Solution;
Given;
R=0.324kJ/kgK
Cv=0.84kJ/kgK
m=2.25kg
V1=1.15m3
P1=P2=7bar
T2=28°C=301K
Value of Cp can be calculated as;
Cp=Cv+R
Cp=0.84+0.324=1.164kJ/kgK
Value for γ can be calculated as;
γ=CvCp=0.841.164=1.386
(a)change in internal energy;
dU=mCv(T1−T2)
From the relation;
PV=mRT
V2=7×1052.25×324×301=0.3134m3
For ideal gases;
T1V1=T2V2
T1=0.31341.15×301=1104.5K
dU=2.25×840×(1104.5−301)
dU=1518.615kJ
(b)change in enthalpy;
dH=mCp(T1−T2)
dH=2.25×1164×(1104.5−301)
dH=2104.366kJ
(c)Heat transferred;
For isobaric process;
dH=dQ=2104.366kJ
(d)Work done;
W=pΔV
W=7×105(1.15−0.3134)
W=585.620kJ
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