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 · ideal Rankine cycle
    Thermodynamics / power cycles

    Rankine cycle

    Four processes: isentropic pumping, constant-pressure heat addition, isentropic expansion, and constant-pressure rejection. The ideal efficiency is bounded by Carnot between the saturation temperatures, and the gap is almost entirely the irreversible expansion — which is why turbine efficiency, not pressure ratio, separates a good plant from a mediocre one.

    QualityWetBack work0.3%Boiler10.0 bar
    Thermal efficiency
    17.0%
    Carnot bound 21.8%
    Net work
    414 kJ/kg
    415 turbine less 1.2 pump
    Steam rate
    8.69 kg/kWh
    mass flow per unit of output
    Exhaust quality
    0.877
    below the 88% erosion limit
    The four states
    Enthalpy at each point, with the isentropic reference for the expansion.
    17.0%
    The four states
    Stateh (kJ/kg)Note
    1 — turbine inlet2777saturated vapour at the boiler pressure
    2s — isentropic exit2288the ideal expansion
    2 — actual exit2362at 85% turbine efficiency
    3 — condenser exit341saturated liquid
    4 — pump exit341.7at 80% pump efficiency
    Turbine work415 kJ/kgh1 − h2
    Pump work1.19 kJ/kgv ΔP / η
    Net work414 kJ/kgturbine less pump
    Heat input2435 kJ/kgh1 − h4
    Efficiency17.0%net work over heat input
    Carnot21.8%between the saturation temperatures
    Exhaust quality0.877dryness fraction at state 2

    The turbine exhaust is 87.7% dry, which is below the 88% usually allowed — the last stages would erode. Reheating or a higher condenser pressure fixes it.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Efficiency against boiler pressure
    Rises, then flattens — and the exhaust gets wetter, which is the constraint that stops you simply raising it.
    Efficiency against boiler pressure0.0004.619.2313.8418.4523.072 bar2 bar5 bar5 bar10 bar10 bar15 bar15 bar20 bar20 barEfficiency %
    Work against losses
    The pump is a rounding error against the turbine — until you look at the back-work ratio.
    Work against losses0.00098.04196.08294.12392.16490.20415.43TurbineTurbine1.19PumpPump414.24NetNetkJ/kg
    Canonical equations
    The exact relationships implemented by this engine.

    w_t = h₁ − h₂ · With η_t on the isentropic drop.

    w_p = v_f ΔP / η_p · Incompressible.

    η = (w_t − w_p)/(h₁ − h₄) · Net over heat input.

    1 − T₂/T₁ · The bound.

    engineering methodology

    A clear path from inputs to results.

    Four processes: isentropic pumping, constant-pressure heat addition, isentropic expansion, and constant-pressure rejection. The ideal efficiency is bounded by Carnot, and the gap is almost entirely the irreversible expansion.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Set the boiler and condenser pressures.
    2. Enter the turbine and pump efficiencies.
    3. Read the efficiency, the net work and the exhaust quality.
    4. Check the quality against the 88% erosion limit.

    The formula

    η=(h1−h2)−(h4−h3)h1−h4\eta = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}

    Reference: Saturated steam properties from IAPWS-IF97; the cycle is the simple ideal Rankine with isentropic efficiencies applied to the endpoints.

    Worked example

    10 bar to 0.5 bar, 85% turbine

    Inputs
    • Boiler at 10 bar, condenser at 0.5 bar.
    • Turbine 85%, pump 80%.
    Calculation path
    1. 1

      Saturated vapour at 10 bar has h₁ = 2,777 kJ/kg and s₁ = 6.585 kJ/kg·K.

    2. 2

      Expanding isentropically to 0.5 bar gives a dryness of 0.845 and h₂s = 2,288 kJ/kg.

    3. 3

      At 85% efficiency the actual exit is 2,362 kJ/kg.

    4. 4

      The pump adds 1.2 kJ/kg, so the heat input is 2,435 kJ/kg.

    5. 5

      Net work is 414 kJ/kg, an efficiency of 17.0% against a Carnot bound of 21.8%.

    Result

    414 kJ/kg · 17.0% · exhaust 87.7% dry.

    Interpreting the result

    The exhaust quality is the constraint that stops you simply raising the boiler pressure. Above about 15 bar the expansion ends below 88% dry, and the last turbine stages erode — which is why reheat exists, and why a real plant is a two-stage cycle rather than a simple one.

    Assumptions & limitations

    • Saturated vapour at the turbine inlet, no superheat.
    • Isentropic efficiencies applied to the turbine and pump.
    • No pressure drop in the boiler, condenser or pipework.
    • No reheat, regeneration or feedwater heating.

    Frequently asked questions

    Why is the efficiency so far below Carnot?

    Because the heat is added over a range of temperatures rather than at one, and the expansion is irreversible. Carnot assumes isothermal heat addition, which a boiler cannot do.

    How does reheat help?

    It expands the steam a second time after reheating to the same temperature, which raises the mean temperature of heat addition and dries the exhaust — solving both problems at once.

    What pressure should the condenser run at?

    As low as practical — 0.05 to 0.1 bar in a large plant. Below that the vacuum is harder to hold than the efficiency is worth.

    What does the exhaust quality mean?

    The dryness fraction after expansion. Below about 0.88 the water droplets erode the last-stage blades, so it is a mechanical limit rather than a thermodynamic one.

    Why is the pump work so small?

    Because liquid water is nearly incompressible: v ΔP is a few kJ/kg against a turbine work of hundreds. The back-work ratio is under 1%, unlike a gas turbine where it is 40%.

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