Question

A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 10.62 kg/s. If 30 W of the supplied power is turned into heat, determine the minimum power that must be supplied to the fan.

Answer 1

Answer:

369.84 W

Explanation:

$\dot m$ = Mass flow rate = 10.62 kg/s

v = Velocity of wind = 8 m/s

Power is given by

$\dot W=\dfrac{1}{2}\dot mv^2\\\Rightarrow \dot W=\dfrac{1}{2}\times 10.62\times 8^2\\\Rightarrow \dot W=339.84\ W$

The power supplied is 339.84 W

If 30 W of this power is wasted that means 339.84+30 = 369.84 W is required