Rankine Cycle Efficiency Calculator

Find the ideal Rankine steam power cycle thermal efficiency from the enthalpies at its 4 state points, eta = (Wturbine - Wpump) / Qin.

♨️ Rankine Cycle Efficiency Calculator
kJ/kg
kJ/kg
kJ/kg
kJ/kg
Thermal efficiency (η)
Net work (Wnet)
Turbine work
Pump work
Heat input (Qin)
Step-by-step working

♨️ What is the Rankine Cycle Efficiency Calculator?

This Rankine cycle efficiency calculator finds the ideal thermal efficiency of a steam power cycle directly from the specific enthalpies at its 4 state points: pump inlet (h1), pump outlet/boiler inlet (h2), boiler outlet/turbine inlet (h3), and turbine outlet/condenser inlet (h4). It computes turbine work, pump work, net work, boiler heat input, and the resulting efficiency, all shown step by step.

This calculator intentionally takes enthalpies as direct numeric input rather than pressures and temperatures. In real-world practice, engineers look up h1 through h4 from steam tables for their actual boiler pressure, condenser pressure, and superheat temperature, then use those looked-up values to evaluate the cycle. That steam-table lookup step is a separate, exact tabulated-data problem, so this tool starts from the enthalpies you already have.

Mechanical and power engineering students use this calculator for Rankine cycle coursework, verifying hand calculations of steam power plant efficiency. Power plant and energy engineers use it as a fast sanity check when comparing different boiler pressures, superheat temperatures, or condenser vacuum levels against each other.

A common point of confusion is that Rankine efficiency, though it looks similar in spirit to Carnot efficiency, will always come out lower than the Carnot efficiency between the cycle's peak and lowest temperatures, since real heat addition happens over a temperature range rather than at a single fixed peak temperature, and real cycles include irreversibilities that Carnot's idealization ignores entirely.

📐 Formula

η  =  (Wturbine − Wpump) / Qin
Wturbine = h3 − h4 (turbine work output, kJ/kg)
Wpump = h2 − h1 (pump work input, kJ/kg)
Qin = h3 − h2 (boiler heat input, kJ/kg)
Example: h1=200, h2=205, h3=3000, h4=2000 kJ/kg (illustrative round-number enthalpies): η ≈ 35.60%.

📖 How to Use This Calculator

Steps

1
Enter the pump inlet and outlet enthalpies - h1 (condenser exit) and h2 (boiler inlet), in kJ/kg.
2
Enter the turbine inlet and outlet enthalpies - h3 (boiler outlet) and h4 (condenser inlet), in kJ/kg.
3
Read the thermal efficiency - See the cycle's thermal efficiency along with turbine work, pump work, and heat input.

💡 Example Calculations

The two worked examples below use clean, round illustrative enthalpy values chosen to demonstrate the formula clearly. They are not tied to any specific real steam condition or pressure, real analysis requires looking up actual enthalpies from steam tables for your boiler and condenser pressures.

Example 1 - Illustrative baseline cycle

1
h1=200, h2=205, h3=3000, h4=2000 kJ/kg (illustrative enthalpies)
2
Wturbine = 3000 − 2000 = 1000 kJ/kg, Wpump = 205 − 200 = 5 kJ/kg
3
Wnet = 1000 − 5 = 995 kJ/kg, Qin = 3000 − 205 = 2795 kJ/kg
4
η = 995 / 2795 = 35.5993%
η = 35.5993%
Try this example →

Example 2 - Deeper turbine expansion (lower condenser pressure)

1
h1=200, h2=205, h3=3000 kJ/kg (same as Example 1), h4=1800 kJ/kg (lower, deeper expansion)
2
Wturbine = 3000 − 1800 = 1200 kJ/kg, Wpump = 205 − 200 = 5 kJ/kg
3
Wnet = 1200 − 5 = 1195 kJ/kg, Qin = 3000 − 205 = 2795 kJ/kg
4
η = 1195 / 2795 = 42.7549%, noticeably higher than Example 1 purely because h4 is lower
η = 42.7549%
Try this example →

