Question #137272
A piston cylinder contains gas initially at 3500 kPa with a volume of 0.03 cubic meter. The gas
is compressed during a process where pV raise to 1.25 = C to a pressure of 8500 kPa. The heat
transfer from the gas is 2.5 kJ. neglecting changes in kinetic and potential energies determine the nonflow work in kJ.
1
Expert's answer
2020-10-08T15:10:41-0400

The polytropic process is one in which the pressure-volume relation is given as pVn=constantpV^n=constant

pV1.25=constant    p1V11.25=p2V21.25pV^{1.25} =constant\implies p_1V_1^{1.25}=p_2V_2^{1.25}

V2=V1(p1p2)11.25=0.03(35008500)11.25=0.01475m3V_2=V_1(\frac{p_1}{p_2})^{\frac{1}{1.25}}=0.03(\frac{3500}{8500})^{\frac{1}{1.25}}=0.01475 m^3

The work done during the polytropic process is found by substituting the pressure volume

relation into the boundary work equation. The result is W=12pdV=p2V2p1V11nW= \int_1^2pdV=\frac{p_2V_2-p_1V_1}{1-n}

W=8500×103×0.014753500×103×0.0311.25=81500JW=\frac{8500×10^3×0.01475-3500×10^3×0.03}{1-1.25}=-81500 J

Then, according to the first law of thermodynamics, ΔU=QW=2500+81500=79000J=79kJΔU=Q-W=-2500+81500=79000 J=79kJ


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