Volume / Pressure Conversion

so, what is your point?

All,

I know this is not necessarily PCDMIS related but it is job related.

The note on the print says:

Max leakage at 5.5 bar: 5 cubic mm per second at atmospheric pressure.

I have a tester that can input 80 psi and I can measure the drop in pressure. For instance I can watch for 1 min and see that it drops to 70psi. Is there a way to convert this into something I can use to satisfy the print callout? Any help would be greatly appreciated.

My point is that 10 psi does not equate to one volume. Earlier you said that the op needs to know the air volume of 10 psi, which to me indicates that 10 psi only equates to one volume, and this is in no way the case.

The volume here is a confined space of compressed air, the same amount of air in a smaller space the greater the air pressure would be.

The escaped air has volume (space), since the pressure of compressed air is greater than atmospheric pressure, that air would expand until it reaches atmospheric pressure which is ~1 bar or ~14.5037738 psi a, equivalent to 0 psi g, depends on where you at. So, his question was; what is the actual cubic mm per second (volume/second) of escaped air at atmospheric pressure of 10psi/minute. which is impossible to answer.

No *****, and if you remove some of the mass of air within the smaller volume you can equate the 2 pressures meaning two different volumes at the same pressure.



NO IT'S NOT. The ideal gas law will tell you the mass of escaped air. Using this mass you can then go back and recalculate for standard atmospheric conditions and find the volumetric flow rate. I am an engineer and this is very simple fluid dynamics. I have done these calculations many many times in the past. All he needs is the volume of the chamber he is pressurizing which should either be given or calculated, and if he has a CAD model then he can just get it off of there, and the state variables for the air.

OK engineer man, whatever you said... You were right about other elements should take into consideration for this calculation, you can't do it without volume, this was taught in college chemistry, but I'm just an inspector not an enginneer. So, you win, have a good day.

Did I not say you need the volume of the chamber? Please re read carefully. You were talking about some "air volume of 10 psi" (which makes absolutely no sense whatsoever) not the volume of the chamber.

Right which means there is a pressure differential of 10 psi and the volume HAS NOT CHANGED. You use the ideal gas law to find the volume of the evacuated air which is then used to calculate the volumetric flow rate.