1. Annual MW.

hr generation of a 120 MW power plant is 770, with a maximum peak load of 105
MW. If the annual average load is 99.66 MW, find the annual load factor.

Solution:

Annual Average Load

Annual Load Factor =
Annual Peak Load
Thus,

99.66 MW
Annual Load Factor =
105 MW
Annual Load Factor =0.9491
2. Data of a 150,000 KW power plant: average load – 87,000 KW, operating hours – 16, determine
the use factor.

Solution:

kw−hrs generated ∈a period

Plant Use Factor=
( Actual no . of hours )(Rated Capacity)
Where,

kw−hrs generated /day

Average Load=
24 hrs/day
Thus,

kw−hrs generated per day=Average Load 24 ( hrs

day )
kw−hrs generated=(87,000 kW )(24 hrs)
kw−hrs generated=2,088,000 kW −hrs
Hence,

2,088,000 kW −hrs
Plant Use Factor =
( 16 hrs ) ( 150,000 kW )
Plant Use Factor =0.87
3. The plant utilization factor of a plant is 80% and a capacity factor of 75% and average load of
20,000 kW. Determine the maximum demand of this plant.

Solution:

Maximum Demand
Utilization Factor=
Rated Capacity
Also,
 Ave . Load
Capacity Factor=
Rated Capacity
Thus,

Ave . Load
Rated Capacity=
Capacity Factor
20,000 kW
Rated Capacity=
0.75
Rated Capacity=26666.6667 kW
Finally,

Maximum Demand=(Utilization Factor)( Rated Capacity)

Maximum Demand=0.80(26666.6667 kW )
Maximum Demand=21333.3333 kW
4. Given load factor 0.87, installed capacity 35,000 KW, reserve over peak 10,000 KW, hours out of
service per year 400. Find the capacity factor.

Solution:

Ave . Load
Load Factor =
Peak Load
Where,

Rated Capacity=Reserve Load+ Peak Load

Peak Load=Rated Capacity−Reserve Load=35,000−10,000
Peak Load=25,000 kW
Thus,

Ave . Load=Load Factor x Peak Load=0.87 (25,000 kW )

Ave . Load=21750 kW
Then,

Ave . Load 21,750 kW

Capacity Factor = =
Rated Capacity 35,000 kW
Capacity Factor=0.6214
 5. A Power station supplies the following loads to the consumer

Time in 0-6 6-10 10-12 12-16 16-20 20-22 22-24

hours
Load in 30 70 90 60 100 80 60
MW

What is the load factor of a standby equipment of 80 MW capacity if it takes up all loads above
45 MW?

Area of CLC= ( 2 ) [(
y 0+ y 1 '
x1 +
y 1 + y 18
2 ) ]
+ y 2+ y 3+ y 4 + y 5 + y 6 + y 7 + y 8 + y 9 + y 10 + y 11+ y 12 + y 13+ y 14+ y 15+ y 16 + y 17 x

Solving for x1’:

By similar triangles,

70−45 70−30
=
x1' 1

x 1' =0.625 hr

Area of CLC= ( 0+25

2 ) [ 2 )+25+ 25+ 25+45+ 45+15+15+15+ 15+55+55+55+55+35+35+ 15] 1
0.625+ (
25+15

Area of CLC=550 MW −hr

Then,

Area of CLC 550 MW −hr

Ave . Load= = =22.9167 MW
24 hrs 24 hr
Thus,

Ave . Load 22.9167 MW

Load Factor= =
Peak Load 55 MW
Load Factor =0.4167

6. The peak load on a power plant is 25 MW. The loads having maximum demands of 30 MW, 20
MW, 10 MW and 14 MW are connected to the power plant. The capacity of the power plant is
80 MW, and the annual load factor is 0.78. Determine the average load on the power plant.

Solution:

Annual Average Load

Annual Load Factor =
Annual Peak Load
Thus,

Annual Average Load=( Annual Load Factor ) ( Annual Peak Load )

Annual Average Load=0.78(25 MW )
Annual Average Load=19.5 MW
7. A 50,000 KW plant delivers an annual output of 238,000,000 KW.hr with a peak load of 30,000
KW. What is the capacity factor?

