๐ Rankine Cycle
Steam power cycle analysis with optional reheat and regeneration. Turbine work, pump work, thermal efficiency, and T-s diagram.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
Thermodynamics
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๐ฆ Fortran Code
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Released
Feb 2024
โฑ๏ธ Latency
< 1 ms
Vapor Power Loop Schematic
The Rankine thermodynamic cycle represents the ideal working model of a steam turbine power plant. Reheating steam between turbine stages prevents excessive moisture accumulation, protecting the turbine blades from erosion.
โข Simple Cycle: Standard 4-component loop (Boiler, Turbine, Condenser, Pump).
โข With Reheat: Expands steam in HP stage, returns to boiler to reheat, expands in LP stage.
โข With Reheat: Expands steam in HP stage, returns to boiler to reheat, expands in LP stage.
๐ Configuration
Rankine Cycle States:
1โ2: Pump (isentropic compression)
2โ3: Boiler (constant P heat addition)
3โ4: Turbine (isentropic expansion)
4โ1: Condenser (constant P heat rejection)
ฮท_th = W_net / Q_in
BWR = W_pump / W_turbine
1โ2: Pump (isentropic compression)
2โ3: Boiler (constant P heat addition)
3โ4: Turbine (isentropic expansion)
4โ1: Condenser (constant P heat rejection)
ฮท_th = W_net / Q_in
BWR = W_pump / W_turbine
๐ Results & Visualization
Configure the inputs and click Calculate to see results.
โน๏ธ About the Rankine Cycle
The Rankine cycle is the fundamental thermodynamic cycle for steam power plants. It converts heat into work using water as the working fluid.
Components:
โข Boiler: heats water to superheated steam
โข Turbine: expands steam to produce work
โข Condenser: rejects heat, condenses steam
โข Pump: pressurizes liquid water
Reheat improves efficiency by reheating partially expanded steam.
The Rankine cycle is the fundamental thermodynamic cycle for steam power plants. It converts heat into work using water as the working fluid.
Components:
โข Boiler: heats water to superheated steam
โข Turbine: expands steam to produce work
โข Condenser: rejects heat, condenses steam
โข Pump: pressurizes liquid water
Reheat improves efficiency by reheating partially expanded steam.
๐ Calculation Methodology
Mathematical Model & Theory
The Rankine cycle thermal efficiency ($\eta_{th}$) is determined by turbine work, pump work, and heat input rate:
$$\eta_{th} = \frac{W_{net}}{Q_{in}} = \frac{W_{turb} - W_{pump}}{Q_{in}}$$
$$W_{turb} = \eta_t (h_3 - h_{4s}), \quad W_{pump} = \frac{v_1 (P_2 - P_1)}{\eta_p}$$
Assumptions & Idealizations
- Isentropic expansion in turbines with efficiency $\eta_t$
- Isentropic compression in water pumps with efficiency $\eta_p$
- Negligible friction and pressure losses in piping loops
- Constant boiler and condenser pressures
References & Literature
- Cengel, Y. A., & Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill.
- Moran, M. J., Shapiro, H. N., Boettner, D. D., & Bailey, M. B., Fundamentals of Engineering Thermodynamics, Wiley.
Worked Engineering Example
Problem Statement:
An ideal Rankine cycle operates between boiler pressure 8 MPa and condenser pressure 10 kPa. Steam leaves the boiler superheated at 500ยฐC. Calculate thermal efficiency. (Enthalpy values: $h_1 = 191.8$, $h_2 = 199.9$, $h_3 = 3399.5$, $h_4 = 2100$ kJ/kg).
Step-by-step Solution:
1. Calculate net work:
$$W_{turb} = h_3 - h_4 = 3399.5 - 2100 = 1299.5 \text{ kJ/kg}$$ $$W_{pump} = h_2 - h_1 = 199.9 - 191.8 = 8.1 \text{ kJ/kg}$$ $$W_{net} = W_{turb} - W_{pump} = 1291.4 \text{ kJ/kg}$$ 2. Calculate heat input:
$$Q_{in} = h_3 - h_2 = 3399.5 - 199.9 = 3199.6 \text{ kJ/kg}$$ 3. Calculate efficiency:
$$\eta_{th} = \frac{1291.4}{3199.6} = 0.4036 \quad (40.36\%)$$
Final Result:
Thermal efficiency is 40.36%.
An ideal Rankine cycle operates between boiler pressure 8 MPa and condenser pressure 10 kPa. Steam leaves the boiler superheated at 500ยฐC. Calculate thermal efficiency. (Enthalpy values: $h_1 = 191.8$, $h_2 = 199.9$, $h_3 = 3399.5$, $h_4 = 2100$ kJ/kg).
Step-by-step Solution:
1. Calculate net work:
$$W_{turb} = h_3 - h_4 = 3399.5 - 2100 = 1299.5 \text{ kJ/kg}$$ $$W_{pump} = h_2 - h_1 = 199.9 - 191.8 = 8.1 \text{ kJ/kg}$$ $$W_{net} = W_{turb} - W_{pump} = 1291.4 \text{ kJ/kg}$$ 2. Calculate heat input:
$$Q_{in} = h_3 - h_2 = 3399.5 - 199.9 = 3199.6 \text{ kJ/kg}$$ 3. Calculate efficiency:
$$\eta_{th} = \frac{1291.4}{3199.6} = 0.4036 \quad (40.36\%)$$
Final Result:
Thermal efficiency is 40.36%.