Hydraulic Power (P = ρgQh)

P = \rho g Q h

P =

ρ =

Q =

h =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Constant used — built into this formula, no need to enter

g_{n} = 9.80665\ \text{m/s}^{2}

Standard acceleration of gravity · exact

Learning zone

Each second a pump moves ρQ kilograms of fluid and lifts them h meters, so the energy per second is ρgQh — the water power before any efficiency losses. Pumping 100 L/s of water up 20 m takes 1000 × 9.80665 × 0.1 × 20 ≈ 19.6 kW at the water; with a 75%-efficient pump the motor draws about 26 kW. Run the formula the other way and it is hydroelectricity: the same Q and h falling through a turbine yield the same power.

Hydraulic Power (P = ρgQh)

P = \rho g Q h

Where

$P$ = Hydraulic power
$\rho$ = Fluid density
$Q$ = Flow rate
$h$ = Head

Missing one of these? Work it out first, then come back

Hydraulic power — Pump Efficiency from Hydraulic and Shaft Power, Power (P = W/t)
Fluid density — Dynamic Pressure (q = ½ρv²), Buoyant Force (Archimedes' Principle)
Flow rate — Volume of Water Over a Period, Pump Water Horsepower
Head — Pump Water Horsepower, Pump Brake Horsepower

Part of this formula set

Pump power and efficiency

7 formulasWater horsepower, brake horsepower, pump efficiency, NPSH available and specific speed — sizing the motor and keeping the impeller wet.