# Wetting on the outside of a glass cylinder

• guv
In summary, The conversation discusses difficulties in analyzing the shape of a water surface, specifically in terms of the unknown variables ##r(y)## and ##F_{surface \; tension}##. The idea of using the minimization of potential energy and the variational principle is brought up, as well as the use of force balance and energy minimization principles. The conversation then moves onto the estimation of the meniscus on a flat plate and the cylindrical problem, with a proposed approach involving a 2nd order ODE. The conversation ends with a detailed explanation of the force balance equations and the Young equation in terms of the parameters involved.
guv
Homework Statement
As in the figure, when water wets on the outside of the glass cylinder, it follows a shape as shown. Is it possible to derive or estimate how ##H## depends on ##g##, water density ##\rho##, radius of the glass tube ##r_0##, surface tension ##\sigma##, wetting angle ##\theta## at the top?
Relevant Equations
##F_{surface \; tension} = \int_0^H \pi (r(y)^2 - r_0^2) \rho g dy##
There are two difficulties, first ##r(y)## is not known, the surface tension force ##F_{surface \; tension}## is not known either. We can write net surface tension force as
##F_{surface \; tension} = \int_0^H 2 \pi r (\sin \arctan \frac{dy}{dr(y)}) dy ##

Is there something else we could use to analyze the shape of the water surface ##r(y)##?

A related question as I am thinking about this is how to apply minimization of potential energy? Maybe variational principle could yield the ideal curve ##r(y)##? This then reminds me how we generally approach wetting inside of a glass tube, there we simply use a force balance and I don't see how we could use energy minimization principle for the internal wetting setup.

Looking for ideas how you might approach this, thanks!

#### Attachments

• surface_tension.png
3.8 KB · Views: 44
Do you know how to solve this for meniscus on a flat plate?

For the cylindrical problem, I would arc length s measured downward from the top as the independent variable and the contact angle ##\phi(s)## as the dependent variable. Then $$\frac{dz}{ds}=\cos{\phi}$$and $$\frac{dr}{ds}=\sin{\phi}$$where z is measured downward from the contact point.

There are going to be differential equations with respect to s.

guv
The force on the LHS of your equation appears to be the vertical component only.
Can you rewrite it in terms of ##\sigma## and ##r'##, and, instead of an integral, just address the element between ##r## and ##r+dr##?

guv
Thanks for the ideas, I wasn't sure if I needed to consider Young's equation and the multi-phase interactions to come up with an reasonable estimation. It looks like people are still (in 2022) trying to work out the meniscus for the water surface inside a cylindrical tube. So this is not a trivial problem. However, maybe as an extremely rough estimation, from the equations above, we can say the weight of the fluid is ##\propto H^3 \rho g##, while the surface tension force is ##\propto 2 \pi r_0 \cos \theta##, therefore
## H \propto \sqrt[3]{ \frac{ \sigma r_0}{\rho g} }##

The problem with this estimation I can see is proportionality on ##r_0##, capillarity suggests the smaller the ##r_0##, the greater the ##H##.

guv said:
Thanks for the ideas, I wasn't sure if I needed to consider Young's equation and the multi-phase interactions to come up with an reasonable estimation.
This would definitely be the approach that I would use. If you would like me to flesh out the analysis, please let me know. My current assessment is that it will involve be a 2nd order ODE, split boundary value problem solving the the liquid reduced pressure value at the contact location.

Chestermiller said:
This would definitely be the approach that I would use. If you would like me to flesh out the analysis, please let me know. My current assessment is that it will involve be a 2nd order ODE, split boundary value problem solving the the liquid reduced pressure value at the contact location.
Definitely would be interested to see how you might approach this with more details. Thanks!

guv said:
Definitely would be interested to see how you might approach this with more details. Thanks!
OK. The unit vector in the s direction is $$\hat{i}_s=\cos{\phi}\hat{i}_z+\sin{\phi}\hat{i}_r$$If we do a differential force balance for the "window" of interface between s and ##s+ds## and between ##\theta## and ##\theta+d\theta##, we obtain $$\sigma\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta+\sigma\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta+(p_a-p_l)rd\theta ds(-\hat{i}_n)=0$$or$$\sigma\frac{\partial (r\hat{i}_s)}{\partial s}-\sigma\hat{i}_r=(p_a-p_l)r\hat{i}_n\tag{1}$$where ##p_l## is the pressure on the liquid side of the interface: $$p_l=p_a-\rho g(h-z)\tag{2}$$with h representing the (unknown) height of the contact point above the pool surface. The unit vector ##\hat{i}_n## is the unit normal vector directed from the water side of the interface to the air side of the interface, and ##\sigma## is the surface tension.$$\hat{i}_n=\cos{\phi}\hat{i}_r-\sin{\phi}\hat{i}_z$$If we combine Eqns. 1 and 2, and dot the resulting relationship with ##\hat{i}_n##, we obtain $$\sigma \frac{d\phi}{ds}-\sigma \frac{\cos{\phi}}{r}=\rho g (h-z)$$This is the Young equation in terms of our parameters.

