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post #65 of Old 02-02-2013
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Re: DIY - discussion on building a DIY hookah diving compressor

One way to measure the volume is how long it takes to empty an inverted container, corrected for a pressure of 15 psi,
the pressure at 32 feet deep, above atmospheric pressure at the surface of the water. One foot of water is about 1/2
psi.(Actually, it's closer to 14.7/32=0.459 psi per foot of water depth)
The absolute pressure(Pabs.) would be:
Pabs=14.7(sea level) + 1/2d, where d=the depth of water to the discharge depth
of the compressor hose.
(P1)(V1)(T1)=(P2)(V2)(T2) (gas law), where pis pressure, V is volume and T is temperature
in degrees absolute.
V2=(P1)(V1)/(P2) where T1=T2 and the pressures are absolute pressures.
Let P2=pressure of water at the discharge depth, and let P1 = 14.7
P2= (14.7 + 0.459d), so the equation becomes:
V=(14.7)(V1)/(14.7 + 0.459d)
So, if your compressor produces 200 cuft/hr at the surface, at a depth of 20 feet,
it will produce:
V(@20 ft)=(14.7)(200)/(14.7 + 0.459 x 20)
=2940/(14-7 + 9.18)
=120(approx) cuft.
If there is any heating of the air, during compression and subsequent cooling
of the air, in the diver's supply line, then temperature must be taken into
consideration and the T2 = the discharge air temperature (measured) and
T1=exit air from the compressor output port. I don't think that the difference
is too significant to the calculation, pressures are the overriding
factors, since on the absolute temperature scale (459 degrees Absolute, Rankine,
= zero Fahrenheit, so (example) 60 degrees F =519 degrees R).
The minimum needed at 32 feet is 100-120 cuft/hour, but 150-180 cufh would


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