# Height of a stable droplet on a perfectly wetting surface, shape of a water drop

I would like to clarify that this is not a homework question, but a problem that I chose to want to solve to satisfy my curiosity. I would like to understand whether the physical concepts I have applied and the physical situation I have modeled are reasonable, as I am really afraid that they are not consistent with the result I have obtained, a frankly unsolvable ODE.

The ceiling of a steam cabin is made of a perfectly wetting material. Since the temperature is slightly below the dewpoint, water droplets are forming on the horizontal ceiling. What is holding these droplets up there? Determine the height $$h$$ of the largest stable (just-not-dripping) droplet. Give this height $$h$$, that is, the distance between the lowest point of the droplet and the ceiling, as a function of the water density $$\rho$$, surface tension $$\alpha$$ and gravitational acceleration $$g$$. Calculate $$h$$ numerically.

First consideration) Perfect wetting means the contact angle is zero. The drop won't have constant mean curvature because the pressure inside at the bottom will be more than at the top.

Second consideration) Considering the forces acting on the portion of the droplet below some level, we have gravity, surface tension (where the surface of the lower portion meets the surface of the upper portion) and air pressure. So, if we take the portion of the droplet below a cut as the body acted on, there are four forces: pressure from air below pressure from water above weight of portion surface tension, but surface tension will contribute to the pressure in the droplet.

Third consideration) The steepest gradient, i.e. the slope of the surface, will be horizontal at the bottom and top (perfect wetting). It reaches a maximum somewhere in between. When the drop becomes unstable, the steepest gradient will be $$\frac{\pi}{2}$$ and it cannot accommodate any more weight. 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. But we need to find the general equation of shape, so I will consider a horizontal slice at an arbitrary height, so now $$\theta$$ can be any value from $$0$$ to $$\frac{\pi}{2}$$. I would 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 $$s$$ and the contour angle $$\theta(s)$$.

All derivatives are w.r.t. distance $$s$$ along the surface from lowest point unless otherwise stated. We consider a horizontal slice through the droplet. $$\theta$$ is the angle of the surface to the horizontal, $$p$$ is pressure within the droplet, $$p_a$$ is atmospheric pressure. Using the Laplace-Young equation: $$p=\alpha \left(\frac{1}{r}+\theta'\right)+p_a$$.

$$F$$ is the force exerted by adjacent layers: $$F=p\pi r^2$$ T is the vertical component of the force due to tension in the surface $$T=\alpha 2\pi r\sin(\theta)$$ The gravitational downward force on an element is $$\pi r^2\sin(\theta)\rho g\Delta s$$ The upward vertical force due to air pressure on an element is $$p_a2\pi r\Delta r$$ Force balance gives: $$\pi r^2\sin(\theta)\rho g\Delta s+\Delta T+\Delta F=p_a2\pi r\Delta r$$ $$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$ $$r'=\cos(\theta)$$ $$T'=2\alpha\pi (\cos(\theta)\sin(\theta)+r\cos(\theta)\theta')$$ $$p'=\alpha(r^{-2}\cos(\theta)+\theta'')$$ $$F'=p'\pi r^2+p2\pi r\cos(\theta) = \alpha\pi(\cos(\theta)+r^2\theta'')+(p_a+\alpha(1/r+\theta')) 2\pi r\cos(\theta)$$ $$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$

From these equations, I get: $$r^2 \rho g \theta + 2 \alpha (\theta + r \theta') + \alpha (1 + r^2 \theta'') + 2 r (p_a + \alpha(1/r + \theta')) =2 p_a r.$$ We can also replace $$r$$ with $$s$$, producing an ODE in $$\theta(s)$$: $$\frac{\rho g}{\alpha}s^2\theta+2\theta+4s\theta'+s^2\theta''+3=0.$$

Ok, this produces an ODE, but there are too many variables. It has both $$r(s)$$ and $$\theta(s)$$. In principle, that can be resolved using $$r'=\cos(\theta)$$, but eliminating theta will result in $$\arccos \theta$$ terms, and to eliminate $$r$$ we would have to differentiate the equation so that it only involves derivatives of $$r$$, not $$r$$ itself. For these reasons, I suspect the problem intends we take the portion below the point where the surface is vertical as being a hemisphere. It won't be far wrong. But even then, it will only let us get a height for that portion. To find the entire height we would still need to solve the shape equation.

