Pump power calculation: hydraulic, shaft and motor power
Sizing a pump motor is three calculations, not one: the hydraulic power the fluid actually needs, the shaft power after pump losses, and the motor rating after drive losses and a margin. Get the order wrong and you either trip the overload or pay for an oversized motor for 20 years.
Why pump power calculation matters
A pump that is undersized cannot hold the duty point and runs hot. A pump that is oversized runs throttled, wastes energy, and shortens seal life. In a plant running a 7.5 kW pump 6,000 hours a year, a single efficiency point is worth roughly ₹30,000 a year in electricity. The calculation below is the difference between a motor that lasts and one that fights its own curve.
Step 1 — Hydraulic (water) power
Hydraulic power is the useful power delivered to the fluid. It depends only on flow, head and fluid density:
Total head H is the sum of static lift, friction losses and any pressure head at the discharge — not just the height you are pumping to. Underestimating friction loss is the most common reason a "correctly sized" pump misses its duty point.
Step 2 — Shaft (brake) power
The pump is not 100% efficient, so the shaft must deliver more than the hydraulic power. Divide by pump efficiency:
This shaft power is what you would read on a brake test — hence "brake horsepower" (BHP). Always size the shaft from the worst-case duty point, not the rated point on the nameplate.
Step 3 — Motor power and margin
The motor must cover the shaft power plus transmission losses, then carry a service margin so it never runs at 100% load:
Standard IEC motor sizes (kW) are 0.37, 0.55, 0.75, 1.1, 1.5, 2.2, 3, 4, 5.5, 7.5, 11, 15, 18.5, 22 and so on. Always round up to the next standard size.
Worked example
Pumping water, Q = 50 m³/h, total head H = 30 m, centrifugal pump η = 70%, direct coupled.
- Hydraulic power: Phyd = 50 × 30 / 367 = 4.09 kW
- Shaft power: 4.09 / 0.70 = 5.84 kW
- Motor power: 5.84 / 1.0 × 1.2 = 7.01 kW → round up to a 7.5 kW motor
Skip the efficiency steps and you would wrongly pick a 5.5 kW motor that trips under load. Add too much margin and you would pay for an 11 kW frame that runs at half load and poor power factor.
Quick reference: motor size by duty
Indicative direct-coupled motor sizes for water, pump efficiency 70%, 20% margin:
| Flow (m³/h) | Head (m) | Hydraulic kW | Motor size |
|---|---|---|---|
| 10 | 20 | 0.55 | 1.1 kW |
| 25 | 25 | 1.70 | 3 kW |
| 50 | 30 | 4.09 | 7.5 kW |
| 100 | 40 | 10.9 | 18.5 kW |
| 200 | 50 | 27.2 | 45 kW |
Use these as a sanity check only — your real duty point, fluid and pump curve decide the final size. Run your own numbers in the pump power calculator to get hydraulic, shaft and motor power in one screen.
Common mistakes
- Using static lift as total head. Friction and discharge pressure can double the head on a long pipe run.
- Forgetting the efficiency division. Hydraulic power is not motor power — pump and drive losses are real.
- Assuming water density for every fluid. Slurry, oil and hot water change the answer.
- Sizing at the rated point, not the worst case. The system runs at the duty point, which may be off the best-efficiency point.
- No service margin. A motor at 100% load has no headroom for voltage dips or fouling.
One more practical note for anyone procuring pumps against a drawing: before you release the casting or impeller for machining, balloon the GD&T on the print so inspection is unambiguous. CadNexa's auto-ballooning tool turns a PDF pump drawing into a numbered inspection sheet in the browser — handy when you are qualifying a new vendor. You can also grab a ready inspection template to record the results.
Frequently asked questions
What is the formula for pump power?
Hydraulic power in kW equals ρ · g · Q · H / 1000, where ρ is fluid density in kg/m³, g is 9.81 m/s², Q is flow in m³/s and H is total head in metres. For water with flow in m³/h, it simplifies to Q · H / 367. Shaft power is hydraulic power divided by pump efficiency, and motor power adds drive losses and a margin.
What is the difference between hydraulic, shaft and motor power?
Hydraulic power is the useful power delivered to the fluid. Shaft (brake) power is higher because it includes pump losses. Motor power is higher still because it includes drive-transmission losses plus a service margin so the motor never runs fully loaded.
How do I size a pump motor in kW?
Calculate hydraulic power, divide by pump efficiency to get shaft power, divide by drive efficiency, multiply by a 1.15–1.25 margin, then round up to the next standard IEC motor size such as 5.5, 7.5 or 11 kW.
Does fluid density change pump power?
Yes. Power is directly proportional to density. The Q · H / 367 shortcut assumes water at about 1000 kg/m³; for slurry, brine or oil use the full formula with the actual density, which can raise the required power by 30% or more.