Pressure of 20 Ft. Head of air

[jrf1]

Can anyone tell me how I would convert 20 Ft. Head of air into psi. Is this the same as water, 2.31.
Thanks,
John

 

[quark]

It is very difficult to give one single answer, to your first question, without knowing the context about which you are speaking. First, how the 20ft. head of air is created? Secondly, does the air column has any pressure or vacuum acting upon it?

You should mind that when a liquid head creates pressure on its bottom, there is atmospheric pressure acting on the vapor liquid interface. But for gases, it is only atmospheric pressure.

Answer to your second question is 'No'.

If I ignore what I said above, approximately, you can calculate Pressure = (head x density of air)/2.31

Regards,

 

[TBP]

For all practical purposes, you can ignore elevation change for compressible fluids.

 

[DRWeig]

Hmmmmm....

I've never seen the term "feet of head" applied to air. Please tell us where the data came from. I suspect that the application will reveal that the 20 ft of head (air) may be better stated as "air pressure equal to 20 ft of head of water" in which case the answer to the second question is "yes".

Otherwise, Quark and TBP gave you the right path.

Old Dave

 

[MintJulep]

The METHOD of conversion would be same as converting feet of water to psi, but the constant would be different. The conversion factor includes the density of the fluid, and probably 144.

 

[lilliput1]

For water psi = Ft water x (62.34 lb water/CF)/144 = Ft water x 0.433

The 2.31 is = 1/.433 = ft of water /psi


Thus with air at 75°F & 50% RH density = 1/13.7 = 0.073 lb air/CF:

psi for air = Ft air x (0.073 lb/CF)/144 = Ft air x .000507 = 20x.000507 = 0.1014

 

[CuriouslyGeorge]

Don't forget, for any liquid, it's really 1 psi = 2.31/S.G.

 

[ChasBean1]

I get:

P = gamma * z (gamma = rho * g)
P = rho * g * z
P = (.073 lb/ft3)(32.2 ft/s2)(20 ft)(lbf•s2/32.2 ft•lbm)(ft2/144 in2)

P = 0.01 psi

 

[25362]

DRWeig is right, it is not customary to express pressures in ft of air.
In fan engineering, standard air is considered to be air with density 0.075 lbm/ft³ when U.S. customary units are used.
The measuring conditions being pressure: 29.921 in. Hg; temperature: 70^oF; RH: 0% (dry air). Or moist air with the same density is taken at 68^oF, RH: 50%, and the same pressure.

Customary pressure units in fan engineering are:

1 psi = 2.3106 ft wg = 2.036 in. Hg = 27.728 in. wg = 144 psf = 68948 dy/cm² = 6894.8 Pa = 704.28 mm wg = 51.715 mm Hg = 0.06805 atm


wg @ 68^oiF or 20^oC
wg @ 62.32 lbm/ft³ or 998.279 kg/m³

Hg @ 32^oF or 0^oC
Hg @ 848.714 lbm/ft³ or 13595.1 kg/m³

The estimate given by ChasBean1 is OK.

 

[ChasBean1]

I agree with all that the units are not the norm, but you could size a fan in fathoms of jello if you want as long as you know density, the base equation, and how to convert.

 

[lilliput1]

I missed the decimal point on the final answer. But the terms & derivations are correct.

 

[quark]

Now I knew where I went wrong. It should be the ratio of densities rather than the density of air i.e (head x density of air)/(density of waterx2.31)

In anyway, these calculations give us wrong values unless we consider density changes(which are very significant) with respect to altitude.

The common figure, we all got, indicates the atmosphere to be of 14.7*20/0.01 = 29400ft only.

PS: I came across same type of problem in my college Thermodynamics textbook.

 

[Slugger926]

I have this problem in some of my engineering books at home. It was common to have Feet Head of Air as units when caculating air pressures for grain drying in bins.