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Otto & Diesel Cycle Efficiency Calculator

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Understanding the Otto and Diesel cycle efficiency calculator — concept and principle

The Otto and Diesel cycles are the idealised models of petrol (spark-ignition) and diesel (compression-ignition) engines. This calculator gives the theoretical thermal efficiency of each from the compression ratio, the heat-capacity ratio γ and, for the Diesel cycle, the cut-off ratio.

Otto and Diesel cycle efficiency at compression ratio 10At r = 10 and γ = 1.4 the Otto cycle reaches 60.19%; the Diesel cycle with cut-off ratio 2 reaches 53.39%.Same squeeze, two ways to burn101ratio r = 10Otto (spark)60.19%Diesel · r꜀ 253.39%workheat lost in exhaustOtto: 1 − 1 / 10⁰·⁴ = 1 − 0.398 = 60.19%Diesel: 1 − 0.398 × 1.171 = 53.39%
At the same compression ratio, burning during the stroke (Diesel) costs about 7 points of efficiency.

How it is calculated

r is the compression ratio (cylinder volume at bottom dead centre ÷ volume at top dead centre). γ = c_p/c_v, 1.4 for cold air. The cut-off ratio r_c is the volume at the end of fuel injection divided by the volume at its start.

The principle

Both are air-standard cycles: the working fluid is treated as air, an ideal gas, compressed and expanded adiabatically and reversibly. In the Otto cycle heat is added instantly at constant volume, as if the spark burned all the fuel at top dead centre. In the Diesel cycle heat is added at constant pressure while the piston moves down and fuel keeps burning. Because temperature after adiabatic compression rises by the factor r^(γ−1), higher compression means heat is added at a higher temperature, which raises efficiency, as the Carnot principle predicts. The Diesel bracket term (r_c^γ − 1)/(γ(r_c − 1)) is always above 1, so at the same compression ratio a Diesel cycle is less efficient than an Otto cycle.

Worked example

  1. Inputs: r = 10, γ = 1.4, r_c = 2.
  2. r^(γ−1) = 10^0.4 = 2.51189, so 1 / 2.51189 = 0.398107.
  3. Otto: 1 − 0.398107 = 60.19%.
  4. Diesel bracket: (2^1.4 − 1) / (1.4 × 1) = (2.63902 − 1) / 1.4 = 1.17073.
  5. Diesel: 1 − 0.398107 × 1.17073 = 1 − 0.466074 = 53.39%.
How Otto efficiency changes with γHow Otto efficiency changes with γ-400-300-200-10001000.350.71.051.41.752.12.8Example: 60.19%γ
Each point is the calculator re-run with only “γ” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
γOtto efficiency
0.35-346.68%
0.7-99.53%
1.0510.87%
1.460.19%
1.7582.22%
2.192.06%
2.898.42%

How to read the result

These are upper limits for the idealised cycle. Real engines lose heat to cylinder walls, burn fuel over finite time, have friction and pumping losses, and use a gas whose γ falls at high temperature, so their actual efficiencies are much lower. The comparison is still informative: diesel engines run without knock and can use much higher compression ratios, so in practice they beat petrol engines. With r = 18 and r_c = 2, the ideal Diesel efficiency rises to about 63%.

Common mistakes

Keep in mind: Air-standard efficiencies are teaching models that ignore heat loss, friction, combustion timing and real gas properties. Engine design and emissions work require measured data, detailed simulation and qualified engineers.

Frequently asked questions

Why can't petrol engines simply use higher compression?

Higher compression heats the fuel-air mixture more, and beyond a limit it ignites on its own (knock), which damages the engine. Fuel octane rating sets that limit.

What is a typical compression ratio?

Petrol engines commonly run around 9 to 13, diesel engines around 15 to 22.

Sources: Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, McGraw-Hill · J. B. Heywood, Internal Combustion Engine Fundamentals, McGraw-Hill

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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