time = 5 hours.
Area of reservoir = ( 2.5 x 1000 x 100) sq. m.
- 2.5x 1000100Ii_1P14±
x
Discharge/sec. - 5x3600 - 72
1000 x 104 x h x 80
Work done/sec. = 72
1000 x 10 4 x hx8O x 0.8 x 0.9
H.P. developed = 72 75
1000x10xax80x0.80 x 0.9 x 0.746
kW produced = 72 x75
= 1.05 x 10 5 x h
But kW produced = 20,000 (given)
20,000 = 1.05 x 10 5 x h
h = 0.2 metre. Ans.
Example 6.7. Fora hydroelectric POWerPlaltt the following data
is supplied
Annual rainfall = 1000 mm.
Catchrnent area = 120 sq. km .
Effective head = 250 rn.
Load factor = 40%
Yield factor to allow for run-off and evaporation, loss 50%
Efficiency of power plant = 70%
Determine the following
(a) Average power produced
(b) Capacity of the power plant.
Solution. Volume of water available per year
= Catchment area x Annual rainfall x Yield factor,

Now catchment area 120 sq. km . = 120 x (1000) m



HYDRO-ELECTRIC POWER PLANT 463

Annual rainfall = 1000 metre

10 100
Volume of water available per year = 120 x (1000)2 x 1 x 0.5
Volume of water available per second
- 120 x (1000) 2 x05
' 8760x60x60

H.P. developed -
75
where 0= water available/sec.
- 120x(1000)2x0.5
- 8760x60x60
w = specific weight of water = 1000 kg/rn3
H = head = 250 metre
= efficiency = 0.7
k = yield factor = 0.5
• H. P. developed = 0.7 x 0.5
8760 x 60 x 60 x 75
35 x 106
876 x 9 4328
•. Average Power = 4328 x 0.746 = 3228 kW
Avepower
Load factor
Maximum demand
Maximum demand
- Average power - 3228
= 8070 kW.
- Load factor - 0.4
The capacity of the power plant can be taken equal to maximum
demand
Capacity = 8070 kW.
Example 6.8. Ca/cu/cite the power that can be deteloped from ci
hydro-electric pou'er station hating the following data
Catchment area = 100 sq. km .
Ateragi' rain fall = 120 cm
Run-off =
Available head = 300 rn
Overall eftcu'ncv o/the power station = 75'.


464 POWER PLANT

Solution. Run-off available ()

- 100 x 10 3 x 1.2 x 0.8 = 3.4 cu-rn/sec.
- 365 x 24 x 60 x 60
0) = 1000 kg
(as 1 cum of water weight = 1000 kg)
H = Available head = 300 in
Tj = Efficiency 0.75

(1)qHxfl = 1000x3.4x300
H.P. developed - x 0.75
75 75
1000 x 3.4 x 300 x 0.75 x 0.746
Power in kW =
75
1000 x 3.4 x 300 x 0.75 x 0.746
Power in MW = 75 x 1000
- 3.4 x 300 x 0.75 x 0.746
- 75•
= 7.48 MW. Ans.
Example 6.9. Determine the H.P. that can be developed from a
hydroelectric power plant with following data
Catchment area of reservoir = 3 1618 sq. metre
Mean head =40m
Annual average rainfall = 110cm
Efficiency of turbine 80%
Efficiency of generator = 85%
Load factor . = 45%
Assume that 25% of the rainfall is lost due to evaporation and
4% head lost in penstock.
Solution. Quantity of water available per annum
Q = 3 x 103 x 1.10 x 0.75 = 2.475 x 10 8 cu-m.
Quantity of water available per second
2.475 x 108
Q -------- = 7.85 cu-m/sec.
365 x 24 x 60 x 60
Head H=40m
Overall efficiency () = Turbine efficiency x Generator efficiency
x Penstock efficiency
= 0.8 x 0.85 x 0.96 = 0.65

HYDROELECTRIC POWER PLANT 465

- wQHr1
H.P. developed
- 75 x load factor
1000 x 40 x 7.85 x 0.65
= = 6032. Ans.
75 x 0.45
Example 6.10. (a) In a hydropower plant the reservoir is 200 in
above the turbine house. The annual replenishment of reservoir is
40 x 1010 kg. Calculate the energy available at the generating station
bus bars if the loss of head in the hydraulic system is 20 in the
overall efficiency of the station is 80%.
(b) Also determine the diameter of two steel penstocks if maxi-
mum demand of 50 MW is to be supplied.
Solution. (a) n = Efficiency = 0.8
h' = Loss in head = 20 rn
h = Actual head available
=200-h 1 =200-20= 180
E = Energy available at the turbine house
= mgh = 40 x 1010 x 9.81 x 180
= 70.63 x 1013 J
- 70.63 x iø' -
16.84 x 107 kWh.
- 36x105 -
(As 1 kWh = 36 x io J)
Energy output E X 1 = 16.84 x 107 x 0.8
= 13.48 x 107 kWh. Ans.
(b) Kinetic energy of water
= mV = Loss of potential energy of water

,mV2=mgh

V='K='I2x9.81x180=59.4 rn/sec.
= Energy available from a mass in kg of water when it flows
with the velocity 59.4 rn/sec.
= mV = I in x 5942 J/s or Watts.

But energy to be supplied = 50 x 106 watts.

in = 22,700 kg.
 POWER PLANT
466
- co = Specific weight of water = 1000 kg/n3.
A = Total area of two penstocks (in 2)
A x V x ti = m
A x 59.4 x 1000 = 22,700
A = 0.38 m2
= Area of each penstock

=A
-- = 0.38 = 0.19 m 2

Let D = Diameter of penstock

= 0.19

D=O.242rn. Ans.
6.43(a) Hydro Power Plants in India
Some of the hydro power plants installed in our country are
mentioned in Table 6.4
6.43.1 Hydro Power in India
Hydro power represents a renewable source of energy which
enjoys man y intrinsic advantages. Our country has large
hydropotential.
It is essential to produce cheap power for airound development
of the country. Our fossil fuel reserve is limited and we are still at
infancy indeveloping nuclear power and hence as far as possible the
long term planning for power development should be hydropower
,oriented. As no fuel is required to produce power therefore, power
produced will be cheaper than other conventional means in the long
run. In India the scope of Miter power development is tremendous.
Only a part of the easily available water power potential has been
developed so far Hydro power plants do not pollute the atmosphere.
Hence in order to keep the atmosphere free from harmful pollution,
water power is the ideal source of power. Being endowed by nature
with a variety of sources its tropical geographical position, its rivers
and wind power mark out for India much hydro power potential.
Hydro power being a renewable source of energy must undoubtedly
receive a high priority in our energy development programme. Total
hydro potential in our country is estimated to be equivalent to about
75,400 MW at 60% load factor of which only 11 to 12% has been
exploited so far. The assumption that hydel generation despite its
high capital costs and long gestation period, is cheap and clean has
rarely been questioned.
HYDRO-ELECTRIC POWER PLANT 467

Table 6.4

S.No. Name of the project River and the State Type of dam Power
develo
1. Srisai!arn Hyd! Project Andhra Pradesh IMasonry and 770
gravity
2. Upper Sileru H y dro Andhra Pradesh Masonry Division 120
Electric Project Weir
3. Umiarn Hyde! Power Urnaim River, Assam Concrete Earthen 39
Project
4. Subarnarokha Hyde!- Subarnarekha, Bihar Masonry Earthen 120
cum-Water Supply
Project
5. Ukar Project Tapi, Gujarat Masonry Earthen 160
6. Chonani Hyde! Project Tawi, Jammu and Masonry Division 37
Kashmir We,ir
7. Sala! Hyde! Project Chenab Kashmir Concrete Gravity 960
8. Iddiki 1-lyde! Project Iddiki, Kerala Concrete Arch
Concrete Gravity 800
Masonry Gravity
9. P Pamba and Kakhi Concrete Gravity
I river. Kerala

Masonry 300
Masonry
10. 1 Chainbal Valley Project Chamba!, river
(i) Gandhi Sagar Madhya Pradesh Concrete Gravity 115
(Li) Jawahar Sagar, Chamba! (M.P.) " 99
Kotah Darn
(iii) Rana Pratap Sagar 128
Dam
11. Kodayar Hydro-Electric Kodayar, Tamil Nadu - 100,
Project
12 Kundah Basin Develop- On different rivers, - 425
merit Project. Kundah Basin, Tamil
Nadu
13. Mettur Ilydro-Electric Mettur Tamil Nadu Concrete Gravity 200
Project
14. jPeriyar Hydro-Electric Periyar, Tamil Nadu Forebay Dam 140
Project
15. Koyria Hydro-Electric Koyna, Maharashtra Rubble Concrete 900
Project
16. Sharavathi Hydro-
Electric Project
(ii Linga Makki Dam Sharavathi, Ka r- Masonry Earthen 891
nataka
(ii) Talakala!a Dam
17. Tungabhadra Project Turigabhadra, Kar- MasonrY ('irav ,_[126
 468 POWER PLANT

iHiaiimeia Hydro-Electric Silera. Orissa .- 480

- Project
19. Hirakund Project Mahanadi, Orissa Concrete Gravity 427
Earthen
20. Seas Dam at Pong Bess, Punjab- Gravel 360
Himachal
21. Beas Sutlej Link Project
22. Bhakra Nangal Project Beas, Punjab. Concrete Gravity 633
Himachal Pradesh
(i) Bhakra Dam Sutlej, Punjab Concrete Gravity 1156
(ti) Nangal
23. Doab Canal Hydro Punjab . 45
Project
24. Orba Hydro-Electric Rehand, V.P. - 99
Project
25. Ram Ganga Project Ram Ganga, V.P. Earth 165
26. Rihand Hydro-Electric Rihand, U.P. Concrete Gravity 300
Project
27. Yamuna Hydro-ElectricYamuna, V.P. Concrete Gravity 444
Project
28. Jaldala Hydro-Electric Jaldala, W. Bengal Concrete 36
Schemes

So far as river-wise potential is concerned the two great rivers

namely Ganga and Brahmaputra have an unexploited power poten-
tial of 5000 MW and 15000 MW respectively. The river Godavari
and its tributaries have a power potential of 6000 MW, which is still
to be exploited.
The gestation period of hydro power plants is more. To ensure
timely implementation of such projects the following steps are
required
(i) A better assessment of geological and environmental fac-
tors.
(ii) Strengthening of construction agencies in terms of
(a) organisation (b) equipment
(c) skilled man power (d) technical know-how.
(iii) Making available optimal equipment both qualitatively
and quantitatively.
(iv) Regular flow of funds.
6.43.2 Preventive Maintenance of Hydro Electric Power Plant
Preventive Maintenance is based on schedule inspectionF of
plant and equipment. Its purpose is to minimise break down .nd
excessive depreciation resulting from neglect. The various par s of
a hydro plant using reaction turbine are inspected and maint, ned
monthly, quarterly, half yearly and yearly are as follows:

HYDRO-ELECTRIC POWER PLANT 469

(a) Monthly inspection and maintenance is carried out for the

following parts:
(1) Turbine cover parts such as drainage holes, leakage
unit, servomotor connections, turbine shaft and cover,
oil pump its auxiliaries and ejector cabinet.
(ii) Guide vane mechanism.
(iii) Operating ring of turbines.
(b) Quarterly maintenance is carried out in case of following
parts:
(i) Governor oil system
(ii) Ejector cabinet
(iii) Servomotor
(iv) Feedback system.
(c) The following parts are inspected half yearly:
(i) Gauges.
(ii) Grease pumps for guide vanes and guide bearings.
(iii) Governor mechanism.
(iv) Grease pipes connected to grease pumps.
All gauges are recalibrated once in two years.
(d) Yearly maintenance is camed out in case of following
parts:
(i) Turbine auxiliaries such as turbine guide bearing,
and water both, oil pressure tank, and turbine instru-
ments.
(ii) Scroll casing runner with cone guide vanes.
(iii) Draft tube.
(iv) Emergency slide valve.
(u) Runner blades checked for cavitational effects, cracks
and we aring out.
(vi) Pit liner.
6.43.3 Electrical and Mechanical Equipment
The electrical and mechanical equipment in a hydro power plant,
consists of the following:
(a) Electrical equipment.
(i) Generators
(ii) Exciters and voltage regulators
(iii) Transformers (iv) Switcgear.
(v) Control room equipment including switch boards.
The generators used in hydro power plants are usually three
phase synchronous machines. The generators have a speed range of
70-1000 R.P.M. Generators have either a vertical shaft arrange-
ment or horizontal shaft arrangement. But vertical shaft arrange-
ment is preferred. The generator cooling can be achieved by air
circulation through the stator ducts. Cooling by water cooled heat
470 POWER PLANT

exchangers in common. The out put power of a three phase alter-

nator is given by
= Vv.i. p. x 10MW
where P = Power
V = Voltage in volts
I = Current in amperes
p = Power factor = 0.9 to 0.95
Thrust bearing is used to carry all the axial loads of the machine
including the weight of runner, the turbine and generator shaft and
rotor weight.
Transformers may be of three phase or single phase type. The
transformers are oil-filled for insulation purposes as well as for
cooling purposes. The generated voltage is usually below 18 kV
while transmission voltage may be as high as 400 kV. This is
achieved by using step up transformers.
Switch gear is used to make and break the circuits. It consists
of switches, isolators, surge arrestors and circuit breakers. For
generated voltage it is preferred to locate switch gear indoors
whereas out door location is used for transmission voltage. Both
transformers and switch gears particularly located outside should
be provided with adequate lighting protection.
The control room equipment is used to perform the following
functions
(i) Machine starting and stopping
(ii) Generator and system voltage control
(iii) Machine loading control
(iv) Frequency contrdl
(v) Hydraulic control
"(vi) Machine running control
(b) Mechanical equipment
(i) Shaft, couplings, bearings etc.
(ii) Compressors and air ducts
(iii) Braking equipment for the generator
(iv) The oil circuits and pumps
(v) Ventilation and cooling systems
(vi) Cranes and other lifting equipment
(vii) Equipment for power houses lighting
(viii) Equipment for water supply and drainage.
6.43.4. Ilydel-Thermal Mix
A judicious combination of both hydro and thermal power is the
optimum solution to meeting the increasing power demand and to
reduce the cost of electrical power. Hydro-power represents a renew-
HYDRO-ELECTRIC POWER PLANT 471

able source of energy which enjoys many intrinsic advantages as

compared to thermal power. Although the cost of construction of a
hydro power plant is nearly same as that of a coal based steam power
plant in terms, of investment for MW, but hydro-power plant uses
water for power generation which is available in abundance in
nature. Our country's full hydro potential is estimated to be 135000
MW at 40 per cent load factor.
While the hydel thermal mix stood at 51: 49 in 1962-63 it went
down to 40 : 60 at the end of.Fifth plan, declined at 34 : 66 at the
end of Sixth plan and declined to 30 : 70 at the end of Seventh plan.
This ratio may be 20: 80 at the end of Eighth plan if proper measures
are not taken to make use of hydro power properly. it is observed
that to produce electrical power economically the hydel thermal mix
ratio must be 40 : 60 by the end of Ninth plan.
Central Electricity Authority (CEA) has projected that by the
end of Ninth Plan period the hydro component will be 34% thermal
60.2 ncl nuclear 5.8'. Without adequate hydel backup the over
al. cost of meeting the power is more expensive. This is further
supported b y the fact that the hydel generation is based on renew-
able source of energy and is pollution free.
In our country's power scenario coal still occupies the prime
position accounting for 68% of commercial energy generation. This
trend has led to the distortion of hvdel-thermal mix ratio in the
installed capacity over the years. In view of the availability of
natural resources, a generation mix of 40% hyde! and 60% thermal
is considered fairly good for Indian systems.
In our country the present ratio of hydrothermal mix is poor
particularly in respect of Eastern and Northern regions (which are
endowed with large economically harnessable hydro-potential).
Priority should be given for tapping of the same.

6.44 Economic Loading of Hydro-Power Plants

For economic loading of hydro power plants, following data is
needed
(i) Efficiency curve of each unit.
(ii) Input and output curves showing total rate of flow of water
against the load.
(iii) Incremental rate curve showing the change in the rate of
flow required per unit change in load.
Capacity of each unit. The use of hydra power plant as
base-load plant or peak load plant depends on storage
available at the plant.

472 POWER PLANT

6.44.1 Run-of-river plant in combination with steam power

plant
In a run-of-river plant the quantity of water available is not
steady throughout the year. The run of river plant can be used as
base load plant during the rainy season when enough water is
available and thermal power plant is used to take up the peak load.
During dry season the thermal power plant should be used as base
load plant and run-of-river plant can be used to take peak load.
In a combined system it is always desirable to use the
hydropower plant to the best extent possible when enough water is
available. Load duration curve for the combined system should be
plotted to know the power to be supplied at different intervals of
time (duration) similarly power available curve for run-of-river
plant should also be plotted for the same duration. By combining
these two curves it can be found out during which period of the given
duration the run-of-river plant will act as base load power plant.
6.44.2 Pump storage plant in combination with steam power
plant
The pumped storage power plant is useful in an inter-connected
system to supply sudden peak loads of shorter duration. In pump
storage power plant water is pumped from tail water pond to head
water pond.
The various advantages of inter-connecting pumped storage
plant with steam power plant are as follows
(i) During off-peak period the extra energy available from
steam power plant is used to pump water from tail water
pond to head water pond. This stored hydraulic energy is
used during peak load period.
(ii) provides added power to meet the peak load and thus:
It
(a) helps in reducing the installed capacity of the steam power
plant.
(b) allows the steam power plant to run at high load factor.
This will reduce power generation cost.
Example 6.11.A hydel power plant produces 18 x 1W k Wander
a head of 14 metres. Determine the following:
(a) Type of turbine.
b) Synchronous speed of generator.
Assume overall efficiency of plant 70%.
Solution. (a) H = head = 14 metres.
P = Power 18 x iO kW
= 18 x 103 x = 2.447 x iO Fl. P.


HYDROELECTRIC POWER PLANT 473
=
75
where = Specific we ,.' ht of water = 1000 kg/M3
Q = Discharge
ii = Efficiency = 0.7
2.447 x io =.1000 x Q x 14 x 0.7
75
Q = 187.3 m3/sec.
As the head is low and discharge is high so a propeller type of
turbine should be used.
(b) N = Specific speed
1150
= (Approx.) 1150 1150
14 1/4 =
=596
N = Speed of rotation
N. . H.'-44 596x14'4
- 'IP- '12.447 xiO

= 103 R.P.M.
For generator N =
p
where f= Frequency in cycles per second.
= 50 cycles/sec
• p = Number of poles
103_120x50
60
p = 58.3 = 60 (say) ...
As the number of poles is necessarily an even number.
• Again N==.°<5° = 100 R. P.M.
p 60
Example 6.12. A steel penstock carries 12 m 3/sec of water under
a head of 240 rn The penstock diameter is 2.8 Tn and the allowable
stress in penstock material is 1400 kg/M 2 . Determine the following:
(a) Water hammer pressure assuming velocity of pressure
wave 1425 rn/sec.
(b) Thickness ofpenstock.
(c) Describe water hammer.
—32


POWER PLANT
474

Solution.
(a) Q = Discharge = 12 m3/sec.
d Diameter of penstock = 2.8 m
H= Head =240m
V1 = Velocity of pressure wave = 1425 mlsec.
V2 = Velocity of water
Q = V2 x d2

12 = V2 x x (2.8)2

V2 = 1.95 rn/sec.
= Water hammer pressure
- Vi . V2 - = 283 m.
g 9.81

(b) f= Safe stress in penstock material

=1400 kg/cm2
h = Total pressure head = H +
= 240 + 283 = 523 in
p = Total pressure = w.h. = 1000 x 523 kg/m 2
= Thickness of penstock
t
pd 1000 )< 523 x 2.8
=
2[. =2x1400x104
(c) Water hammer. When the load on the generator is suddenl:
reduced the valves admitting water to the turbines are to be close
suddenly. This sudden stoppage brings the water near the valve t
stand still but the along column of water in the pipe line is stil
moving. The momentum of this water causes sudden increase ii
pressure in pipe section. This sudden rise of pressure above norrn
is called water hammer. Surge tanks help in releasing and dissipal
ing the excessive water hammer pressure built up in a penstock.
Example 6.13. For a hydro-electric power station the followin
data is available
Head =380m.
Discharge = 4 ,n3/sec.
Efficiency of tzsrb(ne = 80%
HYDR O-ELECTRIC POWER PLANT
475
Generator frequency = 50 c/s
Determine the following:
(a) Output (b) Type of turbine
(c) Speed of turbine.
Solution. Q = Discharge = 4 m3lsec
N=Head =380 in
w = Unit weight of water = 1000 kg/M3
n = Efficiency of turbine = 0.8
Output = 10 x 4 x 380x08
75 75
= 16,200 metric H.P.
Pelton turbine should be used for a head of 380 m.
N= Actual speed of turbine.
N = Specific speed of turbine.
Choosing an approximate specific speed of about 26 (metric
unit).
N- N .H514- 26x3805'4
P"2 - 16200'2
= 342 R.P.M.
p = Number of poles
1= Generator frequency = 50 cycle/sec.
[120x50
N - 342
= 17.54 = 18 (say)
Corrected speed = 120f = 120x50
p 18
= 333 R.P.M.
Example 6.14. (a) What is a flow duration curve?
(b) The mean weekly discharge at a hydropower plant site is
as follows:


POWER PLANT
476

11100
(j
V.

600

200

2
- Weeks
Fig. 6.46

(i) Draw the hydrograph and find average discharge avail-

able for the whole period.
(ii) Develop the flow duration curve and plot it.
Solution. (a) A flow duration curve is used to determine the
available power at the site. It indicates the daily, weekly, or monthly
flows available as ordinates plotted against percentage of time.
3/sec.) and
(b) The hydrograph is plotted between discharge (m
number of weeks as shown in Fig. 6.46.
Average discharge =Totaliscare = 580 m3/sec
Total weeks
6960 -
12
In order to draw flow duration-curve it is essential to find the
lengths of time during which certain flows are available. This
information is indicated in table shown below.
Total number of jicek I Percentage time

more) 11 91.7
200 (and more)
10 83.3
300 (and
8 - 66.7
400 (and more)
7 58.3
600 (and more)
5 41.7
700 (and more)
3 25
900 tand more)
1000 (and more) 2 ____ 16.7 T
Fig. 6.47 shows a flow duration curve. When selecting a suitable
site for a hydro power plant the flow data for a number of.years are
collected and hydrograPhs and flow duration curveEr the various
periods are determined.
 HYDRO-ELECTRIC POWER PLANT
477

2 9U1

10.00

800

lop 600

'.- 00
C
U
U,
200

Fig. 6.47

Example 6.15. At a site for a hydro power plant a flow of 80

m 3/sec is available at a head of 120 m. If turbine efficiency is 90%
and generator effi ciency is 94% determine the following:
(a) Power that can be developed
(b) No. of units required and their capacities.
Solution. Q = Discharge = 80 m3/sec
H= Head 12Om
Tj = Efficiency of turbine = 0.9
P = Power developed
_wQHr 100 x 80 x 120 x 0.9
- 75 - 75
= 115,200 Metric H.P.
Two turbines each of 115200 = 57,600 metric H.P. capacity
2
should be used
= Generator efficiency = 0.94
Tw

Generator capacity of each unit = 57,600 x 0.94 x 0.736

= 39,850 kW.
 POWER PLANT
478

Total power generated by generator

= 39,850 x 2 = 79,700 kW
= 79.7 MW.
river power plant
Example 6.16. (a) It is observed that a run of
operates as peak load plant with a weekly load factor of 24% all this
capacity being firm capacity. Determine the minimum flow in river
so that power plant may act as base load plant. The following data
is supplied
Rated installed capacity of
generating plant = 12 MW
Operating head = 18m
Plant efficiency = 85%
(b) Calculate the daily load factor of the plant if the stream
flow is 17 cu mec.
Solution.
. Average load
(a) Load factor = Maximum load
Averageload = Load factor x Maximum load
= 0.24 x 12,000 = 2880 kW
E = Total energy generated in one week
= Average load x Time
=2880x24X7 = 48.4 x 104 kWH
Let Q = Minimum flow rate (m3/sec)
=
P = Power developed

where = Specific weight of water

= 1000 kg/m3
If Head = 18 in
il = Efficiency of plant = 0.85

(b) P = 1000 Q!11 = 13.3 Q 11 11 (HP)

75
= 0.736 x 13.33 QHi1 (kW)
= 9.8 x QHri (kW)
=9.8xQx18XO.85 kW
= 149.9 Q kW
E 1 = Total energy generated in one week
- =Px Time =149.9QX24<7


HYDRO-ELECTRIC POWER PLANT 479

= 2.5Q x 104 kWh

Now E=E1
48.4 x 104 = 2.5 Q x iO
- Q = 19.36 cumec.
P i = Powcr developed when stream flows
is .7 curnec.
= 149.9 x 17 = 2548.3 kW
E2 = Energy generated per day
= P 1 x Time = 2548.3 x 24
61,159 kWh.

Daily load factor =

Max. load

= 61,159 __=
0.212 or 21.2%.
24 x 12,000
Example 6.17. (a) What is pondage factor c
(b) Determine the firm capacity of a rix . f. rioer hydro power
plant to be used as 9 hours peaking plant assuming daily flow in a
river to be constant at 16 m31sec.
Also calculate pondage factor and pondage if the head of the
plant is 12 m and overall efficiency is SO%.
Solution.
(a) Pondage factor is the ratio of total in flow hours in a given
period to the total number of hours plant running during
the same period.
(b) P = Firm capacity without pondage
w.Q.H.r1
- 75
where w = Specific weight of water

= 1000 kg/m3 i6. Iz

Q = Discharge = 16 m3/s
H = Head = 12 m
Overall efficiency = 0.8

P = = 2048 H.P.

480 POWER PLANT

P.F. = Pondage factor = t2

where tj = Total hours in one day = 24

t2 = No. of hours of power plant running = 9
P.F. = = 2.67

Q 1 =9 hours flow = 16 x 24
9
= 42.67 m3/s
P1 = Firm power with pondage
=2048x2.67
= 5468 H.P.
S Magnitude of pondage
= (24 - 9) = 15 hours flow
=15x60x60x.16
= 864 x 103 m3.
Example 6.18. The following data is supplied for a hydropower
plant.
Catchment area = 2200 sq km
Annual average rain fall = 150 cm
Available head = 130 m
Turbine efficiency = 0.86
Generator efficiency = 0.91
Percolation and evaporation losses = 18%
Determine
(a) Power development in kW taking load factor as Unity.
(b) Suggest type of turbine to be used if the speed of runner is
to be maintained below 260 R. P.M.
Solution.
A = Area = 2200 sq km
= 2200 x 106 m2
h = Annual average rainfall = 150 cm
150
= 1.5 m

HYDRO-ELECTRIC POWER PLANT 481

y = Percolation and evaporation losses

= 0.18
H = Head available = 130 in
Qi = Quantity of water available for power
generation per year
=Axhx(1-y)
= 2200 x 106 x 1.5(1 -0.18)
= 2.7 x 108 m3
Q = Quantity of water available per second
- 2.7x108
- 365x24x60x60
= 8.56 m3/sec.
= Turbine efficiency = 0.86
= Generator efficiency
.O.91
P Power developed
=
=
WQH
X TIt X flj

= 1000 x 8.56 x 130 x 0.86 x 0.91

= 11612 H.P. = 11612 xO.736

= 8546 kW = 8.546 MW
N = 260 R.P.M
NS - NJ ' 260 118546
H514 - 130
= 54.75.
Single Pelton turbine with 4jcts can be used. Further since head
available is large and discharge is low, Pelton turbine will work
satisfactorily.
Example 6.19. The available discharge for a hvdroelrctric
power plant is 320 ,,z 3/sec under a head of 28 metre.The turbine
efficiency is 90 and the generator is directly coupled to the turbine.
The frequency of generation is 50 cycles /sec. and number of poles
used are 26. Find the number of machines required when
(a) A francis turbine with a specific speed of 280 is used.


482 POWER PLANT

(b) A Kaplan turbine with a specific speed of 500 is used.

Solution. N = Speed of generator

p
f= frequency
50 cycles/sec.
p Number of poles 26
120 x 50
N = = 230 R.P.M.
Since the generator is directly coupled to the turbine the speed
of turbine used must be equal to the synchronous speed of the
generator.
w.Q.H
P= Power = XY

where U) = 1000 kg/m3

Q = Discharge =320 m3/sec.
H = Head = 28 m
Tj = Efficiency of turbine = 0.9

1000 x 320 x 28 x 0.9

P = = lOi x 103 H.P.
(a) Let P1 = Power capacity of each Francis turbine
- N'JP
ti

280 =

P 1 = 6400
Number of Francis turbines required
= P= 107x103 = 17. Ans.
(b) Let P2 = Power capacity of each Kaplan turbine
-- N.'P2

230
_fP
00 -
2812"
P2 196 x 102
Number of Kaplan turbines required

HYDRO-ELECTRIC POWER PLANT 483

_f_107x103.6 Ans.
- P2 - 196 x 102
Example 6.20. A pump storage power plant has a gross head of
300 ni. The hedd race tunnel is 3.8 in and 650 in The
in sec. and friction factor is 0.017. if the overall
flow velocity is 7
efficiency, of pumping and generation are 84% and 89% respectively
determine the plant efficiency. The power plant discharges directly
in the lower reservoir.
Solution. f = Friction factor = 0.17
qp = Pumping efficiency = 0.84
= Generation efficiency = 0.89
L = Tunnel length = 650 in
H = head = 300 in
V = Flow velocity = 7 m/s
D = Diameter of head race tunnel
= 3.8 in
him= Frictional head loss
f.L.V 0.017x650x72
- 2gD - 2x9.81
= 7.26 in
Now h1=kH
7.26=kx300
k=0.024
Plant efficiency ('1) is given by
i-k X Tip X Tl

= 1-0.024 x 084 x 0.89

1+0.024
= 0.712 = 71.2.
Example 6.21. A run off river hvdro-power plant with an
effective head of 20 ni and plant efficiency 78- supplies a variable
load as shown below

15


484 POWER PLANT

8-12 28
12-16 32
16-20 42
20-24 50

Draw load duration curve and determine

(a) Flow required for average load
(b) Load factor.
Solution. The load duration curve is shown in Fig. 6.48.
H = head = 20 m
= Efficiency = 0.78
E Energy supplied during 24 hours

TIME (HCURS)-.--
Fig. 6.48

= (8 + 15 +28 + 32 + 42 + 50) x4x 10

= 700 x 103 kWh

Average Load = 700 x

24
=29x 103 kW
Average Load
Load factor
Maximum demand
29 x
= 50 x iO
= 0.58
= 58%
Let Q = Flow (m3/sec)
Average Power (kW) = x y x 0.736
HYDROELECTRIC POWER PLANT

29 x 10 = 1000 x x 20 x 0.78 x 0.736

Q = 189 m3/seçond.
Example6.22. A penstock supplies water at a head equivalent
to 18 kg/cm2. There is a possibility of 15% increase in pressure due
to transient conditions. The design stress and efficiency ofjoint may
be taken as 1000 kgkm 2 and 88% respectively, if internal diameter
of penstock is 1.3 m find wall thickness of penstock.
Solution. p = Total pressure
=18+0.15x18 -
= 20.7 kg/cm2
- .,-
D = Internal diameter
=1.3m
R = Internal radius
=0.65m.
=65cm
1= Design stress
il = Efficiency
= 0.88
t = Wail thickness of penstock
It is given by the following formula . .......-
= pR
+ 0.15 cm
f.t-0.6p
- 20.7 x 65 +0.15
- 1000x0.88-0.6x20.7
= 1.7 cm.
6.45 Power House Planning
The basic requirement for power house planning is functional
efficiency coupled with aesthetic beauty. Two types of power plants
may be used.
(i) Surface power plants. The building of such power plants
is located above the ground.
(ii) Underground power plants. The building of these power
plants is situated in caverns excavated below the ground. Surface
power plants are quite commonly used.
486 POWER PLANT

After the choice of the type of power plant is made, the following
decisions may then be taken.
(i) The size of power plant and the arrangement of various
units.
(ii) The equipment to be located indoors and outdoors. For
example transformers and inlet valves can be located
either inside or outside the power plant.
6.46 Surface Power Plants
They have less space restrictions than the underground power
plants. The following parts of the power plant should be properly
designed.
1. Power house structure. Power house structure can have
following three divisions:
(i) Sub structure (ii) Intermediate structure
(iii) Super structure.
The sub-structure is that part of power house structure which
is situated below the turbine level. The sub-structural transmits the
load of the structure to the foundation. Sub-structure is usually
below the ground level and houses the following
(i) Draft tube (ii) Tail water channel
(iii) Drainage pipes of waste water
(iv) Drainage galleries (u) Grout galleries
The intermediate structure extends from the top of the draft
tube to the top of the generator foundation. It houses the following:
(i) Turbine including its casing
(ii) Galleries for the auxiliary machines
(iii) Governor servo-motor system.
The turbine flour is below the generator floor and is accessible
through stairs from the generator floor.
The super structure is the portion extending from the generator
floor called th main floor up to the rooftop.
It houses the following
(i) Generators (ii) Governors
(iii) Control room (iv) Exciters
U') Auxiliary equipment. Such as needed for
(a) ventilation (b) cooling
(vi) Main travelling gantry crane at the roof level.
(vii) Control room (viii) Offices and stores.
2. Power house dimensions. The super structure of the power
plant has the following three bays.
(i) Machine hail or unit bay (ii) Erection or loading bay
(iii) Control bay.
Machine hail size depends on the number of units, the distance
between the units and size of machines. The machine hall must have
HYDRO-ELECTRIC POWER PLANT 487

a height which will enable the cranes to lift the rotor of the generator
or runner of the turbine clear off the floor.
The loading bay is used to load and unload the vehicles and
where dismantled parts of the machines can be placed and where
small assembling of the equipment can be done.
The control bay houses the main control consisting of control
switches, panels etc. The control bay should be quite spacious.
3. Cables and Bus Bars. High voltage and low voltage cables
should be carried separately cables and bus bars are placed in cable
ducts provided in the floor of the generator or placed in the bus bar
galleries.
4. Operation room. The operat m room may be located inside
the turbine room. The instructions are sent to the operation room
from the control room. The operation room is the centre from where
the machines are controlled and switch gear operated.
5. Miscellaneous equipment. Proper planning for the follow-
ing should be carried out
(i) Transformers (ii) Service cranes
(iii) Lighting (iv) Ventilation.
5.47 Under ground power plants
They are safe against earth tremors, rock slides and snow
avalanches. The various factors which affect the choice are as
Ibilows
(i) Rock quality (ii) Tunneling ease
(iii) Over all economy.
An aesthetically significant advantage offered by an under-
ground power plant is the preservation of the natural land scape
features. The surface power plants would necessitate the deforesta-
tion, interference with natural landscape on account of the material
transportation and constructional activities.
6.48 Components of underground power plant
The various components of an underground power plant are as
follows
(i) Machine hail It houses the turbines and generators
(ii)Transformer hail
(iii)Control room
(iv) Erection bay : It is used for the assembly of rotor stator. It
is also used to serve as repair bay.
(v)Cable gallery
(vi) Valve c/iarnbt'r: It is used to house control valves

488 POWER PLANT

(vii) Ventilation tunnel

(viii) Access gallery.
6.49 Types of underground power plants
The layout of an underground power plant depends on the
following factors
(i) Head and tail race water levels
(ii) Turbines (iii) Generators
(iv) Transformers (u) Control valves
(vi) Control room (vii) Access gallery
(viii) Ventilation shaft (ix) Topography and geology.
The underground power plants required a good sound rock.
Heavy expenditure is incurred in supporting penstocks and on rock
supports and rock bolting various types of underground power
plants are as follows:
(i) Upstream power plant
(ii) Downstream power plant
(iii) Intermediate station development.
The upstream power plant also known as head development is
located close to the intake and thus water is directly fed from the
head pond to the generating units.
In down stream power plant also called tail race development
has a long ar(d nearly horizontal pressure tunnel together with
pressure shafts and a short tail race tunnel. The intermediate power
plant has a long head race tunnel and a long tail race tunnel.
6.50 Largest Underground Power Plant (Nathpa Jhakhri
Hydel Power Project)
Nestled amidst the snow-capped mountains of Himachal
Pradesh and surrounded by an inhospitable terrain, the country's
largest undergtound power house, the Nathpa Jhakhri hydel power
project, promises to sole the northern region's power problem.
The hydel project, estimated to produce an aggregate capacity
of 1500 MW of hydel power in a single underground cavern, is
situated in the Kannaur and Shimla district. The cavern is to house
six units of 250 MW each.
While the first unit is to be commissioned by June, 1998 the
subsequent second, third, fourth, fifth and sixth units have been
scheduled for completion by December 1998.
The Nathpa Jhakhri project envisages to harness the
hydropower potential in the upper reaches of river Satluj in the form
of "largest run-of-the river underground development in the south-
west Himalayas".
 HYDRO-ELECTRIC POWER PLANT 489
The reservoir at Nathpa has virtually no impoundment of water
and related problems associated with ecology displacement of
population and deforestation.
Some of the unique features of this project, one of the largest
underground complexes in the world is that it has 301-metre-deep
surge shaft to produce hydel power. This will be the deepest
surge shaft in the world.
The ca vern six of this power house alone is
216 M x 20 M x 49 M and is to utilise a design discharge of 405
cubecs of water with a design head fall of 425 metres atJhakhri.
The Nathpa Jhakhri Power Corporation (NJPC) proposes a
60.40 metre-high concrete dam on Satluj river at Nathpa to divert
405 cubecs of water through four intakes. Tunnelling and drilling
work relating to this diversion dam is under completion.
6.51 Advantages and Dis-advantages of Underground
Power Plant
Various advantages of under ground hydro power plant are as
follows
(1) Short conduit reduces cost and head losses. It also
reduces pressure developed due to water hammer; (ii) The
power plant construction period is less ; (iii) The under-
ground power house is safe against air-raids; (iv) Govern-
ing of turbines is easier ; (u) The plant is free from land
slides ; (vi) Regular maintenance cost is low.
The various disadvantages are as follows:
(i) Construction cost is more ; (ii) Special ventilation and air
conditioning is needed ; (iii) Construction of air ducts and bus
galleries increase the cost (iv) Operational cost is more.
Example 6.23. A pelton turbine is to satisfy the following
requirements.
Head =Tm
Power = 9.5 x 101k1V
Speed = 740 RPM
Jet dtamter = of wheel diameter
Overall effici ency = 87%
Determine (a) Wheel diameter
(b) Jet diameter
(c) Number ofjets required
Given : Speed ratio .0. 48

—33

490 POWER PLANT

A Cu = 0.986
Solution.
H = Head = 340 m
P = Power = 9.5 x 10 kW
C, = Coeff. of velocity = 0.986
- V1 = Velocity of jet
= Cu 'J
= 0.986 Ix
9.81x 340 = 80.53 mIs
V2 = Velocity of wheel
= Speed ratio x
= 0.48 x 12-x - -8- 9.
1 x3 4 0 3
= 9. 2 mIs
N= RPM 740
D = Wheel diameter
itDN
V2
60
- it xD x 740

D = 1.01 m
d = Jet diameter

= x D = x 1.01

=0.168m
n = Number of jets required
q = Discharge throughout jet
= x d2 x V1 = x (0.168)2 x 80.53

= 1.78 m3/sec
TI = Overall efficiency = 0.87

P
Ti
Water
- 9500
W.Q.H
where w = Specific weight of water


HYDRO-ELECTRIC POWER PLANT 491

= 9.81 kNfm3
Q = Total discharge
9500
087 -
9.81xQx340
Q = 3.273 M3 /see
3.273= nxq=nxl.78
n = 1.838 = 2 (say)
Example 6.24. For a river having a discharge of 230 litres/sec
and available head of 43 metres.
(a) Calculate specific speed of turbine.
(b) Suggest type of turbine for the river.
Take efficiency of turbine as 80% and speed as 480 RPM.
Solution.
H = Head = 43 m

Q = Discharge = 230 litres/sec = 230 = 0.23 m3/s

N = Speed = 480 RPM
= Turbine efficiency = 0.8
P = Power developed
w = Weight density of water = 9.81 x 10 3 N/rn3
P
- W.Q.H.

08 P
9.81x0.23x43
P = 77.6 kW
MS = specific speed
N.'48Ox'fi
- - 435/4 -38.4

Therefore Pelton turbine is suitable.

Example 6.25. A hydro-electric power plant is to be built across
a river having a discharge of 320 m 3/s and head of 30 m. The
generator is directly coup! ed to the turbine. The frequency of gener-
ator is 50 iz and n.imber ofpoles used are 24. The turbine efficiency
is 82%. Fin, he number of turbines required if
(a) A Fra a specific speed of 250 is used.
(b) A Ka r', specific speed of 710 is used.

492 POWER PLANT
4
Solution.
f= Frequency = 50 Hz
p = Number of poles = 24
(a) H= Head =30m
- Q = Discharge = 320 m3/s
= Efficiency of turbine =0.82
N= 120x50 = 250 RPM
p 24
P
Tj where P = Power
= W.Q.H.

0.82 = 9.81 x 320 x 30

P = 77224 kW
NVP
= specific speed where P i = Power of each Francis
= H514
turbine
- 2 50
2z0 =

P1 =4512kW -
Number of Francis turbines required
P 77224
P 1 4512'
= 17,,1 = 18 (say)
(b) For Kaplan turbine
VP
N3-N
H514
710 - 250VP
- 30
P2 = 32618 kW
Number of Kaplan turbines required
77224
32618 = 2.36 = 3 (say).

Example 6.26. Determine the firm capacity of a run of river

hydropower plant to be used as 7hours peaking load plant assuming
daily flow in the river to be constant at 16 in3/sec.
 HYDRO-ELECTRIC POWER PLANT 493
If head of power plant is 12 m and over all effi ciency is 80%
calculate
(i) Pondage factor
(ii) Pondage.
Solution.
H= Head =12m
w = specific weight of water
= 9.81 kN/ni3
Q = Discharge = 16 m3/s
ii = Efficiency = 0.8
P = Firm capacity without pondage
=lxW.Q.H.
= 0.8 x 9.81 x 16 x 12
= 1506.8 kW
ti
Pondage factor =
where ti = Total hours in a day = 24
t2 = Number of hours for which plant remains in operation = 7

Pondage factor = 24= 3.43

Q = Q x 3.43
= 16 x 3.43 = 55 m3/s
P 1 = Firm power with pondage
= P x Pondage factor
= 1506.8 x 343 = 5168.3 kW
Pondage = (24 - 7) 17 hours flow
= 17 x 60 x 60 x Q
17 x 60 )< 60 x 16
= 9.8 x io m3.
Example 6.27. (a) A run of river plant is used as a peak loud
plant with weekly load factor of 26% all the capacity being firm
capacity. Determine the nzini,nuni flow in river so that power plant
may act as base load plant when rated installed capacity is 12 MW
and operating head is 17 in with plant ef ficiency 85%.

494 POWER PLANT

(b) Also determine the laity load factor when stream flow is
16 m3/sec.
Solution.
(a) C = Capacity of plant = 12 MW
H = Head = 17 m
r = Plant efficiency = 85%
Load factor = Average load
Maximum demand
0.26 = Average load
12 x 1000
Average load = 3120 kW
E = Electrical energy generated in one week
= Average load x 24 x 7
=3120x24x7=52x 104 kWh
P
Now ii =
where P = Power developed
P
0.85
P = 141.75 Q kW
E 1 = Tot! energy generated in one week
=Px24x7
= 141.75 x 24 x 7 x Q
= 23814 x Q kWh
Now E=E1
52 x iO = 23814 x Q
Q = 21.8 m3/s
(b) When Q = Flow rate
= 16 m3/s
P 1 = Power developed = 141.75 x 16
= 2268 kW
= Energy generated per day
= P1 x 24
= 2268 x 24

HYDRO-ELECTRIC POWER PLANT 495

= 54432 kWh
Average load
Daity load factor -
- Maximum load
- 54432
= 0.19 = 19%.
- 24 x 12 x 1000

PROBLEMS

6.1. (a) What are the essential features of a hydroelectric power

plant? Describe the various types of dams used in such plants.
(b) Discuss the working principle of hydro power plant.
6.2. Name the different types of hydroelectric power plants. Describe
the pumped storage power plant and high head hydroelectro
power plant.
6.3. What is the function of draft tube ? Describe the various types
of draft tube.
6.4. How a surge tank helps in reducing water hammer effect ?
Describe the various types of urge tanks.
6.5 (a) Classify the water turbines. What type of water turbine is
used in high head hydro electric plant and why? Discuss the
various factors to be considered while selecting a water
turbine.
(b) Sketch an impulse turbine installation.
6.6. What are the different factors to be considered while selecting
the site for hydroelectric power plant?
6.7. State the advantages and disadvantages of hydro-electric power
plant. Compare it with steam power plant.
6.8. Write short notes on the following:
(a) Penstock
(b) Low head hydro-electric power plant
(c) Francis turbine
(d) Hydraulic Accumulator
(e) Drum Gate
(1) Fore bay.
6.9 Explain the terms 'Hydrology'. Describe Hydrologic cycle.
6.10. \Vat is meant by waterway ? Describe the various types of
tunnels and flumes.
6.11. What is the function of a gate in hydro-power station? Describe
Rolling Gate and Tainter Gate.
6.12. Write short notes on the following:
(a) Cavitation. (b) Cost of h y dro Power Plant.
(c) Stoney Gate. (d) Needle valve.
(c) Turbine governing.
(f) Micro and mini hydro-power plant.
496 POWER PLANT

)g) Trash rack. (h) Butterfl y valve

6.13. What is the function of a spillway ? Discuss the various
types of spillways commonly used.
6.14. Tick mark the correct answer. Francis turbine is used for
(a) Low head. (b) Medium head.
(c) High head.

