Pump Efficiency from Hydraulic and Shaft Power
Worked example: 15 kW delivered from 20 kW at the shaft → eta = 0.75 — press Try an example to run it live, then adjust anything.
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Pump Efficiency from Hydraulic and Shaft Power explained
Everything a pump fails to put into the water it puts into heat, noise and vibration. A unit delivering 15 kW of hydraulic power while absorbing 20 kW at the shaft is 75% efficient, and the missing 5 kW is warming the casing and the liquid. Large water-supply pumps reach 85–90%; small close-coupled circulators struggle to pass 40%; and any positive-displacement chemical metering pump is efficient in a different sense entirely.
The trap is comparing efficiencies measured at different boundaries. Pump efficiency is hydraulic over shaft; wire-to-water efficiency multiplies that by the motor and drive efficiencies and is the number that matters on the electricity bill. A 75% pump on a 92% motor behind a 97% VFD is 67% wire-to-water — so if you are calculating operating cost from a "75% efficient" pump, you will underestimate the annual energy by about 12%.
Pump Efficiency from Hydraulic and Shaft Power formula
- = Pump efficiency
- = Hydraulic power (W)
- = Shaft power (W)
Missing one of these? Work it out first, then come back
- Pump efficiency — Pump Brake Horsepower, Fan Brake Horsepower
- Hydraulic power — Hydraulic Power (P = ρgQh), Pump Affinity Law — Power vs Speed
- Shaft power — Steam Turbine Power Output, Pump Affinity Law — Power vs Speed