Question

If \(\gamma\) refers to the ratio of specific heats, the air-standard efficiency of an Otto cycle is

• A. \(\; 1 - \dfrac{1}{(\text{Compression ratio})^{\gamma}} \)
• B. \(\; 1 - \dfrac{1}{(\text{Pressure ratio})^{\tfrac{\gamma - 1}{\gamma}}} \)
• C. \(\; 1 - \dfrac{1}{(\text{Compression ratio})^{\gamma - 1}} \)
• D. \(\; 1 - \dfrac{1}{(\text{Pressure ratio})^{\gamma - 1}} \)

Hint:

Remember: Otto cycle efficiency depends only on compression ratio and \(\gamma\), not on pressure or cut-off ratios.

Answer 1

The Correct Option is C

Step 1: Otto cycle basics.
The Otto cycle consists of two isentropic (compression, expansion) and two isochoric (heat addition, heat rejection) processes. The air-standard efficiency is given by: \[ \eta_{otto} = 1 - \frac{1}{r^{\gamma - 1}} \] where \(r = \dfrac{V_{max}}{V_{min}}\) is the compression ratio.

Step 2: Why compression ratio matters.
During isentropic compression: \[ T_2 = T_1 r^{\gamma - 1} \] and the temperature ratio (hence efficiency) depends only on \(r\) and \(\gamma\), not on pressure ratio.

Step 3: Elimination of options.
(A) Incorrect – has wrong exponent (\(\gamma\) instead of \(\gamma-1\)).
(B) Incorrect – pressure ratio is relevant for Brayton cycle, not Otto cycle.
(C) Correct – matches derived formula.
(D) Incorrect – again involves pressure ratio. Final Answer:
\[ \boxed{\eta_{otto} = 1 - \frac{1}{r^{\gamma - 1}}} \]