Height of a stable droplet on a perfectly wetting surface

[haruspex]

I think you are going to have to set up and. solve the differential force balance on the surface (Young Laplace equation) in cylindrical coordinates and solve this in conjunction with the hydrostatic equilibrium equation within the drop. I would work in terms of axial contour length along the drop s ($ds=\sqrt{(dr)^2+(dz)^2}$) and the contour angle $\phi(s)$.

All of which I attempted in post #44. I would appreciate your reviewing it.

 

[Chestermiller]

All of which I attempted in post #44. I would appreciate your reviewing it.

Oops. Sorry, I missed that. I assume that a prime' represents differentiation with respect to contour length s, right?

 

[haruspex]

Oops. Sorry, I missed that. I assume that a prime' represents differentiation with respect to contour length s, right?

Yes.

 

[TSny]

The profile of a hanging drop can be described by some function $r(z)$ where $r$ and $z$ are as shown

A differential equation for $r(z)$ is $$\frac{r’’}{\left(1 + r’^2 \right)^{3/2}} - \frac{r'}{r\sqrt{1+r’^2}} = z - \frac{2}{R_0}$$ See this link under the section "Axisymmetric equations".

This equation assumes that distances are measured in units of the “capillary length” $\large \sqrt{\frac{\alpha}{g \Delta \rho}}$. Here $\alpha$ is the surface tension of the liquid and $\Delta \rho$ is the difference between the density of the liquid and the density of the surrounding medium. Assuming air is the surrounding medium, then $\Delta \rho$ can be taken to be the density of the liquid for all practical purposes. For water, the capillary length at room temperature is about 3 mm.

The parameter $R_0$ in the differential equation is the radius of curvature of the drop at the lowest point $z = 0$, measured in units of capillary length.

I used Mathematica to numerically solve the differential equation for specified values of $R_0$ and for a contact angle at the top of approximately zero degrees. Here are some results for $R_0$ decreasing from 3 to 0.5. The plots are labeled by the value of $R_0$ and the total volume $V$ of the drop (in units of capillary length cubed). The value $R_0 = 0.785$ corresponds to the case where the profile just achieves verticality (at about $z = 1.5$).

Note that as $R_0$ decreases from 3 to about 1.25, the volume increases. This is what we would expect as the drop accumulates more liquid. However, as $R_0$ continues to decrease from 1.25 to about 0.7, $V$ decreases. So, we have a local maximum of $V \approx 14$ near $R_0 = 1.25$. For water, this volume is $V \approx (14)(3 \, mm)^3 \approx 0.4$ cm³. My interpretation of this is that once the drop accumulates enough liquid for the volume to become about 14, there are no nearby equilibrium profile shapes that have greater volume. I think this implies that the drop must begin to “drip” at this point as it accumulates more liquid. The profiles shown for $R_0$ greater than 1.25 are equilibrium profiles, but they can’t be obtained by starting with $R_0 < 1.25$ and letting the drop slowly accumulate more liquid.

Here is a video that I made that shows a drop slowly growing on the bottom of a plate until it drips. (The shaking in the video is my hand shaking, not the plate.) Of course, the contact angle between the water and the plate is not zero as we assume in our problem. But, it appears to be pretty small. Does it look like the drop begins to drip when it achieves a shape roughly like that of the diagram for $R_0 = 1.25$? Or is that just wishful thinking on my part? It's not easy to tell when the drip actually begins.