Well, making all the same assumptions: frictionless, circular hole of negligible length, constant pressure differential, incompressible fluid (we all know that air is compressible, but for the purpose of fluid dynamics calculations it can be assumed to be incompressible if the velocity is below Mach 0.3 or so), laminar flow and so forth, we have
Q = K * A * sqrt((2 * P ) / rho)
A = pi * 0.0254^2 = 2.026829916E-3
P = 0.75 psi = 5171.067968 Pa
Now, the density of air is a variable with a wide range, given that it will be different at 0.75 psi above ambient than outside at ambient pressure, and is also subject to variation due to temperature and moisture content. This variance is significant, so you will need to determine a value for rho which best approximates your conditions (i.e. find the partial pressure of water vapour at your likely temperature, etc.), and then use an appropriate average between the density inside the hull and the density outside. Just to illustrate the example, though, I'm going to use the density of dry air at IUPAC standard temperature and pressure (0 deg C and 100 kPa), which is 1.2754 kg / m^3.
Q / K = (2.026829916E-3) * sqrt((2 * 5171.067968) / 1.2754) = 0.182515276 m^3/s, at a speed of about 90 m/s.
Assuming a K value of about 0.62, this gives you about 113 liters per second. You might think this is high, and you'd be right - airflow at that speed, in the presence of any piping friction or restriction whatsoever, will not be laminar, but rather turbulent. Accurate calculation of flow will require data on the flow path (valves, piping, etc.) to determine the Reynold's number and calculate from there, but in any case this gives you some idea of the order of magnitude.
-Sean
Okay, having done the water calcs, how about air? I'm working on those pneumatic mushroom valves we were talking about last month. The throat on these particular units is 2" and the side vents in the valve body equate to slightly over that. So, if the valve is considered frictionless and driven by...oh, call it 0.75 pis at the surface, how fast will I vent 5 cubic feet or so?
Vance
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