Engineering question
How much power does a pump duty require, and which efficiency values belong in hydraulic, shaft and electrical calculations?
Hydraulic power is the rate of energy delivered to the fluid. Shaft power is higher because the pump is not perfectly efficient, and electrical input is higher again after motor and drive losses. Keeping these three quantities separate avoids undersizing and makes an energy balance easier to audit.
The calculation requires flow and total dynamic head at the actual operating point. Static lift alone is not total head; friction, pressure requirements and fluid conditions matter. Pump efficiency should come from the pump curve near the duty point, not from a generic fixed assumption used for final selection.
Calculation basis
Formulas and units
Hydraulic power
Phyd = ρgQH
Use kg/m³, m/s², m³/s and metres to obtain watts.
Shaft power
Pshaft = Phyd / ηpump
Pump efficiency must correspond to the selected operating point.
Electrical input
Pelec = Phyd / (ηpump × ηmotor × ηdrive)
Include drive efficiency only when a drive is in the power path.
Worked example
Apply the formula
Water at 100 m³/h and 30 m total head; pump 75%, motor 90%, VFD 97% efficient.
- 1Q = 100/3600 = 0.02778 m³/s.
- 2Phyd = 1000 × 9.81 × 0.02778 × 30 ≈ 8.175 kW.
- 3Pshaft = 8.175/0.75 = 10.90 kW.
- 4Pelec ≈ 8.175/(0.75 × 0.90 × 0.97) = 12.49 kW.
Result: The screened electrical input is about 12.5 kW at this duty. Final motor and VFD selection needs curve and service data.
Open Pump Power CalculatorDetermine the real duty point
- Build the system curve including static and friction head.
- Plot the duty against the pump curve and allowable operating region.
- Check NPSH available against NPSH required with margin.
- Review minimum flow, maximum flow, runout and shutoff conditions.
- Correct for density, viscosity, temperature, solids and gas content.
- Check motor power across the full operating envelope.
Control and energy use
A VFD can reduce throttling losses in variable-flow systems, but savings depend on the system curve and operating schedule. The affinity laws are useful for the same pump near similar efficiency: flow varies approximately with speed, head with speed squared and power with speed cubed. Static head and changing efficiency reduce the accuracy of a simple cube-law estimate.
Common mistakes
- Using m³/h without dividing by 3600
- Using static lift as total head
- Multiplying by efficiency instead of dividing
- Selecting the motor only from one nominal duty point
Troubleshooting checks
- Measured power high: verify flow/head, efficiency, impeller, valve position and mechanical condition.
- Low flow: compare rotation, system resistance, suction condition and pump curve.
- Cavitation signs: review NPSH, suction losses, temperature and air ingress.
- VFD savings disappoint: separate static head from friction head and review the speed schedule.
Assumptions
- Steady incompressible flow
- Total dynamic head is known
- Efficiencies represent the actual duty point
Limitations
- Not a pump, motor or VFD selection
- NPSH, transients, viscosity and solids are not calculated
- Affinity-law energy estimates require a system-specific review
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