Page 121 - Advanced Thermodynamics for Engineers, Second Edition

108 CHAPTER 5 RATIONAL EFFICIENCY OF POWER PLANT

Work done by turbine,
w T ¼ h 4 h 6 ¼ 2799 2200:8 ¼ 598:2kJ=kg K:

w net w T þ w P 598:2 2:0
th
Thermal efficiency of the cycle is h ¼ ¼ ¼ ¼ 0:243
q in h 4 h 2 2799 342:0
The maximum net work available can be evaluated from the change of exergy between state
points 4 and 2. Hence,
b w net ¼ b 4 b 2 :
It is necessary to define a dead-state condition. This is arbitrary, and in this case will be taken as the
condition at 1. Hence, p 0 ¼ 0.5 bar; T 0 ¼ 81.3 þ 273 ¼ 354.3 K.
Thus,
b 4 ¼ h 4 T 0 s 4 ðh 0 T 0 s 0 Þ¼ 2799 354:3 6:340 a 0 ¼ 552:7 a 0 kJ=kg;
b 2 ¼ h 2 T 0 s 2 ðh 0 T 0 s 0 Þ¼ 342:0 354:3 1:091 a 0 ¼ 44:5 a 0 kJ=kg:
Hence,
b w net ¼ 552:7 ð 44:5Þ¼ 597:2kJ=kg:
Thus, the rational efficiency is

w net w net 598:2 2:0
R
h ¼ ¼ ¼ ¼ 1:00:
b 4 b 2 b w net 597:2
In this case the rational efficiency is equal to unity because the turbine and feed pump have
isentropic efficiencies of 100%, and it was assumed that the temperature of the working fluid in the
condenser was equal to the dead-state (ambient) temperature. Hence, although the cycle is not very
efficient, at 24.3%, there is no scope for improving it unless the operating conditions are changed.
It is possible to evaluate the rational efficiency of a steam plant operating on a Rankine cycle in
which the condenser temperature is above the ambient temperature. If the dead-state temperature in the
previous example was taken as 20 C, rather than 81.3 C then the following values would be obtained.

b 4 ¼ h 4 T 0 s 4 a 0 ¼ 2799 293 6:340 a 0 ¼ 941:4 a 0 kJ=kg;
b 2 ¼ h 2 T 0 s 2 a 0 ¼ 342:1 293 1:091 a 0 ¼ 22:4 a 0 kJ=kg:

Hence,
b w net ¼ 941:4 22:4 ¼ 919:0kJ=kg:
Thus, the rational efficiency is

w net w net 598:2 2:0
R ¼ 0:649
h ¼ ¼ ¼
b 4 b 2 b w net 919:0
This result shows that irreversibilities in the condenser producing a temperature drop of 81.3–20 C

would reduce the potential efficiency of the power plant significantly. Basically this irreversibility is
equivalent to a loss of potential work equal to the area of the T–s diagram bounded by the initial dead-
state temperature of 81.3 C and the final one of 20 C and the entropy difference, as shown in Fig 5.4,

Þðs 4 s 2 Þ, which is equal to (354.3 293) (6.340 – 1.091) ¼ 321.8 kJ/kg.
i.e. ðT 0 1 T 0 2