Ans. Medium head)

6.15. Fll in the blanks
(i) The overall efficiency ofa Francis turbine is....
(ii) Pelton turbine is used for .... heads.
(iii) In axial flow turbines the water flows through the vanes....to
the axis of the runner shaft.
(iv) The maximum cone angle of a straight divergent draft tube
is.....
(r) The running cost ofhydro-electric power plant is....
[Ans. (i) 85 to 90% ; (ii) High (iii) Parallel (iv) 8 (v) Low)
6.16. State the head of water for which the following types of turbine
are used
(a) Pelton turbine (b) Francis turbine
(c) Kaplafi turbines.
jAns. (a) 150 to 300 rp (b) 60 to 150 in (c) up to 60 in
6.17. A turbine develops 30,000 H.P. under a head of 28 m when
running at 150 R.P.M. Determine the specific speed of the
turbine, and specify the type of turbine to be used.
6.18. What is unit power? Unit discharge and Unit Speed ? Derive
the expressions for these quantities and explain their impor-
tance in determining the performance of a turbine.
6.19. A proposed hydroelectric power station has catchment area 520
sq. km with an average annual rainfall of' 480 cm. The average
head is 460 metres. Assuming 40 1. energy losses and 50" load
factor, estimate the installed capacity of the power station.
6.20. Describe the various controls usi. h ydro power plants.
6.1. Write short noteson the following:
(a) Hydro-steam inter-connected system.
(b) Mass curve.
6.22. Write a short note on future of hydro power in India.
6.23. Dell tic firm power and secondary powerofa h dro-electi-ic power
plant.
6.24. Write short notes on the following:
(a) Water hammer.
(b) Synchronous speed of a turbine.
(c) Methods of coupling turbine with generator. Sketch and discuss
turbine generator coupling for a low head power plant.
6.25. For a hydro power plant the following data is available.
Head = 180 in
Discharge = 2.5 m2/sec.
Efficiency of turbine = 85%
Generator frequency = 50 cycles/sec.
HYDRO-ELECTRIC POWER PLANT ri.i
Determine the following:
(a) Output (b) Type of turbine
(c) Speed of turbine.
6.26. A steel penstock carries a discharge of 24 m 2/sec under a head
of 48in. The velocity of water is 4 mlsec. Theefficiency ofjoints
is 88% and the permissible stress in penstock material is 1100
kg/cm 2 . Determine the following
(a) Diameter of penstock.
(b) Thickness of penstock.
6.27. A hydro-electric power plant has a capacity of 40 x 13 kW. The
cost of development is Rs. 1400 per kW. The fixed cost is 10%
per year and operation and maintenance cost is Rs. 12 per kW
per year. If the transmission liability is Rs. 16 per year and load
factor is 65% determine the cost per kilowatt hour.
6.28. Run-off rate of400 m 3/sec and head of45 mis available at a site
proposed for hydroelectric power plant. Assuming the turbine
efficiency of 90% and speed of50 r.p.m. find out the least number
of machines, all of equal size required if Francis turbine not
greater than 2000 specific speed is used.
IA.M.I.E. 19811
6.29. (a) What are 'hydrographs' and flow duration curves, and what
are their uses? Describe unit hydrograph.
(b) Name the different types of water turbines and explain the
basis ofselection of turbines for hydroelectric power stations.
(C) The catchment area for the reservior of a hydro-electric power
plants is 14 x 108 sq in an average rain-fall per annum
of 125 cm 70% of the rain-fall is available for power generation
and the mean operating head is :30 metre. Calculate the
maximum plant capacity h.p. to be installed. Assume that 5%
of the head is lost is penstock, the turbine efficiency is 85%,
the generator efficiency is 90% and the annual load factor is
50%. IA.M.I.E. 19791
6.30. Describe a typical layout of a hydro power plant.
6.31. (a) Flow are specific speed, capacity and head related in a
turbine. Explain the significance of specific speed in select-
ing the type of hydroelectric turbine in a hydro-electric power
station.
(b) It is proposed to develop 2000 h.p. at a size where 150 m of
head is available. What type of turbine would bp 'nitable if
.it had to run at 300 r.p.m.?
(c) Draw a line sketch of a typical high pressure intake head
works for a hydro-electric power station Explain briefly its
features. IA.M.1.E. 19751
6.32. Compare a Pelton wheel and Francis turbine.
6.33. (a) Define runoff. Discuss the factors affecting run-off.
(b) Describe the methods to find run-off.
6.34. Write short notes on the following
(n) Flood run-off (it) Types of catchment areas.
(iii) Hydrologic cycle.
 POWER PLANT
498
6.35. Write short notes on the following:
(a) Pondage factor
(b) Rain-fall and its measurement.
6.36. Discuss the investigations to be earned out while selecting the
site for hydro-power plant.
6.37. Discuss electrical and mechanical equipment of a hydro-power
plant.
6.38. Write short notes on the following:
(a) Site selection for a dam.
(b) Advantages of pumped storage plants.
(c) Hydel—Thermal mix.
6.39. The average weekly discharge at the site of a hydri plant is
240 m 3/sec. If the head at the installation is 28 m and the over
all efficiency of the hydraulic turbine generator unit is 88% find
the maximum average power in MW which can be developed.
6.40. Write short notes on the following:
(a) Functions of anchor blocks used for penstocks.
(b) Selection of number of penstocks.
6.41. Name the main component parts of a high head hydro-power
plant.
6.42. Discuss economic loadingof hydro-power plant.
6.43. Write short notes on the following
(i) Power house planning (ii) Surface power plant
(iii) Underground plant.
6.44. Describe the salient features of Nathpa Jhakhri Hydro-power
plant.
6.45. Write short notes on the following
(i) Run-of-river plant in combination with steam power plant.
(ii) Pump storage plant in combination with steam power plant.
(iii) Speed and pressure regulation in turbines.
 7
Gas Turbine Power Plant

7.0 Introduction
Gas turbine power plant has relatively low cost and can be
quickly put into commission. It requires less space. This plant is of
smaller capacity and is mainly used for peak load service.
Gas turbine power plants are very promising for regions where
liquid or gaseous fuel is available in large quantities. Gas turbine
installations require only a fraction f water used by their steam
turbine counterparts Gas turbine has made rapid progress during
the past decade due mainly to the large amount of research. The size
of gas turbine plants used in a large System varies normall y from
10 to 25 MW and the largest size used is about 50 MW. The thermal
efficiency of gas turbine plant is about 22% to 2511.
In our country it may be expected that by 1995 installation of
medium and large size combined cycle plants (CCP) will pick up and
gas turbines will record a faster growth rate. In C.C.P. plants
atmospheric pollution by fly ash will be corresponding lower, cooling
water requirement will be reduced combined cycle gas based power
plants are more economical than coal based power plants as coal
transportation was exorbitantly expensive. Moreover the gas based
power plant has the shortest gestation period as it could be put on
stream in barely two years. In any other system it takes at least five
years to commission a project. More gas power plants would ensure
utilisation of natural gas.
7.0 (A) : Classification of Gas Turbine Plants
Gas turbine plants-may be classified according to the following
criteria
1. Type of load:
(a) Peak load plants (b) Standby plants
(c) Base load plants.

500 POWER PLANT

2. Application:
(a) Aircraft (b) Locomotive
(c) Marine (d) Transport.
3. Cycle.
(a) Open cycle plants (b) Closed cycle plants.
4. Number of shafts:
(a) Single shaft (b) Multi-shaft.
5. Fuel:
(a) Liquid (b) Solid
(c) Gas.

7.1 Elements of a Simple Gas Turbine Plants

A simple gas turbine plant is shown in Fig. 7.1. It consists of
compressor combustion chamber and turbine. When the units runs
the atmospheric air is drawn into the compressor, raised to static
pressure several times that of the atmosphere. The compressed air
Uen flows to the combustion chamber, where the fuel is injected.
The produces of combustion, comprising a mixture of gases at high
temperature and pressure, are passed through the turbine where
they expand and develop motive force for turning the turbine rotor.
After expansion the gases leave the turbine at atmospheric pres-
sure. The temperature of the products of combustion is nearly
1000° to 1500F. The temperature of the exhaust gases is in the
range of 900 to 1100°F. The compressor is mounted on the same
shaft as that of turbine. Major portion of the work developed in the
turbine is used to drive the compressor and the remainder is
available as net power output.

OIL

M1P..JI
Fig. 7.1
 GAS TURBINE POWER PLANT
501
Turbine. Turbine drives the compressor and the load. Both
impulse and reaction turbines can be used in gas turbine plants. As
compared to steam turbines gas turbines have few stages because
they operate on smaller pressure drops.
Axial flow type turbines are commonly used. The various re-
quirements of turbines are as follows:
(i) Light Weight
(ii) High Efficiency
(iii) Reliability in operation
(iv) Long working life.
Combustion Chamber. In the combustion chamber, combus-
tion of fuel takes place. The combustion process taking place inside
the combustion chamber is quite important because it is in this
process that energy, which is later converted into work by the
turbine, is supplied. Therefore, the combustion chamber should
provide thorough mixing of fuel and air as well as combustion
products and air so that complete cbmbustion and uniform tempera-
ture distribution in the combustion gases may be achieved. Combus-
tion should take place at high efficiency, because losses incurred in
the combustion process have a direct effect on the thermal efficiency
.of the gas turbine cycle. Further the pressure losses in the combus-
tion chamber should be low and the combustion chamber should
provide sufficient volume and length for complete combustion of the
fuel.
Initially the temperature developed in combustion chamber is
too high. The difficulty is avoided by adding a satisfactory amount
of air to maintain stable combustion conditions and then the
products of combustion are cooled to a temperature suitable for use
in gas turbine by introducing secondary air. The sum of primary and
secondary air supplied is total air needed for combustion. Fig. 7.2
shows the combustion chamber. In combustion chamber used for
aircraft engines a large quantity of air is used to keep the tempera-
ture of combustion chamber to about 650°C. The air fuel ratio may
be of the order of6O: 1 in this case.

74-
SECONDARY AIR

AIR

M8USTIOW
/ CHAMBER
FVEL Oil.

Fig. 7.2
52 POWER PLANT

The requirements of a combustion chamber are as follows:

(i) Low pressure loss
(ii) High combustion efficiency
(iii) Good flame stability
(iv) Low weight
(v) Thorough mixing of cold air and hot products of combus-
tion to generate uniform temperature
(vi) Reliability
(vii) Low carbon deposit in turbine, and combustion chamber.
Compressor. The various compressors used are reciprocating
compressors, centrifugal compressors and axial flow compressor.
The reciprocating compressors are not preferred due to the friction
in sliding parts, more weight, less speed and inability to handle large
volumes of air. For a gas turbine power plant of high output and
efficiency generally pressure ratios of 10 : 1 or more is used. It is
observed that when a single compressor with a pressure ratio not
more than 4 : 1 is required the centrifugal compressor is the most
suitable. It is quite rugged in construction, can operate more effi-
ciently over a wide range of mass rate of flow of air than a com-
parable axial flow compressor. Centrifugal compressor is mainly
used in super chargers and in jet aircraft plants, where lowr
pressure ratios and small volumes of air is needed.
For higher pressure ratios multi-stage centrifugal compres
does not prove to be as useful as an equivalent axial flow compressoi.
Therefore, when high pressure ratios are needed, axial compressor
is advantageous and is always used for industrial gas turbine
instaI1tions. Further it is desirable that more than one compressor
should be used when thepressure ratio exceeds 6 : 1. Although the
axial flow compressor is heavier than the centrifugal compressor but
it has higher efficiency than the centrifugal compressor.
It is important that air entering the compressor should be free
from dust. Therefore, air should be passed through a filter before it
enters the compressor. Air filters are not needed in the closed cycle
system.
7.2 Terms and Definitions
(i) Work ratio. It is defined as the ratio of network output to the
total work produced in the turbine.
(ii) Thermal efficiency. It is defined as the ratio of network
output the total fuel energy input. The higher the working tempera-
ture of the working medium the higher the thermal efficiency of the
turbine.
GAS TURBINE POWER PLANT 503

iii) Air ratio. It is defined as the amount of air (in pounds or kg)
entering the compressor inlet per unit of network output of the
turbine.
The size of gas turbine plant is dependent on rate of flow of air
in relation to the useful horse power output. Lower the air rate
smaller will be the size of plant. Inter cooling and re-heating reduces
the air rate. An increase in the compressor and turbine efficiencies
will decrease the air rate. An increase in compressor inlet tempera-
ture will decrease the net output of the turbine and hence will
increase the air rate.
(iv) Pressure ratio. It is defined as the ratio of absolute pressure
at the compressor outlet to the absolute atmospheric pressure at
compressor inlet.
(u) Compression efficiency or machine efficiency. This term is
related to the compressor, and is defined as the work required for
ideal compression to the actually required by the compressor for a
given pressure ratio.
Actual work required by the compressor is always more than
the ideal work.
Air rate or Air ratio is a criterion of the size of the plant, i.e. the
lower the air rate the smaller the plant. From the mechanical and
metallurgical point of view the lowering of the air rate results in
turbines of smaller physical dimensions with a more a nearly
uniform temperature distribution. The work ratio acts as a guide in
the determination of the size of gas turbine. If the work ratio is high
the variations in the compressor and turbine efficiencies will have
less effect on the thermal efficiency of the cycle than ifthe work ratio
is low. A plant with a high work ratio will have higher part load
performance efficiency than a plant with a low work ratio.
7.3 Engine Efficiency or Turbine Efficiency
It is defined as the ratio of net output of the turbine to the power
that should be ideally produced by the turbine. Network output of
the turbine is always less than ideal output.
Ideal conditions mean : (1) that compression and expansion
processes are isentropic ; (2) that no losses occur in combustion
chambers, heat exchangers, inter-coolers, and pipes connecting the
different components (3) that heat transfer in heat exchangers is
complete, that working fluid behave as perfect gas with constants
specific heats ; and (4) rise in temperature on cold side is equal to
drop in temperature on hot side.

U
 04 POWER PLANT

7.4 Starting of Gas Turbine Power Plant

This plant is not self-starting and for its initial starting the
various devices used are as follows
Electric motor energrised by the batteries for an isolated plant
whereas stationary plants may he started by the electric energy
supplied by the plant bus bars. Smaller reciprocating internal
combustion engines are also used for starting purposes. First of all
the internal combustion engine is started by hand cranking and then
it is used to start the gas turbine plant.
7.5 Fuels
Various fuels used by gas turbine power plants are liquid fuels
gaseous fuels such as natural gas, blast furnace gas, producer gas,
coal gas and solid fuels such as pulverised coal. Care should be taken
that the oil fuel used should not contain moisture and suspended
impurities.
The different types of oils used may he distillate oils and residual
oils. The various paraffins used in gas.turbine are Methane, Ethan,
Propane, Octane (gasoline) and Dodecane (kerosene oil). Out of
these gasoline and kerosene or blend of the two are commonly used.
7.5.1 Qualities of Fuel
Some of the important properties to be considered while select-
ing the fuel for gas turbine are as follows
1. Volatility. The properties has a major effect on starting and
combpstion efficiency of the engine particularly at low temperature
and other adverse conditions. The-volatility of the fuel should be
'such that it is conductive to a quick and successful restart blowout
of flame.'Highly volative faels are also not desirable as they have
the following disadvantages:
(i) They are more susceptible to fire (although they have less
tendency to explode).
(ii I They are conductive to vapour lock and to excessive loss
of fuel during flight because of evaporation of certain
lighter hydrocarbons. Therefore, in case of aircraft gas
turbines in which the quantity of fuel used is sufficiently
high, the fuel wastage will also be more if the fuel is highly
volatile.
2. Combustion products. The products of combustion should
not be in the form of solids because they tend to deposit on the
combustion chambers, turbine blades and vanes and cause a loss in
efficiency.
3. Energy contents. Fuel should have greater heating value so
that fuel consumption may be less.
GAS TURBINE POWER PLANT 505

4. Lubricating properties. The fuel should provide a certain

amount of lubrication of friction surfaces of fuel pumps.
5. Availability. The fuel selected should be available in large
quantities so tWat it is cheaper.
7.6 Comparison of Kerosene Oil and Gasoline
Kerosene is quite commonly used in aircraft gas turbines. It is
not as volatile as gasoline and, therefore, there is less possibility of
vapour lock and fuel loss. But its combustion efficiency is low
compared to gasoline. The lubrication properties of gasoline are
poorer. About 5 to 20% of a barrel of crude may be refined kerosene
whereas 40 to 50% of a barrel of crude oil may be refined into
gasoline which shows that gasoline can be available in large quan-
tities.
7.7 Air Fuel Ratio .
Air fuel ratio in the gas turbines is nearly 60: 1.
7.8 Gas-Turbine Cycles
The working of gas turbine power plant is mainly based on the
two cycles mentioned below: -
1. Open Cycle.
2. Closed Cycle.
7.8.1 Open Cycle
A simple open cycle plant
is shown in Fig. 7.3. In this pj LOAD
case fresh atmospheric air is
drawn into by the compressor
EAdAusr
(c) continuously and heat is AIR COfrf.ÔU.S7I0N
added by combustion of fuel CIff,F.;BER
in the working fluid (air) it-
self. The products of combus- Fig. 7.3
tion are expanded through
the turbine (T) and exhausted to atmosphere. For a given flow of
working fluid there is a limit to the amount of fuel which can be
burnt in the combustion chamber and the maximum amount of fuel
wlich can be added to combustion chamber is governed by the
working temperature of highly stressed turbine blades which should
not be allowed to exceed to certain limit. This limiting value of
temperature is dependent upon two factors namely creep strength
of the material used in the construction of the turbine blades and
the working life of the system.
Combustion of fuel takes place either at constant volume or at
constant pressure. Theoretically the thermal efficiency of the con-
stant volume cycle is more than the constant pressure cycle. But
there are some practical difficulties in the case of constant volume
—34 .
....
 POWER PLANT

cycle such as valves are necessary to isolate the combustion cham-

bers from the compressor and the turbine. Thus combustion is
intermittent and does not allow the smooth running of machine. Due
to these defects this cycle has not been widely accepted. In case of
constant pressure cycle, combustion is a continuous process and
valves are not necessary. This system has proved to be quite satis-
factory.
7.8.2 Refinement of Simple Open Cycle
Refinement of open cycle is done by providing regenerator
intercooler atd reheating combustion chamber (Fig. 7.5 (a)).

6i4SOUT . GAS /N

Fig. 7.4

In simple open cycle system the heat of the turbine exhaust

gases goes as waste. To make use if this heat a regenerator [Fig.
7.4] is provided. In the regenerator the heat of hot gases coming from
the turbire is utilized in preheating the air entering the combustion
chamber. Air circulates all around the tubes of regenerator and the
exhaust gases pass through the tubes. This arrangement saves some
amount of fuel because less heat will be required in combustion
chamber (C.C.) and consequently thermal efficiency is improved.
However, there will be small reduction in power output because of
pressure losses in this working fluid (air) while passing through the
regenerator but it is negligible against the improvement in thermal
efficiency. For further improvement of the cycle, an intercooler is
placed between low pressure compressor and high pressure com-
pressor. In this case the compressor will have to do less work to
compress the air (because of reduction in volume). Fig. 7.5 (b) shows
an intercooler. In the inter-cooler water is passed through the tubes
and air circulated all round the tubes. The output can be further
improved by providing a reheating combustion chamber between
high pressure turbine and low pressure turbine. In reheating corn-


GAS TURBINE POWER PLANT 507

bustion chamber fuel is added to reheat the exhaust gases of high
pressure turbine to upper temperature limit.
'COMPRESSOR TURBINE
LOAD
PC ARC. PT

AIR
INTER I REGENERATOR
COOLER
Fig. 7.5 (a)

AIR I N N,,-• AIR OUT

B AF FLES •
"AM COOLANT
OUT

1Ia•u_ _I.t
"TUBE PLATE
1'T(J8E5 COOLANT
END COVER
IN
Fig. 7.5 (b)

By adding the above refinement equipment, i.e. inter-cooler

reheating combustion chamber and regenerator, the efficiency of the
plant is improved but at the same time the apparatus becomes
bulky; and somewhat complicated. These additions raise the ef-
ficiency of gas turbine plant to over 30%. The other arrangements
used are straight compound units and cross compounded units. In
straight compound units (Fig. 7.6) the high pressure compressor
(H.P.C.) is driven by the H.P. turbine and low pressure compressor
(L.P.C.) is driven by low pressure turbine (L.P.T.). In this system
compressot, are mechanically independent as, they are driven by
separate t' Inc. i n this arrangement the two shafts are capable of
adjusting tt, mtational speeds independent. The turbo-compres-
sor unit may op " ' needs while turbo-generator unit
can be maintain' 'd.

508 POWER PLANT

Fig. 7.6

7.8.3 Closed Cycle

In the closed cycle (Fig. 7.7) the same working fluid (air or some
other stable gas) is constantly circulated. The fuel is burnt in the
combustion chamber and the heat is transferred to the working
medium through heat transfer surfaces. Thus the working medium
does not mix with the products of combustion. The working medium
is cooled in the water cooler beforeit enters the compressor. This
minimises compressor work.

LOAD

WATER
COOLER

AREEERATOR
COMBUSTION
CHAMBER
Fig. 7.7

Some of the gases used in closed cycle gas turbine plant are as
follows
(i) Helium (He) (ii) Krypton (Kr)
(iii) Hydrogen (112) (iv) Oxygen (02)
(v) Nitrogen (N2) (vi) Carbon-dioxide (CO2)
(vii) Ammonia (NH3) (viii) Methane (CH4)
(ix) Ethylene (C2114)
7.9 Some Other Possible Arrangements
Some other arrangements of open cycle system areas follows:
(i) Simple Open Cycle—Twin Shaft. In this arrangement air
enters the compressor (C) and after being compressed it flows to
combustion chamber. The gases then expand in the turbines (T).
The exhaust of burnt gases takes place in the atmosphere (See Fig.
7.8).
4

GAS TURBINE POWER PLANT 509

Fig. 7.8

(ii) Inter Single Shaft Open Cycle. The air from low pressure
compressor (L.P.C.) flows to inter-cooler and then to high pressure
compressor (H.P.C.). The high pressure compressor delivers the air
to combustion chamber (C.C.). The gases then expand in theturbine
(Fig. 7.10).

WATER
Fig. 7.9

(iii) Reheat Single Shaft Open Cycle. In this arrangement

reheating is achieved by providing a combustion chamber between
the two turbines (Fig. 7.10).

Fig. 7.10

(iv) Regenerative Single Shaft Open Cycle. In this system

the heat of exhaust gases from turbine (7') is utilized in heating the
compressed air (Fig. 7.11).
 510 POWER PLANT
I

Fig. 7.11

CHAMBER

Fig. 7.12

7.10 Semi-closed Cycle Gas Turbine

The'system is shown in Fig. 7.12. This system is combination of
open cycle and closed cycle. The air enters the low pressure com-
pressor (L.P.C.) and then flows through pre-cooler where its
temperature is lowered. Then it is compressed in high pressure
compressor (H.P.C.) and after that it is heated in combustion cham-
ber. It then expands in the turbine (Ti ) and some amount of gases
leaving this turbine is expanded in turbine (T2) which drives the low
pressure compressor (L.P.C.). A portion of gases is cooled in the
pre-cooler. In this system the heat exchanger has been omitted for
simplicity.
7.11 Compressors
The various ty pes of compressors used in gas turbine are as
follows
1. Centrifugal Compressor. 2. Axial Compressor.
Centrifugal Compressor. It consists of stationary casing and
rotating impeller. Impeller is provided with blades. When the im-
GAS TURBINE POWER PLANT 511

poller rotates the air enters axially and leaves radially. When the
impeller rotates the pressure in the region R falls and, therefore,
the air enter through the eye. The air then flows radially outward
through the impeller blades. After that the air flows through the
converged passages of diffuser blades and finally the air flows to
compressor outlet (Fig. 7.13).

DAPFM
8.A DES

4fPEW
PLAT1

IMPELLI
BLADE.-

Fig. 7.13

Axial Compressor. This compressor (Fig. 7.14) is quite com-

monly used in gas turbines. It consists of stator which encloses rotor
(R). Both stator and rotor are fitted with rings of blades (RB—Rotor
Blades, SB—Stator Blades). In this compressor the air flows in afi
axial direction from inlet to outlet. Air entering at one end as shown
flows through the alternatively arranged rings and gets compressed
successively.
7.12 Air Rate
Air rate is the amount of air in kg needed per horse power hour.
7.12.1 Factors affecting air-rate
1. Effect of turbine inlet temperature and pressure ratio
on the air rate. With increase of turbine inlet temperature the air
rate is decreased. As the pressure ratio is increased the air rate
firstly decreases to a minimum value and then starts increasing.
2. Effect of compressor inlet temperature on air rate. With
the increase of compressor inlet temperature the compressor work
is increased and the net turbine output is decreased and therefore,
air rate is increased.

512 POWER PLANT

Fig. 7.14

3. Effect of regeneration, inter-cooling and reheating on

air rate. If only regeneration process is added it increases the air
rate to small rate. Addition of inter-cooling and reheating in open
cycle reduces air rate especially at high pressure ratios.
4. Effect of compressor and turbine efficiencies on air
rate. Air rate is decreased with increase in compressor and turbine
efficiencies.
7.13 Free Piston Gas Generators Turbine System
The system shown in Fig. 7.15 combines the thermal efficiency
of the diesel cycle with the simplicity of the turbine in which
expansion of hot gases takes place. The fuel is injected in the centre
of diesel cylinder where the combustion of fuel takes place. The
expansion of combustion gases forces the pistons and causes out-
"AE
PO#?75
/ 4j wZqx(

N Cw-esso'i
I AIR
INTAKE

I
CYUNDER

RECEVtR

TURBINE

EXW.WS1

Fig. 7.15
GAS TURBINE POWER PLANT 513
ward moment of pistons. Each piston acts as a single stage air
compressor on its inner face and as an air bounce cushion on its
outer face. During outward movement of pistons some energy is
stored in bounce cylinder. The energy is utilised for causing inward
movement of pistons which compress the air in air cylinder and
diesel cylinder. Air at a pressure of about 75 to 100 p.s.i., flows from
compressor cylinder into central air space and then it enters the
diesel cylinder through intake ports. The mixture of air and com-
bustion gases at about a temperature of 1000°F leave through
exhaust ports to the turbine where these gases are expanded to
atmospheric pressure.
The advantages of the system are:
(i) As there are no unbalanced forces and no side forces on
cylinder wall, the engine is vibration free.
(ii) The system is smaller and lighter than a diesel engine of
the same output. Thermal efficiency higher than that of
diesel engine and nearly 40% is obtained in this system.
(iii) As compared to open cycle gas turbine system the size of
this system is about one third the size of open cycle plant.
(iv) Air rate is lower in this system.
The main disadvantages of the system are being its difficult
starting and control.
7.14 Advantages of Closed Cycle Gas Turbines
The various advantages of closed cycle gas turbines are as
follows:
1. As the working fluid does not mix with the products of
combustion, therefore, it is possible to use a gas of higher
density and higher specific heat than air such as krypton,
argon, xenon etc. This will reduce the size of various
components. Helium has been successfully used as work-
ing medium in closed cycle gas turbines. The specific heat
of helium at constant pressure being nearly five times that
of the air so heat drop and hence energy dealt per kg of
mass flow is about five times in turbines using helium as
working medium, as compared to gas turbine using air.
Heat exchanger used in gas turbines using lum as
working medium has a surface area of about I that of heat
exchanger used in gas turbines using air as working
medium. Therefore, the size of helium unit is comparative-
ly smaller.
2. As the working medium does not mix with the product of
combustion so there is no accumulation of carbon deposits
on blades and nozzles of the turbine. The compressor
remains free of dust as the working medium may be
514 POWER PLANT

cleaned. Therefore, the periodic cleaning of components is not

needed.
3. The system has improved part load efficiency as the out-
put can be varied by withdrawing or admitting more
working medium.
4. External heating can be done by inexpensive solid fuel
such as cal.
Disadvantages
1. As the system is under an initial high pressure with a
working medium other than air, therefore, the system
should be gastight. This increase the cost of the system.
2. A large air heater is needed and this air heater is not as
efficient as the combustion chamber used in open cycle.
3. As the system needs .cooling water so the system cannot
be used in aeronautical engines.
7.15 Advantages of Open Cycle
The advantages of open cycle are as follows:
1.Simplicity. The combustion chamber is lighter in weight and
smaller in size with a high rate of heat release. Secondly the ignition
system is simple only a spark is required for a short period to start
the burning after which the combustion continues. The combustion
chamber may be designed to burn almost any of the hydrocarbon
fuels ranging from gasoline to heavy diesel oil including solid fuels.
2. V ibrationless. In this system the moving or rotating parts
being the rotor (consisting of turbine and compressor connected by
a shaft) and the gear trains that drive the other auxiliaries. There
being no unbalanced forces and therefore the engine is vibrationléss.
3. Codling Water. Cooling water is not needed except in those
turbines using inter-cooler.
4. Low Weight and Size. In this cycle the turbine has a lower
specific weight and requires lesser space per horse power output
(Specific weight is the weight of engin per H.P. output). This
property of producing more power output in a small space and low
weight is quite useful in aviation engines.
5. Warm up period. The warm up period of the engine is
negligible because after the engine has been brought up to the speed
by starting motor and fuel ignited the engine then can accelerate
from a cold start to a full load without warm up time. This property
is quite advantageous in marine, aviation etc.

GAS TURBINE POWER PLANT 515

Disadvantages
1. Part load performance is low. This can be improved by
using inter-cooler and reheater.
2. Reduction in component efficiencies lowers the thermal
efficiency of the cycle.
3. The gas turbine in open cycle requires a large quantity of
air.
7.16 Relative Thermal Efficiency at Part Load
The relative thermal efficiency at part loads for various cycles
assuming full load ther. 1 o fficiency of each cycle is 100% as shown
in Fig. 7.16.

ac

tOADY. -
Fig. 7.16

A—closed cycle with regenerator, inter cooler and reheater.

B—open cycle with regenerator, inter-cooler and reheater.
C—open cycle with regenerator (twin shaft system).
D—simple one cycle—twin shaft.
E—simple one cycle—single shaft.
7.17 Jet Propulsion
Gas turbines are quite commonly used for the propulsion of
aircrafts. The aircraft units are classified as follows:
1. Turbo-jet unit.
2. Turbo-propeller unit.
Turbo-jet Engine. Its principle is based on open cycle gas
turbine. It Consists of a diffuser, compressor, combustion chamber,
turbine and exit nozzle. The air enters the diffuser at a velocity equal
to that of aircraft. The pressure of air rises above atmospheric
516 POWER PLANT

2.
pressure in the diffuser. The air is then compressed in the compres-
sor to a pressure of about 3.5 kg/cm The compressor used is
generally radial or axial type. The air is then supplied to the
combustion chamber. The liquid fuel is injected into the combustion
chamber where its combustion takes place. The hot gases expand in
the turbine. The turbine extracts enough energy to drive the com-
pressor and the necessary auxiliary equipment. The hot gases are
then expanded through exit nozzle and they leave the nozzle at high
velocity and cause a momentum due to which the engine exerts a
forward thrust (Fig. 7.17.).
COMPRESSOR WMtJ I
CHAMBER

AIR
00000000
fti11 iiI i_0 0 0 00 000
- -
_- EXHAUST
0 0 0 0 000 GASES
000000

FUEL NOZZLE
DIFFUSER
Fig. 717

7.18 Specific Thrust

The specific thrust of a turbo-jet is defined as the ratio of force
to air flow-rate (sometime it is defined as ratio of thrust force to
fuel flow-rate).
7.19 Applications of Gas Turbine
1. Gas turbine plants are used as standby plants for the
hydro-electric power plants.
2. Gas turbine power plants may be used as peak loads plant
and standby plants for smaller power units.
3. Gas turbines are used in jet aircrafts and ships. Pul-
verised fuel fired plants are used in locomotive.
7.20 Advantages of Gas Turbine Power Plant
The economics of power generation by gas turbines is proving to
be more attractive, due to low capital cost, and high reliability and
flexibility in operation. Quick starting and capability of using wide
variety of fuels from natural gas to residual oil or powdered coal are
other outstanding features of gas turbine power plants. Major
progress has been made in three directions namely increase in unit
capacities of gas turbine units (50-100 MW), increase in their
efficiency and drop in capital cost (about Rs. 700 per kW installed).
Primary application of gas turbine plant is to supply peak load.
However gas turbine plants now-a-days are universally used as
peak load, base load as well as standby plants.

GAS TURBINE POWER PLANT 517

1. It is smaller in size and weight as compared to an

equivalent steam power plant. For smaller capacities the
size of the gas turbine power plant is appreciably greater
than. a high speed diesel engine plant but for larger
capacities it is smaller in size than a comparable diesel
engine plant. If size and weight aréthe main consideration
such as in ships, aircraft engines and locomotives, gas
turbines are more suitable.
2. The initial cost and operating cost..of the plant is lower
than an equivalent steam power plant. A thermal plant of
250 MW capacity cost about Rs 250 erores. Presently
whereas a gas turbine plant of that same-size cost nearly
70crores.
3. The plant requires less water as compared to a condensing
steam power plant.
4. The plant can be started quickly, and can be put on load
in a very short time.
5. There is no standby losses in the gas turbine power plant
whereas in steam power plant these losses occur because
boiler is kept in operation even when the turbine is iot
supplying any load.
6. The maintenance of the plant is easier and maintenance
cost is low.
7. The lubrication of the plant is easy. In this plant lubrica-
tion is needed mainly in compressor, turbine main bearing
and bearings of auxiliary equipment.
8. The plant does not require heavy foundations and build-
ing.
9. There is great simplification of the plant over a steam
plant due to the absence of boilers with their feed water
evaporator and condensing system.
Disadvantages
1. Major part of the work developed in the turbine is used to
derive the compressor. Therefore, network output of the
plant is low.
2. Since the temperature of the products of combustion be-
comes too high so service conditions become complicated
even at moderate pressures.
7.21 Layout
Fig. 7.18 shows the layout of gas turbine power plant. The air
filter is used to clean the air. The air leaving the air filter flows to
the low pressure compressor. The compressed air then flows
through the inter-cooler and then it enters the high pressure com-
pressor. The air leaving the high pressure compressor tws through
 18 POWER PLANT

the heat exchanger and the hot air flows to the combustion chamber.
Products of combustion are expanded in the high pressure turbine
and then in low pressure turbine.
7.22 Advantages of Gas Turbine Over Steam Turbine
The various advantages of a gas turbine over a steam turbine
are as follows

L.P. TUP.SI,VE

AIR FILTER r
INTER
COOLER

-
'TARTIAn
MOTOR ALTERNATOR H.P
If-
- H.P I COMBU.snOlv
COMPRESSOR URBINE I CHAMBER
-. Fig. 7.18
(i) Its operation is simple and it can be quickly started
(ii) Its initial and maintenance costs are low.
(iii) It requires few parts and their design is simple.
(iv) It has low weight-power ratio.
(u) Its lubrication cost is low.
(vi) Maximum operating pressure is about 6 kg/cm 2 , therefore
the material is not subjected to heavy stresses. However
due to high temperature of gases (nearly 900C) the
material used should be able to withstand it.
(vii) In gas turbine the inlet temperature is about 806C
whereas it is about 500 C C in case of steam turbine. There-
fore, thermal efficiency of gas turbine is more than steam
turbine other things being equal in both cases.
7.23 Gas Turbine Cycle Efficiency
Gas turbines may operate either on a closed or on an open cycle.
The majority of gas turbines currently in use operate on the open
cycle in which the working fluid, after completing the cycle is
exhausted to the atmosphere. The air fuel ratio used in these gas
turbines is approximately 60: 1.
The ideal cycle for gas turbine is Brayton Cycle or Joule Cycle.
This cycle is of the closed type using a perfect gas with constant


GAS TURBINE POWER PLANT 519

specific heats as a working fluid. This cycle is a constant pressure
cycle and is shown in Fig. 7.19. On P-V diagram and in Fig. 7.20 on
T - diagram. This cycle Consists of the following processes
The cold air at 3 is fed to the inlet of the compressor where it is
compressed along 3-4 and then fed to the combustion chamber
where it is heated at constant pressure along 4-1. The hot air
enters the turbine at 1 and expands adiabatically along 1-2 and is
then cooled at constant pressure along 2-3.

Constant
Adiabatic I
pressure
I-I
LA
I'.,

Vol urn e Entropy

Fig. 7.19 Fig. 7.20

Heat supplied to the system = Kp (T i - T4)

Heat rejected from the system = Kp (T2 - T3)
where Kp = Specific heat at constant pressure,
Work done = Heat supplied - Heat rejected
=K(T1 -T4)-K(T2-T3)
Thermal efficiency (fl) of Brayton Cycle
Work done K1 [(T i - T4) - ( 7'2 - T3)]
= Heat Supplied K (Ti - T4)
T2 - T3
11 =1
T1-T4
For expansion 1-2
/
T1 1P1
T2
-
Ti = T2
( P2
For compression 3-4
(Y-1Vy (.. hT
T41P4"i P1
T3 P3 P2)
(y - 1)/y
T4 ' Pj
(P2)


520 POWER PLANT

j Substituting the values of T1 and 7'4 in equation (1), we get

T2-T3
(y-i)/y (y-1)/y
1
T2 - -7'3
(P2) (P''
T2-T3
1 (y-i)/y
(P1'l
(T2-T3)
P2 V
2
7.24 Effect of Blade Friction
In a gas turbine there is always some loss of useful heat drop
due to frictional resistance offered by the nozzles and blades of gas
turbine thus resulting drop in velocity. The energy so lost in friction
is converted into heat and, therefore, the gases get reheated to some
extent. Therefore, the actual heat drop is less than the adiabatic
heat drop as shown in Fig. 7.2 1, where 1-2' represents the adiabatic
expansion and 1-2 represents the actual expansion.
Actual heat drop = K (Ti - T2)
Adiabatic heat drop = K,, (Ti -
Adiabatic efficiency of turbine
- Actual heat drop
- Adiabatic heat drop
- K(Ti -Tz) -
- K (Ti - 7'2) - Ti - 7'2'
For adiabatic process 1-2
t
I T2 = (P2 (Y 1)^Y
T1 pi^
In the compressor also
reheating takes place which
causes actual heat increase to
be more than adiabatic heat in-
crease. The process 3-4 repre-
Fig. 7.21
sents the actual compression
while 3-4' represents
adiabatic compression.
Adiabatic heat drop = K,., (T - 7'3)
Actual heat drop = K,, (7'4 - T3)
Adiabatic efficiency of compressor
 N
GAS TURBINE POWER PLANT
521
= K (7 T3) - 7'4 - T3
-
K'(T4-7'3)
7.25 Improvement in Open Cycle
The open cycle for gas turbine is shown in Fig. 7.21. The fresh
atmospheric is taken in at the point 3 and exhaust of the gases after
expansion in turbine takes place at the point 2. An improvement in
open cycle performance can be effected by the addition of a heat
exchanger which raises the temperature of the compressed air
entering the turbine by lowering exhaust gas temperature which is
a waste otherwise. Less fuel is now required in the combustion
chamber to attain a specified turbine inlet temperature. This is
called a regenerative cycle (Fig. 7.22).
This regenerative cycle is shown on T = diagram in Fig. 7.23.
where 0 = entropy.

.m
our
a
GHAIA1D

COMPRESS"

Fig. 7.22

'T I
PRESSURE
LIMCS

HeAT

Fig. 7.23

Heat supplied = K,, (T1 - 7'3 ) = K,, (Ti - 7'2)

Heat rejected =Kp(Ts-Ts)K,,(T4_T3)
() Thermal efficiency of theoretical regenerative cycle


POWER PLANT
522

1 - K(Ti-Ts)
For isentropic compression and isentropic expansion thermal
efficiency is given by

r I: K(Tl-T5)

Example 7.1. In a gas turbine power plant working on Joule

cycle, air is compressed from 1 kg/cm 2 and 17C through a pressure
ratio of 6. It is then heated in the combustion chamber to 700C and
expanded back to a pressure of 1 kg 1cm 2 . Calculate the following:
(a) Cycle efficiency
(b) Work ratio
(c) Specific work output of the plant.
Solution. The cycle is shown in Fig. 7.24 on T = Q diagram
T3 = 273 + 17 = 290'C
71' T1=273+WO=973C

t T4 = (F4
T 3 T3Pa
y = 1.4 for air
Fig. 7.24 14 - 11/i4

74
= T3[]= 29O[
= 5OOK
(y-1y
T2 P2
Ti Pi
L2 - 1)/i [ 1.4 - I)/1.4

, T2 T, [ 973
T i 6
= 554'K
Compressor work = K (T4 - Ts) = 0.24 (500 - 290)
= 50.4 kcal
(Assuming K = 0.24)
Work done by turbine = K (T i - T2)
= 0.24 (973 - 554) = 100.56 kcal
Net work done = 100.56 - 50.4 = 50.16 kcal
 GAS TURBINE POWER PLANT
$23
Heat supplied = K (Ti - 7'4) = 0.24 (973 - 500)
= 0.24 x 473 = 1135 kcal.
Cycle efficiency ('1) = Net work done -
Heat supplied - 113.5
= 0.44 = 44%
Work ratio = Net work done
----- = 50.16 0.49
Work done by turbine 100.56
Specific work output = 50.16 kcal
= 50.16 x 426.94 = 21,415 kgfm/kg.
as (1 kcal = 426.94 kgfm)
7.26 Combined Working of Gas Turbine Plant and
Steam Power Plant
When a load is to be supplied by both steam power plant and
gas turbine power plant then the base load is supplied by steam
power plant and peak load is supplied by gas turbine power plant.
The size of gas turbine plants used in a large system varies normally
from 10 to 25 MW and largest size used is about. 50 MW. A gas
turbine plant can be started quickly and has a short starting time
as compared to steam power plant. Gas turbine plants are par-
ticularly useful and economical where cost of gis is not excessive.
The capital cost of gas turbine power plants is less as compared to
steam power plant of equal size. The fixed charges of gas turbine
power plants are also lower than those of steam power plant of same
size.
7.27 Gas Turbine Power Plants in India
Gas turbine power plants are quite helpful in meeting the
growing power demand of the country with the minimum capital
cost-
Some of the gas turbine power plants in our country are as
follows
(a) Gas turbine power plant in Namrup. This power plant
is located at Namrup in Assam. it is of 70 MW capacity.
(b) Uran gas turbine power plant. This power plant is of 240
MW (60 x 4) capacity and is located at Uran in Maharashtra.
(c) Auraiya Gas power plant. This is India's largest combined
cycle module having 652 MW capacit y . It supplies power to the
northern gid. It is situated in Uttar Pradesh. It has been con-
structed by NT.P.C. India's largest public sector power utility in the
field of power generation and transmission. Th Mitshubishi Heavy
 POWER PLANT
524
;Industry of Japan had collaborated with N.T.P.C. in the construc-
tion of this prestigious project.

PROBLEMS

7.1. What are the essential components of a simple open cycle gas of
turbine plants ? How intercooling and regeneration help in
improving thermal efficiency of the plant?
7.2. Describe a closed cycle gas turbine plant. What are the ad-
vantages of closed cycle?
7.3. What are the advantages and disadvantages of gas turbine
power plant? Discuss the applications of such plant.
7.4. What are the various(actors to be considered while selecting the
site for gas turbine power plant?
7.5. Give the layout plan of gas turbine power plant.
7.6. What methods are used to improve the efficiency of gas turbine?
7.7. What are the various compressors used in gas turbine ? State
their relative advantages and disadvantages?
7.8. Explain free piston gas generator turbine system?
7.9. What are the requirements (Qualities) of fuel used in gas tur-
bines'? Compare kerosene and gasoline as fuels.
7.10. State the principle ofjet propulsion. Describe a turbojet engine.
7.11. Write short notes on the following:
(a) Pressure ratio
(b) Semi-closed cycle
(c) Specific thrust of turbo-jet engine.
(d) Relative thermal efficieicy at part loads of various ga tur-
bine cycles.
7.12'(a) Define work ratio.
(b) Which is better, a high or low work ratio? Why?
7.13. Compare a gas turbine with a steam turbine?
7.14. Define air rate. Does a low air rate increase or decrease the size
of an engine?
7.15. List the five operating variables that strongly effect the thermal
efficiency of an open cycle g isturbine.
Solution. The thermal efficiency of an open cycle gas turbine is
heavily dependent upon the following operating variables:
(i) Pressure ratio
(ii) Turbine efficiency
(iii) Turbine inlet temperature
(iv) Compressor efficiency
(u) Atmospheric or compçessor inlet air temperature.


