Diesel cycle

Example

Take $n$ moles of an ideal gas with constant molar heat capacities $c_V$ and $c_p = \gamma c_V$, where $\gamma > 1$ is the heat capacity ratio. Let $V_2 < V_3 < V_1$ be volumes, with compression ratio $r = V_1/V_2$ and cut-off ratio $\rho = V_3/V_2 > 1$. The air-standard Diesel cycle has four steps, with temperatures $T_1, T_2, T_3, T_4$ at their start points:

1. isentropic compression from $V_1$ to $V_2$;
2. isobaric heat addition from $V_2$ to $V_3$, absorbing $Q_H = n c_p (T_3 - T_2)$;
3. isentropic expansion from $V_3$ to $V_1$;
4. isochoric heat rejection at $V_1$, rejecting $Q_C = n c_V (T_4 - T_1)$, back to the initial state.

The gas returns to its initial state, so this is a thermodynamic cycle. From the isentropes and the ideal gas law: $T_2 = T_1 r^{\gamma-1}$, $T_3 = \rho T_2$, and $T_4 = T_3 (V_3/V_1)^{\gamma-1} = T_3 (\rho/r)^{\gamma-1} = T_1 \rho^{\gamma}$. Hence the thermal efficiency is $$\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^\gamma - 1}{\gamma(\rho - 1)}.$$ Since $\rho \mapsto \rho^\gamma$ is strictly convex, $\rho^\gamma - 1 > \gamma(\rho - 1)$ for $\rho > 1$, so for the same compression ratio the Diesel cycle is less efficient than the Otto cycle. For $r = 18$, $\rho = 2$ and $\gamma = 1.4$, $\eta \approx 0.63$.