Pump Efficiency
Comparing the hydraulic work a pump does with the power it consumes.
Why this matters
Where it occurs
Maintenance teams compare pump power draw with flow and pressure to judge whether a pump is healthy, worn or throttled.
Why calculate it
Efficiency converts electrical bills into a machine-health signal: a falling ratio means internal wear, not cheaper power.
What decision it supports
Overhaul the pump? Re-trim the impeller? Or is the pump simply oversized for its duty?
What happens if it is wrong
Running a worn pump wastes energy month after month, while a throttled oversized pump is burning money by design.
The concept
A pump's job is hydraulic power: moving fluid against pressure. Hydraulic power is Q (m³/s) x ρ x g x h (head in m).
Efficiency compares that useful power with what the drive consumes: η = P_hydraulic / P_input. Typical pump-plus-motor combinations run 65-75%.
A falling efficiency at the same flow means the pump is working harder internally - the signature of worn impellers and open clearances.
Formula
Pump Efficiency
eta = P_hydraulic / P_input
- η = Efficiency (fraction or %)
- P_hydraulic = Hydraulic power delivered (kW)
- P_input = Power consumed (kW)
Worked example
Problem: A pump delivers 5 kW of hydraulic power while drawing 7.5 kW. What is its efficiency?
- Use η = P_hydraulic / P_input.
- Substitute: η = 5 / 7.5.
- Calculate: η = 0.667.
- Express as a percentage: 66.7%.
66.7% - healthy territory for a pump with its motor.
Quick solve: practice it
A pump moves 0.02 m³/s of water against 30 m of head. What hydraulic power does it deliver?
Quick-solve habit: calculate, then ask yourself - does this answer make sense in real units?
Games that use this mathematics
Apply the calculation inside a workplace simulation.