Otto cycle

Example

Take $n$ moles of an ideal gas with constant molar heat capacity $c_V$ at constant volume and constant heat capacity ratio $\gamma > 1$. Let $V_1 > V_2$ be two volumes, with compression ratio $r = V_1/V_2$. The air-standard Otto 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. heat addition at constant volume $V_2$ (isochoric), absorbing $Q_H = n c_V (T_3 - T_2)$;
3. isentropic expansion from $V_2$ back to $V_1$;
4. heat rejection at constant volume $V_1$, 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. On the isentropes $TV^{\gamma-1}$ is constant, so $T_2 = T_1 r^{\gamma-1}$ and $T_3 = T_4 r^{\gamma-1}$. Hence $T_3 - T_2 = r^{\gamma-1}(T_4 - T_1)$ and the thermal efficiency is $$\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{1}{r^{\gamma - 1}}.$$ For $r = 8$ and $\gamma = 1.4$, $\eta = 1 - 8^{-0.4} \approx 0.56$. This is less than the Carnot value $1 - T_1/T_3$ for the extreme temperatures $T_1$ and $T_3$ of the cycle, since $T_1/T_3 < T_1/T_2 = r^{1-\gamma}$.