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Pump Hydraulic Power Calculator

Enter the flow rate, head and pump efficiency to find the hydraulic and shaft power.

Result

9.807 kW
14.010 kW
70.0%

How it is calculated

Flow is entered in litres per second, head in metres and efficiency in percent. The flow is first converted to cubic metres per second, Q = flow ÷ 1000. Hydraulic power is P = ρ · g · Q · H, with ρ fixed at 1000 kg/m³ (water) and g = 9.80665 m/s², giving watts, shown in kW. Shaft power is Pshaft = P ÷ (efficiency ÷ 100), the power the motor must put into the pump shaft. The density is that of water, so the result is not valid for oils, slurries or other fluids of different density. Head should be the total head, including friction and fittings. All three inputs must be above zero. Results are rounded to three decimals in kW. The tool does not add a safety margin for the motor.

Worked example

A pump moves 25 L/s against a total head of 40 m and has an efficiency of 70%. Flow: 25 ÷ 1000 = 0.025 m³/s. Hydraulic power: 1000 × 9.80665 × 0.025 × 40 = 9,806.65 W, which is 9.807 kW. Shaft power: 9.807 ÷ 0.70 ≈ 14.010 kW. So the motor must supply about 14 kW at the shaft, and in practice a motor of the next standard size above that would be chosen.

Try this example →

Common mistakes

  • Entering the flow in m³/h or L/min. The field is in L/s, so 90 m³/h must be converted to 25 L/s first.
  • Using only the static height difference as head. The head must include pipe friction and fitting losses, or the power is too low.
  • Entering efficiency as a fraction such as 0.7. The field is in percent, so type 70; 0.7 would make the shaft power 100 times too large.
  • Using the result for a fluid that is not water. Density is fixed at 1000 kg/m³, so a heavier liquid needs proportionally more power.
  • Treating shaft power as the motor rating. The motor also needs a margin above it, and the tool does not add one. Also note it accepts an efficiency above 100%, which is not physically possible.

About this calculator

The hydraulic power a pump delivers to a fluid is P = ρ·g·Q·H, where ρ is fluid density, g is gravitational acceleration, Q is the volumetric flow rate, and H is the total head (the height the pump effectively lifts the fluid against, including friction losses). This is the useful power actually transferred to the water or other fluid.

Because no pump is perfectly efficient, the shaft power the motor must actually supply is higher: hydraulic power divided by the pump's efficiency. Mechanical and civil engineers use this pair of numbers to size pump motors correctly for water supply systems, irrigation networks, and industrial fluid transfer, making sure the driving motor has enough margin above the theoretical hydraulic requirement.

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Comments (1)

  • Editorial team

    For plain water there is a handy sanity check: hydraulic kilowatts are roughly flow in litres per second times head in metres, divided by 102. With the default 50 L/s and 30 m that gives about 14.7 kW, matching the tool. The shaft figure still has no motor margin or starting conditions in it, so compare it with the motor's rated output. If you pump a fluid heavier than water, post what you used and how you adjusted for its density.

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