Question

A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3 /s. determine the minimum power that must be supplied to the fan. take the density of air to be 1.18 kg/m3

Answer 1

Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate

The minimum power required (at 100% efficiency) is
$\frac{1}{2} (8 \, \frac{m}{s} )^{2}*(9 \, \frac{m^{3}}{s} )*(1.18 \, \frac{kg}{m^{3}}) = 339.84 \, W$

The actual power will be higher because 100% efficiency is not possible.

Answer: 339.8 W (nearest tenth)