Solution;
Let subscripts 1,23... Represent various stages of the cycle;
At state 1,90bar,530°c;
h1=3462kJ/kg
s1=6.7553kJ/kgK
Since process 1-2 is isentropic expansion;
s1=s2=6.7553kJ/kgK
At P=10bar;
h2=2856.92kJ/kg
At state 3;10bar,480°c;
h3=3435.8kJ/kg
s3=7.7075kJ/kgK
Process 3-4 is isentropic expansion;
s3=s4=7.7075kJ/kgK
At 0.6 bar;
h4=2719.06kJ/kgK
At state 5,0.6bar,480°c;
s5=9.0166kJ/kgK
h5=3446.6kJ/kg
Process 5-6 is isentropic expansion;
s5=s6=9.0166kJ/kgK
At P6=0.04bar;
h6=2745kJ/kg
h7=121.40kJ/kg
h8=h7+uΔp
h8=121.40+(90−0.04)×102×1.0063×10−3=130.396kJ/kg
Hence;
Qa=(h1−h8)+(h3−h2)+(h5−h6)
Heat added;
Qa=(3462−130.396)+(3435.8−2856.92)+(3446.6−2719.061)=4638.024kJ/kg
Efficiency;
W=(h1−h2)+(h3−h4)+(h5−h6)
W=(3462−2856.92)+(3435.9−2719.06)+(3446.6−2745)=2023.22kJ/kg
η=QaW=4638.0242023.22=0.4362
η=0.4362 or 43.62%
For output of 20,000kW;
P=m˙W
m˙=2023.2220000=9.885kg/s
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