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

[Ito]

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?

 

[A.T.]

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.

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.

 

[Baluncore]

Welcome to PF.

Anyone an idea how much water one would (theoretically) indeed lose in such a situation?

Assuming the liquid is water, and the fluid is aerated water.

Since the hydrostatic pressure in the tube depends on the liquid level in the barrel, the air will be more or less compressed in the tube. That will change the volume of air, not the mass of air.

You drop the liquid into the barrel, which will release air from the open tube, but also aerate and foam the liquid in the barrel. That aerated liquid has a lower density than the pure liquid, so the level will rise in the barrel.

At the same time, the temperature of the fluid will be changing while it moves through the tube, as it exchanges heat with the environment. That means the volume of the fluid in the system may change during operation.

As you inject air into the fluid, the air changes temperature and pressure. The efficiency and rate of air injection will depend on pressure and temperature. Gasses from the air and/or CO₂ will dissolve in, or be emitted, from the water. That will depend on temperature.

If the air is injected through small pinholes, with a high air manifold pressure, then the liquid will be cooled by the expanding air as it is injected into the tube. The liquid may, possibly, freeze solid.

If you want to know the changing density of the liquid-gas mix that you are pumping around the tube, then you need to weigh the barrel, measure the depth, and temperature of the fluid.

Is the liquid water?
What is the liquid, and why must it be stirred?
How long is the tube?
Why pump the liquid around when you could stir the barrel?

 

[Lnewqban]

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?

No considering any possible effect of air, the volume of pumped liquid should decrease as the level in the container decreases.