Thermodynamic Cycle Analyzer
Thermodynamic Cycle Analyzer computes key parameters of a Rankine cycle for a given working fluid, turbine inlet conditions, and condenser pressure.
Formulas Used in Thermodynamic Cycle Analyzer
The analyzer computes parameters for a simple Rankine cycle using the following equations:
Turbine Work (\\(W_t\\)):
\\[ W_t = h_1 – h_2 \\]Pump Work (\\(W_p\\)):
\\[ W_p = v_f \cdot (P_1 – P_4) \cdot 1000 \\]Heat Input (\\(Q_{in}\\)):
\\[ Q_{in} = h_1 – h_4 \\]Thermal Efficiency (\\(\eta\\)):
\\[ \eta = \frac{W_t – W_p}{Q_{in}} \cdot 100 \\]Where:
- \\(h_1, h_2, h_4\\): Enthalpies at cycle states (kJ/kg)
- \\(v_f\\): Specific volume of saturated liquid (m³/kg)
- \\(P_1, P_4\\): Pressures (MPa, kPa)
- \\(W_t, W_p\\): Turbine and pump work (kJ/kg)
- \\(Q_{in}\\): Heat input (kJ/kg)
- \\(\eta\\): Thermal efficiency (%)
Example Calculation
Example: Water, T = 500 °C, P₁ = 10 MPa, P₄ = 10 kPa
\\[
h_1 \approx 3373.7 \, \text{kJ/kg}, \, h_2 \approx 2115.3 \, \text{kJ/kg}, \, h_4 \approx 191.8 \, \text{kJ/kg}
\\]
\\[
W_t = 3373.7 – 2115.3 = 1258.4 \, \text{kJ/kg}
\\]
\\[
W_p = 0.00101 \cdot (10 – 0.01) \cdot 1000 \approx 10.1 \, \text{kJ/kg}
\\]
\\[
Q_{in} = 3373.7 – 191.8 = 3181.9 \, \text{kJ/kg}
\\]
\\[
\eta = \frac{1258.4 – 10.1}{3181.9} \cdot 100 \approx 39.2 \, \%
\\]
Result: \\(W_t \approx 1258.4 \, \text{kJ/kg}, W_p \approx 10.1 \, \text{kJ/kg}, Q_{in} \approx 3181.9 \, \text{kJ/kg}, \eta \approx 39.2 \, \%\\).