Height of a stable droplet on a perfectly wetting surface

[Frabjous]

My question come from Rudolf Ortvay Competition in Physics. I really don't think elliptic integrals, complex functions, etc. should be involved... I think reasonable approximations are sufficient: this is an advanced competition for high school students, certainly not for university students with solid, advanced knowledge.

That is not accurate at all.
https://ortvay.elte.hu/2023/E23szab.pdf

 

[Hak]

That is not accurate at all.
https://ortvay.elte.hu/2023/E23szab.pdf

OK, fine, but the file I took the problem from should be for students with advanced high school knowledge, or for university students in their first years. It certainly doesn't involve elliptic integrals or solving problems that don't seem to have been solved yet. I would be really curious to understand how to solve the ODE set up by @haruspex, by any possible method. If anyone knows how to do this, their help would be crucial. Thank you.

 

[Frabjous]

OK, fine, but the file I took the problem from should be for students with advanced high school knowledge, or for university students in their first years. It certainly doesn't involve elliptic integrals or solving problems that don't seem to have been solved yet. I would be really curious to understand how to solve the ODE set up by @haruspex, by any possible method. If anyone knows how to do this, their help would be crucial. Thank you.

One has 10 days to solve a subset (10 out of 32 for 2023) of the problems using external resources. It is designed for university students or graduates. You are still mischaracterizing things. These are not problems that will have trivial solutions.

FYI, as your half-truths add up, fewer people will respond to your threads.

 

[Hak]

One has 10 days to solve a subset (10 out of 32 for 2023) of the problems using external resources. It is designed for university students or graduates. You are still mischaracterizing things. These are not problems that will have trivial solutions.

FYI, as your half-truths add up, fewer people will respond to your threads.

I do not understand the point of these sterile and inconclusive polemics. I am not mischaracterising things or stating half-truths. This problem is taken from the 2013 competition, in which (actually, in all editions) there are many problems repurposed from 200 More Puzzling Physics Problems, a book of interesting problems written by the same authors who propose this competition. Usually this book is suitable for advanced knowledge of high school or university students at the beginning of their studies. Not all problems have non-trivial and extremely complex solutions. From the results of 2013, according to some people who participated, this problem turned out - according to the authors themselves - to be simpler than others. I was initially wrong to generalise by saying that this is a competition for high school students, but I am still not underestimating everything. If I have proposed this problem, it is because I know for a fact that someone from outside the university has come up with the right solution.

 

[haruspex]

From your equations, I get:

$$r^2 \rho g \theta + 2 \alpha (\theta + r \theta') + \alpha (1 + r^2 \theta'') + 2 r (p_a + \alpha(1/r + \theta')) =2 p_a r.$$

How should I solve the ODE to which this equation leads?

Yes, that follows from my equations, but you can also replace r with s, producing an ODE in $\theta(s)$:
$\frac{\rho g}{\alpha}s^2\theta+2\theta+4s\theta'+s^2\theta''+3=0$
The constant 3 strikes me as highly suspicious, and that $4s\theta'$ term came from adding two $2s\theta'$ terms. Changing one sign would make those two terms cancel and turn the 3 into a 1:$$r^2 \rho g \theta + 2 \alpha (\theta + r \theta') + \alpha (1 + r^2 \theta'') + 2 r (p_a - \alpha(1/r + \theta')) =2 p_a r.$$
$\frac{\rho g}{\alpha}s^2\theta+2\theta+s^2\theta''+1=0$

Please try to check my equations in post #44.

 

[Hak]

Yes, that follows from my equations, but you can also replace r with s, producing an ODE in $\theta(s)$:
$\frac{\rho g}{\alpha}s^2\theta+2\theta+4s\theta'+s^2\theta''+3=0$
The constant 3 strikes me as highly suspicious, and that $4s\theta'$ term came from adding two $2s\theta'$ terms. Changing one sign would make those two terms cancel and turn the 3 into a 1:$$r^2 \rho g \theta + 2 \alpha (\theta + r \theta') + \alpha (1 + r^2 \theta'') + 2 r (p_a - \alpha(1/r + \theta')) =2 p_a r.$$
$\frac{\rho g}{\alpha}s^2\theta+2\theta+s^2\theta''+1=0$

Please try to check my equations in post #44.

Thank you very much @haruspex, but I did not understand how we should calculate $h$ from this equation. Thanks again.

 

[Hak]

Using the Laplace-Young equation: $$p=\alpha(1/r+\theta')+p_a$$

Why? Should not it be $p = \alpha \pi x + p_a$?

The gravitational downward force on an element is $$\pi r^2\sin(\theta)\rho g\Delta s$$

Why? Shouldn't one do the integration, which I submitted to you a few posts ago?

In general, I do not understand why we are considering $\theta$. Didn't we say that we imposed $\theta = \frac{\pi}{2}$, which maximises the tension force? I don't understand.

 

[haruspex]

Thank you very much @haruspex, but I did not understand how we should calculate $h$ from this equation. Thanks again.

As I noted in post #44, I do not see how to get the full height without solving for the true shape. The purpose of modelling my equations, in a spreadsheet say, is to see if it indicates an error in them. If so, maybe we can find the error(s). Once we have equations that produce reasonable shapes in the spreadsheet, perhaps they will be solvable.

 

[haruspex]

I do not understand why we are considering $\theta$. Didn't we say that we imposed $\theta = \frac{\pi}{2}$, which maximises the tension force?

We are asked to consider a drip that is as large as it can be without becoming unstable. I reason that instability is when some part of the surface reaches vertical. At that point, the vertical component of the tension force has reached its limit. So we are considering a shape in which the gradient just reaches vertical, then tips out again, forming an S.

At that point, $\theta = \frac{\pi}{2}$. But we need to find the general equation of shape, so in post #44 I considered a horizontal slice at an arbitrary height, so now $\theta$ can be any value from 0 to pi/2.

 

[Chestermiller]

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 $\phi(s)$.

 

[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.