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Pump Power Calculator

Calculate the power a pump puts into the water, the shaft power it needs, the electricity its motor draws and what that costs in a year.

Method reviewed October 26, 2026. Version 1.0.0. How it works

Your inputs

Unit system
Pump head given as

From the pump curve at the operating point.

Result

Pump brake horsepower
1.804hp
Electrical input
1.495kW
Energy cost per year
$897

1.8 hp at the shaft, 1.49 kW from the supply — about $897 a year.

Water (fluid) horsepower
1.263hp
Head
50ft
Energy per year
5,978kWh
Pump heat added to the fluid (approx.)
4,590BTU/h
Show the working
  1. Fluid horsepower = GPM × head × SG ÷ 3,960
    100 × 50 × 1 ÷ 3,960 = 1.2626 hp
  2. Brake horsepower = fluid hp ÷ pump efficiency
    1.8038 hp
  3. Electrical input = bhp × 0.7457 ÷ motor efficiency
    1.4945 kW

Pump power, step by step

  • Fluid (water) horsepower = GPM × head in feet × specific gravity ÷ 3,960.
  • Brake horsepower = fluid horsepower ÷ pump efficiency.
  • Electrical input = brake horsepower × 0.7457 ÷ motor efficiency, in kW.

The 3,960 is the unit conversion the DOE pumping sourcebook uses. If you know the pressure rise in psi instead of head, the calculator converts it: head in feet ≈ psi × 2.31 ÷ specific gravity for water.

Where the savings are

Pump energy depends on how much flow and head the system really needs. Oversized pumps throttled by valves, or pumps running at full speed all year, waste energy. Reducing speed with a drive saves energy roughly with the cube of speed, as the affinity laws describe.

Worked example

These are the example values the calculator opens with, and the result the engine calculates for them.

Inputs

Flow rate
100 gpm
Pump head given as
Feet of head
Total head
50 ft
Specific gravity of the fluid
1
Pump efficiency
70 %
Motor (and drive) efficiency
90 %
Hours of operation per year
4,000 h
Electricity price
0.15 $/kWh

Result

Pump brake horsepower
1.804 hp
Electrical input
1.495 kW
Energy cost per year
$897
Water (fluid) horsepower
1.263 hp
Head
50 ft
Energy per year
5,978 kWh

1.8 hp at the shaft, 1.49 kW from the supply — about $897 a year.

Assumptions

  • 3,960 converts gpm × ft of water to horsepower (DOE pumping sourcebook).
  • psi to head uses water at 62.37 lb/ft³ (about 60 °F) and the specific gravity you enter.

Limitations

  • Pump efficiency changes along the pump curve; use the value at your actual flow and head.
  • The heat figure assumes all shaft power ends up in the fluid.

Frequently asked questions

What is pump head?

The energy a pump adds per pound of fluid, expressed as the height of a column of that fluid it could lift, in feet. In a closed hydronic loop it is the friction loss around the loop, not the building height.

What pump efficiency should I use?

The value from the pump curve at your flow and head. Small circulators are often well below large pumps.

Does pump heat matter?

In a closed loop the shaft power ends up as heat in the fluid. It helps a heating loop slightly and adds load to a chilled-water loop.

Sources

  1. Improving Pumping System Performance: A Sourcebook for Industry (2nd ed.) — U.S. DOE (with the Hydraulic Institute). Fluid power (hp) = flow (gpm) × head (ft) × specific gravity ÷ 3,960; pump efficiency = fluid power ÷ shaft (brake) horsepower.
  2. NIST Guide to the SI (SP 811), Appendix B.9 — conversion factors — National Institute of Standards and Technology. 1 ton of refrigeration = 12,000 BTU/h; 1 BTU/h = 0.2930711 W; 1 atm = 101.325 kPa.

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