About Hydroelectric Power Calculator
The hydroelectric power calculator returns the power a hydro scheme can generate from a given flow of water falling through a head. It applies the fundamental hydropower equation P = rho g Q H eta, where the available water (hydraulic) power equals the density of water times gravity times the volumetric flow rate times the net head, and the electrical output is that hydraulic power multiplied by the overall turbine-and-generator efficiency. It distinguishes gross head from net head by subtracting penstock and friction losses, since only the net head does useful work.
Enter the flow rate in cubic metres per second, the gross (static) head in metres, the penstock head loss, and the overall efficiency (typically about 0.85 for a modern small-hydro plant). The tool reports the net head, the hydraulic power, the electrical power output in kilowatts, and, with a capacity factor, the annual energy in megawatt-hours. A chart shows how output rises with net head so you can compare sites or assess the cost of penstock losses.
How It Works
- Enter flow rate Q (m^3/s), gross head (m), penstock head loss (m), and the overall turbine-plus-generator efficiency (0-1, typically ~0.85).
- The calculator finds the net head H_net = gross head - head loss; only the net head produces power.
- Hydraulic power P_hyd = rho g Q H_net with rho = 1000 kg/m^3 and g = 9.81 m/s^2; electrical output = P_hyd x efficiency.
- Optionally add an annual capacity factor to estimate annual energy = electrical power (kW) x 8760 hours x capacity factor.
Worked Example
A run-of-river scheme passes Q = 10 m^3/s through a gross head of 20 m with negligible penstock loss, so the net head is 20 m. The hydraulic power is rho g Q H = 1000 x 9.81 x 10 x 20 = 1,962,000 W = 1.962 MW. With an overall efficiency of 0.85 the electrical output is 1.962 x 0.85 = 1.668 MW (1668 kW). Running at a 50% capacity factor the annual energy is 1668 x 8760 x 0.5 = 7.30 million kWh, about 7305 MWh per year.
Formulas
- Net head
H_net = H_gross - H_loss- Hydraulic (water) power
P_hyd = rho * g * Q * H_net- Electrical power output
P_el = P_hyd * eta- Annual energy
E = P_el(kW) * 8760 * CF
Standards & References
- Fundamental hydropower equation P = rho g Q H eta
- ESHA — Guide on How to Develop a Small Hydropower Plant
- IEC 60041 (field acceptance tests of hydraulic turbines)
- IEC 62006 (hydraulic machines acceptance tests for small hydropower)
Frequently Asked Questions
What is the difference between gross head and net head?
Gross head is the total vertical drop between the intake water level and the turbine. Net head subtracts penstock friction and fitting losses; only the net head produces power, so P = rho g Q H_net. Higher flow or longer penstocks increase the loss and reduce output.
What overall efficiency should I use?
Overall efficiency is the product of the turbine and generator (and transformer) efficiencies. Modern plants reach about 0.85-0.92 at the design point; small or part-load schemes may be 0.70-0.85. Use the manufacturer hill chart value at your operating flow when available.
Why is the power proportional to flow and head but not their squares?
Hydraulic power is the rate of change of potential energy: mass flow rate (rho Q) times g times head H. Both flow and head enter linearly, so doubling either doubles the power, unlike wind power which scales with the cube of speed.
How does the capacity factor affect annual energy?
The capacity factor is the ratio of actual annual energy to the energy if the plant ran at full power all 8760 hours. Run-of-river schemes often see 0.4-0.6 because flow varies seasonally; multiplying rated power by the capacity factor and 8760 gives a realistic annual yield.