Question

For the given same compression ratio and the same heat input the air standard efficiency of otto, diesel and dual combustion cycles are in the order of:

• A. $\eta_{otto} \gt \eta_{diesel} \gt \eta_{dual}$
• B. $\eta_{otto} \gt \eta_{dual} \gt \eta_{diesel}$
• C. $\eta_{diesel} \gt \eta_{otto} \gt \eta_{dual}$
• D. $\eta_{dual} \gt \eta_{diesel} \gt \eta_{otto}$

Hint:

If you see "Same Compression Ratio," always put Otto first. Heat addition at constant volume is theoretically the most "efficient" way to get work out of an engine if you don't have to worry about mechanical stress limits.

Answer 1

The Correct Option is B

1. Analyzing the Condition (Same $r$ and Same Heat Input): When we maintain the same compression ratio ($r$) for all three cycles:

• Otto Cycle: Heat is added at

constant volume. This is the most efficient way to add heat because the pressure and temperature rise most rapidly.

• Diesel Cycle: Heat is added at

constant pressure. This results in the lowest peak temperature and pressure for a given heat input, leading to the lowest efficiency.

• Dual Cycle: Heat is added

partly at constant volume and partly at constant pressure. Naturally, its performance falls in between the Otto and Diesel cycles.

2. Mathematical Reasoning: The efficiency of the Otto cycle is purely a function of the compression ratio ($\eta = 1 - 1/r^{\gamma-1}$). For the Diesel and Dual cycles, the addition of the "cut-off" factor in the denominator (due to heat addition at constant pressure) always reduces the efficiency relative to the Otto cycle when the compression ratio is identical.

3. Order of Efficiency: Under these specific conditions (Same $r$, Same heat input): $$\eta_{otto} \gt \eta_{dual} \gt \eta_{diesel}$$ Note: In practice, Diesel engines are often more efficient than gasoline engines because they can operate at much higher compression ratios without pre-ignition (knocking), which is a different set of conditions.