I am curious about determining how much vacuum a vessel can see during draining.

My case would be a vessel 90% full of liquid being drained with nothing else open. Is it as simple as P1*V1=P2*V2, where I would solve for P2 when I have 100% empty vessel?

Does the rate of draining impact the amount of vacuum developed? It seems that if it took 5 days to drain this vessel, developing a vacuum would be more difficult than if it was drained in 5 minutes, but perhaps this logic is incorrect?

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you have water filled up to the tank at 90% where the 10% air space is atmospheric air.  the level starts to drop.  as soon as the bottom of your tank sees about 34' (14.7 psi converted to head) of head, you'll start getting air bubbles sucked into the tank outlet to equalize pressure.  similar to if you ever bought one of those 2 gallon water jugs and tried to fill a glass without punching out the vent port

like the above posts say, what is left is the vapor pressure inside the vessel and the pressure of the air at the new expanded volume (assuming you don't consider the air coming into the outlet).  At a max with water at 20 deg C youll see about 20 mmHg from the vapor pressure + the final pressure of the 10% air if you have a check valve at the tank outlet.


if this is a tank, id expect this to have PVSV on the top to make sure you didnt pull too much vacuum on the vessel, but assumptions above, you can pull pretty close to complete vacuum.

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