Discussion
If \(\gamma\) refers to the ratio of specific heats, the air-standard efficiency of an Otto cycle is
Answer: C
The efficiency is \(1 - 1/(\text{Compression ratio})^{(\gamma - 1)}\).
In the air-standard Otto cycle heat is added and rejected at constant volume, and the two other processes are isentropic between the same two volumes. With compression ratio \(r = V_{1}/V_{2}\):
\[ \begin{aligned} \eta &= 1 - \frac{T_{4} - T_{1}}{T_{3} - T_{2}} = 1 - \frac{T_{1}}{T_{2}} \\ \frac{T_{2}}{T_{1}} &= r^{\gamma - 1} \Rightarrow \eta = 1 - \frac{1}{r^{\gamma - 1}} \end{aligned} \]
The exponent is \(\gamma - 1\), not \(\gamma\), because the isentropic temperature-volume relation is \(TV^{\gamma - 1}\) = constant. The forms in terms of a pressure ratio raised to \((\gamma - 1)/\gamma\) describe the Brayton (gas turbine) cycle, where heat is added at constant pressure; the pressure ratio with exponent \(\gamma - 1\) matches neither cycle.
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