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We have a funnel that is thin enough to keep a drop of liquid inside it as shown in this figure.

Horizontal thin funnel

Assuming that the funnel is placed on a horizontal table, will the drop flow from the left side to the right side?

Here is why I think it may move. Each side of the drop experiences the same pressure, however, side A has a larger surface, so the force applied upon it is accordingly larger. Side B has a smaller surface, hence smaller force is exerted upon it. Is this considered an imbalance that will cause the drop to slide? And is this related to the capillary action?

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For the case that you have drawn, the behavior of the drop is actually the exact opposite of what you mention: it will move from right to left.

This is caused by surface tension and the curvature of the droplet caps which creates a larger pressure in the drop at side B than at side A.

To make it more quantitative. Let's assume that the funnel is axisymmetric such that the radius of the tube $R$ and the slope of the funnel $\beta$ are the only geometric parameters. The air pressure on the outside of the droplet is atmospheric $P_{atm}$ and the pressure at the inside of the droplet at points A and B are $P_A$ and $P_B$ respectively.

If we calculate the capillary pressure jump across the interfaces $A$ and $B$ we get: $$P_A-P_{atm}=\Delta P_{c,A}=\frac{2 \gamma}{R_A} \tag{1}$$ and $$P_B-P_{atm}=\Delta P_{c,B}=\frac{2 \gamma}{R_B} \tag{2} $$ where $\gamma$ is the liquid-gas surface tension and $R_A$ and $R_B$ are the radii of the circular gas-liquid interfaces.

If we now eliminate $P_{atm}$ by subtracting $(1)$ and $(2)$ we find: $$P_A-P_B=2\gamma \left(\frac{1}{R_A}-\frac{1}{R_B}\right) $$

Because $R_A>R_B$ (as is clear from your picture) this means that $P_A-P_B<0$ which will result in a liquid flow from the right to the left that drives the droplet. A larger change in $R$ from $A$ to $B$ will, i.e. larger $\beta$, will result in a steeper pressure gradient.

Note that I have assumed here that the situation is as drawn in your figure: with positive curvatures from the viewpoint of the liquid. If the contact angle is such that the droplet has negative curvatures (i.e. the center of the circle describing the interface is in the gas), as shown below, then the droplet will move to the right, towards B.

enter image description here

Coincidentally, this week a paper came out in Langmuir that describes exactly this case, for both wetting and non-wetting droplets. Unfortunately it is behind a paywall, but for those with a university login: Luo et al. 2014, Langmuir, Behavior of a Liquid Drop between Two Nonparallel Plates

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  • $\begingroup$ You should calculate $R_A$ and $R_B$, I believe you have been to quick and get the wrong sign for their difference. $\endgroup$ – Joce Jul 2 '14 at 15:17
  • $\begingroup$ @Joce As it is sketched and according to the Laplace-pressure, this is exactly right. Obviously, both $R$'s are positive, and nothing is wrong with this derivation. Maybe with different contact angle behavior this changes, but that really depends on the material of the funnel. $\endgroup$ – Bernhard Jul 2 '14 at 18:14
  • $\begingroup$ @Joce, indeed as Bernhard mentions: the radii should be positive when the center of the circle fitting the interface is inside the phase you are considering. $\endgroup$ – Michiel Jul 2 '14 at 18:20
  • $\begingroup$ @Joce, I do agree that the contact angle matters in this case and that a contact angle <90 degrees (measured through the liquid) would result in the droplet moving to the right (towards B), but the droplet is drawn with the contact angle >90 degrees and that is what I wrote the answer for $\endgroup$ – Michiel Jul 2 '14 at 18:21
  • $\begingroup$ I was a bit quick last week I admit, sorry about that. Your answer is correct for completely non-wetting drop, but for partly non-wetting as drawn there is an equilibrium point. $\endgroup$ – Joce Jul 6 '14 at 15:31
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Consider this container in pressurized air but zero-gravity (and ignore surface tension, which would make the liquid ball up).

enter image description here

If your guess were right, the liquid would squirt out the small hole on the right, but that ignores the role of the wall on the right, which counters the pressure on the left. Think of the liquid as a collection of horizontal cylinders separated by tissue paper, some having a wall on the right, and one having a hole on the right. The ones having a wall on the right will not move, and the one having a hole on the right will not move either because it has the same force on both sides.

Also, the shape of the wall on the right makes no difference. It can be slanted any way you like. It won't force liquid between cylinders, because the fluid pressure everywhere is equal to the air pressure.

(On the other hand, capillary action has everything to do with surface tension, not with changing diameter of the tube. Capillary action is how plants drink against gravity.)

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There is no quick answer, except if the droplet is completely non-wetting or if it is at least partly wetting.

  • If it is completely non-wetting, it will be move towards the wide side of the funnel until it is a spherical drop touching only its wall.

  • If it is at least partly wetting, it will move to the narrow side until it reaches its apex (if air is allowed to flow out of course!)

  • If it is as you draw it, the answer is it depends. Then there is indeed an optimum distance $d$ of $B$ from the apex which can be approximately given for small $\beta$ by $$ d = \frac{L_0}{\beta} \frac{\cos \theta}{2 \sin \theta + \cos \theta} $$ where $\pi/2 \leq \theta \leq \pi$ is the contact angle, and $L_0$ the height of a cone having the volume of the drop and the angle $\beta$. The drop will move towards this equilibrium from its initial position, so it will be either to the left or right depending on where it starts.

This result is obtained by writing the pressure difference between the two menisci, $$ \delta p = \frac{\cos(\theta-\beta)}{h_B} - \frac{\cos(\theta+\beta)}{h_A} $$ and that the heights at point B, $h_B \simeq d \beta$, and A, $h_A \simeq h_B + L \beta \simeq h_B + L_0 - \beta d^2$. Then one can expand this for small $\beta$ and find the equilibrium.

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