Solution:

Ave . Load
Capacity Factor=
Rated Capacity
Also,

kw−hrs generated/ yr
Ave . Load=
8760 hrs / yr
238,000,000 kw−hrs / yr
Ave . Load= =27168.94977 KW
8760 hrs / yr
Thus,

Ave . Load 27168.94977 KW

Capacity Factor = =
Rated Capacity 50000 KW
Capacity Factor=0.5434

8. A 28 MW power station has the following annual factors: use – 45.2%, capacity – 46.2%, load –
58.5%. Find the reserve load.
Solution:

Ave . Load
Load Factor =
Peak Load
Also,

Ave . Load
Capacity Factor=
Rated Capacity
Then,

Ave . Load= (Capacity Factor )( Rated Capacity )

Thus,

( Capacity Factor )( Rated Capacity )

Load Factor =
Peak Load
Where,

Rated Capacity=Peak Load+ Reserve Load

Hence,

( Capacity Factor )( Rated Capacity )

Load Factor =
Rated Capacity−Reserve Load
(Capacity Factor )( Rated Capacity )
Reserve Load=Rated Capacity−
Load Factor
0.462(28 MW )
Reserve Load=28 MW −
0.585
Reserve Load=5.88772 MW
9. Determine the minimum base length and base width in mm of a concrete foundation for a 100
kW slow speed diesel-generating set applying the following design data and parameters: Over-
all weight of the genset – 48,000 kg; Over-all dimensions of the bedplate – 5m x 2m; Efficiency
of the generator – 85%; Depth of the foundation – 3.5 m; Specific weight of the foundation reqd.
– 480 kg/bhp; Safe load bearing capacity of site – 20 tons/m 2; Density of concrete – 2.5 tons/m3;
Top edges of foundation should not exceed 2 m from the bedplate of the machines.

Solution:

W F =Weight of the foundation per bhp x Rated capacity

(
W F = 480
kg
bhp)( 100 KW ) ( 1 hp
0.746 kW )
=64343.1635 kg

Also,
 WF
V F=
ρF
Where,

(
ρ F= 2.5
tons
m3 () 1000 kg
1 ton ) =2500
kg
m 3

Thus,

W F 64343.16354 kg 3
V F= = =25.7373 m
ρF kg
2500 3
m
Assuming Rectangular Foundation,

Since,

Ltotal=L+2 C=5+2 (C )=5+ 2C

W total =W +2C=2+ 2 ( C )=2+2 C

Since,

h=3.5 m
V
A=
h
3
( 5+2 C ) ( 2+2 C )= 25.7373 m
3.5 m
By Quadratic Equation,

C=¿
For Soil Bearing Capacity,

20,000 kg (643431.635+ 48,000)kg

=
m
2
A
Where,

A=L x W
A=(5+2 C)(2+2 C)
Thus,

20,000 kg (643431.635+ 48,000)kg

=
m
2
(5+2 C)(2+2 C)
 2
10+10 C+ 4 C +4 C =34.5716
By quadratic Equation,

C=1.284 m
Thus,

Ltotal=L+2 C=5+2 (1.284 )=7.568

W total =W +2C=2+ 2 ( 1.284 )=4.568

V 25.7373 m3
h= = =0.745 m
A 7.568 x 4.568

10. What is the required base area of the foundations to support an engine with specified speed of
1200 rpm and weight of 1200 kg? Assume bearing capacity of soil as 100 KPa. Use e = 0.11

Solution:

W F =e W e √ n

W F =( 0.11 )( 1200 kg ) √ 1200=4572.6141 kg

Then,

W F+ W e
Sbd =
A
WF
A= =¿¿
S bd

A=0.5663 m2
11. A steam turbine operating at 5 MPa, 400°C (h = 3195.7 KJ/kg) has an exhaust pressure of 100
KPa (h = 2475 KJ/kg). It is made of two-row, velocity compounded stage at the vapor region of
expansion with a blading efficiency of 65 %. For an 80 % rotational and leakage allowance, find
the efficiency of this stage.

Solution:

ActualWork =( h1 −h2 ) x e blading x rotational∧leakage

kJ kJ
ActualWork =( 3195.7−2475 ) x 0.65 x 0.8=374.764
kg kg
The

kJ
374.764
Actual Work kg
e stage = =
( h1−h 2) kJ
( 3195.7−2475 )
kg
e stage =0.52∨52%

12. A large regenerative steam turbine receives steam at 8.7 MPa, 510 °C (h = 3419.9 KJ/kg) and
exhaust at 100 KPa (h = 2517 KJ/kg). The blades of the turbine are made of pressure
compounded stages for both the superheated and saturated stages. The blading efficiency after
the steam become saturated is x % and the moisture effect of this stage is y %. Find the
efficiency of this stage. Assume a rotation and leakage loss of 3%.
13. A water-tube boiler with 250 m2 heating surface was found to evaporate y kg of steam per hour
at 1.8 MPa (h = 2797.1 KJ/kg). The feedwater is 66°C (h = 276.3 KJ/kg). Determine the rated
boiler horsepower.

