R=25.8lb⋅Rft⋅lbk=1.09=cvcpv=15cuftp=75psia
a. For one mole of ideal gas we know that
cp−cv=Rcvcp−1=cvRk−1=cvRcv=k−1Rcv=1.09−125.8=286.6667BTU/F/lbcp=cvkcp=286.667×1.09=312.4667BTU/F/lb
b. For ideal gas we know
pV=mRTm=RTpVm=in2×ft275lb×144in2×25.8ftbl×(80+459.67)R15ft3×lbRm=25.8×539.6775×144×15lbm=11.635lb
c. Q = 30 BTU
T1=80°FQ=mcvΔTQ=mcv(T2−T1)mcvQ+T1=T2T2=286.67×11.63530+80T2=80.009°F
For ideal gas we know
pV=RT
at constant volume we know
Tp=constantT1p1=T2p2p2=T1p1×T2p2=8075×80.009p2=75.0084psia
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