Question

Unit mass of a certain fluid is contained in a cylinder at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law PV2=Constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original position; heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 20 bar. Calculate the net work done by the fluid, for an initial volume of 0.05 m3.

Answer 1

Answer:

61750

Explanation:

P1​V12​=C=P2​V22​⟹P2​=25×(2×0.050.05​)2=6.25bars

W1−2=∫v1v2Pdv=∫v1v2Cv2dv=0.065[1v]0.050.1=0.65barm3=65000NmW_{1-2}=\int_{v_1}^{v_2} Pdv=\int_{v_1}^{v_2} \frac{C}{v^2}dv=0.065[\frac{1}{v}]_{0.05}^{0.1}=0.65 barm^3=65000NmW1−2​=∫v1​v2​​Pdv=∫v1​v2​​v2C​dv=0.065[v1​]0.050.1​=0.65barm3=65000Nm

W2−3=∫v2v3Pdv=P2(V3−V2)=65000(0.05−0.1)=−3250NmW_{2-3}=\int_{v_2}^{v_3} Pdv= P_2(V_3-V_2)=65000(0.05-0.1)=-3250 NmW2−3​=∫v2​v3​​Pdv=P2​(V3​−V2​)=65000(0.05−0.1)=−3250Nm

W3−1=∫Pdv=0W_{3-1}=\int Pdv=0W3−1​=∫Pdv=0

Net work =W1−2+W2−3+W3−1=65000−3250+0=61750NmW_{1-2}+W_{2-3}+W_{3-1}=65000-3250+0=61750 NmW1−2​+W2−3​+W3−1​=65000−3250+0=61750Nm