Height of a stable droplet on a perfectly wetting surface

[Hak]

Homework Statement

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. Show all the passages conducting to symbolic formula.

Relevant Equations

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I would assume that the droplet on the ceiling is spherical, since it is the shape that minimizes the surface energy for a given volume. The droplet is held by the surface tension force, which acts along the contact line between the droplet and the ceiling and is balanced by the weight of the droplet, which acts downward. Therefore:

$$2\pi \alpha r=\frac{4}{3}\rho g\pi r^3$$,
where $r$ is the radius of the droplet.

Solving for $r$, I get:

$$2\pi \alpha r=\frac{4}{3}\rho g\pi r^3 \Rightarrow \ \alpha = \frac{2}{3}\rho g r^2 \Rightarrow \ r^2 = \frac{3}{2}\frac{\alpha}{\rho g} \Rightarrow \ r = \sqrt{\frac{3}{2}\frac{\alpha}{\rho g}}$$

The height $h$ of the droplet would be the distance from the lowest point of the droplet to the ceiling, which is equal to $r \cos \theta$, where $\theta$ is the contact angle. Since the ceiling is perfectly wetting, the contact angle is zero, so I get:

$$h = \sqrt{\frac{3}{2}\frac{\alpha}{\rho g}}$$Plugging in the values of $\alpha=0.0728 \ N/m$, $\rho=1000 \ kg/m^3$, and $g=9.81 \ m/s^2$, we get:

$$h \approx 2.7×10^{-3} \ m$$.

This reasoning seems too simplistic and incorrect to me. Where do I go wrong? Do you have any hint or insights? Thanks.

 

[Hill]

Homework Statement: 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. Show all the passages conducting to symbolic formula.
Relevant Equations: /

Therefore:

2παr=43ρgπr3,
where r is the radius of the droplet.

The left side of this equation tells me that the droplet is held by its equator. Then the droplet is not a sphere. But the right side is volume of a sphere. --?

 

[Hak]

The left side of this equation tells me that the droplet is held by its equator. Then the droplet is not a sphere. But the right side is volume of a sphere. --?

Yes, you are right. Do you have any hints on this?

 

[Hill]

Yes, you are right. Do you have any hints on this?

IIRC the surface of the droplet will make an angle with the horizontal surface of the ceiling.

 

[Hak]

IIRC the surface of the droplet will make an angle with the horizontal surface of the ceiling.

I cannot understand this. If the material is perfectly wetting, shouldn't the angle between the surface of the droplet and the horizontal surface of the ceiling be zero?

 

[Hill]

I cannot understand this. If the material is perfectly wetting, shouldn't the angle between the surface of the droplet and the horizontal surface of the ceiling be zero?

I understand that it would be zero if there were no weight of the droplet pulling it off.

 

[Hak]

I understand that it would be zero if there were no weight of the droplet pulling it off.

So, I followed your advice by trying various approaches, but they all turned out to be very time-consuming and unlikely. They involve third- and fourth-degree equations that cannot be solved canonically, according to traditional methods: we need Cardano formulas, etc... If you can confirm to me that you all are sure that there is a simpler and more reasonable approach that does not involve these types of equations, I look forward to some advice to try other methods. Thank you.

 

[haruspex]

I understand that it would be zero if there were no weight of the droplet pulling it off.

Perfect wetting means the contact angle is zero, no exceptions. The shape will be like an upturned bell. It won't have constant mean curvature (https://en.wikipedia.org/wiki/Mean_curvature) because the pressure inside at the bottom will be more than at the top. (Edited)
Here, however, you only have to consider a droplet about to fall. When a droplet falls from, say, a dripping tap, what does it look like just before it becomes unstable?

@Hak, for the first part, you mention surface tension and gravity. No other forces you can think of?

 

[Hak]

@Hak, for the first part, you mention surface tension and gravity. No other forces you can think of?

