Discussion
The efficiency of diesel cycle is given by :
(Where γ = specific heat ratio, r = compression ratio, ρ = cut-off ratio)
Answer: D
The Diesel cycle efficiency is \(1 - \frac{1}{\gamma r^{\gamma - 1}} \frac{(\rho^{\gamma} - 1)}{(\rho - 1)}\).
Heat is added at constant pressure and rejected at constant volume, so
\[ \begin{aligned} \eta &= 1 - \frac{c_{v}(T_{4} - T_{1})}{c_{p}(T_{3} - T_{2})} \\ &= 1 - \frac{1}{\gamma} \cdot \frac{T_{4} - T_{1}}{T_{3} - T_{2}} \end{aligned} \]
Writing every temperature in terms of \(T_{1}\) using the compression ratio r and the cut-off ratio \(\rho\) gives \(T_{2} = T_{1}r^{\gamma - 1}\), \(T_{3} = \rho T_{2}\) and \(T_{4} = T_{1}\rho^{\gamma}\). Substituting yields the expression above.
The bracket \(\frac{\rho^{\gamma} - 1}{\gamma(\rho - 1)}\) is always greater than 1, which is why a Diesel cycle is less efficient than an Otto cycle of the same compression ratio. Option C has this ratio upside down, and options A and B put \(\gamma\) and \(r^{\gamma - 1}\) in the wrong places.
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