Pump Power Calculator
From flow and head to watts — the hydraulic power the water receives, then the efficiency tax that sizes the motor.
Pump Power Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Steady duty point; no VFD or throttling model
- Pump efficiency entered, not derived
In short: Lifting 10 L/s of water through 20 m of head asks 1,962.000000 W of hydraulic power — ρgQH = 1,000 × 9.81 × 0.010000 × 20. At a 70% pump efficiency the shaft must supply 2,802.857143 W: the 840.857143 W difference never reaches the water — it becomes heat, noise and wear inside the pump, and the motor nameplate is sized on the larger number.
Formula
P_hyd = ρ·g·Q·H ··· P_shaft = P_hyd/η
Hydraulic power is the rate of gravitational work — density times gravity times the flow you lift times the height you lift it — the one line every pump curve secretly prices. Efficiency then splits the shaft's bill: everything the water does not receive is spent inside the pump, which is why the motor is always bought bigger than the water's side of the ledger.
Worked Example
- Enter the duty flow and total dynamic head.
- Enter the fluid density.
- Enter the pump efficiency at that duty point.
- Read hydraulic watts, then the shaft watts.
Defaults: 10 L/s, 20 m, water, 70% → hydraulic 1,962.000000 W, shaft 2,802.857143 W. The halved efficiency check: at 50% the shaft pays 3,924.000000 W for the same water — efficiency is the second-most-expensive metre in pumping.
Strengths & Limits Of This Model
Where this engine is strong
- Hydraulic and shaft watts both shown
- The efficiency tax stated in watts
Where it stops
- No motor efficiency stage
- No affinity-law scaling
Practical Use Cases
Motor selection
nameplate from the duty point
Energy audits
the kWh per cubic metre
Teaching
where the watts actually go
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Pump Power Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
What is the difference between hydraulic and shaft power?
Hydraulic power is what the WATER receives — ρgQH, the pure gravitational bill at 100% — while shaft power is what the COUPLING delivers, hydraulic power divided by the pump's efficiency. The gap becomes heat and noise inside the pump. Motors are sized on the shaft number; energy audits that quote the hydraulic number flatter the system by exactly the efficiency.
Which H belongs in the formula?
Total dynamic head — static lift plus pipe friction plus any pressure the destination sits under. Using the physical lift alone under-promises: the 12 m balloon to 16.7 m in the classic worked example once friction and discharge pressure climb aboard. The pump head page assembles this H from its parts; this page only prices it once it exists.
Why does efficiency swing the answer so much?
Because it divides: at 70% the shaft pays 1.43× the hydraulic bill, at 50% it pays double. And efficiency is a CURVE, not a constant — the value at the duty point matters, not the catalogue's best-case peak. A pump run far from its best efficiency point wastes watts in exact proportion to how far off-peak it sits.
Does the motor's efficiency stack too?
Yes — motor efficiency divides AGAIN if you want the electrical input: wall watts = hydraulic watts ÷ (pump η × motor η). The page stops at the shaft because that is the pump's boundary; the utility meter keeps going. Chain enough efficiencies and the overall figure explains why pumping is the world's favourite electricity bill.
What about fluids heavier than water?
Enter the real density and the formula handles it — power scales directly with ρ. The head, notably, does not: a given pressure reads as FEWER metres on a heavy fluid. That asymmetry (head is energy per weight, power is energy per time) is why pump curves in metres work for any fluid but power curves do not.
Why is g = 9.81 used here?
It is the engineering convention the pump industry standardises on — close enough to standard gravity that catalogue arithmetic and field measurements agree within their own tolerances. The exact standard is 9.80665; the difference is under 0.04%, far below pump-curve uncertainty. Consistency matters more than the fourth decimal.
How do I price the energy bill from this?
Shaft watts ÷ motor efficiency = electrical watts; run hours × kilowatts = kWh. The page's shaft figure is the honest start of that chain — starting from hydraulic watts silently deletes the pump's inefficiency from the bill, which is the oldest optimistic error in pumping. Watts at the shaft, then the meter takes its share.
Is a bigger pump ever the cheaper answer?
At the duty point, efficiency matters more than nameplate: an oversized pump throttled back to duty runs off its best efficiency curve and pays for it every hour. Two pumps with identical hydraulic output can differ by a quarter of their energy bill on efficiency alone. The page prices one duty point honestly — system curves are the next conversation.