GAS TURBINE POWER PLANT

525
7.16. Discuss combined working of gas turbine plant and steam power
plant.
7.17. Describe any two gas turbine power plants in India.
7.18. State the requirements of a combustion chamber for a gas
turbine.
7.19. Discuss the factors affecting air-rate in gas turbine.

-.1
•

• I ••tst.'

- -.

% •1' -
a
LIJ

Instrumentation

8.0 Introduction
In every power station different t ypes c,f instruments are used.
They measure pressure, temierature and flow etc. in the plant. The
various functions of such instruments can be summarised as follows:
(i) They provide guidance to operate the power plant effi-
ciently and economically.
(ii) They can be used to checkthe internal conditions of the
equipment and thus provide maintenance guidance.
(iii) The y enable economical supervision.
(iv) The y -hell) in plant performance calculations.
(v) They help in costs accounting and cost allocations.
8.1 Classification of Instruments
The various type of instruments can be divided in t o two
categories as follows:
(i) Electrical Instruments.
(ii) Mechanical Instruments.
Elec'trical Instruments. Electrical instruments include am-
meters, voltmeters, wattmeters power factor meters, ground detec-
tor and reactive voltampere meters.
Mechanical Instruments. Mechanical instruments used for
various purposes are of the following types.
8.2 Measurement of Pressure
Instruments known as gauges are used to measure pressures
The various gauges used are as follows:
(i) Barometers
(ii) Manometer gauges
(iii) Vacuum gauges
(iv) Pressure gauges.
A barometer is a device for measuring atmospheric pressure.
Manometers are used to measure pressure below or above atmos-
pheric pressure. ti-tube manometers used for measuring pressure

INSTRUMENTATION 527

above atmospheric pressure are known as

Pa
manometer gauges. Gauges used to measure
pressure less than atmospheric are known as
vacuum gauges. Such gauges are used to
measure condenser vacuums. Manometers
contain fluid such as mercury, water etc. and
read pressure in centimeters of mercury and
water respectively.
Fig. 8.1 shows a U-tube manometer used
for measuring pressure below atmospheric
(vacuum). One leg of U-tube is connected to
the vessel in which pressure is less than
atmospheric, the other leg of the tube is open
Fig.8.1 to atmospheric. In this manometer,
Ii. = -P
where Pat m = atmospheric pressure
p = pressure of gas inside the vessel.
Fig. 8.2 shows a U-tube manometer used to measure pressure
above atmospheric pressure. Here

Gauges used for measuring pressure greater than atmospheric

pressure are called pressure gauges. Pressure indicated b y such
gauges is called gauges pressure. Fig. 8.3 shows a Bourdon pressure
gauge. The pressure of steam in the boiler is generally measured by
this gauge.

Pat

Fi

Fig. 8.2 Fig. 8.3

0
 528 POWER PLANT

8.3 Temperature Measurements

Temperature measuring instruments are of two types as fol-
lows:
(i) Thermometers. (ii) Pyrometers.
Pyrometers. Pyrometers or thermocouple thermometers are
used to measure very high temperatures of furnaces, flue gases and
super heated steam. Fig. 8.4 shows a thermocouple pyrometer. It
consists of thermocouple, an indicating device and connecting wires.
The hot junction is placed in an iron tube to protect it against
damage and cold end is buried in ground so to keep its temperature
constant. Due to difference in temperatures of two junctions an
electromotive force is generated. The indicating instrument is a
potentiometer or a millivoltmeter. The millivoltmeter is so
calibrated as to read in degrees by comparison with a standard
thermometer. This pyrometer can measure temperatures from 300
to 3000SF.
Copper lead s

zo;'^
box
—0— _^ ^-^ I Indicator

I I Flexible
_ jocouie
II auxiliary
couple
Hot
junction
Furnace

Ground
Pipe

L Cold junction
Fig. 8.4

Therrnwnett'r . '"hey are used to measure ordinary temperature.

(i) Gas fillea and tube thermometers. They can measure
temperatures of gases or liquids up to 500C.
(ii) Vapour pressure thermometers. They can measure tempera-
ture up to 250 C.


INSTRUMENTATION
52S
(iii) Glass tube mercury thermometers. They are used to measure
temperature of feedwater, condensate, circulating water and bear-
ing oil etc.
8.4 Flow Measurement
The various flow meters used measured flow rate areas follows:
(i) Steam flow meters. They are used to measure steam output
of boilers and turbine supply etc.
(ii) Water flow meters. These are used to measure feed water
condensate and pump discharge.
(iii)Airflow meters. Rate of flow of water is generally measured
by a venturi meter. Fig. 8.5 shows a venturi meter, a simple device
operating on the Bernoulli's principle. It consists of two tapering
lengths of pipe A and B interconected by a cylindrical part C. The
diameter of the wide ends is equal to that of the pipe in which the
rate of discharge is being measured. On the cylindrical parts
piezometric tubes Ti and T2 are fixed. As the water flow from end 1
to end 2 a difference between water level (h) in the PiezometerS is
obtained.

1i Y21
I
hf
1 I .p - _
I
rç11
A iC 8

Fig, 8.5 Venturimeter,

The rate of water discharge in the pipe is given by

Q Rate of discharge

= Cd A 1 metre per sec

Al

where C = Discharge coefficient

A 1 = Area at section Y, Y-2 =

A 2 = Area at section Y2Y2 = d2

 530 POWER PLANT
4
h = Difference between water levels in
the piezometers.
8.5 Fuel Measurement
The various instruments needed to measure the quantity of coal,
gas and fuel oil consist of belt conveyor, weighers, or coal volume
measuring meters, gas meters and oil meters respectively.
8.6 Speed Measurement
Vibrating reed tachometer, stroboscope, clock t ype tachometer
and revolutions counter are various instruments used for measuring
speed.
8.7 Level Indicators
They are used to record the level in boilers, tanks etc.

Compression
cylinder

U,
0
E

To drain

Fig. 8.6
 INSTRUMENTATiON
531

8.8 Gas Analysis.

The equipment used for the gas analysis may be Orsat ap-
paratus or CO 2 meters, oxygen meters. Orsat apparatus is used to
determine the .volumetric analysis of products of combustion.
Fig. 8.6 shows Hay's CO2 recorder which is used to record only
ç02 in flue gas. Flue gas enters, at regular intervals and CO 2 gets
absorbed and its percentage is indicated by the pointer moving
across the chart. Water falling from aspirator draws the flue gases
from the measuring burett-'. (M.B.) and enters the stand pipe. When
the level of water reaches V, a iample of gas is trapped. Rising water
sends the extra gas to the atmospheric tube. When the water level
rises to the point W burette contains an accurate sample of gas at
atmospheric pressure. Rising water pushes the gases from burette
through the capillary tube (C.T.) into the absorption chamber (A.C.)
where it comes in contact with liquid cardisorber. Some of the
cardisorber is pushed by the gas which seals the compression
chamber of the caustic tank and simultaneously the bellows are
operated to move the pointer. Water in stand pipe reaches the top
of machine at the end of cycles. The water than starts a syphon
which empties the machine of water and a fresh sample of gas is
drawn.
8.9 Calorimeters Fuel and Steam Meters
8.9. (a) Gong alarms give warning about the higher tempera-
tures of generator or transformer coil of high water in hot well or of
low water in boiler feed tank and lubricating oil temperature.
8.9. (b) Other instruments used are: indicating, and recording.
Indicating instruments provide operating guidance whereas record-
ing instruments help in performing calculations.
8.10 Selection of Instruments
The instruments used should be accessible, accurate and easy
to handle. Although size and importance of a power station are the
deciding factors in selecting the instruments but it is desirable to
use as far as possible minimum number of instruments. This saves
money and avoids unnecessary complication of the power stations.
As compared to hydro power station and diesel power station, steam
power station requires more number of mechanical instruments.
The electrical instruments are about the same for all plants of
comparable capacity. Some of the essential instru inents for a steam
power station include pressure gauge, boiler water gauge, a ther-
mometer on the feed water main and on steam line, feed water
meter, a vacuum gauge on condenser, lubricating oil thermometer,
a feed water meter, a steam flow meter, fuel meter, flue gas ther-
mometer, switch board voltmeter, ammeter and watt hur meter.

532 POWER PLANT

; 8.11 Electrical Instruments

The electrical instruments are used to measure current voltage
electric resistance, power and energy.
(i) Electric current is measured by ammeters and voltage is
measured by voltmeters.
(ii) Electrical bridges are used to measure electric resistance.
(iii) Wattmeters are used to measure power.
(iv) Energy is measured by means of watthour meter.
8.12 Instrumentation and Controls in Steam
Power Stations
In a steam power station the major controls provided are as
follows:
(a) Boiler Instrumentation:
(i) Combustion control
(ii) Feed water control
(iii) Steam flow control
(iv) Main steam temperature system
(v) Furnace safe guard supervisory system
(vi) Typical measurement of flue gas analysis, chemical and
physical analysis of feed water.
Combustion control regulates the flow of fuel and air so as to
generate the required amount of heat in accordance with the steam
load. Fuel feeds is controlled by varying the speed of coal feeder and
air flow is controlled by positioning the damper in the primary air
inlet ducts and inlet vanes of forced draft fans.
Feed water control is used to regulate the flow of feed water to
the boiler. The quantity of feed water flowing to the boiler will
depend upon the steam requirements. The feed water flow is
measured by a nozzle fitted in the feed water discharge pipe.
Flow of the steam through a stationary turbine is usually
regulated so as to produce constant rotative speed in the presence
of variable load demand. Control is exercised by varying the quan-
tity and pressure of steam flowing through the turbine. The main
steam temperature at the super heater outlet is controlled.
(b) Turbine instrumentation
(i) Turbine supervisory instrumentation
(ii) Typical measurement and controls in turbine like conden-
sate level controls. Condensate re-circulation controls,
vacuum controls etc.
Modern thermal power stations are becoming of higher capacity
and require coordinated master control hand manual facility, sur-
veillance of the control system, redundancy in critical areas, and
 INSTRUMENTATION
533

more exact turning of control loops. In fact there is a trend towards

greater automation. Reliability of instrumentation and control in
thermal power plants is of vital importance. To ensure that the
sophisticated electronic equipment control and instrumentation
system works with maximum efficiency it is essential to provide the
specified environment such as air conditioning and dust free atmos-
phere in control room.
Fig. 8.6 (a) shows a typical central control room layout. The
positioning of the instruments depends on their shape, size and
number. It is desirable to group together the various controls in
order to have easier regulation and adjustment either automatic or
manual.

ELECTRICAL
CONTROL

TURBINES
AND
cONTROl. _f L / I 1
AUXiLIARIES._1-[IIJ LI: [IIJ4OPERATORS

AUXIUARI

BOILER CONTROL BOILER CONTROL

Fig. 8.6 (a)

8.13 Instruments and Controls Arrangement

In steam power stations a grouped arrangement of instruments
and controls is used. Two systems generally used are as follows:
(i) Area system
(ii) Centralised system.
In area system all the controls and instrument for boiler and its
auxiliaries are placed in one room or area in the boiler room.
Similarly all the controls and instruments for the turbine and its
auxiliaries are arranged in one roothbr area in the turbine room and
all the controls and instruments for generator and its auxiliaries are
arranged one room are area in the switch house. In centralised
control system the instruments and controls for all equipment of
power station are placed in the same room.

-
 534 POWER PLANT

8.14 Orsat Apparatus

This apparatus is used to carry out the analysis of flue gas from
boiler. The amount of CO2. 02, CO and N2 present in flue gases can
be determined.
K

Fig. 8.7
This apparatus as shown in Fig 8.7 consists of levelling bottle
A, measuring burrette B and three absorbing pipettes, C, D and E.
The pipettes C, D and E contain respectively solutions of potassium
hydroxide for absorption of CO2 , pyrogallic acid for absorption of
02 and curous chloride for absorption of CO. These pipettes
provided with stop cocks, K1 , K2 and K3 respectively. The
connects the apparatus to the flue gas.
100 cc. of flue gas sample is filled in the measuring burrette
lowering the levelling bottle A. The stop cock Ki is opened and by
• adjusting the bottle A the flue gas sample is transferred to pipette
C wherç CO2 is absorbed. The remaining gas is brought back to
measuring burette B and Volume measured. The difference between
original volume (100 c.c.) and this volume is the volume of CO2.
Again flue gas is transferred to pipette D where 02 is absorbed. The
remaining gas is brought back to B and volume measured. Dif-
ference of two volumes is the volume ofO 2 . Finally gas is transferred
to pipette E where CO is absorbed. The gas is brought back to
measuring burette B and volume measured. Difference in volumes
is the volume of CO. when percentage of CO2, 02 and CO are known,
the remainder of the gas is assumed to the nitrogen.
During a test of boiler performance flue gas samples are
analysed periodically to find the contents of CO2, 02, CO and N2.
8.15 Oxygen Meter
The presence of dissolved oxygen in feed water for boiler is
responsible for the corrosion of boiler tubes and impairs the perfor-
mance of condenser. The deaerating system is included in the plant
INSTRUMENTATION 535

to reduce the concentration of oxygen. It is desirable that the value

of oxygen should be kept below 0.1 cc. per litre or a value of 0.007
ppm is considered the limit for a plant operating in the pressure
range of 100 to 170 kg/cm2.

-- Meter

Cool In
water
g1 Cat0nt
head dtvice To
T d.c.
mains
H

ritice_'L.,,

Drain
Fig. 8.8
Fig. 8.8 shows a dissolved oxygen meter. In this meter the
difference in thermal conductivity of pure hydrogen gas is compared
with that of a mixture of hydrogen and oxygen. Hydrogen enters
from the right chamber and displaces some of the ox ygen in the
sampled water. Consequently the required conditions for instru-
ment based on the principle mentioned above are fulfilled.
8.16 Impurity Measuring Instruments
The feed water used in a boiler to generate steam may consist
of the following impurities
(i) Calcium sulphate (CaSO4) and other sulphates
(ii) Sodium chloride and other chlorides
(iii) CO 2 from evaporators
(iv) Metallic pieces picked up by steam while passing through
pipes
(v) Silica from dust etc.
(vi) Oxygen from air in condensers etc.
These impurities should be kept to the minimum.
Fig. 8.9 shows dionic water purity meter. The principle used in
the meter is that the electrical conductivity of an electrolyte dis-
solved in water depends on the amount of salt in solution i.e. the
extent of impurity. The resistance between the opposite faces of a
cube of standard water in the instrument is compared with the
resistance when the water contains more impurities.

536 POWER PLANT

Temp. c
bimetal

Insulating
tube

CAM m ow
Collar
(-v.a)

flow at water
sample
Fig. 8.9

In this meter a sample of water is allowed to flow through it.

The automatic correction for temperature variation is effected by
the plunger. The insulating plunger vanes water cross section
automatically by temperature compensator to give correction to
20C. Conductivity is measured df two water columns in between
rings ad G.M. coilars. The conductivity per cm 3 indicated is the
reciprocal of the resistance measured. The standard of reference is
the resistance of one meg-ohm between the opposite face of a cm 3 of
solution. The conductivity being the reciprocal of resistance equals
one micro mho per cm3.
Dionic readings of condensate and feed water at various points
of flow are brought to one control point, connections to the meter
being provided by special piping. This arrangement enables to locate
the source of contamination speedily.
8.17 Measurement of Smoke and Dust
The amount of smoke and dust present in the flue gases going
out of chimney should not exceed a predetermined level in order to
avoid pollution of atmosphere. The main pollutants from the ther-
mal power plants are dust and gases like CO, CO2. S021 NO2 etc.
 INSTRUMENTATION
537
scns'vity
Gas Control Urit Control
Project 6rF ih Reciever (/

[nit

I Gas I I! UL6 To
- IL__..JI Rccord
—rr----- cr
A C.Mairis
Fig. 8.10

8.17.1 Photo cell-type smoke meters

Fig. 8.10 shows photo cell principle for the smoke density meas-
urement. In this arrangement a focused light beam through the
chimney is passed on a photo cell and the variations in the signal of
the photo cell circuit are measured. The variation of the obscuration
of light source due to the smoke and dust in the gases is measured.
8.17.2 Reflected light dust recorder
In this dust meter shown in Fig. 8.11, the reflected light from
he dust particles is measured. Light from the lamp is beamed into
e dust through a suitable opening. Some of this light is reflected
ck on the photo cell. The light reflected varies with variation of
st quantity in the flue gas.

Photoc11
Duct qr
Power PcordQr

f:)jy TT ArnptItQr
Lamp tor
U
Fig. 8.11

PROBLEMS

8.1. (a) What are the functions of instruments in a power station?

(b) Name any five electrical and any five mechanical instru-
ments used in a power station.
8.2. State and explain the instruments used to measure the follow-
ing:

-36
538 POWER PLANT

(a) Pressure
(5) Temperature
(c) Steam flow.
8.3. Write short notes on the following:
(a) Hay's CO2 recorder
(b) Selection of instruments.
8.4. Explain area system and centralised system for the arrange-
ments of instruments and controls in a steam power plant.
8.5. Describe any three controls used in steam power plant.
8.6. Explain how a venturi.meter is used to measure rate of water
discharge in a pipe.
8.7. Describe Orsat apparatus used for flue gas analysis.
8.8. Describe oxygen meter.
8.9. Sketch and describe the feed water impurity measuring instru-
ment.
8.10. Sketch and describe the photo-cell type smoke meter.
8.11. Sketch and describe the reflected light dust recorder.
 Miscellaneous Problems

9.1 Magneto-hydro Dynamic (MHD) Generator

Principle. The principle of a magneto hydro dynamic (MHD)
generator is based on Faraday's law of electromagnetic induction
which states that a changing magnetic field induced an electric field
in any conductor located in it. This electric field while acting on the
free charges in the conductor causes a current to flow. As in case of
conventional electric generator conductor crosses the fine of the
magnetic field and a voltage is induced. Similarly in a magneto
hydrodynamic generator when an ionised gas flows across the lines
of magnetic field a voltage is induced. The ionised gas acts like an
electrical conductor. The gas used may have a temperature between
2000'-3000°K.
M.H.D. generator is a highly efficient heat engine which directly
converts thermal energy into electricity. It is the latest technique of
advanced method of power generation where efficiency as high as
60% can be achieved as compared to about 35% efficiency of conven-
tional thermal power stations. A M.H.D. generator requires a
suitable working fluid which is electrical conducting. The working
fluid is a partially ionised gas. The concepts of M.H.D. generation
depends much more on the conductivity of the gas. The conductivity
of the gas is a function of temperature. Gases become conducting
when their atoms of molecules are stripped of one or more electrons
thermally, electrically or by using radiations. However to achieve
thermal ionization of the products of combustion of fossil fuel or
inert gases extremely high temperatures are necessary. Reasonable
ionisation and hence reasonable value of electrical conductivity is
obtained at temperature around 2000 to 3000'K when the gases are
seeded with additives of easily ionising materials (alkali metals).
This m€ - .od ca power generation will reduce environmental pollu-
tion considerably.
The initial cost of setting up of an M.H.D. power plant is
anticipated to be slightly higher than that of conventional thermal

540 POWER PLANT

power station but this would be offset by factors like higher efficien-
cy and improved cycle of operation.
Working. The various components of MHD generator are
shown in Fig. 9.1. The hot ionized
T SONISED
gas passes between the poles of an
6A electro-magnet and induces a
1 potential difference between a
pair of electrodes which are at
j right angles of magnetic field and
a current starts flowing in the
MAG LOAD resistive load connected between

ELECTRODES electrodes. The ionised gas has

high temperature and pressure
Fig. 9.1
. and after passing through MHD
generator it is passed through a
generator where energy is added to it and it is then recirculated in
the MI-ID generator. This system is simpler and has large power and
temperature handling capacity. Having no moving parts it has
high-reliability, MHD power plants tan operate as base load, peak-
ing or semi peaking units and along with a large load variations
without significapt loss in efficiency.
The technology of MHD power generation is poised for a big leap
and as a major contender for future power plant schemes. In India
the department of science and the technology has sponsored a
research and development programme in the field of coal based
MHD generator. The programme aims at studying various aspects
of MHD with a plasma obtained from coal as fuel. The programme
• is being carried out by Bhabha Atomic Research Centre in collabora-
tion wit} Bharat Heavy Electricals Limited.
The Indian MHD programme emphasis on the following:
(i) Developing the necessary competence in areas of as-
sociated technology which will form the basis of sophisti-
cated and commercially viable equipment for MHD power
generation.
(ii) Creation of a suitable base of R and D work in the field of
MHD power generation, setting up a 5 MW—MHD ex-
perimental plant with a provision to increase the thermal
rating to 15 MW with suitable modifications and develop-
ing scientific data.
(iii) Considering the abundance resources of coal and com-
pared to natural gas and oil, the MHD research
programme will be based on coal technology. Initial effort
in the development of technology will be through coal


MISCELLANEOUS PROBLEMS 541

gasification route, specially because of the considerable experience

available in clean fuel MHD generators.
In our country an MHD power plant of thermal capacity 5 MW
is being set-up at Tiruchirapally. Later this capacity will be in-
creased to 15 MW. This power plant is expected to be commissioned
in 1983.

9.2 Fuel Cell 4N V +C4TUOOe

It is an electrochemical device
which converts chemical energy
directly in electrical energy. Fig.
9.2 shows hydrogen oxygen fuel
cell. This fuel cell uses hydrogen ES 401

(or hydro-carbon) as a fuel and

oxygen (or air) as an oxidiser.
There are two chambers. In one
chamber hydrogen in introduced
and in other chamber oxygen is
introduced. The gases are at high
pressure. iI I—,-H20
The two chambers are
separated by an electrolyte, which F'9 92
may be solid or liquid. The various electrolytes used are Potassium
hydroxide. Zirconia oxide porous ceramic and solid Polymers. When
the temperature is high the electrolyte material acts as sieve and
the hydrogen ions can migrate through the material. The electrical
load is connected between anode and cathode. Hydrogen ions are
produced by the dissociation of hydrogen molecules at the anode
electrolyte interface. The reaction being as follows:
21 12 -* 4W 4 4e
The electrons so formed return to fuel cell at cathode leaving a
positive charge at anode. The hydrogen ions diffuse through
electrolyte and when they reach cathode they combine with
electrons and oxygen molecules and form water. The reaction being
as follows
4W + 4e +02 — 2H20
In this chemical reaction the energy representing the enthalpy
of combustion of fuel is released and a part of it is available for
conversion into electrical energy.
At the present time the fuel used in fuel cells is usually either
hydrogen or a mixture of gaseous hydrocarbons and hydrogen. The
oxidizer is usually oxygen. However, current development is
directed toward the production of fuel cells that use hydrocarbon
542 POWER PLANT

(uels and air. Although the conventional (or nuclear) steam power
plant is still used in large-scale power-generating systems, and
conventional piston engines and gas turbines are still used in most
transportation power systems, the fuel cell may eventually become
a serious competitor. The fuel cell is already being used to produce
power for space and other special applications.
Advantages -
1. It is simple.
2. It has high power to weight ratio.
3. Theoretical efficiency as high as 90 (7 can be expected but
it is possible only at light loads.
Disadvantages
1. Its cost is high.
2. It has relatively short life particularly at high tempera-
tures.
3. It is very essential to select proper materials for com-
ponents so that the reaction cannot attack them.
The hydrogen and oxygen for operating the cell are stored in
liquid form to minimise the volume occupied. The Hydrox (H 2, 02)
cells are of two types.
(i) Low temperature cell. In this cell the temperature of
-1 rtrolyte is 90 ' C. The electrolyte may be pressurised up to four
atmospheres.
(ii) hg" pressure cell. In this cell the temperature of
electrolyte is anout 300'C and it is pressurised up to 45 atmospheres.
Depending upon the type of fuel used the fuelcells are classified
as follows : (a) Hydrogen (H 2) fuel cell ; (b) Hvdrozine (N2 H4) fuel
cell (c) Hydrocarbon fuel cell ; (d) Alcohol (Methanol) fuel cell.
Fuel cells are particularly suited for low voltage and high
current applications.
9.3 The Indian Electricity Act 1910
Some extracts of the Indian Electricity Act are as follows:
Definitions
(i) Area of supply. It means the area within which alone a
licensee is for the time being taken authorised by this licence to
supply energy.
(ii) Consumer. Consumers means any person who is supplied
with energy by a licensee or the government or by any other person
engaged in the business of supplying energy to the public under this
Act or any other law for the time being in force and includes any
person whose premises are for the time being connected for the
MISCELLANEOUS PROBLEMS •543

purpose of receiving energy with the works of a licensee, the govern-

ment or such other person as the case may be.
(iii) Distributing main. It means the portion of any main with
which a service line is or is intended to be immediately connected.
(iv) Electric supply line. It means a wire, conductor or other
means used for conveying, transmitting or distributing energy.
(v) Energy. Energy means electrical energy : (a) generated.
transmitted, or supplied for any purpose or (b) used for any purpose.
except the transmission of message. . . . -.
(vi) Licensee. Licensee means person licensed to supply energy
(explained under the heading 'Supply Energy'). .
(vii) Main. It is an electric supply line through which energyis
or is intended to be supplied to the public.
(viii) Service line. Service line means any electric supply line
through which energy is intended to be supplied.
(a) to a single consumer either from a distributing main or.
immediately from the supplier's premises, or
(b) from a distributing main to a group of consumers on the
same point of the distributing main.
(ix Works. Works include energy supply line and any building
plant, machinery, apparatus and any othei thing of whatever
description required to supply energy to the publi and to carry into
effect the object of licence or sanction granted under this Act or other
law for the time being in force.
Supply of Energy
1. Grant of Licence. This State Government may on applica-
tion made in the prescribed form and on payment of the prescribed
fee (if any) grant after consulting the State Electricity Board a
licence to any person to supply energy in any specified area and also
to lay down or place electric supply lines for the conveyance and
transmission of energy:
(a) where the energy to be supplied is to be generated outside
such area from a generating station situated outside such
area to the boundary of such area or
(b) where energy is to be conveyed or transmitted from any
place in such area to any other place therein across inter-
vening area not included therein across such area.
2. Revocation or Amendment of Licences. The State
Government may, if in its opinion the public interest so requires,
revoke a licence in any of the following cases:
 544 POWER PLANT

(i) where the licensee breaks any of the terms or condition of

his licence the breach of which is expressly declared by
such licence to render it liable or revocation
(ii) where the licensee in the opinion of the State Government
makes willful and unreasonably prolonged default in
doing anything required of him by or under this Act
(iii) where thi licensee fails within the period fixed
(a) to show to the satisfaction of the State Government that
is in a position fully and efficiently to discharge the duties
and obligations imposed on him by his licence or
(b) to make the deposite or furnish the security required by
his licence.
Provision as to the opening and breaking up of stress railways
and tramways.
The licensee from time to time but subject always to the terms
and conditions of his licence may
(ci) open and break up the soil and pa y ment ofany street, railway
or tramway; (b) open and break up any sewer, drain or tunnel in or
under any street, railway or tramway; (c) lay down and place electric
supply lines and other works ; (d) repair alter or remove the same.
Criminal Offences
(a)Theft of energy. Whoever, dishonestly abstracts consumes
or uses any energy shall be deemed to have committed theft within
the meaning of Indian Penal Code.
(b) Penalty for maliciously wasting energy or injuring
works. Whoever maliciously causes energy to be wasted or diverted
or with intent to cut off the su of energy cuts or injures or
attempts to cut or injure any electn . rply line or works shall be
punishable with imprisonment for a term wiich may extend to two
years or with fine which may extend to one thousand rupees or with
both.
(c)Penalty for interference with meters. Whoever connects
any meter, indicator, or apparatus with any electric supply line
through which energy is supplied by a licensee or disconnects the
same from any such eleclricsupply line, maliciously injures any
meter, indicator or apparatus or wilfully alters the index or any such
meter shall be punishable with fine which may extend to five
hundred rupees and in the case of continuing offence with a daily
tine which may extend to fifty rupees.
(d) Penalty for extinguishing public lamps. Whoever mali-
ciously extinguishes any public lamp shall be punishable with
 MISCELLANEOUS PROBLEMS
$45
imprisonment for a term which may extend to six months or with
fine which may extend to three hundred rupees or both. -
9.4 Indian Electricity Rules, 1956
Some extracts of Indian Electricity Rules are as follows
Some Important Definitions
(i) Conductor. Conductor means any wire, cable, bar tube, rail
or plate used for conducting energy and so arranged as to be
electrically connected to a system.
(ii)Voltage. Voltage means the difference of electrical potential
measured in volts between any two conductors or between any part
of either conductor and the earth as measured by a suitable
voltmeter.
(iii) Low Voltage. It means the voltage which does not exceed
250 volts under normal conditions.
(iv) Medium Voltage. It means the voltage which does not
exceed 650 volts under normal conditions.
(u) High voltage. It means the voltage which does not exceed
33,000 volts under normal conditions.
(vi) Extra-high Voltage. It means the voltage which exceeds
33,000 volts under normal conditions.
Inspector
Qualifications of Inspector. No person shall be appointed as
an inspector unless
(a) he possesses a degree or diploma in electrical engineering
from a recognised university or college.
(b) he has been regularl y engaged for a period of at least eight
years in the practice of electrical engineering of which not
less than two years have been spent in an electrical or
mechanical engineering workshop or in generation trans-
mission or distribution of electricity or the administration
of the Act and the Rules made thereunder in a position of
responsibility.
Inspection
(i) Any inspector or any officer appointed to assist an inspec-
tor may enter, inspect and examine any place and carry
out tests.
(ii) Every supplier, consumer, owner and occupier shall afford
at all times all reasonable facilities to any such inspector
or officer to make such examinations and tests.
(iii) An inspector may serve an order upon any supplier, con-
sumer, owner or occupier calling upon him to comply with

.546 POWER PLANT
d
any specified rule and person so served shall thereupon comply with
the order within the period named therein and shall report in
writing to the inspector when the order is complied with.
General Safety Precautions
(i) All electric supply lines and apparatus shall be sufficient
in power and size and shall be of sufficient mechanical
strength and shall be constructed, installed, protected
worked and maintained in such a manner as to prevent
danger.
(ii) The consumer shall take precautions for the safe custody
ofthe equipment in his premises belonging to the supplier.
(iii) The consumer ensure that the installation under his con-
trol is maintained in a safe conditions.
(iv) The supplier shall provide suitable cut-out in each conduc-
tor of every service line other than an earthed or earthed
neutral conductor.
(v) Where more than one consumer is supplied through a
common service line each oonsumer shall be provided with
an independent cut-out at the point of junction to the
common service.
Accessibility of Bare Conductors. Where bare conductors
are used in a building the owner of such conductors shall
(i ensure that they are inaccessible;
(ii) provide, in readily accessible position, switches for render-
ing them dead wherever necessary.
Caution Notice
The owner of every medium, high and extrahigh voltage instal-
lation shall affix permanently a caution notice on
(a) every motor, generator, transformer and other electrical
plant and equipment together with apparatus used for
regulating the same;
(b) all supports of high and extra-high voltage overhead lines.
Provision applicable to protective equipment
(a) Fire buckets filled with clean, dry sand and ready for
immediate-use for extinguishing fire shall be kept in all
generating stations, enclosed switch station in convenient
situations.
(b) First-aid boxes or cupboards shall be provided in every
generating station, enclosed sub-section and enclosed
switch station so asto be readily accessible.
 MISCELLANEOUS PROBLEMS
547
Sealing of Meter and Cut-outs
(a) The supplier may affix one or more seals to any Cut-out
and any meter and no person other than the supplier shall
break any such seal.
(b) The consumer shall ensure that no such seals is broken
otherwise than by the supplier.
Overhead Lines
(a) Materials and strength. (i) All Conductors of overhead
lines shall have a breaking strength of not less than 317.5 kg.
(ii) Where the voltare is low and span is of less than 15.24
metres and is on u consumer's or owner's permises a
conductor having an a tual breaking strength of not less
than 126.08 kg may be used.
(b) Joints. Joints between conductors of overhead lines shall be
mechanically and electrically secured under the conditions of opera-
tion.
(c) Clearance above ground. No conductor of an over-head
lines shall be erected across a steel at a height less tha'
(i) for low and medium voltage-5.791 metres; •
(ii) for high voltage line-6.096 metres.
Overhead line should not be erected along any street at a height
less than
(i) for low medium and high voltage line upto and including
11,000 volts if bare-4.6 metres,
if insulated-4 metres;
(ii) for high voltage lines above 11,000 volts-5.182 metres.
For extra-high voltage lines the clearance above ground shall
not be less than 5.182 metres plus 0.305 metres for every 33,000
volts or part thereof by which the voltage of the lines exceeds 33,000
volts.
Penalty for Breaking Seal. The persons breaking the seal
shall be punishable with fine which may extend to two hundred
rupees.
Payment of Bills
(i) Bills should be paid at the licencee's local office within 15
days from the date of their presentation.
(ii) Any complaints with regard to the accuracy of the bills
shall be made in writing to the licensee and the amounts
of such bills shall be paid under protest within the same
period of 15 days.
9.5 Energy Cycles
Thermodynamic cycles which are generally used are of two
types:
 POWER PLANT
548

(i Gas power cycles.

These may be classified as follows
(a) Spark ignition (S.I.) cycle.
(h) Compression ignition (C.!.) cycle.
(c) Gas turbine cycle.
(ii) Vapour power cycles.
Various power cycles are as follows
1. Carnot cycle 2. Rankine cycle
3. Regenerative cycle 4. Reheat cycle
5. Regenerative Reheat cycle
6. Binary Vapour cycle.
9.5.1 Carnot Cycle
This cycle is of great value to heat power theory although it has
not been possible to construct a practical plant on this cycle. It has
high thermodynamic efficiency.
It is a standard of comparison for all other cycles. The thermal
efficiency ( ft) of Carnot cycle is as follows
T1-T2
11 -
Ti
where Ti = Temperature of heat source
= Temperature of receiver.
9.5.2 Rankine Cycle
Steam engine and steam turbines in which steam is used as
working medium follow Rankine cycle. This cycle can be carried out
in four pieces of equipment joint by pipes for conveying working
medium as shown in Fig. 9.3. The cycle is represented on Pressure
Volume P-V and S-T diagram as shown in Figs. 9.4 and 9.5 respec-
tively.
Efficienc y of Rankine cycle
H 1 -- H2
- H i -H..

where = Total heat of steam at entry pressure

H2 = Total heat of steam at condenser pressure
(exhaust pressure)

qt
TURBINE

PUMP .. 7E",
Fig. 9.3
 MISCELLANEOUS PROBLEMS 549

H.2 = Total heat of water at exhaust pressure.

P RATION
—lc_ f
7,
I \EPAN5I01V

4-
V —a-
S —'-
Fig. 9.4 Fig. 9.5
9.5.3 Reheat Cycle
In this cycle steam is extracted from a suitable point in the
turbine and reheated generally to the original temperature by flue
gases. Reheating is generally used when the pressure is high say
above 100 kg/cm'. The various advantages of reheating are as
follows

Fig. 9.6
(i) It increases dryness fraction of steam at exhaust so that
blade erosion due to impact of water particles is reduced.
(ii) It increases thermal efficiency.
(iii) It increases the work done per kg of steam and this results
in reduced size of boiler.
The disadvantages of reheating are as follows:
(i) Cost of plant is increased due to the reheater and its long
connections.
(ii) It increases condenser capacity due to increased dryness
fraction.
Fig. 9.6 shows flow diagram of reheat cycle. First turbine is high
pressure turbine and second turbine is low pressure(L.P.)- turbine.
This cycle is shown on T.S (Temperature entropy) diagram (Fig.
9.7).

550 POWER PLANT

-,- S
Fig. 9.7

(Hi -H2)+(H3-H4)
Efficiency
- H 1 + (H3 - H2) - HL4
where = Total heat of steam at 1
Hi
H2 = Total heat of steam at 2
H3 = Total heat of steam at 3
H4 = Total heat of steam at 4
= Total heat of water at 4.

9.5.4 Regenerative Cycle (Feed Water Heating)

The process of extracting steam from the turbine at certa
points during its expansion and using this steam for heating for fee
water is known as Regeneration or Bleeding of steam. The arrange-
ment of bleeding the steam at two stages is shown in Fig. 9..

HOT
WELL.

Fig. 9.8

Let rn2 = Weight of bled steam at a per kg of feed water heated

rn 3 Weight of bled steam at b per kg of feed water heated
H 1 , H, 1 = Enthalpies of steam and water in boiler
 MISCELLANEOUS PROBLEMS
551
H4 , H 3 = Enthalpies of steam at points a and b
t2, 6 = Temperatures of steam at points a and b

H 4 , H. 4 =Enthalpy of steam and water exhausted to hot well.

Work done in turbine per kg of feed water between entrance and

a
H i - H2
=

Work done between a and b = (1 - mz) (H2 - H3)

Work done between b and exhaust
= (1 - m2 - m3) (H3 - H4)
Total heat supplied per kg of feed water = H1 - H2
Efficiency it
_(Hi -H2)+(1-m2)(H2-H3)+(1_m2_m3)(H3_H4)
- (H1 - H2)
9.5.5 Binary Vapour Cycle
In this cycle two working fluids are used. Fig. 9.9 shows Ele-
ments of Binary vapour power plant. The mercury boiler heats the
mercury into mercury vapours in a dry and saturated state. These
mercury vapours expand in the mercury turLine and then flow
through heat exchanger where they transfer the heat to the feed
water, convert it into steam. The steam is passed through the steam
superheater where the steam is super-heated by the hot flue gases.
The steam then expands in the steam turbine.
STeAM
5UPERKATER

RCLcv
TLIR5INE

S TEAM
TURBINE

MERCUR',
SOiL ER

TER
I EXCHANGER
Fig. 9.9
552 POWER PLANT

; Advantages of Mercury as a vapour cycle fluid

1. It has high density and, therefore, it is easier to separate
vapour from liquid in the boiler.
2. It is an element and hence stable.
3. At higher fluid temperature it has moderate vapour pres-
sure.
4. The efficiency of binary vapour plant using mercury is
more than steam power plant of same capacity.
Disadvantages
1. In mercury vapour plant there is danger due to poisonous
fumes if any leakage of mercury vapour occurs.
2. The investment cost per kW of binary vapour cycle power
plant is more than a steam power plant. Therefore, it is
desirable to use bipary vapour cycle plant as base load
plant because of its higher thermal efficiency tojustify the
cost.
3. Cost of mercury is high.
4. Mercury has toxic qualities.
9.5.6 Reheat-regenerative cydle
In steam power plants using high steam pressure reheat
regenerative cycle is used. The thermal efficiency of this cycle is
higher than only reheat or regenerative cycle. Fig. 9.10 shows the
flow diagram of reheat regenerative cycle. This cycle is commonly
used to produce high pressure steam (90 kg/cm2) to increase the
cycle efficiency.
Ti irh,n ?C

Heaters
Fig. 9.10

Example 9.1. (a) Explain the process of feed water heating by

bleeding. How bleeding improves the efficiencies of steam power
plant?
 MISCELLANEOUS PROBLEMS

(b) Calculate theoretical tht!r,nU/ ef/ ictenc y ole plant work n

b'twee,z 20 4,g/em 2 ((zbs..) dr y and saturated and 0.03
1,glf-1112.

(i) 11iit/K)Ut bleeding.

(ii) What correct iteight of steam is bled at 2.5 kg1cm 2
(abs)
Solution. i F'rom ?lollier diagram
Heat drop from 20 kg/cm 2 to 0.08 kg/cm2
IL
197 kcal/kg.
1'roin steam tables
Heat at 20 kg/cm 2, H i = 668.7 kcal/kg
Total heat of water at 0.08 kg/cm2.
= -11.1 kcalfkg.
Therefore, theoretical thermal efficiency
.197
= 668.7-41.1= 0.29 Ans.
(ii) When steam is bled at 2.5 kg/cm2
From Moltier diagram
Heat drop from 20 to 2.5 kg/cni 2 = 88 kcal/kg = II
Heat drop from 2.5 to 0.08 kg/cm2
= 109 kcaVkg
Dryness fraction q = 0.87
Let in = Weight of bled steam per kg of feed water
m (112 - 1-I) = 11,-, H,,
H,
In =
llk2

where 14,, 2= Total heat of water at 2.5 kg/cni2

= 127.2 kcaLfkg
112 = q2L
L 2 = Latent heat of steam at 2.5 kg/cm)
(0.87 x 522.3 # 127.2) kcaVkg

iO
127-41.1
=0.870.16 kg.

l'herefore. weight of steam left in the turbine to expand from 2.5

kg/cm 2 to 0.0$ kg/cm 2 is 1 O. 16 - 0. 8. 1 kg.
—37
 POWER PLANT
554

Work done per kg of feed water

= lid1 x 1 Hd2 x 084
= 88 x + 109 x 0.84 = 179.56 kcal.

Heat supplied by fuel

= H1 ii r2 = 668.7 - 127.2 = 541.5 kcal.

Theoretical thermal efficiency

179.56 -
=. = 0.3a1

Hence there is an increase in theoretical thermal efficiency due
to bleeding.
15 kg/cm 2 (abs.) and
Example 9.2. Steam it a pressure of to a pressure of
t.'!nperature of 250 C is expanded through a turbine
at constant pressure to a tempera-
/rtO 2 (abs.). it is then reheated
it completes its expansion through the
ii ,' of 200 C after which
'IF1 ii tO (U? ('110 U 't p I- c-su re
ufO, 1kg/e1,71 2 (abs). Calculate theoreti'
t' 'ncV
Taking r('h,t'a(iFIg into (ICCOU at.
exhaust pressure
It lht st'ui? Was 'xp(lnd('d direct
h to
w i thout r'he(iting.
Solution. (a) From Mollier diagram
and 250'C
11 Total heat of steam at 15 kg/cm 2
698 kcallkg
II: Total heat of stream at 5 kg/cm 2
.l6kcalfkg
v' am is relit:it '0 to 200 C at. constant pressure
I hat iii this stage = 682 kcal/kg
'[hen t( 0, is expanded to 0.1 kg/cmY
1. Heat at this stage 553 kcal/kg

j1,, I T otal heat of water at 0.1 kg/cm2

.15.4 kcal/kg
'I'Iiei,i .t I Cii ('ttlCIeflL'\
41 1 l/..i
III ' 1I l!.) hI.
MISCELLANEOUS PROBLEMS 555

(698 646) (682 - 53:3)

- 698 (642 646) - 45.4

= 0.293 or 29.3. Ans.

