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
Thanks,
John
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Pressure of 20 Ft. Head of air
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RE: Pressure of 20 Ft. Head of air
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
RE: Pressure of 20 Ft. Head of air
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
RE: Pressure of 20 Ft. Head of air
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
RE: Pressure of 20 Ft. Head of air
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
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
RE: Pressure of 20 Ft. Head of air
RE: Pressure of 20 Ft. Head of air
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