Question

Your boss ask you to size a fun to transfer 50000 SCFM of air to ventilate a building 500 ft from the intended location of the fun. However, your boss has found an old fan and would like to reuse it if the specifications are compatibles with the project need. The identification plate shows: "design flow: 50000 SCFM and motor power 50 kW" Unfortunately, the value of the static pressure is unreadble because the identification plate is worn. Assuming that the friction loss in the pipe is one inch of water column, what would be the required power of the fun? mechanical efficience: 65% and 90% - fun and motor.

Answer 1

Given,

Q = 50000 ft³/min = 23.6 m³/s

h = 500 ft = 152.4 m

Power of old fan P_o = 50 kW

h_L = 1 inch of water = 0.0254 m of water

= 65 %

= 90 %

Assuming density of air = 1.225 kg/m³

For Old Fan and Motor:

Input power to the old fan = * P_o

                                         = 0.9 * 50 = 45 kW

Output power to the old fan = * Input power to the old fan

                                            = 0.65 * 45 = 29.25 kW                                           ......(i)

Power required as per project:

Head loss is given in terms of in of water. It has to be converted in m of air or simply m.

Hydraulic Power is given by

H = total head required = h + h_L

Required fan Power,

Required power is greater than the Power output of the old fan.

Hence, the old motor and fan can't be used for the project.