Solution:

HS 250
Rated Boiler Horsepower= =
0.91 0.91
Rated Boiler Horsepower=274.7253 Bo . Hp
14. The percent rating of a 250 m2 water-tube boiler is 100 % receiving steam at 70°C (h = 292.98
KJ/kg) and evaporating it at 2 MPa, 350°C (h = 3137 KJ/kg). Find the developed boiler
horsepower.

Solution:

Developed Boiler Hp
Percent Rating=
Rated Boiler Hp
Where,

HS 250
Rated Boiler Horsepower= = =274.7253 Bo . Hp
0.91 0.91
Thus,

Developed Boiler Hp=Percent Rating x Rated Boiler Hp

Developed Boiler Hp=1 ( 274.7253 Bo . Hp )
Developed Boiler Hp=274.7253 Bo . Hp
15. A steam boiler was found to have a factor of evaporation of 1.75 producing 4950 kg/hr of steam.
If the fuel consumption of this boiler is at 20 kg/min, determine the equivalent specific
evaporation.

Solution:
 Equivalent Specific Evaporation=Bo . Economy (Factor of Evap.)
Where,

kg
4950
m hr
Bo. Economy = s = =4.125
mf
20 (
kg 60 min
min 1 hr )
Thus,

Equivalent Specific Evaporation=4.125(1.75)

kg steam
Equivalent Specific Evaporation=7.2188
kg fuel
16. The mass of fuel requirement of a steam boiler is 1000 kg/hr to produce 4950 kg of steam per
hour at 2 MPa, 300°C (h = 3022.1 KJ/kg) from feedwater at 75°C (h = 313.93 KJ/kg). Determine
the boiler economy.

Solution:

ms
Boiler Economy=
mf
4950
Boiler Economy=
1000
kg steam
Boiler Economy=4.95
kg fuel
17. A steam boiler requires 150 liters per hour of fuel to produce 500 kg of steam per hour at 2.3
MPa, 340°C (h = 3107.8 KJ/kg). Feedwater is at 50°C (h = 209.33 KJ/kg) and fuel oil used is at 20
API. Find the boiler efficiency.

Solution:

kJ
q h=41130+ 139.6 (° API )=41130+139.6 ( 20 )=43922
kg
Thus,

ms (h s−h f )
e bo=
mf x q h

Where,

141.5
° API = −131.5
SG @15.6
 141.5
20= −131.5
SG 15.6

SG 15.6=0.934

Then,

mf =V f x ρf =V f x SG15.6 ( ρh 20 )

( )( ) ( )
3
150 L 1m kg kg
mf = ( 0.934 ) 1000 3 =140.1
hr 1000 L m hr

Finally,

e bo=
ms (h s−h f )
=
(500 kghr )(3107.8−209.33) kJkg
140.1 ( 43922 )
mf x q h kg kJ
hr kg
e bo=0.2355∨23.55 %

18. Determine the slope of the Willian’s line for a 600,000 KW straight-condensing turbine
operating at 6 MPa, 510 °C (h = 3446 KJ/kg) and exhaust at 10 KPa (h = 2520 KJ/kg). No load
steam factor is 2.19 x 106 KJ/hr and combined engine efficiency is 85 %.

Solution:

Wk
η gen=
Wt
600,000 kW
0.85=
Wt

W t =705882.3529 kW

W t =ms ( h1 −h2 )

KJ KJ
705882.3529 =ms∗( 3446−2520 )
s kg

ms =762.292 (
kg 3600 s
s 1 hr )
kg
ms =2744251.048
hr
For Point 2,
 P2 ( x , y )=P2 ( W t , ms ) =( 705882.3529, 2744251.048 )

For Point 1,
P1 ( x , y )=( 0,2190000 )

y 2− y 1 2744251.048−2190000
slope=m= =
x 2−x 1 705882.3529
kg
slope=0.7852
kW −hr

19. A 5000 KW steam plant has a full load steam rate of 12 kg of steam per KW.hr. No load steam
consumption may be taken as 10% of the full load consumption. Calculate the hourly steam
consumption at 50 % of the rated load.

