Solution;
Given;
r=13.5
k=1.35
T1=140+460=600°R
V1=1cuft
Cv=k−1R=778(1.34)53.34=0.2016Btu/lb°R
Cp=kCv=1.34×0.2016=0.3602Btu/lb°R
P2=P1(r)k−1=14(13.5)0.34=33.833psia
P2=P3
V2=13.5V1=13.51=0.074ft3
TT2=T1rk−1=600×13.50.34=1454°R
V3=V2+0.06VD
V3=0.074+0.06(1−0.074)=0.1298ft3
T3=T2(V2V3)k−1
T3=1454(0.07410.1297)0.34
T3=2445°R
T4=T3(V2V3)k−1=2545(10.1297)0.34
T4=1271°R
P4=P3(V4V3)k
P4=457.9(10.1297)1.34
P4=29.7psia
Qr=mCv(T1−T4)
Qr=0.063×0.2016(600−1271)
Qr=8.52Btu
Wnet=Qa−Qr
Qa=mCp(T3−T2)
Qa=0.063×0.2702(2545−1454)=18.57Btu
Wnet=18.57−8.52=10.05Btu
η=QaW=18.5710.05=0.5412Pm=V1−V2W=(1−0.0741)×14410.05×778
Pm=58.64psi
Wr=42.410.05×1000=237hp
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