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truckpoor (Mechanical)
26 May 10 20:24
I'm hoping I can get some help here, as I'd like someone to reassure me this is as easy as it seems.  We've got a gage that we're having manufactured that has to be leak tested at 5000 psig for not less than 3 minutes and the air equivalent leak rate not to exceed 5x10-5 atm cc/sec.  The manufacturer, in the interest of speeding the process up, would prefer to do a helium leak test, measured in mbar*l/sec.  To convert from cc/sec to mbar*l/sec, you just multiply by 1.013, which comes out to 5.1x10-5 mbar*l/sec.  With this, they're willing to test to 1x10-6 mbar*l/sec, which exceeds our requirement.

I was thinking I need to convert the air leak rate to a helium leak rate, which according to my books equates to dividing the air leak rate by .374 for molecular flow, and that comes out to 1.3x10-4.  Either way, what they're willing to test to seems to exceed our requirement just based on these basic conversions.  

What I haven't been able to figure out is if this an equivalent/acceptable conversion?  To me, it sounds like the original method is a pressure decay, fill whatever volume they have to 5000 psig and after 3 minutes, check the pressure to see if it has dropped enough to exceed the leak rate.  I've seen helium leak detectors in operation, but is it equivalent to a pressure decay leak rate?  Thanks for any insights into this.
BigInch (Petroleum)
27 May 10 5:09
It would help if you posted the relationships you've found in your book, but in the meantime

What you're doing doesn't seem right, but again I don't know what method you're using.

First, anything short of an outright explosion will be a "pressure decay".

Secondly, leak rates depend on pressure; the higher the pressure, the faster the leak rate and the faster the pressure drop for any given escape path.

Third: The leak rate depends on the viscosity and specific gravity of the gas.

Lastly, If you want to talk about volumes, the volumes should be corrected to standard pressure and temperature.

A method for correction leak rates should include terms for specific gravity of the gas in relation to air, probably something like (1/SG)^0.5 and for viscosity.

This method might work for you. Have a look at thas and see if it makes sense,

"Being GREEN isn't easy" ..Kermit

JR97 (Mechanical)
27 May 10 9:54

I agree with BigInch.  Leak rate depends on pressure (actually pressure differential across the pressure boundary) and the gas used.

Per ANSI N14.5, the first thing you need to do is calculate the leakage hole diameter for the specified air test then, given that hole diameter, back-calculate to a corresponding leak rate for helium at whatever pressure they will be using.  To be accurate you need to consider both laminar and molecular flow although, at least for the air test portion, you can probably assume molecular flow.  Depending on your parameters one of the leak rates will be insignificant.

The formulas to use are:

Fc(continuum or laminar flow) = (2.49E6 x D^4)/ (a x u)
Fm(molecular flow) = (3.81E3 x D^3 x (T/M)^.5) / (a x P)

Fc = continuum or laminar flow leak rate (in cc/atm sec)
Fm = molecular flow leak rate (in cc/stm sec)
D = leakage hole diameter (in cm)
a = leakage hole length (okay to assume 1 cm)
u = viscosity (air=0.0185, He=0.0198 centiPoise)
T = Test Temperature in Kelvin
M = Molecular weight (air=29.0, He=4.0 g/gmol)
P = pressure in atm, absolute

truckpoor (Mechanical)
27 May 10 16:29
Thanks JR97, but that's kind of what I've been struggling with.  How do you calculate a leakage hole diameter when you don't want one in the first place?  I could back-calculate how much mass of air is lost according to the leak rate.  This is the requirement verbatim:  Each gage shall be leak tested at 5000 psig for not less than 3 minutes duration.  The air equivalent leak rate shall not exceed 5x10-5 atmospheric cubic centimeters per second.

I guess I need to figure out how they were testing these in the first place.  The only way that makes sense to me is a pressure decay, since I have no clue how they'd measure an air leak any other way.  But there is no volume requirement, so I'd imagine they could play this any way they wanted since it's not going to leak as much at lower pressures.

As far as the relationships I was using, they come from a Nondestructive Testing Handbook.  For molecular flow, flow rate for one gas may be compared for any other gas by the following equation:
where M is the molecular weight.  So comparing air to helium, Q(air)=((28.966/4.02)^-.5)*Q(Helium), or Q(air)=.374*Q(helium), or Q(helium)=Q(air)/.374.  Actually running the numbers myself, I come up with .373, but the book says .374 so close enough.

I'll take a look at your link BigInch and see if it helps.  I'll keep you posted and if you have any more insights, please share.  Thanks!
BigInch (Petroleum)
27 May 10 16:50
Forget about a leak rate.  Leak rate through a hole won't work for you, as you have realized, you have no hole size and you don't care anyway.  You're looking for a finite volume lost over 3 minutes, through any size of hole, which must be < your allowable leak rate.

So, you do need to look at the link I provided and work with finite pressures and volumes.  Then correct those volumes at 5000 psig pressure to atmospheric pressure and standard temperature, using the proper compressibility factor and actual temperature, to get the volume actually leaked, which should be <= 5E-5 cc/s.  