Solution:

kg
Full load steam consumption=12 ( 5000 kW )
kW −hr
kg
Full load steam consumption=60,000
hr
For Point 2,
P2 ( x , y )=( 5000 , 60000 )

For No load Steam Consumption,

Noload Steam Consumption=0.10 60,000 ( kg

hr )
kg
Noload Steam Consumption=6000
hr
For Point 1,
P1 ( x , y )=( 0,6000 )

Finding the slope,

kg
60000−6000
y 2− y 1 hr
m= =
x 2−x 1 5000−0 KW
kg
m=10.8
kW −hr
Also,
 y− y1 =m ( x−x 1 )

y−6000=10.8 ( x−0 )
y=10.8 x +6000(eqn. 1)
x at 50% load,
x=0.5 x 5000 kW
x=2500 kW
Finally,
y=10.8 x +6000=10.8 ( 2500 )+6000
kg
ms =33,000
hr

20. The steam consumption of the two units of a steam plant in kg/hr are: Unit 1 (6000 KW) –
27,800 at 50% load and 51,100 at full load; Unit 2(9000 KW) – 41,500 at 50% load and 75,000 at
full load. Find the total mass of steam used in kg/hr if the power plant carries a system load of
5000 KW divided equally between the units.

Solution:

For Unit 1:
For Point 2,
P2 ( W 2 ,m2 )

P2 ( x , y )=( 75000 , 51100 )

@50% Load,
kg
m1=27,800
hr
W 1=0.5∗( 6000 )=3000 kW

For Point 1m
P2 ( x , y )=( 3000 , 27800 )

ms − y 1=slope ( W −x 1 )

51100−27800
ms – 27800= ∗( W −3000 )
7500−3000
ms =5.1778W + 12266.6667(eqn . 1)
For Unit 2,
For Point 2,
P2 ( W 2 ,m2 )

kg
m2=75,000
hr
W 2=9000 kW

P2 ( x , y )=( 9000 , 75000 )

@50% Load,
kg
m1=41,500
hr
W 1=0.5∗( 9000 )=4500 kW

P2 ( x , y )=( 4500, 41500 )

ms − y 1=slope ( W −x 1 )

75000−41500
ms – 41500= ∗( W −4500 )
9000−4000
ms=6.7 W +11350(eqn.2)
For the Values at the System Load,
System Load=5000 kW
Since,
5000 kW
W 1=W 2= =2500 kW
2
For Unit 1,
ms 1=5.1778 W +12266.6667

(
ms 1= 5.1778
kg
kW . hr )
x 2500 kW +12266.6667
kg
hr
kg
ms 1=25211.1667
hr
For Unit 2,
ms2=6.7 W +11350
 (
ms 2= 6.7
kg
kW . hr
x 2500 kW + 11350 )
kg
hr
kg
ms 2=28100
hr
So,
mstotal =ms 1+ ms 2

kg kg
mstotal =25211.1667 +28100
hr hr
kg
mstotal =53311.1667
hr

21. The heat generated by fuel is 2500 KW. If the jacket water loss is 30 %. Determine the mass of
water circulated in the engine if the temperature rise is 10°C.

Solution:

q j=m j c pj ∆ T

Where,

q j=0.3 ( 2500 KW )=750 KW

Since not given,

kJ
c pj =4.187
kg−K
Thus,

qj 750 kW
mj= =
c pj ∆ T
( 4.187
kJ
kg−K )
( 10° C )

kg
m j =17.9126
s
22. An engine working on ideal Diesel cycle with pressure of 100 KPa and temperature of 37°C at the
beginning of compression process. The peak temperature of the cycle is not to exceed 1200 K
and a compression ratio of 15. Determine the cycle thermal efficiency.

Solution:

T 1=37+273=310 K

T 3=1200 K
For process 1-2,

( )
k−1
T2 V 1
=
T1 V 2

( )
k −1
V1 1.4 −1
T 2=T 1 =310 K ( 15) =915.7949 K
V2

Also,

V3 T3
=
V2 T2

V 3= ( ) ( )( )
T3
T2
V 2=
T3
T2
V1
15

V3
= (
T3
=
1200
)
V 1 T 2 x 15 915.7949 x 15
=0.08736

Since,

V 1=V 4

V3 V 3
= =0.08736
V1 V 4

For Process 3-4,

( )
k−1
T4 V3
=
T3 V4

( )
k−1
V3 1.4 −1
T 4=T 3 =1200(0.08736)
V4

T 4=452.5827 K

Thus,

QL T 4−T 1
e=1− =1−
QH k (T 3−T 2)
452.5827−310
e=1−
1.4 (1200−915.7949)
e=0.6417∨64.17 %
23. A 39.37 cm x 55.88 cm x 327 rpm, 16 cylinder, 4-cycle stationary diesel engine is connected to a
3125 KVA (80% power factor) generator. It also drives a 10 KW exciter. Assume a generator,
efficiency of 80% determine the bmep at rated load.

