EngiMetric Arena / Spot the Error
A 15 kW motor for a 120 m3/h pump: hydraulic or input power?
Competency Quantitative reasoning primary Systems and balances supporting
What you should take away: Separate hydraulic power, shaft power and motor input power, and recognise that motor sizing must include both efficiencies.
The situation
What is happening
A pump delivers 120 m3/h against a total dynamic head of 42.9 m. The pump efficiency is 75 % and the motor efficiency is 92 %. The reviewer specifies a 15 kW motor because the hydraulic power is about 14 kW. What calculated motor input power does the review establish before selecting a standard motor?
Given
| Quantity | Value | Unit | Note |
|---|---|---|---|
| Flow rate | 120 | m3/h | Design flow. |
| Total dynamic head | 42.9 | m | Static, friction, minor and residual head. |
| Pump efficiency | 75 | % | Hydraulic-to-shaft efficiency. |
| Motor efficiency | 92 | % | Shaft-to-electrical input efficiency. |
| Reviewer selection | 15 | kW | Based on hydraulic power only. |
Assumptions this challenge makes
- The stated total dynamic head already includes static, friction, minor and residual pressure head.
- Pump and motor efficiencies are operating-point values, not catalogue maxima.
- The answer is the calculated motor input, not the next standard motor rating.
Your call
Your answer
Hints
- Compute hydraulic power from flow, head and water density first.
- Divide hydraulic power by pump efficiency to get shaft power. Do not stop there.
- Divide shaft power by motor efficiency to get the motor input. The answer is before standard-size selection.
Your answer Engine result
Where that comes from
Each limit, checked
Why
- The engine confirms a total dynamic head of 18 + 7.5 + 2.1 + 15.2958 = 42.8958 m.
- Hydraulic power is about 14.02 kW. Dividing by 0.75 gives a shaft power of about 18.70 kW.
- Dividing shaft power by 0.92 gives a motor input of about 20.32 kW.
- The reviewer's 15 kW is below even the shaft power, so it would overload the motor. A standard 22 kW motor is the next selection above 20.32 kW, but that selection is not the calculated input.
- The engine reports hydraulic, shaft and motor power separately so the three bases cannot be confused.
What this challenge does not establish
- The efficiencies are single operating-point inputs; real pump and motor efficiency vary with load.
- The calculation does not model viscosity, solids, suction conditions or motor service factor.
- Motor starting, voltage dip, altitude derating and duty cycle are separate checks.
- The result is a calculated input, not a guarantee that a particular motor will deliver the duty.