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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,

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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

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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 $\rho gh+P_0$.

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  • $\begingroup$ 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
    Commented Nov 30, 2016 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
    Commented Nov 30, 2016 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
    Commented Nov 30, 2016 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
    Commented Nov 30, 2016 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
    Commented Nov 30, 2016 at 15:22

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