Solution:

WK
eg =
Wb
Where,

W K =( Pf ) ( KVA ) +W ex

W K ( P f ) ( KVA )+W ex 0.8 ( 3125 )+10

W b= = = =3137.5 kW
eg eg 0.8

Then,

W b =Pbmep x L AN

Wb 3137.5 kW
Pbmep= =
L AN
( π
( 0.5588 m ) x 0.39372 m2 x
4 2) (
1 327 rev
min
x 16 )(
1 min
60 s )
Pbmep=1057.8404 KPa

24. Specification of a 4-cylinder, 4-cycle diesel engine are 10.16 cm x 15.24 cm x 900 rpm.
Mechanical efficiency, 82%. Find the KW output of a direct-connected generator of 90%
efficiency when the indicated mep is 10 kg/cm2.

Solution:

W I =Pmep L AN

Where,

N=ns x n c x n

1
n s= ( For 4−cylinder engine )
2
1
N= ( 900rpm ) ( 4 )
2
N=1800 /min
Thus,

W I =Pmep L AN
 (
W I = 10
kg f
cm
2 ) ( 15.24 cm ) ( π4 x 10.16 ) cm ( 1800
2 2
min ) ( 9.81 N
1 kg
f
) ( 1000 N 100 cm 60 s )
1 kN
)( 1m
)( 1 min

W I =36.3624 kW

Hence,

Wb
e m=
WI

W b =e m x W I =0.82(36.3624 kW )
W b =29.8172 kW

Finally,

WK
eg =
Wb

W K =e g x W b =0.9(W b =29.8172 kW )
W K =26.8355 kW

25. A 2,000 KW diesel electric plant has a brake thermal efficiency of 85 %. If the heat generated by
fuel is 9,000,000 KJ/hr. What is the generator efficiency?

Solution:

WK
eg =
Wb
Also,

Wb
e tb =
Ec

(
Ec = 9,000,000
kJ
hr )( 3600
1hr
s)
=2500 kW

Then,

W b =e tb x E c =0.85 ( 2500 kW )=2125 kW

Hence,

W K 2000 kW
eg= =
W b 2125 kW
e g =0.9412∨94.12 %
26. Determine the brake power of an engine having a brake thermal efficiency of 83% and uses 25°
API with fuel consumption of 200 kg/hr.

Solution:

q h=41130+ 139.6 (° API )=41130+139.6(25)

kJ
q h=44620
kg
Also,

Ec =mf x q h= 200( kg
hr )( 3600
1 hr
s )(
44620 )
kJ
kg
Ec =2478.8889kW

Then,

Wb
e tb =
Ec

W b =e tb x E c =0.83(2478.8889 kW )
W b =2057.4778 kW

27. A diesel cycle operates with a compression ratio of 5 and with a cut-off occurring at 6% of the
stroke. One cu.ft. of air at 14 psia and 120 °F is used. Determine the cycle thermal efficiency.

Solution:

T 1=120+460=580 R

For process 1-2,

( )
k−1
T2 V 1
=
T1 V 2

( )
k −1
V1
T 2=T 1 =580 R(5)1.4 −1=1104.1193 R
V2

Also,

V1
=5
V2
V 1 1 f t3
V 2= =
5 5
3
V 2=0.2 f t
Then,
V s =V 1−V 2
3 3
V s =1 f t −0.2 f t

V s =0.8 f t 3

Hence,
V 3=0.06∗V s +V 2

V 3=( 0.06∗0.8 f t ) + 0.2 f t

3 3

V 3=0.248 f t 3

Since,
V3
rc=
V2

0.248 f t 3
rc=
0.2 f t 3
r c =1.24

Thus
k
1 r c −1
e=1− k−1
x
k∗r k r c −1
1.4
1 1.24 −1
e=1− 1.4 −1
x
1.4∗5 1.24−1

e=0.4506∨45.06 %

28. What ideal thermal efficiency is possible from a Diesel having a compression ratio of 5 air. Fuel
ratio is 40:1 and the heating value of fuel is 30,000 KJ/kg? Initial temperature is 16°C with Cp =
1.0467 KJ/kg.K.