I don't want to ask the question at all in terms of checking my algebraic calculations - I have rechecked them many times. Rather, I would like to understand whether it is my assumptions of the physical situation that lead me to these apparently wrong calculations. Thanks.

Edit: 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 $$1$$: $$r^2 \rho g \theta + 2 \alpha (\theta + r \theta') + \alpha (1 + r^2 \theta'') + 2 r (p_a - \alpha(1/r + \theta')) =2 p_a r.$$ $$\frac{\rho g}{\alpha}s^2\theta+2\theta+s^2\theta''+1=0$$.

I still cannot figure out what my errors are in the physical assumptions and related concepts. Could you help me point them out? The issue regarding the shape of a falling drop seems very difficult to fix.

• From Wikipedia's definition of contact angle en.wikipedia.org/wiki/Contact_angle, contact angle 0 means the droplet is flat, and there's no shape to be discussing about, maybe you mean something else by your first consideration? Commented Nov 10, 2023 at 18:21
• Thanks for information. I had assumed the contact angle to be 0 because the material is perfectly wetting. Apparently this is wrong, but I knew it was so by definition.
– Bml
Commented Nov 10, 2023 at 19:31

If I understand your elaborations correctly, your problem is this: Since the surface is perfectly wetting¹, the contact angle in the typical sense does not matter since there is no point at which the portion of the liquid forming the droplet is in contact with your actual surface. The first thing your condensing liquid does is to coat the entire ceiling. Before that has happened, no droplets will form. Afterwards, the layer of liquid coating the surface will just stay as it is, and below droplets will form as goverened by a balance of surface tension and gravity. These droplets will not interact with the actual ceiling but only with the layer of liquid coating it².

Here is how I would approach this problem: First fix the volume of your droplet. Consider functions $$f: ℝ⁺→ℝ⁺$$ that describe half the vertical cross-section of your droplet. These completely determine the droplet’s shape due to rotational symmetry. (Functions may not suffice here such that you need curves instead.) Use variational calculus or some numerical method to find that function $$f$$ that minimises the energy of your droplet taking into account the surface tension (as energy per surface of your function) and the gravitational energy of the water descending from the surface. You do not have to consider the adhesion to the ceiling as that concerns the unchanged layer that perpetually wets it. Finally find the volume at which this function becomes singular in a way corresponding to a tearing droplet, and determine the corresponding height.

¹ I here assume that the tiny difference made by gravity doesn’t change that.
² Mind that this is only an approximation based on the assumption that molecules only interact with direct neighbours. If you take into account further ranging molecular interactions, things become a bit more complicated, but then you’ll certainly know these details about your molecular interactions.

• Good answer. In what sense do you say "First fix the volume of your drop. Consider functions [...] that describe half the cross-section of your droplet. These completely determine the droplet’s shape due to rotational symmetry."? I would have other things to think about, partly stemming from your reply. If you and @Jono94 want, we can start a discussion at the following link: chat.stackexchange.com/rooms/149635/…
– Bml
Commented Nov 10, 2023 at 18:31
• @Bml: Can you specify your question? I am happy to clarify, but right now I can only say that I mean everything as it is written or paraphrase in the hope that this addresses your concern. Commented Nov 10, 2023 at 18:37
• Of course. I had already calculated the volume of the drop, but apparently my method is not correct. I cannot understand how the two methods (mine and yours) differ. Could you then paraphrase your words that I quoted in the previous comment? I didn't really understand what the shape of the drop should be: bell-shaped, hemispherical, spherical? Also, I would be immensely grateful if you would like to start a discussion at the link I attached to the previous comment, I have written some thoughts on which I would greatly appreciate a follow-up. Thanks.
– Bml
Commented Nov 11, 2023 at 9:50
• The shape of the droplet would be roughly bell-shaped, but not pre-determined. Also see my edit. Commented Nov 11, 2023 at 20:48
• Thank you very much. Did you happen to see the updates I set up in the room whose link I posted? Let me know.
– Bml
Commented Nov 11, 2023 at 21:09