Fan Brake Horsepower

BHP=QSP6356ηBHP = \frac{Q \cdot SP}{6356 \, \eta}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Air power is flow times pressure rise, exactly as hydraulic power is flow times head — and the fan's efficiency turns that into shaft power. The trade constant 6356 converts cfm × inches of water gauge into horsepower (1 in w.g. is 5.202 lbf/ft², and 33 000 ft·lbf/min is 1 hp). Move 2 m³/s against 500 Pa through a 65%-efficient fan and you need 2 × 500 / 0.65 ≈ 1.54 kW at the shaft. This page evaluates Q·Δp/η in SI, so any pressure and flow units work.

Two subtleties bite. First, use fan total pressure, not static, if you want honest total efficiency — static-pressure numbers paired with total-efficiency values overstate the fan. Second, the constant assumes standard air at 1.2 kg/m³; at 5000 ft of altitude air is about 17% thinner, so a fan moving the same cfm develops proportionally less pressure and needs proportionally less power, which is why altitude derating tables exist for every rooftop unit sold.

Fan Brake Horsepower
BHP=QSP6356ηBHP = \frac{Q \cdot SP}{6356 \, \eta}
Where
  • BHPBHP= Fan brake power
  • QQ= Airflow
  • SPSP= Fan total pressure
  • η\eta= Fan efficiency
Missing one of these? Work it out first, then come back