Solution:

T 1=16+273=289 K
k−1
T 2=T 1 r k

1.4−1
T 2=289 K (5 )
 T 2=550.156 K

mf
Q a=¿Q h x
ma

kJ 1 kg fuel
Q a=30,000 x
kg−fuel 40 kg air
kJ
Qa=750
kg air

Qa=cpa ( T 3−T 2 )

750
kJ
kgair (
= 1.0467
kJ
kg−K )
( T 3−550.156) K

T 3=1266.6937 K

T3
rc=
T2
1266.6937 K
rc=
550.156 K
r c =2.3024
k
1 r −1
e=1− x c
kxr
k−1
r c −1

1 2.3024 1.4−1
e=1− x
1.4 x 51.4−1 2.3024−1
e=0.3621∨36.21%

29. A 4-stroke, 8-cylinder diesel engine with bore and stroke of 9 in. x 7 in., respectively and speed
of 1000 rpm has a brake mean effective pressure of 100 psi. Determine the engine brake power.

Solution:

W b =Pbmep L AN

Where,

N=ns x n c x n

1
n s= ( For 4−cylinder engine )
2
 1
N= ( 1000 rpm ) ( 8 )
2
N=4000/min
Thus,

W b =Pbmep L AN

(
W b = 100
lb f
¿
2 )¿
W b =449.819 hp

30. An 8-cylinder diesel engine, 4-stroke cycle, 550 rpm, 400 mm x 500 mm is directly coupled to a y
KW AC generator 13,800 V, 3-phase, 60 cycles and 93% efficiency. Calculate the brake
horsepower of Diesel Engine.

Solution:

Wk
e gen =
Wb
1 hp
13,800 kW x
W 0.746 kW
W b= k =
e gen 0.93

W b =19891.0317 hp

31. A gas turbine working on a simple ideal Brayton cycle has a temperature of 300 K at the inlet of
the compressor and 1200 K at the inlet of the turbine. The cycle produces a net power of 15 MW
with a pressure ratio of 5. Determine the mass flow rate of air in kg/s.

Solution:

T 1=300 K

T 3=1200 K

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1 1.4 −1
P2 k 1.4
T 2=T 1 x =300 K ( 5 ) =475.1489 K
P1

For Process 3-4

 ( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P4
()
1.4−1
k 1
T 4=T 3 x =1200 K 1.4
=757.662 K
P3 5

Then,

W net =W t−W c

Where,

W t =ma C p 34 (t 3−t 4 )

W c =ma c p 12 (t 2 −t 1 )
Thus,

W net =ma C p 34 ( t 3−t 4 ) −m a c p 12 (t 2−t 1 )

Assume:C p 34 =C p 12=0.24 (
kcal 4.187 kJ
kg−K 1 kcal
=1.005 )
kJ
kg−K
W net 15,000 kW
m a= =
C p [ ( t 3−t 4 )−( t 2−t 1 ) ]
(1.005 kg−kJ K ) [( 1200−757.662)−( 475.1489−300 ) ]
kg
ma=55.8607
s
32. Consider an ideal Brayton cycle with a maximum and minimum temperatures of 1100 K and 320
K. The pressure ratio of the cycle is 4, determine the backwork ratio.

Solution:
T 1=320 K

T 3=1100 K

T 2−T 1
BWR=
T 3−T 4

Solving for T2,

( )
k−1 1.4−1
P2 k 1.4
T 2=T 1 =320 K ( 4) =475.5182 K
P1

Solving for T4,

 ( )
k−1
P3 k 1
1.4−1
T 4=T 3 =1100 K ( ) 1.4
=740.2451 K
P4 4

Thus,
T 2−T 1
BWR=
T 3−T 4
475.5182−320
BWR=
1100−740.2451
BWR=0.4323

33. A simple gas turbine has a temperature of 300 K at the inlet of the compressor and 1200 K at the
inlet of the turbine. The cycle produces a net power of 30 MW with a pressure ratio of 5.
Determine the mass flow rate of air in kg/s if both the compressor and turbine have 85%
isentropic efficiency.

Solution:

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1 1.4 −1
P2 k 1.4
T 2=T 1 x =300 K ( 5 ) =475.1489 K
P1

For Process 3-4

( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P
()
1.4−1
1
T 4=T 3 x 4 k
=1200 K 1.4
=757.662 K
P3 5

Then,

' ' ' Wc

W net =W t−W c =e s W t −
es
'
W net =e s ¿

30,000
kJ
s (
= 1.005
kJ
kg−K )[ 0.85 [ m ( 1200−757.662 ) K ]− m ( 475.1489−300
0.85
)K
]
 kg
m=175.6652
s
34. The pressure ratio of an air-standard Brayton cycle is 5 with a minimum and maximum
temperatures of 530 R and 1230 R, respectively. Determine the cycle thermal efficiency.

