One for the STW Sci...
 

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[Closed] One for the STW Science Dept.

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CAUTION: I don't know the answer so close this thread if you're likely to get frustrated!

L = PD X 6,000
SP X M

L = Leak rate in % volume loss per hour
PD = Pressure decay in Pascals
SP = 101,325 + starting pressure in Pascals
M = Test time in minute

This is the formula for calculating % volume loss from a separative device, e.g. a pharmaceutical isolator. The device is pressurised to say +250 Pa above room pressure and left for 5 mins. If the pressure decay is 84 Pa, ie. +166 Pa residual pressure, the %age volume loss can be calculated thus:

L= 84 x 6000 = 0.99 call it 1%
101575 x 5

What is the constant 6,000

And how can %age volume loss be calculated without knowing the volume of the separative device?

>scratches chin again<


 
Posted : 25/08/2020 1:01 pm
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Can I just check I've got the equation right? Is it

L = (PD X 6,000)/(SP X M)

I would say 6000 is a time constant (60 mins to the hour).
The equation is calculating percentage loss so presumably you don't need the actual volume.


 
Posted : 25/08/2020 1:06 pm
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Your equation is correct.

An hour is 3,600 seconds so I can't see why 6,000 pertains to an hour? The time variable is the test duration in mins.


 
Posted : 25/08/2020 1:11 pm
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6000 = 60 x 100

60 because you have a leak rate in hours but a test time in mins.

That would then give a ratio, x 100 to convert to a %


 
Posted : 25/08/2020 1:12 pm
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6000 = 60 x 100

60 because you have a leak rate in hours but a test time in mins.

That would then give a ratio, x 100 to convert to a %

This


 
Posted : 25/08/2020 1:29 pm
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If it's a pressurised vessel, there's no volume loss. the volume stays the same, but the pressure is reduced. Do you mean % mass loss of contents?


 
Posted : 25/08/2020 1:35 pm
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Yeah, that.

If it’s a pressurised vessel, there’s no volume loss. the volume stays the same, but the pressure is reduced. Do you mean % mass loss of contents?

You lose a volume of gas, but the remaining gas expands to fill the vessel at a lower pressure. It's ambiguous, for sure, but guessing that since this is an engineering situation that particular outcome is desired for a reason.


 
Posted : 25/08/2020 1:36 pm
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This is just a rehashing of PV=nRT (the ideal gas equation).

Assuming V (volume of vessel), T (temperature) and R (a constant) stay constant, then:

P~n
P is proportional to n
also written as P = k n
where k is (another) constant.

Change in pressure △P = k △n

The rest of your formula is to calculate the change over time a specific format. But since I doubt the pressure decays linearly* it's a bit basic.

* Meaning that I suspect more gas leaks out when the pressure differential is greatest I.e. when the vessel is at highest pressure (WRT atmospheric pressure) it'll leak more quickly.


 
Posted : 25/08/2020 3:51 pm
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Meaning that I suspect more gas leaks out when the pressure differential is greatest I.e. when the vessel is at highest pressure (WRT atmospheric pressure) it’ll leak more quickly.

It's rather complex in the case of a significant leak compared to the mass because as the pressure in the hole increases past some point the flow becomes turbulent so the relationship between pressure differential and flow is complex. But I'm not a fluid dynamics expert 🙂


 
Posted : 25/08/2020 4:04 pm
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So is the previous explanation of 6,000 correct?

Thanks chaps.


 
Posted : 25/08/2020 5:30 pm

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