A centrifugal Blower delivers 6m^3 of air per second against a head of 55mm of alcohol of relative density 0.79 when running at 400 RPM. Calculate the power required if the efficiency os 0.7. Estimate the discharge of a blower of similar design of twice the diameter when discharging against a head of 55mm of alcohol. Calculate the operating speed and the expected power. Air Density is 1.17Kg/m^3

speed is 200 rpm

power is 14614.1 W

Explanation:

given data

discharge Q = 6 m³

head H = 55 mm

running s = 400 rpm

relative density R = 0.790 Kg/m³

efficiency e = 0.7

Air Density = 1.17 Kg/m³

to find out

speed and the expected power

solution

we will apply here 1s power formula that is

power = Q * R * g * H / e ... 1

put here all value

power = 6 * 790 * 9.8 * 0.055 / 0.7

power = 3653.52 W

and we know discharge is directly proportional to square of diameter

so

Q1/Q2 = D1²/D2² ... 2

here we know blower design twice the diameter so D2 is 2D

6/Q2 = D² / (2D) ²

so Q2 = 24 m³/s

so speed is

D1 * S = D2 * S1

D * 400 = 2D * S1

S1 = 200

speed is 200 rpm

and

power is

P1/P2 = Q1/Q2

3653.52 / P2 = 6 / 24

P2 = 14614.1

so power is 14614.1 W