Chapter 1-5

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
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
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.
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
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

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-
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
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.

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

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
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:
(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:

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

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%

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?

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.
(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
(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
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
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.
(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
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.
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.
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.

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.
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
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
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.

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.

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.

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

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

- Diversity factor
20 t 1±J2. = 40 MW.
-- 15

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
= 0.45 x 8 = 3.6 MW = 3600 kW
=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.
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.
= 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
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
= (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
= 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

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

_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.
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
(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

=0.25x700= 175 kW.

= 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.
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
= 10 x 0.70 + xO.746 x 0.75 + 4 x 0.70
= 7 + 4 + 2.8 = 13.8 kW
Diversity factor = Simultaneous maximum demand
•. Simultaneous maximum demand on transformer 1

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

= 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.

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
= 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..
(a) Reserve capacity

(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
= 0.6 x 125 = 75 MW = 75 x 10 kW.
= 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
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
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.
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.

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.
= 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

(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
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
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
(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:
Capacity factor = 55%
Installed capacity of the plant = 200 MW
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:
(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
- -
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

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


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
= 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
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
= 30 MW
= 130 x 106 kWh
M = Maximum demand
= 25 MW.
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
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
= 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
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
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 --
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)


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%
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 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- -
(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;
(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
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.
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.
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
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)

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.
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
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

(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
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

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)

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.
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
(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

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

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
[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).

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
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)
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

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(


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.
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

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

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 --
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.
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

- 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.
• _________
-
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

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.

"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.

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.
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.

(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-
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.

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.

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.
(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
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.
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.
(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.
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.

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
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
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
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.
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
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)
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.
..
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

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.

(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.
(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.
(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.

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.
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
(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 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.
(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
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
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
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.

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.
(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.
(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.
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.
(a) Steam turbine performance
(b) Steam turbine generators.
3.35. Describe environmental control of steam power plants.
(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.
(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.
(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.
(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
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.

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.
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.

(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


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
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

.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
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
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
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
(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
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
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.


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.
(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


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.

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.
(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'


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.
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.
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
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
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).
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)

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'
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. -
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


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 -

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.
(a) Boiling water reactor (B.W.R.)
(b) Pressurised water reactor (P.W.R.)
(c) Multiplication factor.
(d) Fertile and fissionable material.
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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1

As tchnology is advancing the consumption of power is steadily

plants or b y thermal power plants depending upon the relative

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

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

sunlight is abundant for a major part oft l i

Perhaps the great advantage of solar power is that the system

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

(i) Gulf of Cambay

Disadvantages. The various disadvantages of tidal power

sector COUSUWOS nearly 379r, of total coal produced aH

1.4 Calorific Value of Fuel

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

where Ki . constant 0.35 for Bituminous coal

= 0.63 for BituminoUs coal

1.6 Types of Coal

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

Table 1.1 (A)

Fixed curb- Volatile Fixed carb•

Calorific Value, kcaJkg

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

AFuelsp^xjmate Ana1YSS.% Witrrute Analysts,%

5 1.5 05 11 7'00 7700

volatile _ J 1 7 6750 7100

an thrac ite 1 7500 7700

1.7 Liquid Fuels

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

1.7.1 Oil Properties. Petroleum or any of its derivatives is

1.9 Gaseous Fuels

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

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

(d) ProducerGas. Producer gas is produced tliirint incomplete

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

10 Advantages of Gaseous Fuels over Solid Fuels

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

ll€-.. fuel 0.95 86.1

1.11 Comparison of Sources of Power

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

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

where the user is benefitted from assured uninterrupted supply and

Year _J__. Generation in Billion k

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

I 1951 575 1.261 - 1.836

To improve power generation the generation capacity of power

change power between inter connected systems. Use of UHV and

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

petroleum producing countries. The world is mostly depending on

ethanol programmes using maize in the USA and sugarcane in

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

If the amount of oxygen supplied is not sufficient the combustion

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

1.16 Combustion of Gaseous Fue!s.

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

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

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

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

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

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

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