Skip to main content
PM Assist

Pump power calculator

Hydraulic, shaft and motor input power from flow, head and density. Both efficiencies are yours to enter — a default would hide the thing you are probably looking for.

Hydraulic power

2.724 kW

Work done on the fluid

Shaft power

3.892 kW

Absorbed at the pump shaft

Motor input

—

Enter a motor efficiency

Method and sources

Equations used

P_hydraulic (kW) = ρ × g × Q × H ÷ 1000

ρ in kg/m³, g in m/s², Q in m³/s, H in metres. The useful work done on the fluid.

P_hydraulic (kW) = Q × H × ρ × 9.806 65 ÷ 3 600 000

The same equation with flow in m³/h, which is the unit most schedules use.

P_shaft = P_hydraulic ÷ η_pump

Absorbed at the pump shaft. η from the pump curve at the duty point.

P_motor input = P_shaft ÷ η_motor

Electrical input. η from the motor nameplate.

Constants assumed

SymbolValueSource
g9.806 65 m/s² (exact)Standard acceleration of free fall, 3rd CGPM (1901); NIST SP 811 App. B.9
ρEntered by youNot assumed — 1000 kg/m³ is a cold-water figure and is nearly 3% out at 80 °C
η_pump, η_motorEntered by youPump curve and motor nameplate. Deliberately not defaulted.

Sources

  • •NIST Special Publication 811, Appendix B.9 — standard acceleration of free fall.
  • •The hydraulic power relationship is first principles (weight flow × head), not an empirical correlation.

What this calculator deliberately does not do

No efficiency is assumed if you leave the field blank.

Shaft and motor power are shown as unavailable rather than estimated. Pump efficiency varies by more than a factor of three across real installations and collapses off the best efficiency point, so a default would introduce more error than the calculation removes — and would mask the oversizing that is usually the reason for doing the sum.

This is the power at one duty point. It does not model the system curve, NPSH available, variable-speed operation across a year, or the extra losses in a pump running off its best efficiency point. For anything load-bearing, work from the manufacturer's curve for the specific pump.

Frequently asked questions

How do you calculate pump power?
Hydraulic power is P (kW) = ρ × g × Q × H ÷ 1000, with density in kg/m³, g as 9.80665 m/s², flow in m³/s and head in metres. With flow in m³/h the same thing is Q × H × ρ × 9.80665 ÷ 3,600,000. Shaft power is hydraulic power divided by pump efficiency, and motor input power is shaft power divided by motor efficiency.
Why does it not assume an efficiency for me?
Because the assumption would carry more error than the calculation. Pump efficiency ranges from around 25% on a small circulator to over 85% on a well-selected end-suction unit, and it falls away sharply either side of the best efficiency point — which is exactly the condition someone doing this calculation is usually investigating. A built-in default would be a guess about your pump wearing the authority of a calculated result, and the commonest real finding, that a pump is running well off its curve because it is oversized, is the one a default would hide. Both figures are on the pump curve and the motor nameplate.
What head should I use?
Total dynamic head at the duty point — the pressure the pump actually develops against the whole circuit, including pipe and fitting losses, terminal and control valve authority and any static lift. Not the static lift alone, and not the closed-valve head from the top of the curve, both of which are common substitutions that give an answer with no relationship to what the pump is doing.
What density should I use?
1000 kg/m³ is a reasonable figure for cold water and is right to within 0.2% at chilled water temperatures. For LTHW it is not: water is about 972 kg/m³ at 80 °C, so using 1000 overstates the power by nearly 3%. For a glycol mixture use the manufacturer's figure for that concentration and temperature — this calculator takes density as an input precisely so you can supply the right one rather than have a wrong one assumed.
Why is my measured motor power higher than this says?
Usually because the real efficiencies are lower than the figures entered, and most often because the pump is operating away from its best efficiency point. A pump selected with generous margin runs further right on its curve at lower efficiency, drawing more power for the same useful work. The gap between this calculation and a clamp-meter reading is a reasonable first indication of how far off its design point a pump is running.