Leakage rate units

  • Thread starter Runei
  • Start date
  • #1
193
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Hello,

At my work I'm working with testing the leakage rate of a gas system, and right now im trying to gain some understanding of it all. However, there are some things which are very confusing to me, in regards to these rates.

Searching the internet I find several different ways to describe the leak rate, some of mL/min and some are mbar*mL/min.

To me, the first one seems to only be usable in systems with incompressible liquids, and the second is usable in systems using gas.

My intuition so far tells me that using the ideal gas law, the change in pressure over time (due to leakage) will be equal to

[itex]\Delta p \cdot V = \Delta n RT[/itex]

The temperature can be approximated to be constant, and the volume of the system is constant (its an aluminium gas string with valves and such).

Using this, we correct leakage rate units would be

[itex]\frac{\left[p\right]\left[V\right]}{\left[t\right]}[/itex]

Is this all there is to the story, or is there something I am missing?

Thank you,
 

Answers and Replies

  • #2
2,081
104
All leak rate units consist of three components:

1) Pressure (mbar, atm or Pa)
2) Volume (l, cc or m³)
3) Time (s or m).

from that you can see a leak rate is two possible scenarios
-a changing volume (2) in a given time (3) at a certain pressure (1)
-a changing pressure (1) in a given time (3) at a certain volume (2).

so your scenario is the second one.

The pressure drop can be converted to a leak rate by multiplying the pressure drop with the inner volume of the part and by then dividing the result by the measurement time needed for the pressure drop.

which is the same as your equation

so you got it.

just make sure you look at the pressure units, for proper conversion. for example a pascal is
newton per square meter. a bar is 100,000 Pa
 
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