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

[Supercazzola]

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.

Quote

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

ρ

, surface tension

α

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

π2

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

θ

can be any value from

0

to

π2

. 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

θ(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.

θ

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=α(1r+θ′)+pa

.

 

F

is the force exerted by adjacent layers:

F=pπr2

T is the vertical component of the force due to tension in the surface

T=α2πrsin(θ)

The gravitational downward force on an element is

πr2sin(θ)ρgΔs

The upward vertical force due to air pressure on an element is

pa2πrΔr

Force balance gives:

πr2sin(θ)ρgΔs+ΔT+ΔF=pa2πrΔr

πr2sin(θ)ρg+T′+F′=pa2πrr′

r′=cos(θ)

T′=2απ(cos(θ)sin(θ)+rcos(θ)θ′)

p′=α(r−2cos(θ)+θ′′)

F′=p′πr2+p2πrcos(θ)=απ(cos(θ)+r2θ′′)+(pa+α(1/r+θ′))2πrcos(θ)

πr2sin(θ)ρg+T′+F′=pa2πrr′

 

From these equations, I get:

r2ρgθ+2α(θ+rθ′)+α(1+r2θ′′)+2r(pa+α(1/r+θ′))=2par.

We can also replace

r

with

s

, producing an ODE in

θ(s)

:

ρgαs2θ+2θ+4sθ′+s2θ′′+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 would like to understand whether it is my assumptions of the physical situation that lead me to these apparently wrong calculations. 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. Thanks.

Edited November 8, 2023 by Supercazzola

[exchemist]

38 minutes ago, Supercazzola said:

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.

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

π2

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

θ

can be any value from

0

to

π2

. 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

θ(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.

θ

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=α(1r+θ′)+pa

.

 

F

is the force exerted by adjacent layers:

F=pπr2

T is the vertical component of the force due to tension in the surface

T=α2πrsin(θ)

The gravitational downward force on an element is

πr2sin(θ)ρgΔs

The upward vertical force due to air pressure on an element is

pa2πrΔr

Force balance gives:

πr2sin(θ)ρgΔs+ΔT+ΔF=pa2πrΔr

πr2sin(θ)ρg+T′+F′=pa2πrr′

r′=cos(θ)

T′=2απ(cos(θ)sin(θ)+rcos(θ)θ′)

p′=α(r−2cos(θ)+θ′′)

F′=p′πr2+p2πrcos(θ)=απ(cos(θ)+r2θ′′)+(pa+α(1/r+θ′))2πrcos(θ)

πr2sin(θ)ρg+T′+F′=pa2πrr′

 

From these equations, I get:

r2ρgθ+2α(θ+rθ′)+α(1+r2θ′′)+2r(pa+α(1/r+θ′))=2par.

We can also replace

r

with

s

, producing an ODE in

θ(s)

:

ρgαs2θ+2θ+4sθ′+s2θ′′+3=0.

 

Ok, this produces an ODE, but there are too many variables. It has both

r(s)

and

θ(s)

. In principle, that can be resolved using

r′=cos(θ)

, but eliminating

θ

will result in

arccosθ

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 would like to understand whether it is my assumptions of the physical situation that lead me to these apparently wrong calculations. 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. Thanks.

I can't help you with the maths, but physically I think it may not be correct to assume that the surface of the droplet becomes vertical when the maximum mass it can support is reached. You have a constant surface tension (of 70 dynes/cm or so) and there will come a point at which the weight of the water exceeds the force from the vertical component of this tension. But it is not immediately clear what maximum angle from horizontal the surface will attain when that point is reached. It could well be that the weight becomes too much when the angle is only π/4, or some other angle.  I think this has to be calculated. 

Edited November 8, 2023 by exchemist

[Supercazzola]

18 minutes ago, exchemist said:

I can't help you with the maths, but physically I think it may not be correct to assume that the surface of the droplet becomes vertical when the maximum mass it can support is reached. You have a constant surface tension (of 70 dynes/cm or so) and there will come a point at which the weight of the water exceeds the force from the vertical component of this tension. But it is not immediately clear what maximum angle from horizontal the surface will attain when that point is reached. It could well be that the weight becomes too much when the angle is only π/4, or some other angle.  I think this has to be calculated. 

