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# estimating fan pressure rise

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### David Walden

Sep 4, 2018, 2:27:08 PM9/4/18
to

Hello,
I am attempting to estimate the pressure rise generated by a centrifugal fan by first estimating the head (air column) produced by using equations for estimating head of a centrifugal water pump and substituting the density of air for that of water and plugging in other fan parameters found in Example 11.1 of Fluid Mechanics, 2nd Edition by Frank M. White, McGraw-Hill, pp. 640 - 641.
The pressure rise was estimated to be 51.53 psi, which seems awfully high to me.
Am I doing this correctly?

The power is computed thus:

P-sub-a = ρ*Q*u-sub-2*V-sub-t2 = (0.00237 [slugs/ft^3])*(32.34 [ft^3/s])*(324.63 [ft/s])*(11.84 [ft/s])
= 294.6 [ft-lbf/s]

Where

P-sub-a = power [ft-lbf/s]
ρ = density of air; std. atm. sea level = (0.00237 [slugs/ft^3])
Q = volume flow rate = (32.34 [ft^3/s])
u-sub-2 = outlet_tip_speed = (324.63 [ft/s])
V-sub-t2 = tangential component at outlet = (11.84 [ft/s])

H ≈ P-sub-a/ρgQ = 294.6 [ft-lbf/s]/((0.076314 lbf/ft^3)*(32.34 [ft^3/s])) = 119 ft

Where

g = 32.2 ft/s^2

The head can be converted to pressure thus:

p = 0.433 h SG = 0.433 * 119 * 1.000 = 51.53 psi

where

p = pressure (psi)

SG = specific gravity of the fluid = 1.000

David

### Wally W.

Sep 5, 2018, 12:11:54 AM9/5/18
to
On Tue, 4 Sep 2018 11:27:07 -0700 (PDT), David Walden wrote:

>
>
>
>Hello,
>I am attempting to estimate the pressure rise generated by a centrifugal fan by first estimating the head (air column) produced by using equations for estimating head of a centrifugal water pump and substituting the density of air for that of water and plugging in other fan parameters found in Example 11.1 of Fluid Mechanics, 2nd Edition by Frank M. White, McGraw-Hill, pp. 640 - 641.
>The pressure rise was estimated to be 51.53 psi, which seems awfully high to me.
>Am I doing this correctly?
>
>The power is computed thus:
>
>P-sub-a = ?*Q*u-sub-2*V-sub-t2 = (0.00237 [slugs/ft^3])*(32.34 [ft^3/s])*(324.63 [ft/s])*(11.84 [ft/s])
> = 294.6 [ft-lbf/s]
>
>Where
>
>P-sub-a = power [ft-lbf/s]
>? = density of air; std. atm. sea level = (0.00237 [slugs/ft^3])
>Q = volume flow rate = (32.34 [ft^3/s])
>u-sub-2 = outlet_tip_speed = (324.63 [ft/s])
>V-sub-t2 = tangential component at outlet = (11.84 [ft/s])
>
>
>H ? P-sub-a/?gQ = 294.6 [ft-lbf/s]/((0.076314 lbf/ft^3)*(32.34 [ft^3/s])) = 119 ft
>
>Where
>
>g = 32.2 ft/s^2
>
>The head can be converted to pressure thus:
>
> p = 0.433 h SG = 0.433 * 119 * 1.000 = 51.53 psi
>
>where
>
> p = pressure (psi)
>
>
> SG = specific gravity of the fluid = 1.000
>
>
>
>David

51.53 psi can be reasonable for a pump.

Ideally, a centrifugal pump is a constant-head machine.

119 ft of air would be a pressure of rho*g*h
= 0.075 lbf/ft^3 * 119 ft = 8.925 psf

8.925 psf * psi / 144 psf * 2.307 ft wc/psi * 12 in / ft = 1.7 " w.c.,
which seems like a reasonable fan pressure.

### David Walden

Sep 5, 2018, 7:14:05 PM9/5/18
to
"51.53 psi can be reasonable for a pump.

Ideally, a centrifugal pump is a constant-head machine.

119 ft of air would be a pressure of rho*g*h
= 0.075 lbf/ft^3 * 119 ft = 8.925 psf

8.925 psf * psi / 144 psf * 2.307 ft wc/psi * 12 in / ft = 1.7 " w.c.,
which seems like a reasonable fan pressure"

Outstanding! Thank you Wally W. I'm new to the forum so please forgive my lack of understanding the reply process.

It appears that I was off by a factor of 144! Thank you. 0.06 psi sounds (feels) better.

Thanks again!

David
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