❓ Frequently Asked Questions

What is the Rankine cycle?+
The Rankine cycle is the ideal thermodynamic cycle used to model steam power plants. Water is pumped to high pressure, boiled and superheated into steam, expanded through a turbine to produce work, then condensed back to liquid before the cycle repeats. Its thermal efficiency is eta = (Wturbine - Wpump) / Qin.
What is the formula for Rankine cycle efficiency?+
eta = ((h3-h4) - (h2-h1)) / (h3-h2), where h1 through h4 are the specific enthalpies in kJ/kg at the pump inlet, pump outlet, turbine inlet, and turbine outlet respectively. The numerator is net work output and the denominator is heat added in the boiler.
What do h1, h2, h3, and h4 physically represent?+
h1 is the enthalpy at the pump inlet, condenser exit, saturated liquid. h2 is the enthalpy at the pump outlet, boiler inlet, after the feedwater pump raises the pressure. h3 is the enthalpy at the boiler outlet, turbine inlet, usually superheated steam at high pressure and temperature. h4 is the enthalpy at the turbine outlet, condenser inlet, after expansion.
Where do I get real values for h1, h2, h3, and h4?+
In practice, engineers look these enthalpies up from steam tables (or a steam property calculator) using the actual boiler pressure, condenser pressure, and superheat temperature of their system. This calculator intentionally takes enthalpies as direct input rather than pressures, since the steam-table lookup step is a separate, exact tabulated-data problem outside the scope of this tool.
Why is Rankine efficiency always lower than the Carnot efficiency between the same two temperatures?+
Two reasons. First, real cycles have irreversibilities (friction, non-ideal expansion) that a Carnot cycle assumes away entirely. Second, and more fundamentally, heat addition in the boiler happens over a range of temperatures as water heats, boils, and superheats, not entirely at the single highest temperature the way an idealized Carnot cycle requires, so the average temperature of heat addition is lower than the peak temperature.
How does lowering condenser pressure improve Rankine efficiency?+
Lowering condenser pressure lowers the turbine outlet enthalpy h4, which increases turbine work (h3-h4) without changing the heat input Qin much, directly raising net work and efficiency. This is exactly why real power plant condensers operate under a vacuum, typically well below atmospheric pressure, rather than at atmospheric pressure or above.
Why is pump work so much smaller than turbine work?+
Pumping a liquid (nearly incompressible) to higher pressure takes far less work than the pressure-volume work extracted from an expanding, low-density gas (steam) in the turbine. Pump work is often under 1 percent of turbine work, though this calculator still accounts for it explicitly since h2 and h1 are both required inputs.
What is a superheated Rankine cycle versus a saturated one?+
In a saturated cycle, the boiler outlet (state 3) is saturated steam right at the boiling point for that pressure. In a superheated cycle, the steam is heated further past saturation before entering the turbine, raising h3 and generally increasing both net work and efficiency while also keeping the turbine exhaust drier (less moisture damage to turbine blades).
Does the Rankine cycle use water specifically, or any working fluid?+
Steam power plants overwhelmingly use water as the working fluid because it is cheap, non-toxic, and has well-characterized thermodynamic properties, but the Rankine cycle itself is a general model. Organic Rankine cycles use other fluids with lower boiling points for waste-heat recovery and geothermal applications where source temperatures are too low for water to be practical.
What happens if I enter h3 less than or equal to h2?+
The calculator shows an error, since h3 must exceed h2 for the boiler to actually be adding heat to the working fluid (Qin = h3-h2 must be positive). A Rankine cycle with zero or negative heat input is not physically meaningful.
Can Rankine cycle efficiency exceed the Carnot limit?+
No. The Rankine cycle operates between the same peak boiler temperature and condenser temperature, and the second law of thermodynamics guarantees the Carnot efficiency between those two temperatures is the absolute ceiling. Any correctly computed Rankine efficiency from valid steam-table enthalpies will always come out lower than the Carnot efficiency for the corresponding temperature range.

