Hi crvz, thanks for the additional details. It sounds like you have a BX cylinder of oxygen or similar high pressure cylinder that is supplying a constant flow of oxygen to your tank that is remaining at 0.3 atmospheres. In order to maintain constant flow into this tank, I'm assuming the cylinder has a regulator on the outlet. Is that right?
The tank at 0.3 atmospheres is then being maintained at a constant temperature (70 F) using an 'independent cooling system'.
You then want to control the flow out of this 0.3 atmosphere tank using an orifice. Also, this orifice is passing the oxygen into a vacuum. Did I get that all correct? If so, you can apply the second equation I gave you. (Note: I made a mistake in my other post and said the first equation was the choked flow one, that should be the second equation.) The flow through the orifice is choked with the inlet pressure of 0.3 atmosphere. Make sure you use absolute pressure in the equation. Also, if you use a drilled orifice (ie: a flat plate that has a drilled hole in it) I'd suggest using a discharge coefficient of 0.80. You may need to play around with this a bit, or you might simply put a metering valve on the outlet. Seems like a metering valve would be much easier. If you want a valve recommendation, let me know what tubing you're using.
Clausius, regarding the thermodynamics of this, the thermo isn't needed to determine the orifice size. Also, I'm unclear on what you're suggesting. I believe the first law for this would reduce as follows.
Since we have constant pressure and constant temperature inside the tank (the one at 0.3 atmospheres) the internal energy change inside the tank is zero (dU = 0). Thus the first law applied to the 0.3 atm tank would reduce to:
dU = 0 = Hi - Ho + Qi
Note that for no change in pressure or temperature, the density also remains constant and thus the mass flow in must equal mass flow out. (I know - I was drunk as a skunk yesterday and said enthalpy in equals enthalpy out, my bad.) So now we have:
Qi = ho mdot - hi mdot = mdot (ho - hi)
If I plug some values in, the first thing I do is look at the source. Let's say the bottle pressure is high (~ 5000 psig) and also assume there is little heat transfer between the regulator outlet and this 0.3 atm tank. As the oxygen expands across the regulator (from 5000 psi) and enters the tank, it cools to roughly -55 F. Internal energy is converted to PV energy as the molecules rearrange themselves. Once inside the 0.3 atm tank, heat must be added to get the oxygen back up to 70 F. So actually, assuming I'm corrrect in how this system is set up, the "independent cooling system" is actually warming the gas inside the tank in order to maintain a constant temperature and pressure, not cooling it.
What is happening also, is that as the pressure inside the high pressure cylinder drops, the oxygen is cooling and heat from the atmosphere is going to warm the contents. It won't cool too much at these flow rates because the thermal mass of the bottle is relatively large and heat transfer is sufficient to maintain a roughly constant temperature.
Also, as the high pressure cylinder drops in pressure, the enthalpy of the oxygen increases (ie: it picks up heat from the atmosphere/thermal mass of the cylinder) so that there is less and less heat needing to be added to the 0.3 atm tank. At first, the heat flux is relatively high but as the cylinder pressure decays, the amount of heat that must be added to the 0.3 atm tank also drops.