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Dam-Based Hydropower Calculator

Dam-Based Hydropower Calculator picks turbine, power, and energy with losses from head, flow, efficiencies, penstock, and storage for dam energy planning.

Formulas Used in Dam-Based Hydropower Calculator

The calculator uses the following formulas and logic to select a turbine and estimate power and energy:

Flow Velocity:

\\[ v = \frac{Q}{\pi (D/2)^2} \\]

Head Loss:

\\[ h_{\text{loss}} = \left( f \cdot \frac{L}{D} \cdot \frac{v^2}{2g} \right) + \left( K \cdot \frac{v^2}{2g} \right) \\]

Net Head:

\\[ h_{\text{net}} = h_{\text{gross}} – h_{\text{loss}} \\]

Power Output:

\\[ P = \eta_{\text{turbine}} \cdot \eta_{\text{generator}} \cdot \rho \cdot g \cdot h_{\text{net}} \cdot Q \\]

Power in Kilowatts:

\\[ P_{\text{kW}} = \frac{P}{1000} \\]

Energy Output:

\\[ E = P \cdot t \cdot \frac{V_{\text{storage}}}{Q \cdot t_{\text{max}}}, \quad t_{\text{max}} = \frac{V_{\text{storage}}{Q} \\]

Turbine Selection:

  • Kaplan: Head 2–20 m, Flow > 1 m³/s
  • Francis: Head 10–100 m, Flow 0.5–50 m³/s
  • Pelton: Head > 100 m, Flow 0.01–10 m³/s
  • Crossflow: Head 2–40 m, Flow < 1 m³/s

Where:

  • \\( v \\): Flow velocity (m/s)
  • \\( Q \\): Flow rate (m³/s)
  • \\( D \\): Penstock diameter (m)
  • \\( h_{\text{loss}} \\): Head loss (m)
  • \\( f \\): Friction factor
  • \\( L \\): Penstock length (m)
  • \\( K \\): Minor loss coefficient
  • \\( g \\): Gravitational acceleration (9.81 m/s²)
  • \\( h_{\text{net}} \\): Net head (m)
  • \\( h_{\text{gross}} \\): Gross head (m)
  • \\( P \\): Power output (W)
  • \\( \eta_{\text{turbine}} \\): Turbine efficiency
  • \\( \eta_{\text{generator}} \\): Generator efficiency
  • \\( \rho \\): Water density (kg/m³)
  • \\( E \\): Energy output (kWh)
  • \\( t \\): Operating time (hours)
  • \\( V_{\text{storage}} \\): Reservoir storage volume (m³)
  • \\( t_{\text{max}} \\): Maximum depletion time (s)

Example Calculations

Example 1: Medium Dam

Input: Head = 50 m, Flow = 10 m³/s, Turbine Eff. = 0.85, Generator Eff. = 0.90, Density = 1000 kg/m³, Penstock Length = 500 m, Diameter = 2 m, Friction = 0.02, Minor Loss = 0.5, Storage = 1,000,000 m³, Time = 24 h

\\[ v = \frac{10}{\pi (2/2)^2} \approx 3.18 \ \text{m/s} \\] \\[ h_{\text{loss}} = \left( 0.02 \cdot \frac{500}{2} \cdot \frac{3.18^2}{2 \cdot 9.81} \right) + \left( 0.5 \cdot \frac{3.18^2}{2 \cdot 9.81} \right) \approx 3.09 \ \text{m} \\] \\[ h_{\text{net}} = 50 – 3.09 = 46.91 \ \text{m} \\] \\[ P = 0.85 \cdot 0.90 \cdot 1000 \cdot 9.81 \cdot 46.91 \cdot 10 \approx 3520750.5 \ \text{W} \\] \\[ P_{\text{kW}} = \frac{3520750.5}{1000} \approx 3520.75 \ \text{kW} \\] \\[ t_{\text{max}} = \frac{1000000}{10} = 100000 \ \text{s} \approx 27.78 \ \text{h} \\] \\[ E = 3520.75 \cdot 24 \cdot \frac{1000000}{10 \cdot 100000} \approx 84498.0 \ \text{kWh} \\]

Turbine: Francis (Head 10–100 m, Flow 0.5–50 m³/s)

Result: Turbine = Francis, Power = 3520.75 kW, Energy = 84498.0 kWh, Net Head = 46.91 m

Example 2: Small Dam

Input: Head = 10 m, Flow = 2 m³/s, Turbine Eff. = 0.80, Generator Eff. = 0.85, Density = 1000 kg/m³, Penstock Length = 200 m, Diameter = 1 m, Friction = 0.03, Minor Loss = 0.3, Storage = 200,000 m³, Time = 12 h

\\[ v = \frac{2}{\pi (1/2)^2} \approx 2.55 \ \text{m/s} \\] \\[ h_{\text{loss}} = \left( 0.03 \cdot \frac{200}{1} \cdot \frac{2.55^2}{2 \cdot 9.81} \right) + \left( 0.3 \cdot \frac{2.55^2}{2 \cdot 9.81} \right) \approx 2.29 \ \text{m} \\] \\[ h_{\text{net}} = 10 – 2.29 = 7.71 \ \text{m} \\] \\[ P = 0.80 \cdot 0.85 \cdot 1000 \cdot 9.81 \cdot 7.71 \cdot 2 \approx 102915.2 \ \text{W} \\] \\[ P_{\text{kW}} = \frac{102915.2}{1000} \approx 102.92 \ \text{kW} \\] \\[ t_{\text{max}} = \frac{200000}{2} = 100000 \ \text{s} \approx 27.78 \ \text{h} \\] \\[ E = 102.92 \cdot 12 \cdot \frac{200000}{2 \cdot 100000} \approx 1235.0 \ \text{kWh} \\]

Turbine: Kaplan (Head 2–20 m, Flow > 1 m³/s)

Result: Turbine = Kaplan, Power = 102.92 kW, Energy = 1235.0 kWh, Net Head = 7.71 m

Example 3: High-Head Dam

Input: Head = 150 m, Flow = 5 m³/s, Turbine Eff. = 0.90, Generator Eff. = 0.95, Density = 1000 kg/m³, Penstock Length = 1000 m, Diameter = 1.5 m, Friction = 0.015, Minor Loss = 0.7, Storage = 5,000,000 m³, Time = 48 h

\\[ v = \frac{5}{\pi (1.5/2)^2} \approx 2.83 \ \text{m/s} \\] \\[ h_{\text{loss}} = \left( 0.015 \cdot \frac{1000}{1.5} \cdot \frac{2.83^2}{2 \cdot 9.81} \right) + \left( 0.7 \cdot \frac{2.83^2}{2 \cdot 9.81} \right) \approx 4.79 \ \text{m} \\] \\[ h_{\text{net}} = 150 – 4.79 = 145.21 \ \text{m} \\] \\[ P = 0.90 \cdot 0.95 \cdot 1000 \cdot 9.81 \cdot 145.21 \cdot 5 \approx 6091467.8 \ \text{W} \\] \\[ P_{\text{kW}} = \frac{6091467.8}{1000} \approx 6091.47 \ \text{kW} \\] \\[ t_{\text{max}} = \frac{5000000}{5} = 1000000 \ \text{s} \approx 277.78 \ \text{h} \\] \\[ E = 6091.47 \cdot 48 \cdot \frac{5000000}{5 \cdot 1000000} \approx 292390.6 \ \text{kWh} \\]

Turbine: Pelton (Head > 100 m, Flow 0.01–10 m³/s)

Result: Turbine = Pelton, Power = 6091.47 kW, Energy = 292390.6 kWh, Net Head = 145.21 m

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