688.6
(b) Without reheating
Theoretical efficiency =
h i-
= Heat of steam at 0.1 kg/cm-' (From Mother diagram) if
steam was expanded from 25 kg/cm 2 to 0.1 kg/cm2(abs)
510 kcallkg
Total heat of water at 0.1 kg/cm2
= 45.4 kcal/kg
Theoretical efficiency
698- blO
0.282 28.2. Ans.
- 698 45.4
9.6 Installation of Power Plant
Some of the factors to be considered while installing a power
plant are as follows
(a) Selection of plant system,
(h) Location of plant,
(C) Building layout,
(ci) Selection of prime mover,
(e) Selection of operating conditions
(f) Cost,
g) Selection of units.
1. Selection of Power System. The type of power plant to be
installed depends upon the source of energy. A hydropower plant
should be installed where sufficient head of water is available
whereas a steam power station is suitable near coal mines. A nuclear
plant should be installed near a source of water for example lake,
river etc. Diesel plant is preferred for smaller loads. Further choice
about the number of generating units should be made. For example
if a steam power plant of 200 MW capacity is to be installed then it
is to be decided whether a single boiler will be supplying steam to a
single steam turbine or there will be two units each of 100 MW
capacity or four units of 50 MW each or there can be single boiler
supplying steam to four turbines each of 50 MW capacity. Unit
system in which a single boiler supplies steam to single turbine as
preferable.
The points to be considered while choo'sing the t y pe of genera-
tion are as follows
 56 POWER PLANT

ui T y pe of fuel available
( it) Type of load to be supplied
(iii) Reliability of operation
(iv> Cost of land
v) Cost of fuel transportation
(j) Availability of cooling water
(vii) Cost of power transmission.
2. Location of Power Plant. It should be installed near the
load centre so that the cost for transmitted power is reduced. It
should be nearer to source of fuel and sufficient amount of water
should he available near the power station. The soil should be such
that special and costly foundation is not required. The plant site
should he accessible by road or railway line so that the transporta-
tion of equipment is easy.
3. Building for a Power Plant. The power plant building
should be simple, rugged and should have pleasing appearance. The
size, arrangement and shape of power plant depends upon the type
of power plant. The roofs are made usually flat. The roof deck can
be carried on reinforced concrete s1a)s and beams. The floors may
be made up of concrete or tiles. Concrete floors as preferred. The
structure should be fire proof. While laying out the various equip-
ment allowari.ces should he made for sufficient clearance and
walkway s. Some space should be left for future expansion. The
rooms of the building should be spacious, uncrowded, well lighted
and clean.
4. Selection of Prime Mover. Depending on the load and type
of power stations, the selection ofsteaun turbine, hydraulic turbines,
.boilers, diesel engines, gas turbiies should be made. The units
installej should have capacity more than the peak load and some
provisions should be made Tor future expected load. For example the
peak load of one city is 36 MW. Then two units each of 20 MW or
four units each of 10 MW could be used keeping in view the future
load.
Heat transfer is the primary function of the boiler. It should be
capable of utilising the heat of the combustion of fuel to a great
extent. In central power stations water tube boilers are preferred.
Coal fired boilers have higher efficiency, than boilers using oil or gas
as fuel because hydrogen loss is more in gaseous fuels. Pulverised
fuel is preferred in case of low ranking fuels such as lignite etc. The
cost of boiler depends upon operating pressure, operating tempera-
ture, type of firing and efficiency desired. Higher pressure requires
forced circulation. For higher temperature superheaters are needed
and, therefore, cost becomes high. The system used for coal tiring in
the boiler also influences the cost of boiler. Stoker firing is cheaper
than pulverised fuel firing. The efficiency of boiler is increased by
 MISCELLANEOUS PROBLEMS
557

adding heat recovery equipment such as economisers and air adding

heat recovery equipment such as economisers and air preheater.
Economiser improve the efficiency of boiler by about 4 to iofl;
whereas preheater further improves the efficienc y by about is 6 to
8•. The temperature of exhaust flue gases should be below than
150 C because otherwise the condensation of moisture ma y take
place and in combustion with sulphur dioxide it (moisture) will
produce sulphuric acid dilute solution which is harmful for equip-
ment. In selecting it boiler its fixed cost and operating cost should
be considered. The working pressure and capacity ofniodern boilers
are quite high. In modern boilers working pressure has reached up
to about 360 kg/cm 2 and evaporative capacity of about 2000 tonnes
per hour.
5. Selection of Operating Conditions. The power plant
should supply the varying power demand. To supply varying power
the supply of fuel, air, water, etc. should he varied accordingly, In
order to keep the power plant efficient, the older and inefficient units
should be replaced by new and efhcienct units.
6. Cost. The cost of power plants consists of fixed cost, operating
cost, depreciation charges etc. Cost of' the power station should be
kept as low as possible. Some of the factors, which affect the cost art'
as follows
to The cost can he kept low by installing the power Station
near the load centre.
(b ) The cost can be reduced b y selecting prime movers of
proper capacit. The cost of the power station is incieitsi'd
if the prime movers installed are of too high capacity a
corn pared to t h in ax i in urn expected ttwe r dein and
c) As considerable area is rLqU ii't'd for it
power stat jim so, 1
cost of land should be reasonable. As far as positW.' it,t'
suit-soil conditions should h such hat pilin or },fast lug
is not required and it founditjwi ciii be iii,h' at
reasonable depth
t/i Water should he available in lar'ne quantit y ill case of
steaiii power station, nu.'t'ar power' ' . t.ljon and diesel
powe r stat ion
7 Selection of Generating Units. The generrtii1 on
hou Id fulfill the following requirements.
The n u in be r of units selected should not be too iii an v hit
the y should not be Less than two. In case two units ar.'
selected each should be capable of supplying the maxi
mum load.
 POWER PLANT
558

(,I 'l'he units selected should be capable to supply the maxi-

mum load. The future demand extension should be kept
in view while deciding the capacity of the units.
(ill) A reserve unit should be installed to supply the load if
there is a break clown of highest capacity.
The units should operate at high load factors and should
I) high efficiency and their cost should be below.

9.7 Blo-Gas
The most important sources of organic waste for the production
of bio gas in India is cow-dung and agricultural wastes. The cow-
clung and agricultural waste are used extensively as direct sources
of e tiergv in rural meas. The development of the technology for
hio . gas production from organic wastes to preserved their manurial
value and at the same timQ to provide the rural area with a
substitute source of energy has been recognised. Indian has about
237 million cattle population and if the entire dung field from the
cattle population is utilized for the hio gas production in an opt nuin
manner, the annual gas availability would be about 66,000 million
in s . Agricultuial wastes have the capcity to produce a lot more gas
than cow-dung A kilogram of paddy straw can produce about six
cubic feet of a ga as against a kilogram of cow-dung which call
t)rodli('e about 1.3 cubic teed of gas. in a 1)10-gas plant apart from the
gas that is produced, enriched mnamuire is also obtained.
A number of institutionS have played a significant role in thd'
development (it technology of hio-gas production. The Khadi and
village Industries Commission has played It ke y role in popularmang
the installation of Gobar Gas plants . i n the country.
The program inc of hio-gas as being i in plemented now in I ud i a
is oriented towards establishing mairill, small size plants to serve
the needs of individual families. To achieve maximum econoniv as
also for accelerating the development of bio-gas potential it is
considered 1imt large size ofbio-gas plants should be established for
serving the whole village community in a village. Some of the
institutionS carr ying out research on bio-gas technology are as
follow -
(I (obar Gas Research centre of Khadi and Village In-
dustries Commission.
I ridian Agricultural Research Institute, New Delhi.
National Environmental Engg. Institute, Nagpur.
One of the main advantages ot'street lighting or home lighting
with bio-n is that the entire process will he simple with absolutely
no recurri ig expenditure Also in case of bio-gas system failure the
villagers themselves will be able to rectify the fault. Further the
same gas th. is used for cooking can after a simple pro cess of
MISCELLANEOUS PROBLEMS 559

conversion be used to light a bulb. The maintenance costs on a

community type gubar gas plant are eas y to hear as it is run by a
group of persons.
Gobar gaA plants are cheap, simple eas y to handle and can be
made locally byusing indigenous materials. The plant not only
produces enough gas for the kitchen and oil engines but also makes
available fertilisers to increase food production. Bin-gas plants have
proved to be a boon to the farmer because the manure from bin-gas
plants is richer in nitrogen content and more in quantit y than that
obtained by conventional composing. The gohar gas plants also
reduce firewood consumption.
At present about 60,000 bio-gas plants are operating in our
country as against air potential of about 18 million family
size and half million communit y size plants. In the sixth five year
plan about one million hio-gas plants will be installed.
In the gobar gas plant the cow-dung is fermented to yield a
combustible gas which can be used fer cooking as well as lighting
purposes. The residue becomes available for use as manure.
A gobar gas plant consists of two main parts namely digester-
and gas holder. The digester is a sort of wall, coiitructed ofinasonrv
work below the surface of ground. The gas hold, r is built with bricks
and cement as dome cover of the well of the u!ester into which a
mixture of vanous organic wastes mostl y cow and buffalo clung and
water is fed at regular intervals. When the gas is formed b y fermen-
tation of cow-dung in the digester it ascends towards the top ofcbnme
and pushes the slurry down.
The displaced level ofsiurry provides necessar y pressure for the
release of the gas U)) to the burners of kitchens or other outlets. The
manure obtained from bio-gas plants is richer in nitrogen content
and more in quantity than obtained by conventional composting.
Bio-gas, a mixture of methane 55-65., carbon dioxide (CO2)
30-40'. and other impurities is produced during decomposition The
materials for generation of the gas should possess the following
characteristics
(a) Proper carbon to nitrogen (C/N) ratio.
(b) Appropriate Volatile-solid concentration.
(c) Particle size as small as possible.
Material from which bio-gas is produced retains its nutrient
value and can be used as fertilizer.
Bio-gas is produced by digestion or h ydrogasitication process.
Digestion is biological process that occurs in the absence of oxygen
and in the presence of anaerobic organisms at ambient pressure and
temperature 350 C.
560 POWER PLANT

Biogas produced by fermentation of organic wastes essentially

contains methane and carbon dioxide in large proportion as indi-
cated below

- _:-------_ Volume --ii_iii

Methane 50-60
Carbon-dioxide 30---15
ri0
L
----------------------------------
Nitrogen 05-0.7
_Traces
Methane is the only combustible portion the gas. Biogas is a
flammable gas. The main products of the bio-gas Plant are fuel gas
and organic manure.
The gobar gas plant should be near the kitchen, in an open place
arid awa y from an y wall or tree so as to be under sunshine as much
as possible. This will ensure better fermentation and better gas
production. This plant helps to keep the environment clean.
In the face of global energ y crisis and depleting forest resources
in the country bio-gas energy assumes great importance in
strengthening rtiral economy. Toda y along with other items like
food, shelter and clothing energy also forms part of the basic needs
of the rural household.
Fuel from gobar gas plant has a very high thermal efficiency of
about 60'; whereas efficienc y of cow-clung is only 11%.
l3io-gas is a flammable gas. Fhe combustible portion is onl
methane in the gas. One in gas may be thought of equivalent to 0.7
kg of gasoline for power (utput. A hio-gas lamp of a luminosity of
about 60 watts equivalemt electrical light can be functioning for six
to seven hours if one m gas is available. (hie horse power engine
can work for miearlv two hours with a in of gas. The gas has a flame
temperature of about 500 C.
The major constraints in the wide spread adoption of bio-gas
technology are as follows
Ii) The high cost of gas plants with the model.-; now in use.
(ii) The constant attention required in operation at these
plants.
The target of setting up of over half a million bio-gas plants
during the Seventh Plan is expected to be exceeded. The rapid
progress in the last few years in this direction has been due to the
package of measures taken by the Department of Non-conventional
Energy Sources DNES.
 MISCELLANEOUS PROBLEMS
561

C3 1

CL

cj IZ
Io
I-

a.
Q
U,
C
.-

0
0
x 0)

-I U-

a
a.

9.7.1 Biogas manure plant

Fig. 9.10 (a) shows a precast R.C.C. biogas manure plant. This
plant is easy to operate and is simple in construction. The raw
material of the material is mixture of dung and water which flows
down through the inlet pipe to the bottom of digester. A pipe is
provided at the top for flow of gas for usage.
Fig. 9.10(b) shows the flow sheet of community Bio-gas plant.


POWER PLANT
562

House Lighting

Better -- Cooking
[SanitationJ
BioBiogas Water -- pumping
Plant jEnergy

Water heating

Manure - Misc uses

Fig. 9.10(b)

9.8 Combined Working of Different Types of Power

Plants
rfhere is alwa y s a trend to investigate the CCOilOtfltCS of the
combined working of electric power plants. Hydro power plants can
be developed only at sites where a large quantity of water is
available at sufficient head. In considering the economy of power
development the transmission liability of !ivdi-electric projects
should be taken into account. Steam power plants can be located at
or near the load centre and can be used as base load plant. When a
number of power plants are worked iii combination to supply
electric power system they are all connected together and the system
is called inter-connected s y stem. Before using inter-connected svs-
tern it is necessar y to draw the annual load duration curve of the
area to which power is to he supplied and then to fit the various
tvpesf power plants iflt4) the area under the curve. Base load plants
which run at a high load factor should be used to supply the load on
the base portion of load curve and peak load plants which work at
a low load factor should he used to supply rest of' load on the top
portion of load curve. Steam power plants can he used with ad-
vantage in combination with h y dro electric power plant to obtain
economy from the mixed s y stem. Steam power plants can he used
on any portion of the load curve although it is not economical to Lkse
them as peak load plants.
In combined working of hvdro power plant and nuclear power
plant the nuclear power plant should be used as base load plant and
hydropower plant can supply the variable load. Gas turbine plants
are the cheaper type in some situations when they are used as peak
load plants in combination with base load plants.
During combined working of hydro power plant and steam
power plant the hydro power plant with ample water storage should
MISCELLANEOUS PROBLEMS 563

be used as base load plant and steam power plant should be used as
peak load power plant. If the amount of water available is less at
hydro plant then the steam power plant should supply the base load
and hydro-plant should act as peak load again. Diesel power plants
are generally used as as peak load plants.
When pumped storage plant is used in combination with steam
power plant then pumped storage plant is used to supply sudden
peak loads of short duration. The advantages of pump storage plant
in an inter connected system are as follows
(a) It stores the energy using off peak energy of thermal plant
and the same is supplied during demand.
(b) It minimises wastage of off-peak energy of steam power
plant.
(c) Use of pumped storage plant helps in loading economically
the steam power plant. During coordination of hvdro
power plant and gas turbine power plant the gas turbine
power plant is used peak load plant. The high working cost.
of gas turbine plant is compensated b y lower fixed cost and
lower operating and maintenance cost.
Inter connected power system can provide large savings both in
capacit y and fuel cost at the same time ensuring reliability and
contmuit y ot'jmwcr supply. The main purpose oiiter-connection IS
to distribute the load among the interconnected power plants system
ill to achieve the overall economy.
The various advantages of combined working of different types
ot power plants are as follows
(a) There is a reliability of power supply to consumers be-
cause in the event of power failure at one of the power
plants the system can be fed from other powers plant to
avoid complete shut down.
(b) The spinning reserve required in a power system in
reduced.
Combined working of different types of power plants
reduces the amount of generating capacity required to be
installed as compared to that which would he required
without inter-connection.
The inter-connection of various power plants helps in
reducing the amount of generating capacity to he in-
stalled.
' In an inter-connected system the overall cost of energY is
less
54 POWER PLANT
4

3:
4

K
Fig. 9.11

Fig. 9.11 shows a typical annual . load duration curve between

percentage ofycar (K) and load (W). The various loads which should
be supplied by different power plants are indicated in Table 9.1.
Table 9.1

Power plant

-
A
-__
- - Nr
Ru n off river plant
power plant
H ydro,werp!tnt with sut1icient stor ag-
.. -

.... 1) ._. St m pwerplant

.111f .d- ropow erplant -with limited storage
..I)ieeIpowcr plant or Gas turbin werplant
Following Factors should be considered while deciding the load
to be shared among different t ypes of power plants
Capacit y of l)OVer plant
(ii) Degree of reliability
(tii )Probable load far or
(U; Cost of fuel and transport facilities
c) Initial cost and operating cost
(vi) Load between different power plants should he so (leCi(led
that overall econom y is achieved.
vii Time required to start the plant from cold condition
I

9.9 Economy of Operation

The factors direct IV a 11cc) ing the econom y of o)u.ra t intl of va riou
power plants are as follows
 MISCELLANEOUS PROBLEMS
56!
(i) Cost of generating units
(ii) Normal and emergency equipment ratings
(iii) Fuel cost
(iv) Labour cost
(v) Reserve requirements
(vi) Voltage limitations
(vii) Characteristics of prime movers
(viii) Transmission losses.
9.10 Efficiency of Power Plants
The efficiency of a power plant depends on load and time period
Fig. 9.12 shows the efficiency of various power plants with respect
to load. The increase in efficiency of hydro power plants is higher as
. compared to other power plants at high load.

Uj

Fig. 9.12

The efficiency of power plants is reduced with increase in timc

period. Fig. 9.13 shows effect of time period in years on efficiency ol
power plants.

IOC

Fig. 9.13

566; POWER PLANT

PROBLEMS

9.1. (a) What is M.H.l). generator?

(h) Discuss the working principle ofan MIII). generator.
(c) State the advantages of M.H.D. power generator.
9.2. Write short notes on the following:
(a) Fuel cell'
(h ) Bio-gas
() Carnot cycle,
9.3. Describe the factors to be considered while installing a power
plant.
9.4. A steam power plant uses dry steam at a pressure of 38 7
kg/cm 2 absolute and exhausts it at 0.06 kg/cm 2 . If the plant
works on perfect regeneration cycle determine work done per kg
of steam. Calculate also.the ideal efficiency of the cycle. 4
9.5. Describe following cycle
(a) Regenerative cycle
(h) Reheat cycle
(c) Binary vapour cycle
(d) Ranking cycle.
9.6. State some extract of Indian electricity Rules 1956.
9.7. Discuss the combined working of different t y pe of power plants.
9.8. State the advantages of combined working of different type of
power plants.
9.9. State the factors to be considered while distributing load among
different t y pes of power plants.
9.10. Sketch and describe reheat regenerative cycle.
9.11. State factors affecting economic operation of power plants
9.12. Discuss the variation of efficiency of power plants with respect
to load and time period.
 10

Major Electrical Equipment in

Power Plants

10.1 Introduction
A power station is equipped with electrical equipment for power
generation and supply of ekctrca1 power. The major electrical
equipment in a power plant consists of the following
(i) Generators (ii) Excitors
(iii) Transformers (it') Reactors
'v) Circuit breakers (w) Switch board
(tti) Control board equipment.
10.2 Generator
It is an important part of a power plant. All modern type of
alternating current generators (alternators) essentiall y consist of
fixed stator and revolving rotor.
Stator mainly consists of following three parts
(i Stator frame
(ii Stator core
(iii ) Stator windings.
The stator frame is of circular in shape and is made of welded
steel plates. The core is made of stampings of high permeability, low
hysteresis and eddy current losses. The rotors are generall y built in
cylinderical form and diameter of rotor is limited to 100 to 110 cm.
Large number of deep slots are machined in the rotor to accoin-
modate the field windings which will carr y field currents. There
should be a in plc passage in the rotor through which cooling fluid
can he circulated freely.
Generators coupled directly to the steam turbines are called
turbo-generators and those (lircctiv coupled to water turbines are
called \vat er wheel generators.
The fiw.nv at which generator operates is given by

568 POWER PLANT

f -- p_fl
/ 120
where f= frequenc y in cycles per second
ii = speed in R.P.M.
p = number of poles of the generator.
The frequency is the standard one used in the particular
country. In India it is 50 cycles per second. The number of poles is
necessarily an even number.
The generators used with diesel engines are of salient pole type
having large diameters and short lengths. The number of poles
varies between 4 to 28 as the speed range of diesel engines driving 4
the generator varies between 1500 to 214 R.P.M. A common range
of capacities is from 25 to 5000 WA, the power factor rating being
0.8 lagging. A 3 phase 50 cycles per second suppl y is to be obtained
from generators and a common voltage rating is . 1 . 1() V. Generator
efficiency varies between 92 to 95.
A common voltage rating for turbo-alternators is 11 kV, The
turbo-alternators are generall y rated at 0.8 power factor lagging
most of turbo-alternators are 2 pole generators with a speed of H)()
R.P.M. Their efficiency may be taken between 98 1Y, and 99g..
The water wheel generators are built with salient (projecting)
poles as they require large number of poles and run at comparatively
low speeds. Therefore water wheel generator rotors are much
greater in diameter than turbo-generator rotors. The water wheel
generators usually run at 62.5 to 125 R.P.M.
In a power plant several generators are run in parallel with each
other. The parallel opetation of generators has the following ad-
vantages
1 There exists a high reliability of power supply to the
consumers.
(ii) It becomes possible to achieve more stable frequenc y and
voltage under conditions of varying load.
(iii) Better economy in running of power stations and their
equipment is obtained.
10.3 Exciter
The generator needs an exciter to build up the necessary voltage
on no load arid then to keep it constant on load, when greater
excitation will be required. The exciter is a D.C. generator shunt or
compound. The capacit y of the exciter depends on the speed, on the
voltage rating of ac. generator and on the voltage required to
compensate for the drop due to load on generator.
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 569

The rating of exciter is 0.3 to 1% of the generator output.

Exciters usually operate at a voltage between 115 volts and 400
volts.
10.4 Generator Constants
The following details of generator should be known to enable to
determine their performance in the power system under different
loading conditions:
(a) kVA rating
(b) number of phases
(c) frequency
(d) voltage
(e) power factor
(1) temperature rise limits
(g) star connections of the stator.
10.5 Generator Cooling Methods
Generator losses appearing as heat must be continuously
removed in order to keep the temperature of various parts ofthe
generator and winding insulation with in limits. If the heat is not
removed properly then the generator gets overheated and insulation
is damaged. Thus every generator requires contipuous cooling
during its operation.
The two methods commonly used for cooling of generator are as
ibilows:
(i) Air Cooling
(ii) Hydrogen Cooling.
Air Cooling. Generators of smaller capacity (about 50 MW) are
cooled by this method. In air cooling the air circulates through ducts
in the stator and along the air gap and then passes through a set of
water cooled coils. The air is passed over the surfaces of the water
cooled coils for transfer of heat to the water.
Hydrogen Cooling. Hydrogen (112 ) is used for cooling gener-
ators of larger capacities. This method offers following advantages
over air cooled generators:
(i) The windage losses of the rotor turning in hydrogen are
only 10% of those in air.
(ii) The noise is considerably reduced due to lighter cooling
medium and lower friction.
(iii) Life of insulation is increased because of absence of
oxygen.
(iv) The higher thermal conductivity of hydrogen allows gen-
erators to develop about 25% more output than with air
ceoliilg for the same physical dimensions.
—38 -

570 POWER PLANT
4
(v) The cooling suti.c required for H 2 cooling is considerably
smaller than that needed for air cooling due to high heat
transfer rates.
The main difficulty in hydrogen cooling is that it is reuired to
ensure full safety against an y possibility of explosion. This is due to
the reason that when hydrogen is mixed with air in certain propor-
tions the mixture becomes explosive. To avoid this following precau-
tions are taken
(a) It is necessary to use pure hydrogen. The generator
enclosure must be gas tight so that air cannot penetrate
into the generator.
(b) The pressure of hydrogen is kept at least 0.035 kg/cm2
higher than outside atmospheric pressure to prevent the
leakage of air into the system.
(c) Special labyrinth type oil gland seals on the bearings are
used to help the hydrogen leakage to a safe minimum.
The efficiency of large turbo-generators can be taken as 98%
with air cooling and 99% with hydrdgen cooling.
10.5.1 Parallel running of alternators
The method of connecting an alternators in parallel with
another alternator or with bus bars to which a number of alternators
are connected is called synchronising. In order to achieve proper
synchronising of the alternator of the following conditions should be
fulfilled:
(i) The frequency of the two systems must be identical.
(ii) Thp phases of the inconling alternator must be identical
with the phases of the bus bars.
(iii) The voltage of the incoming alternator must be the same
or approximately same as that of the alternator with
which it is to be run in parallel or with that of bus bar.
10.6 Power Transformers
They are used to convert alternating current of one voltage to
alternating current of some other desired voltage. The power trans-
formers used in power stations are either single phase transformers
banks of three each or 3-phase transformers.
Three phase transformers can be used with advantages in place
of banks of three single phase transformers. The connections most
widely used in 3 phase transformers are as follows:
(i) Delta-star (ii) Star-delta
(iii) Delta-delta,
Star-star connections are not normally used.
The size of transformers depends on the size of generator units.
The general practice is to use the unit system of operation i.e. one
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 571

transformer for one generator. In such case the kVA capacity of the
3-phase transformer must then be the same as that of generator to
which it is connected.
While selecting the type of transformer, the significant operat-
ing characteristic of the transformers as well as their initial cost
should be considered. The losses in the transformer are also an
important criterion for their selection. A 3-phase transformer is
preferable compared with a bank of three single phase transformer
as the core loss is about 20% less in 3-phase transformer.
Following factors should be considered in deciding the
suitability of particular type of transformer for a particular applica-
tion:
(a) Load factor (6) Losses
(c) Initial cost (d) Cost of losses
(e) Continuity of service (J) Methods of cooling
(g) Percentage impedance voltage
(h) Floor area (i) Weight
(j) Regulation of voltage.
Different methods of cooling the transformers are ü'sed. Two
commonly used methods used are a s' self-oil cooled transformers or
forced oil cooled transformer. Self-oil cooled transformers do not
need any external equipment and are self-contained whereas forced
oil cooled transformers require pumps etc.
Fig. 10.1 (a) shows a power transformer. The various parts are
as follows
1. Tank 2. Magnetic circuit (core)
3. Radiators 4. Windings
5. Transformer oil 6. Oil gauge
7. Relay 8. Vent pipe
9. Oil conservator 10. Terminal bush
11. Terminal bush
12. Temperature indicating thermometer
13. Tank supporting frame.
The tank of transformer is made of steel. The core and windings
comprise an assembly which is placed with in the tank. The core
serves as the magnetic circuit for the flux set up by the windings.
The windings are made of copper wire. The transformer oil which is
a special grade of mineral oil serves two functions as follows:
(i) it serves as medium for transfer of heat generated in the
windings and core to the air surrounding the transformer.
(iij to insulate the current carrying parts from each other and
from the earthed tank.

Radiators increase the surface available for cooling the oil of

transformer. The conservator ensures that the tank is always filled

572 POWER PLANT

with oil. The breather pipe permits the

air to be taken in or expelled out as the
level of oil in the conservator varies
with change in temperature. Relay
protects the transformer in the event
of internal faults. The losses irl the
transformer are an important
criterion for the selection of a trans-
former. A 3-phase transformer is
preferable compared with a bank of
three single phase transformers as the
core loss is about 20% less in 3-phase
Fig. 10.1 (a) transformers.
Further a 3-phase transformer
has the following advantages as compared to three single phase
transformers for ground surface power plant.
(i) Its cost is less (ii) Its weights less
(iii) It occupies less space (iv) It is more compact.
10.7 Reactors
Reactor help in protecting the electrical system. In large
capacity instllations short circuit current may attain very high
values. This can be reduced considerably by the use of current
limiting reactors and by their judicious placing.
A reactor should be so arranged that while there is no voltage
drop across it under normal conditions of operation, it will prevent
a large short circuit current being fed by the generators into the
fault, thereby limiting the breaking currents of the circuit breakers.
10.8 Location of Reactors
The various ways by Aich reactors are located are as follows:
(a) Reactors in generator circuits. In this system the reac-
tors are connected in the lines between the generators of the power
plant and generator (G) bus bars. Fig. 10.1 shows this system. The
disadvantage of this system is that if a short circuit occurs on one
feeder the voltage at the common bus bars drop to a low value and
the synchronous machines connected to the other feeders may fall
out of step.
(b)Reactor in series with feeders. Fig. 10.2 show this arran-
gement. In this arrangement when a short circuit occurs on one of
the feeders, most of the voltage drop occurs across its reactor and
bus bar voltage does not drop to any consideraM extent.
(c) Bus bar reactors in ring system. Fig. 10.3 shows this
system. In this system under short circuit cortditions the faulty
feeder is mainly fed by one generator. The other generators feed the
MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS

REACTORS

FEEDERS

Fig. 10.1 (b)

REACTOR
a

FEEDERS

Fig. 10.2

faulty feeder only through the reactors which thus limit the fault
current.
(d) Bus bar reactors in Tie-bar system. Fig. 10.4 shows this
system. This system helps in reducing fault currents in the same
way as the ring bus bar reactor system but in addition it has the
advantage of limiting the currents.

4Reactors

FEEDERS
Fig. 10.3

10.9 Circuit-breakers
Their function is to break a circuit when various abnormal
conditions arise and create a danger for the electrical equipment in
an installation. There are two types of circuit breakers.


574 POWER PLANT

BUS BAR

REACTORS

FEEDERS
Fig. 10.4

(a) Air circuit breakers (b) Oil circuit breakers.

In air blast circuit breakers compressed air is stored in a tank
and released through a nozzle to produce a high velocity jet and this
is used to extinguish the arc. Fig. 10.5 shows an axial air blast circuit
breakers and Fig. 10.6 shows a cross Ilir blast circuit breaker.
C = Stationary finger contacts
- A = Are restraining insulating barrier
M = Movable blade contact.
Oil circuit breakers are quite commonly used. In these circuit-
breakers oil plays an important role in the interrupting process. The
rating range of these circuit breakers lies between 25 MVA at 2.5
kV to 5000 MVA to 230 W. The use of oil in these circuit breakers
offers following advantages. -
(a) it has high dielectric strength.

AIR

Fig. 10.5 . Fig. 10.6

 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 575

(b) Surrounding oil in close proximity to the arc presents a

large cooling surface.
(c) It acts as an insulator between live part and earth.

10.10 Earthing of a Power System

A power system is earthed at a suitable point. The various
advantages of earthing are as follows
(a) It provides lightning protection.
(b) It reduces operation arid maintenance cost.
(c) It improves service reliability.
(d) Greater safety of power s ystem is achieved.
(e) It provides greater safety to electrical equipment against
over current.
The various methods commonly used for earthing a power
system are as follows

Fig. 10.7

1. Solid earthing. Fig. 10.7 shows this method ,f earthing. In

this method neutral of a generator, power transformer or earthing
transformer is connected direct to the station earth. This method of
earthing does not make a zero impedance circuit.
2. Resistance earthing. Fig. 10.8 shows this method ofearthing.
In this method neutral is connected to the earth through one or more
resistors. This method has the following advantages:
a) It reduces the effect of burning and melting of faulted
electrical equipment.
(b) It reduces electric shock hazard to personnel caused by
stray earth fault current on the earth return path.
tnl It reduces mechan i cal stresses in circuits carrying fault
currents.
3. Reactance earthing. Fig. 10.9 shows this method ot'earthing.
In this method a reactor is connected between the machine neutral
and earth, In order to minimise transient over voltages the earth
fault current should not he less than 25'. of the 3-phase thult
Current.

576 POWER PLANT

10.11 Layout of Electrical Equipment

The layout of electrical equipment of a power station consists of
the following:

XG RN

Fig. 10.8

XGXN

Fig. 10.9

(a) Arrangements of bus bars at the generator voltage.

(b) Arrangements of circuit breakers and switches.
(c) Location of transformers.
(d) Controlling switch board arrangement.
The various system used for L . of electrical equipments are
as follows'
(ci) Single bus bar system. Fig. 10.10 shows this system. In a
power plant having a number of generators and a si• '- hiis bar
arrangement, the bus bar is sectionalised by circuit breaker it has
the advantage that fault on one part of the bus bar or system does
not completely shut down the whole station. However the use of such
a large number of circuit breakers is out of date and now-a-days
there is a tendency to use fewer.
(b) Double bus bar system. Fig. 10.11 shows this s ystem. In
this system both low voltage and high voltage bus bars are dupli-
cated, any one of the bus bar Sections can be used as desired. A bus
G = Generator
CB Circuit breaker
S = Isolator or switch


MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 577

CB 0 CB 0 C8 CG C8 CB
4

I
Ic,
s4S4
CSCBCO CB3 C8 CB
-

s4 s,L s4
CB C8 CB1 CB C8 CS

TO TRANSMISSION LINES
LVB = low voltage bus bar
HVS = High voltage bus bar
T = Step up transformer. III
Fig. lOiG -

TO CTATIAKI
AU XII. IA RI.

CE

LVB

live BC

10 1RANSMSSION LINES

Fig. IOu

578 POWER PLANT

bar coupling switch is provided to transfer operation from one bus

bar to the other. This system has the advantage that it is possible
to have one bus bar "live" and to carry out repairs on the other when
required.
(c) Ring bus bar system. Fig. 10.12 shows this arrangement.
In this system each line is served by two circuit breakers. The
advantage of thi system is that there are always two parallel paths
to the circuit and failure of one section does not interrupt the service
completely.
10. 12 Protective Devices for a Power Plant
INF

1
T

LINE
Fig. 10.12

In powir plant the main equipment used consists of gener-

ators. transformers, bus'bar section, circuit breakersand circuits
going out of power plant. The function of circuits breakers is to make
the circuits connections when required and to trip or disconnect the
circuits under faulty or ab normal conditions. The fault y conditions
ma\' be due to short circuits either at power plant itself'or elsewhere
in the s ystem. In order to enable a circuit to be isolated from the
s ystem onl y under faults conditions, it is essential to ensure correct
operation of the circuit breakers by means of protective relays.
10.13 Characteristics of Relays
The main characteristics of protective relaying equipment are
(I) Sensitivity
(ii) Selectivity
thE) Speed.
The relaying equipm!1t should he sufhcientiv nsitiv to
operate reliabl y and it should be able to select between corI(lltwns
MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS

under which prompt operation is required and those under which

no operation is required. A protective relay must operate at the
required speed and must be reliable.
The speed with which relays and circuit breakers operate have
direct bearing on the following:
(i) Quality of power supplied to the consumer.
(ii) Stability of the system.
(iii) Amount of power supplied.
The relay has to operate in case of heavy over-loads, short
circuits, reversals and such abnormal occurrences. The selection of
relay is made on the basis of time grading of the system, the various
relays being given appropvi time settings.
10.14 Types of Relays
Protective relays avoid damage to the equipment. The function
of the relaying system is to recognise a short circuit and to initiate
the operation of devices to isolate the defective element with the
minimum of disturbance to the normal service of the power system.
The main principle used in the operation of relays is either
electromagnetic attraction or electromagnetic induction. The
various types of relays commonly used are as follows:

R ELECTRO
GNET
PRIMARY
WINDING

CLOSED
SECONDARY
WINDING

LOWER
ELECTROMAGNET
TAPPED PRIMARY
WIN DINS

LUR. LUKP(LNT TERt.t.NALS

Fig. 10.13
580 POWER PLANT

(i) Electromagnetic attraction type relay

(ii) Induction type directional overcurrent relay
(iii) Induction type overcurrent relay
(iv) Differential relay
(v) Electromagnetic current balance relay.
Fig. 10.13 shows an induction type over current delay. Where
as electromagnetic current balance relay is shown in Fig. 10.14.
10.15 Voltage Regulation of Transmission Lines
CONTRC*.
SPRIN6 PIVOT ARMATURE

1A
H91!B
10 TRIP
CIRCUIT
OF LINE A
Fig. 10.14

Constant voltage at the alternator terminal is essential for the

satisfactory operation of the main supply or any grid. The following
factors cause the deviation of terminal voltage from the normal
value.
(i) Change in the speed of alternator
(ii) Variation of the load-of the alternator
(iii) Change of power factor of load
(iv) Variations in temperature.
Voltage regulation is aimed at maintaining consumer's voltage
constant.
There is voltage drop due to load. It is, therefore, essential to
regulate voltage of the system at various points within the
prescribed limits. The various methods used for voltage regulation
are as follows
(i) Booster transformers
(ii) Tap changing transformers
(iii) Phase-angle control
(iv) Brown-Boveri regulators.
Voltage regulators should be fast acting. There are two type of
voltage .gulators.
(a) Electro mechanical regulators
(b Static voltage regulators.
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 581
Transformers with on load tap changers are widely used in
electric power stations and power system sub-stations. Built in tap
changers increase the cost of transformers.
Fundamentally there are two systems by which the electrical
energy can be transmitted.
(i) High voltage D.C. system
(ii) High voltage A.C. system.
Various advantages of D.C. transmission are as follows:
(i) Only two conductors are used for transmission as com-
pared to three conductors in A.C. system.
(ii) No stabilisers are required when power is to be trans-
mtted over longer distances.
(iii) Inductance, capacitance, phase displacement and surge
problems are eliminated in D.C. transmission.
(iv) In D.C. system the potential stress produced on the insula-
tion is about 70% of A.C. effective voltage of same value.
The various advantages of A.C. transmission are as follows:
(i) It is possible to generate voltage as high as 33 kV as
compared to 11 kV in D.C. system.
(ii) In A.C. system the alternattng voltages can be efficiently
stepped up by transformers which is not possible in D.C.
system.
(iii) It is easier to maintain A.C. substation.
(iv) The transforming sub-stations are much efficient than the
motor-generator sets used in D.C. system.
The various disadvantages ofA.C. system are as follows
(i) In A.C. system the amount of copper tied is more.
(ii) The inductance and capacitance of the line effects the
regulation of the line.
(iii) The speed of generator and alternators is not economical
variation of these speeds must be controlled within very
low limits.
(iv) Construction of transmission lines is not easy.
10.16 Transmission of Electric Power
The electric power may be transmitted by either underground
cables or overhead lines.
Transmission by underground cables is quite suitable in densely
populated areas. However underground transmission is about two
times as much costly as over head system of transmission. There-
fore, under normal circumstances an overhead tflsthjsjon system
is preferred.
The main parts of the transmission system are as follows

Dad POWER PLANT

(i) Supports
(a) Poles (b) Towers
(ii) Conductors (iii) Insulators
(iv) Transformers (v) Protective devices
(vi) Control devices.
Poles or towers support the structures which support the con-
ductors above the ground. Both poles and towers should ful fill the
following requirements
(i) High mechanical strength
(ii) High durability
(iii) Light weight
(iv) Low maintenance cost
(v) Low initial cost.
Poles may be made of materials like wood, steel, cement concrete
etc. Usually RCC poles are used.
Fig. 10.15 shows RCC pole for single circuit and Fig. 10.15 (a)
shows RC. C pole for doub}e circuit.
Towers made of steel are mechanically sturdy and durable and
are commonly used. Fig. 10.16 and Fig. 10.17 show steel towers. The
steel towers are generally four legged, each leg anchored properly.

Pole

Fig. 10.15(a)
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 583
Conductors are generally made of copper, aluminium, copper
clad steel and steel cored aluminium conductors are used to carry
electrical power from one end to the other for transmission and
distribution. Conductors should ful-fill the following requirements.

tor

Fig. 10.16 Fig. 10.17

(i) Good electrical conductivity
(ii) Low weight per unit volume
(iii) High tensile strength
(iv) Low cost
(v) High corrosion resistance.
Insulators are made of porcelain, glass or stealite. An insulator
should have following properties
(i) High mechanical strength
(ii) High insulation resistance
(iii) High dielectric strength.
Various types of insulator commonly used are as follows:
(i) Suspension type (ii) Pin type
(iii) Shackle type (iv) Strain type.
Fig. 10.18 shows a suspension type insulator.
Fig. 10.19 shows pin type insulator. In this insulator P is steel
pin.
 584 POWER PLANT

s Shackle insulators are used for low voltage distribution lines.

Fig. 10. 19.1 shows shackle insulator mounted in a clamp.
Conductor tied down
with soft copper wire

II
Sheds

Fig. 10.18 Fig. 10.19

Fig. 10.19.2 shows a strain insulator. They are used for high
voltage transmission step up transformers are used at the generat-
ing end and step down transformers are used at the receiving end.
DUCT ER

CLA
ULATOR

Fig. 10. 19.1

Fig. 10. 19.2

Protective devices such as circuit breakers break the circuit

when a fault occurs. Relays protect the circuit by energising the trip
coils of the circuit breakers at the proper time.
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 585
The voltage of the transmission line is maintained constant or
within limits b y voltage control devices such as
(i) Synchronous compensators at the receiving end of the line
(ii) Regulators for the generators
(iii) Indudtion regulators
(iv) Tap changing transformers V

(u) Line drop compensators.

The s ystem control apparatus mainly consists of
(z) Quick-response excitation control apparatus for improv-
ing the stability of the system.
(ii) Governors and frequency control apparatus for controlling
the load division between different stations in system.
10.16.1 Transmission Lines
Transmission lines are used to transmit power from the power
station to load centre. The size of transmission lines, i.e. number of
conductors and their cross-sectional area depends on the following
factors
(i) Amount of power to be transmitted.
(ii) Distance our which power is to be transmitted.
The various requirements of goo.d and efficient electricatservice
of a power system are as follows:
(i) Constant voltage
(ii) Dependability
(iii) Balanced voltage
(iv) Minimum annual cost
(u) Constant frequency.
The voltage of transmission depends on the following factors
(z) Power to be transmitted.
(ii) Distance over which power is to be transmitted.
(iii) Specific performance requirements regarding permissible
regulation, losses etc.
(iv) Cost of transmission line equipment.
The standard voltages used for transmission in India are 11, 22,
33, 66, 110, 132, 166 and 230 W.
Transmission lines are classified as follows
(i) short (ii) medium (iii) long.
Transmission lines having a length less than 80 km are
regarded as short transmission lines. Medium transmission lines
have lengths between 80 and 240 km. Long lines have lengths more
than 240 km. Performance of transmission line depends on length
of line and power to be transmitted.
Percentage regulation. The percentage regulation of a trans-
mission line is calculated as follows
—39

586 POWER PLANT

3 r
xlOO
V
,.
1
where V = Sending end voltage
Vr = Receiving end voltage
Regulation
Efficiency. Efficiency of transmission line is given by
Pr
S

where Pr = Receiving end power iT L•

P = Sending end power S

ii
= Efficiency
Power loss. Power loss in a 3-phase transmission line is given
by
Power loss =3JR
where I = current in line
R = Resistance of each phase.
The spacing of the transmission line between conductors is
chosen considering the voltage between phases and the reactance
and capacitance constants.
Capacitance (c). Capacitance of transmission line is defined
as the ratio of charge to voltage

C= V
where Q = charge j
V = voltage
I UI I ;3 F.
Capacitive reactance (Xc). It is given by
- 1
c_ 21c -
where f= frequency
Charging current (!). It is defined as the ratio of voltage to
capacitive reactance
S
5

--S

Xr S..

t5

MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS

587

10.17 Systems of Electrical Energy Transmission

Fundamentally there are two systems by which the electrical
energy is transmitted.
I. High voltage DC system. 2. High voltage AC system.
DC systm
(i) DC two wire
(ii) DC two wire with mid point earthed
(iii) DC three wire. 1 b .4- 1ki mii

AC System )J lYD)A
(a) Single phase AC system
U) Single phase two wire ).
(ii) Single phase two wire with mid-point earthed
(iii) Single phase three wire.
(a) Two phase AC system
(i) Two phase three wire
(ii) Two phase four wire
(c) Three phase AC system
(i) Three phase four wire.
However, three phase alternating currents system Is mostly
used. When transmission is being-carried on over-head lines great
care is taken to insulate the conductors from the cross arms and
support towers.
The single phase system has its own limitations and therefore
has been replaced by poly-phase system.
Poly-means many (two or more than two) and phase means
windings or circuits each of them having a single alternating voltage
of the same magnitude and frequency. Hence a polyphase system is
essentially a combination or two or more than two voltages. These
voltages have same magnitude and frequency but are displaced from
one another by equal electrical angle called phase difference.
- 360 electrical degrees
- - N
Aq
where 0 = Phase difference 11
N = Number of phases
In two phase s y stem there are two equal voltages of same
magnitude and frequency having a phase difference of 90 degrees
electrical
In three phase system there are three voltages of same mag-
nitude and frequency and having a phase difference of 120 degrees
electrical.
588 POWER PLANT

The three phase system is most commonly used polyphase

system and is exclusively used for generation, transmission and
distribution of electric power.

10.18 A.C. Power Distribution

Electric power is generally generated and distributed in the
form of alternating current. The advantage of using A.C. system is
that voltage can. be conveniently regulated by means of trans-
formers.
A.C. power distribution system is of two types
(i) Primary distribution system.
According to this system the bulk consumers are generally
supplied power at high voltage of 11 kV, although voltage may also
be 6.6 kV or 3.3 W. The voltage is stepped down to 400 V by
step-down transformers. Three phase three wire system is preferred
for distribution due to economic considerations.
Fig. 10.19 (a) shows a typical distribution system.

Bulk
consumer

11 kV
Primary distributor
33kV 1
3W.r
P VVjtI
Tr
Substation L1
Factory Distribution
Sub-station
Fig. 10.19 (a)

(ii) Secondary distribution system. This system is used to

supply power to small consumers at low voltage (400/230 V). Three
phase-4 wire system is preferred the voltage is stepped down from
11 kV to 400 V by transformers installed at distribution sub-station.
Fig. 10.19 (b) shows a secondary distribution system.


MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 589

11 k V/400 V
Transformer
11 kV I 1 400v12301, 34', 4 Wire
.Ph ¶

Distributior
Substation

Hou5e I I I I House

-
ory Mouse

Fig. 10.19(b)

10.19 Practical Working Voltage

Following voltages are generally adopted
(i) For generation 6.6, 11 or 33 W.
(ii) For main or primary transmission 66, 132, 220, 275, 330
and 400 kV.
(iii) For secondary transiission 11 or 33 kV.
(iv) For distribution 6.6, 11 or 33 kV.
It is not doubt economical to use higher voltages for transmis-
sion but there is a limit to which the voltage can be increased. Since
an increase in transmission voltage introduces insulating problems
as regards supporting of conductors and clearance between them.
10.20 Over-head and Under Ground Power
Transmission Systems
Both over-head and underground systems are used for trans-
mission of electrical power.
The over-head transmission system has the following ad-
vantages:
(i) It is cheaper to transmit power by over-head lines than
underground cables.
(ii) Over-head lines can be easily repaired.
(iii) Insulation of over-head lines i s easier.
The advantages of underground cable power transmission are
as follows
(i) It is easier to transmit power in densely populated areas.

590 POWER PLANT

(ii) Maintenance cost is low.

(iii) Underground cable is susceptible to less number of faults.
10.21 Conductors
Conductors are used to carry electrical power from one end to
the other for transmission and distribution. The conductors used for
over head lines are stranded in order to increase the flexibility.
Stranded conductors usually have a central wire around which there
are successive layers of 6, 12, 18,24 wires. A conductor having seven
strands is shown in Fig. 10.20.

