Compute hydraulic power, shaft power, and overall efficiency for pumps. Understand performance curves, energy savings, and best practices for centrifugal and positive displacement pumps.
Pump efficiency is a measure of how effectively a pump converts mechanical energy (from a motor) into hydraulic energy (flow and pressure). It is a critical factor in energy consumption, operating cost, and system sustainability.
Phyd = ρ · g · Q · H
η = Phyd / Pshaft
where:
Hydraulic power represents the energy per unit time imparted to the fluid. It depends on flow rate and head. For water (ρ = 1000 kg/m³), a common simplification is: Phyd (kW) = Q (m³/s) × H (m) × 9.81. When using mixed units, conversions are essential – this calculator handles unit conversions automatically.
Shaft power is the mechanical power delivered to the pump shaft, usually from an electric motor or engine. The difference between shaft power and hydraulic power represents losses inside the pump (hydraulic, mechanical, and volumetric losses).
Overall efficiency is the product of three internal efficiencies:
| Pump Type | Typical Efficiency Range | Best Efficiency Point (BEP) |
|---|---|---|
| Centrifugal (end suction) | 60% – 85% | ~80% |
| Centrifugal (split case) | 70% – 90% | ~85% |
| Axial flow | 70% – 85% | ~80% |
| Positive displacement (gear) | 75% – 90% | ~85% |
| Positive displacement (piston) | 85% – 95% | ~90% |
| Submersible pump | 60% – 80% | ~70% |
Note: Efficiency varies with operating point. Operating a pump far from its Best Efficiency Point (BEP) can significantly reduce efficiency and increase wear.
The blue curve shows head vs. flow; the orange dashed curve shows efficiency. The red circle marks the Best Efficiency Point (BEP). Operating within the shaded region ensures optimal performance.
A water treatment plant uses a centrifugal pump rated at 100 m³/h, 20 m head, with a measured shaft power of 7.5 kW. The calculated hydraulic power is 100 m³/h = 0.02778 m³/s → Phyd = 1000×9.81×0.02778×20 = 5449 W ≈ 5.45 kW. Efficiency = 5.45/7.5 = 72.7%. If the pump operates 4000 h/year and electricity costs $0.10/kWh, the annual energy cost is 7.5×4000×0.10 = $3000. Improving efficiency to 80% would reduce required shaft power to 5.45/0.8 = 6.81 kW, saving $276 per year. Over the pump’s life, this justifies impeller trimming or variable speed drive investment.