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Pressure of 20 Ft. Head of air

Pressure of 20 Ft. Head of air

Pressure of 20 Ft. Head of air

(OP)
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

RE: Pressure of 20 Ft. Head of air

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,

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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.

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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: 70oF; RH: 0% (dry air). Or moist air with the same density is taken at 68oF, 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 @ 68oiF or 20oC
wg @ 62.32 lbm/ft3 or 998.279 kg/m3

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

The estimate given by ChasBean1 is OK.

RE: Pressure of 20 Ft. Head of air

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.

RE: Pressure of 20 Ft. Head of air

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

RE: Pressure of 20 Ft. Head of air

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.

RE: Pressure of 20 Ft. Head of air

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.

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