Given:
P
s
= 2.5 MW
h = 1981.2 m
Required:
Power
Solution:
P
act
= 29.2 
h
1000
P
act
= 29.2 
1981.2 m
1000
(
3.281 ft
1m
)
P
act
= 23.4196828 in Hg
Power = P
s
(
Pact
29.92
)
Power = 2.5 MW
(
23.4196828 in Hg
29.92
)
PROBLEM 9
A 373 kW (500Hp) internal combustion engine has a brake mean effective pressure
of 551.5 kPa at full load. What is the friction power if mechanical efficiency is 85%.
Power = 1.956858523 MW
P
b
= 321.1040678 kPa
Given:
Brake Power = 373 kW = 500 Hp
IP = 551.5 Hp
ƞ
m
= 85%
Required:
Friction Power (FP)
Solution:
ƞ
m
=
BP
IP
0.85
=
500 Hp
IP
IP = 588.2352941 Hp
FP = IP
–
BP = 588.2352941 Hp
–
500 Hp
REVIEW PROBLEM NO. 23
–
GAS TURBINE POWER PLANT/ HYDRO
ELECTRIC POWER PLANT
Mea, Jane May M.
1455383
Problem No. 1
Water is being heated by the exhaust gases from a gas turbine. The gases leave the
gas turbine at 648
̊
C and may be cooled to 148
̊
C. The water enters the heater at 9
3̊
C. The
rate of gas flow is 25 kg/s and the water flow is 31.5 kg/s. Assume that the mean specific
FP = 88.23529412 Hp
heat of the gas and the water are respectively 1.088 and 4.27 kJ/kg
̊
C. What is the available
energy removed from the hot gases in kW? Take available sink temperature as 311
̊
K.
Given:
T
gt
= 648
̊
C
T
c
= 148
̊
C
T
h
= 95
̊
C
m
gf
= 25 kg/s
m
w
= 31.5 kg/s
MEH
g
= 1.088kJ/kg
̊
C
MEH
w
= 4.27kJ/kg
̊
C
Sink Temperature= 311
̊
C
Required:
Energy removed from hot gases, kW
Solution:
ΔQ = Q
gas
–
Q
water
Q
gas
= m
g
CΔT
Q
gas
= (25 kg/s)(1.088 kJ/kg 
̊
C)(648148)
̊
C
Q
gas
= 73600 kW
Q
water
= m
w
CΔT
Q
water
= (31.5 kg/s)(4.27kJ/kg 
̊
C)(31193)
̊
C
Q
water
= 7397.7558 kW
ΔQ = 13600 kW –
7397.7758 kw
ΔQ = 8345.6 kW
Problem No. 2
In a gas turbine plant, the mass flow rate is 6.2 kg/s, the enthalpy at the combustor
entrance is 250 kJ/kg and the enthalpy at the exit is 980 kJ/kg. What is the capacity of the
combustor in kW?
Given:
m= 6.2 kg/s
h
1
= 250 kJ/kg
h
2
=980 kJ/kg
Required:
Capacity of the combustor, kW
Solution:
W= Δh = m(h
2

h
1
)
= 6.2kg/s (980 kJ/kg  250 kJ/kg)
W= 4526 kW
Problem No. 3
A Brayton cycle has a compressor rated power of 451.3kw. If compressor
efficiency is 78%, find the ideal compressor work.
Given:
η
c
= 78%
W
T
= 451.3 kw
Required:
Ideal compressor work, Wc
Solution:
η
c
= W
c
/W
t
0.78 = Wc/ 451.3 kW
Wc = 352.014 kW
Coro, Corinne S.
14
–
54200
Problem No. 4
A hydroelectric generating station is supplied from a reservoir of a capacity
6000000 m3 at a head of 170 m. Assume hydraulic efficiency of 80% and electrical
efficiency of 90%. The fall in the reservoir level after a load of 15 MW has been supplied
for 3 hours, if the area of the reservoir is 2.5 sq. km is closest to:
Given:
h = 170 m
Ƞ
T
= 0.80
Ƞ
G
= 0.90
Generating Output = 15 MW
t = 3 hours
A = 2.5 km
2
Required:
Fall in the reservoir, H
Solution:
A = 2.5 km
2
A = 2.5 x 10
6
m
2
Generating Output = (w Q h) (
Ƞ
T
) (
Ƞ
G
)
15,000 kW = (9.81
kN
m
3
x Q x 170 m) (0.80)(0.90)
Q = 12.4922548
m
3
sec
After 3 hours,
Q = 12.4922548
m
3
sec
(3 hours x
3600 sec
1 hour
)
Q = 134,916.3519 m
3
Volume = Area x Height
134,916.3519 m
3
= (2.5 x 10
6
m
2
) (H)
H = 0.05396654074 m
Problem No. 5
In hydroelectric plant having 50 sq. km reservoir area and 100 m head is used to
generate power. The energy utilized by the consumers whose load is connected to the
power plant during a fivehour period is 13.5 x 10 to the 6th power kWhr. The overall
generation efficiency is 75%. Find the fall in the height of water in the reservoir after 5
hour period.
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 Combustion, Internal combustion engine, Diesel engine, Gas turbine