# Leak rate / Hole size question

1. Oct 5, 2005

### CAF

Hi everyone, this is my first post here. It looks like this is a great community so I am hoping for some help in either a better explanation or different approach to my problem: How can you determine the size of a hole in a container that is filled with gas given the starting pressure, ending pressure, time interval between starting and ending, density of the gas, and volume of the container? I found a reference but after playing around with the equations I couldn't get it to work out to the same answer the authors did. I don't know if there is an error in their work or if I am making a mistake figuring it out (probably the latter). Anyway, here is the reference where I got my equations from: http://www.piug.org.uk/IUGNewsletter3-2.pdf [Broken] , pages 2 to 3. They get an answer of 472 microns, when I calculate it I get 459 microns. I am trying to set it up in excel so I can enter different values and have it calculate the hole size for any given leak rate.

I hope I explained this clearly, thanks in advance for any and all suggestions.
CAF

Last edited by a moderator: May 2, 2017
2. Oct 6, 2005

### quark

The procedure given in the above link is OK except that velocity calculation should be based on mean pressure which is (P1+P2)/2. The author of the paper knows it, as he used it in his sample calculation, but mistakenly cited it as the half of differential pressure.

You should note that this calculation only holds good for isothermal condition. This can be used, without any problem, for checking leak rates at low pressures. Leakage from high pressure containers is, generally, adiabatic in nature and there are different sets of equations for different types of flow. Search for "flow through orifices".

This link gives you the details of calculating leak rates http://www.air-dispersion.com/feature2.html

For all practical purposes, I strongly suggest getting your hands on Crane Technical Paper 410.

3. Oct 6, 2005

### CAF

Thanks Quark!

I think I will stick with the equations in the reference since the pressures I am dealing with are no where near those in the link you posted, typically less than 250 Pa. I just have a couple of basic math questions about the equations.

First, when I try to duplicate the calculations done in the sample, I get n=27.27 , judging by the filled in example on the top right column of page 3 the author somehow got 0.03.

Next, I get confused when the equation is simplified to the next one down where it starts with "2,000". Did they just cancel a bunch of stuff out or what?

When I try to just plug all my numbers into the top right equation on page 3 I get a=159, even when I use n=0.03 as the author does.

Lastly, how is it that pi suddenly appears in the last equation on page 3 when it is nowhere to be seen in the preceding equations?

Thank you for your patience, I know these answers are probably obvious to you guys.

4. Oct 6, 2005

### quark

Note that the pressures should be absolute. You calculated the leak rate as n = (100*140/110)-100 which is wrong. You should rather calculate (100*100140/100110)-100. The calculation approximates the atmospheric pressure as 100000 Pa where as it is actually 101325 Pa (but the error is not significant).

Area = Pi*d^2/4, so d = 2*(a/Pi)^1/2. If you observe, area is given in sq.mm. So calculated diameter from the above formula will be in mm. 1mm = 1000 microns. That explains 2000 = 2*1000

Regarding a = 159, do the calculation carefully.

5. Oct 6, 2005

### CAF

Sweet

Quark,
Your explanation of it being in absolute pressure makes a ton of sense, I couldn't figure out why one of the other equations had 100125 in it, it makes sense now. To get the equations to work though, I had to adjust the mean pressure by subtracting 100000 from it. Why is it that you don't use the absolute pressure for the mean pressure also, as you do for the initial and final? Anyway, I think I am all set now, thank you very much, your help was invaluable.

6. Oct 6, 2005

### quark

The equation for the velocity is a simplified and modified version of the Bernoulli's principle. It is assumed that the potential energy is converted into kinetic energy totally. At no flow conditions, the absolute pressure inside is Po+P1 where Po is ambient pressure and external pressure is Po. As the gas flows out, the mean pressure is considered and its absolute value is Po+(P1+P2)/2. The differential pressure that is causing flow is internal pressure - external pressure (i.e Po+Pm-Po) = Pm which is the mean gauge pressure.

The main drawback with this calculation is that, the coefficient of discharge is considered as 1 and this is not a practical case. But with lower pressures, the error may not be significant.

Just curious, are you working with Isolators?

7. Oct 7, 2005

### PerennialII

Would 2nd this, trying to find an accurate discharge coefficient can be quite a task in itself and the small error which is done in this case likely doesn't matter.