Hydraulic Power (P = ρgQh)

Each second, a pump moves $\rho Q$ kilograms of fluid and raises them $h$ metres. The work done on that mass is $\rho Q g h$ joules, and since it happened in one second, that is the power. There is nothing more to the derivation than $mgh$ with the mass supplied at a steady rate, which is why the equation is exact rather than empirical. What it gives is the water power — the energy actually landing in the fluid, before any machine is asked to deliver it.

Pumping 100 L/s up 20 m: $P = 1000 \times 9.80665 \times 0.1 \times 20 = 19.6$ kW at the water. Now walk it back through the machinery. A pump running at 75% efficiency needs 26.2 kW at the shaft, and a motor at 92% draws about 28.5 kW from the panel. Run it 4000 hours a year at ten cents a kilowatt-hour and that pump costs roughly $11 400 annually in electricity — a figure that usually dwarfs the capital cost and is the real reason efficiency points are worth arguing over.

Run the equation backwards and it is hydroelectricity. The same $Q$ and $h$ falling through a turbine yield the same $\rho gQh$, less the turbine's own efficiency, which is why a run-of-river plant's output can be estimated from a flow gauge and a contour map before anything is built. It is also the honest reply to most perpetual-motion schemes involving pumped water: the energy to lift it is exactly the energy available when it falls, and every real component takes a cut in both directions.

The mistake that undersizes more pumps than any other is putting the physical lift into $h$. The $h$ this equation wants is the total dynamic head: the static lift, plus the friction loss through every metre of pipe, elbow, valve and heat exchanger, plus any residual pressure the discharge has to be delivered against. On a long or restricted run, friction can exceed the geometric lift several times over. Sizing on height alone gives a pump that produces its rated flow into an open pipe and much less into the real system.

The reverse error appears in closed loops and is just as common. A hydronic circuit's return leg gives back everything the supply leg spent lifting, so the static height cancels entirely and $h$ is friction alone — a pump for a thirty-storey closed loop may need only a few metres of head. Height sets the fill pressure; friction sets the pump. Three more points of arithmetic. $Q$ belongs in cubic metres per second, and the usual slips are a factor of 1000 from litres per second or 60 000 from litres per minute. Efficiency is not a fixed property of a pump but a curve, and a unit operating far from its best efficiency point can fall to 40% or worse, which is why throttling a valve to reduce flow is such an expensive way to control a system. And ρ is the density of what you are actually pumping — 40% glycol at 1045 kg/m³ costs about 4.5% more power for the same head and flow, before its higher viscosity adds friction on top.