Question

If an ideal air-standard Otto cycle and an ideal air-standard Diesel cycle operate on the same compression ratio, then the relation between thermal efficiencies ($\eta_{th}$) of the cycles is:

• A. $\eta_{th,Otto} = \eta_{th,Diesel}$ and $\eta_{th,Otto}<1$
• B. $\eta_{th,Otto}>\eta_{th,Diesel}$
• C. $\eta_{th,Otto}<\eta_{th,Diesel}$
• D. $\eta_{th,Otto} = \eta_{th,Diesel} = 1$

Hint:

At equal compression ratio, Otto cycle is always more efficient than Diesel cycle, because Diesel has additional irreversibility due to finite cut-off ratio.

Answer 1

The Correct Option is B

Step 1: Otto cycle efficiency.
\[ \eta_{Otto} = 1 - \frac{1}{r^{\gamma -1}} \] where $r =$ compression ratio, $\gamma =$ specific heat ratio.
Step 2: Diesel cycle efficiency.
\[ \eta_{Diesel} = 1 - \frac{1}{r^{\gamma -1}} . \frac{\rho^\gamma -1}{\gamma (\rho-1)} \] where $\rho =$ cut-off ratio $(>1)$.
Step 3: Comparison.
Since $\rho>1$, \[ \frac{\rho^\gamma -1}{\gamma (\rho-1)}>1 \] Thus, \[ \eta_{Diesel}<\eta_{Otto} \text{for the same } r \] Final Answer: \[ \boxed{\eta_{th,Otto}>\eta_{th,Diesel}} \]