Question

1kg of a fluid expands reversibly according to a linear law from 5.2bar to 2.4 bar. The initial and final volumes are 0.005m3 and 0.03m3. The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law pv= constant back to the initial conditions of 5.2bar and 0.005m3. Calculate the work done in each process and the net work of the cycle. Sketch the cycle on a p-v diagram.

Answer 1

$W_{1\to2} = ?;\ W_{2\to3} = ?;\ W_{3\to1} = ?$

$\begin{aligned} W_{1\to2} &= \frac{1}{2}(P_1+P_2)(V_2-V_1)\\ &= \frac{1}{2}((5.2+2.4)×10^5)(0.03-0.005)\\ &= 9500J \end{aligned}$

$\begin{aligned} W_{2\to3} &= P_2(V_3-V_2)\\ \textsf{Accord}&\textsf{ing to Boyles Law}\\ V_3 &= \frac{P_1}{P_3}V_1 = \frac{5.2}{2.4}×0.005 = 0.0108m^3\\ W_{2\to3} &= 2.4×10^5(0.0108-0.03)\\ &= -4608J \end{aligned}$

$\begin{aligned} W_{3\to1} &= P_1V_1\ln\frac{V_1}{V_3}\\ &= 5.2×10^5 × 0.005 × \ln0.463\\ &= -2002J \end{aligned}$

$\therefore$ The net work of the cycle is;

9500 - 4608 - 2002 = 2890J