I'm not sure, but I believe we have also to consider gravitational force and buoyant force. Is it correct?

 

[haruspex]

I'm not sure, but I believe we have also to consider gravitational force and buoyant force. Is it correct?

You already consider gravity, correctly. Can you be more specific about buoyant force?

 

[Hak]

You already consider gravity, correctly. Can you be more specific about buoyant force?

I gave a spur-of-the-moment answer. But, maybe, thinking it through, there is a downwards force from the pressure difference $p_{\text{fluid}}-p_{\text{air}}$ on the area of the horizontal cut.

 

[haruspex]

I gave a spur-of-the-moment answer. But, maybe, thinking it through, there is a downwards force from the pressure difference $p_{\text{fluid}}-p_{\text{air}}$ on the area of the horizontal cut.

That's not quite how I would put it. 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.

When a drip is soon to fall, what can you say about the shape?

 

[Hak]

That's not quite how I would put it. 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.

When a drip is soon to fall, what can you say about the shape?

Which has a spherical shape?

 

[haruspex]

Which has a spherical shape?

That is rather past the point at which it becomes unstable. Or did you mean hemispherical?

 

[Hak]

That is rather past the point at which it becomes unstable. Or did you mean hemispherical?

Sorry for expressing myself wrongly. I meant inverted spherical cap.

 

[haruspex]

Sorry for expressing myself wrongly. I meant inverted spherical cap.

It won’t be exactly a spherical surface. As I posted, it will be more bell-shaped. I'll see if I can figure out what it will be, but maybe spherical is near enough. More importantly, how big a part of a sphere? I.e. what do you think the steepest gradient will be?

 

[Hak]

It won’t be exactly a spherical surface. As I posted, it will be more bell-shaped. I'll see if I can figure out what it will be, but maybe spherical is near enough. More importantly, how big a part of a sphere? I.e. what do you think the steepest gradient will be?

What do you mean by "steepest gradient"?

 

[haruspex]

What do you mean by "steepest gradient"?

The slope of the surface. It will be horizontal at the bottom and top (perfect wetting). It reaches a maximum somewhere in between. When the drop becomes unstable, what do you think the steepest gradient will be? Draw some pictures, study a slowly dripping tap.

 

[Hak]

It won’t be exactly a spherical surface. As I posted, it will be more bell-shaped. I'll see if I can figure out what it will be, but maybe spherical is near enough. More importantly, how big a part of a sphere? I.e. what do you think the steepest gradient will be?

This is the process I had adopted before your advice. I took a coordinate system with $x$ pointing to the right and $y$ pointing down and imagined the drop hanging from the $y=0$ plane. "Cutting" the drop horizontally at $y$, the circumference of the bell at the cut is $2\pi x$ and the total force on the cut part is $2 \pi x \alpha$, where $\alpha$ is the surface tension per unit length (a constant). The surface tension is parallel to the tangent and the total upwards force on the cut part has the component
$$F_y = -2\pi x \alpha \sin(\theta),$$

where $\theta$ is the angle (a negative value) of the tangent at $(x,y)$.
What do you think? Obviously this is not the complete process, but I got here.

I was also thinking of employing energy considerations, but I don't think I have suitable knowledge here.

 

[Hill]

When a droplet falls from, say, a dripping tap, what does it look like just before it becomes unstable?

A teardrop?

 

[haruspex]

This is the process I had adopted before your advice. I took a coordinate system with $x$ pointing to the right and $y$ pointing down and imagined the drop hanging from the $y=0$ plane. "Cutting" the drop horizontally at $y$, the circumference of the bell at the cut is $2\pi x$ and the total force on the cut part is $2 \pi x \alpha$, where $\alpha$ is the surface tension per unit length (a constant). The surface tension is parallel to the tangent and the total upwards force on the cut part has the component
$$F_y = -2\pi x \alpha \sin(\theta),$$

where $\theta$ is the angle (a negative value) of the tangent at $(x,y)$.
What do you think? Obviously this is not the complete process, but I got here.

