Pumping liquid around and draining a part of it, how much is lost?

Ito
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I have a question about a collector/barrel connected to a tubular system pumping around a liquid.

Imagine you have a barrel that has at the bottom an opening connected to a tubular system. There is also a pump that actively pumps the liquid around next to the barrel. The liquid is as such pumped around and enters back in the barrel at the top, above the liquid level). In the barrel you have 500L. In the tubes you have X L (unknown) of the liquid + air because you also inject air in the tubes (after the barrel/pump) to keep the liquid mixed well.

Now imagine you open a drain in the barrel, while keeping the pump and air injection on, and you keep removing liquid until you are about half of the original level (approximately 250 L left) in the barrel.

How much liquid did you actually lose? An easy and quick estimation would be that you effectively lost 250 L given you went from 500L to 250L in the barrel and assuming that the liquid volume in the tubes remained the same given you keep pumping the liquid around in the system. I assume in the tubes the liquid (and air) volume will stay the same. However, is this actually the case?

If there was no pumping of liquid going on (and no injection of air), this would a situation of communicating barrels and you would also have lost water in the tubes (the same amount) as well, totaling a loss of approximately 500L in total.

However, in this situation you pump the liquid around (in combination with air injection in the tubes) so I would assume the total volume in the tubes stays the same, but perhaps this is a bit too simplistic as the communicating barrels itself might also still play a role? And perhaps because there is less liquid in the barrel to start with, perhaps there is less 'power' (pressure) from the liquid causing the liquid to be less 'strongly' pumped around and there will be more air in the tubes?

Anyone an idea how much water one would (theoretically) indeed lose in such a situation?
 
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Ito said:
If there was no pumping of liquid going on (and no injection of air), this would a situation of communicating barrels and you would also have lost water in the tubes (the same amount) as well,...
If the water volume in the tube is unknown, why would you assume that without the pump the volume loss from the tube is the same as from the barrel? Same water level doesn't imply same volume.

Ito said:
However, in this situation you pump the liquid around (in combination with air injection in the tubes) so I would assume the total volume in the tubes stays the same,
If the pump keeps the tube filled to the top and the amount of injected air stays constant, this assumption is OK.
 
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