Discussion
If an ideal air-standard Otto cycle and an ideal air-standard Diesel cycle operate on the same compression ratio, then the relation between the thermal efficiencies (\(\eta_{th}\)) of the cycles is
Answer: B
For the same compression ratio, \(\eta_{th,Otto} \gt \eta_{th,Diesel}\).
\[ \begin{aligned} \eta_{Otto} &= 1 - \frac{1}{r^{\gamma - 1}} \\ \eta_{Diesel} &= 1 - \frac{1}{r^{\gamma - 1}}\left[\frac{r_{c}^{\gamma} - 1}{\gamma(r_{c} - 1)}\right] \end{aligned} \]
For any cut-off ratio \(r_{c} \gt 1\), the bracketed factor exceeds 1, so the Diesel cycle loses more. Physically, the Otto cycle adds all its heat at the highest temperature, at constant volume, while the Diesel adds part of its heat at constant pressure as the piston moves away, at a lower effective temperature.
The two would be equal only in the limit \(r_{c} \to 1\), and no heat engine reaches an efficiency of 1. In practice Diesel engines are more efficient because they run at much higher compression ratios, which this comparison excludes.
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