Total Dynamic Head: the combined static head and system losses the pump must overcome.

How does efficiency affect power?

Lower efficiency means higher input power for the same hydraulic output.

How do I convert kW to hp?

Approximately 1 kW = 1.341 hp, and 1 hp = 0.746 kW.

Can this result be used directly for final design?

No. CivilCalc is intended for preliminary sizing and checking; final design must follow applicable standards, authority criteria and engineering review.

Pump Power Calculator | Flow, TDH, kW & HP

Pump Power Calculator: Flow, TDH, Efficiency and Motor Power

Pump sizing is not based on flow and static elevation alone. Static head, pipe friction, minor losses and required discharge pressure combine to form total dynamic head (TDH). CivilCalc connects hydraulic power, pump shaft power and approximate motor input power in one traceable workflow.

Engineering reference updated: 19 September 2026

Core power equations

P_h = ρ × g × Q × H

Shaft and motor input power increase as pump and motor efficiency decrease.

P_shaft = P_h/η_pump • P_motor = P_shaft/η_motor

Building TDH

TDH = static head + friction loss + minor loss + discharge pressure head

Static elevation difference is not automatically the required pump head. Pipe diameter, length, fittings, flow and terminal pressure all affect the operating point.

Worked pumping example

Q = 20 L/s (317 gpm), static head = 24 m (78.7 ft), friction = 5 m, minor loss = 1 m → TDH = 30 m (98.4 ft). Hydraulic power is about 5.89 kW. At 75% pump efficiency, shaft power is about 7.85 kW; at 90% motor efficiency, electrical input is about 8.72 kW (11.7 hp).

How flow changes required power

Illustrative comparison with TDH = 30 m (98.4 ft), pump efficiency 75% and motor efficiency 90% held constant.

5 L/s

2.18 kW

10 L/s

4.36 kW

15 L/s

6.54 kW

20 L/s

8.72 kW

This is not a pump curve. In a real piping system TDH normally changes with flow, so the duty point is found from the intersection of the system curve and pump curve.

Efficiency versus input power

Overall efficiency	Input at 20 L/s, 30 m
80%	7.36 kW
70%	8.41 kW
60%	9.81 kW
50%	11.77 kW

Energy use

Energy = motor input power × operating time

An 8.72 kW input operating 12 hours/day represents about 104.6 kWh/day. Actual consumption varies with duty point, VFD control, level changes and efficiency.

Six engineering checks before selecting a pump

1. System curve
Evaluate system head across the expected flow range.

2. BEP
Check how close the duty point is to the best efficiency region.

3. NPSH
Compare NPSH available with manufacturer NPSH required using an appropriate margin.

4. Motor margin
Do not match the motor nameplate directly to theoretical kW.

5. Operating range
Check minimum/maximum flow, wet-well levels and parallel pump cases.

6. Surge
For long force mains, review starts, stops, check valves and transient pressure.

Frequently asked questions

Is static elevation enough to size pump head?

No. Add friction, minor losses and any required discharge pressure to establish TDH.

Should an 8.7 kW result use an 8.7 kW motor?

Not directly. Standard motor sizes, pump curves, efficiencies, service conditions and manufacturer data must be checked.

If flow doubles, does power double?

If TDH and efficiency stay constant, yes. In real pipe systems friction rises with flow, so TDH may also increase.

Why is NPSH checked separately?

NPSH addresses suction-side cavitation risk and cannot be determined from motor power alone.

Can a VFD reduce energy use?

It can in variable-demand systems, but savings depend on the system curve and operating profile.

Pump Power Calculator