Page 127 - Advanced Thermodynamics for Engineers, Second Edition
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114 CHAPTER 5 RATIONAL EFFICIENCY OF POWER PLANT
This term is shown on Fig 5.6(a), and is energy which is unavailable for the production of work.
If the efficiencies of the turbine and compressor are not 100% then the T–s diagram is shown in
Fig 5.6(b). It can be seen that the rejected energy is larger in this case.
Q5. Gas turbine cycles
For the gas turbine cycle defined in Q4, calculate the effect of (a) a turbine isentropic efficiency of
80%; (b) a compressor isentropic efficiency of 80% and (c) the combined effect of both inefficiencies.
Solution
(a) Turbine isentropic efficiency, h T ¼ 80%.
This will affect the work output of the turbine in the following way
¼ 0:8 693:2 ¼ 554:6kJ kg:
T
w T ¼ h w T isen
Hence
w net ¼ w T þ w C ¼ 554:6 þð 407:9Þ¼ 146:7kJ=kg;
and
146:7
w net
th ¼ 0:296:
h ¼ ¼
q 2s3 496:1
The temperature after the turbine is T 4 ¼ T 3 h T DT isen ¼ 1200 0.8 690.1 ¼ 647.9 K and the
T 4 p 4 647:9
entropy at 4, related to 1, is s 4 s 1 ¼ c p ln R ln ¼ 1:0045 ln ¼ 0:7734 kJ=kg K:
T 1 p 1 300
Hence, the rejected maximum net work is
T 4 647:9
¼ b 4 b 1 ¼ c p ðT 4 T 1 Þ T 0 ln ¼ 1:0045 ð647:9 300Þ 300 ln
b w net rejected
T 1 300
¼ 117:4kJ=kg:
This is a significant increase over the rejected potential work for the ideal Joule cycle. However, in
addition to the inefficiency of the turbine increasing the maximum net work rejected, it also increases
the unavailable energy by the irreversibility, T 0 (s 4 s 4s ), as shown in Fig 5.6(b). This is equal to
T 0 ðs 4 s 4s Þ¼ 300 ð0:7734 0:5328Þ¼ 72:2kJ=kg
Since the maximum net work is the same for this case as the ideal one then the net work is given by
T 0 s 4 s 4s ¼ 336:3 117:4 72:2 ¼ 146:7kJ=kg:
w net ¼ b w net ½ b w net rejected
This is the same value as found from the more basic calculation.
w net 146:7
R ¼ 0:436.
The rational efficiency of this cycle is h ¼ ¼
b w net 336:3
(b) Compressor isentropic efficiency, h C ¼ 80%.
The work done in the compressor is
w C isen
¼ 509:9kJ=kg
w C ¼
h C
