Velocity Pressure 4005

[wufi2004]

I know Pv = (V/4005)^2 in English Unit.

Who can tell me how to get the 4005? I need the process of transform.

 

[MintJulep]

Pv is in inches of water

V is in feet per minute.

Density of air = .075 lb/ft^3 is built in. More general formula is

V = 1097 * Sqrt(Pv/d)

where d is density.

Throw enough 12's and 60's at it and you're there.

 

[wufi2004]

Thank! But where's 1097 from?

 

[imok2]

The 1097 is the average velocity at the traverse plane. In order to use this constant one has to meet certain qualifacations, otherwise the final results could be bogus!
Example: For a duct with a 28-by-20-in. airway, the Log-Tchebycheff method calls for a five-by-five grid
of unequally spaced measurements, while the Equal Area method requires a five-by-four grid with the distance between measurements no more than 6 in. To read more goto

http://www.hpac.com/member/archive/pdf/2001/0301/Klaassen.

pdf

 

[quark]

thread100-92174

 

[lilliput1]

The velocity pressure expressed in Ft of the fluid flowing =
V^2/2g (Equation 1)

Where V = Ft/sec
g = 32.1722 Ft/Sec^2

But Ft of Air = ((in. wg) x 62.3215 ) / (12 x .075) (Equation 2)

Where 62.3215 lb/ft^3 is density of water

12 = inches/ft

.075 = lb/ft^3 density of air

and in. wg. = inches of water gauge

Equate Equation 1 to Equation 2 & solve for In. wg.
and get 4005 = 60 x((2 x 32.1722 x 62.3215) / (12 x .075))^0.5

 

[25362]

To lilliput1, excellent presentation. I think, however, the final V = 4005 [Δ]P^0.5 is in feet per minute, not fps. Please advise.

 

[wufi2004]

Really helpful! Thank you all!

 

[lilliput1]

The 60 in the term converts FPM to FPS.
Velocity pressure in Inches wg = (FPM velocity/4005)^2