I was also thinking of employing energy considerations, but I don't think I have suitable knowledge here.

Good.
Of course, that is just the force on that portion of the drop due to surface tension. There's the weight of the portion and air pressure from below.
Now, when it becomes unstable, there is a y value at which surface tension is no longer able to hold the drop in place. What does that suggest about $\theta$ at that latitude?

 

[Hak]

There's the weight of the portion

The volume $V$ of this part is the integral $V = \int_y^{y_{\text{tip}}}\pi x^2dy$, where $y_{\text{tip}}$ is the

y value at which surface tension is no longer able to hold the drop in place.

The mass of this volume is $\rho V$, where $\rho$ is the density of water, and the force of gravity thus $W = \rho gV$. Now, $V$ is an integral of $x^2$ versus $y$ and $\tan(\theta) = \frac{\mathrm{d}y}{\mathrm{d}x} = y'$, so

$$\sin(\theta) = \frac{y'}{\sqrt{1+y'^2}}$$.

We have:

$$F_y = - 2\pi \alpha x\frac{y'}{\sqrt{1+y'^2}}$$.

What does that suggest about $\theta$ at that latitude?

I am not sure I understand this question, nor do I have in mind how to calculate air pressure from below. But are you sure that there isn't a downwards force from the pressure difference between air and fluid on the area of the cut?
Thanks.

 

[haruspex]

I am not sure I understand this question

Look at your expression for $F_y$. If you increase the weight of the drop, how will $\theta$ change so as to try to hold things together? At what point will that no longer work?

how to calculate air pressure from below

Air pressure acting over a (horizontal) area.

a downwards force from the pressure difference between air and fluid on the area of the cut?

Ok, I think I see your point. 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 I've just had a nasty thought… surface tension will contribute to the pressure in the droplet. I'll need to think about that more.

 

[Hak]

Look at your expression for $F_y$. If you increase the weight of the drop, how will $\theta$ change so as to try to hold things together? At what point will that no longer work?

Perhaps what you mean is that the angle must vary according to $- \frac{\pi}{2} < \theta < 0$ because if $\theta = 0$, then the force cancels and the weight maximises? Let me know if this is correct or if I still don't understand. Thank you very much.

Ok, I think I see your point. 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 I've just had a nasty thought… surface tension will contribute to the pressure in the droplet. I'll need to think about that more.

Yes, that is what I meant. Regarding your last statement...it is absolutely true. I really wouldn't know how to approach this kind of complication. That's why I thought of energy considerations. For this point I haven't the faintest idea.
Thank you again.

 

[haruspex]

Perhaps what you mean is that the angle must vary according to $- \frac{\pi}{2} < \theta < 0$ because if $\theta = 0$, then the force cancels and the weight maximises? Let me know if this is correct or if I still don't understand. Thank you very much.

No. According to your expression for $F_y$, what angle maximises its magnitude?

 

[Hak]

No. According to your expression for $F_y$, what angle maximises its magnitude?

$\frac{\pi}{2}$, I think.

 

[haruspex]

$\frac{\pi}{2}$, I think.

Right. When the steepest gradient reaches that the droplet becomes unstable. It cannot accommodate any more weight. So take the horizontal cut at that point.
If the radius there is R, what is the pressure in the water at that level?

 

[Hak]

If the radius there is R, what is the pressure in the water at that level?

Maybe $p = \frac{\alpha}{R}$?

 

[haruspex]

Maybe $p = \frac{\alpha}{R}$?

If you mean gauge pressure, yes. So what force does the water above exert on the lower portion?

 

[Hak]

If you mean gauge pressure, yes. So what force does the water above exert on the lower portion?

