š Steam Rankine Cycle Calculator
Analyze steam power plant cycles ā simple and reheat configurations with efficiency and power output calculations.
š Configuration
Rankine Cycle States:
1ā2: Pump (isentropic compression)
2ā3: Boiler (constant P heat addition)
3ā4: Turbine (isentropic expansion)
4ā1: Condenser (constant P heat rejection)
Ī·_th = W_net / Q_in
BWR = W_pump / W_turbine
1ā2: Pump (isentropic compression)
2ā3: Boiler (constant P heat addition)
3ā4: Turbine (isentropic expansion)
4ā1: Condenser (constant P heat rejection)
Ī·_th = W_net / Q_in
BWR = W_pump / W_turbine
š Results & Visualization
Configure the inputs and click Calculate to see results.
ā¹ļø 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:
ā¢ Boiler: heats water to superheated steam
ā¢ Turbine: expands steam to produce work
ā¢ Condenser: rejects heat, condenses steam
ā¢ 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:
ā¢ Boiler: heats water to superheated steam
ā¢ Turbine: expands steam to produce work
ā¢ Condenser: rejects heat, condenses steam
ā¢ Pump: pressurizes liquid water
Reheat improves efficiency by reheating partially expanded steam.
š Calculation Methodology
Mathematical Model & Theory
The Rankine cycle is the thermodynamic cycle for steam power plants. The cycle thermal efficiency ($\eta_{th}$) is determined by turbine work, pump work, and heat input:
$$\eta_{th} = \frac{W_{net}}{Q_{in}} = \frac{W_{turb} - W_{pump}}{Q_{in}}$$
$$W_{turb} = \eta_t (h_3 - h_{4s}), \quad W_{pump} = \frac{v_1 (P_2 - P_1)}{\eta_p}$$
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 500Ā°C. 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 500Ā°C. 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%.