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Please refer to the attached drawing! After watching some videos about the U-tube effect in drilling, a question came to me.

Imagine one fluid of heavy density being pushed down the drilling pipe by another lighter fluid, with the help of a pump (the drawing illustrating the concept shows 10 bar for the pump but this was actually randomly put). The heavy fluid at some point will start flowing up the annulus until reaching a certain height. When the pump can no longer do the work (a static condition is then reached), we close/seal the annulus with a "packer" (the crossed rectangles you see on both sides in the drawing).

When the seals are placed, we shut down the pump, the pressure which was then applied on top of the lighter fluid to push down the heavy fluid becomes $P_{atm}$ (say 1 bar).

This is where I need help. What happens at this point ? The heavy fluid will push the lighter one back up the drilling pipe (middle tube), but a void will be created under the seals, therefore pulling back the fluids, am I correct ?

So my question is, when the pressure below the seals reaches a value close to 0 bar, how much fluid can it support ?

U-tube effect in drilling

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  • $\begingroup$ Might I suggest that you annotate the vertical dimensions in your picture - in particular the difference in height between the light/heavy interface and the top of the annulus of liquid, and the distance from the interface to the top of the light liquid. The pressure due to these two must be the same, so must the product of $\rho_i \cdot h_i$ where the $i$ suffix refers to either of the two liquids and their respective heights. There may be a few additional meters due to the atmospheric pressure over the central column (is the outer truly void?) but that is a small effect $\endgroup$ – Floris Jun 26 '15 at 2:58
  • $\begingroup$ Hello Floris, thank you for your contribution. I did not fully annotate the drawing in the first place because it was not meant to describe any particular case, but rather in a general purpose. But let us say for clarity that the lighter and heavier fluids interface in the drilling pipe is down 3700m, and as shown in the drawing, the packers are 10m below the surface. In this configuration, the pump being technically set at the surface level, the distance from the interface to the top of the lighter fluid would also be 3700m. $\endgroup$ – SimoMJ Jun 26 '15 at 16:01
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Let us assume that the volume of the heavier fluid remains constant (as we always do), and that the drilling pipe and the accompanying setup is vertical. In the absence of the packer, buoyant forces would take over and push up on your pump. But in the presence of your packer, the heavier fluid is restricted from creating vacuum under the packer due to the presence of the force of vacuum on it.

If the heavier liquid were to exert buoyant force on the pump, one of the ways to do it is that it will have to fill the space left behind by the lighter liquid-pump system. Hence, it will have to give up the space it has filled below the packer. But it cannot because of the reason we mentioned earlier. Therefore, the only way is to expand its volume. But again that is not possible as we have assumed the volume of the heavier liquid is constant.

Therefore, we have proven any change of state of the system after insertion of the packer is impossible since it demands a change in volume of liquid.

We conclude by stating that the system will remain in static equilibrium after the insertion of the packer.

Any failure of equilibrium is caused by either of the following factors:

  1. Failure of the packer: If the grip of the packer against the walls of the drain is poor, it may be first to cause failure. This also includes the contact between your pump and the lighter fluid. In short, as you mention, mechanical constraints.
  2. The setup not being air-tight, along with the presence of of froth-creating contaminants: This is the next main culprit I can think of. Drains contain various industrial wastes enough to create froths such as these:

enter image description here

Presence of small pockets of air along with these contaminants and the application of vacuous pressure is enough to trigger bubble formation, if your system is large enough.

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  • $\begingroup$ First of all, thank you for your reply, much appreciated. The details you provided were very useful. But there must be a limit at which the system will break down, I mean if the height of the column of fluids becomes too significant, the equilibrium cannot be maintained at some point, am I right ? Is the limit related to the mechanical resistance of the walls/packer ? Thank you again ! $\endgroup$ – SimoMJ Jun 26 '15 at 15:54
  • $\begingroup$ @SimoMJ Take a look at the edit. $\endgroup$ – Gaurav Jun 26 '15 at 16:47
  • $\begingroup$ Thanks a lot for the follow up, you put serious effort into this and I truly appreciate your help ! So if I get this right, when the system is air-tight, so long as the packer holds, the counter force generated by the vacuum can "pull" any volume of fluid ? $\endgroup$ – SimoMJ Jun 26 '15 at 23:11
  • $\begingroup$ @SimoMJ Yes, exactly. $\endgroup$ – Gaurav Jun 27 '15 at 4:35
  • $\begingroup$ Ok great, I think I've got what I needed. Thank you again Gaurav. $\endgroup$ – SimoMJ Jun 27 '15 at 10:46
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Even if your seals are perfect, you have to consider that all liquids start to boil as your pressure approaches a perfect vacuum. Some liquids do better at this than others, but they all eventually boil because they will constantly adjust to achieve a gas/liquid equilibrium for the pressure. Thus you will find that you can never actually get down to a perfect 0 bar. You'll always have some residue left from evaporation/boiling of the heavy liquid, depending on your choice of liquid.

Mercury actually does pretty well at this sort of game. It might be worth considering. Also, if you could find a heavy liquid which is denser than mercury (!!!), you could use a "cap" of mercury floating on the drilling fluid to cut down on evaporation issues.

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