Otto and Diesel cycle calculator
Calculate the thermal efficiency of the ideal Otto (constant-volume) and Diesel (constant-pressure) reference cycles. Choose the compression ratio, the isentropic exponent and, for Diesel, the cut-off ratio – the calculator returns efficiency, corner states, heat balance, mean effective pressure and the p-V diagram, live with every input.
Cycle calculator (Otto/Diesel)
Model: ideal reference cycles with reversible isentropes and ideal gas (constant properties). Without friction, wall heat, incomplete combustion or gas exchange; real efficiencies are lower.
Results
Calculating …
Formulas and fundamentals
The ideal Otto cycle consists of two isentropes and two isochores. Its thermal efficiency depends only on the compression ratio ε = V1/V2 and the isentropic exponent κ = cp/cv: η = 1 − 1/ε^(κ−1). A higher compression ratio raises the efficiency but is limited by knock in real spark-ignition engines. Example: ε = 10 and κ = 1.4 give η = 0.60189, about 60 percent in the ideal case.
The ideal Diesel cycle replaces the constant-volume heat addition with a constant-pressure one. It has the cut-off ratio φ = V3/V2 as an extra parameter: η = 1 − (1/ε^(κ−1))·(φ^κ − 1)/(κ·(φ − 1)). The factor containing φ is always greater than 1, so for the same ε the Diesel cycle has a lower efficiency than the Otto cycle. Because Diesel engines compress much more, however, their real efficiency is higher. Example: ε = 18 and φ = 2 give η = 0.63158.
From an intake state (pressure p1, temperature T1) the ideal-gas state equations give all corner states: the end-of-compression temperature T2 = T1·ε^(κ−1), the pressure p2 = p1·ε^κ and from these the states after heat addition and expansion. The specific heats are formed consistently from κ and the gas constant Rs (cv = Rs/(κ−1), cp = κ·cv). The mean effective pressure p_m = w/(v1 − v2) relates the net work to the swept volume and allows the comparison of engines of different size.
Worked example
A spark-ignition engine compresses air (κ = 1.4) with ε = 10 from the intake state 1 bar and 300 K. Isentropic compression leads to T2 = 300·10^0.4 = 753.6 K and p2 = 1·10^1.4 = 25.1 bar.
For a peak temperature of 2000 K after constant-volume combustion the thermal efficiency is η = 1 − 1/10^0.4 = 0.60189. The heat added and rejected and the net work follow from the temperature differences of the isochores.
A Diesel engine with ε = 18, φ = 2 and the same medium reaches η = 1 − (1/18^0.4)·(2^1.4 − 1)/(1.4·(2 − 1)) = 0.63158. Despite the less favourable shape factor the Diesel efficiency exceeds that of the Otto example thanks to the higher compression.
Frequently asked questions
Why does the Diesel cycle have the lower efficiency at the same compression ratio?
Because heat is added at constant pressure (and thus increasing volume) rather than at constant volume. The extra factor (φ^κ − 1)/(κ·(φ − 1)) is always greater than 1 and lowers the efficiency relative to the Otto cycle. In practice, however, Diesel engines compress far more, so their real efficiency is still higher than that of spark-ignition engines.
What does the Otto efficiency depend on?
Only on the compression ratio ε and the isentropic exponent κ, not on the amount of heat added. A higher ε raises the efficiency but is limited by the knock resistance of the fuel in real engines.
What is the cut-off ratio φ?
The ratio of the volumes after and before the constant-pressure combustion, φ = V3/V2. It describes over which fraction of volume the heat is added at constant pressure in the Diesel cycle. As φ → 1 the Diesel efficiency approaches the Otto efficiency.
How realistic are the results?
These are ideal reference cycles: reversible isentropes, ideal gas with constant properties, no friction, no wall heat transfer, complete combustion, no gas exchange. Real engines reach considerably lower efficiencies. The comparison nevertheless shows the fundamental relationships and the upper bound.
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