Pump Power Calculator

Calculate hydraulic power from density, gravity, flow, and head using P_hyd = ρgQH. Then divide by pump efficiency to obtain required pump shaft input. If you need motor electrical input, also divide by drive and motor efficiencies. Keep all power definitions separate so hydraulic, shaft, and electrical power are not confused.

For an incompressible liquid, hydraulic power is P_hyd = ρgQH. Pump shaft input is P_shaft = P_hyd/η_pump. If motor and coupling losses are included, electrical motor input can be estimated as P_elec = P_hyd/(η_pump η_drive η_motor).

Convert flow to m³/s and head to metres, use fluid density in kg/m³, then calculate hydraulic power with ρgQH. Divide by the expected pump efficiency to estimate pump shaft power. Use the manufacturer curve at the actual operating point for final engineering checks.

Calculate power in watts or kilowatts first, then divide watts by 745.699872 to convert to mechanical horsepower. The result may represent hydraulic horsepower or pump shaft horsepower depending on which power stage you convert. Always state the power definition with the horsepower value.

Calculate hydraulic power in watts from ρgQH and divide by 1000 for kW. For pump shaft kW, divide hydraulic power by pump efficiency before converting to kW. If electrical motor input is required, also account for motor and any drive efficiency.

Use P_hyd = ρgQH for an incompressible liquid, with density in kg/m³, gravity in m/s², flow in m³/s, and head in metres. The result is watts of useful power transferred to the fluid. It does not include pump, motor, or drive losses.

Hydraulic power is useful power delivered to the fluid. Pump shaft power is mechanical input supplied to the pump. Because pumps are not perfectly efficient, shaft power is normally greater than hydraulic power. Pump efficiency is the ratio P_hyd/P_shaft.

Pump shaft power is mechanical power at the pump shaft. Motor electrical input is electrical power consumed upstream of motor losses. If a coupling or drive is present, motor shaft power can also differ from pump shaft power. The calculator separates these stages instead of treating them as one number.

For fixed hydraulic duty, lower pump efficiency requires more shaft input power because P_shaft = P_hyd/η_pump. A pump operating away from its best-efficiency region can therefore require more input power to deliver the same flow and head than a more efficient operating point.

At a fixed flow rate and fluid density, yes: hydraulic power is proportional to head through P_hyd = ρgQH. In a real pump, however, changing head generally changes the operating point and possibly the flow and efficiency, so the pump curve should be consulted.

At fixed head and density, hydraulic power rises in proportion to flow. In a real system, increasing flow also changes friction losses, pump head, and efficiency, so actual shaft power should be checked at the corresponding point on the manufacturer performance curve.

Yes. You can convert US GPM and feet to SI and use ρgQH, or use the conventional US water-horsepower relationship when appropriate. Be certain whether GPM means US gallons per minute, because Imperial gallons are a different volume and must not be mixed.

For water-like service, first combine flow in US GPM, head in feet, and specific gravity to estimate water horsepower. Divide by pump efficiency to estimate brake or pump shaft horsepower. Final shaft power should still be compared with pump manufacturer data at the actual operating point.

Water horsepower is the hydraulic power transferred to the liquid expressed in horsepower. A common US customary estimate is WHP = Q × H × SG / 3960, with Q in US GPM and H in feet. It does not include pump-efficiency losses.

Brake horsepower is commonly used for mechanical input power at the pump shaft. For a simplified pump calculation, BHP = WHP/η_pump. It is therefore greater than water horsepower when pump efficiency is below 100%.

Shaft horsepower is mechanical power transmitted through the pump shaft, expressed in horsepower. It is the appropriate input power basis for pump efficiency. Motor electrical horsepower or nameplate horsepower should not automatically be treated as the same quantity.

WHP represents useful hydraulic power delivered to the fluid. BHP represents mechanical power required at the pump shaft. Pump efficiency connects them: η_pump = WHP/BHP. Motor losses occur upstream of BHP and are therefore not included in pump efficiency.

One mechanical horsepower is approximately 0.745699872 kW. Other horsepower definitions exist, such as metric horsepower, so the conversion should state which definition is being used. This calculator uses mechanical horsepower by default.

You can calculate hydraulic power from flow, head, and density without knowing pump efficiency. To calculate required pump shaft input power, you need pump efficiency. Without it, you cannot determine how much shaft power is lost inside the pump before reaching the fluid.

Use the pump efficiency from the manufacturer performance curve at the actual operating flow, head, speed, and impeller configuration whenever possible. A generic efficiency assumption can introduce substantial error, especially when the pump operates far from its best-efficiency region.

Yes. For the same volumetric flow and head, hydraulic power is proportional to density. A denser liquid requires more hydraulic power than a lighter liquid at the same Q and H. Use density at the actual temperature and composition.

Yes. Specific gravity is a relative-density measure and can be used to estimate fluid density for pump-power calculations. In US water-horsepower formulas, SG appears directly. When more accurate density data are available, use those project-specific values rather than a simplified reference conversion.

Temperature changes water density and viscosity and can also change system and pump performance. Density directly affects hydraulic power, while viscosity can influence the pump curve and efficiency. For accurate work, use fluid properties and manufacturer data corresponding to the actual operating temperature.

Yes, it is well suited to preliminary centrifugal-pump power calculations. Final shaft and motor power should be checked at the actual flow-head operating point on the manufacturer curve, because pump efficiency and power can vary substantially across the performance range.

