p1=160kPa=1.6×105N/m2,V1=0.8m3,n=1.4
C=p1V1n
C=1.6×105N/m2(0.8m3)1.4≈1.17070×105N⋅m2.2
Work done=W=∫V1V2pdV,where p=VnC
W=∫0.8m30.6m3V1.41.17070×105N⋅m2.2dV=
=1.17070×105N⋅m2.2(−0.41)[V0.41]=−0.6m30.8m3≈
≈1.17070×105N⋅m2.2(−0.333353m1.2)≈−39025.649J Simpson’s rule
Work done=W=∫V1V2pdV,where p=VnC
W=∫0.8m30.6m3V1.41.17070×105N⋅m2.2dV=
=1.17070×105N⋅m2.2∫0.8m30.6m3V1.41dV Evaluate ∫0.8m30.6m3V1.41dV using Simpson’s rule.
∫abf(V)dV≈≈3ΔV(f(V0)+4f(V1)+2f(V2)+...+2f(Vn−2)+4f(Vn−1)+f(Vn))
ΔV=nb−a,n=4,a=0.6,b=0.8,ΔV=80.8−0.6=0.05
f(V0)=f(0.6)=0.61.41≈2.044505
4f(V1)=4f(0.65)=0.651.44≈7.311076
2f(V2)=2f(0.65)=0.71.42≈3.295283
4f(V3)=4f(0.75)=0.751.44≈5.983761
f(V5)=f(0.8)=0.81.41≈1.366703
∫abf(V)dV≈
≈601(2.044505+7.311076+3.295283+5.983761+1.366703)≈
≈0.333355
W≈1.17070×105N⋅m2.2(−0.333355m1.2)≈−39025.923J
Work done≈−39026J
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