The Rankine cycle is the model for steam power plants. Water is pumped to high pressure, boiled and superheated, expanded through a turbine, and condensed back to liquid. Four steady-flow energy balances, with states read from steam tables, give the work, heat and thermal efficiency.
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The four processes
Number the states the usual way. 1 → 2: the pump raises saturated liquid from condenser to boiler pressure. 2 → 3: the boiler adds heat at constant pressure, turning the liquid into superheated steam. 3 → 4: the steam expands through the turbine, producing work. 4 → 1: the condenser removes heat at constant pressure until the fluid is saturated liquid again.
In the ideal cycle the pump and turbine are isentropic (reversible and adiabatic), so s₂ = s₁ and s₄ = s₃. That is where the second law and entropy come in: the turbine exit state is found by matching entropy, not by a guess.
Energy balances
Each component is a steady-flow device with one inlet and one outlet. Neglecting kinetic and potential energy changes, the energy balance per unit mass is q − w = h_out − h_in. So the turbine work is h₃ − h₄, the pump work is h₂ − h₁, the heat added is h₃ − h₂, and the heat rejected is h₄ − h₁.
Liquid water is nearly incompressible, so the pump work is simply v₁·(P₂ − P₁). It is small next to the turbine work, which is the reason steam cycles pump a liquid instead of compressing a gas.
Real turbines and pumps are rated by isentropic efficiency: η_t = (h₃ − h₄a)/(h₃ − h₄s) and η_p = (h₂s − h₁)/(h₂a − h₁). Raising boiler pressure or superheat temperature, lowering condenser pressure, and adding reheat or regeneration all push efficiency up.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Steady-flow energy balance
q − w = h_out − h_in
per unit mass (kJ/kg), ΔKE and ΔPE neglected
Çengel & Boles, Thermodynamics: An Engineering Approach
Pump work (liquid)
w_p = v₁·(P₂ − P₁)
v₁ = specific volume of saturated liquid (m³/kg), P in kPa gives kJ/kg
Moran et al., Fundamentals of Engineering Thermodynamics
Thermal efficiency
η = (w_t − w_p) / q_in = 1 − q_out / q_in
w_t = h₃ − h₄, q_in = h₃ − h₂
Çengel & Boles, Thermodynamics: An Engineering Approach
Quality of a wet mixture
x = (s − s_f) / s_fg, h = h_f + x·h_fg
f = saturated liquid, fg = liquid-to-vapour change
Moran et al., Fundamentals of Engineering Thermodynamics
Isentropic turbine efficiency
η_t = (h₃ − h₄a) / (h₃ − h₄s)
a = actual, s = isentropic exit state
Çengel & Boles, Thermodynamics: An Engineering Approach
Worked examples
Example: Ideal Rankine cycle for a small plant
Given
An ideal Rankine cycle runs between 4 MPa in the boiler and 15 kPa in the condenser. Steam enters the turbine at 400 °C.
Steam properties from IAPWS-IF97: at 4 MPa, 400 °C, h₃ = 3214.4 kJ/kg and s₃ = 6.7712 kJ/(kg·K). Saturated at 15 kPa: h_f = 225.9, h_fg = 2372.4 kJ/kg; s_f = 0.7548, s_fg = 7.2523 kJ/(kg·K); v_f = 0.001014 m³/kg.
The plant must deliver 5 MW of net power.
Find
The pump and turbine work.
The thermal efficiency.
The steam mass flow rate.
Solution
Pump work and pump exit enthalpy, with h₁ = h_f at 15 kPa.w_p = 0.001014 × (4000 − 15) = 4.04 kJ/kg, h₂ = 225.9 + 4.04 = 230.0 kJ/kg
Turbine exit: isentropic, so s₄ = s₃. Find the quality, then the enthalpy.x₄ = (6.7712 − 0.7548) / 7.2523 = 0.830, h₄ = 225.9 + 0.830 × 2372.4 = 2194 kJ/kg
Turbine work and boiler heat.w_t = 3214.4 − 2194 = 1020 kJ/kg, q_in = 3214.4 − 230.0 = 2984 kJ/kg
Net work and efficiency.w_net = 1020 − 4 = 1016 kJ/kg, η = 1016 / 2984 = 0.341
Mass flow for 5 MW net.ṁ = 5000 / 1016 = 4.92 kg/s
Answer: The pump needs about 4 kJ/kg and the turbine delivers about 1020 kJ/kg, so the pump uses under 0.5% of the turbine output. The efficiency is about 34%, and 5 MW needs about 4.9 kg/s of steam.
Example: The same cycle with real turbine and pump
Given
The cycle in the first example, with the same states, now has a turbine isentropic efficiency of 85% and a pump isentropic efficiency of 80%.
Find
The actual turbine exit state.
The thermal efficiency and steam flow for 5 MW net.
Solution
Actual turbine work and exit enthalpy.w_t = 0.85 × 1020.4 = 867.3 kJ/kg, h₄a = 3214.4 − 867.3 = 2347 kJ/kg
Exit quality. It is higher than in the ideal case: the losses leave the steam drier.x₄a = (2347 − 225.9) / 2372.4 = 0.894
Actual pump work and boiler heat.w_p = 4.04 / 0.80 = 5.05 kJ/kg, h₂a = 225.9 + 5.05 = 231.0, q_in = 3214.4 − 231.0 = 2983 kJ/kg
Net work and efficiency.w_net = 867.3 − 5.05 = 862.3 kJ/kg, η = 862.3 / 2983 = 0.289
Mass flow for 5 MW net.ṁ = 5000 / 862.3 = 5.80 kg/s
Answer: Efficiency drops from 34.1% to about 28.9%, and the plant needs about 5.8 kg/s of steam instead of 4.9 kg/s, 18% more fuel for the same output. Nearly all of the loss is in the turbine; the pump barely matters.
Common mistakes
Reading the turbine exit from tables by temperature alone. In the wet region, temperature and pressure are not independent. Use s₄ = s₃ and the quality to find h₄.
Leaving pressure in MPa in the pump work. v·ΔP gives kJ/kg only with P in kPa.
Applying isentropic efficiency the wrong way round. Turbines: actual work = η × isentropic work. Pumps: actual work = isentropic work ÷ η.
Neglecting pump work in the boiler heat. q_in = h₃ − h₂, not h₃ − h₁. The difference is small but it changes the answer.
Comparing efficiency with Carnot at the wrong temperatures. Heat in a Rankine boiler is not added at one temperature, so a Carnot limit using the peak steam temperature will always look much higher.
In industry
Frequently asked questions
Why is a Rankine cycle less efficient than a Carnot cycle?
Heat is added over a range of temperatures, from cold feedwater up to superheated steam, so the average temperature of heat addition is well below the peak. A Carnot cycle adds all its heat at the peak temperature.
Why superheat the steam?
It raises the average temperature of heat addition, which improves efficiency, and it keeps the turbine exhaust drier. Water droplets erode turbine blades, so exit quality is usually kept above about 0.9.
What do reheat and regeneration do?
Reheat expands the steam part way, returns it to the boiler, then expands it again, which raises efficiency and dries the exhaust. Regeneration bleeds some steam to preheat the feedwater, which raises the average temperature of heat addition.
Where do the steam property values come from?
Steam tables and property software are based on the IAPWS-IF97 formulation. Textbook tables round the values slightly differently, so expect small differences in the last digit.