Hong Life CalculatorsUnderstanding the pump power calculator — concept and principle
Every pump has to lift a flow of liquid against a head, and the power that takes is fixed by physics; the motor then has to supply extra to cover the pump's own losses. This calculator gives the hydraulic (water) power, the shaft power after pump efficiency, and the pressure equivalent of the head. It is used to choose motors for water supply, irrigation, cooling and process pumps.
How it is calculated
- Hydraulic power: P_h = ρ g Q H, with ρ in kg/m³, g = 9.80665 m/s², Q in m³/s (m³/h ÷ 3,600) and H in m; result in W, ÷ 1,000 for kW.
- Shaft power: P_s = P_h / η, where η is pump efficiency as a decimal.
- Pressure rise: Δp = ρ g H, ÷ 10⁵ for bar.
The principle
Head is energy per unit weight of liquid. Raising a mass m through height H needs m g H joules, so a mass flow ρQ per second needs ρ Q g H watts. Head also includes friction losses in pipes and any pressure difference between suction and discharge, all expressed as an equivalent height of the liquid being pumped. No pump converts shaft work perfectly, because of internal leakage, friction and turbulence, so the shaft must deliver P_h divided by the efficiency. For oil-hydraulic pumps the same idea is usually written as P (kW) = p (bar) × Q (L/min) / 600.
Worked example
- Inputs: 20 m³/h, head 30 m, water 1,000 kg/m³, efficiency 70%.
- Q = 20/3,600 = 0.005556 m³/s.
- P_h = 1,000 × 9.80665 × 0.005556 × 30 = 1,634.4 W = 1.634 kW.
- P_s = 1.634/0.70 = 2.335 kW.
- Δp = 1,000 × 9.80665 × 30 / 10⁵ = 2.942 bar.
| Flow (m³/h) | Hydraulic power (kW) |
|---|---|
| 5 | 0.4086104167 |
| 10 | 0.8172208333 |
| 15 | 1.22583125 |
| 20 | 1.634441667 |
| 25 | 2.043052083 |
| 30 | 2.4516625 |
| 40 | 3.268883333 |
How to read the result
Shaft power is what the motor must deliver at the duty point. Motor input power is higher again because the motor has its own efficiency, and motors are normally chosen one standard size above the shaft power for margin, which here would point to a 3 kW class motor depending on the pump curve and supplier advice. For water, about 10.2 m of head equals 1 bar. Efficiency varies widely: small pumps are often well below large ones, and the best efficiency occurs near the design flow, so running far off the curve wastes power.
Common mistakes
- Entering Q in m³/h without converting to m³/s (the calculator does this for you).
- Using only the static lift and ignoring pipe friction in the head.
- Forgetting the liquid's density; brine, oil or slurry change power in proportion.
- Treating shaft power as electrical power.
Frequently asked questions
Does a viscous liquid need more power?
Yes. Viscosity lowers centrifugal pump efficiency and head, so manufacturers apply correction factors.
How do I convert head to pressure for other liquids?
Multiply by density: Δp = ρ g H. Light oil gives less pressure for the same head than water.
Why does the motor need margin?
Real flow can drift to the right of the curve where power rises, and motors run hot if overloaded.
Sources: ISO 9906, Rotodynamic pumps — Hydraulic performance acceptance tests · Çengel, Y. A. and Cimbala, J. M., Fluid Mechanics: Fundamentals and Applications, McGraw-Hill