Fan power calculation: enter a complete duty
Enter the required airflow in litres per second and the fan pressure rise in pascals. Choose a fan efficiency that uses the same pressure basis: total pressure with total efficiency, or static pressure with static efficiency. Enter motor and drive efficiencies as percentages. The result changes as you edit the inputs.
The starting values are an arithmetic example, not a Fumeflow performance rating. If you have a complete measured wire-to-air efficiency, air power divided by that efficiency gives input power directly. Do not apply motor and drive losses again when they are already included.
Illustrative inputs. This calculation does not select equipment or establish a safe operating limit.
Air, shaft and electrical power
Air power is the useful rate of energy transfer represented by airflow multiplied by pressure. In SI units, cubic metres per second multiplied by pascals gives watts. Divide L/s by 1,000 before using this equation.
- Air power: Pair = Q × Δp.
- Fan shaft power: Pshaft = Pair ÷ ηfan.
- Electrical input: Pinput = Pshaft ÷ (ηmotor × ηdrive).
Each efficiency in the equations is a fraction, so 60% becomes 0.60. A drive allowance can include the transmission or controller losses relevant to the chosen arrangement. AMCA's fan power map distinguishes the fan, transmission, motor and control boundaries and uses efficiencies at the required speed and load.
Fan shaft power calculation: a worked example
For an illustrative duty of 500 L/s at 300 Pa, Q is 0.5 m³/s. Air power is 0.5 × 300 = 150 W. With 60% fan efficiency, shaft power is 150 ÷ 0.60 = 250 W. At 85% motor efficiency and 95% drive efficiency, electrical input is about 310 W, or 0.310 kW.
A motor nameplate rating describes its rated output capability under specified conditions. It is not a measurement of the electricity used at this duty. Selection also checks the fan's power requirement across its permitted operating range, starting arrangements and the motor's rating conditions. Our fan-curve guide explains where the proposed duty sits.
Energy use and specific fan power
For a constant operating point, energy in kWh equals electrical input in kW multiplied by running hours. The example above uses about 3.10 kWh over ten hours. Multiply by your own electricity tariff to compare operating costs. A variable-duty system needs each operating mode calculated separately and added together.
For the same example, dividing 310 W by 500 L/s gives about 0.62 W/(L/s). That is a specific-fan-power comparison for the stated boundary, not a minimum safety airflow or a compliance result. Use our specific fan power guide when comparing complete system duties.
What to check before choosing a motor
Pressure losses from ductwork, bends, silencers and other components all affect the duty. Installation details can also change the achieved performance. Confirm the required airflow, air density and temperature, chemical compatibility, pressure basis, rotation and connection arrangement with the designer.
A speed change needs a fresh operating-point check. The fan affinity laws can help compare corresponding points, but they do not replace the manufacturer's performance curve. For unusual gas conditions or significant compression, use the applicable engineering calculation rather than this low-pressure air estimate.
Your questions
Can I calculate fan power from airflow alone?
You also need the pressure duty and the relevant efficiency. Two systems moving the same airflow can require different power because their resistance differs.
Is calculated input power the motor size?
No. Motor selection also considers the fan's operating range, starting conditions and the motor's permitted duty. Use the calculation as one input to that selection.