Hydroelectric Power
Power from falling water: flow rate times head.
Formula
Variables
Example
Q=10, H=50, η=0.9: P ≈ 4.4 MW
How it works
This formula estimates the electrical power a hydroelectric system can produce from moving water. It comes from the fact that falling water loses gravitational potential energy, and the rate of energy transfer depends on how much water passes each second, how far it falls, and how effectively the turbine-generator converts that energy into electricity. The density of water and gravity set the energy available per cubic metre, while efficiency accounts for real-world losses such as turbulence, friction, and electrical losses.
When to use it
Use this formula when you know the river or reservoir flow rate, the vertical drop across the plant, and the turbine efficiency. It is commonly used in hydropower design, feasibility studies, and estimating output from small or large hydroelectric installations.
Worked example, step by step
Given: ρ = 1000 kg/m³, g = 9.81 m/s², Q = 25 m³/s, H = 30 m, η = 0.85 P = ρgQHη P = 1000 × 9.81 × 25 × 30 × 0.85 P = 6,251,250 W P ≈ 6.25 MW
Common mistakes to avoid
- Forgetting to multiply by efficiency, which makes the power too high
- Using head in the wrong units, such as centimetres instead of metres
- Mixing up flow rate and water volume, or using mass flow without converting correctly
Did You Know?
Hydropower supplies about 15% of the world's electricity — the largest renewable source by far.
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