Hydropower Energy Yield Calculator
Hydropower Energy Yield Calculator estimates the annual energy yield of a hydropower plant using flow duration curve data, gross head, head loss, and turbine efficiency, with a detailed breakdown and interactive chart.
Formulas Used in Hydropower Energy Yield Calculator
The calculator uses the following formulas to estimate annual energy yield:
Net Head:
\\[ h_{\text{net}} = h_{\text{gross}} – h_{\text{loss}} \\]Power Output for Flow \\( Q_i \\):
\\[ P_i = \eta \cdot \rho \cdot g \cdot Q_i \cdot h_{\text{net}} \\]Energy Contribution for Flow \\( Q_i \\):
\\[ E_i = \frac{P_i}{1000} \cdot 8760 \cdot \frac{\Delta P_i}{100} \\]Total Annual Energy Yield:
\\[ E_{\text{total}} = \sum E_i \\]Where:
- \\( h_{\text{net}} \\): Net head (m)
- \\( h_{\text{gross}} \\): Gross head (m)
- \\( h_{\text{loss}} \\): Head loss (m)
- \\( P_i \\): Power for flow \\( Q_i \\) (W)
- \\( \eta \\): Turbine efficiency (0 to 1)
- \\( \rho \\): Water density (1000 kg/m³)
- \\( g \\): Gravitational acceleration (9.81 m/s²)
- \\( Q_i \\): Flow rate (m³/s)
- \\( E_i \\): Energy for flow \\( Q_i \\) (kWh)
- \\( \Delta P_i \\): Time fraction (%) for flow \\( Q_i \\)
- \\( E_{\text{total}} \\): Total energy yield (kWh/year)
Example Calculations
Example 1: Small-Scale Site (Full FDC)
Input: FDC = [(1.5,7.69), (1.3,15.38), (0.8,50.00), (0.2,92.31)], Gross Head = 10 m, Head Loss = 0.5 m, Efficiency = 0.8
For \\( Q_1 = 1.5 \\), \\( \Delta P_1 = 7.69\% \\):
\\[ P_1 = 0.8 \cdot 1000 \cdot 9.81 \cdot 1.5 \cdot 9.5 \approx 111,834 \ \text{W} \\] \\[ E_1 = \frac{111834}{1000} \cdot 8760 \cdot \frac{7.69}{100} \approx 75,316 \ \text{kWh} \\]Similarly for other segments…
\\[ E_{\text{total}} \approx 75,316 + 149,112 + 279,072 + 17,208 \approx 520,708 \ \text{kWh/year} \\]Result: Net Head = 9.5 m, Total Energy Yield = 520,708 kWh/year
Example 2: Medium-Scale Site (Percentiles)
Input: Q0 = 3.5, Q10 = 3.1, Q50 = 1.9, Q90 = 0.6, Q100 = 0.5, Gross Head = 50 m, Head Loss = 3.54 m, Efficiency = 0.85
Segment 0–10% (\\( Q = 3.3 \\), average of 3.5, 3.1):
\\[ P = 0.85 \cdot 1000 \cdot 9.81 \cdot 3.3 \cdot 46.46 \approx 1,279,141 \ \text{W} \\] \\[ E = \frac{1279141}{1000} \cdot 8760 \cdot \frac{10}{100} \approx 1,120,528 \ \text{kWh} \\]Similarly for 10–50%, 50–90%, 90–100%…
\\[ E_{\text{total}} \approx 1,120,528 + 3,848,640 + 1,058,112 + 98,496 \approx 6,125,776 \ \text{kWh/year} \\]Result: Net Head = 46.46 m, Total Energy Yield = 6,125,776 kWh/year
Example 3: Large-Scale Site (Full FDC)
Input: FDC = 30 daily flows (e.g., 25.0 to 5.0 m³/s), Gross Head = 100 m, Head Loss = 3.3 m, Efficiency = 0.9
For \\( Q_1 = 25.0 \\), \\( \Delta P_1 = 3.23\% \\):
\\[ P_1 = 0.9 \cdot 1000 \cdot 9.81 \cdot 25.0 \cdot 96.7 \approx 2,135,708 \ \text{W} \\] \\[ E_1 = \frac{2135708}{1000} \cdot 8760 \cdot \frac{3.23}{100} \approx 604,404 \ \text{kWh} \\]Sum over 30 segments…
\\[ E_{\text{total}} \approx 37,589,664 \ \text{kWh/year} \\]Result: Net Head = 96.7 m, Total Energy Yield = 37,589,664 kWh/year