Rankine Cycle
The Thermodynamic Cycle Behind Steam Power Plants — Components, TS Diagram, Formula & Solved Problems
Last Updated: March 2026
📌 Key Takeaways
- The Rankine cycle is the standard thermodynamic cycle for steam power plants — it describes how thermal, nuclear, geothermal, and solar-thermal plants generate electricity.
- Four components: Pump → Boiler → Turbine → Condenser.
- Four processes: Isentropic compression (pump) → Constant-pressure heat addition (boiler) → Isentropic expansion (turbine) → Constant-pressure heat rejection (condenser).
- Efficiency: η = (Wturbine − Wpump) / Qboiler, calculated using enthalpy values at each state point.
- Efficiency is improved by superheating, reheating, and regeneration.
- Rankine cycle efficiency is always less than Carnot efficiency for the same temperature limits, because heat addition occurs over a range of temperatures rather than at a single TH.
1. What is the Rankine Cycle?
The Rankine cycle is the thermodynamic cycle that models the operation of steam power plants. It describes how water is pumped, heated into steam, expanded through a turbine to produce work, and then condensed back into water to repeat the cycle. Virtually every coal-fired, nuclear, geothermal, biomass, and concentrated solar power plant in the world operates on some version of the Rankine cycle.
Named after Scottish engineer William John Macquorn Rankine, this cycle is the practical engineering counterpart of the Carnot cycle. While the Carnot cycle is the theoretical ideal, the Rankine cycle accounts for the fact that real power plants use water/steam as the working fluid, which undergoes phase changes (liquid to vapour and back) during the cycle.
The Rankine cycle is one of the most heavily tested topics in GATE ME — questions typically ask you to calculate efficiency, turbine work, or the effect of modifications like superheating and reheat.
2. The Four Components
| Component | Function | Energy Interaction | Process Type |
|---|---|---|---|
| Pump | Compresses liquid water from condenser pressure to boiler pressure | Work input (Wpump) | Isentropic compression (ideally) |
| Boiler | Heats compressed water into superheated steam using fuel combustion or nuclear reaction | Heat input (QH) | Constant-pressure heat addition |
| Turbine | Expands high-pressure steam to produce shaft work (drives the generator) | Work output (Wturbine) | Isentropic expansion (ideally) |
| Condenser | Condenses exhaust steam back into liquid water by rejecting heat to cooling water or air | Heat rejected (QL) | Constant-pressure heat rejection |
The cycle is a closed loop — the same water circulates repeatedly through all four components. The pump, boiler, turbine, and condenser are connected in series, and each serves a specific thermodynamic function.
3. The Four Processes — Step by Step
Process 3→4: Isentropic Compression (Pump)
Liquid water from the condenser (state 3, low pressure) is compressed by the pump to boiler pressure (state 4). Since water is nearly incompressible, the pump work is small compared to the turbine work. Ideally, this process is isentropic (s&sub3; = s&sub4;).
Wpump = h&sub4; − h&sub3; ≈ v&sub3;(P&sub4; − P&sub3;)
Where v&sub3; is the specific volume of liquid water at the condenser exit.
Process 4→1: Constant-Pressure Heat Addition (Boiler)
High-pressure liquid water enters the boiler and is heated at constant pressure. Three stages occur: the liquid is preheated to saturation temperature (subcooled → saturated liquid), then evaporated (saturated liquid → saturated vapour), then superheated beyond the saturation temperature (saturated vapour → superheated steam).
Qboiler = h&sub1; − h&sub4;
Where h&sub1; is the enthalpy of superheated steam leaving the boiler and h&sub4; is the enthalpy of compressed liquid entering the boiler.
Process 1→2: Isentropic Expansion (Turbine)
Superheated steam at high pressure and temperature enters the turbine and expands, producing shaft work that drives the electrical generator. The steam exits the turbine at condenser pressure as a wet mixture (liquid + vapour) or low-pressure steam. Ideally, this process is isentropic (s&sub1; = s&sub2;).
Wturbine = h&sub1; − h&sub2;
This is the gross work output of the cycle.
Process 2→3: Constant-Pressure Heat Rejection (Condenser)
Wet steam from the turbine enters the condenser, where it is cooled and condensed back into saturated liquid by rejecting heat to a cooling medium (river water, cooling tower). Pressure and temperature remain constant during condensation.
Qcondenser = h&sub2; − h&sub3;
This is the heat rejected to the environment — the “waste” energy that cannot be converted to work.
4. TS Diagram
On a Temperature-entropy (TS) diagram, the ideal Rankine cycle appears as follows:
- 3→4 (Pump): A nearly vertical line moving upward from the saturated liquid line — entropy is approximately constant, and the temperature rises very slightly due to compression.
- 4→1 (Boiler): A line moving rightward at constant pressure — first along the saturated liquid line (preheating), then horizontally through the two-phase dome (evaporation at constant T), then curving upward and rightward into the superheated region (superheating).