A pressure force proportional to the depth of the water and the density of the water? Also, according to which you said, I think that the cohesive force between the upper part and lower part of the cutting boundary should be subtracted from the other downward forces, but I don't know how. Maybe as a constant multiplied by the area of the cut? Thanks.

 

[haruspex]

A pressure force proportional to the depth of the water and the density of the water?

As discussed, it might not be as simple as that. But you don't need to go that route. You know the pressure there from the radius, so what is the force?

 

[Hak]

As discussed, it might not be as simple as that. But you don't need to go that route. You know the pressure there from the radius, so what is the force?

Thanks. I would say $F= \pi x^2 \frac{\alpha}{R}$. Correct?

 

[haruspex]

Thanks. I would say $F= \pi x^2 \frac{\alpha}{R}$. Correct?

How is x different from R?

 

[Hak]

How is x different from R?

I don't understand, what do you mean?

 

[haruspex]

I don't understand, what do you mean?

Aren't they both the radius of the cut disc?

 

[Hak]

Aren't they both the radius of the cut disc?

I don't think so. If we use Laplace-Young equation, we have $\Delta P = \alpha \left(\frac{1}{R_1} +\frac{1}{R_2} \right)$, where $\Delta P$ is the pressure difference across the interface and $R_1 = \frac{\mathrm{d}x}{\mathrm{d}(\sin \theta)}$ and $R_2 = \frac{x}{\sin \theta}$ radii of curvature normal to the surface.
Am I wrong in thinking that, and is the matter simpler than that? Perhaps yes, it is as you say, $F = \pi x \alpha$.

 

[haruspex]

I don't think so. If we use Laplace-Young equation, we have $\Delta P = \alpha \left(\frac{1}{R_1} +\frac{1}{R_2} \right)$, where $\Delta P$ is the pressure difference across the interface and $R_1 = \frac{\mathrm{d}x}{\mathrm{d}(\sin \theta)}$ and $R_2 = \frac{x}{\sin \theta}$ radii of curvature normal to the surface.
Am I wrong in thinking that, and is the matter simpler than that? Perhaps yes, it is as you say, $F = \pi x \alpha$.

We agreed that at the cut disc the surface is vertical, so forms a cylinder. The radii are infinity and R.

 

[Hak]

We agreed that at the cut disc the surface is vertical, so forms a cylinder. The radii are infinity and R.

Yes, you are absolutely right. The $F$ force will be $F = \pi x \alpha$. Now what?

 

[haruspex]

Yes, you are absolutely right. The $F$ force will be $F = \pi x \alpha$. Now what?

As I mentioned, the pressure you found in post #28 is gauge pressure, i.e. the extra pressure above atmospheric.

 

[Hak]

As I mentioned, the pressure you found in post #28 is gauge pressure, i.e. the extra pressure above atmospheric.

Okay, I see, so what should I find?

 

[haruspex]

Okay, I see, so what should I find?

Proceed with finding expressions for the other forces on the portion.
You will need to assume a shape for it. For now, take it to be hemispheric.

 

[Hak]

Proceed with finding expressions for the other forces on the portion.
You will need to assume a shape for it. For now, take it to be hemispheric.

I have already found the gravitational force and the upward force $F_y$, now I only need to find the force due to air pressure? What other forces are you talking about, if any?

 

[Hak]

You will need to assume a shape for it. For now, take it to be hemispheric.

I'm not sure I understand what you are saying. That is, I didn't understand how I should assume in advance that the shape of the drop is hemispherical. I would equalize upward forces and downward forces, differentiate both members of the equation, and get an ODE from which to derive the shape of the drop. Assuming the latter to be hemispherical (as I understand it, that's not really correct) right off the bat, I just wouldn't know how to do it.

 

[haruspex]

I'm not sure I understand what you are saying. That is, I didn't understand how I should assume in advance that the shape of the drop is hemispherical. I would equalize upward forces and downward forces, differentiate both members of the equation, and get an ODE from which to derive the shape of the drop. Assuming the latter to be hemispherical (as I understand it, that's not really correct) right off the bat, I just wouldn't know how to do it.

