Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.

Dismiss

4 views

Skip to first unread message

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])

The head is estimated thus:

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)

h = head (ft)

SG = specific gravity of the fluid = 1.000

Source: https://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

Thanks In Advance.

David

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])
>

>

>

>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:

>

> = 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])
>

>Where

>

>P-sub-a = power [ft-lbf/s]

>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])

>

>The head is estimated thus:

>

>H ? P-sub-a/?gQ = 294.6 [ft-lbf/s]/((0.076314 lbf/ft^3)*(32.34 [ft^3/s])) = 119 ft
>u-sub-2 = outlet_tip_speed = (324.63 [ft/s])

>V-sub-t2 = tangential component at outlet = (11.84 [ft/s])

>

>The head is estimated thus:

>

>

>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)

>

> h = head (ft)

>

> SG = specific gravity of the fluid = 1.000

>

>Source: https://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

>

>Thanks In Advance.

>

>David

51.53 psi can be reasonable for a pump.
>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)

>

> h = head (ft)

>

> SG = specific gravity of the fluid = 1.000

>

>Source: https://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

>

>Thanks In Advance.

>

>David

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.

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.
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"

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

Thanks again!

David

0 new messages

Search

Clear search

Close search

Google apps

Main menu