How to Estimate Air Leakage into a Vacuum Vessel?

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To estimate air leakage into a vacuum vessel, the pressure rise during a leak interval is crucial. With an external pressure of 1020 mBar and an internal pressure of 10 mBar, a 10 mBar rise indicates approximately 48 liters of air leakage over 6 hours, translating to about 8 liters per hour. Adjustments to the pumping speed, now confirmed at 130 m3/hour, suggest a revised leakage estimate of approximately 1.01 m3. It’s important to consider pump efficiencies, as rough pumps may not achieve optimal vacuum levels. Additionally, outgassing from the vessel's materials can also impact overall leakage measurements.
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Question:
How can I estimate the amount of air that leakes into a vacuum vessel. I know the pressure outside the vessel (1020 mBar), the pressure achieved in the vessel (10 mBar), the pumping speed of the vacuum pump (400 cubic metres per hour) and the duration of the leaking (6 hours). Volume of the vessel = 4800 litres.
 
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The amount of leakage is measured by the pressure rise during the 6hr leak interval.
If you had a 10mBar rise during that period, you had about 1/100th of the volume leak in, so about 48 liters in 6 hrs, or about 8 liters/hr.
 
etudiant said:
The amount of leakage is measured by the pressure rise during the 6hr leak interval.
If you had a 10mBar rise during that period, you had about 1/100th of the volume leak in, so about 48 liters in 6 hrs, or about 8 liters/hr.

Thank you for answer. I should have made clear that there was a stable pressure of 10 mBar during the 6 hours. During this time several pumps were operating. Turns out I overestimated the pumping speed of the system. It is in fact 130 m3/hour. I now believe the estimate to be: ((pumping speed)*(duration))/100 so (130*6)/100≈1*101 m3.
 
Zrq said:
Thank you for answer. I should have made clear that there was a stable pressure of 10 mBar during the 6 hours. During this time several pumps were operating. Turns out I overestimated the pumping speed of the system. It is in fact 130 m3/hour. I now believe the estimate to be: ((pumping speed)*(duration))/100 so (130*6)/100≈1*101 m3.


In practice, you want to be pretty cautious about pump efficiencies.
Rough pumps that are used initially to pump down the installation can't produce a good vacuum, although they should get to maybe 10**-2 mB.
A 4800 liter vessel is good size and may have elements that outgas in a vacuum, which might become a factor.
 
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