Fig. 10.20

Conductor materials. The codductor materials should pos-

sess the following properties:
(i) High conductivity i.e. low resistivity so that area of cross-
scctmn can be reduced.
(ii) High tensile strength.
(iii) High ductility so that the conductor can be easily drawn
in shape of wires.
(iv) Low specific gravity so that low weight per unit volume is
used.
(u) Low cost.
The most commonly used materials are as follows
(i) Copper (ii) Aluminium
(iii) Steel cored aluminium (iv) Galvanised steel.
10.21.1 Disposition of Conductors
The disposition of conductors in transmission or distribution of
electrical energy has much bearing on the performance of the line.
The line may be either of single circuit or double circuit system.
Single circuit and double circuit system can be compared as follows:

- Single Crcurt Destgjj _J Double Gcttilik,agn

1TfThere is less danger at the time of 1. fuanger always exists from the
^air ther alive circuit

2. I This design is subjected to lessi 2 There is more wind pro.ur

wind pressure and structure and This requires heavier struc-
therefore, it requires lighter tures and weight of to—r
towers nnil wdiuni ftjoiJ ..... -
MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS 591

3. Larger spacing of conductors is 3. Lesser spacing in conductors is

4. lHigher reactance as spacing be-1 4 Low reactance as spacing be-
Itween conductors is mo4_
5.This design is not reliable. far 5 This design is quite reI Ille as
as continuity of power supply is regards continuit y wer
tppIy.
10.22 Electric Power System Stability
Whenever the speed of an alternator changes with respect to its
synchronous speed it gives rise to variations in voltage and frequen-
cy. These variations are further transmitted to the lines and feeders
connected to the alternators thereby setting up disturbances in the
connected equipment.
Following measures help in providing stability to the electric
power system
(i) The stability can be improved by using lighting arrestors
for protection of lines, by using quick acting circuit
breakers and with relays having minimum time lag.
(ii) Voltage regulators connected on the line should be quick
acting.
(iii) The governors attached to the turbines driving alter-
nators should be as quick acting as possible so that the
generator input to the load is quickly adjustable.
(iv) Stability is also improved mechanically by connecting the
synchronous motors to heavy flywheels.
10.23 Control Room
The control room of a power plant should be properly located.
Following points should be kept in mind while locating the control
room:
(i) It should be near the switch house to reduce the lengths
of connecting cables.
(ii) It should be well ventilated and lighted.
(iii) It should he located away from the sources of noi s-
Following equipment is fitted in the control room
) control equipment like
(a) voltage regulators
(h) frequency stabilisers .. ...
(c) load distributors. , ,f
(ii) protective devices like
(a) circuit breakers ,..
(5) relays l..
(c) emergenc y trippers.
ifi!'fiIi (pci I

J h911 jl' ?IlJ.Ot .I

ci-, ..• u .i

. .


592 POWER PLANT

(iii) Indicating devices such as instruments for indicating the

load, voltage, frequency, power factor and winding
temperatures.
10.24 Layout of Power System
The generator, transmission and distribution of electric power
is called power system.

POWER 1l; 11 II V
PLANTt
'1/400 OR 200
OR 137 K STEP UP
\TRANSFoRMER

10 SUB STATION
(220KV)
PRIMARY 720166 0R335V
STEP DOWN
TRANSMISSION 'r•" TRANSFORMER
I
TO SUB STATION
(66kv)
SECONDARY I

TRANSMI59014JI l6/11 NV
. STEP DOWN
TRANSFORMER
T
10 SUB STATION
(11KV) I
PRIMARY
DISTRIBUTION 11NV140V
STEP DOWN
TRANSFORMER

TO LOW VOLTAGE
DISTRIBUTOR
Fig. 1 fl 21 -

Fig. 10.21 shows line tliagrain oi ;er svsteni. Generall y a

power system Consists of the following stage.
(i) Power station
(it) Primary transmission
(iii) Secondary transmission
(it') Primary distribution
(t') Secondary distribution.
At the power plant the electric power is generated. The genera-
tion voltage ma y he 3.3, 11 or 33 kV. However 11 kV generation
voltage is quite commonl y adopted. Then the generation voltage of
11 kV is stepped up to 132, 220 or 400 kV at the power plant.
The power is then transmitted by primary and secondar y trans-
mission s y stem using three phase three wire system. The electric
power is finall y supplied to the consumers by primary and secondary
distribution system.
 MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS
593
10.25 Factors Affecting Power Generation and
Distribution
Electric power is generated by means of electric machines called
generators when power generated is D.C. and Alternators when
power generated is A.C. These machines are essentially convertors
which convert mechanical energy into electrical energy i.e. these
machines must be mechanically coupled to prime-movers.
Three phase system should be used for generation, transmission
and distribution of electric power because of its superiority over
single phase system.
Three phase three wire A.C. system is invariably employed for
transmission of electric power because in this case power at high
voltage can be generated and the voltage can be stepped up and
stepped down easily and efficiently by using transformers.
To transmit the same amount of power over a fixed distance at
a given voltage, 3-phase system requires about 3/4 of weight of
conducting material of that required by single phase system.
Electric power generated at the power station is conveyed to the
const ners through a network of tr ansmission and distribution. The
part power s y stem by which electric power is distributed among
Consumers is called distribution system. The electrical energy
generated at the power plants is conveyed to the sub-stations from
where the electric power is convoyed to the bulk consumers by high
voltage distribution s y stem whereas it is conveyed to the small
consumers b y low voltage distribution system.
The various factors which affect electric power generation and
distribution are as follows
,) Type of phase System
(a) Single phase system
b Three phase system.
(ii)Type of fuel used. Hydro-electric power plant is best suited
where water is available in large quantities and at sufficient head.
Steam power should be installed near coal mines. However steam
power plant or nuclear power plant can he used where neither coal
nor water is easily available. For small loads diesel power plant or
gas turbine power plant ma y he used.
(iii) Transmission Voltage. It is desirable to use highest
possible voltage for transmission of electric power. Economic trans-
mission voltage is one for which cost of transmission (cost ofconduc-
tors. insulators, t ransformers, switch gear and other terminal
apparatus like lightning arrestors etc.) is minimum.
(ir) Cost of transmission and distribution.

594 POWER PLANT

(v) Type of transmission. The electric power is generally

4ransmitted and distributed by over head lines because of economy.
(vi) Type of substations:
(a) Indoor substations.
(b) Out door substations. Outdoor substations are preferred
over indoor substations.
(vii) Type of ldad to be taken by the power station.
(viii) Capacity of power plant.
(ix) Running cost of power plant. The running cost of hvdro
power plant is less than steam or nuclear power plant.

PROBLEMS

10.1. What is the electrical equipment in a power plant.

10.2. Write short notes on the following:
(a) Generator (b) Excitor (e) Factors affecting power generation
and distribution.
10.3. Describe the various generator cooling methods.
104. Write short notes on the following:
(i) Power transformer
(ii) Voltage regulation
(iii) Transmission of electrical power.
10.5. Discuss the factors to be considered while deciding the
suitability of a transformer,
10.6. What is a circuit breaker? What are the different types olcircuit
breakers that are employed in typical power stations ?
10.7. What is protective relay? Describe any one relay system tiscd to
,protect the equipment in a power plant.
10.8. Give a list of various electrical protective equipment used in a
power station. A.M.LE. 1979)
10.9. Describe the various inethod.- controlling the voltage at the
consumer terminal used in p>&r supply system.
10.10. What are the different methods ofearthing in pocer system?
Explain in details.
10.11. What is the function of a reactor? Describe the various arran-
gements used for location of reactors,
10.12. Write short notes on the following
(a) Over-head and underground power transmission systems
(b) System of electrical energy transmission
(c) Practical working voltage
(c/i Disposition otconcluctors
(e) Characteristics of relays
(f) I 'a ral Ic! running of alternators


MAJOR ELECTRICAL EQUIPMENT IN POWER PLANTS

595
(g) Control room
(h) Poles and towers
(i) Types of insulators.
10.13. Discuss electric power system stability.
10.14. Sketch and describe a power transformer.
10.15. State the advantages and disadvantages of A.C. and D.C.
systems of power transmission.
10.16. What are the properties ofrnaterials used forconductors Name
the materials used for conductors.
10.17, Describe A.C. power distribution systems.
10.18. Sketch and describe layout of power system.

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0' 1IJU ij.4d .0)7 Ui Jfl f QI(l 11i-..1flJlI. lfl')l'Il . ii -
.- ' yy1IJ1)i. iJII'',flC)-rI(,fl 'jtdi'iiiji hri wn b r1tfl-.'.1A.)
ILUJ1 r1Lf 1)111 IIj (U ',rT9to4 ')l'C/ i J0U11 ,'i 1i:)
.'!91'IfI011
¶)I110 In OLtrl!;'Ij)f 1.(0
 11

Non-Conventional Sources of
Energy

11.0 Introduction
In our country power is needed in plenty to sustain our in-
dustrial growth and agricultural production. Existing sources of
energy such as coal, oil, water and uranium may not be sufficient to
meet the increasing energy demands This requires that scientists
and engineers should undertake research work which would help in
exploring the possibilities of harnessing energy from several non-
conventional sources of energy. The various non-conventional sour-
ces of energy ae as follows
(i) Solar energy (ii) Wind energy
(iii) Bio-gas and bio-mass (iv) MHD generat9r
(v) Geo thermal (vi) Tidal power
(vii) Fuel cells (viii) Thermo-electric generation
(ix) Thermionic converter.
It is recognised that an early, smooth and successful transition
from fossil fuel (coal, oil, water) to new and renewable sources of
energy depends on a strong, integrated and purposeful R and D
programme. In our country the Commission for Additional Sources
of energy is responsible for the development of new and renewable
sources of energy. The technology to obtain power from renewable
sourccs should be so developed that power generated is cheaper. The
power plants working on renewable sources of energy should have
low capital cost, there should be no pollution of atmosphere and
running cost should be less.
The Energy Ministry proposes to give a bigger thrust to the
exploitation of new and renewable (non-conventional) sources of
energy to meet a variety of energy needs in urban and far flung rural
areas in a decentralised manner.
The main advantages of non-conventional sources ofenergy are
as f0llows
 NON-CONVENTIONAL SOURCES OF ENERGY
597

(i) They are widely available.

(ii) They are non-polluting.
(iii) They are well suited for decentralised use a distinct ad-
vantage in a large country like India. Owing to technologi-
cal developments and improvements in systems designs it
has been observed that renewable sources of energy can
now be shown to be more attractive in many situations
and locations. For example calculations of the economics
of bio-gas plants show that the pay back period is two and
half years in terms of value of fuel saved and manure
produced. According to energy experts the non-conven-
tional sources of energy can be used with advantage for
power generation as well as other applications in a large
number of locations and situations in the country. It is
hoped that these alternative energy sources will be able
to meet considerable part of energy demand in coming
future. Solar energy, however is not free from problems.
It is now available at night and when local weather con-
ditions obscure the sun. Moreover solar energy is diffused
in nature and is at a low potential. Consequently if solar
energy is to be economically competitive it must be con-
verted into a usable form of energy with maximum effec-
tiveness to reduce capital cost of solar power plants. -
111 Solar Energy
Among the various non-conventional sources of energy the solar
energy seems to hold out the greatest promise for the mankind. It
18 free, in exhaustible, non-polluting and devoid of political control.
Rapid depletion of fossil fuels and combined crisis of pollution and
of steep rise in oil prices have brought about an upsurge of interest
in solar energy.
Solar radiations can be converted directly or indirectly in the
form of energy such as heat and electricity. Solar power would
eliminate most of the serious environmental problems associated
with fossil fuel and nuclear power. Energy is released b y the sun as
electromagnetic waves of which 99% have wavelengths between 0.2
to 4.0 p. where, p. is micrometer. Solar energy reaching the top of
earth's atmosphere consists of about 8% ultra violet radiation (short
wave length less than 0.39 p.), 46% visible light (0.39) to (0.78 p.) and
46% infrared radiation (long wave length more than 0.78 p.).
Solar radiation that has not been absorbed or scattered and
reaches the ground directly from the sun is called Direct Radiation.
It is the radiation which produces a shadow when interrupted by an
opaque object. The radiations received after scattering is called
Diffuse Radiation. Diffuse radiation comes to earth from all parts of


598 POWER PLANT

the sky. The total solar radiation received at any point on the earth's
surface is the sum of direct and diffuse radiation.
Fig. 11.1 (a) shows solar energy storage.

0)

0) o
CJ)

00
LU
ÜILU
'U

0 0
a)
0 0
0 00
0D 0 a. -=
0) E Eu

1 211 F-P.
0
0) n
= 0
w E
a) a)
0
Cn

CL

ID 0
>. t, IIUl
0)
;flirIl9
j1a,c
uJ jJjQJ

Fmfl,) :)Jg
U) oh
rim
C
0
I L9
F; 60-

The thrust of the research efforts in our county in the soiar

energy has been directed both towards solar thermal applications
and direct conversion of solar electricity. Utiliation of solar energy
is of great importance to our country as it lies in tropical climate
region where sun light is abundant for a major part of the year. The
target for power generation from non-conventional energy sources
have been upgraded to 200 MW from 600 MW for eighth plan. The
most promising and fast moving solar technology today is that of
solar cells, flat metallic blue chips made of highly pure silicon that
 NON-CONVENTIONAL SOURCES OF ENERGY

can convert sun light into electricity. These photo-voltaic cells are
being used in rural areas and isolated locations for a variety of
applications such as () water pumping for irrigation and drinking
water supply.
(ii) Community and street lighting.
(iii) Power supply for micro wave repeater station.
(iv) Communication equipment, radio and television
receivers.
(v) Solar water heaters.
(vi) Solar refrigeration.
11.2 Solar Radiation Measurement
Following two type of instruments are used for solar radiation
measurement
W P vrheliotneter. It collimates the radiation to determine the
-
beam intensity as a function of incident angle.
(ii) Pyronorneti'r. It measures the total hemi-spherical solar
radiation. The pyranometer is quite popular.
11.3 Solar Constant -
It is the rate at which solar energy arrives at the top of atmos-
phere. This is the amount of energy received in unit time on a unit
area perpendicular to the sun's direction at the mean distance of the
earth from the sun. The rate of arrival of solar radiations vary
throughout the year. According to National Aeronautics and Space
Administration (NASA) the solar constant is expressed in following
three ways.
(i) 1.353 kilowatts per square metre or 1353 watts per square
metre.
(ii) 429.2 Btu per square foot per hour.
(iii) 1164 kcal per sq. m. per hour.
nocv;f f
ui
11.4 Solar Energy Collectors
Solar energy can be exploited in various ways as follows
(i) By direct conversion to a fuel by photosynthesis. 4

(ii) By direct conversion to electricity by photo.Istltaic.

(iii) By conversion to electricity via thermo-ctric power
system. However thernio-electric system is common-
ly used to as other two systems are still far away from
acceptble limits.
Following thetmo-electric systems are presently used for power
generation.
(i) Low tcmpt.raturc cycles Using flat plate collectors.
600 POWER PLANT

(ii) Concentrator collector for medium and high temperature

cycles.
(iii) Power concept for power generation.
11.5 Flat Plate Collectors
A flat plate collector is shown in Fig. 11.1. Flat plate collectors
are made in rectangular panels from about 1.7 to 2.9 sq. in area
and are relatively simple to construct and erect. Flat plates can
collect and absorb both direct and scattered solar radiation they are
thus partially effective even on cloudy days where there is no direct
radiation.

Solar.
radiation
\\% Absorbin
Surf oc
TrQnsporcr
cover DItS

Hot water

Cold water
Fig. 11.1

The solar rays pass through transparent covers and fall on

absorbing surface. The absorbing surface which is usually made of
copper, aluminium, or steel coated with a heat resistant black
(carbon) point intercepts and absorbs the solar radiation enorgy.
Radiation energy is converted into heat and water flowing through
the tubes gets heated. It isiot possible to generate steam with plate
collectors so this system can not be used directly to run the prime
mover. So some other organic fluid such as Freon-14, 150 butane
etc. which evaporate at low temperature and high pressure by
absorbing heat from heated water. The vapours formed can be used
to run the prime mover (turbine or engine) to generate power.
Insulation is used to prevent loss of heat from the absorber and
heat transporting fluid. The insulating materials commonly used
are fibreglass or styrofoam.
Flat plate collectors are also called non-concentrating type.
Collectors are classified as low temperature collectors because they
can generate temperature less than 90 and have a collection 'hf-
ficiency of about 30 to 50%.
NON-CONVENTIONAL SOURCES OF ENERGY 601

11.6 Concentrating Collectors

('oitcen t r.i I ri or f tcu i rtg colk'ct rs a fl of two tvfls
p Lill( . tut'usuilg 1 lcctors
(ii Point focusing collector,
Focusi'i collectors collect solar energY with high inte'natv of
solar radiatuout on the energy ahsorluuig urlice. Such collect'r
generally uu eutucal Ystt'lfl in the firm of reflectors or rtfracturs
and can heat t h fit uk oil) to about UUU C. An important uli tIe rt I Ice
between collectors of riuu-t'ocusuuug antI lousing type e- that the
latter conceult rate only direct radiation coming from a specific d t red-
ti on.
11.7 Line Focusing Collectors
l"ur. 11.2 liows ,i parabolic trough collector. In this collector the
solar radiation couluuutg from a parttctuliti'directioti are collected over
the area ott he cell cutittg -irface a uid is concentrated at the kc ti s (F)
of parabola.lotlyc yl i ndrical parabolic concentrators are used in
which absorber i. placed along foctisfocus axis. Fig. 11 :3 shows a typical
cy lindrical parabolic collector. Itconsi,tsotparjboltcuyhulerretlCc
tor to concentrate sun light On to it collecting pipe The retlector is
steered during cia time to keep still focused on the collector.
'[his t y pe of concentrator produces much higher teniperitLire than
flat plate collectors. The dimensions of parabolic trough collector or
parabolic c y lindrical collector can vary over a wide range, the length
of a reflector unit may be above 3 to S in and width about 1.5 to 2.4
In. Ten or More suit units may he connected end to end in a row,
several rows being connected in parallel. Parabolic trough reflectors
may be made From polished il Uflil ut iu In, silvered glass or a thin til iii
of highly aluntinised plastic on a firm base.

Stay
rois
C)'
Mr r or
strips \
\ t'"_

0 -Absorber
rettec tir tube

II 2 1' 3

--40
602 POWER PLANT

11.8 Point Focusing Collector

Fig. 11.4 shows a paraboloidal dish collector which bring solar
radiation to a focus at a point. In this collector a dish 6.6 metre in
diameter made from about 200 curved mirror segments forming a
paraboloidal surface is used. The absorber located at the focus is a
cavity made of a zirconium-copper alloy with a black chronic selec-
tive coating. The heat transport fluid flows into and out of absorber
cavity through pipes bonded to the interior. The dish can he turned
left and right and up and down so that sun rays can he focused
properly.

Absorb
er

dish I

Fig 11.4.

11.9 Advantages and Disadvanfages of

Concentrating Collectors
Concentrating collectors have the following advantages over flat
collectors: -
() Reflecting su!faces are structurally simpler and need less
material.
(ii) Cost of collecting system per unit area is low.
(ii) The absorber area of concentration is small and therefore,
solar energy concentrated can produce more heat and
therefore, working fluid can attain temperature for the
same solar energy falling on the concentrator.
(iv) Since the temperature that can be attained with con-
centrating collector system is higher, the amount of heat
which can be stored per unit volume is larger and conse-
quently the heat storagcosts are less for concentrator
systems than for flat plate collectors.
(t') They have more efficiency.
Disadvantages
hi Diffused solar radiations can not be focused and is lost.
ii Initial cost is high.
Costly orienting system for reflector to track the sun is
rt. iu red.
NON-CONVENTIONAL SOURCES OF ENERGY 603

11.10 Solar Thermal Power Generation

Solar thermal power generation uses power cycles which are
broadly classified as
(i) Low temperature cycle.
(ii) Medium temperature cycle.
(iii) High temperature cycle.
Low temperature cycles usuall y can be used up to 100 C and use
flat plate collectors whereas medium temperature cycles are used
for temperatures between 150 to 300 C and high temperature cycles
are used for temperatures above 300 C. Two principal forms of
energy into which solar radiations can be converted for practical
applications are as follows
(i) Heat
(ii) Electricity.
Heat is obtained when solar radiation is absorbed b y a black
surface. The heat then may be used in two ways
(a) Direct thermal applications such as water heacing, drying,
cooking, distillation etc.
(b) In-direct conversion of solar energy into electricity using
photo voltaic cells, thermo electric thermionic and photo
chemicals. Photo voltaic effect is quite popular.
The mechanical power production is called solar thermal power
production system. So t'---r as conversion of solar energy into electri-
cal energy is concerned it ma y be clone either by' solar therival power
oroduction route or solar radiations can he directl y converted into
electrical power.
11.11 Low Temperature Thermal Power
Generation
Fig. 11.5 shows a low temperature thermal election power
generation scheme using solar pond. Hot water from pond enters an
evaporator where the organic working fluid is vaporised. Then the
vaporised organic fluid at high pressure enters a turbine and there-
by expanding through the turbine wheel to produce power.
The vaprn'r now passes through a condenser where it is con-
densed to a liquid. This liquid is pumped back to the evaporator
where cycle i repeated.
Another type of low temperature solar power plant is shown in
Fig. 11.6. This system uses an array of flat plate collectors to heat
water to about 70C and then in heat exchanger the heat of water is
used to boil butane. The butane at high pressure is made to pass
through a turbine. This scheme is quite commonl y used for lift
in it ion purposes.
 604 POWER PLANT

Hot water
np Cooling
lower

Con denser Cold water

Solar
radiations
E vaporator
boiler

Solar
çGenerotcr Pump pon d
Organic Hot
workmg brine
fluid

Turbine

Fig. 11.5

Water circuit

Solar

Array of
collectors Hot
water

Butane
r
Bu tan: ro p
boiler ectors

PumpT

1
I 1urbne

Water for •
[1
irrigation purposeWell
:^
L Condenser
Fig ll.

High Temperature Systems

Two basic arrangements for converting solar radiation into
electrical energy systems are as follows.
11.12 Tower Concept Type Solar Power Plant
This type of power plant uses art of plane mirrors or
heliostats which are individually controlled to reflect radiation from
the sun into it mounted on a tall (ahout 500 in high) tower.
The steam is generated irl the boiler. The steam mav attnin a
teUlI PriItUlU up to 2000 K. The steam s. rodiiccd is ued t drive
NON-CONVENTIONAL SOURCES OF ENERGY

a turbine coupled to a generator. hg. 11.7 shows a tower concept

t y pe solar power plant.

Railer Incident
solar rays

To 9r
z $tots

Fig 11.7

Another type of solar power plant based on similar principle is

shown in Fig. 11.8. It uses an array of heliostats guided mirrors to
focus sun light into a cavity t y pe boiler near the ground to produce
high tcm1Rrature high pressure steam which is used to drive a
steatil t whine. The solar ra y s striking the mirrored faces of helios-
t.ats modules are reflected and concentrated in the cavity of hailer.
sclor
S
Porot,oIc
rcflector

fliT )T

- Boner

Fig. 1 1 .8

11.13 Photovoltaic (PV) Cells or Solar Cells

These cels din-CH cuvert sIar energy to D.C. power. These
made
cells are m of se nu i com
' i ductors thiitginerute uketricity when they
absorb light. Solar cells made of single crystal silicon are cuiflmuOI1ly
used as its theoretical efhciencv is about. 2-1 . Hut commercially
available cells have anefficienc y of about 10 to 12? Gallium
arsenide is the another solar cell material. Cells oft his material ulay
achieve an efficiency of 20 to 25"; cells made of gimllmuiii
arsenide can retaineffmciencyat much higher temperature tliaui tells
made of silicon.
The silicon cell consists of a single cr y stal Of silicon into which
a doping material is diffused to form a semi-conductor. The best
known application of photovoltaic cells for electrical power genera-
606 POWER PLANT

tion has been space craft for which silicon solar cell is the most
highly developed type.
Various advantages of solar cells are as follows
(i) They need little maintenance
(it) They have longer life
(iii) They do not create pollution problem
(iv) Their energy source is unlimited
(u) These are easy to fabricate
(vi) They call made from raw materials which are easily
available in larger quantities.
The disadvantages of solar cells are as follows
(i) Compared with other sources of energy solar cells produce
electric power at very high cost.
(ii) Solar cell output is'lot constant and it varies with the time
of day and weather.
(iii) They can be used to generate small amount of electric
power.
Solar cells have also been usea to operate irrigation pumps,
vigational signals, highway emergency cell systems, rail road
cro- rw signals_etc.
The mosr common configuration of a typical solar cell to form a
p . n junction semi-conductor is shown in Fig. 11.9.
Solar
rays

Load

Current
collection grid
(metal

P - fleion
Base material Difttjsed
layer
Metal
conductor
Fig. 119

11.14 Solar Pumping

The various parts of solar pumping system are as follows
(1) Solar coD ('ctOls SUC Ii as


NON-CONVENTIONAL SOURCES OF ENERGY 607

(a) Flat plate collectors

(b) Stationary concentrators.
(ii) Heat exchanger or Boiler
(iii) Heat engine such as
(a) Brayton cycle as turbine
h) Stirling hot gas engine
(c) Rankine engine
(it) Condenser
(v) Pump such as
(a) Centrifugal pump
(b) Reciprocating pump
(c) Rotary pump.
Heat engine

orrog"

rganic
fluid
T^ 7
un
Condenser

Circulating Feed
pump pump
Fig. 11.10

The power generated by solar energy is used for water pumping

useful for irrigation purpose. Fig. 11.10 shows a typical solar energy
power water pumping system. The primary components of the
system are an array of flat plate collectors and Rankine engine with
an organic fluid as the working substance. In this systems a heat
transfer fluid flows through the collector arrays. The fluid get heated
due to solar energy. The fluid (water) is then made to flow through
a heat exchanger (boiler) where it transfers its heat to other fluid in
the boiler. This other fluid then gets evaporated and expands in the
engine before reaching the condenser. Some of the working fluids
used in cycle are toluno (CP-25), monochloro benzene i MC B, and
trifluoro ethanol. The major obstacle to the increased use of solar
irrigations system is its relatively high capital cost. 1-I.E. in figure
indicates heat exchanger.
11.15 Wind Energy
Energy of wind can be used for the generation of electrical
energy . The potential of wind energy as a source of power is large.
Wind energy which is an indirect source of energy can be utilised to
run wind mill which in turn drives a generator to produce electricity.
Wind energy is one of America's greatest natural resources. Accord-
ing to NASA study the U.S. has enough harvestable wind to
 608 POWER PLANT

• 1 1 110 rate 1 .2 trillion kWh ofeleiric;rl c I w Ijgy per y ear. Wind mills
have been usi of for several cent urie in cmintrie like Denmark,
Netherland China. Europe. Persia etc fit Itigli wind speeds
are obtainable in coastal aias of Sot ir.ishtra Western Rajasthan
and snow parts of (errl India Iii these areas there could be a
poilnlity (f uing nirditini and lal 6 iicd wind mills for genera-
tioi of elCt rRit \
4ome of the haractere-tics ol wind (rIrIg\ ale as I'ollows
\'irul power s y stems are ii n olhit inn
(it) It is it renewable source of energy
iii The eni rgy generated by wind-power sv'tvms is cheaper
when produced on small scale. On largesraic it is confliCt!-
tive v lth conventronal pOWCF gvnerw ing systems.
Wind energy svtrriis avoid fuel prove-ion and transport.
however tilowing prribRiiis are a,ociated with wind energy
I I \\i;id cfl(rgv is fliictiiiting in nature

li-ause of its irregularit y it needs storage clivices

(n) Uoi- is sufficient mini-c 1ioducid h\ \\ mud eiiergv power
gener. it irIg sv-t eiiI-.
In oil-country a niuimlierof in (I mill svtonis br wate r 1111n)p111g
and for j )rolucti)n ofsinall iimflotint oleiectncal powei- ha'. e been rt
up. Soon- o! the dvelopinviit s ire-tolled are as follows
Melurai wind mill at Madural ('l'.N.)
hr S}oii:ipur wind null at Sholapur
(iii) C._ virid mill at Jodhpur (Rajasthan)
ut - Ml' I all wind null at NAL Bangalori
Wire! ills are Cemitral.Salt and Marine (lwmicais. Be-
sear Inst i tute, lthiivnagar i(ujarat
Atpri-amit about 43 JlV aggregate ' md pwcr capacity his
hecn (-,tahie-lied In our , country. It is propo-ed to t--tat ' ie' li addition-
al 1IHI MV capacity in the Eighth five \ear pLo 1-11uplee-is would
ho to -et up imidigetiously produced willd Electric Gemieiiitor (\VEG).
As ii conservative assessnient wind power 1 i0(tr16.11 in tilt' country
is abe-mt 2000 0 MW.
11.16 Wind Mills
eel energ y is used to roll winch mill which in turn drives a
geerati
n t' reduce clictiicit A wind mill converts the kinetic
energy of riroving air uitu nie--hanical inotiomu that can he either used
directly to run the machine or to turn the generator to produce
electricity \anous t y pes of wind nnlI are ii-- toiiuw -
a llniizorital axis wind mills
in I Multi blade type wind iiiill
I I'
I -'ui I type \VII1II 111 111
NON-CONVENTIONAL SOURCES OF ENERGY 609

• Propel lo! tvp( \V ad irnl I.

ti \'erticai axis wind mills
ionIU.- t y pe viiid inifl
(l) 1)arrieus t wind
ifld rail!.
\'ertical aX_is wind fflulls are ofsirnple design as coi 1ind to the
liorizoiittl aNi s . horizontal axis wind mills may IX' single l,lnd•d
double hlid&'d OF multi bladed.
Wind,d en rgv conversion devices are commonly known s win
turbine, because the y convert the energy of the wind stream into
energy of rotation : the component which rotates is called rotor. The
e F in turbine and rot or ire however often regarded as 1wilip
svno!lvIumH An electric generator is coupled to the turbine to
produce el ' ctrc nw er. The combiatio
n nof the Wind turbine alld
generator is sometime called as an Aero Generator.
ihe fr i ('lion of free fl o ,,% , wind power than can be extracted I
rotor is called the O\\ er coefficient.
1 Power coefficient
I, /I
where P 1 •.- Power of wind rotor
P ' Power available in the wind.
The in,Ixiinum theoretical power coefficient is 0.593.
The iis Iai rs calculated as follows
• Amount of air passing in f .iilt time
-A x
where A Area through which air passes
V \elocitv of air
- Mass of air traversing through area A swept
b y the rotating blades of' wind null t y pe generator.
Al = p A. V
where p = Density of air
K E. - Kinetic energy of moving air

= MV'2

= pA.V.V2 = pAV3

-Available wind power

Kinetic energy

POWER PLANT
610

= p.A.V3

The physical conditions in a wind turbine are such that onl y a

fraction of available wind power can be converted into useful work.
The power available in the wind increases rapidly with the speed
and hence Wind Energy Conversion (WECmachines should be
located preferably in areas where winds are strong and persistent.
Wind turbine generators have been built having capacity rang-
ing from a kilowatt or so to a few thousand kilowatts. Wind power
has been successfully used for cooling of homes, space heating, for
operating irrigation ; navigational signals and for offshore drilling
operations.
Fig. 11.11 shows the various parts of a wind-electric generating
power plant. They are
(i) Wind turbine or rotor
(ii) Wind mill head
(iii) Generator
(iv) Supporting structure.
The wind mill head supports the rotor housing the rotor bear-
ings. The moving air makes the blhdes to rotate and the electricity
is produced at the generator. Part A indicates transmission, Speed
increaser. Drive shaft and bearing brake clutch and coupling.

Sb

Wnd
turbine

Fig. 1.11

Fig. 11.12 shows schematic arrangement of a horizontal axis

type wind mill and Fig. 11.13 shows vertical axis type wind mill.
 NON-CONVENTIONAL SOURCES OF ENERGY
611

-.--Wind

Wind mill
head

'RotOr
Supportirg
Structure

Fig. 11.12

Gent

Fig. 11.13

11.16.1 Basic Components of a Wind Energy Conversion

System (WECS)

CONTROL COUPLING

POWER
PITCH
CONTROL GENERATOR
J
SPCEO
& TS
OtJTPUl
TEMPERATURE

POWER
ROLLER

Fig. 11.13(a)
 POWER PLANT

Fig. 11 13 a shows the basic components of a wind energy

conversion system.
Fig. 11.13.1 shows a horizontal axis multiblade t y pe wind mill.
The blades are made of metallic sheets. The y have high starting
torque and are economical.
Fig. 11.13.2 shows a horizontal axis wind mill I)utch type. The
blades are made of wood.

Fig. 11.13.1 Multiblade windmill.

Fig. 11.13.2 Horizonta l axis, Dutch type windmill,

NON-CONVENTIONAL. SOURCES OF ENERGY 613

Aeroturbines convert the Wind energ y to rotar y mechanical

energy. A mechanical interface consisting of a Step up gear and a
coupling transmits the rotar y mechanical energy to an electrical
generator. The output of generator is connected either to load or
power rid. The purpose of controller is to sense wind speed, wind
direction, shafts speeds and torques, output power and generator
temperature.
11.17 Site Selection for Wind Mill Units
Following factors should be considered while locating Wind
Energy Conversion Systems (WECS).
(i) Wind energy conversion machines should be installed at
sites where winds are strong and persistent.The most
suitable sites for wind turbines would be found where the
annual average wind speeds are known to be moderately
high. It is desirable to have average wind speed of about
3.5 - 1.5 in/sec which is the lower limit at which \VECS
generators start turning. An ideal site will be one where
a ,nooth steady wind flows all the time.
(ü) It is desirable to instal WECS at higher altitudes because
the winds tend to have higher velocities at higher a!-
i tudes
(iii) The ground conditions at the site shoLild he such that the
Raindations for WECS are secured. The land cost should
be low.
(iv) Icing problem, salt spray or blowing dust should not be
present at the Site as the y affect aeroturbing blades.
() The .site selected should he near to the users of generated
electric energy.
(vi) The site should be near to the road or railway facilities.
The best site for wind energy s ystems are found off' shore and
the sea coast and at mountains.
11.17.1 Performance of Wind Machines
The wind electric plants should make use of wind energ y in the
best possible method. The overall efficiency of an aero-generator is
calculated as follows

where q = overall conversion (' ihciencv of art aero-generator.

vjg Efficienc y of gearing
Efficiency of coupling
= Efficienc y of gencrattr
Efficienc y of aeroturbirie

614 POWER PLANT

- Useful shaft power output -

-
Wind power input
where Cp = coefficient of performance
The coefficient of performance of an aero-turbine is 0.593 for
horizontal axis wind
machines. - kA

Wind speed plays an im-

portant role in the power out-
C
put. The efficiency of wind
generator depends on design
of wind rotor and rotation
speed expressed as

where VT = Blade tip speed.

0 1 2 3 45 67
V = wind speed. VT
-
The value of VT is given by
Fig. 11.13(b)
VT = 71 DN rn/sec
where D = Diameter of rotor (metres)
N =lRotation frequency or rotation per second.

Fig. 11.13 (b) indicates the variation of coefficient of

mance (Cp) and the ratio YT- for some of the wind machi.
m

The curve A shows ideal efficiency for propeller type wind i.

The curve B shows the variation of coefficient of performance i
high speed two blade wind mill and curve C is for Dutch four am,
type 'wind mill.
11.18 Biomass
As a result of energy shortages in the years to come, interest in
the alternative fuel sources has increased considerably. Bioinass is
one of the such sources being considered. Biomass as a source of
energy has several advantages over dwindling fossil resources:
(i) It is renewable
(ii) It is environmentally clean
(iii) It is easily adaptable.
Biomass is produced in nature through photosynthesis achieved
by solar energy conversion.
Solar energy - Photosynthesis —* Biomass - Energy genera-
tion.
Energy from Biomass is obtained in following three ways
NON-CONVENTIONAL SOURCES OF ENERGY 615

(i) Biomass in its traditional solid mass (wood and agricul-

tural residue).
The biomass is burnt-directly to obtain the energy.
(ii) Biomass in non-tradition form (converted into liquid
fuel).
In this case the biomass is converted into ethanol and methanol
to be used as liquid fuels in engines.
(iii) To ferment the biomass anaerobically to obtain a gaseous
fuel called bio-gas.
Biomass resources are as follows
(i) Concentrated wastes like
(a) industrial wastes
(b) municipal solid
(c) manure at large lots.
(ii) Dispersed waste residue like cro esIdue, disposed
manure.
(iii) Harvested bio-mass, standing bio-mass.
11.19 Biomass Gasification
Biomass gasification is used to produce a gas for burning pur-
poses. Gasification is a process which converts a solid or liquid into
a gaseous fuel without leaving a carbonaceous residue. Gasification
is carried in a gasifier which is an equipment that can gasify a
variety of biomass like wood waste, agricultural wastes. Biomass
gets dried, heated pyrolysed, partially oxidised and reduced. The
gas produced in the gasifier is a clean burning fuel. A gasifier can
be easily operated, is reliable and its maintenance cost is low.
11.20 Tidal Power
Tide is periodic rise and fall of water level of the sea. In about
24 hours there are two high tides and two low tides. These are called
as semi-diurnal tides. The rise and fall of wall level follows a
sinusoidal curve as shown in Fig. 11.14 with point M indicating the
high tide point and point N indicating the low tide point. The
difference between high and low water levels is called the range of
the tide.
Tides occur due to the attraction osea water by the moon. These
tides can be used to produce electrical power which is called tidal
power. World's first tidal power plant was commissioned at Rance
in France. This plant is of 240 MW capacity.
Following are the important points for the selection of location
of tidal power plant:
 POWER PLANT
616

4
M
I +

Sn,

I He

____12h.25rnir. H
Fig 11 4

The tidal range at the (lesireil locattoti 511001(1 he vlc(11int(

throughout theVIar.
I'he Site seIect(d . for tom1 j)OWCr plant sh(flhl(l I) free fr
the wave attack of sea.
410 I There ShOUld be 1)0 appreciable ciiiiige Ui tidal pattern at
the proposed site
or The site at which tidal 1 ower plant is to he 1,cateil hould
riot have excessive Seilim' nt load.
11.20.1 Tidal Power l'Iaiits
(ho Iae Cale up and (Iowa nlovcili(flt olson watci , !r(o(1tS
an unhiiuted source of energ y . It soilie part of this U'I'V can
Iieconverted j I to elect rical energy it wouldheofall important ource
of power.
follows.
'Flu. three iii iti collwollellt oLt tidal pOIVO r plant are
(i) The d y ke to form basic or basin
ii) Sluice wa y s from the basin to the s&'tL 111(1 LLLC L''TSTi
ij) The power house.
'The turbine, electric generators and other auxiliary equip'
nicniL are the main equipments of power house.
11.21 Classification
'lid al power plants are classified on the basis ol' n urtiher ot basin
used for the power gunerati()n1. The y are further sulnitviil'ol a,
way or two way svt'm as per the c y cle of operation for power
ge iieration.Variou s tv pes of tidal power plants are as follows
(t) Single basinsvstenl
Vi One w•iv sYstem
i/i Two wa y v1'i:
( C I [lvi) WI'.' '.l ith pump t('i'ilg&'.
ill r)OUi)l,' t)0.StI1 % stylli
Sinijili' double i 1,an svtinl
i/i I) ui l I .oi Ii WIth p0111 pIng
 NON-CONVENTIONAL SOURCES OF ENERGY 617

OykQ

Power
Basin

523

slut's

Fig. 11.15

Fig. 11.15 shows a single basin one way tidal power plant. In
this plant a basin is allowed to get filled during the flood tide and
during the ebb tide, the water flowing from the basin to te sea
thruugh the turbine and generates power. The power is iilahle for
a short duration during ebb tide.
In single basin two way tidal power plant the power is generated
hoh duing flood tide as well as ebb tide. The direction of flow
through the turbines during the ebb and flocd tides alternates but
machine acts as a turbine for either direction of flow.

Dyhe
Low basin
High
basin

Sluice\,<
Sea

Fig. 11.16

Fig. 11.16 shows a double basin one way tidal power plant. In
this plant one basin is intermittently filled by flood tide and ocher
is intermittently drained by ebb tide. For some more details of tidal
power see Chapter 1.
11.22 Advantages and Disadvantages of Tidal Power
Advantages
Various advantages are as follows:
(i It is free from pollution.
(it) It is ir.exhaustible and does not depend on rain.
(iii) Tidal power plants do not require large area of valuable
land because they are located on sea shore.
—41

618 POWER PLANT

(iv) Tidal power has a unique capacity to meet peak power

demand effectively when it works in combination with
hydropower plant or thermal power plant.
Disadvantages
(i) The output varies because of variation in tidal range.
(ii) The power transmission cost is high because the tidal
power plants are located away from load centres.
(iii) Sedimentation and silteration of basins are the problems
associated with tidal power plants.
(iv) Because of variable tidal range the turbines have to work
on a wide range of variable head.
(v) Capi t al cost of the plant is high.
11.22.1 Regulation of Tidal Power Supply
The tidal power plants generate unregulation power from tides.
Some of the methods which help to generate regular power from
tidal power plants are as follows
(i) Combining two or more tidal schemes with different tidal
phases.
(ii) Providing two basins tcñe high and one low) having inter-
connection with each other and with the sea.
(iii) Tcrinfer-onnect tidal power plant with (a) steam power
plant (b) pumped storage hydro power plant (c) hydro
power plant.
11.23 Biogas
Biogas is produced by the decomposition of animal wastes, plant
wastes and human wastes. It is produced by digestion, pyrolysis or
hydro-gasitication. Digestion is a°biological process that takes place
in the pbsence of oxygen and in the presence of an aerobic organisms
at ambient pressures arid temperature of 35-70C. The container
used for digestion process is called digester. There are two sig-
nificant, temperature zones in an aerobic digestion. It is observed
that two types of micro-organisms mesophilic and thermophilic are
responsible for digestion at the two temperature ranges. The op-
timum mesophilic temperature is around 35°C while optimum ther-
mophilic temperature is about 55C. The gas temperature falling
very steeply when the temperature goes below 20°C. It is easier to
maintain the temperature of the digester at the mesophilic range
rather than at the thermophilic range. See some more details in
Chapter 9.
11.24 Classification of Bio-gas Plants
Various types of bio-gas plants as follows
G) Continuous and batch type
( ii) The dame and drum type
NON-CONVENTIONAL SOURCES OF ENERGY 619

(iii) Different variations in the drum type.

Continuous types. The continuous process may be completed
in a single stage or separated into two stages. Fig. 11.17 shows a
single stage continuous process type digester. In the digester the
entire process of conversion of complex organic compounds into
biogas is completed in a single chamber. This chamber is regularly
fed with raw materials while the spent residue keeps moving out.

Scum layer
lays,

inlet Actively digesting

slurry

sludg :Sludge
Outlet

Fig. 11.17

Fixed dome digester. Fig. 11.17 (a) shows circular fixed dome
digester plant. The fixed dome is made of masonry structure. The
digestion takes place in the masonry well. The gas generated is
taken out from the top. Are movable man hole cover sealed with clay
is provided.
GAS PIPE
N HOLE JLOOSE COVER

INLET
PIPE
DISPLACEMENT
UL

O
, UTLET
SLURRY
Fig. 11.17(a)

11.24.1 Factors affecting bio-digestion or generation of

gas.
The generation of gas in bio-gas plants depends upon the follow-
ing factors
pH or hydrogen ion concentration. A pH value between 6.5
and 8 is best for fermentation and gas formation.
(ii) Temperature

620 POWER PLANT
(iii) Loading rate
(iv) Seeding
(v) Solid contents of the feed material
(vi) Type of feed stocks
(vii) Nutrients
(ciii) Pressure
(ix) Stirring and mixing of the contents of the digester
(x) Acid fothing and methane forming bacteria.
Flexible bag digester. This digester is made of plastic
material and can be easily installed. Howeverthe life of this digester
is small. Fig. 11.17 (b) shows a flexible bag type combined digester.
GAS
PIPE GAS

LEVELLE
SURFACE

SLURRY
Fig. 11.17(b)
11.24.2 Thejmionic Generation
This method of power generation utilises the thermionic emis-
sion effect which means emission of electrons from heated metal
surface. The energy required to extract an electron from the metal
is called work function of the metal and depends on the nature of
metal and its surface condition.
11.24.3 Thermionjc Converter
The thermionic converter utilises thermionjc emission effect. It
consists of two metal (or electrodes) with different work functions
sealed into an evacuated vessel on heating one electrode, the
electrons are emitted which travel to opposite colder electrode called
collector or anode.
The hot electrode (emitter or cathode) emits electrons and
acquires a positive charge whereas colder electrode collects
electrons and becomes negatively changed. A voltage (or clectro
motive force) thus develops between the electrodes and a direct
current starts flowing in the load connected as shown in Fig. 11.17
(c).
The electrode with the large work function is maintained at
higher temperature than one with the smaller work function.
NON-CONVENTIONAL SOURCES OF ENERGY 621

C.VTAT ED
1101 SF-L.
ELE.CTROI

COLD
ELECTRObE

RON ,

ECIERNAL D.C. LOAD

Fig. 11.17 (c

The thermionic converter will continue to generate electric

power as long as heat is supplied to the emitter and a temperature
difference is maintained between it and the collector.
By thermionic converter the currents that can be produced are
extremely small except in special case of metals at high tempera-
tures. To achieve a substantial electron emission rate and hence a
significant current out put as well as a high efficiency the emitter
temperature in a thormionic converter containing cesium should be
at least 1000 C. The efficiency is then about 10 percent. Higher
efficiency can be obtained by operating at still higher temperatures.
(P) is given by
Electric power
P E.I.
where E = voltage
I current.
11.25 Types of MHD Generators
An MUD generator is a device for converting heat energy
directly into electrical energy without a conventional electric gener-
ator. The major advantage of MILDgenerator (converter) is that it
can take better advantage of high temperatures attained in the
combustion of a fossil fuel.
MHD generator are of two types:
(i) Open cycle generators
(ii) Closed cycle generators.
In open cycle the working fluid after generating electrical energy.
is discharged to the atmosphere through a stack. In closed cycle the
working fluid is continuousl y recirculated.

