Question

A fan draws air from the atmosphere through a 0.30-mdiameter round duct that has a smoothly rounded entrance. A differential manometer connected to an opening in the wall of the duct shows a vacuum pressure of 2.5 cm of water. The density of air is 1.22 kg/m3 . Determine the volume rate of air ow in the duct in cubic feet per second. What is the horsepower output of the fan?

Answer 1

Answer:

V = 50 ft³/s

H.P = 0.466

Explanation:

Given that

Diameter of the duct, D = 0.3 m

Vacuum Pressure of the duct, Z = 0.025 m

P(w) = pressure of water, 1000 kg/m³

P(a) = pressure of air, 1.22 kg/m³

To find the pressure change we use the formula

ΔP = P(w) * g * Z

ΔP = 1000 * 9.8 * 0.025

ΔP = 245 Pa.

We need the area, do we find that too

A = πd²/4

A = π * 0.3² * 1/4

A = 0.071 m²

Recall the energy equation to be

1/2v² = ΔP/p(a) , so that if we rearrange, we have

v² = 2ΔP/p(a)

v = √(2ΔP/p(a)), on substituting the values, we have

v = √(2 * 245)/1.22

v = √490/1.22

v = √401.64

v = 20.04

The volume flow rate is then equal to

Velocity * Area.

V = 20.04 * 0.071

V = 1.42 m³/s, converting to ft³/s, we have 50 ft³/s

Horsepower output is gotten using

P = ΔP * V

P = 245 * 1.42

P = 347.9 w, converting this to HP, we have 0.466 HP