- 1→2 (Turbine): A vertical line dropping downward — entropy is constant (isentropic), temperature decreases. The line may end inside the two-phase dome (wet steam at turbine exit).
- 2→3 (Condenser): A horizontal line moving leftward at constant temperature and pressure — the wet steam is condensed into saturated liquid.
The area enclosed by the cycle on the TS diagram represents the net work output. The area under the top curve (4→1) represents Qboiler, and the area under the bottom line (2→3) represents Qcondenser.
Exam tip: When asked to sketch a Rankine cycle on a TS diagram, always show the vapour dome (saturation curve) and mark all four state points clearly. Indicate the direction of processes with arrows.
5. Efficiency Formula & Derivation
Rankine Cycle Efficiency
η = Wnet / Qboiler = (Wturbine − Wpump) / Qboiler
η = [(h&sub1; − h&sub2;) − (h&sub4; − h&sub3;)] / (h&sub1; − h&sub4;)
Since pump work is usually very small (1–3% of turbine work):
η ≈ (h&sub1; − h&sub2;) / (h&sub1; − h&sub3;) (approximate, neglecting pump work)
To calculate efficiency, you need enthalpy values at all four state points. These are obtained from steam tables or the Mollier diagram (h-s diagram) using the known pressures, temperatures, and the isentropic condition (s&sub1; = s&sub2; for ideal turbine, s&sub3; = s&sub4; for ideal pump).
| State Point | How to Find Enthalpy |
|---|---|
| State 1 (turbine inlet) | Superheated steam table at boiler pressure and superheat temperature → h&sub1;, s&sub1; |
| State 2 (turbine exit) | Use s&sub2; = s&sub1; (isentropic). At condenser pressure, find quality x = (s&sub2; − sf) / sfg, then h&sub2; = hf + x·hfg |
| State 3 (condenser exit) | Saturated liquid at condenser pressure → h&sub3; = hf |
| State 4 (pump exit) | h&sub4; = h&sub3; + v&sub3;(P&sub4; − P&sub3;), or h&sub4; ≈ h&sub3; when pump work is neglected |
6. Improving Rankine Efficiency
The basic Rankine cycle has lower efficiency than the Carnot cycle because heat is added over a range of temperatures (not at a single TH). Several modifications improve performance:
Superheating
Heating steam beyond the saturation temperature increases the average temperature of heat addition, improving efficiency. It also reduces the moisture content at the turbine exit, protecting turbine blades from erosion. Modern plants superheat to 500–600°C.
Reheating
After partial expansion in a high-pressure turbine, steam is sent back to the boiler to be reheated to the original superheat temperature, then expanded again in a low-pressure turbine. This increases net work output and keeps the steam dryer at the turbine exit. Efficiency gain: 3–5%.
Regeneration (Feedwater Heating)
Steam is extracted from intermediate stages of the turbine and used to preheat the feedwater before it enters the boiler. This raises the average temperature of heat addition (since less heat is needed to warm cold water in the boiler), improving efficiency. Two types exist: open feedwater heaters (direct mixing) and closed feedwater heaters (heat exchangers). Efficiency gain: 2–4% per stage.
Increasing Boiler Pressure
Higher boiler pressure raises the average temperature of heat addition, increasing efficiency. Modern supercritical plants operate at pressures above 22.1 MPa (critical pressure of water), where the liquid-vapour distinction disappears.
Lowering Condenser Pressure
Lower condenser pressure (deeper vacuum) means the turbine can expand to a lower back pressure, increasing net work. However, this is limited by the ambient cooling water temperature.
| Modification | Effect on Efficiency | Typical Improvement |
|---|---|---|
| Superheating | Increases avg. temperature of heat addition | +2–4% |
| Reheat | More work from same heat input, drier exhaust | +3–5% |
| Regeneration (per stage) | Raises feedwater temperature before boiler | +2–4% |
| Higher boiler pressure | Higher Tavg during heat addition | +1–3% |
| Lower condenser pressure | More expansion, more work output | +1–2% |
7. Worked Numerical Examples
Example 1: Basic Rankine Cycle
Problem: An ideal Rankine cycle operates with a boiler pressure of 10 MPa, condenser pressure of 10 kPa, and turbine inlet temperature of 500°C. Using the following steam table data, calculate the cycle efficiency.
Given data: h&sub1; = 3,374 kJ/kg, s&sub1; = 6.597 kJ/kg·K, h&sub3; = hf at 10 kPa = 191.8 kJ/kg, sf at 10 kPa = 0.649 kJ/kg·K, sfg at 10 kPa = 7.502 kJ/kg·K, hfg at 10 kPa = 2,392.8 kJ/kg, v&sub3; = 0.00101 m³/kg.
