How to use this calculator

Choose the cycle. For an Otto (gasoline engine) cycle enter the compression ratio and the peak temperature; for a Diesel cycle, the compression ratio and the cutoff ratio; for a Brayton (gas turbine) cycle, the pressure ratio and the turbine inlet temperature. Then give the inlet state. The defaults use air with k = 1.4 and R = 287 J/kg·K. Every result is per kilogram of working gas.

Formulas

Otto: compression, constant-volume heat, expansion, constant-volume rejectionη = 1 − 1r^k−1 q_in = c_v(T₃ − T₂)

Diesel: heat added at constant pressureη = 1 − 1r^k−1 · r_c^k − 1k(r_c − 1) q_in = c_p(T₃ − T₂)

Brayton: steady-flow compressor, combustor, turbineη = 1 − 1r_p^(k−1)/k q_in = c_p(T₃ − T₂)

Isentropic steps and MEPT₂T₁ = r^k−1 c_v = Rk − 1 MEP = w_netv₁ − v₂

Worked example

Otto cycle, r = 10, 300 K and 100 kPa inlet, 1,600 K peak (the default)

c_v = 287/0.4 = 717.5 J/kg·K.
Compression: T₂ = 300 × 10^0.4 = 753.6 K and P₂ = 100 × 10^1.4 = 2,512 kPa.
Heat in: q_in = 0.7175 × (1600 − 753.6) = 607.3 kJ/kg. Peak pressure is 5,333 kPa.
Expansion: T₄ = 1600/10^0.4 = 637.0 K, so q_out = 0.7175 × (637.0 − 300) = 241.8 kJ/kg.
w_net = 365.5 kJ/kg and η = 365.5/607.3 = 60.2%, matching 1 − 1/10^0.4. MEP = 472 kPa.

Comparing the three cycles

Otto and Brayton efficiencies depend only on the compression or pressure ratio, so raising it is the main way to improve them. Gasoline engines are held to r ≈ 8 to 12 by knock, which is why Diesel engines, compressing air alone to r ≈ 14 to 22, are more efficient in practice even though a Diesel cycle is less efficient than an Otto cycle at the same r. Gas turbines run pressure ratios of 10 to 40, and their back work ratio, often 40 to 60%, shows how much turbine output the compressor consumes.

Assumptions and limits

Cold-air standard: the working fluid is air, an ideal gas with constant specific heats evaluated near room temperature.
All processes are internally reversible; compression and expansion are isentropic.
Combustion and exhaust are replaced by heat addition and rejection, so there is no change of gas composition.
Constant specific heats overpredict efficiency at high temperature, because c_p rises and k falls as air gets hotter. Real engines also lose heat, friction and pumping work, so actual efficiencies are much lower.

Frequently asked questions

Why doesn’t the peak temperature change Otto efficiency?

In the ideal Otto cycle every temperature ratio across compression and expansion is r^k−1, so the heat rejected is always the same fraction of the heat added. A hotter peak gives more work per kilogram, not a higher efficiency.

What is mean effective pressure?

The constant pressure that, acting on the piston over one stroke, would produce the same net work. It lets you compare engines of different sizes: higher MEP means more work from the same displacement.

Which k should I use?

1.4 is the room-temperature value for air and is standard for cold-air-standard problems. Some textbooks use k ≈ 1.35 to approximate hot combustion gases.