Answer to Question #177771 in Mechanical Engineering for Sana ullah

Question

Unit mass of a certain fluid is contained in a cylinder at an initial pressure of 25 bars. The fluid is allowed to expand reversibly behind a piston according to a law pv 2 =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 25 bars. Calculate the net work done by the fluid, for an initial volume of 0.05m3 & Sketch the process on P-v Diagram.

Answer 1

For position 1 to 2

$P_1V_1^2=C=P_2V_2^2 \implies P_2=25 \times (\frac{0.05}{2 \times 0.05})^2=6.25 bars$

$W_{1-2}=\int_{v_1}^{v_2} Pdv=\int_{v_1}^{v_2} \frac{C}{v^2}dv=0.065[\frac{1}{v}]_{0.05}^{0.1}=0.65 barm^3=65000Nm$

$W_{2-3}=\int_{v_2}^{v_3} Pdv= P_2(V_3-V_2)=65000(0.05-0.1)=-3250 Nm$

$W_{3-1}=\int Pdv=0$

Net work =$W_{1-2}+W_{2-3}+W_{3-1}=65000-3250+0=61750 Nm$