Answer:
speed is 200 rpm
power is 14614.1 W
Step-by-step 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