Modern Engineering Thermodynamics

13.10 Modern Steam Power Plants 485
a. The isentropic thermal efficiency of this Rankine cycle power plant is given by Eq. (13.15) with (η s ) pm1 = (η s ) pm2 =
(η s ) p = 1.0 as
where the numerator in this equation is
and the denominator is
then, the isentropic thermal efficiency is
ðη T Þ s
= ðh 1 − h 2s Þ + ðh 3 − h 4s Þ + ðh 5 − h 6s Þ − v 7 ðp 8s − p 7 Þ
ðh 1 − h 8s Þ + ðh 3 − h 2s Þ + ðh 5 − h 4s Þ
Numerator = ð1530:8 − 1316:9Þ + ð1505:9 − 1343:8Þ + ð1526:4 − 955:1Þ
144
− 0:01606ð5000: − 0:400Þ = 932:4 Btu/lbm
778:16
Denominator = ð1530:8 − 55:76Þ+ ð1505:9 − 1316:9Þ+ ð1526:4 − 1343:8Þ
= 1847 Btu/lbm
ðη T Þ s =
932:4 Btu/lbm
1847 Btu/lbm
b. The isentropic efficiency of the complete power generating unit is
= 0:505 = 50:5%
ðη s Þ = ð _W out Þ net actual
turbine-generator
ð _W out Þ net isentropic
where the actual net power output is given as
ð _W out Þ net actual = 325 MW = ð325,000 kWÞð3412 Btu/kW . hÞ = 1:11 × 10 9 Btu/h
and the isentropic net power output can be calculated from
Then,
ð _W out Þ net isentropic = _W turbine 1 + _W turbine 2 + _W turbine 3 − j _W j pump
= _mfðh 1 − h 2s
Þ+ ðh 3 − h 4s
Þ+ ðh 5 − h 6s
Þ− v 7 ðp 7 − p 8s Þg
= ð1:50 × 10 6 lbm/hÞfð1530:8 − 1316:9Þ+ ð1505:9 − 1343:8Þ
+ ð1526:4 − 955:1Þ− 0:01606ð5000: − 0:400Þð144/778:16Þg Btu/lbm
= 1:40 × 10 9 Btu/h
ðη s Þ = 1:11 × 109 Btu/h
turbine-generator
1:40 × 10 9 = 0:793 = 79:3%
Btu/h
The isentropic efficiencies calculated in parts a and b are somewhat low because we choose to omit the eight regenerator
units in this analysis.
Exercises
22. Determine the isentropic thermal efficiency of the Eddystone Power Plant discussed in Example 13.8 if the boiler
outlet state temperature is increased from 1200.°F to 1500.°F. Assume all the other variables remain unchanged.
Answer: (η T ) s = 52.1%.
23. If the second reheat pressure in Example 13.8 is increased from 300. psia to 600. psia, what would be the new
isentropic thermal efficiency of the power plant? Assume all the other variables remain unchanged. Answer:
(η T ) s = 50.2%.
24. If the operating conditions of the power plant in Example 13.8 remain unchanged, but the generator power output
drops from 325 MW to 315 MW, determine the new isentropic efficiency of the power generating unit. Answer:
(η s ) turbine-generator = 76.8%.
The introduction of nuclear power in the 1960s added a new facet to heat source technology. Nuclear safety
restrictions require the use of a double-loop heat transfer system to keep the radioactive reactor cooling fluid
from entering the turbine, and this effectively limits the maximum nuclear power plant secondary loop temperature
and pressure to around 1000°F and 800 psia. This has the effect of limiting nuclear plant thermal efficiencies
to the low 30%s, while the thermal efficiencies of fossil fueled plants reached the low 50%s.