Hydropower calculator
Estimate the installed capacity and yearly energy of a hydropower scheme from its flow, head and efficiency — using the standard P = ρgQHη relation.
Built around Nepal's run-of-river context, with a capacity-factor input and an optional PPA tariff to gauge annual energy revenue. A planning-level tool, computed in your browser.
Scheme parameters
Discharge through the turbines.
Vertical drop from intake to turbine.
Friction in headrace & penstock — typically 3–10%.
Turbine × generator × transformer — usually 80–90%.
Share of full output over a year. Nepal run-of-river ≈ 50–60%.
Enter a rate to estimate annual energy revenue.
Installed capacity
39.61 MW
39,608 kW at the generator terminals
Annual energy
190.83 GWh
Net head
190.0 m
Annual energy
190.83 GWh
Homes powered
1,27,220
≈ per home/yr
1,500 kWh
| Power equation | P = ρ · g · Q · H_net · η |
| Net head (H_net) | 200 m gross − 5% loss = 190.0 m |
| Hydraulic power | 1000 × 9.81 × 25 × 190.0 × 0.85 = 39.61 MW |
| Annual energy | 39,608 kW × 8,760 h × 55% = 190.83 GWh |
A planning-level estimate. Real designs account for the flow-duration curve, varying head, turbine efficiency curves, downtime and dry/wet-season tariffs. The “homes powered” figure assumes ≈1,500 kWh per grid-connected home per year and is illustrative only.
From falling water to kilowatt-hours
Power depends on how much water falls (flow) and how far it falls (head); annual energy then depends on how steadily the plant runs through the year.
Net head
Start from the gross vertical drop, then subtract headrace and penstock friction losses (usually 3–10%).
Power
P = ρ·g·Q·H·η. In handy units, P (kW) ≈ 9.81 × flow × net head × efficiency.
Annual energy
Multiply power by 8,760 hours and the capacity factor — the share of full output the plant achieves over a year.
Hydropower, answered
How is hydropower output calculated?+
Hydraulic power is P = ρ · g · Q · H · η, where ρ is water density (1000 kg/m³), g is gravity (9.81 m/s²), Q is the design flow in m³/s, H is the net head in metres and η is the overall efficiency. Dividing by 1000 gives the power in kilowatts: P (kW) ≈ 9.81 × Q × H × η.
What is the difference between gross head and net head?+
Gross head is the full vertical drop from the intake water level to the turbine. Net head subtracts the friction losses in the headrace and penstock (typically 3–10%). Power is driven by the net head, which this calculator works out from the gross head and your loss percentage.
What efficiency should I use for a hydropower plant?+
Overall efficiency combines the turbine, generator and transformer. Modern plants reach 85–92%; micro-hydro schemes are often 60–80%. The default here is 85%, which is reasonable for a planning estimate.
What is a capacity (plant) factor and what value is typical in Nepal?+
The capacity factor is the share of full-rated output a plant actually produces over a year. Most of Nepal's plants are run-of-river, whose flow falls sharply in the dry season, so their annual capacity factor is typically around 50–60%. Storage and peaking-reservoir plants differ.
How is annual energy estimated?+
Annual energy (kWh) = installed power (kW) × 8,760 hours × capacity factor. For example a 10 MW plant at a 55% capacity factor generates about 10,000 × 8,760 × 0.55 ≈ 48 GWh per year.
Sources & data note
Based on the standard hydropower relation P = ρgQHη and annual energy = power × 8,760 h × capacity factor. Efficiency, loss and capacity-factor defaults reflect typical run-of-river practice. This is a planning estimate, not a substitute for a detailed feasibility study.