Pump Power Calculator

Calculate pump power, hydraulic head, and efficiency for fluid pumping systems.

10 L/s = 0.01 m³/s
Please check your inputs and try again.

From Hydraulic Power to Motor Input Power

A pump must add mechanical energy to a flowing fluid, and the required hydraulic power is set by flow rate and total head. For an incompressible fluid, Phyd=ρgQH. The product ρgH is an energy-per-volume or pressure-like term, while Q is volume per time, so their product has units of watts. Doubling either flow rate or required head doubles ideal hydraulic power.

Total head is an energy-per-unit-weight measure and can include elevation change, pressure difference, velocity-head change, and piping or fitting losses. It is not always the same as vertical lift. Pump efficiency then relates hydraulic output to shaft input: Pshaft=Phyd/ηpump. If a motor drives the pump, its electrical input is larger again: Pmotor,in=Pshaft/ηmotor.

Phyd=ρgQH,   Pshaft=Phyd/ηpump,   Pmotor,in=Pshaft/ηmotor
SymbolMeaningWhy it appears / units
ρFluid densitykg/m3; denser fluid requires more power for the same Q and H.
QVolumetric flow ratem3/s; volume delivered per second.
HTotal head addedm of fluid; represents mechanical energy per unit weight.
ηpumpPump efficiencyFraction or percent; accounts for hydraulic and mechanical losses in the pump.
ηmotorMotor efficiencyFraction or percent; converts electrical input to shaft output.

Real pump selection also depends on the pump curve and system curve. The operating point occurs where the two intersect, and efficiency varies with flow. Designers also check net positive suction head, cavitation risk, motor service factor, starting behavior, fluid properties, and expected operating range rather than sizing from one power number alone.

Worked Examples

Example 1: Water pump: Q=10L/s, H=20m, η=75%
P_hyd=1000×9.81×0.01×20=1962W
Result: P_shaft=2616W, P_motor≈2.9kW at η_motor=90%
Select 3kW motor
Example 2: Building pump: Q=5L/s, H=50m
P_hyd=1000×9.81×0.005×50=2453W
Result: 2.5 kW hydraulic power
For multi-story building pressurization
Example 3: Including both pump and motor efficiency
Water, Q=0.020m3/s, H=15m, ηpump=80%, ηmotor=92%
Result: Phyd=2.943kW, motor input≈4.00kW
The electrical input must exceed hydraulic power because losses occur in both the pump and the motor.
Example 4: Pumping light oil
ρ=850kg/m3, Q=0.005m3/s, H=30m, ηpump=70%, ηmotor=90%
Result: Phyd≈1.251kW, motor input≈1.985kW
For the same Q and H, lower-density fluid gives lower hydraulic power, but efficiency losses still determine the required input.

Common Mistakes

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Entering liters per second as cubic meters per second

Convert using 1L/s=0.001m3/s. Missing this factor makes the power 1000 times too large.

⚠️
Treating vertical elevation as the entire total head

Total head can also include pressure, velocity, and friction losses. A long pipe can require substantial additional head even when elevation change is small.

⚠️
Multiplying by efficiency instead of dividing

Input power must be greater than useful hydraulic output when efficiency is below 100%, so divide by the efficiency fractions.

Frequently Asked Questions

Head vs pressure?
H = P/ρg. 1 bar = 10.2 m water head. Head is pressure expressed in height of liquid — useful because it's fluid-independent in comparisons.
Pump types by specific speed?
Conventional pump specific speed requires rotational speed as well as flow and head, commonly in a form proportional to N√Q/H3/4. Its numerical ranges depend on the unit convention. Because this page does not request rotational speed, it should not assign a pump type from Q and H alone.
What is the difference between pump head and pressure?
Head expresses mechanical energy per unit weight as an equivalent fluid-column height. Pressure is force per area. For a static fluid, ΔP=ρgH, so the same head corresponds to different pressure changes for fluids with different densities.
Why is actual motor power larger than hydraulic power?
Hydraulic power is the useful rate of energy delivered to the fluid. The pump loses energy through hydraulic, leakage, bearing, and mechanical effects, and the motor has electrical and mechanical losses. Dividing by both efficiencies gives the required electrical input in the idealized chain.
Does pump efficiency stay constant at every flow rate?
No. Centrifugal pumps have efficiency curves and usually perform best near a best-efficiency point. Moving far away from that operating region can reduce efficiency and increase vibration, recirculation, heating, or mechanical loading. Manufacturer pump curves are needed for actual equipment selection.
Why must cavitation be checked separately from pump power?
A motor can provide enough power while the pump still cavitates if suction pressure falls too low. Cavitation forms vapor bubbles that collapse inside the pump, causing noise, performance loss, erosion, and damage. Net positive suction head requirements address this separate limitation.

Formula Explorer connections

Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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