Jet Engine Simulator

Tools · Thermodynamics

Brayton cycle calculator

Enter the compressor pressure ratio and the turbine inlet temperature. The calculator works out the cycle temperatures, the work, the thermal efficiency and the back work ratio, and draws the cycle on a temperature–entropy diagram.

Calculator

K
Standard day at sea level: 288.15 K (15 °C)
kPa
Sea level: 101.325 kPa
Early jets 5–10, modern airliner engines 40–50
K
Limited by turbine materials and cooling
%
100 % for the ideal cycle
%
100 % for the ideal cycle
Air: 1.4
J/kg·K
Air: 1005 J/kg·K
Thermal efficiency43.3 %ideal cycle at this pressure ratio: 57.5 %
Net specific work334.9 kJ/kgper kg of air through the engine
Compressor exit, T2731 K (458 °C)
Compressor exit pressure, P22,027 kPa
Turbine exit, T4724 K (450 °C)
Compressor work, wc445.4 kJ/kg
Turbine work, wt780.3 kJ/kg
Heat added, qin772.5 kJ/kg
Back work ratio, wc/wt57.1 %
Pressure ratio for most work (ideal cycle)17.9
4008001200Entropy (relative)Temperature, K1234
Temperature–entropy diagram: 1→2 compression, 2→3 heat added at constant pressure, 3→4 expansion in the turbine, 4→1 heat rejected.

How to use it

The calculator follows one kilogram of air around the cycle. State 1 is the compressor inlet, 2 the compressor exit, 3 the turbine inlet and 4 the turbine exit. The two numbers that matter most are the pressure ratio and the turbine inlet temperature; the component efficiencies turn the ideal cycle into a realistic one.

The equations

T2s = T1 · PR(γ−1)/γ    T2 = T1 + (T2s − T1) / ηcCompression. T2s is the temperature a perfect (isentropic) compressor would reach; a real one, with efficiency ηc, gets hotter for the same pressure rise.
T4s = T3 / PR(γ−1)/γ    T4 = T3 − ηt · (T3 − T4s)Expansion through the turbine, with turbine efficiency ηt.
wc = cp(T2 − T1)   wt = cp(T3 − T4)   qin = cp(T3 − T2)
η = (wt − wc) / qin    ηideal = 1 − 1 / PR(γ−1)/γWork, heat and thermal efficiency. For the ideal cycle the efficiency depends on the pressure ratio alone.
PRmax work = (T3 / T1)γ / (2(γ−1))The pressure ratio that gives the most net work per kilogram in the ideal cycle.

Worked example

With the default inputs (a standard day, pressure ratio 20, turbine inlet 1500 K, an 88 % compressor and a 90 % turbine), the compressor heats the air to about 731 K and takes 445 kJ for every kilogram. The turbine gives back 780 kJ, leaving 335 kJ/kg of net work from 773 kJ/kg of heat: a thermal efficiency of 43.3 %. The ideal cycle at the same pressure ratio would reach 57.5 %, and the compressor alone uses 57 % of the turbine's work, which is why component efficiency matters so much.

What the numbers show

Assumptions and limits

This is the textbook "cold-air standard" cycle: air is a perfect gas with constant cp and γ, there are no pressure losses in the combustor or ducts, and the mass of the fuel is ignored. Real engines use hotter gas with a lower γ, lose a few percent of pressure in the combustor, and bleed air for cooling, so their numbers are somewhat lower. In a jet engine the net work does not leave as shaft power: it speeds up the exhaust, which the jet thrust calculator turns into thrust. The game's own engine model is more detailed; the About page explains how.