622 POWER PLANT

11.25.1 Open cycle generator

Fig. 11.18 shows schematic arrangement of open cycle MHD
generator. In this generator the fuel (oil, coal, natural gas) is burnt
in the combustion chamber. Preheated air is supplied to the com-
bustion chamber to burn the fuel. The hot gases from the combustion
chamber is then seeded with a small amount of an ionised alkali
metal (cesium or potassium) to increase the electrical conductivity
of the gas. The g eed material which is generally potassium car-
bonate is injected into the combustion chamber, the ionisation of
potassium taking place by hot combustion gas at temperature of
about 2300-2700C. The hot pressurised working fluid leaves the
combustion chamber and passes through a convergent divergent
nozzle. The gases coming out of nozzle at high velocity then enter
the MHD generator. The hot gases expand in the generator sur-
rounded by powerful magnets. The MHD generator produces direct
current which can be convefted into alternating current with the
help of an inverter.
Iftertff

Uok..up Hd

Fig. 11.18
11.25.2 Closed cycle systems
-J
There are two types of closed cycle MHD generator systems
(i) seeded invert gas system
(ii) liquid metal system.
Fig. 11.19 shows a liquid metal system. In this system a liquid
metal is used as working fluid. The liquid potassium coming out of
breaker reactor at high temperature is possessed through a nozzle
to increase its velocity before passing to MHD generator. The liquid
potassium coming out of MHD generator is passed through a heat
exchanger (boiler) to ue it q remaining heat to system to run a
turbine and then pumped back to the reactor. There are lot of
operational and constructional difficulties in this system. MUD-
NON-CONVENTIONAL SOURCES OF ENERGY 623

GEN indicates MHD Generator, Inv. means invertor and S.T. indi-
cates steam turbine.
Condenser

Magnin F..6
wIhr
140
Gen
Nozzle Seperolor
eoCiOI

Gen.roor
Steam

Pump Liquid

Fig. 11.19

11.25.3 Advantages of MHD Systeni

The various advantages of MI-ID Steam power plant are as
follows:
(i) The size of the plant is consiaerably smaller than conven-
tional fossil fuel plants.
(ii) It can be started and put on line within few seconds.
(iii) It has high thermal efficiency (50-55%).
(iv) It proves instantaneous stand by power. It can be used
most economically as peak load plant.
(v) Closed cycle produces power free from pollution. Some
more details are given about MHD system in Chapter 9.
11.25.4 Combination of 4iID power plant and steam
power plant
There is substantial improvement of thermal efficiency when
MHD power plant an steam power plant operate together.
The cost of a combined MI-ID and conventional power plant ui'
be too high and is not economical at low power generation. But
higher capacities, the cost will not he more than 20% that of a
conventional coal-fired thermal power plant. The open cycle coal-
fired MHD system, when combined with a conventional steam
power plant, can develop efficiencies upto 50 17r. Closed cycle MHD
may have the potential to approach the efkiciency of open cycle.

.., ..,..'.,. . . - .--

 624 POWER PLANT
It is economical to generate electric power by a combined MHD
steam power plant. The potential of MHD to operate at higher
efficiencies substantially reduces the thermal waste discharge and
thus stresses the need for development of more practicable MHD
devices. The high power density feature and rapid start up
capability enable the MilD for situationswhere compact electrical
sources are required.
MI-ID power-plant can also be operated in combination with
(i) gas turbine power plant
or (ii) 11 UL power plant

The '- triotis hurdles in the progress of MHD power plant are as
follows
(i) Development of super conductor magnet.
(ii) Materials to withstand high temperatures.
(iii) Efficient conversion of 1)C to AC crosioiik,ss electrodes.
(ic) Heavy losses in the power electrodes.
11 26 Fuel Cell
l"til rell details are discussed in Chapter 9.
11.27 Thermo-Electric Power
This method of producing electrical energ y uses see-beck effect
according to whici an is produced when in a loop of two
dissimilar mçtals the juncttons arc kept at different temperatures
The efficienv of thermo-electri 'r( , nerator depends upon the
temperatures ul hot and cold ju'.ctions.

Material-i
iot coi
Junctjøt, Junction
14 Matericil-2
cm
Fig. 11.20

Fig. 11.20 shows basic principle ofthermo electric powergenera-

tion. If temperature difference is maintained between the hot and
cold junctions an electric current will flow round the loop. The
magnitude of the current produced depends on the following fctors:
(i) Materials used
(ii) Temperature difference of the junctions.
If the circuit is brokenz, pen circuit voltage (V) appears across
the terminals of the break as shown.
Thermo e.rn.f V produ.t the device is given l)V

- a. x
NON-CONVENTIONAL SOURCES OF ENERGY
625
where = Seebeck coefficient
AT = Temperature difference
=T2-
where T2 = temperature of cold junction
= Temperature of hot junction.

+Ve

cts 0

—v e

Ternperoture(°K)
Fig. 11.21

Fig. 11.21 shows variation ofseebeck coefficient with tempera-

ture for the following materials.
(ci) Metal
(b) P-t ype semi conductors
(c) N-Ty pe semi conductors
The coefficients is positive for P-type semi conductors and nega-
tive for N-type semi-conductors.
11.28 Thermo-electric Power Generator
Fig. 11.22 shows a simple arrangement for utilising the thermo
couple arrangement for power generator.

Hot
A
Cold

Fig. 11.22


626 POWER PLANT

The materials A and B are joined at the hot end. An electric

voltage (or electromotive force) is then generated between the cold
ends. A direct current will flow in a circuit or load. Connected
between the ends. For a given thermo couple the voltage and electric
power output are increased by increasing the temperature dif-
ference between the hot and cold ends.
To increase voltage and power several thermo-couples are con-
nected in series in a thermoelectric power generator as shown in
Fig. 11.23.
Hot Hot Hot

P:E
B

D.C. Load
Fig. 11.23

The direct current generated can be converted into alternating

current by an inverter. -
11.29 Thermo-electric materials
The conmonly used materials in thermoelectric power gener-
ator are as follows
(i) Lead telluride (ii) Bismuth telluride
(iii) Bismuth sulfide (it') Zinc antimonick
(v) Cesium sulfide.
An index used in rating thermo-electric material is calkd
figure of merit.
Tle figure of merit (z) depends on properties of materials. A high
value of figure of merit i g obtained by using materials of
(i) large seebeck coefficient
(ii) small thermal conductivity
(iii) small electrical resistivity.
Table 11.1 indicates figure of merit for sonic of thermo electric
materials.
Table 11.1

B i smuth telluride 4 . 10_

I.e.(I telluride 1 II)

Ctsiun uIflde i • yr '

(e,iniuiitIurdt
Zinc •,1iiI,1onId - 10 3
NON-CONVENTIONAL SOURCES OF ENERGY
627

Example 11.1. At a tidal site the observed difference between

the high and low water tide is 9 in. The basin area is 0.6 sq. km .
which can generate power for 3 hours in each c ycle. The average
available head is 8.5 m. and over all efficiency of generation is 75(7(
calculate the yearly power out put. Specific weight of sea uciter is
1025 kg/m:3 and there are 705 full tidal cycles in a year.
Solution.
A Area of basin
= 0.6 sq. km . = 0.6 x 10 6 rn 0
h = Difference in water tide = 9 in
v = volume = A x h
= 0.6 x 106 x 9 = 5.4 x 10 6 m3
T = Time = 3 hours
Q = Average discharge
V

=o500 m3IS
to = specific weight of water = 1025 kg/rn3
P = Power
=X
75 T1

where H = Average available head = 8.5 ni

II = Efficiency = 0.75

1025 x 500 x 8.5

P = x 0.75

= 4.3 x 104H.P.
E = Energy per cycle
= P x 0.736 x T
= 4.3 x 10 x 0.736 x 3 = 9.3 x 10 kWh.
N = Number of tidal cycle per year 705
Total out put per year
= E x N = 9.3 x 10'x 705
= 6.4 x 107 kWh.

628 POWER PLANT

Example 11.2. For a thermo-electric power generator following

para meters are given
Temperature of sink = 280K
Temperature of hot reservoir of source = 620K
Figure of merit for the material = 1.8 x iO k'
Calculate -
(a) Efficiency of thermo-electric power generator
(b) Carnot efficiency of generator.
Solution.
TH = Temperature of hot source = 620°K.
Tc = Temperature of sink = 280K
Z = Figure of merit
Figure of merit depends on the properties of thermo-electric
materials used. The high val' of figure of merit can be obtained as
follows
(i) by using materials of large seebeck coefficient
(ii) small thermal conductivity
(iii) small electrical resistivity
Z=1.8x103k'
Tj = Efficiency of thermo electric generator
F TH- Tcl M-1
L
TH
TC

wher M=[1 +(TJi+TC)]

F 1.8x103
(620+280)j = 1.32
2
Tcl I M
[TL^
l Mi-
--
Tc
[620-2801 1 1.32-1
- 620 i 132+
280
0.12 = 12
 NON-CONVENTIONAL SOURCES OF ENERGY
629

Tic = carnot cycle (reversible) efficiency

i - TC 0-280
X 100 = 62 100

= 52%.

11.30 Methods for Maintaining Biogas Production

The techniques usually suggested for maintaining the biogas
production are briefly summarized as follows:
1. Insulating the gas plant. To reduce heat losses from the
digester, the external surface of the digester is adequately insulated
using different materials like mineral wool, aluminium cladding,
fibre glass, straw etc.
2. Composting. In this process the heat released in aerobic
composting of agriculture residues around the annular ring in the
upper part of the digester could be utilized to raise the digester
operating temperatures. It is observed that a temperature rise of
8-10°C above ambient temperature even during the coldest season,
takes place as a result of heat released from composting manure and
straw. Under optimum conditions of moisture, composting f com-
plete in 3-4 vees and the released heat varies with time markedly.
Small quantity of water in the straw around the digester is added
later when the drop in temperature occurs and therefore the
temperature of the composting material and digester contents can
be kept relatively at constant temperature.
Geothermal Sources
Geothermal energy is the energy which lies. According to
various theories the earth has a molten core. The fict that volcanic
action takes place in many places on the surface of the earth
supports these theories. The steam and hot water comes naturally
to the surface of the earth in some locations of the earth.
Various geothermal energy re-sources are as follows
1. Hydro thermal systems
(a) Vapour dominated or dry steam fields
(b) Liquid dominated system or wet steam fields
(c) Hot water fields
Hydrothermal systems are those in which water is heated by
contact with the hot-rock, as explained earlier.
In Vapor-dominated systems, the water is vaporiz' I into
steam that reaches the surface in a relatively dry condition a bout
200°C and rarely above 7 kg/cm 2 (8 bar). This steam is the st
suitable for use in turbo electric power plants.
 630 POWER PLANT
I
In Liquid-dominated systems, the hot water circulating and
trapped underground is at a temperature range of 175 to 315°C.
When tapped by wells drilled in the right places and to the right
depths, the water flows naturally to the surface or is pumped up to
it..
2. Geopressu red Systems. These resources occur in large,
deep sedimentary basins. The reservoirs contain moderately high
temperature water (or brine) under very high pressure. They are of
special interest because substantial amounts of methane CH4
(natural gas) are dissolved in the pressurized water (or brine) and
are released when the pressure is reduced. Geopressured water is
tapped in much deeper underground acquifers (it is a water-bearing
stratum of permeable rock, gravel or sand), at depths between about
2400 to 9000 M.
3. Petro-thermal or Hot dry rocks (HDR). This represents
by for the largest resource of geothermal energy of any type, as it
accounts for large per cent of the geothermal resource. Much of the
HDR occurs at moderate depths.
4. Magma resources. Thee resources consist of partially or
completely molten rock, with temperatures in excess of650°C, which
may be ecoóntered at moderate depths, especially in recently
active volcanic regions. These resources have a large geotheri
energy content, but they are restricted to a relatively few 10
5. Volcanoes.

PROBLEMS

,11.1. What are non-conventional sources of energy?

11.2. Write short notes on the following:
(a) solar energy
(b) solar radiation measurement
(c) solar constant.
11.3. Describe various types of solar collectors.
11.4. State the advantages and disadvantages of current rating type
solar collectors.
11.5. Describe solar thermal power generation.
11.6. Write short notes on the following:
(a) Low temperature solar thermal power generation
(b) Tower concept type solar power plant
(c) PV cells
(d) Solar pumping.
11.7. What is wind energy? State the characteristics of wind energy.
11.8. (a) What is a wind mill? What are various types of wind mills?
(b) Describe a horizontal axis and vertical axis wind mill.
NON-CONVENTIONAL SOURCES OF ENERGY
631

11.9. State the factors to be considered while selecting the site for wind
mill.
11.10. Write short notes on the following:
(a) Bto mass
(b) Bib gas
(c) Fuel cell
(d) Power coefficient of wind mill.
11.11. (a) Describe MI-ID generator principle.
(b) Describe an open cycle and a closed cycle MHD power gener-
ator.
11.12. What is tidal power? What is tidal range? State advantages
and disadvantages of tidal power.
11.13. (a) Name the main components of tidal power plant.
(b) State the points to be considered while selecting a site for
tidal power plant.
11.14. I-io.v do you classify tidal power plants. Describe
(a) single basin one way
(b) double basin one way tidal power plant.
11.15. How do you classify bio-gas plants? Describe a continuous type
bio gas plant and fixed dome type plant.
11.16. Describe various parts of a wind electric generating plant.
11.17. State the advantages of MHD p ower generation system.
11.18. Write short note on the following:
(i) Thermo-electric power
(ii) Thermo-electric power generator
(iii) Materials for thermal electric power generator
(iv) Figure of merit.
Cv) Basic components ofa wind energy conversion system.
11.19. Write short notes on the following:
(i) Performance of wind machines
(ii) Overall efficiency of aero-generat.or
(zn) Efficiency of aero-turbine
(iv) Methods of regulation of tidal power.
11.20. Write short notes on the following
(i) Methods for maintaining bio-gas production
(ii) Geothermal energy resources.

. :
 12

Environment Pollution and its

- Control

12.1 Introduction
The electric power demand is continuouLy increasing. The
electric utilities and faced with simultaneous demand for increased
power as well as problem of harmful imp.rities which are ejected
into the environment surrounding the utilities. The power engineer-
ing specialists should develop and implement the effective means in
the field of environmental protection from harmful effluents of
power plants. The effects of power plant pollutants on the environ-
ment are raifily on the air and water and to a lesser extent on the
land.
Most of the pollution in the cities can be avoided if power plants
are located outside the city boundary. The pollution from the nuclear
power plants is radio-active waste in the form of gases, liquids and
solids whose radio-active property may retain number of years. The
dumping and leakage of these wastes are major problems in ouclear
power plants.
dombustion generated pollution is by far the largest man-made
contributor to atmospheric pollution. The principal pollutants
emitted by fossil fuel tired boilers in large power plants are products
of incomplete combustion such as combustible particulate matter
CO, NOx, SOx and fly ash and particulates formed due to the use of
fuel additives.
The methods of control of pollutants include:
(1) low excess air combustion (excess air 2%).
(ii) staged combustion with heat removal between stages.
(iii) flue gas re-circulation.
(it') gasification of coal and residual fuel oil.
(v) fluidised bed combustion.
ENVIRONMENT POLLUTION AND ITS CONTROL 633

unproved atmms.ttion and aeration 01 the tUI sprLv iii

case of gas turbine combustors.
12.2 Stean. rower Plant Pollutants
The thornia l power plants burning conventional fuels (ml, oil,
gas I contribute to air pollution in a large measure. The influence of
thermal power plants on the surroundings is determined b y their
ejection of flue gases, heat and contaminated waste watrs. Though
thermal power plants are not among tliu worst contitmint,its of
water basins in terms of scope and composition (It tutu' liquid
wastes, their discharge into water basins call great harm if
protRr nems iire nit taken for water protection
The ma in pollutants front power plants are fdlnws
U I ('0 (ii) CO2
Ut') Nitrogen oxides such as
ii I Nitrogen dioxide (NCh) (b) Nitric oxide NO
ci l)ust (v z) Fl y ash
It Ii incomplete combustion of fuel in furnaces, carbon
niunoxide ' ( ) 1, hydro carbons C1 4 , C 2 1 L etc. are produced 'Fin' CO
is injurious to human health as it combines with hacoinglobin ill the
red bit tod corpuscles and thus interfere tvi t h their norm a'tii uct ion
of supplying ox y gen to the blood tissues.
The thermal power plants contribute substantiall y to ('02 etnis-
sions CO has vt'rv harm ful efb .ct on atopht'nic
in climate which
could turn- b.'rtilt' land into deserts. Therefore the implications and
control oFCO2 nOCCI pnioritvstudv. Sulplairdioxide(SO ) is the main
pollutan t from steam power plant The priin.-urv source ot SO in the
atiflus re is the combustion of bituminous coal and residual oil
fuel. Vegetables are most Sensitive to the content ofSU .' gas ill
atmo " phere The toxic effect of SO 2 gas is associated with deteriora-
tion of the surfaces of leaves or needles due to destruction of their
chloroph y ll. Nitrogen dioxide (NO) and nitric oxide (NO) are t,ften
ni'ferrcd collectively as nitrogen oxides.
Nitrogen oxides are toxic and produce a sharp irritatia effect.
People living in NO 2 contaminated areas suffer front duced
respiratory function and have a higher incidence of re- iratorv
diseases.
Acid rains is another menace caused ijv the t h,'rnct ever
plants. Three main constituents of flue gases which mainly
acidit y of rains are SO 2 and nitrogen oxide.,. In the atmo ire
SO ' is fairl y readily converted into sulphuric acid (H 2SO4 1 wheri.'iis
the nitrogen oxides get converted into nitric acid (FINO. 1 ). During
rain y season the acid formed in the atmosphere falls on the ground
in the form of rain called acid rain. The effect of this rain is o
encase the acidit y of lake, well water and water of t1o\vin river.
634 POWER PLANT

Ili ginraI SO. citrilniti's about 60Y? of the acidit y whereas

nitrogen oxdes cijOtrlute 3.5; rH (liOXi(l( (CO ' )also cause the
rain to be acidic but to a very small extent.
Maximum l)ernnsIl)le limit if nitrogen oxide is 005 to 0.1 ppm.
The further detrimental cffeut of acid runs is the reduction of
ground t'rtilitv and crops void smoke, dust and fl y ash carried by
flue gases also produce inirius effects oil health. The
i I uantmtv of ash (Q carried ott lv flue gases per kt. ofhiel burnt and
taki Into accotirit unhurnt ('a rhori i.- haincl b y the fI lowing for-

j) 1
Q = A . lt)i) lt)0 1'
where A. = Fraction (if ' solini particles carried off from furnace with
flue gase-..
W = Ash content ofworkiiig Itlass of fuel in c
1'= Content of conil,astili . le in fl y ash in
The ash is also a problem as it also emits heat to the atmosphere
as well as small parto:l'. of ash are carried Lv the air. Large
quantity of heat d,sc}iiig'd to the at iiosphrn- also is a cause of
pollution. 'Toxic substance, contained in flue oases discharged from
stacks iii' ther.nial po'.'. or pLiiits (an produce harmful effects on the
Whole complex of living nature or the bio-sphere The bin-sphere
comprises the atmosphere la y er near the earth's surface and upper
la y ers of soil and water basins.
Depending on the t y pe of fuel and capacit y of boiler the ash
collection from (hernial power plamits can be effected b y the following
devices
Ash'collecter (it) Fl y ash scrubbers
(iii) Electro static pr.'cipit.at.ors.
Fly ash, cinders, various gases and smoke discharged from the'
stack become atmospheric contaminants. Gases diffuse in all direc-
tions. The path followed b y the flue gases depends upon the thermal
and dynamic properties of gases and wind flow past the stack
The various variables affecting the area over which flue gases
ronstitme:its will settle out are as follows
(i) Stack height
(ii) Stack exit gas velocity
(iii) Wind velocity
(im.') Gas temperature
(m.) Particle size
(m.'i ) Surrounding topography.
ENVIRONMENT POLLUTION AND ITS CONTROL 635

istible content of stack dust is important in pollution

S I'S settlecloser to tile plant. Flue gases dust quantity
and qualo upend oil t y pe of fuel and burning equipment -
The '. tifistanues like CO, CO. SO., SO S , NO, ash, hydro
carin)us etc contained in the waste Vases may be harmful fr
v ) -i-tation. imals, water basins and people even In \'iV
co1fltriltiis i)I above the specified safe limits. It is therefore
desirable to remove th -e toxic substances from flue' g;mses as tar as
1)0 ii I e
The control of tIn' atniisphere at steam power plain Is iiiamlv
aimed at illililinising the di'charge of t'xe, s htaiic'&.'s into the
atillospliel k . .
Enviroiinu-ntal issues in power sector are of major importance
III our cociutrv ilii.' ti the signitieince of electric power ui the
(c()flOiIied&'V&'ln'piiU'Iit process. Sevent y percent of powergi.'nei ,itioi
Ili our coulitmi i coal based, The ('nVirOflfl)enti)l imupii.'t of powir
production iii general and coal based power production in pai'ticular
are sei. jnus in imrrns of human health.
Coal based power gi-iu'rati in affl'cts air, land and wat ('rr&e)cIr
ces. Emission of particulate matter, sulphur dioxide and oNldt. of
nitrogen causes air pollution. Accumulation cit ash at powt'rstations
pre-i.- iupts land and endangers both ground and surface water. The
asnclt id increase in coal production to meet the additional
demand can degrade more land, deplete water resources and cause
water jmllut Ion.
12.3 Control of Pollutants
The control oft 1w atmosphere at thermal power plants is mainly
ainii'd at milillIusilIg t iW discharge of toxic substances into the
atmosphere. The task of preserving the purity of atmosphere and
water basins is of national significance. Thermal power plants
consume about more than 1/3 of all the fuel produced and thus can
significantl y affect the local environment and the whole biosphere.
Large condensation plants are among the greatest sources of heavy
ejections of contaminants into the atmosphere. The effects of par-
ticulate matter in the atmosphere surrounding steam power plant
are man y and varies Adverse effects on health, climate and water
basin are quite serious Adverse health effects are associte(l with
SO concentrations Acid fall out in the form of acid rain is one of
the more seriou environmental hazards of increased concentra-
tions of sulphur and nitrogen oxides in the atmosphere.
As regards tllirifl.iI power plants the state of environment
around t Item depuiui. tot the fellowing
i i i kind (It fuel used
636 POWER PLANT

(ii) organisation of fuel combustion

(iii) operation of dust collecting and gas cleaning plants
(iv) devices used for ejection of flue gases into the atmosphere
ft ) the influence of thermal power plants oil surroundings
is also determined by their ejection of flue gases, heat and
contaminated waste waters.
During high temperature combustion o'gaseous or liquid fuel
the pollution of the atmosphere by solid particles, Co and SO 2 can
be kept minimum b y suitable organisation of the following
(1 corn I) U S lion process

(ii) burning gas with excess air

(iii) choice of proper length and diameter of furnace chamber
(iv) corrc't stabilisation of flame.
It is necessary to minimise the emission ofSQ into the atrnos-
phere from thermal power plants as its contribution to pollution is
maximum.
Large amounts otair pollutants are Pr0d1ced with coal combus-
tion than an y other fuei Therefore research is continued on convert-
ing coal to cleaner and more convenient gaseous and liquid fuels.
The permissible niaximum concentrations ofS() at ground level
are 0.05 to 0.08 ppin for 24 hours, 0.12-0.2 ppm for 1 hour and
0.1-05 ppm for live minute . The maximum permissible limit of
nitrogen oxidis 0.05 to 0.1 ppm,
12.4 Control of Particulate Matter
Steam power plants generally use the following mechanical
arrestors for the removal of solid particles
(a) Fabric filters b. Electrostatic precipitators.
It is economically feasible to obtain a high degree of pollution
control over particles which are larger than 2 to 3 micruns. About
95 17c of fly ash under 2 microns in diameter are difficult and costly
to be removed. The particulate removal is the major problem (so far
as cost is concerned) in power plants using pulverised fuel. Irrespec-
tive of all steps taken to remove the particulates from gases before
going to stack, about 1% is always discharged to the atmosphere.
The particulates effects can be reduced b y using the hllowing
(a) coal cleaning
(b) using improved electrostatic precipitator design
(c) to control the (lust within allowable limit. It can he done
by increasing the height of chimney so that the dispersion
will be on the Larger area thus reducing concentration,.
Electrostatic precipitators are used to remove the dust particles
from flue gases. Details of electrostatic precipitator are given in
Chapter 3. A combination of mechanical and electrostatic
precipitators cm r'iv e more than 99.5% of the particulate matter
ENVIRONMENT POLLUTION AND ITS CONTROL 637
from the effluent gases. Fl y ash scrubbers are used to remove fly
ash. A fl y ash scrubber is explained in Chapter 3.
Cinder catcher and cyclone dust collector are explained in Chap-
ter 3.
Furnaces burning coal in suspension (pulverised coal burners
and spreader stokers)throw (lust in the form of fly ash and collectors
should be installed in the breeching to remove ash particles.
Smok y atmosphere is less healthful than smoke free atmos-
phere. Smoke has deadly effect oil vegetation principally be-
cause of sulphur products it carries. Smoke Corrodes metals and
darkens paints. Fuels should be burnt completely to reduce quantity
of dust particles in the flue gases.
12.5 Control of S02
Solid fuels contain sulphur in the following three forms.
(i) as inclusions of pvrite (FeS2)
(ii .) sulphur in molecules of organic mass of fuel
(iii) sulphate sulphur (in sulphur salts of calcium and alkali
metals).
502 is one of the principal toxic component which ma y pollute
the atmosphere Substantially.
Following methods are used to reduce the quantit y of sulphur
dioxide (SO 2 ) produced during combustion of fuel
(a) Desulphurisation of fuels. Decreasing the sulphur con-
tent in fuel is called desulphurisation process. This process can
remove a substantial amount of sulphur from fuel.
Following three methods are used to remove sulphur from
coal:
(i ) Chemical treatment (ii) Froth flotation
(iii) Magnetic Separation.
These processes leave the coal unchanged in form. In chemical
treatment coal is leached with an aqueous solution of ferric sulphate
at temperature in the range of 90-130 'C. In froth flotation process
the coal is suspended in water through which air is bubbled. The air
bubbles tend to attach themselves to the coal particles rather than
to the mineral matter. The mineral waste falls to the t)uttom and is
discharged.
In magnetic separation the fine] vru s hed coal is passed through
a strong magnetic Field which removes pyrite (Fe52) from coal. Coal
itself is lion-magnetic.
Ihe sulphur from liquid petroleum fuels is generall y rerucived
b y reaction with h y drogen gas in the presence of a catal st at i
 b38 POWER FLANT

iiitcl III ll I% arid pressur ilnu

ur i.
; o hvlrgic ulplude which is then removed.
I) Ti ue low sulphur fuels. To ue low -cii phu r content fulls
I.-; the coinlnercilllv proved nwans to cntiol St) erriiswu into
itn 105 pIer
Use of tall stakes. To prevent air 1coltutcirc with St ) tall
chiri III ey are UCd to clisper'e flue gases OVI'I larger au-u
(C/) Cl caning of flue gases. Commonly thc thu. ncctlud. used
to remove SO from the flue gases are as fotIow
1. Wet scrubbing 2. Solid absorbent
3. Catal ytic oxidation.
It is observed that to remove SO 2 from flue gases is more
economical its compared to removing sulphur from coal. Methods
used to prevent air pollution with SO 2 are different for gaseous,
liquid and solid fuels. It is advisable to remove 1125 from natural gas
before burning it
The sulphur content of liquid fuels can be reduced b y following
wa y s subjecting the fuel to a high temperature either with the use
of (;xidant- gasification) or, without them (Pyrolysis).
The process of gasification is effected at a high temperature
9OO 130 (') with it admission of oxygen.
Pyrolvsis.- of fuel is carried out It 700 1000 C witf out an
oxidant. l'vrulvsis is (ltd-c ted b y contacting atornised oil directl y with
a heat carrier which may he Ili (other it stationar y or moving state.
The combui.tible gas tlitu produced i Purified front com-
tuncis arid other harmful ircipuritres and used as pure power fuel
l'rivi, offuel oil, crude petroh-uiri and heav y petroleum residues
can also he made hN using liquid heat carriers as fused salts, slags
etc
12.5.1 Wet Scrubber
Wet scrubber al ciH'-d wet, flue gas desulpliurisation systuill
ues lime stone fit forici of an aqueous slurr y . This slurr y when
brought into contact with the flue gas ahsurhs SO lit Fig 12.1
shows schematically the wet lime stone scrubbing process. In this
scrubber SO of exliaiu-t gases is absorbed and reacts chemically
with waterarici liiiestori hi foot fit products that are transferred from
scrubberr to tank Ili rtrs lion tank chemical reactions take place
rculting in disposable precipitates. Make up slurr' N added to the
tank arid scrubbi ig liquid N sent hack to tIR cr'ul lie!' The thicknec
rrucIv-s if rm.\tUl'C (t ' , to In'1 uspeiided illlmd.
water which are
ciiniciritratc-d 1v edirncntattori and rerimneud to it poicil in lucid fill
ENVIRONMENT POLLUTION AND ITS CONTROL 639

E x h us
cc3s

ler
Rac tic
tank

0
waste
Fig . 12 1
In the scrubber following reactions take place.
Ca (011)2 SO2 -. CaSO3 I 110
CaCO 3 - SO 2 CaS0 3 CO2
The c'iticiuin sulphite (CaSO3) SO produced gets partially
oidiscd to calcium sulphate. Iii most of the cases products of
neutr:iltsatw i.are not utilistd hut go to waste. Lime stone SCrul)herS
are capat)li. removing up t )O'; Of' So, from the entering
the ciuhher wiuch may have 0.2 to 0.3 502.
\'irtous ilvantages of wet scrubber are as follows
'i ) I ugh efficiency
Ui) 1 .usv flue gas energ y requirement
(iii Good reliability.
The disadvantages are as follows
(i) 11 igh capital and operating cost
ii) ('ostiv disposal problem for the waste material which is
water logged sludge.
12.5.2 Catalytic oxidation
It used to produce sulphuric acid. from dilute 502 in the flue
gas. The sulphuric acid is separated from tlue gases.
12.5.3 Magnesium oxide scrubbing
In this process magnesium sulphate and sulphite salts are
regenerated. producing a concentrated stream of 50 1 and mag-
nesium oxide ( \Ig( )) for reuse in the scrubbing loop.
Mg() S0 2 MgSO.i
 640 POWER PLANT
Thi . iu:i riuiiu iphitiS( f, )1 11 d react.- further . () and
at&r hi jerIII lZiiesiLllfl hi -ilpht
I -( )_• I I_( I lg l Is( .c

'l'hi.• latter is iiccitralisid tin adLtion of \IgO.

1It ) \lg() 2 \lgSOc I120
?d agnesi U II r u ph te e caic ned at 800 900 C and thus decoin-
pl) .d therm int, the original product as follows
Mg() t S) 2
MgO is ret uriied into the process whereas SO 2 can he
ri-p ts-t-ed j ich, sulphuric acid. The priiicip:il (lisacivantilge of this
process I, that it iuutes liumc-ruu procedures VItII solid SUl)tan-
It c:ius- ahrasi,ri wear of equipment and formation of much
du5t Further much heat is needed for the dr y ing of cr ystals and
, t'hYdriitc moisture.
12.6 Control of NO2
\iti -ogen oxides possess high biological act ivitv. Nitrogen oxides
are po)rlv soluble
ill and for that reason can penetrat ;cep
into lung nd tin cause harmful diseases.
People hying ill N() 2 contaminated areas ma y suffer from fl-
lowing 1-
reduc-d respirator y tuimrt,on
(ii }cilu r mntulcnct- of respiratory diseases
uxlcliit certain changes in the peripheric blood
Lower cn--ntritiem 'f NI)2 though being apparentl y harmless
for plants can iim1ihit their growth.
The ci 'ill !nistloll of fossil fuels in air is accompanied f) N, the
hrnra I nitric- oxide (NO) which is partl y oxidised to nitrogen
dioxide i No - resulting mixture of variable composition is
represicitic! b y the NOx where x has ;I between I arid
.1
o.\iiles N() and NO-c) are present in flue gases produced
b y burning coal, oil and gas, in exhaust gases from internal combus-
ion engines and gas turbines.
i- hirnu-d ill high ti'Inperature Zones of rUiUh)ci.-tion chain-
her from N and () of air. It is observed that usual' p to about
1000C the formation of NO from air is negligible nut above
lilt))) C time aln,)unt increases with increasing temperature.
The various methods coni mon lv usc-cl to reduce fit(- emission of
N( Ix from steam power plants and gas turbine power plants are as
ful lows
ENVIRONMENT POLLUTION AND ITS CONTROL 541

(1 I Reduction ill in COlfll)UStic)n Zulu'. liv adjust-

ing the combustion conditions to minimise the formation
of nitric oxide. The obvious requirements are low combus-
tion temperatures and use of low nitrogen fuels it possible.
Althou g h bigher temperature during combustion of' fuels
produce less amount of'Co but higher temperatures also lead to the
formation of nitric oxide (NO,. Therefore the combustion tempera-
ture should he so adjusted ti'at minimum amount ofNO and CO are
formed.
(it) Reduction of residence time ill zone.
This is the' most promising method because reducing the time
of residence of combustion products in high temperature zone not
onl y reduces the formation of NO.v but also produces minimum
amount of C( )", " ;02 and hydrocarbons.
Increase in the equivalence ratio in the combustion Zone.
BY carrying out combustion using equivalence ratio of 1.6 to 1.8
the amount of NOx produced call reduced,
12.7 Control of Waste Waters from Steam Power Plant
The waste waters discharged from steam power plant pollute
the water basin if the waste waters ae not ' pHv handiNi. 'the
waste water discharge into the basins may be int lie following forms.
(i) Single
( it) Periodic
(tit) Continuous with constant flow rate
t I Continuous with variable flow rate
(i - ) Occasional.
The most favourable mode of waste water discharged is that at
which the maximum permissible (safe) concentration Ml'('i of
impurities in the basin is not exceeded
Steam Power Plants are the sources of following t ypes i,l'waste
waters
i) Cooling waters which mainly cause thermal contanunu-
tion.
(it) Waste waters of water treatment plants.
(iii) Waters from hydraulic ash disposal systems.
(ic) Used water after h y draulic cleaning of fuel conveying
system.
Ic) Rain water collected on the territor y of power plant.
Cooling waters of steam power plant carr y : tit amount
of beat into water basins waste waters of water treatment plants
contain various neutral salts acids and alkalis which ma y affect
water basin by changing p11 value of water and b y varying the salt
concentration in water. The waste waters of hydraulic ash disposal
svteins should be discharged into water basin only when they
contain no coarse particle substances.
642 POWER PLANT

Since %t,- waters usuall y COIILa111 a number of lIutatit-

therefore vastc waters should he purified before being di
into water bjsins. Use of contaminated waste waters ma y hirru
agriculture, fishing industry etc.
Purification of waste waters is carried b y number of ri.etlic,ds.
Some of the nietliod are as follows
Methods for direct separation of [-I tics
liitering
ii) ( ntrifL.ging
(iii) Flotation
(it Settling and clarifying
Micro straining through fine nets
lil Coagulation
(1) ) B io-Ch e in ica I methods.
(c) Methods of impurity sep.ration with a change in the
liase state of Water or impurity.
Methods based on transformation of impurities.
Steam power plants should use stacks of proper design. These
stacks sliuld he able to disperse harmful substances of flue gases
in tilt atinosphere so as to reduce their concentrations to the
spccitmd safe limits. Taller stacks are preferred stacks having
heights 100 in to 300 in and even more may he used In a steam
power pla4it Use (1 poor quality fuels which are characterised by
jijEiem uncentriitiumls of ash and sulphur should be avoided.
Pollution of biosphere by gaseous contaminants can be avoided
b y the following
it BY using fuel of proper quality
lit B y cleaning fuel
(iii B Y cleaning flue gases
[lv using stack of sufficient height for proper (lispersion of
efflients.
The global concern for the environment protection and pollution
control has male it mandator y for most combustion plants to con-
in uouslv monitor flue gases for pollutants like CO. NO, SO 2 , and
also measure (.) 2, temperature, smoke densit y or dust concentration.
Anal y ser used for CO. NO and SO 2 and for the reference variable
ma y be micro processor-controlled which fi.aturc autoniat.Lc
status monitoring.
12.8 Pollutants from Nuclear Power Plants their Effects and
Control
A nuclear reactor produces a and 1 3 particles, neutrons and y
quanta which call the normal functioning of living or
ganismils. The radioactive isotopes which form in nuclear reactors


ENVIRONMENT POLLUTION AND ITS CONTROL b43

have a hi g h hxtcitV and their effect oil organisms ma y be

accumulative. For thi reason the problem of disposal transport and
torage of ii . I and liquid radi oacti v e wastes arc extremely impor-
tant.
At atomic power 1dirtt there are three main ources of radio
active contarnitlation ol air

fission ot riticlit it solid or ,':iseous nuclear f uel.

(11 DLW to the eftc t of neutron fluxes (-It heat carrier in
th priln:Lry cooling s y st&fl1 :LIt(f ott the atiThierit air.
'Ihi third source air contamination is (lanniage of shells
of turf eleI1lents Potential souftes of radioactivit y itt
:ltoiitii_' f)o\'('i' pl:lIlts are also various :tuxiliaiv strticttires
and iltinents such as cooling poiiiis, reactor below off,
vStti1l tanks for collecting radioactive leakage etC. They
ma y liberate radintetivi' inert gases etc.
One of the nitior prohhniis at niilear power plants is the
disposal (,t vaste fuOIILU'tS which IN ltigIIl\ ;tiliia(ti'; LINY emit
lartz(' di tnt it ie raYs
' File SOlids 11 (11 11 ( 1 aII(f gsodis rii(lIoact:w ;IsN- IN priciuued
at different stages of nonchar fuel cYcle. 'Hnt radjaitive \vastes
silo tdl)edis o'edoff'niisuch awa y thatt.lir c :: ari1 tollinulin
and plant life special Nrc is taken to pr( cut I -akige of I 1cnds
containing radioactive substances into the g round in til l area
r
ground tilt' 1)OW(t' plant Lii1tiiil w:isti art tivallv lied ii st a less 1

steel hulks mounted in concrete cl i ffs with bottoms. Th t . -iitanks are

buried in th &grnuni till their dectv il r;nlitariivitv. Solid w:iSti
are buried iii the g-round The waSte gases Ire mixed with iii active
air di charged from venmtil ill nui ystinN :tnml ifrir j:tSiI1g througli
tile filters ar. released through high stacks. 'lii' radioactive g:iis
ma y also in colleetcil :iinl stored 111:1 tank Iorrritd in the ground and
disposed of to the atmosphere when rill !niictivitv level is SIflicIOIitiv
low The waste gaos should ia j irsd tinougli it luinl up plant to
remove radioactive iodine \vlIIt'll cnnst it Lit es the illajor gast'ru
hazard ('leaning of gases front ra(liactivi rolinc is imuli by using
adsorhin g filters with activated carbon its iidson'hrrrt
Waste waters of iltoiflic power plant ille contafltiIl:tted with
radioactive itupuritii's. Titesi' impurities ninist he separated before
discharging waste waters i nto water basins Flic radioactivit y of
waste waters can be lowered b y ',evimal tirrits of 111,1 '^ _'llitUdC by
treating them in evaporators. Tin livel of r:iilmoactivity oil tue
territory surround the atoulltc power plant slaulit la ch. ed pin
tidicall y 'l'his level l:uuld be kept below tia- piinnthle lvtl
 644 POWER PLANT

; 12.9 Noise Pollution and Noise Control

Heavy noise environnient has extremely unplease effect on
people exposed to it. Continuous exposure reduce hearing ability.
The main sources of noise in a power plant are turbo-alternators,
fans, power transformers etc. Complete sound proofing and com-
plete casing of turbo-alternators can reduce noise level.
12.10 Standardisations for Environmental Pollution
Stand ardisations in the field of atmospheric protection are the
rating for exhaust emissions in air depending on the composition
and concentration of emissions. The quantities of contaminants
discharged with effluents should not exceed the permissible limits.
The standardisation varies from country to country as well as from
place to place. Standardisation depends on the following factors
(i) Type of fuel used
(ii) Surrounding air velocity
(iii) Temperature
(it . ) Humidity
(u) Population density
(li) Atmospheric dispersion characteristics.
The rules and regulations regarding the emission of pollutants
to the atmosphere should be properly followed. The standards
specify the general requirements for the protection ofatinosphere.
Water .basins, land and living organisms. The standards define
the maximum permissible emission (MPE) for each of the atmos-
pheric pollutant.
Table 12.1 indicates t y pical values of Maximum Permissible
concentration (MPC) of some of harmful substances in air of popu-
lated areas.
Table 12.1 -
__•___.-,-. - -.-
-- .. . XfP
-- Acemge
- ...—.
-. 005
L__4._ IlMhest Sin.......
- 015
Non toxic dust 0.15 1 05
-c:. 1---- -- ano -
:Nitr'nd,oxLde _._J__._928 5 .............
SuhurAnhdriJc .... - 005105
L_ Ilvdruf
1 en sulphide0OOS - ).Ou - -.

12.11 Thermal Pollution

Thermal power plants like fossil, nuclear, solar plant reject low
temperature heat tn the environment. All the heat energ y used in
human activities evtrittiallv becomes a heat input iiitoenvironmt'rit
aiiiiost all t'vliic}i ciitvi', the atmosphere and Can affect climate.
 ENVIRONMENT POLLUTION AND ITS CONTROL 545
It is observed that in thermal power plants heat added and heat
rejected are functions of plant efficiency.
il = Efficiency
_w_ w
HA W+HR

HR
I +

where W = Working output

HA = Heat added = W + HR
HR = heat rejected
HR 1
w-
H11 1
i,v - -
This shows that an increase inefficiency will reduce the amount
of heat rejected.
12.12 Cleaning of Ventilation Air at Atomic Power Stations
The air in various rooms of atomic power stations may contain
radioactive aerosols and gases, including iodine, which ma y have
evolved with heat-carrier leakages or formed due to neutron a'ctiva-
'tion of air. The flow rate of ventilation air at a 1000-MW station mar
be of an order of hundreds of thousands cubic metres per hour. It is
not needed to purify such a large volume of air from radioactive
noble gases, since their concentration in the ventilation air is not
dangerous for the environment. On the other hand, these gases must
be cleaned from aerosols.
The methods for air cleaning from aerosols are much like the
common methods of air Purification from inactive dust.
At present, filtration is the principal method forcli'ariiiu the air
from radioactive aerosols. This is usuallydone b y using tiI• . hist'cl
on special tine-fibrous materials.
The resistance of the filter fabric at various air speeds can be
found by the formula
V
Pa

where V is the nominal air speed, m/s, and p is the standard

resistance of the fabric, Pa.
646 POWER PLANT

12.13 Fuel economy in furnaces of boilers

Fuel econunv in furnaces leads to uiuInie iperatiori if tb
furnaces, in it tlu large t losses are C)IltaIIll'd Ill liii \\iSte
flue gases and radiation fion furnace walls.
Ihe heat carrR'd away b y flue gases can be reduced by
providing sufficient air for combustion
adequT'ite mixing of t\iel and all-
(fit) preventing air infiltration
ti t -) 1 >rovidingheatexchaiigers in the path iffl tie gases leaving

the boiler.
Recover of waste heat escaping through the boiler chimne y by
use of superheater, economiser and combustion air preheater
reduces ill" conuiflpti)n in boiler. Heat lost in boiler chimney
flue gases at various teln } )erilture for various excess air levels have
h(-(-n shown in Fig 12 12.
I. Preheating of Solid Stocks for 11CIt lk'(oVerV
This technique is also used for recovery of heat f'i'on flue gases
going out of the s y stem examples are Multi-chambered Ring Fur-
naces used in brick, refractory/ceramic industries, carbon electrode
industr y , Rotary kilns in cello nt industr y In all these appliance the
charges are, prhcatc-d b y the counter current contact of the flue
gases which-are consequently exiled (luring this process before it is

Il

tA

tce. (VI i.) -

Fg
 ENVIRONMENT POLLUTION AND ITS CONTROL

ixIunjtid tlirocig}: the chinnev. IIIU COililiustion;ill 1, prulle:lted

b y cx I ract mg h,:it from !I iv hot products coining out of fl I maces.
I! Li rig it(&tiilt(' amount of it
t itilnn:atic proportioning of fuel and air improves
fuel ecoi1om I .i ace effic RflcV and process cojitiol. The use of
ox ygen au dvstr the products of combustion leds to ii proved
fuel econuluIv and (ItICICIICV.
0ne ni the most iiflportaiit factors influencing the efficient use
of t}uc' luck is the r stock loss i.e. that proportion of the futI energy
icIat which is cariued out of the operating furnace as the sensible
hc;it Of the flue gusts. The important factors governing this
paranleter are iiorinallv the temperature of the flue gases leaving
tb furnace and their ox y gen content twhiirh is related to the
prodwtnui f , aur tiod for combustion above strict stuichionn.tric

to

LB
to
to
0
fl
6

()
P

u 20 40 60 80 100
1. Eocess or--
Fig 12 3 Rci . cnship of ". oxygen in waste gases and '. excess air.
eqwnunitut. .• exces5 air:. The general relationship between
the uxv,:i c ' :i:t. and 'xc&ss air Idr both low ('V, fill
lulat furnace g;u and rich fuel coke oven gas. heav y fuel oil
ett. I> 'va in Fig l d. 'ftc• It:uu that the fcirim uong
Ii niitI bed e.inuHitiuu •,! rid hi more (flleii:leV as compared to
f:irniccs using puiveried coal. Iii the figure curve A is for blast
lui, uetiu gu. thuc cur ve P m i c:nvcrt'r gas mid curve C is for
fuel oil and coke oven gas

648 POWER PLANT

Thus through the above means of heat recovery systen, t he

overall thermal efficiencies of boiler furnaces may be brought to a
level of S5-90 or even more.

