Comparing ideal-cycle efficiencies: Why is Otto cycle efficiency higher than Diesel cycle efficiency for the same compression ratio and heat input?

Introduction / Context:
The Otto and Diesel cycles are idealized models for spark-ignition and compression-ignition engines. For the same compression ratio, the Otto cycle typically attains higher thermal efficiency because of the way heat is added. Understanding this distinction is central to thermodynamic comparison of engine cycles.

Given Data / Assumptions:

• Air-standard analysis with constant specific heats.
• Equal compression ratio and equal total heat input for both cycles.
• Reversible compression/expansion (isentropic) in both cycles.

Concept / Approach:
In the Otto cycle, heat addition occurs at constant volume, raising pressure and temperature more steeply at top dead center for a given heat input. The Diesel cycle adds heat at constant pressure over a finite cut-off, leading to a lower average temperature of heat addition and thus lower efficiency at the same compression ratio. The key differentiator is the constant-volume heat addition of the Otto cycle, not the isentropic legs (which both cycles share).

Step-by-Step Solution:

Verification / Alternative check:
Standard efficiency expressions show η_Otto = 1 − 1/r^(γ−1), while Diesel efficiency contains the cut-off ratio; for identical r, Diesel's η decreases as cut-off increases.

Why Other Options Are Wrong:

• Isentropic legs: Present in both cycles; not the differentiator.
• Maximum temperature is higher: A consequence of constant-volume addition, but the fundamental reason is the heat-addition mode.
• Lower heat rejection: Not inherently guaranteed; it is the higher average temperature of heat addition that drives higher η.

Common Pitfalls:
Attributing differences to compression alone; forgetting the role of cut-off in Diesel efficiency.

Final Answer:
combustion is at constant volume