Otto Cycle Efficiency Calculator
Enter the compression ratio and specific-heat ratio to compute the ideal Otto cycle efficiency η = 1 − 1/r^(γ−1).
Input Data
Results
At a glance:The ideal Otto cycle models a spark-ignition (gasoline) engine: isentropic compression 1→2, constant-volume heat addition 2→3, isentropic expansion 3→4, constant-volume heat rejection 4→1. Its thermal efficiency depends only on the compression ratio r = V_max/V_min and the specific-heat ratio γ: η = 1 − 1/r^(γ−1). For air γ≈1.4, so r=8 gives η≈1−8^(−0.4)≈56%; r=10 gives ~60%. Higher compression raises efficiency but risks knock (pre-ignition) and needs higher-octane fuel. This tool returns η (fraction and percent) and the temperature ratio T_max/T_min = r^(γ−1).
Formula
η = 1 − 1/r^(γ−1)
Temperature ratio: T_max/T_min = r^(γ−1)
$$r = \frac{V_1}{V_2}$$$$\eta = 1 - \frac{1}{r^{\gamma - 1}}$$How to Use
- Enter the compression ratio r.
- Enter the specific-heat ratio γ (≈1.4 for air).
- The calculator returns η and the temperature ratio.
Case Studies
Gasoline engine
r = 10, γ = 1.4.
η = 1 − 10^(−0.4) ≈ 0.602.
≈60% ideal; real engines ~30–35% due to losses.
FAQ
Why does higher compression improve efficiency?
More compression extracts more work from the heat added (larger expansion ratio), raising η = 1 − 1/r^(γ−1). Real gains are limited by knock and mechanical stress.
What is γ?
The ratio of specific heats C_p/C_v. For diatomic air γ≈1.4; it sets how much pressure rises with compression.
Related Tools
References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.