Skip to content

    Under active development. New calculators and features ship every week. Everything currently runs free while the site matures — pricing pages are a preview, checkout is not open yet.

    Search ⌘K
    Calculates locally in your browser · Brayton cycle
    Thermodynamics / gas turbines

    Brayton cycle

    A gas turbine is four processes in air: compression, constant-pressure heat addition, expansion, and constant-pressure rejection. The ideal efficiency depends only on the pressure ratio — the most quoted result in power engineering and the most misleading, because the real cycle is dominated by the compressor and turbine efficiencies and by how much exhaust heat a recuperator can recover.

    Net workPositiveBack work53%TIT1500 K
    Thermal efficiency
    38.5%
    ideal 50.8%
    Net work
    323 kJ/kg
    690 turbine less 367 compressor
    Compressor exit
    665 K
    isentropic would be 610 K
    Heat rate
    9355 kJ/kWh
    fuel energy per unit of output
    The four states
    Cold-air-standard analysis: constant cp, constant k, no pressure loss.
    38.5%
    The four states
    StateTemperatureNote
    1 — compressor inlet300 Kambient
    2s — isentropic exit610 Krp^((k−1)/k) = 2.034
    2 — actual exit665 Kat 85% efficiency
    3 — turbine inlet1500 Kthe firing temperature
    4s — isentropic exit737 Kthe ideal expansion
    4 — actual exit814 Kat 90% efficiency
    Compressor work367 kJ/kgcp (T2 − T1)
    Turbine work690 kJ/kgcp (T3 − T4)
    Net work323 kJ/kgturbine less compressor
    Heat input839 kJ/kgcp (T3 − T2), or the recuperated equivalent
    Efficiency38.5%net work over heat input
    Ideal efficiency50.8%1 − 1/rp^((k−1)/k)

    Efficiency is 38.5% against an ideal 50.8%; the gap is the compressor and turbine irreversibility, which is where most of the real loss lives.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Efficiency against pressure ratio
    Rises steeply then flattens; the ideal curve keeps climbing, which is why real engines stop around 15–20.
    Efficiency against pressure ratio0.00013.9227.8541.7755.6969.622:12:14:14:18:18:112:112:116:116:120:120:130:130:1RealIdeal
    Work split
    A gas turbine spends a large share of its turbine work driving its own compressor — 40–60% is normal.
    Work split0.000162.77325.54488.30651.07813.84366.74CompressorCompressor689.69TurbineTurbine322.95NetNetkJ/kg
    Canonical equations
    The exact relationships implemented by this engine.

    T2 = T1 + (T2s − T1)/η_c · T2s = T1 rp^((k−1)/k).

    T4 = T3 − η_t (T3 − T4s) · T4s = T3 / rp^((k−1)/k).

    (w_t − w_c)/q_in · Net over heat added.

    1 − 1/rp^((k−1)/k) · The most quoted result.

    engineering methodology

    A clear path from inputs to results.

    A gas turbine is four processes in air: compression, constant-pressure heat addition, expansion, and constant-pressure rejection. The ideal efficiency depends only on the pressure ratio — the most quoted result in power engineering and the most misleading.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Set the pressure ratio and the compressor inlet temperature.
    2. Enter the turbine inlet temperature and the gas properties.
    3. Add the compressor and turbine efficiencies.
    4. Read the efficiency, the work and the back-work ratio.

    The formula

    ηideal=1−1rp(k−1)/k\eta_{ideal} = 1 - \frac{1}{r_p^{(k-1)/k}}

    Reference: Cold-air-standard Brayton analysis; the isentropic relations and the efficiency definitions are standard.

    Worked example

    Pressure ratio 12, firing at 1,500 K

    Inputs
    • rp = 12, compressor inlet 300 K.
    • Turbine inlet 1,500 K, k = 1.4, cp = 1.005.
    • Compressor 85%, turbine 90%.
    Calculation path
    1. 1

      rp^((k−1)/k) = 2.034, so the isentropic compressor exit is 610 K.

    2. 2

      At 85% efficiency the actual exit is 665 K, and the compressor work is 367 kJ/kg.

    3. 3

      The isentropic turbine exit is 737 K; at 90% efficiency the actual exit is 814 K.

    4. 4

      The turbine work is 690 kJ/kg, leaving 323 kJ/kg net.

    5. 5

      Heat input is 839 kJ/kg, so the efficiency is 38.5% against an ideal 50.8%.

    Result

    323 kJ/kg · 38.5% · back work 53%.

    Interpreting the result

    The gap between 38.5% and 50.8% is entirely irreversibility in the compressor and turbine. That is why a real engine spends its engineering effort on blade cooling and aerodynamics rather than on the pressure ratio, and why the firing temperature is the number that has driven gas turbine progress for fifty years.

    Assumptions & limitations

    • Cold-air-standard: constant cp and k, air as a perfect gas.
    • No pressure loss in the combustor or ducts.
    • Isentropic efficiencies applied to compressor and turbine.
    • No cooling flow, no reheat and no intercooling.

    Frequently asked questions

    Why does the ideal efficiency depend only on pressure ratio?

    Because with constant cp the work terms and the heat input all scale with temperature, and the temperatures themselves are fixed ratios of the pressure ratio. It is a peculiarity of the ideal cycle.

    When does a recuperator help?

    Only when the exhaust (T4) is hotter than the compressor exit (T2). At high pressure ratios T2 overtakes T4 and the recuperator does nothing.

    Why not just raise the pressure ratio?

    Because the firing temperature limits the work, and beyond a point the extra compression costs more than the extra expansion returns. Most engines sit between 15 and 20.

    Why is the real efficiency lower?

    Compressor and turbine irreversibility, plus pressure losses. A real engine loses 2–5% of pressure in each component, which costs several points of efficiency.

    What is the back-work ratio?

    Compressor work over turbine work. It is 40–60% for a gas turbine, against under 1% for a steam plant, which is why a gas turbine is sensitive to compressor efficiency.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.