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Pump Hydraulic Power Calculator

Convert cubic metres per hour to cubic metres per second, calculate rho-g-Q-H hydraulic power, then move upstream through pump and motor efficiencies to shaft and electrical input power.

Volumetric flow (m³/h)Total dynamic head (m)Fluid density (kg/m³)Pump efficiency (%)Motor efficiency (%)Operating hoursEnergy rate per kWh

Flow in m³/s (m³/s)-

Hydraulic power (kW)-

Pump shaft input power (kW)-

Electrical motor input (kW)-

Combined pump and motor efficiency-

Energy over entered hours (kWh)-

Energy cost over entered hours-

Pump and motor losses (kW)-

Decision view

Pump and motor energy-flow losses

Pump and motor energy-flow lossesElectrical input separates into motor loss, pump loss, and useful hydraulic output on one proportional flow.

Exact scenario comparisonTotal dynamic head (m) changes while all other entered assumptions remain constant.

Total dynamic head (m)	Flow in m³/s (m³/s)	Hydraulic power (kW)	Pump shaft input power (kW)	Electrical motor input (kW)	Combined pump and motor efficiency	Energy over entered hours (kWh)	Energy cost over entered hours	Pump and motor losses (kW)

How to use Pump Hydraulic Power Calculator

Enter volumetric flow, head, and fluid density.
Enter pump and motor efficiencies.
Choose operating hours and energy rate.
Read the conversion chain from electrical input to hydraulic output.

Calculator guide

Understanding Pump Hydraulic Power Calculator

A pump transfers hydraulic power to fluid, while the shaft and electrical input must also cover pump and motor losses. This calculator keeps the flow conversion, hydraulic duty, efficiency stages, energy, and cost traceable.

Convert flow first m3/h must become m3/s.

Output is smallest Losses make upstream inputs larger.

Efficiencies multiply Combined efficiency is eta_p times eta_m.

Cost uses input Electrical kWh drives the bill.

Calculation method

How the calculation works

Apply density, gravity, flow, and head to obtain hydraulic power, then divide by explicit pump and motor efficiencies to reach electrical input. Convert flow to SI seconds, multiply density, gravity, flow, and head for hydraulic watts, divide by pump efficiency for shaft power, and divide again by motor efficiency for electrical input.

Detailed calculation process

Move upstream from fluid duty to electrical input

The defaults move 85 m3/h of 998 kg/m3 fluid through 42 m head at 72% pump and 92% motor efficiency.

General formula: Q_s = Q_h/3600; P_h = rho g Q_s H/1000; P_shaft = P_h/eta_p; P_elec = P_shaft/eta_m; eta_total = eta_p eta_m Hydraulic power is useful fluid output. Each upstream input is larger because it must cover the loss at its conversion stage.

What each symbol means

Q_h / Q_s Flow in m3/h and converted m3/s.

rho Fluid density, measured in kg/m3.

g Standard gravitational acceleration, 9.80665 m/s2.

H Total dynamic head, measured in metres (m).

eta_p / eta_m Pump and motor efficiency fractions.

P_h / P_shaft / P_elec Hydraulic, shaft, and electrical power in kW.

Worked substitution with the default inputs

1. Convert flow: Q_s = 85/3600 = 0.0236111 m3/s The SI power equation requires flow per second.

2. Calculate hydraulic power: P_h = 998 x 9.80665 x 0.0236111 x 42 /1000 = 9.70548 kW This is the useful rate of energy transferred to the fluid.

3. Account for pump loss: P_shaft = 9.70548/0.72 = 13.47983 kW The pump shaft must supply hydraulic output plus pump loss.

4. Account for motor loss: P_elec = 13.47983/0.92 = 14.65199 kW; eta_total = 0.72 x 0.92 = 66.24% The combined efficiency links electrical input directly to hydraulic output.

5. Reconcile period energy and cost: E = 14.65199 x 3000 = 43,955.97 kWh; cost = 43,955.97 x 0.13 = 5,714.28; loss = 4.94651 kW Total modeled loss equals electrical input minus hydraulic output.

The default duty requires 9.705 kW hydraulic output, 13.480 kW shaft power, and 14.652 kW electrical input at 66.24% combined efficiency.

Energy conversion

Expose pump loss and motor loss separately

A proportional energy-flow view shows where input power goes.

Electrical input Power purchased.

Motor loss Difference from shaft power.

Pump loss Difference from hydraulic power.

Fluid duty Useful output.

Worked situations

Practical examples

85 m3/h converts to 0.023611 m3/s.
The hydraulic duty is 9.705 kW.
Combined pump and motor losses total 4.947 kW.

Better inputs

Useful tips

Use total dynamic head at the actual operating point.
Use efficiency at the same flow and head.
Check motor loading as well as annual energy.

Before relying on the result

Limitations and common mistakes

The model assumes steady incompressible flow.
System curve, NPSH, viscosity, solids, impeller trim, and variable-speed behavior are excluded.
Part-load and service-factor decisions require manufacturer data.

Reference

Key terms

Hydraulic power: Useful energy rate delivered to the fluid.
Shaft power: Mechanical input to the pump.
Total dynamic head: Energy per unit fluid weight expressed as metres.

Important note

Calculated from the entered values using the displayed engineering relationship. Confirm design values, load cases, safety factors, standards, and field conditions with a qualified professional.

Frequently asked questions

Why use total head instead of pressure alone?

Head expresses the complete energy rise per unit fluid weight for the pump equation.

Why divide by efficiency?

Input must be larger than useful output when efficiency is below one.

Does higher density increase power?

Yes, at the same volumetric flow and head.

Does this select a pump?

No. A pump curve, system curve, NPSH, materials, and operating envelope are also needed.