Solution;
m˙=1lb/s
r=5.5
k=1.32
P1=15psia
T1=130°F=460+130=590°R
T3=5000°R
State 1:
By the Ideal gas equation;
V1=P1mRT1
V1=15×1441×53.54×590=14.62ft3/s
State 2:
From the relation;
P1V1k=P2V2k
P2=P1(V2V1)k
P2=15(5.5)1.32=142.35psia
Also;
T2=T1(V2V1)k−1
T2=590(5.5)1.32−1=1018.05°R
And;
V2˙=rV˙1=5.514.62=2.66ft3/s
State 3:
Constant volume heat addition;
V˙3=V˙2=2.66ft3/s
T3=5000°R
P3=T3(T2P2)
P3=5000(1018.05142.35=699.13psia
State 4:
Isentropic expansion;
Hence;
T4=T3(V4V3)k−1
T4=5000(5.51)1.32−1
T4=2897.70°R
Ans: T4=2897.70°R
Since process 4-1 is constant volume process;
P4=T4(T1P1)
P4=2897.7(59015)=73.65psia
Ans: P4=73.65psia
(c) Heat rejected;
We know that;
Cv=k−1R=(1.32−0.32)×77853.54
Cv=0.2151Btu/lb°R
Qr=m˙Cv(T1−T4)
Qr=1×0.2151(590−2897.7)
Qr=−496.39Btu/s
(d)Thermal.efficiency;
η=QaW
Heat added;
Qa=mCv(T3−T2)
Qa=1×0.2151(5000−1018.05)=856.52Btu/s
Work;
W=Qa−Qr=856.52−496.39=360.13Btu/s
Hence;
η=856.52360.13=0.4205
Ans: 42.05%
(e) Work net in horsepower,hp;
W=42.4360.13×60=509.62hp
Ans:509.62hp
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