I have been trying to discover the correct shape. Can't find anything online. Here is where I got to with my own efforts:

All derivatives are wrt 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, pa is atmospheric pressure.
Using the Laplace-Young equation: $$p=\alpha(1/r+\theta')+p_a$$
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: $$\pi r^2\sin(\theta)\rho g\Delta s+\Delta T+\Delta F=p_a2\pi r\Delta r$$
$$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$
$$r'=\cos(\theta)$$
$$T'=2\alpha\pi (\cos(\theta)\sin(\theta)+r\cos(\theta)\theta')$$ $$p'=\alpha(r^{-2}\cos(\theta)+\theta'')$$
$$F'=p'\pi r^2+p2\pi r\cos(\theta)$$
$$=\alpha\pi(\cos(\theta)+r^2\theta'')+(p_a+\alpha(1/r+\theta')) 2\pi r\cos(\theta)$$
$$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$

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.
Maybe my equations are wrong.

 

[Hak]

I have been trying to discover the correct shape. Can't find anything online. Here is where I got to with my own efforts:

All derivatives are wrt 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, pa is atmospheric pressure.
Using the Laplace-Young equation: $$p=\alpha(1/r+\theta')+p_a$$
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: $$\pi r^2\sin(\theta)\rho g\Delta s+\Delta T+\Delta F=p_a2\pi r\Delta r$$
$$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$
$$r'=\cos(\theta)$$
$$T'=2\alpha\pi (\cos(\theta)\sin(\theta)+r\cos(\theta)\theta')$$ $$p'=\alpha(r^{-2}\cos(\theta)+\theta'')$$
$$F'=p'\pi r^2+p2\pi r\cos(\theta)$$
$$=\alpha\pi(\cos(\theta)+r^2\theta'')+(p_a+\alpha(1/r+\theta')) 2\pi r\cos(\theta)$$
$$\pi r^2\sin(\theta)\rho g+T'+F'=p_a2\pi rr'$$

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.
Maybe my equations are wrong.

So, what do you recommend?

 

[haruspex]

So, what do you recommend?

I recommend you study my equations to see if you can find an error or a way forward.
I just found https://physics.stackexchange.com/questions/218504/shape-of-a-water-drop; same question but no successful solution.

 

[Hak]

I recommend you study my equations to see if you can find an error or a way forward.
I just found https://physics.stackexchange.com/questions/218504/shape-of-a-water-drop; same question but no successful solution.

OK, thanks. Maybe this link https://www.physicsforums.com/threads/wetting-on-the-outside-of-a-glass-cylinder.1054405/ can be useful?

 

[haruspex]

I note this in that thread:

It looks like people are still (in 2022) trying to work out the meniscus for the water surface inside a cylindrical tube

This suggests the analytical shape is an unsolved problem.
Could try a numerical solution. At the bottom, we can approximate $r=s, \sin(\theta)=\theta, \cos(\theta)=1$. Thereafter, apply the equations from my post.
At the least, if the shape looks unreasonable then we'll know my equations are wrong.

Where does your question come from?

 

[Hak]

I note this in that thread:

This suggests the analytical shape is an unsolved problem.
Could try a numerical solution. At the bottom, we can approximate $r=s, \sin(\theta)=\theta, \cos(\theta)=1$. Thereafter, apply the equations from my post.
At the least, if the shape looks unreasonable then we'll know my equations are wrong.

Where does your question come from?

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.

 

[Hak]

I note this in that thread:

This suggests the analytical shape is an unsolved problem.
Could try a numerical solution. At the bottom, we can approximate $r=s, \sin(\theta)=\theta, \cos(\theta)=1$. Thereafter, apply the equations from my post.
At the least, if the shape looks unreasonable then we'll know my equations are wrong.

Where does your question come from?

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?