(a) percentage decrease
V1V2×100
V1=0.5m3
p2p1=(V1V2)1.4=101=(0.50V2)1.4=V2=0.09653
=0.50.09653×100=19.306%
(b) Air standard efficiency
η=1−T2T1=1
T2=T1(p1p2)1.40.4=(27+273)(10)1.40.4=579.2K
η=1−579.2300=48.2%
(c)mean effective pressure
Pmean=V3W
W=ηQa=10048.2×200kJ=94.47kJ
Vs=V1−V2=0.5−0.09653=0.40347m3
Pmean=0.4034796.41=238.95kPa
(d) Ideal power
P=w×N=96.41×1200=19282KW
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