9-81
9-110 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be
determined.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room
temperature. 3 Kinetic and potential energy changes are negligible.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).
Analysis For the compression and expansion processes we have
T2 s  T1 r p( k 1) / k  (293 K)(8) 0.4/1.4  530.8 K
C 
c p (T2 s  T1 )
c p (T2  T1 )

 T2  T1 
T2 s  T1
 1
T5 s  T4 
 rp

T 




c p (T4  T5 )
c p (T4  T5 s )
1
 (1073 K) 
8
4
qin
1073 K
C
530.8  293
 293 
 566.3 K
0.87
( k 1) / k
T
2s
293 K
1
0.4/1.4
 592.3 K
2
5
3
6
5s
qout
s

 T5  T4   T (T4  T5 s )
 1073  (0.93)(1073  592.3)
 625.9 K
When the first law is applied to the heat exchanger, the result is
T3  T2  T5  T6
while the regenerator temperature specification gives
T3  T5  10  625.9  10  615.9 K
The simultaneous solution of these two results gives
T6  T5  (T3  T2 )  625.9  (615.9  566.3)  576.3 K
Application of the first law to the turbine and compressor gives
w net  c p (T4  T5 )  c p (T2  T1 )
 (1.005 kJ/kg  K )(1073  625.9) K  (1.005 kJ/kg  K )(566.3  293) K
 174.7 kJ/kg
Then,
m 
W net
150 kW

 0.8586 kg/s
wnet 174.7 kJ/kg
Applying the first law to the combustion chamber produces
Q in  m c p (T4  T3 )  (0.8586 kg/s)(1.005 kJ/kg  K )(1073  615.9)K  394.4 kW
Similarly,
Q out  m c p (T6  T1 )  (0.8586 kg/s)(1.005 kJ/kg  K )(576.3  293)K  244.5 kW
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