Pump Power Calculator
Calculate the shaft power a pump must deliver to move fluid through a system against a given total head. Use this to size pump motors or estimate energy consumption in HVAC, water supply, and process engineering.
About this calculator
Pump shaft power (also called brake power) is the mechanical power that must be supplied to the pump shaft to deliver the desired flow. The formula is: P = (Q × H × ρ × g) / (η × 1000), where Q is the volumetric flow rate in m³/s, H is the total head in metres, ρ is the fluid density in kg/m³, g = 9.81 m/s², and η is the pump efficiency expressed as a decimal (e.g. 0.75 for 75%). The factor of 1000 converts watts to kilowatts. Total head includes static lift, friction losses, and any velocity or pressure head changes. A lower pump efficiency means more shaft power — and therefore more electricity — is needed to achieve the same hydraulic output.
How to use
A pump moves water (ρ = 1000 kg/m³) at 0.02 m³/s against a total head of 30 m with an efficiency of 0.75. Step 1 — compute hydraulic power: 0.02 × 30 × 1000 × 9.81 = 5886 W. Step 2 — divide by efficiency: 5886 / 0.75 = 7848 W. Step 3 — convert to kilowatts: 7848 / 1000 = 7.85 kW. This is the shaft power required from the motor. An electric motor with a separate efficiency of 0.90 would draw 7.85 / 0.90 ≈ 8.72 kW from the grid.
Frequently asked questions
What is the difference between hydraulic power and pump shaft power?
Hydraulic power is the useful power actually delivered to the fluid, calculated as Q × H × ρ × g. Pump shaft power (brake power) is the power that must be supplied to the pump's input shaft, and is always larger than hydraulic power because no pump is perfectly efficient. The ratio of hydraulic power to shaft power is the pump efficiency (η). A pump with η = 0.80 wastes 20% of the input shaft power as heat through internal leakage, disc friction, and hydraulic losses within the pump casing.
How does pump efficiency affect electricity costs in a water supply system?
Because shaft power equals hydraulic power divided by efficiency, even a modest improvement in pump efficiency has a significant impact on running costs. For example, replacing a pump with η = 0.65 with one rated η = 0.80 for the same duty point reduces shaft power — and therefore electricity consumption — by about 19%. Over a year of continuous operation, this translates to substantial savings in energy bills and reduced carbon emissions. This is why selecting the correct pump for its best-efficiency point (BEP) on the performance curve is a key engineering and economic decision.
Why do engineers include fluid density in the pump power formula?
Pump power is proportional to the weight of fluid being moved per unit time, and weight depends on density. Water at 1000 kg/m³ requires significantly less power than a dense slurry at 1400 kg/m³ for the same flow rate and head. Conversely, hot water or light hydrocarbons with densities below 1000 kg/m³ need less power. Neglecting to account for fluid density — for example by always assuming water — can lead to undersizing the pump motor, causing overloads, tripped breakers, or premature motor failure in industrial applications.