Solution
Step 1 — Find h&sub2; (turbine exit):
s&sub2; = s&sub1; = 6.597 kJ/kg·K (isentropic expansion)
Quality: x = (s&sub2; − sf) / sfg = (6.597 − 0.649) / 7.502 = 0.793
h&sub2; = hf + x·hfg = 191.8 + 0.793 × 2392.8 = 191.8 + 1897.5 = 2,089.3 kJ/kg
Step 2 — Find h&sub4; (pump exit):
Wpump = v&sub3;(P&sub4; − P&sub3;) = 0.00101 × (10,000 − 10) = 10.09 kJ/kg
h&sub4; = h&sub3; + Wpump = 191.8 + 10.09 = 201.9 kJ/kg
Step 3 — Calculate efficiency:
Wturbine = h&sub1; − h&sub2; = 3374 − 2089.3 = 1,284.7 kJ/kg
Wnet = Wturbine − Wpump = 1284.7 − 10.09 = 1,274.6 kJ/kg
Qboiler = h&sub1; − h&sub4; = 3374 − 201.9 = 3,172.1 kJ/kg
η = 1274.6 / 3172.1 = 0.4018 = 40.2%
Example 2: Effect of Superheating
Problem: If the turbine inlet temperature in Example 1 is increased from 500°C to 600°C (h&sub1; becomes 3,625 kJ/kg, s&sub1; becomes 6.903 kJ/kg·K), what is the new efficiency?
Solution
x&sub2; = (6.903 − 0.649) / 7.502 = 0.834
h&sub2; = 191.8 + 0.834 × 2392.8 = 191.8 + 1995.6 = 2,187.4 kJ/kg
Wturbine = 3625 − 2187.4 = 1,437.6 kJ/kg
Wnet = 1437.6 − 10.09 = 1,427.5 kJ/kg
Qboiler = 3625 − 201.9 = 3,423.1 kJ/kg
η = 1427.5 / 3423.1 = 0.4170 = 41.7%
Superheating by 100°C improved efficiency from 40.2% to 41.7% — a gain of 1.5 percentage points.
8. Rankine vs Carnot — Why the Gap?
For the same temperature limits, the Rankine cycle is always less efficient than the Carnot cycle. The fundamental reason is the shape of the heat addition process.
| Feature | Carnot Cycle | Rankine Cycle |
|---|---|---|
| Heat addition | Entirely at TH | Over a range of temperatures (preheating + evaporation + superheating) |
| Average T of heat addition | TH | Lower than TH |
| Working fluid | Any (theoretical) | Water/steam (undergoes phase change) |
| Practicality | Impossible to build | Standard for real power plants |
| Efficiency | Maximum (benchmark) | Lower, but improvable with modifications |
Reheat and regeneration aim to close this gap by raising the average temperature at which heat is added, making the Rankine cycle’s temperature profile closer to the Carnot ideal.
9. Common Mistakes Students Make
- Forgetting to calculate pump work: While pump work is small, GATE problems often test whether you include it. Neglecting it without stating the assumption can cost marks.
- Using Carnot formula for Rankine efficiency: η = 1 − TL/TH does NOT apply to the Rankine cycle. You must use enthalpy values from steam tables.
- Not checking dryness fraction at turbine exit: If x&sub2; < 0.88 (moisture > 12%), turbine blade erosion becomes a serious concern. This is a common design check in exam problems.
- Confusing state points: Different textbooks number the state points differently. Always draw the cycle and label each state clearly before solving. This guide uses: 1 = turbine inlet, 2 = turbine exit, 3 = condenser exit (pump inlet), 4 = pump exit (boiler inlet).
- Mixing up enthalpy values from steam tables: Be careful to read the correct row (pressure or temperature) and column (hf, hfg, hg) from the tables. hf is saturated liquid, hg is saturated vapour, hfg = hg − hf.
10. Frequently Asked Questions
What is the Rankine cycle?
The Rankine cycle is the thermodynamic cycle that describes steam power plant operation. Water is pumped to high pressure, heated into steam in a boiler, expanded through a turbine to produce work, and condensed back into water in a condenser. This cycle is used in coal, nuclear, geothermal, and solar-thermal power plants worldwide.
What is the efficiency formula for the Rankine cycle?
η = (Wturbine − Wpump) / Qboiler = [(h&sub1; − h&sub2;) − (h&sub4; − h&sub3;)] / (h&sub1; − h&sub4;). Enthalpy values are obtained from steam tables using the known operating pressures and the isentropic assumption for ideal pump and turbine. Typical ideal Rankine cycle efficiencies range from 30% to 45% depending on operating conditions.
Why is the Rankine cycle less efficient than the Carnot cycle?
In the Carnot cycle, all heat is added at the maximum temperature TH. In the Rankine cycle, heat addition starts at a lower temperature (subcooled liquid), passes through evaporation (constant temperature), and continues through superheating. This means the average temperature of heat addition is lower than TH, reducing efficiency. Superheating, reheating, and regeneration are techniques used to raise this average temperature and close the gap.