OK so far?

Last edited:
guv and TSny
Chestermiller said:
OK. The unit vector in the s direction is $$\hat{i}_s=\cos{\phi}\hat{i}_z+\sin{\phi}\hat{i}_r$$If we do a differential force balance for the "window" of interface between s and ##s+ds## and between ##\theta## and ##\theta+d\theta##, we obtain $$\sigma\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta+\sigma\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta+(p_a-p_l)rd\theta ds(-\hat{i}_n)=0$$or$$\sigma\frac{\partial (r\hat{i}_s)}{\partial s}-\sigma\hat{i}_r=(p_a-p_l)r\hat{i}_n\tag{1}$$where ##p_l## is the pressure on the liquid side of the interface: $$p_l=p_a-\rho g(h-z)\tag{2}$$with h representing the (unknown) height of the contact point above the pool surface. The unit vector ##\hat{i}_n## is the unit normal vector directed from the water side of the interface to the air side of the interface, and ##\sigma## is the surface tension.$$\hat{i}_n=\cos{\phi}\hat{i}_r-\sin{\phi}\hat{i}_z$$If we combine Eqns. 1 and 2, and dot the resulting relationship with ##\hat{i}_n##, we obtain $$\sigma \frac{d\phi}{ds}-\sigma \frac{\cos{\phi}}{r}=\rho g (h-z)$$This is the Young equation in terms of our parameters.

OK so far?
I agree with the final equation being the Young's equation in this setup, can you elaborate on how this part ##\sigma\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta+\sigma\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta## is obtained. They are correct, I am wondering if these come from any formal definition such as ##\nabla \cdot \hat n##, in particular, ##\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta## and ##\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta##

guv said:
I agree with the final equation being the Young's equation in this setup, can you elaborate on how this part ##\sigma\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta+\sigma\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta## is obtained. They are correct, I am wondering if these come from any formal definition such as ##\nabla \cdot \hat n##, in particular, ##\frac{\partial (r\hat{i}_s)}{\partial s}ds d\theta## and ##\frac{\partial \hat{i}_\theta}{\partial \theta}ds d\theta##
The in-plane force on the side of our little window at ##s+ds## is $$[\sigma(rd\theta)\hat{i}_s]_{s+ds}$$The in-plane force on the side of our little window at ##s## is $$[\sigma(rd\theta)\hat{i}_s]_{s}$$The in-plane force on the side of our little window at ##\theta +d\theta## is $$[\sigma ds\hat{i}_{\theta}]_{\theta +d\theta}$$The in-plane force on the side of our little window at ##\theta## is $$[\sigma ds\hat{i}_{\theta}]_{\theta}$$The pressure force normal to the window is $$(p_a-p_l)(rd\theta)ds(-\hat{i}_n)$$
What does this give for the force balance on the window?

guv
Thanks for the clarification. What would you suggest how to determine ##h## from here? It looks like we need boundary condition for ##\phi(z=0)## which needs to be determined from Young's equation at the air-glass-water contact point? The other boundary condition ##r(z=0) = r_0## is given. i.e. if we are given ##r_0, \phi(z=0)##, then ##h## should be uniquely determined including ##s(\phi)##. Alternatively, one can also start with ##r_0## and the surface energy numbers of air-glass-water to determine ##\phi(z=0)## and subsequently the other parameters.

guv said:
Thanks for the clarification. What would you suggest how to determine ##h## from here? It looks like we need boundary condition for ##\phi(z=0)## which needs to be determined from Young's equation at the air-glass-water contact point? The other boundary condition ##r(z=0) = r_0## is given. i.e. if we are given ##r_0, \phi(z=0)##, then ##h## should be uniquely determined including ##s(\phi)##. Alternatively, one can also start with ##r_0## and the surface energy numbers of air-glass-water to determine ##\phi(z=0)## and subsequently the other parameters.
The next step is to reduce the equations to dimensionless form. For all length parameters (including radius), I would use the characteristic length scale ##\sqrt{\frac{\sigma}{\rho g}}##.

In terms of the dimensionless parameters, at s = 0, ##\phi## is equal to the contact angle ##\phi_0## (specified), and at infinite s, ##\phi=\frac{\pi}{2}##. The dimensionless h has to be chosen so that, at large s (say s = 10), z =h and ##\phi=\frac{\pi}{2}##. In this split boundary value problem, you would iterate on h until these conditions at the far end are satisfied.

I would start out first solving the problem with no curvature, where the curvature ##\cos{\phi}/r## approaches zero. Then I would use the h value for this case as a starting guess for the problem for a curved rod.

guv
Interesting... My first thought was using matlab ode solver but I can see where your approach is coming from. Thanks!

guv said:
Interesting... My first thought was using matlab ode solver but I can see where your approach is coming from. Thanks!
You can still use the matlab equation solver on your 3 1st order odes

• Introductory Physics Homework Help
Replies
2
Views
762
• Introductory Physics Homework Help
Replies
63
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
864
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
19
Views
846
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
283
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
2K