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Derivation of air flow constant 4005

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Tim A

Mechanical
Joined
Jan 7, 2020
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thread14-92174
From Bernoulli's equation

(P1-P2)/Rho = V2/2g (i.e static pressure difference creates flow)

1 lb/sq.ft = 0.1922 inches of wc
1g = 115820 ft/min2

So V = [(2 x 115820)*dP/(0.075 x 0.1922)]1/2
i.e V = 4008.6 x dP1/2 (there may be some conversion problems for the costant to be high by 3.6)

Q = A x V (it is not aspect ratio here but the product of two sides)

Regards,

Previous thread pasted above
Understanding the 4005 constant
Trying to understand where the .1922 inches of water column of air comes from as air weighs .075 for each cu ft. .075 x .1922 but not explained how you get .1922 inches of wc in a lb of air
V= is the square root of "16,069,372.18175512" or 4,008.662143627861 V2 is from 2 x 115820 is 2g or 231,640 divided by .075 x .19220 or .014415


Thanks
 
The density should be of air as this is standard Bernoulli's Equation. However, you can't measure dP in terms of air column for practical reasons. So, liquid based manometers are used in the past to check dP and get volumetric flowrates from that. That is the reason dP is always to be kept in height of liquid column.

 
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