PROBLEMS

12 I. \V ht is eiivir nmental control at power plant.

12 2 Discuss thc viri)ti pollutants from steam power plant.
12.3. What are the effects Of -S 02 and Nt), on human life and Veg('tt-
tiflhl ?
12.4. Write short note., on the following
it ) Arid rain
tzi1 Control of particulate matter l'rom steam power plant.
12.5. J)escriln the methods used to control the following from a steam
power plant.
S(.)2 ' (ii) NO
12.6. Write short notes on the following
C i nder catcher
ii I (vclone dust collector

t in I Wet Scrubber
Electrostatic Pieci pttitor.
12.7 l)escribt the various t y pes of waste waters from a si in power
plant and the methods for removing impurities frorn waste
vafors.
12 S. l)eciil i the ill utants of atomic power plants uiid their control
1 ,-,g Write short ti,tcs on the following:
Noise control at power plants
.u .i Standardiations for environment pollution.
(III I Iliernial pollution,
12.10. Discuss the cleaning of ventilation au at atomic power plants.
12,11. Discuss Iliet econom y in furnaces of boilers
 13

Direct Energy Conversion Systems

13.1 Introduction
The direct energy conversion systems convert naturally avail-
able energy into electricity there being no intermediate conversion
into mechanical energy.
The various direct energy conversion systems are is follows.
1. Thermo electric conversion system
2. Therrnionic conversion system
3. Electrostatic mechanical gnerators
4. Photo voltaic power system
5. Electro gas dynamic generator (EGD)
6. MHD system
7. Nuclear batteries
8. Fuel cells
13.2 Thermo-electric Conversion System
The direct conversion of heat energy into electric energy (i.e.
without a conventional electric generator) is based on seebeck
Thermo electric effect Fig. 13.1 shows two dissimilar materials
joined together in the form of a loop so that there are two junctions.
If a temperature difference is maintained between hot and cold
junctions an electric current will flow round the loop. The magnitude
of the current will depend upon
(i) temperature difference
and (ii) materials used

Material-1,Cold

Hot
Junction
unction

McI

Fig. 13.1
—43
 POWER PLANT
650

Let T2 = Temperature of hot junction

Ti = Temperature of cold junction
V=e,m.f.

where u see beck coefficient

AT = Temperature difference
=T2-Ti
Fig. 13.2 shows D.C. load connected in simple thermocouple
arrangement.
Hot

Cold Cold

Load

Fig. 13.2

Fig. 13.3,, shows variation of see heck coefficient (t with

temperature for
(a) p-type semi conductor
(b) metal
(c) n-type semi-conductor

Temp ero1ure,K----
Fig. 13.3

For a given thermocouple, the voltage and electric power output

are increased by increasing the temperature difference between the
hot and cold ends. In a practical thermo electric converter, several
couples are connected in series to increase both voltage and power
as shown in Fig. 13.4.

DIRECT ENERGY CONVERSION SYSTEMS 651

Lii
Co r. Cold

Load

Fig. 13.4

A thermoelectric converter is a form of heat engine. H .tt is taken

up at an upper temperature (i.e. the hot junction) and part is
converted into electrical energy ; the remainder is discharged (or
removed) at a lower temperature (i.e. the cold ends). As with other
heat engines, the thermal efficiency of a given thermocouple for
conversion of heat into work (electrical energy) is increased by
increasing the upper temperature and/or decreasing the lower
temperature.
Thermoelectric generators have been built with powerout put
ranging from a few watts to kilowatts.
The source of heat for a therino electric power generator may be
(i) a small oil or a gas burner
(ii) a radio 150 tope
or (iii) direct solar radiation.
13.3 Performance Analysis of Thermo-electric Power Gen-
erator
Fig. 13.5 shows a schematic arrangement of Thermoelectric
convertor. It consists of two blocks A and B of semi conductor
materials connected together by a co:durtor.

Heat Reservoir

f4-A A-.
I I
1-
ii:.i L
L 1 (-B

I.
L - sink

Load
Fig. 13.5
 POWER PLANT
652

The conductor receives heat from a thermal source and the

lower open (not connected) ends of the blocks reject heat to a low
temperature sink. The sides of the blocks are insulated. Hence, heat
flow occurs along the length of the blocks only, supplying heat to
the hot junction causes the electrons in then type block and holes in
the p-type block to flow away from the hot junction, thereby produc-
ing a potential difference between two open ends. If the circuit is
completed at the coldjunction, an electric current will flow through
the load.
Let RL = External Load Resistance
AT = Temperature difference between hot and
cold functions
P = Power fed into external load

RL
=L
where aAB = seebeck coefficient
R = Resistance of thermo couple

The factor is called figure of merit.

13.4 Thermo electric Materials

Semi conductors due to their much higher values of seeback
coefficient (cx) are preferred as compared to metals. The commonly
used materials for thermoelectric elements are as follows.
Lead telluride (Pb Te) in n andp-type forms, bismuth telluride
(Bj2Te3), bismuth sulfide (B 2S3), antimony telluride (Sb2T,
tin
telluride (SnTe), indium arsenide, germanium telluride (GeTe),
cesium sulfide (CeS) and zinc antimonide (Z0 SO.
Lead telluride (Pb Te), a compound of lead and tellurium,
containing small amounts of either bismuth (n-type); or sodium
(p-type), has been commonly used in recent times for thermoelectric
converters.
Table 13.1 shows Figure of merit for some of thermo-
electric materials.
Materials Z( K
Bismuth telluride (doped with Sb or Se) 4 x 10
Lead telluride -___L5
________ 1 5>'
x
1.5 X
Germanium telluride (with bismuth)
Zinc antimonide (doped with silver) 1.5 x 10
Cesium sulfide 1.0>'
DIRECT ENERGY CONVERSION SYSTEMS 653

13.5 Analysis of Thermionic Generator

A thormionic generator works on the principle of thermionic
emission which implies emission of electrons from the metal when
it is heated.
A therniionic generator consists of two metals or electrodes with
different work functions seated into an evacuated vessel. The anode
(cold electrode or collector) has low work function and cathode (hot
electrode or emitter) has high work function.

Evocuc
vessel 0-c.
Load

Heat input
Fig. 13.6

Anode is generally made of barium and strontium oxidy whereas

cathode is made of tungsten impregnated with a barium compound.
The fuel used in thermionic geneator may be fossil fuel, nuclear
fuel or solar energy. The emitter (cathode) temperature in ther-
mionic converter containing cesium should be at least 1000°C.
The emitted current density as given by Fermi Dirac is
= 4n ni e- o/K7
j

= A V e - ilKT amp/cm2
Where A is constant and has the value equal to
120 x 10 4 Amp/m 2 as described in the previous section.
k2
A- 1z3

h = Planck's constant
e = Charge
Q = work function in o
T = Temperature (K)
K = Boltzmann constant J1
m = mass of an electron (kg)
Now, J = Current leaving cathode
 POWER PLANT
654
• e V6)/KT
Jc =A T e-

=A 1?
Ja = electron current leaving the anode

J. =AV VJKT.,
J = net current density flow
J= Jc - Ja
= A 7 e'M - A7e -
V Voltage output
= 01 -0. -OP
where 0, = Cathode work function
= Anode work function
op = Plasna potential drop.
P = Power produced = J.V
(work output)
H = Heat Supplied to the cathosk (cathode heat flux)
QC
- Ac
"
=J(Ø+--)+ M, - 77 )

where T = Temperature of cathode

= Temperature of surroundings
E = Emissivity
o = Stefan Boltzmann Constant
= 5.668 x 10 12 Joule/sec cm k4

= Efficiency of generator =

where I' = JV
Qc

13.6 ElectrO-gas dynamic generator (EGD)

It uses the potential energy of a high pressure gas to carry
electrons from a low potential electrode to a high potential electrode,
thus doing work against an electric field.
DIRECT ENERGY CONVERSION SYSTEMS 655

Fig. 13.7 shows gas duct in EGD convertor. The carona

electrode at the entrance of the duct generates electrons. This
ionised gas particles are carried in the duct with the neutral atoms
and the ionised partiles are neutralised by the collector electrode
at the end of the insulated duct. The working fluid may be either
combustion. gases produced by burning fuel at high temperature or
it may be pressurised rcctor gas coolant. Number of such channels
are connected in seriet or in parallel. The output of EGD is nearly
10 to 30 W per channe'. This system is as good as MHD system of
producing electricity.

Collector
electrode
Load
Attract(
etectro

a
rode

Fig. 13.7

13.7 Power output of MHD generator

The MHD power generation is a method of generating electric
power which utilizes a high temperature conducting plasma moving
through an intense magnetic field.
Fig. 13.8 shows
(i) direction of magnetic field
(ii) lonised gas velocity
and (iii) force
in MHD system (Faraday generator).
The motion of the gas is in x direction, magnetic fid B is in y
direction and force on the particle is in z direction.
Ri. = load resistance
I = current flowing across load resistance
V = voltage across the load
= Electric intensity between the plates
V
d

656 POWER PLANT

where d = Distance between the plates.

fonised go Plates

Fig. 13.8

u = Gas velocity
E' = Total electric field
= E2 + B . u

=-7+Bu
= (B.u.d. - u)

The electrumagnetic field Ez and B acting on the moving gas

produce the same force on the ions as !e electromagnetic fields Ez
and B produce on a gas with zero average velocity.
Thus the open circuit volt.ge (E0 ) is giv& by
E0 = B.u.d.
Let Rg = Internal resistance of the generator
d.

where a = Conductivity of gas

A = Plate area
Fig. 13.9 shows electric circuit of M.H.D. generator

EIclrodes R Lood

Fig. 13.9


DIRECT ENERGY-CONVERSION SYSTEMS 657

RL. = Load resistance
when, Rg = RL the power obtained is maximum power (Pm..
Pmax = Er, I

- =I.Rg.I
=I2Rg
E,,
Rg -

Ell
asRi.=R
4Rg
- B2u2d2
as Eo = Bud.
4Rg

d
Putting the value of Rg as Rg
- oA
B2ti 2d2
we get, niax = ---- GA

=•

Maximum Power per unit volume

=
1 2 -,

Combustion
chamber
Nozzle
- EIctrodes
Load
(,s flow

Generator
channI

Fig. 13.10 shows a simple MHD generator.

An ionized gas is emplo y ed as the conducting fluid. Ionization

is produced either by thermal means i.e. by an elevated temperature
or by seeding with substance like cesium or potassium vapours
which ionize at relatively low temperatures. The atoms of the seed
element split off electrons. The presence of the negatively charged
electrons makes the carrier gas an electrical conductor.
 S.

POWER PLANT
658

O IdLJ1 Load

pressOr

rom Cooler
1

Fig. 13.11 shows over all power cycle for MHD generator.

11 16\tOC3

01

0.
El
01
4 14et rejection

- Entropy, $
Fig. 13.12 shows Temp. entropy diagram for the cycle.

In the overall power cycle, the MHD converter takes the place
of a turbine in a conventional vapour or gas turbine cycle. Still, a
compressor must be used to elevate the pressure, heat is added at
high pressure and the flow is accelerated before entering the con-
verter.
Pig. shows Temp. entropy (T-S) diagram for the cycle.
flth = Thermal efficiency
It is limited by carnot efficiency
Now P1 =P2
Pressure, P3= P4
- work output - ( h 20 - h30 ) - (h 10 - h40)
11th - heat input - ( h20 - h10)
Where the indicated enthalpies are stagnation values which
take inço account the K.E. of the flow. The stagnation enthalpy is
(lCfiflCd as


DIRECT ENERGY CONVERSION SYSTEMS 659

ho =h + V2

where V is the flow velocity. In practical MHD converters, high

velocity ionized gases are usually employed for the conversion
process, so that the K.E. of the flow represents a substantial portion
of the total energy.
Example 13.1 An M.H.D. generator has the following specifica-
tions
Average gas velocity 103rn/s
Plate area = 0.26 m2
Distance between plates = 0.50 m
Flux density = 2 Wb/m2
Gaseous conductivity = 10 Mho/m.
Determine: maximum power out put
Solution. E0 = B u d where E. = open circuit voltage
B = flux density
u = Average gas velocity
d = distance between plates
E0 = 2 x 1000 x 0.50= 1000 volts
R = Generator resistance

Rg =- where A - plate area = 0.26 in

= = 0.19 ohm
10 x 0.26

Maximum power 4Rg

-
012 1252 x 10 3 W.
- 4 x 0.19
13.8 Materials for MHD generator
In IYlHD generators the temperature of plasma in nearly
2700'C which is quite high. Therefore refracting materials are
commonly used in several parts of the generator like electrodes,
channel or duct wall-
Important factors to be considered for the selection of materials
are as follows
(i> Density
(1i) Melting point
(iii) Thermal shock resistance
(it') Electrical conductivity
 POWER PLANT
660

(u) Corrosion resistance

:j) Erosion resistance
(vii) Oxidation reduction resistance
13.9 EIetrode materials
The materials used should be with higher electrical conductivity
so that they contribute towards better charge transfer across the
electrodes/plasma interface. Most of the initial channel wall designs
involved water cooled metallic electrodes due to the oxidising con-
dition. But the cold surface creats cold boundary layer which
generally promote high thermal losses, low conductivity and leads
to considerable voltage drop in this region. The alkali seed conden-
sation and cold slag condensation degrade the properties of
electrode.
The commonly used inattrials for electrodes are as follows.
(i) Zirconia Based Materials. Zirconia is a good refractory
oxide (Melting point 2800'C). However, pure zirconia can not be
used as it undergoes phase transitions at different temperatures.
During these phase transitions, non-linear thermal expansion takes
place Therefore elimination of th: disruptive phase transition is
necessary. This can be achieved the addition of CaO (calcium
oxide), Y 203 (vitrium oxide), MgO, which results in the formation of
cubic phase that is stable over a wide range of temperature and
composition to 27 CaO and 7 to 52% Y 203 are required for
stabilizatic' i ofZrO2.
(ii) Chromite Based Materials. Lanthanum Chromite
LaCrO3 is one of the materials which has been found suitable.
Incorporation of substantial amount of MgO improves its ther-
mochemical propertis without significantly changing its thermo-
mechanical properties and electrical properties as compared to SrO
doped chromite (5-20 MI0). It is a good electronic conductor with
little variation in conductivity with temperature. Materials of an
optimum composition LC 20 M (20% MgO doped LaCr03) is found
to be highly promising for use in MIlD.
(iii) Aluminate Based Materials. Spinel (MgAI204) is a high
resistivity insulator and hence used for inter-electrode insulation on
addition of Fe 3 0 1 it becomes conducting electrodes material. As its
melting point is 1860C, it cannot be used safely above 1750 C. It is
suitable for coal MFI1) generators.
(iv) Nickel Oxide Based Materials
Lithium doped NiO is a good electronic conductor but does not
posses high enough melting point. Hence the most suitable
materials are Cc02 and Zr02 - Ce02 solid solution.
DIRECT ENERGY CONVERSION SYSTEMS 661

13.10 Materials for channel

Ceramics are quite commonly used for channel or duct as
ceramics offer superior chemical stability towards oxidation and
corrosion. This will therefore resist corrosion and erosion caused by
high velocity plasma.
13.11 Batteries
Batteries can play an important role in energy resources con-
servation.
A battery is defined as a combination of individual cells. A cell
is the elemental combination of materials and electrolyte constitut-
ing the basic electro-chemical energy storer.
13.12 Division of batteries
There are two division of batteries.
(i) Primary batteries
(ii) Secondary batteries
Primary Batteries non chargeable, e.g. "drycell" flash light bat-
teries. In primary batteries the chemical reactions are norever- 3,
sible.
Secondary Batteries rchargeablee.g. a lead arid battery. There
are many types of seconlary batteries. The chemical reactions are
reversible in secondary batteries.
Secondary batteries are of chief interest for solar electrIcs (Solar
and wind energy, electrical storage), and therefore secondary bat-
teries or storage batteries are important.
In a storage battery, individual cells are connected in various
ways to provide the desired power output. Since storage batteries
are generally portable, they are common mobile sources of energy.
A major use, for example, is the starting-lighting Ignition (SLI)
system of automobiles. Among the many other applications are
those in the operation of mine locomotives, forklift trucks, golf carts,
road vehicles, and submarines and other underwater craft.
13.13 Battery Principle
A generalized cell consists of two electrodes called the anode and
cathode immersed in a,suitable electrolyte. When an electrical load
is connected between the electrodes charge separation occurs at the
interface between one electrode and the electrolyte, freeing both an
electron and ion. The electron flows through the external load and
ion through the electrolyte, recombining at the other electrode.
The polarity and magnitude of the cell terminal voltage is, in
general, "unction of th electrode materials electro' - fe, c11
tomperat e, etc.
662 POWER PLANT

In the charged cell, electrical energy is stored as chemical energy

which can be recovered as electrical energy when the cell is dis-
charged.
The storage capacity of a battery depends on the discharge rate
(or discharge time). Increase in the discharge rate (or decrease in
discharge time) for a given battery results in a decrease in the
amount of electrical energy that can be delivered. This effect is very
marked in the lead-acid battery commonly used in automobiles.
13.14 Types of batteries
Various types of batteries are ask. fo1lows:
(i) Conventional Batteries, such as lead acid, Nickel cadmium
Nickel-iron, Nickel-zinc. Silver Zinc,, Silver Cadmium, and Zinc-
Bromine.
(ii) Metal-Gas Batteries, such as Iron-Air, Zinc-Air, Zinc-
Oxygen, Zinc-chlorine, Nickel hydrogen, Cadmium-oxygen, Cad-
mium-air, Aluminium-air, Lithium-Sulfur Dioxide, Sodium-air and
Magnesium-air.
(iii) Alkali-metal-High tempera'ture Batteries, such as Sodium-
sulfur, Sodium-Chlorine, Lithium-Sulfur, Lithium-Iron Sulfide,
Lithium-Chlorine, Lithium-Copper, Lithium-Nickel halide.
13.15 Battery Equivalent Circuit Models
There are two such models
(i) First order model. In this model leakage effects are negli-
gible, and it is the one which is most widely used, Fig. 13.13 shows
this model.
R internal

f__W^
voc

Fig. 13.13

VOC = open circuit voltage

(ii) Second order model. In this model. l .akage is the cause of
"run down" and short shelf life on secondary batteries.
Fig. 13.14 shows this model.
DIRECT ENERGY CONVERSION SYSTEMS 663

Rl,0kO, V
oc

Fig. 13.14

13.16 Types of Battery Arrangements

There are generally two types of arrangements
(i) Series arrangement. In this arrangement cells are con-
nected in series with the positive electrode of each cell connected to
the negative electrode of the adjacent cell. The total electromotive
force (emf) or voltage of the battery is then the sum of the separate
voltages. In most automobile batteries, for example, six cells each
with an emf of close to 2 volts are connected in series to provide a
12-volt output. The total current (in amperes) drawn from the series
of cells, is however, the same as that drawn from each cell.,
Fig. 13.15 shows cells connected in series.

+. 1. I I 1
au uc [(I.II
ectrodes Cell-'
Fig. 13.15

•1

HA H

IIIFig. 13.16

(ii) Parallel arrangement. In this arrangement cells are con-

nected in parallel. All the positive electrodes of the individual cells
are connected together and so also are all the negative electrodes.

ii
664 POWER PLANT

The batter y voltage is now the same as that of a single cell, but the
current is the sun of the currents supplied by the individual cells.
Fig. 13.16 shows such arrangement.
13.17 Power of Battery
The power 01 a battery (i.e. the rate at which stored energy is
withdrawn) in it is equal to the product of the ernf in volts and
the current in amperes thus,
Power (w t) EMF (volts) . current (amp(' . .•)
Specific power is the maximum rated power output per kg of
battery can s.ipply. Rating '.arious batteries by specific power per-
mits rapid performance comparison of different kinds of batteries.
13.17.1 Energy efficiency () of battery.
It is given by:
- Useful_enegy out put ('vatt hours)
1 - Re charge energy (watt hours)
13.17.2 Cycle life of battery
The cycle life of battery is Lhe number of times the battery can
be charged and discharged under specified conditions.
13.18 Principal and analysis of H2, 02 (Hydrogen-oxygen)
fuel cell
This cell is quite commonly used. Fuel cells are particularly
suitabh' for low voltage and high current applications. Fuel cells are
chemical devices in which the chemical energy of fuel is converted
directly into electrical energy chemical energy is the free enegy of
the reactants used.
Fig. 13.17 shows a hydrogen oxygen fuel cell.
Load

H 2 in o.
Fig. 13.1 r A Hydi3x (H2, 02 cell).
 DIRECT ENERGY CONVERSION SYSTEMS
665
The main components of a full cell are as follows.
(i) Fuel electrode (anode)
(ii) an oxidant or air electrode (cathode)
and (iiL an electrolyte. The electrolyte is typically 40% KOH be-
cause of its high electrical conductivity and it is less
corrosive than acids.
In most fuel cells, hydrogen (pure or impure) is the active
material at the negative electrode and oxygen (from the oxygen or
air) is active at the positive electrode. Since hydrogen and oxygen
are gases, a fuel cell requires a solid electrical conductor to serve as
a current collector and to provide a terminal at each electrode. The
solid electrode material is generally porous.
Porous nickel electrodes and porous carbon, electrodes are
generally used in fuel cells. Platinum and other precious metals are
being used in certain fuel cells. The porous electrode has a larger
number o f sites, where the gas electrolyte and electrode are in
contact the electro chemical reactions occur at these sites. The
reactions are formall y very slow, and catalyst is included in the
electrode to expedite them. The best electrochemical catalysts are
finely divided platinum or platinum-like metal deposited on or
incorporated with the porous electrode material.
The reactions which take place are as follows
4KOH -* 4K t 4 (OH)
Anode: 2112 - 4(OH) - 41120 + 4e
Cathode: 02 + 2H20 + 4e - 4(OH)
Cell reaction: 21-1 2 + 02 - 21-120
The electrons liberated at the anode find their way to the
cathode through the external circuit. This transfer is equivalent to
the flow of a current from the cathode to the anode. The movement
of electrons constitutes a current passing through an external load.
Thus useful work is obtained directly from the chemical process.
The gases in the hydrogen oxygen cell must be free from carbon-
dioxide, because this gas can combine with the potassium dydroxide
electrolyte to form potassium carbonate. If this occurs, the electrical
resistance of the cell is increased and its Output voltage is decreased.
Consequently, when air is used to supply the required oxygen,
carbon dioxide must first be removed by scrubbing with an alkaline
medium (i.e. lime).
Depending or. the fuel used the main types of fuel cells are as
follows
(i) Hydrogen (1-1 2 ) fuel cell,
(ii) Hydrazine (N 21-14 ) fuel cell
—44
 POWER PLANT
666

(iii) Hydrocarbon fuel cell, and

(iv) Alcohol (Methanol) fuel cell
13.19 Types
Hydrogen fuel cells (Hydrox) are of two types:
In this cell the electrolyte temperature
(i) Low Temperature cell.
is 90°C. It is sometimes pressurised, but not by a great amount,
usually say upto 4 atmospheres.
(ii) High Pressure cell. In this cell pressure is upto about 45
atmospheres and temperatures upto 300°C say. A single "Hydrox"
fuel cell can produce an e.mf. 1 of 1.23 volts at 1 atm and 25°C. By
connecting a number of cells , it is possible to create useful potential
of 100 to 1000 volts and power levels of 1 kW to 100 MW nearly.
13.20 Applications
Fuel cells are quite commonly used for the following applica-
tions.
(i) Automotive vehicles
(ii) Power stations
(iii) Space applications
13.21 Output
- = work output
- =1Q—AH
where 1iQ = Heat transferred to the steady flow stream from the
surroundings
AL! = Change in enthalpy of the flow stream from entranel to
exist
= Efficiency of energy conversion of fuel cell
useful work -AW
=
Heat of combustion of fuel - AH
Now H = Enthalpy
S = Entropy
0= Gibbs free energy = H - TS
i G = All - Tds - SdT
Since temperature of flow stream at both entravncel and exit is
T
AG = All— TAS
AWm = Maximum workdone by flow stream on surroundings.
DIRECT ENERGY CONVERSION SYSTEMS 667
E = E.M.F. of cell
W AG
n.Fn.F
where n = Number of electrons transferred per molecule of the
reactant
F = Faraday's constant
= 96500 Coulombs per gram molecule
Example 13.2. In a hydrogen-oxygen fuel cell the reaction taking
place is
H2 + 02 = 1120

If(iG)2s = - 56690 callgm-mol of H2 or - 237.3 x 10 3 Joule/gm

mole of H2
Calculate the reversible voltage for the fuel cell
Solution.
E = Reversible voltage
zW -iG
n.F n.F asiW=-G
_237.3 x103
- 2 x 96500 = 1.23 volt.
Example 13.3. At thermo electric power generator operator
between the following parameters. Figure of merit of material
= 2.1 x 10 K . Temperature of hot reservoir of source = 610°K
Temperature of sink = 310°K
Determine. (a) efficiency of thermoelectric generator
(b) Carnot efficiency of generator
Solution.
il = Efficiency of thermoelectric generator
(TH - TC M-1 X

) M+
10
T11

where 2',, = Temperature of hot reservoir of source = 610°K

= Temperature of sink = 3100
Z = Figure of merit = 2.1 x 10 °K'

668 POWER PLAN

M = [1+ z(Ti, + TO

=[i + 2.1 x103 (610+310) = 1.4

610-310 1.41 310 xlOO=10.2%
600 1.4 +
Example 13.4. A thermionic generator works on the following
data.
Cathode work function = 2.6 volts.
Anode work function = 2.0 volts.
Temperature of cathode = 2000°K.
Temperature of surroundings = 1000°K.
Plasma potential drop = 0.1 volt.
Emissivity for electrode materials = 0.2, calculate the efficiency
of the generator and also compare with the Carnot efficiency.
• Solutioij.
= Cathode work function
= 2.6 volts.
= Anode work function
= 2.0 volts.
Tc = Temperature of cathode
= 2000°K
Ts = Temperature of surroundings
= 1000°K
Op = Plasma potential drop
= 0.1 volt
V = output voltage
V=o, øAp 2.6 2.0 — 0.1 = 0.5 volt.
J = net current in the generator
= Jcathode - Junode

e = Charge of an electron = 1.6 x 10 19 Coulomb

 DIRECT ENERGY CONVERSION SYSTEMS

K = Boltzmann constant = 1.38 x 10 23Joule/'K

One electron volt = 1.6 x 10 19 Joule
Jc = JCathode
Now J =Ale_T
= (1.20 x 10 6) ( 2000)2
1-(1.6 x 10 19) 2.5
exp. L1.38x 1023) 2000
= 2.4 x 106 amp/M2
Janode = Ar e - eOJkT
= (1.20 x 10 6) ( 1000)2
I -(1.6x 10')2.0
exp. L1.38x 10- 23 i000
= 1.2 x 102 amp/M2
Net current J can be taken = J, as J0 can be neglected as
compared to J..
i.e. J = Jc = 2.4 x 106 amp/r12

= heat supplied to the cathode

AC
2kT
=
JI e
5 + 2(1.38 X 10 ) 20001
= (2.4 < 106) [ .
L L6x1O'J
+ 0. 2, 5.67 x 10 12, 10 -4 (20004 - 1000)
=6x 106 +8.26x 105 x2x105 =7.026x106 wattirn2
Efficiency of the generator
iv

= (4x 106) (Q =
0.139= 13.9% Ans.
7.026x
Carnot efficiency this device
Tr - Ts 2000-1000 0 0
T' - 2000
= 50% Ans.

670 POWER PLANT

PROBLEM

I. Sketch and describe the principle of thermo electric conversion

system.
2. Define figure of merit in case of thermo electric conversion
system.
3. Discuss the principle of thermionic generator.
4. Sketch and describe gas depramic generator.
5. Derive the expression for power Output of MHD generator.
6. Describe materials for
(i) electrodes
(ii) channel in case of MHD generator
7. What is a battery. Discuss various types of batteries for
electricity generation.
8. Discuss various arrangement of batteries.
9. Write short notes on
(i) Power of a battery
(ii) Life cycle of battery
(iii) Energy efficiency of battery
10. Discuss principle and analysis of H2 —02 fuel cell.
 14

Objective Type Questions

1. Load factor of a power station is defined as

load
(a) Maximum demand
(b) Average load x Maximum demand
Maximum demand
C
Average load
2. Diversity factor is always
(a) less than unity (b) more than unity
(c) equal to unity.
3. Load factor of a power station is generally
(a) more than unity (b) less than unity
(c) equal to unity.
4. It is generally seen that load factor of domestic load is
(a) 10 to 15% (b) 60 to 70%
(c) 30 to 40%.
5. The load factor for heavy industries may be taken as
(a) 10 to 20% (b) 25 to 40%
(c) 70 to 80%.
6. The load factor for street lighting on 24 hour basis may be
taken as
(a) 20 to 25% (b) 40 to 50%
(c) 80 to 90%.
7. A steam power station requires space
(a) more than diesel power plant
(b) less than diesel power station
(c) equal to diesel oower station.
8. State whether the following statements are true or false
bi
POWER PLANT

(a) The depreciation rate of diesel power station is more

than steam power station of equal capacity.
(b) The depreciation rate of a hydro-power station is less
than steam power station of same size.
(c) It is economical to use a few generating unit of large
size than a large n umberofsmafl size for the same total
capacity.
(d) A diesel power station produces less noise than a steam
power station of the same capacity.
(e) A diesel power station requires larger quantities of
cooling water than a steam power station of the same
size.
(/) A steam power station needs longer time for starting
and for taking load as compared to diesel engine power
plant.
(g) Diesel power stations are of smaller capacities.
9. Ina diesel engine the heat lost to the cooling water is about
(ci) 30% (b) 70%
(c) 10%.
10. State whether the following statements are true or false
(a) Pulverised coal firing requires high percentage of cx-
ees air.
(b) It is economical to fire coal into furnace by stokers than
hand firing.
(c) A steam station needs more space for storing fuel than
a diesel power station.
11. Economiser is used to heat
(a) feed water (b) air
:(c) flue gases.
12. Steam is supplied to the turbine through
(a) safety valve (b) throttle valve
(C) blow-off valve,

13. The function of a condenser is to

a reduce the back pressure at the steam turbine exhaust
tin increases the hack pressure at the steam turbine ex-
haust
(c) make the stvain Prsure more than atmospheric pres-
sw.e.
14. State whether the following statements are true or false

If
OBJECTIVE TYPE QUESTIONS
673
(a) There is no direct contact between steam and cooling
water in surface condensers.
(b) Secondary air is not supplied in cyclone burner used to
burn pulverised coal.
15. A venturimeter is generally used to measure rate of flow
of
(a) air (b) steam
(c) water.
16. Pipes carrying steam are generally made up of
(a) cast iron (b) steel
(c) copper.
17. Running cost of hydra-electric power plant is
(a) more than running cost of a steam power plant
(b) less than running cost of steam power plant
(c) equal to running cost of a steam power plant.
18. The initial cost of erecting a nuclear power plant is
(a) equal to the initial cost of steam power plant olsame
size.
(b) less than the initial cost ofa steam power plant olsanie
size.
(c) more than the initial cost of a steam power plant of
same size.
19. State whether the following statements are true or false
(a) In fire tube boilers the gases pass through tubes and
water surrounds these tubes.
(b) More heating surfaces is available in bent tube boilers.
20. The modern steam turbines are
(a) reaction turbines (b) impulse turbine
(c) impulse-reaction turbines.
21. The nuclear power plant at Tarapur has
(ci) pressurised water reactors
(b) boiling water reactors
(c) sodium graphite reactors.
22. The boiling water reactor uses
(a) enriched uranium as fuel th plutonium
(c) thorium.
23 Fill in the blanks
Narora atomic power station is located in ... .... (ALP
V.P., Gujarat)


674 POWER PLANT

(b) Kalpakkam atomic power station it located

in..........(Andhra Pradesh, Maharashtra, Tamil Nadu
(c) Neyveli thermal power station is located in...... (M.P.,
Orissa, Tamil Nadu)
(d) Badarpur thermal power station is located
in.........(U.P., Haryana, Delhi).
24. The average thermal efficiency of modern nuclear power
plant is about
(a)80% (b) 60%
(c) 30%.
25. Control rods used in a nuclear reactor are made up of
(a) steel (b) cadmium
(c) copper.
26. Reflector of a nuclear reactor are made up of
(a) steel (b) boron
(c) beryllium.
27. Mechanical efficiency of a diesel engine is defined as
I.H.P.
(a)
B.H.P. -
(b) I.H.P.
c) B.H.P. x I.H.P.
28. For the safety of steam boiler the number of safety valves
fitted are
(a) two (b) three
(c) one.
29. Natural draught in a steam power plant is produced by
(a) chimneys (b) fans
(c) steam jets.
30. Steam turbines commonly used in steam power station are
(a) condensing type (b) non-condensing type
(e) none of the above.
31. The temperature of the combustion gas at the gas turbine
inlet is about
La)900C (b) 715C
() 1200 C.
2. Name two fuels used in a gas turbine.
33. Name three materials used for insulation of-iteain plp(.S

OBJECTIVE TYPE QUESTIONS 675
34. Laminated asbestos is recommended-as insulation
material for a temperature up to
(a) 875C C (b) 150C
(c) 275C.
35. Statethè approximate heads under which the following
types of hydro-power plant work
(a) high head power plant
(b) medium head power plant
(c) low head power plant.
36. State the range of specific speed for the following types of
water turbines.
(a) Pelton turbine (b) Francis turbine
(c) Kaplan turbine.
37. Pelton turbines are mostly
(a) horizontal (b) vertical
(c) inclined.
38. Jet ratio of a Pelton turbine is defined as
Least diameter of jet
a Mean diameter of runner
(b) Mean diameter of runner
Least diameter of jet
Mean diameter of runner - Least diameter of jet
C
Least diameter of jet
39. The empirical relation for determination of number of
buckets (Z) for Pelton turbine in terms of jet ratio (m) is
given by
(a)Z= 15rn + 0.5 (b) Z = 0.5m 15
(c) L = + 15

40. Francis turbine is usually used for

(a) low heads (b) medium heads
(c)high heads.
41. State whether the following statements are true or false
(a) Turbine runner are made of cast steel and then coated
with stainless steel.
(b) Draft tube is not used in Pelton turbine.
(c) Kaplan turbine is an impulse turbine.
(d) Propeller turbine is used for low head of water.

676 POWER PLANT

(e) The main function of water turbine governor is to

maintain constant head when load on the turbine fluc-
tuates.
(1) Surge tanks are needed only for low head water plant
and they are not required for high and medium head-
power plants.
42. Sensitiveness of a water turbine governor is defined as
Mean speed
(a)
Maximum speed - Minimum speed
Maximum speed - Minimum speed
(b)
Mean speed
(c) Mean speed x Maximum speed - Minimum speed.
43. Fig. 14.1 illustrates the variation of water turbine efficien-
cy with load. State which cufve is for Pelton turbine and
which curve is for Francis turbine.

4 of Full Load—..
Fig. 14.1

44. State whether the following statements are true or false

(a) Specific speed of a water turbine is directly pioportion-
al to its rotational speed.
(b) Horizontal shaft arrangement is mostly employed for
large size impulse turbines.
(c) Vertical shaft arrangement is mostly employed for
large size reaction turbine.
(d) A vertical shaft turbine requires deeper foundation
and a high building.
(e) A horizontal shaft turbine does not require greater
floor area.
(1) In axial flow turbines the water flows through the
vanes parallel to the axis of runner shaft.
OBJEC11VE TYPE QUESTIONS 677
(g) In impulse turbines the entire head of water available
is converted in the velocity head by making it to pass
through a n'zzle.
45. Pelton turbine is suitable for
(a) high head high discharge
(b) high head low discharge
(c) low head high discharge.
46. Belt conveyor can be used to transport coal at inclinations
upto
(a)60 (b) 30
(c) 90g.
47. The maximum length of a screw conveyor is about
(a) 60 metres (b) 100 metres
(c) 30 metres.
48. Fill in the blanks:
(a) Bhakhra dam is.......dam (earth, gravity, arch)
(b)The side of the dam to which water from the 1iver or
the stream approaches is called.........and the other is
called.........(down stream, up stream)
49. The annual depreciation of a hydro power plant is about
(a) 10 to 20% (b) 10 to 15%
(c) 0.5 to 1.5%.
50. The function of a moderator in a nuclear reactor is
(a) to start the chain reaction
(b) to transfer heat produced inside the reactor to a heat
exchanger
(c) to slow down the fast moving neutrons.
51. Statethe factors which influence the selection of steam
pressure and temperature in a steam power plant.
Solution. Proper selection of temperature and pressure of
steam is very essential for proper functioning of steam power plant.
For each combination of fuel cost, load factor and capacity factor
there is an economical pressure and temperature of steam which
will result in the lowest total cost of power.
The following factors should be considered while selecting the
proper steam pressure and temperature
(a) plant capacity (b) annual capacity factor
(c) cost of plant (d) annual costs
(i) fixed cost (ii) variable cost
(e) total annual cost (/) energy generated
(i) gross (ii) net.

678 POWER PLANT

(g) kilo calorie consumption per kilo watt hour

(h) cost of fuel per million kilo-calorie.
52. Discuss the factors which influence the selection of
vacuum in condenser for a steam power plant.
Solution. The vacuum obtainable in a condenser is governed
by outlet water temperature which in turn varies with the amount
of condensing water used per kg of steam and its initial temperature.
Air entrainment in the condenser has its effect upon the vacuum.
The addition of air lowers the vacuum. In the condenser absolute
vacuum is neither possible nor economical to maintain. A higher
vacuum in a condenser results in increase the condenser size. It also
results in lower temperature of the condensate and increase the rate
of flow of cooling water and thus power required to drive the cooling
water circulating pumps is increased. For steam turbines the most
profitables vacuum is about 72 cm of mercury.
53. Vacuum efficiency of a condenser is defined as
(b) Hxh

(c)-j (d) none ofthe above.

where h is the actual vacuum in the condenser and H is the
theoretical 'acuum in a condenser.
54. State whether the following statements are true or false
(a) The theoretical vacuum in the condenser' is the vacuum
when no air is present in it.
(b) The efficiency ratio of a steam turbine is defined as the
ratio of the thermal efficiency of actual turbine to that of
a perfect turbine.
55. A nuclear chain reaction is possible when
(a) fission produces more neutrons than absorbed
(b) fission produces less neutrons than absorbed
(c) none of the above.
56. State whether the following statements are true or false
(a) In a homogeneous reactors the fuel is uniformly dis-
tributed in the core.
(b) In a heterogeneous reactor the fuel elements are
separated by a moderator.
(c) In a thermal reactor thermal neutrons are used.
57. When a nuclear reactor is operating at constant power the
multiplication factor (K) is
(a) less than unity (b) greater than unity
(c) equal to unity (d) none of the above.


OBJECTIVE TYPE QUESTIONS 679

58. The critical mass of fuel in a nuclear reactor is the amount

required to make the multiplication factor.
(a) more than unity (b) equal to unity
(c) less than unity.
59. The efficiency of a gas turbine open cycle with regenerator,
intercooler and reheater is about
(a) 59% (b) 29%
(c) 39%.
60. The efficiency of a modern boiler using coal can heat
recovery equipment is about
(a)65to?O% (b) 85to90%
(c) 25 to 30%.
61. The temperature of cooling water leaving the diesel engine
should not be more than
(a) 80C (b) 30'C
(c) 60C.
62. Statements
(a) Nozzles are not provided in a reaction steam turbine.
(b) In a reaction steam turbine fixed blades act both as nozzles
in which the velocity of the steam is increased and as the
means of directing the steam so what it enters the ring of
moving blades without shock.
(c) Modern steam turbines generally have the first stage
velocity compounded and subsequent stages are either
pressure compounded or reaction stages.
63. Compare the following properties of pressure compounded
steam turbine and a velocity compounded steam turbine
(a) efficiency (b) cost.
64. The compression ratio of an I.C. engine is given by
V vc
(a)v (b)
(c) V x V (d) none of the above.
where V is the total volume (clearance vol. + swept vol.)
and V, is the clearance volume.
65. pH value of a neutral solution is
(a)1 (b) 7
(c) 14 (d) none of the above.
66. The conversion ratio (S) of a nuclear reactor is given by
(a) S l (b) Sfl2 flj
fl2
680 POWER PLANT
c) S = n i x 1 12 (d) none of the above.
where ni isthe number of secondary fuel atoms and n2 is
the number of consumed primary fuel atoms.
67. The conversion ratio of a breeder reactor is
(a) Less than unity (b) equal to unity
(c) more than unity.

68. Surge tank in a hydro power plant is placed as near as

possible to the
(a) reservoir (b) tail race
(c) turbine (d) none of the above.
69. The average ash content in Indian coals is about
(a) 5% (b) 20%
(c) 10%.
70. The maximum inclination of belt conveyors for transport-
ing coal is
(a) 5 (b) 10
(c) 20.
71. State five requirements of a good air preheater.
7 Statements
(a) Gas turbine plant has lower thermal efficiency as com-
pared with a diesel plant.
(b) The capital and maintenance costs of a gas turbine plant
are higher than steam turbine plant.
(c) Thermal efficiency of a gas turbine is more than steam
turbine.
(d) Gas turbine usually operates at higher temperature than
steam turbine.
73. The ideal fuel for gas turbine is
(a) natural gas (b) pulverised coal
(c) producer gas.
74. Total cost of a diesel power plant per kW of installed
capacity is less than that of steam power plant by about
(a)70to80% (b) 20to30%
(c) 5 to 10%.
75. Compare steam power plant, nuclear power plant, diesel
power plant and hydro-power plant, as regards
(a) fixed cost (b) operating cost
(c) reliability
(d) ease in planning, design and construction.
 OBJECTIVE TYPE QUESTIONS
681
76. The specific speed of Pelton turbines are around
(a) 12-45 (metric) (b) 150-300 (metric)
(c) 350-400 (metric).
77. In high head hydro power plant the velocity of water in
penstock is about
(a) 2 rn/sec (b) 7 rn/sec
(c) 12 rn/sec (d) none of the above.
78. Pelton turbine is suitable for high head and
(a) low discharge (b) high discharge
(c) both for low and high discharge.
79. In diesel engine power plant the heat lost to cooling water
is about
(a)90% (b) 60%
(c) 30% (d) none of the above.
80. In diesel power plant the maximum allowable difference
between inlet and outlet temperature of cooling water is
normally
(a) 5C (b) 11CC
(c) 18CC.
81. State whether the following statements are true or false
(a) Nuclear power plant should be used as peak load plants.
(b) Steam power plants can be used both as base load and
Peak base load plants.
(c) Peak load plants work at low load factors.
(d) Gas turbine power plants are normally used as base load
plants.
(e) The flow of water in Kaplan turbine is axial.
(I) In Francis turbine the flow of water is radial.
(g) The, annual depreciation of a hydro electric station is
about 0.5 to 1.5% of the capital cost.
(h) The nuclear power plants cannot be operated at varying
load efficiently.
82. State three advantages of super charging in diesel en-
gines.
83. Name four boiler accessories.
84. Name any four boiler mountings.
85. Statements
(a) Water tube boilers are used only for low pressures.
(b) All large power plants use water tube boilers.