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

"Being GREEN isn't easy" ..Kermit


JR97 (Mechanical)
28 May 10 9:26

The problem you have before you is to determine the difference between how air and helium react to a leak (pressure drop) scenario.  Hopefully you do not have an actual hole in your part, but you use a "hole diameter" to account for the different parameters of the two test gases, namely viscosity and molecular weight.  I must respectfully dsiagree with BigInch, the leak rate is important and it is specified as 5E-5 atm cc/sec AIR.  The method to determine what the equivalent leak rate would be for a different gas, Helium in your case, is to bridge between the two gases through using a common hole diameter.  That is, you first calculate what diameter hole would result in a rejectable leak with air as the gas.  The second step is to use the now known "rejectable" hole characteristics and plug in the helium parameters.  The answer will be the resulting leak rate (Helium) that would occur through the "rejectable" hole.  

If you can get your hands on a copy of ASNE N14.5 or Leak Testing volume of the ASNT Nondestructive Testing Handbook they can explain it much more elequently than I can.

BigInch (Petroleum)
28 May 10 11:08
JR, a volume difference over time can be considered a leak rate, if you know the volume left through any type of escape path, not just a hole with a specific known diameter, or assumed diameter, which he doesn't know anyway.  Why do you insist on having a hole diameter?  It is not needed to solve the problem, if you have two known pressures, two times and the volume of a tank.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

"Being GREEN isn't easy" ..Kermit


JR97 (Mechanical)
28 May 10 12:37

Let's look at an extreme, silly, example.  Let's assume his part actually has a 5E-5 leak in it.  Instead of using air, he decides to pressurize with molassas, or heavy mud as our friends are doing in the gulf.  With the greater viscosity, a lower volume of fluid would escape per unit time, resulting in a lower calculated leak rate for the part; accepting a rejectable part.  That is why a specified leak rate is based on what the pressurizing medium is.  If a different pressurizing medium is used, the leak rate must be adjusted accordingly.  Bridging through a hole diameter is the method used by ANSI N14.5 and the method I am familiar with.  There could be other means to achieve the same end.

The reason the vendor wants to use Helium is the relatively small leak rate.  5E-5 atm cc/sec is equivalent to a car tire losing 1 psi of pressure in 2 years.  In order for the target leak to create a large enough pressure change to be reliably measured on the pressure gage the test period must be extended beyond the point where constant temperature can be assumed.  Once temperature is introduced as a variable interpreting the test results is all but impossible.  The vendor will probably use a detector probe in a helium mass spectrometer.  Under carefully controlled circumstances that equipment will detect a 1E-6 atm cc/ sec (Helium) leak in a few minutes.

Hope this helps.

BigInch (Petroleum)
28 May 10 13:23
Yes I perfectly understand all that.  I already said exactly that in my first post.   I just asked why do you want to use an imaginary hole and calcualte the leak rate across it, when you can directly measure the average leak rate from (V2-V1)/time, then correct for meduim and compressibility to arrive at the leak rate at STP.  Perhaps you should look at the link I posted as well.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

"Being GREEN isn't easy" ..Kermit


btrueblood (Mechanical)
28 May 10 16:08
You're both saying the same thing.  JR is doing a two step, and BigInch is already playing with alge's bra.

Sorry, but the lads in the shop have been playing C&W music today.
BigInch (Petroleum)
28 May 10 16:33
"High school days in Big Spring" by ...

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

"Being GREEN isn't easy" ..Kermit


israelkk (Aerospace)
28 May 10 18:18
We do it with weight loss. We fill a vessel to a predefined pressure, 5000psi in your case, weigh it and then weigh it again after a predefined time. The difference in weight is the weight of the leaked gas. Knowing the molecular weight of the gas and that one mole occupies 22.4 liter at standard conditions (STP) we know the average leak. For small leaks the  pressure will change very little therefore you can assume a constant average pressure. However, the 3 minutes seems too short for a noticeable weight loss that you can measure. There are machines that can detect 0.1 gram difference even it the total vessel weight is 10 kg.
hacksaw (Mechanical)
29 May 10 10:57
use a helium leak test and be done with it. your conversion factor is found in any text/web-site that deals with leak detection and appears correct.

This is not flow through an orifice, but a leak and you are going to have considerable difficulty reducing the leak to a hole size. In part because, the leakage path its "effective discharge coefficient" or in some cases the leak process i.e. diffusion, etc. are unknown.  

mass spec is the way to go, but there are other detectable gases that can be used with less expensive detectors.

it would probably be a mistake to use the submersion method and measure the plume size...

good luck
truckpoor (Mechanical)
1 Jun 10 20:08
I was thinking about this over the weekend and I think what they're wanting to do is more than adequate.  They're just converting the air leak to units they can measure.  They're not even worried about the conversion, just treating it as a 1:1 conversion.  

So, by just using a 1:1 conversion, the leak rate they want to measure is 1x10-6 mbar-l/sec, which is a little better than the required 5x10^-5 cc/sec.  Without converting it to helium, this should more than cover our requirement.  

Think about a leak filling a balloon.  A helium leak is going to be considerably larger than an air leak, which would fill a balloon much faster than air would through the same leak.  If they're willing to helium leak test an air leak requirement, I should be well in the positive of my requirement.  Does that make sense?

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