Fan efficiency calculation
Static air power is Q × Δpₛ, where Q is airflow in m³/s and Δpₛ is fan static pressure in Pa. The result is watts. Static efficiency is that air power divided by shaft power, multiplied by 100 for a percentage.
Keep the units consistent. Divide L/s by 1,000 to obtain m³/s, and multiply kW by 1,000 to obtain watts. Use shaft power at the selected operating point, not the motor’s nameplate rating. A motor rated at 1.1 kW does not mean the fan always absorbs 1.1 kW.
The pressure definition matters too. Read static pressure vs total pressure before mixing a duct pressure measurement with a fan curve.
Worked example: static air power and shaft input
Take an illustrative duty of 500 L/s at 400 Pa fan static pressure, with 500 W absorbed at the shaft. Airflow is 0.5 m³/s, so static air power is 0.5 × 400 = 200 W. Static efficiency is 200 ÷ 500 × 100 = 40 per cent.
| Input | Value |
|---|---|
| Airflow | 0.5 m³/s |
| Fan static pressure | 400 Pa |
| Static air power | 200 W |
| Shaft power | 500 W |
| Static efficiency | 40% |
If the electrical input were 625 W at that same duty, the static air-power-to-electrical-input ratio would be 32 per cent. The different result reflects a different measurement boundary, not an arithmetic disagreement.
Fan total efficiency and electrical efficiency
Total efficiency uses fan total pressure in the air-power calculation. Static efficiency uses fan static pressure. Report which one is being quoted and compare like with like.
A complete electrical assessment includes losses in the motor and any drive or controller. Transmission losses also matter where the fan uses a belt arrangement. The useful comparison is the input required to deliver the specified duty under the stated conditions.
Peak efficiency elsewhere on a curve is useful context, but it does not tell you the efficiency at your operating point. Keep the airflow, pressure, speed and power values together rather than selecting a fan from a single headline percentage.
Find an efficient duty without losing required extraction
Plot the design duty on the manufacturer’s performance data. Review the permitted range, power requirement and the likely operating modes. A fan that runs well near one point may behave differently when branches close or controls reduce airflow.
Lower pressure losses can reduce the duty the fan must provide. Suitable duct sizing, transitions and inlet conditions can therefore matter as much as the fan itself. Review fan system effect before expecting a catalogue efficiency to describe an awkward installation.
Reducing airflow simply to improve a power figure is not a design objective for hazardous extraction. The required capture and containment performance comes first. Any control change needs the appropriate assessment and verification.
Estimate energy using input power and operating time
Electrical energy in kWh is electrical input in kW multiplied by operating hours, when the input is reasonably constant. For varying duties, calculate each operating period separately and add the energy. Use measured or selected electrical input, not the shaft-power figure without accounting for the drive chain.
For example, an illustrative 0.625 kW electrical input for 100 hours uses 62.5 kWh. Converting that to cost needs the applicable tariff and billing terms. We do not need an assumed electricity price to compare two designs on the same operating schedule.
Record the assumptions so the estimate can be checked. Our fan affinity laws guide explains why speed changes can materially change shaft power, and where the simple ratios stop being a whole-system prediction.
Questions about fan static efficiency
Can I calculate efficiency from the motor nameplate alone?
No. You need airflow, the correct pressure quantity and actual or selected power at the operating duty. The nameplate rating is not a measurement of absorbed power.
Is the highest peak efficiency always the best selection?
Compare the proposed fans at the duty your system needs. Chemical compatibility, the stable operating range, controls and installation conditions also affect the selection.