—45
 POWER PLANT
682

(c) In water tube boilers water flows through tubes and hot
gases flow outside the tubes.
86. Name two fire tube boilers.
87. State three factors which affect the efficiency of a steam
turbine.
88. State the pressure and temperature ranges commonly
used for boilers.
89. (a) Define speed factors for a diesel engine.
(b) State speed factors for low speed, medium speed, and
high speed diesel engines.
90. The approximate efficiency of a water tube boiler used for
power purposes without heat recovery equipment using
coal is
(0)45% (b) 7517,
(c) 95%.
91. The ratio of piston stroke t'o bore of c y linder for internal
combustion engines varies between
(a) 0.9and-1.9
(b) 0.5 and 0.8
(c) 0.3 and 0.6.
92. State basic characteristics of a boiler unit.
93. Compare a hydro power plant and steam power Plant..
94. State the methods of feeding various types of fuels into
furnace.
95. The velocity of water in the penstock for high head pOWCI
plant is abçut
(a) 12 m!sec (b) 7 in/see
(c) 3 im/sec.
96. The steam consumption in large turbines is aI)OUt
(a) 5 kg per kWh (b) 10 kg per kWh
(r) 15 kg per kWh (d) None of the aove.
97. The most economical average vacuum in a condenser for
steam turbine is about
(cu) 42 cm of mercury (b) 22 cm of mercury
(c) 72 cm of mercury (d) None of the above.
9S. Statements (True or false)

OBJECTIVE TYPE QUESTIONS 683

(a) The mechanical efficiency of a Pelton turbine fails

more rapidly with time as compared to Francis tur-
bine.
(b) The variation in the operating head can be more easily
controlled in Francis turbine than I'elton wheel.
(c) The thermal efficiency of a (lieSCi engine power plant
is less than thermal efficiency of a comparable size
steam power plant.
(d) It is economical to run nuclear, power plants at low
load factors.
99. State three characteristics of low grade coal-
100.Statements
(a) Initial cost of a surface condenser is high.
(b) In surface condenser a slightly better vacuum can be
obtained.
(c) jet condenser requires less floor area.
A
(d) Cooling water need not be cleaned in Jet condenser.
ft ) Moderators have low atomic numbers.
101 Name four moderators commonl y used in nuclear power
plant
102. State the functions of a steam condenser.
103. State the functions of a carburettor used in S.I. engines.
104. Statements
a) In a smoke tube or fire tube boiler the hot gases flow
through tubes.
(b) Locomotive boiler is water tube boiler.
(c) A boiler safety valve prevents the pressure in the boiler
from rising above a certain safe limit.
(d) A pressure gauge fitted on the boiler shows the pres-
sure of feed water in boiler.
105. In locomotive boiler draft produced is by
(a) induced draft fan (b) steam jet arrangement
(c) chimney (d) forced draft fan.
106. Tho proper indication of incomplete combustion of coal in
boiler is
(a) high CO content in flue gases at exist
(b) high CO2 content in flue gases at exist
(C) high temperature of flue gases
(ci) the smoking exhaust from chimney.

684 POWER PLANT

107. The % 02 by weight in atmospheric air is

(a) 20% (b) 23%
(c)79% (d) 77%.
108. The % 02 by volume in atmospheric air is
(a)23% (b) 77%
(c) 21% (d) 79%.
109. Name two types of steam pressure gauges.

110. Economiser is used to heat

(a) ooiler feed water (b) steam used in turbines
(c) condenser cooling water (d) air.
111. Draft is plus if
(a)Pg>Pa (b) Pg<Po
(C)PgPa.
where Pg is gas pressure and P. is atmospheric pressure.
112. In a nuclear power plant
(a) uranium 238 and thorium 232 are called fertile
materials. -
(b) moderator is used to slow the fast neutrons to a ther-
mal speed.
(c) thermal shielding must surround the entire reactor
core to absorb some of the radiations (beta particles,
escaping neutrons and gamma rays) produced by the
fissioning.
(d) the reflector usually completely surrounds the reactor
core with in the shielding.
113. State ,the basic factors considered in the design of a
nuclear power ractor.
114. State the formula for calculating the thickness of steel
penstock used in hydro power plant
115. State the purposes of draft tubes used in hydro-power
plant.
116. What is a dam ? Define following types of dams
(a) Gravity dam (b) Arch dam
(c) Earthen dam.
117. Classify the dams based on material of construction.
118. Classify dams based on method of design and analysis.
 4

OBJECTIVE TYPE QUESTIONS

685
119. The main force trying to stabilise the dam against water
and other disturbing forces is
(a) Foundation reaction (b) Weight of darn
(c) Earth pressure (d) None of the above.
120. The power output from a hydro electric power plant
depends on three parameters.
(a) Head, discharge and efficiency of the system
(b) Head, type of darn and discharge
(c) Type of darn, discharge and type of catchment area
(d) Efficiency of the system, type of draft tube and type of
turbine used.

121. The low head hydropower plants are those which usually
have water head
(a) less than 5 m (b) less than 15 m
(c) less than 30 m (d) None of the above.
122. Air fuel ratio required for the combustion in diesel engine
is about
(a)15:1 (b) 5:1
(c) 10: 1 (d) none of the above.
123. The air-fuel ratio by weight required in gas turbine is
(a)30:1 (h) 60:1
(c) 90: I (d) none of the nhovc
124. Name three gases used in closed cycle gas turbine plant.
125. Turbo alternators are generally rated at
(a) 0.8 power factor lagging
(b) 0.6 power factor lagging
(c) 0.5 power factor lagging
(d) None of the above
126. Define superstructure and sub-structure of a hydropower
plant.
127. Tick mark the correct answers
(A) The function of superheater is
(a) to increase steam temperature sufficiently above
saturation temperature
(b) to preheat the air entering the furnace
(c) to heat the feed water entering the boiler.
(B) The capital cost of gas turbine power plant is about
(a) Rs. 500 to 700 per kW
(b) Rs. 1500 to 1700 per kW
(c) Rs. 2500 to 2700 per kW.
 POWER PLANT
686
128. State five outstanding features of a gas turbine power
plant.
129. State the functions of a super heater in a steam power
plant.
130. State whether the following statements are true or false.
(a) Francis turbine and propeller turbine are reaction
turbines.
(h) Impulse turbine utilises the kinetic energy of a high
velocity jet of water to transform the water energy into
mechanical energy.
(c) Reaction turbine develops power from the combined
action of pressure energy and kinetic energy of water.
(ci) The operating cost of diesel power plant is less as
compared to steam power plant of equal capacity.
(e) Diesel power plants have good over load capacity.
(f) A hydro-plant can he easily started from cold conditions.
(g) Major drawback of a diesel power plant is limited unit
generation capacity.
(h) A diesel power plant has higher thermal efficiency
than a steam power plant of comparable size.
131. State two factors to be considered while selecting generat-
ing nits.
132. Name four effects of impurities present in feed water for
boiler.
133. State the factor considered in selecting the coal handling
s y stem in a steam power plant
134. The heating value of natural gas at ST.I'. is
(a) 7000 kcallm 3 (b) 9000 kcal/in1
(cj) 11000 kcal/in3.
135. The national research centre for research and develop-
ment work in nuclear energy in our country is
(ci) Bhahha atomic research centre (BARC)
(h ) Tata institute of fundamental research
(C) Saha institute of nuclear physics.
136. State four limitations in the use of nucl.ar power pIant.
137. State two limitations in the use or solar energy.
138 Name two materials used for making solar cells.
139. Statements
OBJECTIVE TYPE QUESTIONS 687

tot Tower concept t ype solar power plants uses an array

of plane mirrors or heliostats.
(h) The flat plate solar collectors are classified as low
temperature collectors
140. Regenerators used in gas turbines have an efficienc y of
about
(a)45% (b) 75%
(c) 95(4•
141. Statements:
In a gas turbine
(a) the work ratio is increased as compressor inlet
temperature is decreased.
(b) inter cooling and reheating improve the work ratio.
142. Name three types of compressors used in gas turbine
plants.
143. State the water head for the following water turbines to
be used as reversible pump turbines
(a) Propeller and Kaplan turbines
(b) Deriaz turbines
(c) Francis turbines.
114. C!assifv dams for hydro plants according to construction
materials.
it
145. State two situations under which the gravity dams ca not
be considered feasible.
116. Classify dams according to h ydraulic design.
147. State three basic t ypes ofarch dams.
148. Classify clams according to design.
149. State the turbines used for the following heads
(a) Medium heads
Ib> fUgh heads
c Very high heads.
150. Tick mark the correct answer
(at Francis turbines are radial/Axial flow tuilitns
(b) Propeller turhine are radial/axial flow turbines.
151. Define the bllowing
Runner of a water turbine
(b) Spillways for it dan.

688 POWER PLANT

152. State the actions by which the coal pulverising mills

reduce the coal to powdered form.
153. Natural draft in a steam power plant is produced by
(ci) Fans (b) Steam jets
(c) Chimney.
154. Which of the following coals has the highest calorific value
(a) Peat (/)i Legiiite
(c) Anthracite (d) Bituminous.
155. In a furnace 14 kg of air is required to burn one kg of coal
while the theoretical amount of air needed is 8 kg. Find
the percentage of excess air.
156. The commonly used steam pressure is steam power plants
is
(a) 20 kg/cm2 (b) 40 kg/cm2
(c) 80 kg/cm2 (d) 150 kg/cm2.
157. Capacity (C i ) of steam turbine in kW and capacity of
generator in kW are related as
(a)C1=C 2 xfl (b) C= C2
—
Ti
C)
(c) C 2 = -
Ti
where ri is the efficiency
generator.
of

158 Steam power plant basically works on

(a) Dual cycle (h) Otto cycle
c) Rankine cycle.
159. State two main factors to be kept in view while deciding
the size and number of generating units in a power plant.
160. pH value of water is numbered from
(a)0-7 (b) 0-10
(c) 0-14.
161. Water having pH value 7 is called
(ci) neutral water (b) acidic water
(c) alkaline water.
162. Statements
(a) Sample of water having pH value less than 7 is said to
be acidic.
(b) Sample of water having pH value more than 7 is called
alkaline.
OBJECTIVE TYPE QUESTIONS
689
(c) The operating cost of diesel power plant is less than a
steam power plant of equivalent capacity.
163. Fill in the blanks. Flue gas borne matter from steam
plants
(a) larger than I in diameter is called.......
(b) larger than 100 J.x is called.......
164. Name two of the following equipments used in hydro
power plant
(a) water conveyance structures
(b) energy conversion equipments
(c) Distribution systems.
165. Statements
(a) In a ponstock, longitudinal stresses are much less than
circumferential stresses.
(b) The length of penstock should be maximum as far as
possible.
(c) Higher velocity of fluid n penstock increases friction
loss.
(d) It is easier and economical to regulate power out put
from a hydro-power plant than stn power plant.
(e) Draft tube has gradually increasing area from exit of
turbine to the tail race.
(f Governing of water turbines keeps tL' turbine speed
constant under all working conditions.
166. The air fuel ratio used in gas turbines is
(a)15:1 (b) 50:1
(ci 100: 1 (d) None of the above.
167. The air fuel ratio in diesel engine is
(a) 15:1 (b) 50:1
(c) 100: 1.
168. Without the use of superheater a boiler produces steam of
about
(a) 80%. dryness fraction
(b) 90% dryness fraction
(c) 98% dryness fraction
(d) 88% dryness fraction.
169. How to increase efficiency of a boiler?
170. Classify reaction turbines with reference to direction of
flow of water.
 POWER PLANT
690
171. State the basis on which penstocks may be classified.
172. Name five safety measure in hydro plant.
173. State the difference between base load power plant and
peak load power plant.
174. Name five parts of a reaction turbine
175. Statements
(CL) The capacity of a boiler is expressed in terms of the
thousands of killogrammes of steam that it can supply
per hour at rated pressure and temperature
(b) In steam power plants the efficiency of large turbo-al-
ternators may be taken as 98% with air cooling and
99% with hydrogen cooling.

176. Where following power plants are located?

(a) Auraiya Gas power plant
(b) Badarpur thermal power plant
(c) Balco-captive power plant.
177. Define specific fuel consumption of an I.C. engine.
178. Solar cells are made of
(a) copper (b) Silicon
(c) Aluminium (d) None of the above.
179. State two factor to be considered for site selection ofWECS
(wind energy conversion system).
180. State three advantages and three disadvantages of wind
energy.
181. Power co-efficient (k) for wind power is given by
(cz)K= (b) K=
(c) I? - P i x P 2 (d) None of the above
where P 1 = Power of wind rotor P2 = Power available in wind.
182. Name the raw materials fbr steam power plant.
183. In biogas plants the best p1-I value for fermentation and
normal gas production is
(a) 4 to 5 (h) 6.5 to 8
(c) 9.5 to 11.5 (d) None of the above.
184. State two main advantages of pulverised fuel tiring
method.


OBJECTIVE TYPE QUESTIONS 691

185. The particle size of coal used in pulverised fuel tired

furnaces is limited to
(a) 70-100 microns (b) 10-20 microns
(c)150-180 microns (d) None of the above.
186. Combustion in coal fired boilers usually takes place at
(a) 600 to 800C (b) 1400 to 1700C
(c) 200 to 300C (d) None of the above.
187. In steam power plants the most commonly used device to
remove SO2 fri.. ¶'.ue gases is
(a) Economiser (b) Heater
(c)Scrubber (d) None of the above.

188. In fluidised bed combustion (FBC) system the combustion

of fuel takes place at about
(a) 300 - 400C (b) 800 - 900C
(C) 1500 - 1700C (d) None of the above.
189. Name three materials used as inert materials in fluidised
bed combustion (FBC) system.
190. Statements
(a) The SO 2 produced in fluidised bed combustion (FBC)
system is less as compared to pulvcrsccl fuel firing
method of coal.
(b) In high pressure boilers the tendency of scale forma-
tion is eliminated due to high velocity of water through
the tubes.
tc) In high pressure boilers the steam can be raised quick-
ly to melt the variable load.
191. Match the words of Column A and Column B.
Column A Collin B
(,_JPtltonTurbifle }'eak )d plant
. IIighhcolofwat0r
.. c !_......._4____. .i
Stoker (iit Ikavv. ...'r
lii
)_ - . . ' Eli
- Iotrtor_ -
Run . of- i.rpoWer plant - Low head of ater
it earn Power
- !iJr Turbine

192. Fill in the blanks

(a) l)iversity factor for residential consumers iss flCUIl........
umer, is
(b) Diversit y factor for large industrial con
about......
(et Pumped storage plant is used as.....

692 POWER PLANT

(d) The combustion of fuel in F.B.C. system takes place at

about
Statements
193. (a) During feed water treatment the suspended solids
which can not be removed by sedimentation are removed
with the help of filtration process.
(b) The minimum generating capacity of power plant must
be equal to the maximum demand.
194. Name two types of filters commonly used during filtera-
tion of feed water for boilers.
195. Electric power is generated at the power station by using
number of alternators in
(a) parallel (b) series
(c) both parallel and series.

ANSWERS
1. (a) 2. (ix) 3. (b) 4. (a)
5. (c) 6. (a) 7. (a)
8. (a) True (b) True (c) True (d) False (e) False (/) True (g) True.
9. (a) 10. (a) False (b) True (c) True.
11. (a) 12. (a) 13. (a) 14. (i) True (ii) False
15. (c) 16. (b) 17. 'b) 18. (C)
19. (a) True (&) True 20. (c) 21. (b) 22. (a)
23. (i)U.P. (ii) Tamil Nadu (iii) Tamil Nadu (iv) Delhi
24(c) 25. (b) 26. (c) 27. (b)
28.(a) 29(a) 30. (a)' 31. (b)
32. (z) Natural gas (ii) Blast furnace gas
3. (a) Laminated asbestos (b) Mineral wool (c) Glass wool.
34. (c)
35. (a) 100 in and above (b) 30 to 50 in (c) 25 to 60 m
36. (a) 10 to 35 (b) 60 to 300 (c) 300 to 1000. 37. (a)
38. (a) 39. (b) 40. (b)
41. (a) True (b) True (c) False (d) True (e) False (/) False.
42. (a) 43. (a) Francis turbine (6) Pelton turbine.
44. (a) True (b) True (c) True (d) True (e) False (J) True (g) True.
45. (b) 46. (b) 47. (c)
48. (a) Gravity (6) upstream, down steam.
49.(c) 50.(c) 51(c) 53(c)
54. (a) True (6) True 55. (cx)
56. (a) True (b) True (c) True
57. (c) 58. (b) 59. (b) 60. (b)
61. (c) 62. (cx) True (b) True (c)True
OBJECTIVE TYPE QUESTIONS
693
63. (i) Pressure compounded steam turbine is the most efficient
turbine because the ratio of blade velocity to steam velocity
remains constant whereas velocity compounded turbine has low
efficiency because the ratio of blade velocity to steam velocity is
not optimum.
(ii) Pressure compounded steam turbine has large number of stages
and, therefore, it is most expensive whereas a velocity com-
pounded steam turbine has relatively fewer number of stages
and hence its initial cost is less.
64. (a) 65. (b) 66. (a) 67. (c)
68. (c) 69 (/,) 'in
71. (i) It should have high thermal efficiency.
(ii) Its initial cost should be low.
It should occupy small space.
(iv) It should be easily accessible and serviceable.
W) Its maintenance cost should be low.
72. W True (ii) False (iii) True (iv) True.
73. (c) 74. (b)
r y ___I__ - . .
75. I__suz, Hydro electric power plants involve nvolve large capital expendi-
tures. The capital cost of a diesel power plant is lowest.
(b) Operating cost of steam power plant is more than..a hydro
power plant.
(c) Reliability of hydro plant, steam power plant and nuclear
power is almost equal whereas reliability of diesel power
plant is more than other plants.
(d) Planning, design and construction of a nuclear power plant
is very difficult and takes long time whereas planning design
and construction of steam power plant is easier than hydro-
power plant. Planning design and construction of a diesel
power plant is the easiest.
76(a) 77(b) 78.(a)
79(c) 800)
81. (a) False (b) True (c) True (d) False (e) True (1) True
(g) True (h) True.
82. The ad'antages ofsuper charging in diesel engine areas follows:
(i) It increases the output of the engine.
(ii) It overcomes the effect of high altitudes.
(iii) It reduces the weight of the engine per horse power
developed.
83. (i) Feed pumps (ii) Feed water heater
(iii) Air preheater (it') Draught equipment.
84. (a) Stop valve (b) Safety valve (c) Fusible plug (d) Pressure
gauge.
85. (a) False (b) True (c) True.
86. (i) Locomotive boiler (ii) Cornish boiler
87. (t) Steam pressure and temperature at the throttle valve of
turbine.
(ii) Exhaust steam pressure and temperature.
(iii) Number of bleedings.
 POWER PLANT
694

88. The pressure and temperature ranges for steam generators are
as follows
(1)70 to 140 kg/cm 2 steam temperature range of 450 to 560 C.
(u) 56 to 70 kg/cm 2 steam pressure with temperature range of
440C to 480°C.
(in) 28 to 56 kg/cm 2 steam with temperature range of 400 to
440 C.
(iv) 17.5 to kg/cm 2 saturated at 400'C.
(') 9 to 17.5 kg/C,112 saturated steam superheated to 65 C super
heat.
89. (a) The speed factor (S.F.) of diesel eng,ne is given by
S.F. =-----
3048 x iü
where N = Rotational speed in R.P.M.
V = Piston speed in centimetres per second.
Speed factor is also given by

1524 x
where 1, stroke of piston in centimetres.

)h) The speed factor for low speed engine is less than 1.2 whereas
for medium speed engine it varies between 1.2 and 3.5. For h h
speed engines the speed factor is between 3.5 and 11.
90. (b) 91(a)
92. Boiler units consists of a steam boiler, a furnace a nw
auxiliary device for heating water and air, and draught e,
meritetc. 'l'lie basic characteristics ofa boiler unit are its capacit
and steam parameters. The capacity of boiler unit is the amount
of steam in kilograms produced per second. The parameters
characterising steam are pressure and temperature of super-
heated steam and pressuie and dryness fraction for saturated
steam.
91 Comparison of hydro and steam power plants.
These plants are compared as follows
(i) Capital Cost. The capital cost of hydro power plant is high as compared
to steam power plant of equal size.
(ii) Operating Cost. The operating cost of both the plants is about same.
(iii)Maintenance Cost. Maintenance cost ofstcani power plant is high as
compared to hydro power plant.
(it) Location near load centre. Steam power plant can be easily located
near load centre. Whereas hydropower plant cannot be located near load
ce.ltre.
,t') Sterling period. Hydro power plant can be quickly started from cold
Cofldtti')flS whereas it takes about 5 to 10 hours to put steam power plant
uperition.
OBJECTIVE TYPE QUESTIONS 695

it) Load Sharing. When both steam power plant and hvd ro power plants
are used to suppl y the given toad it is preferable to use hydropower plant
as base load plant if sufficient amount of water is available at power plant
Site and steam power plant is used as peak toad plants However if the
amount of water available at hydro power plant Site is small then it is
desirable to use steam power pla ifl as base load plant and hydro power plant
as peak load plant
94. Solids fuels like coal, cock etc. are burnt on grate in the furnace
stokers are used to feed the fuel automaticall y . Air enters from
bottom of grate and burns the fuel
l'ulverised fuels enter the furnace through burners. The flame in the
furnace ignites the fuel and it burns ii) SUSp'flSiOfl.
Oil fuel is supplied to furnace t rough b.irners which atomise the oil
into in in ute droplets and mix it with air. The mixture ignites and burns
Gas fuel is also supplied through burners. However gas needs no
at o in is a tio ri
95. (b) 96. (a) 97. (c)
98. (I) True (ii) True (iii) False (it) False.
99. t(1) High ash contents (h) Excessive moisture content (c) Low
calorilic value.
100. (a True (II)True (c ) Irue (il) True (0) True.
101, The moderators commonl y used are as follows
(i) Light water (1120)
(ii) Det.urium or heavy water (1)20)
tie) Carbon
(ic') Beryllium.
102 The function ofa steam condensers are as follows
0 1 Exhaust pressure or back pressure of steam is redtice t.herby
the capabilit y of prime mover to work is increased.
(ii) Exhaust steam from prime mover is condensed into water v.,hich
IS re-used as hot feed to boiler. B y doing so saving in watti- cost
in av he substa n tial in large steam power plant.
103. The function of it carburettor is to discharge into air steam the
desired quantity of liquid fuel, to atomise it and to produce a
homogeneous air fuel mixture.
104 1 1;) True ( h ) False )c) True (ii) False.
105 0 106. (a) 107. (h)
lOS. nt
109. There are usuall y two types ofstearn pressure gauges
I (ou rd on pressure gauge
Ut Diaphragm pressure gauge
lit). (z) Ill. ((1) 112. ((iTrup(b)True)elTrueftflTriie
113. The basic factors considered during the design of a nuclear power
reactor are as fell ''vs
(i) Reactor type
i:) Power rating ufrenctom in megawatts
(it,) (...nla ot SyStem
(it ) Control svste in
(I) Rates of neut run production and tit> ion
696 POWER PLANT

(vi) Safety of the reactor.

114. The thickness of a steel penstock depends on head and hoop stress
allowed in the material. it is given b y the following formula
= Thickness of penstock
0.1 HD
2fti
whore H = Head in metres
D = Diameter of penstock in centimetre
1= Permissible stress in kg/cm2
= Joints efficiency
115. The purposes of a draft tube are as follows
(i) To reduce velocity head losses of water and to use head on
reaction turbine to the maximum extent.
(ii) To permit the installation of runner of reaction turbine at a level
above that of water in tail race.
116. Dam. A dam is defined as the structure built across a river to
store water.
Gravity Darn. Depending upon the material of construction a gravity dam
may be either masonry gravity or concrete gravity. A gravity dam resists
the forces mainly water pressure acting to disturb it by its weight, hence
the name is gravity. Therefore, a gravit y dam has to be very massive, so as
to be able to withstand large pressure due to water.
Arch darn. An arch dam is usually made in the shape ofa circular arc and
is made of concrete or masonry. Whole of water load is transferred to the
abutment of.sides due to arching action of darn Section. This type of darn is
specially suited in narrow valleys.
Earth darn. When a darn is made of earth of locally available soil it is
called earth darn. it is used when effective height of dam is not large.
117. Based on material of construction the dams as classified as
follows
(i) Masonry dam (ii) Arch dam
(iii) Earth darn (iv) Rockfill dam
(v) Timber dam , (vi) Steel dam.
118. Based on method of design and analysis the dams are classified
as follows
(t) Gravity darn (ii) Arch dam
(tit) Buttress darn (iL) Others as steel, timber etc.
119.(b) 120. (a) 121. b)
122.(a) 123(b)
124. Helium carbon dioxide and ammonia.
125. (a)
126. The super Structure provides protective housing for the gener-
ator and control equipment as well as structural support tr the
cranes
The sub-structure consists of steel and concrete components
necessary to form draft tube, support the turbine staving and
generator and encase the spiral case.
127. A (a) B (a)
 OBJECTIVE TYPE QUESTIONS
697
128. The outstanding fetures of a gas turbine power plant are as
follows
(i) Low capital cost.
(ti) Quick starting.
(iii) Capability of using wide variety of fuels from natural gas to
residnalojl or powdered coal.
(iv) High reliability and flexibility in operation.
(i') Higher efficiency (about 37%).
129. The functions of a super heater are as follows:
(i) It removes the last traces (1 to 2%) from the saturated steam
coming out of boiler.
(ii) It raises the temperature ofsteam sufficiently above the satura-
tion temperature
(iii) It avoids too much condensation in the last stages of turbine
which avoids blades erosion.
(iv) It raises over all efficiency of the cycle.
130. (a) True (b) True (c) True (d) True (e) True (f)
(h) True. True (g) True
131. (i) A power station should have at least two generators.
(it) The best way of deciding the size and number of generating sets
in a power plant is to select the number of sets in such a way so
as to fit in the load curve as closely as possible so that the
capacity of plant is used efficiently.
132. Feed water impurities may cause
(i) Scale formation (ii) Corrosion
(iii) Carry over (iv) E mbrittlement.
133. Mechanical systems ii a steam power plant move the coal to
storage or to the furna .. Significant factors considered in select-
ing coal handling systns are as follows
(i) Plant fuel rate
(ii) Coal storage area
(iii) Plant location in respect to coal shipping.
134.0) 135.(a)
136. Limitations in use of nuclear power plants are as follows:
(1) High capital Cost of nuclear power plants.
(ii) Limited availability of raw materials.
(iii) Difficulties associated with disposal or radioactive wastes.
(iv) Shortage of well trained personnel to handle the nuclear power
plants.
137. Limitations in the use of solar energy are
(i) Solar energy is not available at night or when local weather
conditions obscure the sun.
(ii) Solar energy is diffused in its nature and is at a low potential.
Consequently if solar energy .s to be economically competitive
then it must be converted into a usable form of energy with
maximum effectiveness to reduce the capital cost of solar plants
138. (1) Silicon (ii) Gallium arsenide. 139. (a) True (b) True
140. (b)
141. (a) True (b) True
-46

•1
 POWER PLANT
698
142. (i) Axial flow (ii) Centrifugal type (iii) Positive displacement.
143. (a) Less than 20 m (b) Less than 150 in (c) Less than 500 in.
144. (i) Stone masonry (ii) Concrete and R.C.0 (iii) Earth (iv) Rock
pieces and fragments.
145. (i) When good rock y foundation strata are not available
(it) When height of darn is more than approximately 250 in.
146. (t) Liver flow dams
(u) Non over flow dams.
147. Arch dams can be described under the following three types.
(i) Constant radius arch dam
(ii) Constant angle arch dam
(iii) Variable radius an6 variable angle arch darn.
148. (i) Gravity darn (ii) Arch dam (iii) Buttress darn (iv) Earthen
dam.
149. (i) Kaplan or Francis turbine
(ii) Francis or Pelton turbine
(iii) Pelton turbine.
150. (i) Radial (ii) Axial.
151. (i) Runner is that portion of a reaction turbine which revolves
and converts the water head into mechanical energy.
(ii) Spillways are the structures provided with darn to allow a
-,I

safe passage of excess water front reservoir to the down

stream side of dam without'ovcr toppling the darn.
152. (i) Impact (ii) Attrition (iii) Crushing.
153. (c) 154. (c)

155. % excess ir =- 14 8 x 100
=75%
156(d) 157(b) 158.ie)
159. (i) A power plant should have at least two generators
(ii) The generating units should be selected in such a wa y so as
to fit in the load curve as cloely as possible in order that tltQ
capacity of the plant is used efficiently.
160. (c) 161. (a)
162. (a) True (b) True (C) Tru.
163. (a) dust (b) cinders.
164. (a) Penstocks, draft tube (b) Turbine, generator (c) Substation.
transformer.
165. (i) True (u) False (iii) True (iv) True (t ) Trite (t) True
166.(c) 167(a) 168, (c)
169. Efficiency of a boiler is increased by the use of
(i) air preheater (it) superheater
(iii) economiser (iv) feed water heater,
170. With reference to direction of flow of water, react to" turbines are
of following types.
(t) Radial flow inward In such turbines the runner receives ater
under pressure in a radially inward direction and discharges it in a sub'
stantially axial direction. Francis turbine is radial flow inward turbine.


OBJECTIVE TYPE QUESTIONS

699
(ii) Axial flow turbines In these turbines the runner vanes are either
fixed or adjustable. Fixed vane type is preferred where head and flow are
substantially constant and where base load operation is possible. Adjus-
table, vane type is preferred where head and flow vary over a very wide
range and the plant is subjected to variable head operation. These are
propellor ty pe turbines.
171. The penstocks may be classified on the basis of
(z) the material of fabrication
(ii) the method of support
(iii) rigidity of connections and supports
172, (i) Surge tanks (ii) Spillways (iii)
Relief valv e u) Trash screen (u)
Sand traps.
173. Base load plants run throughout the year
They operate at high load factors.
Nuclear power plants are used as base load plants and have as high
as 80 and more load factor. F1ydropowr plants with ample storage are
used as base load plants.
Peak load plants run for a few hours in the year and operate at low
load factors. Hydro power plants with limited storage of water are used
as peak load plants. Pumped storage plants are always used as peak load
plants. Steam power plants can be used both as base load plarc a''
peak load plants. Diesel and gas turbine plants are used as pe
plants.
174. (i) Spiral casing (ii) Runner (iii)'
iii) .Runner shaft (iv) Guide wheel (v
Draft tube
17 5.(a) True (b)True
176.(i) Auraiya Gas Power plant: It is located in Uttar Pradesh.
It is Jndi?'S largest combined cycle module which has an in-
stalled Capacity of 652 MW.
(ii) Badarpu- thermal power plant: It is of 720 \lV capacity
and is located at Badarpur in New Delhi.
It is being managed by National thermal Power corporation
(N.T.P.C.) of India.
(c) l3alco Captive power plant is located at Korba in Madhya
Pradesh,
177. Specific Fuel Consumption (S.F.C.)
It is the ratio of amount of fuel used by the engine per hour to the horse
power produced or delivered by the engine.
S.F.C. =
where S = Amount of fuel used by the engine (kg/hr.)
H.P. = House power produced.
It is one of the most important parameters used in comparison of engines
when SFC is based on I.H.P. (Indicated Horse Power) produced it is
termed as Indicated Specific Fuel Consumptio t ISFC) and when S.F C
is based on B.H.P. delivered It is called Brake speeific fuel consumption
(BSF'C).

700 POWER PLAi

178. (b)
179. (i) The best sites for WECS are found offshore and the sea Coast.
(ii) The second best site are in mountains.
180. Advantages:
(i) It is a renewable source of energy
(ii) It is non-polluting
(iii) Low cost.
Disadvantages:
(i) It is dilute and fluctuating in nature.
(ii) It is noisy in operation.
(iii) Large areas are needed.
181. (a) -. 182 Air, fuel and water
183. (b)
184. (i) Pulverised fuel firing method can handle successfully high as)
coals which can not be fired easily by conventional burninl
methods.
(ii) High temperature is the furnace can be achieved.
185. (a) 186. (b) 187. (c)
188. (b)
189. (i) Dolomite (ii) Fused alumina (iii) Zirconia
190. (i) True (ii) True (iii) True.
191. (i) (e) (ii) (0) (iii) (d) (iv) (b) (v) (/) (vi) (c)
192. (a) 5(b) 1.3 (c) Peak toad plant d) 800-900C
194. (a)True(b)True
195. (i) Pressure, filters (ii) Gravity filters
196(a).
 Appendix A
CONVERSION TABLES
(i) Length
- 1 inch = 2.54 cm
1 foot = 30.48 cm
1 yard = 0.9143 metre
1 metre 3.28 feet = 39.37 inches
1 t(micron) = 3.281 x 10- 6 ft
=O.00lmm
(u)Area
1 M2= 10.7639 ft2
1 inch 2 = 6.4516 cm2
(iii) Weight
1 ton = 1.016 tonnes
1 kg = 2.204 lb
llb=453.6gm.
I Imperial gallon of water weights 10 lb
(iv) Volume
1 Cu. ft. = 0.0283 cu. m.
1 Cu. in. = 16.39 cu. cm.
I Imperial gallon = 4.543 litres.
(v) Density
1 kg/rn 3 = 0.062 lb/ft3
1 lb/ft2 = 16.02 kg/M2
(vi) Energy
1 ft. lb = 0.13825 kg.m
1 kcal (kilocalories) = 3.961 B.T.U.
1 H.P. (F.P.S. units) = 746 watts
1 metric H.P. = 735.5 watts = 0.7355 kW
1 H.P. = 2544 B.T.U. per hour
1 kcal/kg = 1.8 B.T.UJIb
1 metric H.P. = 4500 kg-m/minnte
= 10.54 kcal per minute
1 British H.P. = 550 ft. lb per second
1 kWh (kilowatt-hour) = 3413 B.T.U.
702 POWER PLANT

= 860 kcal.
1 erg = 2.78 x iO kWh
1 joule = 107ergs
1 kg-rn = 2.34 x 10 kcal
1 kW = 1 kJ/sec.
1 eV = 1.6 x 1012 erg = 1.6 x 10-10 joules.
(vii) Pressure
1 standard atmosphere = 14.696 p.s.i
= 1.033 kg/cm2
29.92 inches of mercury
= 760 mm of mercury
= 10.332 tn HO
= 33.8985 feet H20
1 cm of Hg = 0.01359 kg/cm3
1 p.s.i. = 0.0703 kg/cm!
1 kg/cm3 = I ata = 754.6 mmHg.
1 mm of'iater = 1 kg/rn2
(viii) Temperature
K=273+C
°R= 460 + 'F
where °K = Degree Kelvin
= Degree Rankiiie
'C = Degre Centigrade.

j4;' •


Appendix B
Properties of Dry Saturated Steam

Entropy
Pressure Snfuru Sensible Latent Total I Specific Liquid Vapour
p kg/sq lion heat hw heat L hct ii ilunie Ol
cm tempera- k(-al/kg kcal/kg kcalikg Vsin'/kg kcal/kg kcal/kg
lure
K
T (C)

0.02 17.1
L.Lj
I 17.1 5.8 6022 6827
H(7)
2.0803
0.0428.6 286 580.3J_ 6809 3546 0.0936 2.0219
0.06 35.8 3 5.8 6114 24.19 0.1232 1.9880
0.08 41.1 41.1 572.9 614.0 18.45 0.1407 1.9642
0.10 45.4 6570.5 6159 1 14.96 0.1539 1.9458
0j2 49M 686 6176 L' 26 ° 0J652 1.9308
0.13 59.6 660 619.6 I 10.22 0A892 1.9126
0.20 59.7 63.6 622.3 7.797 0.2976 1.8892
025 6459k q 6245146.325 2I24 1.8712
0.30 68.7 57.9 -.626.3
- 5.331 0.2242 J8562
0.35 72.3 72.3555.5 627.8 4.614 1 0.2348 1.8444
0.40 75.4 75.4 553.8 62 .2 02439
1.6334
0.50
0.60
80.9
85.5
809
855 f547.0
L550.6 631.5 3.304 LO 2593
1.8156
633.4 L2.785 L0.21.8019
0.70 89.3 89.3 545.8 635.1 2.411 0.2832 1.7889
0.80 93.0 93.0 ..J 543.5 6362 2.128 0.2931
0.90 96.2 .--962
962J 541.4 637.8 1.7693
1.806 L03018
1.0 _3.9.1 i 536.8 639.0 1.727 0.3056 1.7600
104.1 104.1 536.9 641.0 1.457 0.3235 1.7467
1.4 118.7 108.9 533.9 6428 1 261 0.3354 1.7344
2 7 112.9531.4 644.3 1.113 0.3460 1.7283
1.86.3 116.5 529.1 645.7 0.996 0.3554 1.7148
96 1199 527.0 646.9 0 903 0.3639 i
2 6.8 127.2 522.0 649.5 0.7341 0.3823 1.6888
M2.51 2.2 133.4 5182 651.6 0.6180 1 0.3977 1.6748
3.5 1382 138.4'1 514.5 653.4 0.5352 0.4 110 1 6625
4.0 142.9 1417 1 511.2 654.9 0.4418 0.4227 1.6515
4.5 - 147.2 148.1 L £P-±4 33 1 6425
5.0 rMJ15245L .2.42.3825iQ!.6311
55 154.7 155.9 5025 658.4 0.3489
--- -.----
0.4515 16295
1
6.0 158.1 I 159.4 L 499.94 6593 03222 1 0.4596 1.6165
O29O4671l 165
7.0 164.2 165.7 495.2 I 6609 1 -
0 2785
-- -0.1742 1.6075 1
L1 5 168.7 L4930 6617
02609 0.4808 ----I

L60 . [_ 0 .J7j4 - 430.9T6623 1 02111


704 POWER PLANT

i3 (5) I
L8.5 !_i- 1 u -662.9 T 2217
19 50I3
9.0 -14.5 I7'• 4..' 663.4 0.2195 . 04985 15562
10 --1179.0 181.3 4, 1 i 664.40i985 05090 1.5775
11832 147920181:1 i 08615698
12
L_ 187.1 189.9 470.1 665.9 9.1668 0.5335 1.5628
13 190.7 193.9 473.0 666.6 0.1415 0.5358 1 5556
--
L 14 4 4.S' 4i
l 3I 469.4 L 7.o I oi4s8o5336 i5493
15 1974 ,j, 200.7 I 466.7 667.4 0.1346 1 0 5508 1 5432
16 200.4 204.0 163.4 667.8 0,1164 0.5577 1.5375
17 203.4 207.1 461.1 6682 0.1192 j.5643 15321
18 206.2 210.1 1 458.2 1 6683 0128 0.5705 1.5270i
19 L 2088 . 455.5 668.5 0.1070 05764 1.5220
20 211.4 215.8 452.9 668.7 0.1027 9.5821 1,5173
22 216.2 221.0 4479 668 9
-_ L 220.8 6.04 JiI.0 401)350L06026 '15504
268 40.0785 I 1.4 92 1
26 829.0 2250 433.8 668.8 0.0725 0.6205 L4850 -1
30 i 232.6 23111 --429.5
- 668.6 0.06802 0.6287
"t'r - 1.4780
32 2364 243.1 425.2 .4 0.06372 06364 -- 1.4813
344_239MJ 421.1 6685 0.05991 37±L450
L 230.1 417.1 L. 0.0565 1 .
[_38 24.2 -. 254.1 413.0 667 1 005365 9,6573 1.4530 I
F40 i 2492 257.4 409:2 6666 0.05169 0.6637 114474
E 42 252.1 267: 405.666.5 0.04817 0,6698 1 1.4478
44 254.5 26394 401,6 6665 01)4588 1.4365
L 46 257.6 266.9 397.9 6648 0.04378 0.6813 J 1.4314
48 260.2 269.8 394.2J 664.1 0.04185 L_1215
50 2627 272.7 390.7 0.04007 0.6921 I 14215J
55 268.7 279.6 819661.5 0. O3616t' 7046 1.4098
60 - 274.3 286.1 . 373 4 1 6595 003289 0.7162
65 279.6 292.2 . 365.3 0.03009 0.7270 .1. 2211
70 284.5 299.0 359.3 1 6573 0.02469 0.7371 j 1.378J
Ej80 1 293.6 308.8 341.8 650.6 L 0.02374 0.7557 1.3591
90 . 301.1 319._j 326.6 I 6456 L 0.02064 0.7731 _1413
100 1 309.5 3287 318.8 640.5 1 001815 , 07893 13245
150 340.5 374.1 338.8 612.9 001044 1 3.8622 1.2514
200 4272 H728 oTo 9404 1.1715
LP_ 1 _374.0 1 501.1j 501.1L0 00310Q


APPENDIX
705
Properties of Superheated Steam

flT1/ heat llkt,,1'k

SIa r,'
1.
kI?
nfl

Th ion j.fl 20(1 2.51) 300 3.511 4(1(1

(1.01 OOo J195 542 664 5. 6,7 5 7105 734 758 7820 807 822 s 858 s
005 6115 61USj642 664 5 687.5 710.5 734 758 7820

0.10 617 619 642 664.5 687.5 710.1 734 758 782.0 807 832.5 1
834.3
0.51J 6:31.5 641 664 687 710 734 737.5 782 807
1.00 639.3 - 632 663 686.3 710 733.5 757.5 782 806.5 S32 858

500 6573 - - 682 701 7:31 737.3 780.5 805.5 1$37

100 664.4 - - - 676 702.8 728.1 733.3 778.4 804 830- 856
1.5.0 667.4 - - - 608.5 698.6 75 751 776.5 8(12 5 825.8 833
200( 658.5 - - -- 694 721.9j 748.5 871 827.5 854
25.6( 669(1 - - -

th
- 669 718.5 746 772.5 7995 826 853
3001 (108.6 - - -
- 683.5 715J435 771 797.5 824.5 85

b534: 763 791 .5 1 8 19.6 S4761

Properties of Dry Standard Steam

Suiurajjg.j,, Sp.ci/Ie L',i.j h.ot L
Teinp'r,i ,zr. Tojuih..otJf
T( C) V, Af3lkg

W00623 3GO.3 1 597.2 .597.2

(1007 5966

000829
I FZ 108.2
157.3
595.3
5:150
4_js 5
399.0
I 0.1095:3 I
147.2
137.8 8


706 POWER PLANT

7
( 2)
(1 01020
0.01090
_iii.1IiiIJFiII.
129.1
121.0
59:1.2
592,))
600.2
600.7
8
9 0.01170 113.4 592.1 600.1
10 0,01251 106.4 591.6 61)1.6
0.01338 -1 . 602
12 0.01429 93.90 590.5 602.5
13 (1.01556 88.19 589 9 602.9
14 0.0163 82.91 589.4 603.4
15 0.0174 77.90 588.8 603.0
16 0.0185 73.40 588.3 601.3
17 0.0197 69.10 587.7 604.6
18 0.0210 65.10 587.1 6(15 1
19 0.0224 69.35 586.6 605,0
20 0.0238 57.81 586.0 606.6
21 0.0254 54.56 585.5 606.5
22 0.0274 5 1.49 544 9 "06 0
23 0.0286 48.63 584.3 607.3
24 0.0304 45.L4 583.8 6(17.8
25 0,0323 43.41 583.3 608.2
26 0.0342 41.04 582.6 608.6
27 - 0.0263 38.82 582.1 609.5
28 0.0386 36.78 581.5 6(19.5
29 0.0408 34.77 581.0 . (100
30 0.114:32 32.93 581) 4 6(0.4
31 0.0458 :31.30 579.8 610.8
32 0.0485 29.60 579.3 611 .3
33 0.0513 28.05 578.3 611 '7
34 ()0542_, 26.61 578 1_,1___
612
36 0.0606 7 22 57 57711 613.))
613.4
37 0.064)) 22.77 56.4
38 0.0676 21.66 575.9 613 9
39 0,0713 28.56 5751 614.6
40 0.0752 19,55 574 7 614.3
41 0 ((793 18.60 574,1 615,2
42 00836 17.70 573.6 61.5,1;
43 0E01i81 16.85 5730 616.1)
44 0,11928 16.04 572.4 616.1
45 00977 15,28 571.8 611; 8
- 46 0.10S 14.56 571 2 17 2
47 - ().103 1:3.48 570.7
JOTh 13,23 - 570.1
49 120 _,j;1__-, 618.5
APPENDIX

ii ---IIIII1I
J
50 ,__
(2)
0.126 12.03
(4)
569.0 61 'JJJ
707

51 0.160 9.544 566.1 621.0

52 0.2(1.3 7.682 563.2 623.2
53 0.255 6,2(15 560.3 625.2
54 0.318 5.049 567.7 627.3
75 0.398 4.136 554.4 629.3
80 0.383 3.410 551,3 631.3
- 85 0.589 2.830 T4 633.2
90 0.715 2.361 635 1
1.981 542.0 637.0
100 1 -03:32 1.673
538.9 6:18.9
105 1.2318 1.419
535.6 6405
lie, 1.4609 1.210 539.4 642.5
1.7238 329.1 644.3
.0245 525.7 (; 46.0
125 0.770 522.4 617.7
130 0.668 518.9
135 (1.582 515.2 650.8
140 (1,508 511.6
145
652.5
4.237 () 446 558.9
654.0
150 4.854 (1.392
655.5
155 5.540 0,3:164 500.8 656.5
5.302 0.3068 658:1
7.141 '1.2724 492.1
659.6
170 8.078 0.2425
660.9
175 9 101 0.2166 663.1