Question

A fan draws air from the atmosphere through a 0.30‐m‐diameter 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 flow in the duct in cubic feet per second. What is the horsepower output of the fan?

Answer 1

Answer:

The volume air flow rate is $47.92 ft^{3}/s$

The output of the fan is 0.446 hp

Solution:

As per the question:

Diameter of the round duct, d = 0.30 m

Radius of the duct, R = $\frac{d}{2} = \frac{0.30}{2} = 0.15 m$

Pressure of the opening, P = 2.5 cm of water = 0.025 m of water

P = $0.025\times 9.8\times 1000 = 245 Pa$

Density of air, $\rho_{a} = 1.22 kg/m^{3}$

Now, the velocity, v can be calculated as:

$\Delta P = \frac{1}{2}\rho_{a} v^{2}$

$v = \sqrt{\frac{2\Delta P}{\rho_{a}}$

$v = \sqrt{\frac{2\times 245}{1.22} = 19.212 m/s$

Now,

Volume rate of air flow is given by:

$V_{f} = Av = \pi R^{2}\times v$

$V_{f} = Av = \pi (0.15)^{2}\times 19.21 = 1.357 m^{3}/s$

Now,

1 ft = 3.2808 m

$V_{f} = 1.357\times (3.2808)^{3} = 47.92 ft^{3}/s$

Now, the output of the fan in horsepower (hp):

Power output, P' = $\Delta P\times V = 245\times 1.36 = 333.2 W$

Now,

1 hp = 746 W

P' (in hp) = $\frac{P'}{746} = 0.446 hp$