Pump Efficiency Calculator (Electric Motor Driver)
Determine a motor-driven centrifugal pump’s efficiency from motor power, discharge and suction pressure, flow, and motor efficiency. Enter any five of the six variables and the calculator returns the sixth.
The fluid (hydraulic) power a pump adds equals flow × pressure rise, so no specific gravity is needed — the pressure already carries the fluid density. The shaft (brake) power is the measured electrical input to the motor multiplied by the motor efficiency, and pump efficiency is fluid power ÷ shaft power. K = 3960 ÷ 2.31 = 1714 for Q in gpm, ΔP in psi (power in hp), or 367 ÷ 10.2 = 35.98 for Q in m³/h, ΔP in bar (power in kW). Assumes all readings are at the same operating point (flow, pressures, speed), the motor efficiency at that load is known, and no viscosity correction is required.
For preliminary and educational use. Field efficiency depends on accurate, simultaneous readings of power, pressures, and flow at one operating point; verify against the manufacturer performance curve.
You can measure a running pump’s efficiency without ever opening it up — you only need what a power meter and two pressure gauges already tell you. This calculator turns the electrical power drawn by the motor, the pressure the pump develops, and the flow it moves into a pump efficiency, and it runs the relationship in any direction: enter any five of the six variables and it returns the sixth.
How it works
Two power figures bracket the pump. The fluid (hydraulic) power it actually delivers is the flow times the pressure rise, Q × (Pd − Ps) — notice there is no specific gravity term, because pressure already carries the fluid’s density. The shaft (brake) power going into the pump is the electrical power the motor draws multiplied by the motor’s efficiency at that load. Pump efficiency is simply the first divided by the second:
η_pump = [Q × (Pd − Ps) / K] / (motor power × η_motor), where K = 3960 ÷ 2.31 = 1714 for US units (gpm, psi, hp) or 367 ÷ 10.2 = 35.98 for metric (m³/h, bar, kW).
Worked example
A motor drawing 35 hp at 92% efficiency delivers 32.2 hp to the shaft. If the pump moves 500 gpm while raising pressure from 15 to 95 psi (ΔP = 80 psi), the fluid power is 500 × 80 / 1714 ≈ 23.3 hp. Its efficiency is therefore 23.3 / 32.2 ≈ 72%. Solve the other way to size a motor for a target efficiency, or to back out the efficiency implied by a field power reading.
Assumptions and cautions
All readings are taken at the same operating point — the same flow, pressures, and speed.
The motor efficiency at that load is known (part-load efficiency differs from nameplate/full-load).
No viscosity correction is applied; for viscous fluids the pump’s performance shifts and a correction is needed.
A result above 100% is physically impossible and means the inputs are inconsistent — usually the flow, pressures, or motor power are mis-read.