What is the Rankine cycle?

The Rankine cycle is the ideal thermodynamic cycle used to model steam power plants. Water is pumped to high pressure, boiled and superheated into steam, expanded through a turbine to produce work, then condensed back to liquid before the cycle repeats. Its thermal efficiency is eta = (Wturbine - Wpump) / Qin.

What is the formula for Rankine cycle efficiency?

eta = ((h3-h4) - (h2-h1)) / (h3-h2), where h1 through h4 are the specific enthalpies in kJ/kg at the pump inlet, pump outlet, turbine inlet, and turbine outlet respectively. The numerator is net work output and the denominator is heat added in the boiler.

What do h1, h2, h3, and h4 physically represent?

h1 is the enthalpy at the pump inlet, condenser exit, saturated liquid. h2 is the enthalpy at the pump outlet, boiler inlet, after the feedwater pump raises the pressure. h3 is the enthalpy at the boiler outlet, turbine inlet, usually superheated steam at high pressure and temperature. h4 is the enthalpy at the turbine outlet, condenser inlet, after expansion.

Where do I get real values for h1, h2, h3, and h4?

In practice, engineers look these enthalpies up from steam tables (or a steam property calculator) using the actual boiler pressure, condenser pressure, and superheat temperature of their system. This calculator intentionally takes enthalpies as direct input rather than pressures, since the steam-table lookup step is a separate, exact tabulated-data problem outside the scope of this tool.

Why is Rankine efficiency always lower than the Carnot efficiency between the same two temperatures?

Two reasons. First, real cycles have irreversibilities (friction, non-ideal expansion) that a Carnot cycle assumes away entirely. Second, and more fundamentally, heat addition in the boiler happens over a range of temperatures as water heats, boils, and superheats, not entirely at the single highest temperature the way an idealized Carnot cycle requires, so the average temperature of heat addition is lower than the peak temperature.

How does lowering condenser pressure improve Rankine efficiency?

Lowering condenser pressure lowers the turbine outlet enthalpy h4, which increases turbine work (h3-h4) without changing the heat input Qin much, directly raising net work and efficiency. This is exactly why real power plant condensers operate under a vacuum, typically well below atmospheric pressure, rather than at atmospheric pressure or above.

Why is pump work so much smaller than turbine work?

Pumping a liquid (nearly incompressible) to higher pressure takes far less work than the pressure-volume work extracted from an expanding, low-density gas (steam) in the turbine. Pump work is often under 1 percent of turbine work, though this calculator still accounts for it explicitly since h2 and h1 are both required inputs.

What is a superheated Rankine cycle versus a saturated one?

In a saturated cycle, the boiler outlet (state 3) is saturated steam right at the boiling point for that pressure. In a superheated cycle, the steam is heated further past saturation before entering the turbine, raising h3 and generally increasing both net work and efficiency while also keeping the turbine exhaust drier (less moisture damage to turbine blades).

Does the Rankine cycle use water specifically, or any working fluid?

Steam power plants overwhelmingly use water as the working fluid because it is cheap, non-toxic, and has well-characterized thermodynamic properties, but the Rankine cycle itself is a general model. Organic Rankine cycles use other fluids with lower boiling points for waste-heat recovery and geothermal applications where source temperatures are too low for water to be practical.

What happens if I enter h3 less than or equal to h2?

The calculator shows an error, since h3 must exceed h2 for the boiler to actually be adding heat to the working fluid (Qin = h3-h2 must be positive). A Rankine cycle with zero or negative heat input is not physically meaningful.

Can Rankine cycle efficiency exceed the Carnot limit?

No. The Rankine cycle operates between the same peak boiler temperature and condenser temperature, and the second law of thermodynamics guarantees the Carnot efficiency between those two temperatures is the absolute ceiling. Any correctly computed Rankine efficiency from valid steam-table enthalpies will always come out lower than the Carnot efficiency for the corresponding temperature range.