I'm looking into capillary flow in a tube as explained in this link. The system consists of two liquid phases, one wetting and one non-wetting, inside a tube,

enter image description here

What is the relation between the inlet pressure $p_i$ and the outlet pressure $p_r$? I would personally expect that $p_i>p_r$ due to the capillary forces that establish a pressure gradient.

EDIT: In the figure below the point B denotes the tube above and the point A the reservoir that the tube is "attached" to. Say that the velocity in reservoir A is nonzero -- then by Bernoulli's law, should the pressure in reservoir A be lower than that in tube B?

enter image description here


There is no relation.

The purpose of the experiment is to find the unknown capillary pressure $p_c=p_w-p_{nw}$ from measured quantities : pressure difference $\Delta p=p_l-p_r$, flow rate $u$, viscosity $\mu$ and length $d$ and radius $r$ of the tube. The pressures $p_l$ and $p_r$ are variables which you control in the experiment.

$p_r$ would usually be atmospheric pressure $P_0$. $p_l$ is the pressure applied to the liquid by a pump or a constant 'head' of liquid - ie $\rho gh+P_0$.

| cite | improve this answer | |
  • $\begingroup$ But say that this is a pore inside a piece of rock or chalk and that the outlet is, e.g., a reservoir. Then what would the relation be? I'd expect $p_I>p_R$ still in the case of a solid boundary that is water-wet, since that is the only way for the capillary forces to "push" the non-wetting fluid out $\endgroup$ – BillyJean Nov 30 '16 at 14:00
  • $\begingroup$ $p_l$ and $p_r$ are the fluid pressures at the left and right sides of the tube. I do not think it would be easy to apply this experiment to a pore inside a rock. $p_l$ is easy : it is as given in the last sentence of my answer, where $h$ is the depth of the reservoir. But if the air or gas on the other side of the pore is not open to the air, $p_r$ is not atmospheric pressure $P_0$. I think it would be difficult to know what $p_r$ is in that case. If you knew $p_c$ and all other variables, you could deduce $p_r$. But what use would that be? $\endgroup$ – sammy gerbil Nov 30 '16 at 14:09
  • $\begingroup$ If the liquid flows left to right then $p_l>p_r$. But you could have $p_l<p_r$. Then the liquid flows right to left and the interface is concave instead of convex. $\endgroup$ – sammy gerbil Nov 30 '16 at 14:11
  • $\begingroup$ I see, that is a good point. As a thought experiment, I now consider there to be a velocity in the reservoir attached to the outlet (see the OP) - then by Bernoulli's law, can we now state that $p_l>p_r$? $\endgroup$ – BillyJean Nov 30 '16 at 14:53
  • $\begingroup$ Sorry I don't understand how your 2nd figure relates to the 1st. If you know that the liquid is flowing left to right, then it follows that $p_l > p_r$. You don't need Bernoulli's Law to tell you this. $\endgroup$ – sammy gerbil Nov 30 '16 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.