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Fan Brake Horsepower

BHP=Q⋅SP6356 ηBHP = \frac{Q \cdot SP}{6356 \, \eta}

Worked example: 2 m3/s against 500 Pa at 65% efficient → 1538.46 W — press Try an example to run it live, then adjust anything.

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Fan Brake Horsepower explained

QSPBHPη

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 formula

BHP=Q⋅SP6356 ηBHP = \frac{Q \cdot SP}{6356 \, \eta}
Where
  • BHPBHP= Fan brake power (W)
  • QQ= Airflow (L/min)
  • SPSP= Fan total pressure (kPa)
  • η\eta= Fan efficiency

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