Pump Motor Power Calculator

Hydraulic power, shaft power, motor input power, current and running cost from the flow and head your pump has to deliver — in L/min, L/s, m³/h or gpm, with head in metres or feet.

Last updated 22 Sep 2026

Pump motor power

Flow and head → kW, amps and cost per hour

Flow rateWhat the pump actually has to deliver, not the figure on the box — a domestic pump is usually throttled well below its maximum.

Flow unit

Total headStatic lift plus every friction loss in the pipework. This page does not estimate friction — see the note below.

Head unit

Fluid density (kg/m³)1,000 for cold fresh water. Seawater is about 1,025; a slurry or a glycol mix is different again, and the power goes up in direct proportion.

Pump efficiency (%)From the pump curve at the duty point. Small domestic centrifugal pumps are often 40–60%; large well-matched ones reach 80% and more. This default is an example.

Motor efficiency (%)From the motor plate or its IE class. Small motors are much worse than large ones.

Supply

Supply voltage (V)

Power factorFrom the motor plate. A pump motor running well below its rating has a much worse power factor than its full-load figure.

Electricity rate per kWhYour own tariff, in your own currency — the cost below comes out in the same units you type in here.

The supply feeds the motor, the motor turns the pump, and the pump lifts water. Each stage loses something: the motor's own loss first, then the pump's. The dots show the current the supply actually carries, which is set by the electrical input power — the figure at the far left, not the hydraulic power at the far right.

961.4WExample

100 L/min against 30 m of head, water, pump 60% efficient, motor 85%, 230 V single-phase at PF 0.85, 8 per kWh

From water to watts

P_hydraulic = ρ × g × Q × H P_shaft = P_hydraulic ÷ η_pump P_electrical = P_shaft ÷ η_motor I = P_electrical ÷ (√3 × V × PF) for three-phase

ρ: fluid density in kg/m³ — 1,000 for cold fresh water
g: 9.806 65 m/s², standard gravity (pump handbooks round it to 9.81, which is 0.034% high)
Q: flow rate in cubic metres per second
H: total head in metres — static lift PLUS friction losses
η: efficiency as a fraction; the pump’s and the motor’s multiply

Worked example

100 L/min against 30 m of head, water, pump 60% efficient, motor 85%, 230 V single-phase at PF 0.85, 8 per kWh
Q = 100 L/min = 0.001667 m³/s, so the pump lifts 1.667 kg of water every second
Hydraulic power = 1,000 × 9.80665 × 0.001667 × 30 = 490.3 W
Shaft power = 490.3 W ÷ 0.60 = 817.2 W, which is 1.11 metric horsepower — a "1 HP" pump is already working hard
Electrical input = 817.2 W ÷ 0.85 = 961.4 W
Current = 961.4 ÷ (230 × 0.85) = 4.918 A
Running cost = 0.961 kWh × 8 = 7.69 per hour

What common duties need, at 60% pump and 85% motor efficiency

Hydraulic is the water power, Shaft what the motor must deliver, PS the same shaft power in metric horsepower and Electrical in what the supply provides. Both efficiencies are held at the page’s example values across the table, so the columns scale exactly with flow × head. Real pumps vary widely; a badly matched pump at 35% efficiency needs nearly twice the shaft power of a good one at 65%.
Flow	Head	Hydraulic	Shaft	PS	Electrical in
20 L/min	20 m	65.38 W	109 W	0.15	128.2 W
50 L/min	25 m	204.3 W	340.5 W	0.46	400.6 W
100 L/min	30 m	490.3 W	817.2 W	1.11	961.4 W
200 L/min	30 m	980.7 W	1.634 kW	2.22	1.923 kW
500 L/min	40 m	3.269 kW	5.448 kW	7.41	6.41 kW
1,000 L/min	25 m	4.086 kW	6.81 kW	9.26	8.012 kW

What it really costs to lift water

The physics is the easy part. Lifting water takes energy equal to its weight times the height, so the power is the weight lifted per second times the height: ρ × g × Q × H. A hundred litres a minute is 1.667 kilograms a second, and pushing that up 30 metres is 490.3 W of genuinely useful power. Everything after that is loss.

Two efficiencies, multiplied. The pump turns shaft power into water power at 40–80% depending on how well it is matched to the duty, and the motor turns electrical power into shaft power at 70–95% depending on its size and class. They multiply, so a 60% pump on an 85% motor is 51% overall: for 490.3 W of water power you pay for 961.4 W of electricity. That is where the money goes, and it is why an oversized pump throttled down is such an expensive habit — it drops off its own efficiency curve.

