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Pump power

A pump must supply enough energy to overcome the total loss in a piping system, both friction loss and minor losses, as well as any lift in elevation. The hydraulic power is computed from the total loss expressed in meters of fluid column, the density and the volumetric flow rate.

P = \rho gQh_f

hydraulic pump power to overcome a loss

h_f

Symbols

$P$	hydraulic pump power	W
$Q$	volumetric flow rate	m³/s
$h_f$	total loss	m

Example

A pump delivers $Q = 8.0$ L/s of water against a total loss of $h_f = 3.5$ m. $P = 1000\cdot 9.81\cdot 0.0080\cdot 3.5 \approx 275$ W.

The volumetric flow rate must be in m³/s (divide L/s by 1000) before substituting into

P = \rho gQh_f

Practise pipe flow and dimensionless numbers for free →

← Minor losses · Pumps in series and parallel →

Part of Fluid Mechanics: Pipe flow and dimensionless numbers.