Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 35A / d

This is the ideal Diesel cycle. In the $(p,V)$ plane, $A\to B$ is an adiabatic curve running up and left, $B\to C$ is a horizontal constant-pressure segment running right, $C\to D$ is an adiabatic curve running down and right, and $D\to A$ is a vertical constant-volume segment running down. The traversal is clockwise, so the enclosed area is positive work output.

In the $(T,S)$ plane, the two reversible adiabats are vertical because entropy is constant: $A\to B$ rises at $S_A=S_B$, and $C\to D$ falls at $S_C=S_D$. Along $B\to C$, the heat increment at constant pressure issoAlong the constant-volume curve $D\to A$,which givesThus $B\to C$ runs up and right and $D\to A$ runs down and left between the two vertical adiabats.

Heat enters only during the constant-pressure step, so the heat capacity at constant pressure givesThe positive magnitude of the heat rejected during constant-volume cooling isBecause the working gas returns to its initial state, the thermodynamic cycle has zero net change in internal energy and $W=Q_1-Q_2$. Its thermal efficiency is therefore