Total head is not the lift. This is the number most often got wrong. The head the pump has to produce is the static lift plus every friction loss on the way — the pipe itself, every bend, every valve, the foot valve, the filter — plus any pressure the outlet has to hold. KSB’s own formulation writes the system head as the geodetic head plus pressure and velocity differences plus the sum of the losses. On a long run of narrow pipe, friction can equal the lift. This page does not estimate friction, because doing it honestly needs the pipe’s diameter, material, length and every fitting on it. Get the total head from the pipework calculation or from a pressure gauge at the pump, and put that number in.

Reading the answer against a pump you can buy. The shaft power figure is what a motor has to deliver, so it is the one to compare with a nameplate. The example works out at 1.11 metric horsepower, which means a “1 HP” pump is at its limit at that duty and a 1.5 HP one would be comfortable. Remember that Indian and Gulf pump plates marked in HP normally mean metric horsepower (735.5 W), not the American 745.7 W — the HP to kW converter sets the three definitions side by side.

Running cost. The cost line uses your own tariff and your own currency; the page has no opinion about either. A borewell pump running four hours a day at the example duty costs about four times 7.69 a day in whatever unit you typed. For the whole household’s consumption, use the electricity consumption calculator; for the cable and breaker the pump needs, the motor full-load current calculator and then the MCB size calculator. If the pump runs off an inverter or a solar supply, size that from the electrical input figure and from the starting current, not from the hydraulic power — see the inverter size calculator.

Frequently asked questions

How do I calculate the motor power needed for a pump?

Work out the hydraulic power ρ × g × Q × H, divide by the pump efficiency for the shaft power, and divide again by the motor efficiency for the electrical input. For 100 L/min at 30 m with a 60% pump the shaft power is 817.2 W, which is 1.11 metric HP.

What is hydraulic power in a pump?

The rate at which the pump adds energy to the liquid: density × gravity × flow × head. It is the useful output, and it is always less than the shaft power by the pump’s efficiency.

Does the head include friction losses?

Yes — the total head a pump must deliver is the static lift plus all the friction in the pipes, bends and valves, plus any outlet pressure. On a long or narrow pipe the friction can be as large as the lift. This page does not estimate it; take the figure from the pipework calculation.

How much current does a 1 HP water pump draw?

It depends on the motor, not on the pump. A 1 metric HP (735.5 W output) motor at 75% efficiency and 0.8 power factor on 230 V single-phase draws about 5.33 A. Read your own plate, and use the motor full-load current page.

Why is my pump using more electricity than the calculation says?

Usually because the real head is higher than assumed — friction losses are the common culprit — or because the pump is running away from its best efficiency point. A worn impeller, a partly blocked foot valve or an oversized pump throttled at a valve will all do it.

Does pumping seawater take more power than fresh water?

Yes, in direct proportion to density. Seawater at about 1,025 kg/m³ takes 2.5% more power for the same flow and head.

Related calculators

References

KSB. Selecting Centrifugal Pumps (technical information). Pump input power P = ρ·g·Q·H/η in W = ρ·Q·H/(367·η) in kW with Q in m³/h and ρ in kg/dm³, g = 9.81 m/s² and η a fraction “not in %”; and the system head H_sys = H_geo + (p_a − p_e)/(ρ·g) + (v_a² − v_e²)/2g + ΣH_L, the last term being the friction losses.
IEC 60034-30-1 efficiency classes for line-operated AC motors (IE1 standard, IE2 high, IE3 premium, IE4 super-premium), as tabulated in ABB technical note 9AKK107319 EN 05-2018. At 50 Hz and 4 poles the nominal efficiencies are 79.6 / 82.5 / 85.7% at 0.75 kW, 85.5 / 87.7 / 90.4% at 3 kW, 88.7 / 90.4 / 92.6% at 7.5 kW and 91.6 / 93.0 / 94.5% at 22 kW for IE2 / IE3 / IE4.
National Institute of Standards and Technology. Guide for the Use of the International System of Units (SI), Special Publication 811, Appendix B.8, factors for units listed alphabetically: horsepower (550 ft·lbf/s) = 7.456 999 × 10² W, horsepower (metric) = 7.354 988 × 10² W, horsepower (electric) = 7.46 × 10² W, horsepower (boiler) = 9.809 50 × 10³ W.
ABB. Softstarter Handbook, publication 1SFC132060M0201. Direct-on-line starting current “Usually between 6-8 times the rated current, but it can be more than 10 times the rated current”; star-delta: “The resulting current when Y-connected will be 1/3 of the current when delta connected” and the torque “ending up being 33% of the torque available when delta connected”; n = 2 × f × 60 / p; s = (n₁ − n)/n₁; T_n = 9550 × P_r/n_r.