Solution:

T 1=530 R

T 3=1230 R

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1 1.4 −1
P
T 2=T 1 x 2 k
=530 K ( 5 ) 1.4
=839.4244 K
P1

For Process 3-4

( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P4
()
1.4−1
k 1
T 4=T 3 x =1230 K 1.4
=776.6036 K
P3 5

Then,

W net =W t−W c

Where,

W t =ma C p 34 (t 3−t 4 )

W c =ma c p 12 (t 2 −t 1 )
Thus,

W net =ma C p 34 ( t 3−t 4 ) −m a c p 12 (t 2−t 1 )

Also,

Q a=ma C p 23 ( t 3−t 2 )

Assume:C p 34 =C p 12=C p 23
 W net ma C p 34 ( t 3−t 4 ) −m a c p 12 (t 2−t 1 ) ( t 3−t 4 )−(t 2 −t 1 )
e= = =
Qa ma C p 23 ( t 3−t 2 ) ( t 3−t 2 )
( 1230−776.6036 )−( 839.4244−530)
e=
(1230−839.4244)
e=0.3686∨36.86 %

35. A Brayton cycle operating at 500 R and 1300 R minimum and maximum temperatures is found
to produce 3.2 x 108 Btu/hr of work. For a pressure ratio of 5, determine the amount of heat
added.

Solution:

T 1=500 R

T 3=1300 R

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1 1.4 −1
P2 k 1.4
T 2=T 1 x =500 K ( 5 ) =791.9098 K
P1

For Process 3-4

( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P4
()
1.4−1
k 1
T 4=T 3 x =1300 K 1.4
=820.8005 K
P3 5

Then,

Btu
Assume:C p 34=C p 12=C p 23=0.24
lb−R
Hence,

W net =W T −W e

W net =ma C p 34 ( t 3−t 4 ) −m a c p 12 (t 2−t 1 )

W net
m a=
C p¿ ¿
 Btu
3.2 x 10 8
hr
m a=
( 0.24 lbBtu−R ) [ (1300−820.8005 ) −( 791.9098−500) ]
lb 1 hr lb
ma=7119095.889 x =1977.5266
hr 3600 s s
Thus,

Q A =ma C p 23 (T 3 −T 2 )

(
Q A = 1977.5266
lb
s )(
0.24
Btu
lb−R )
( 1300−791.9098 ) R

Btu
Q A =241142.8569
s
36. What is the mass flow rate of a gas turbine power plant operating at a suction pressure of
101.325 KPa, 30°C and a pressure ratio of 5 if 200,000 J/hr of heat is added to the cycle?
Temperature of the flue gas entering the turbine is at 1800 °C.

Solution:

T 1=30+273=303 K

T 3=1800+273=2073 K

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1 1.4 −1
P2 k 1.4
T 2=T 1 x =303 K ( 5 ) =479.8973 K
P1

Then,

Q A =ma C p 23 (T 3 −T 2 )
Assume,

C p 23=0.24 (
kcal 4.187 kJ
kg−K 1 kcal
=1.005
kJ
kg− K)
Thus,
 m a=
QA
=
200,000
J 1kJ
(
hr 1000 J )
C p 23 (T 3−T 2)
( 1.005 kg−K
kJ
) ( 2073−479.8973 ) K
kg
ma=0.1249
hr
37. A simple gas turbine takes air at 100 KPa, 30 °C and is compressed to 850 KPa. The gas leaves
the combustor at 1500 °C. Find the network. Use cpa = 0.24 and cpg = 0.27 kcal/kg.C.

Solution:

T 1=30+273=303 K

T 3=1500+273=1773 K

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1
P2
( )
1.4−1
k 850
T 2=T 1 x =303 K 1.4
=558.4585 K
P1 100

For Process 3-4

Assume P 4=101.325 Kpa

( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P4
( )
1.4−1
k 101.325
T 4=T 3 x =1773 K 1.4
=965.5923 K
P3 850

Thus,

W net =W T −W e

W net =ma C pg ( t 3−t 4 ) −ma C pa (t 2−t 1 )

W net
=C pg ( t 3−t 4 )−C pa(t 2−t 1 )
ma

W net
ma [(
= 0.27
kcal 4.187 kJ
kg−C
x
1 kcal )
( 1773−965.5923 ) − 0.24
kg−C
x
] [(
kcal 4.187 kJ
1kcal
( 558.4585−303 ) ) ]
 W net kJ
=656.0612
ma kg

38. A gas turbine producing 1000 KW of power is operating at an inlet temperature of 1250°C. The
pressure ratio is 10 and the specific heat of the flue gas is 1.215 KJ/kg-°C. For a mass flow rate of
1.1 kg/sec, determine the air-fuel ratio.

