A fluid undergoes a reversible adiabatic compression from pressure 1 mpa bar & volume 0.m3, according to the law pv1.3=constant. Determine the work done
Assume that "p_1=1\\ MPa, \\ V_1=0.8\\ m^3,\\ V_2=0.4\\ m^3,\\ n=1.3"
"W=\\frac{p_1V_1}{n-1}[1-(\\frac{V_1}{V_2})^{n-1}]=\\frac{1\\cdot10^6\\cdot 0.8}{1.3-1}[1-(\\frac{0.8}{0.4})^{1.3-1}]=-172 \\ (kJ)" . Answer
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