The required power depends on flow, head, fluid density, pump efficiency, speed, impeller configuration, and operating point. Calculate the hydraulic duty first, then divide by pump efficiency for shaft power. Use the manufacturer curve to verify the actual expected power demand.

Start with hydraulic power, divide by pump efficiency for pump shaft power, divide by drive or coupling efficiency if applicable, and divide by motor efficiency for electrical input. The resulting operating requirement is not automatically the final motor nameplate selection.

No. Calculated operating power is a duty-point requirement. Motor nameplate selection can require additional consideration of startup torque, service factor, overload conditions, operating range, control method, ambient conditions, electrical standards, and pump manufacturer recommendations.

Motor-sizing margin depends on the pump curve, service factor, expected operating range, startup and transient requirements, drive type, electrical code, and manufacturer guidance. This calculator intentionally does not apply one universal safety factor or automatically recommend a motor size.

Possible causes include lower pump efficiency, higher actual flow or head, higher fluid density or viscosity, wear, a different operating point, motor or drive losses, instrumentation error, speed changes, impeller differences, or system conditions that differ from the assumptions.

Actual power can be lower if the pump is operating at lower flow, lower head, different speed, smaller impeller diameter, lower-density fluid, or a different efficiency than assumed. Confirm the real operating point and compare it with the manufacturer performance curve.

Yes. Changing speed shifts the pump performance curve and can significantly change flow, head, and power. For many rotodynamic pumps, affinity-law relationships are useful within appropriate limits, but manufacturer data should be used for final predictions.

Yes. Changing or trimming impeller diameter alters the pump's head-flow and power characteristics. The exact effect depends on pump design and trim amount. Use manufacturer curves for the actual impeller diameter rather than assuming one fixed proportional change.

Pipe friction increases the system head required at a given flow. If the pump must provide more head to maintain the same flow, hydraulic power and therefore required shaft power increase. System pressure loss can be evaluated separately with the Pressure Drop Calculator.

Pump head is the energy added across the pump, while required system head can include pipe friction. In a design calculation, the pump must provide enough head to overcome the system's static, pressure, friction, equipment, and other energy requirements at the target flow.

The operating point is the flow and head where the pump performance curve intersects the system curve. Pump power, efficiency, and NPSHR should be evaluated at or near this actual operating point rather than using unrelated nameplate or design values.

The Best Efficiency Point is the region of the pump curve where pump efficiency is highest for the specified speed and configuration. BEP is an important reference for performance and reliability, but it is not automatically the maximum-flow point.

Yes. If two pumps have different efficiencies at the same hydraulic duty, the less efficient pump requires more shaft power. Motor and drive efficiencies can create further differences in electrical input even when hydraulic output is the same.

Internal hydraulic losses change as the operating point moves along the pump curve. Recirculation, incidence, leakage, friction, and flow separation vary with flow, so efficiency is not constant. Use the curve value for the actual duty point whenever possible.

Not as a direct substitute for actual pump shaft power. Nameplate horsepower is a motor rating, while the pump may draw less or more mechanical power depending on operating conditions. Use measured or curve-based shaft power when evaluating pump efficiency or hydraulic duty.

Motor electrical input is the power drawn from the electrical supply. Pump shaft power is the mechanical power reaching the pump. Motor and drive losses occur between those two points, so electrical input is normally higher than pump shaft power.

The equations are straightforward, but the result depends on input accuracy: flow, head, density, efficiency, motor and drive efficiencies, and the actual operating point. Field measurements, manufacturer curves, and consistent power definitions are important for high-confidence engineering results.

Common units are watts, kilowatts, megawatts, and horsepower. Internally, this calculator uses watts and converts displayed results. Always identify whether the number is hydraulic power, pump shaft power, motor shaft power, or motor electrical input.

This calculator is designed for positive pumping duty and positive power input. Signed energy-flow conventions can be used in broader hydraulic analysis, but negative pump power may indicate turbine operation, reversed reference direction, regenerative conditions, or a sign-convention issue that requires separate analysis.

At 100% pump efficiency, shaft power would equal hydraulic power in the idealized calculation. Real pumps have losses, so actual efficiency is below 100%. An entered 100% value is mathematically permitted here but should not be assumed realistic for physical pump selection.

Enter water flow, pump head, water density at the operating temperature, and pump efficiency. Calculate hydraulic power from ρgQH, then divide by efficiency for shaft power. The default water density is only an approximate reference near 20°C.

Multiply kilowatts by 1000 to obtain watts, then divide by 745.699872 for mechanical horsepower. The conversion does not change the power definition: hydraulic kW converts to hydraulic hp, while shaft kW converts to shaft hp.

Multiply mechanical horsepower by 745.699872 to obtain watts, then divide by 1000 for kilowatts. If a specification uses metric horsepower or another horsepower definition, use that definition's conversion instead of the mechanical-horsepower factor.

Yes. Viscosity can change the pump's achievable flow, head, efficiency, and required power. The simple hydraulic power equation uses density directly, but predicting an actual pump operating point for viscous fluids may require manufacturer viscosity corrections or more detailed pump modeling.

It estimates power for the entered duty and efficiencies. Actual operating power should be checked against the pump performance curve, system curve, speed, impeller configuration, motor/drive losses, fluid properties, and field measurements. The calculator does not independently solve the full operating point.

No. It can estimate operating hydraulic, shaft, and motor electrical power. Final motor selection requires consideration of nameplate ratings, overload margin, startup, service factor, controls, ambient conditions, code requirements, pump curve, and manufacturer recommendations.