Fan Power
Power a fan needs to move air against a pressure drop.
Formula
Variables
Example
Q=2, Δp=300, η=0.7: P = 857 W
How it works
A fan does work on air by increasing its pressure, and the useful power required is the rate of that work. The airflow rate Q tells you how much air is being moved each second, while the pressure rise Δp tells you how much energy per cubic metre is added to that air; multiplying them gives the air power in watts. Because real fans are not perfectly efficient, the input power must be higher than the air power by a factor of 1/η.
When to use it
Use this when sizing or checking a fan for ductwork, ventilation, exhaust, or cooling systems where a known airflow must be delivered against a specified pressure drop. It is also useful for estimating electrical demand from fan performance data and comparing different fan selections.
Worked example, step by step
Given: Q = 1.5 m³/s, Δp = 480 Pa, η = 0.8 Step 1: Multiply airflow by pressure rise: Q × Δp = 1.5 × 480 = 720 W Step 2: Divide by efficiency: P = 720 / 0.8 = 900 W Final answer: P = 900 W
Common mistakes to avoid
- Using kPa instead of Pa for pressure without converting, which makes the power 1000 times too large
- Forgetting to divide by efficiency, which gives only the ideal air power, not the required input power
- Mixing up airflow units such as m³/s and m³/min without converting first
Did You Know?
Fan power rises with the cube of speed, so trimming airflow 20% can cut power roughly in half.
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