Question Bank • QB_PARAMETRIC_FP_CYC_013MCQ • 1 Mark
Section 3: Farm Power
Thermodynamic Air Cycles: Otto, Diesel & Dual
Subtopic: Air-standard Otto cycle efficiency
Question Statement:
An internal combustion engine operating on the air-standard Otto cycle has a compression ratio of $r = 15$. Taking the adiabatic exponent $\gamma = 1.4$, the air-standard thermal efficiency of the cycle is: (Variation QB_PARAMETRIC_FP_CYC_013)
Options:
(A) $66.15\%$
(B) $60.65\%$
(C) $72.35\%$
(D) $49.61\%$
Answer Key:
A
Step-by-Step Solution & Explanation
The air-standard efficiency of an Otto cycle is given by:
$\eta_{th} = 1 - \frac{1}{r^{\gamma - 1}}$
Given $r = 15$ and $\gamma = 1.4$:
$\eta_{th} = 1 - \frac{1}{15^{0.4}} = 1 - \frac{1}{2.9542} = 1 - 0.3385 = 0.6615$
In percentage: $\eta_{th} = 66.15\%$.