Need help calculating pressure for a real project

[Jeverard]

TL;DR

I am trying to figure out how to calculate pressure of an exhaust line.

We have a product container we purge with nitrogen to create a low oxygen environment The container is roughly 12x12x12 inch. It has an exhaust port with a 1/4 inch poly line that measures 0.17 inch ID. The top of the container is a door with a rubber gasket. We are a little surprised at how little flow the exhaust port can exhaust before it blows the lid gasket and the nitrogen blows out the door seal. With 20slm input flow at 80psi we measure a max of 4.5 slm out of the exhaust line and the rest blows out the door seal. Long story short is the pressure generated in the exhaust line at 4.5slm overcomes the door seal. I would like to know how to calculate the pressure of a line where I know the line length, ID and flow or better yet a chart. And let's just say the line pressure is 5psi, how does that pressure of 5 psi in a 0.17 line translate to the internal force on a 1 cubic foot container?

 

[Baluncore]

Welcome to PF.
It is all a question of balance.

If the inlet from the 80 psi supply to the container is shorter and wider, then the pressure in the container will be close to 80 psi, with the pressure in the exhaust line falling from 80 psi at the inlet to 0 psi at the open exhaust end.

If you used an identical 0.17” line from the 80 psi supply to the container, then the pressure in the container would approximate 40 psi with half the pressure being dropped along each line.

A pressure of 80 psi applied to 12" x 12" will be a force of 5760 11,520 pounds. [corrected].

 

[jrmichler]

I would like to know how to calculate the pressure of a line where I know the line length, ID and flow or better yet a chart.

There might be a chart somewhere for plastic lines of that diameter, but I did not find one with a quick search. The best way is to do it the hard way. Search Moody chart. That search will find numerous good hits that will tell you exactly how to accurately calculate pressure loss in a line of any diameter, length, roughness, fluid, and flow rate.

Another search using terms pressure loss in air hose found this site: https://www.gates.com/us/en/knowledge-center/calculators/air-flow-calculator.html. That calculator might give you what you want at less effort. They assume air, which is 80% nitrogen, so the results should be pretty good.

 

[The Fez]

You have choked flow on your outlet tube so the increased pressure does not have any increase in flow. For most gasses venting to atmosphere, the flow will be choked for any tank pressure above about 25-28 psia, or about 10+ gauge pressure. You need to get a regulator to reduce the pressure to about 10 psi and limit the inflow to ~5 slm. Even then, as Baluncore mentioned above, your door needs to withstand 12"x12"x10 psi = 1440 pounds of force.

 

[The Fez]

To answer your last part of the question, 5 psi will be 5 psi in the entire container, and the pressure will drop only in the short length of the outlet tube, so you just need to multiply the area of the door by the pressure to get the force acting on it.

 

[Jeverard]

Welcome to PF.
It is all a question of balance.

If the inlet from the 80 psi supply to the container is shorter and wider, then the pressure in the container will be close to 80 psi, with the pressure in the exhaust line falling from 80 psi at the inlet to 0 psi at the open exhaust end.

If you used an identical 0.17” line from the 80 psi supply to the container, then the pressure in the container would approximate 40 psi with half the pressure being dropped along each line.

A pressure of 80 psi applied to 12" x 12" will be a force of 5760 pounds.

I understand 12×12=144 and 144 x 80=11,520. But I don't understand derstand why you divided the 11,520 for 5760. Is that because the 3rd dimension divides the pressure?

 

[Lnewqban]

If all you need is sweeping the volume of the container with Nitrogen, why don’t you just slowdown the incoming flow?
With 20 standard liter per minute you are getting a full change of gas volume in 43 seconds.
Could you increase the cross section of the exhaust and/or use vacuum for it?

 

[jrmichler]

Your description is a little unclear. The sketch below show what I think you are doing:

If this is correct, then the pressure in the box is unknown, so you cannot calculate pressure drop in either the supply or discharge tubing.

One cubic foot is 28.3 liters. Your supply flow rate should be determined by the allowable time to purge atmospheric air after opening the box. If the incoming nitrogen enters so as to get good mixing (not aimed straight at the discharge opening), then a good rule of thumb is that it will be 63% purged after one times the box volume and fully purged after five times the box volume of nitrogen flows through. That would be 28.3 X 5 = 140 standard liters of nitrogen. Divide by the allowable time to get the flow rate. If the allowable time is 20 minutes, the required flow rate will be 140 / 20 = 7 SLPM. If you want a more accurate number, search purging calculations.

If the pressure in the box is less than about 3 PSI, an easy way to measure it is by using a water manometer (search the term). There is no need to be fancy. A piece of 1/2" diameter plastic tubing taped to a stick does the job very nicely.

 

[Baluncore]

Is that because the 3rd dimension divides the pressure?

No. I got it wrong. The 5,760 was from the 40 psi example, the 80 psi will be 11, 520 pounds force.
You must reduce the input psi to what the container can physically handle, or you will have a compressed air explosion if/when the exhaust outlet gets blocked.