Hydroelectric Power Output

Weight, drop and what survives. The water passing through the turbine each second has a weight, $\rho g Q$, and it gives up energy as it falls through the head $H$. The product $\rho g Q H$ is the power in the water itself, which is what the hydraulic-power page gives. The overall efficiency $\eta$, the turbine's times the generator's, is the fraction that leaves as electricity, so $P = \eta\,\rho\,g\,Q\,H$ is the power delivered by the generator, not the power the water carries.

Which head, which flow. Use the net head: the gross height minus the friction losses in the penstock, since only the net drop reaches the turbine. $Q$ is the volume of water per second through the turbine, and the density is about 1000 kg/m³ for fresh water. Efficiency is the product of the turbine's and the generator's, for example 85 %.

A scale to keep in mind. At 100 % efficiency, one cubic metre per second falling one metre carries 9.81 kW, a textbook figure and the page's first example. Everything else scales from it: doubling the flow or the head doubles the power. For a hand-worked case, a 5 MW plant at 30 m net head and 88 % overall efficiency, with $g = 9.80665$ m/s², needs $Q = 5\,000\,000 / (0.88 \times 1000 \times 9.80665 \times 30) = 19.31$ m³/s. The same equation solves for the head, the efficiency or the density when the other quantities are known.

Limits. The formula gives the output at the flow and head you enter, not over a year. The capacity-factor page gives the fraction of this rated power a plant actually delivers across a year, and the total-dynamic-head page is the pump's version of the same bookkeeping: the head a pump has to add.