Question 2.13: A fluid is contained in a cylinder behind a piston of specif...
A fluid is contained in a cylinder behind a piston of specific volume 0.2 m³/ kg at 5 bar pressure. The fluid is compressed reversibly to a pressure of 0.5 bar according to the law : pv1.5 = C, where C is a constant. Calculate the work done by the fluid on the piston.
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v2=v1(p2p1)1.51=0.2(0.55)1.51=0.92832 m3/kg
Work done,
w1−2=v1∫v2p.dv=C0.2∫0.92832v1.5dv=C[0.5−v−0.5]0.20.92832=2C[0.2−0.5−0.92832−0.5]=2.39635 C
Now
Cw1−2=p1v11.5=5×105×0.21.5=44.72×103=2.39635×44.72×103=107.168 kJ/kg
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