Solution:

T 3=1250+273=1523 K

( )
k−1
T 4 P4 k
=
T 3 P3

( )
k−1
P4
( )
1.4−1
k 1
T 4=T 3 x =1523 K 1.4
=788.834 K
P3 10

W T =m g C pg ( T 3−T 4 ) =ma (1+f )C pg ( T 3 −T 4 )

WT
f '= −1
ma C pg ( T 3−T 4 )

kJ
1000
s
f '= −1
( 1.1 )(
kg
s
1.215
kJ
kg−C )
(1523−788.834 )

f '=0.0191
1
f=
0.0191
f =52.356
39. An ideal gas turbine power plant takes in air at 30°C and 100 KPa. The compression is up to 850
KPa and produces 500 KJ/kg of work. The compressor utilizes 255 KJ/kg of work. If the plant has
a regenerator with an effectiveness of 55 %, determine the increase of its efficiency. Use cpa =
0.24 kcal/kg-K and fuel-air ratio of 0.025.

Solution:

For Without Regen:

 T 1=30+273=303 K

For process 1-2,

( )
k −1
T 2 P2 k
=
T 1 P1

( )
k−1
P2
( )
1.4−1
k 850
T 2=T 1 x =303 K 1.4
=558.4585 K
P1 100

Solving for T2’,

Wc
=C pa (T 2' −T 1 )
ma

255
kJ
kg (
= 0.24
kcal 4.187 kJ
kg−C
x
1 kcal
( )
T 2' −303 ) K

'
T 2 =556.7616 K
W net =W T −W C

W net =m g C pg ( T 3−T 4 ) −ma C pa (T 2 '−T 1 )

W net =m a (1+ f )C pg ( T 3−T 4 ) −m a C pa (T 2 ' −T 1 )

Where,

(
C pa=C pg= 0.24
kcal 4.187 kJ
kg−C
x
1kcal
=1.005 )
kJ
kg−C

500
kJ
kg (
=( 1+0.025 ) 1.005
kJ
kg−C
( )
T 3−T 4 )− 1.005 ( kJ
kg−C
( 556.7616−303 ) K )
( T 3−T 4 ) =732.9503 K
For Process 3-4

Assume P 4=101.325 KPa

( )
k−1
T 3 P3 k
=
T 4 P4

( )
T3 850
1.4−1
= 1.4
T 3−732.9503 101.325
T 3=1609.4977 K

Thus,
 kJ
500
W net W net kg
e w /o = = =
Q A C pa (T 3−T 2 ')
(1.005 kg−C
kJ
) ( 1609.4977−556.7616 ) K
e w /o =0.4726∨47.26 %

For with regen:

T 3−T 2 '
Reg . eff =
(1+f )(T 5−T 2 ' )

Since,

( T 4 −T 5 ) =732.9503 K
T 5=1609.4977−732.9503=876.5474 K

T 3−556.7616
0.55=
(1+0.025)(876.5474−556.7616)
T 3=737.0408 K

Thus,

e kJ
500
W net W net kg
w /¿= = = ¿
( )
Q A C pa (T 4−T 3) kJ
1.005 (1609.4977−737.0408 )
kg−C

e w /¿=0.5702∨57.02% ¿

Finally,

% Increase=e w/ ¿−e w/o =57.02−47.26 ¿

% Increase=9.76 %
40. A gas turbine plant receives 0.1 liters/min of air producing 8.2 KW of power. Generator
efficiency is 90%. Fuel air ratio is 0.01 and heating value of the fuel is at 45420 KJ/kg. Determine
the plant thermal efficiency.

Solution:

m
ρ=
V
kg
Assume ρ=1.225 3
m
 ( )( )
3
kg L 1m 1 min kg
ma= ρV = 1.225 3
1000 x x =0.02042
m min 1000 L 60 s s

Also,

mf
f=
ma

kg kg
mf =f x ma=0.01 x 0.02042 =0.0002042
s s
Then,

(
Ec =mf x q h= 0.0002042
kg
s )(
45420
kJ
kg )
=9.274 kW

Finally,

WK 8.2
e= =
E C 9.274
e=0.8842∨88.42 %