Thank you for your response. Could someone give me suggestions on how to calculate this angle? Thanks. However, my procedure that should lead to the water droplet shape for the time being still does not assume that the maximum angle is 90°. So, there must also be something else wrong.... I would be grateful if you could help me out.

Edited November 8, 2023 by Supercazzola

[exchemist]

25 minutes ago, Supercazzola said:

Thank you for your response. Could someone give me suggestions on how to calculate this angle? Thanks. However, my procedure that should lead to the water droplet shape for the time being still does not assume that the maximum angle is 90°. So, there must also be something else wrong.... I would be grateful if you could help me out.

This looks to me like quite an advanced applied mathematics question and as such is beyond me, I'm afraid.  I started wondering about catenaries for a moment, but I don't know what I'm talking about. I think @studiot may be a mathematician. I wonder if he has any ideas. 

[Supercazzola]

51 minutes ago, exchemist said:

This looks to me like quite an advanced applied mathematics question and as such is beyond me, I'm afraid.  I started wondering about catenaries for a moment, but I don't know what I'm talking about. I think @studiot may be a mathematician. I wonder if he has any ideas. 

Thanks. Any help is appreciated by me.

[studiot]

This is not a simple standard school problem so it would be helpful to tell us the context in which it set.

As said the contact angle at the ceiling must be zero.

So you then need a 'hump' shaped line that flattens out to parallel to the horizontal at both sides. As in Fig A.

Rotating this line will produce a domed hat like an inverted ww1 or ww2 tommie's hat with a dome and a wide flat brim.

This is modelled as in Fig B.

You can reduce this to a one dimensional solution by assuming radial symmetry. This would be expected by the action of surface tension.

This is shown in Fig C.

You have identified the important features of this problem, including the fact that the brim plays a crucial role in maintaining the adhesion since it contains little or no water  (weight) but presents an large surface area to the atmousphere to press against.

As a first estimate I would suggest that surface tension holds the droplet together but does not hold the droplet against gravity.

A more sophisticated model would include the change in surface energy cause by the droplet breaking away, which could be incorporated in the virtual work equation below.

Your force balance can be conveniently solved by the theorem of virtual work, either using a differential virtual vertical displacement or a differential virtual increase in the weight of the drop.

I don't think Young's formula will help you here, but it was a good idea.

The weight of the drop is derivable from the geometry and in particular the surface area.

This is then an extremal (maximisation) problem which could be attacked by good old trial and error or by the calculus of variations if you are seeking an analytical solution.

 

 

 

Edited November 9, 2023 by studiot

[Supercazzola]

6 minutes ago, studiot said:

This is not a simple standard school problem so it would be helpful to tell us the context in which it set.

 

 

 

I thank you for your response. This problem is taken from Rudolf Ortvay Competition in Physics, 2013. I am not sure what type of solution is best to look for in competitions of this type.

 

7 minutes ago, studiot said:

This is not a simple standard school problem so it would be helpful to tell us the context in which it set.

As said the contact angle at the ceiling must be zero.

So you then need a 'hump' shaped line that flattens out to parallel to the horizontal at both sides. As in Fig A.

Rotating this line will produce a domed hat like an inverted ww1 or ww2 tommie's hat with a dome and a wide flat brim.

This is modelled as in Fig B.

You can reduce this to a one dimensional solution by assuming radial symmetry. This would be expected by the action of surface tension.

This is shown in Fig C.

 

 

 

 

I don't think I fully understood. Are the three figures A, B, C three different ways of approaching the problem, or three physical situations consequential to each other? Thank you.

 

9 minutes ago, studiot said:

As a first estimate I would suggest that surface tension holds the droplet together but does not hold the droplet against gravity.

Did I not make this assumption in my calculations? Excuse my naivety, I can't understand it.

 

10 minutes ago, studiot said:

Your force balance can be conveniently solved by the theorem of virtual work, either using a differential virtual vertical displacement or a differential virtual increase in the weight of the drop.

I am not familiar with this theorem. Could you give me some suggestions? Obviously, I don't want the ultimate solution, but I would like to be led to it through some help. It seems to me to be quite a challenging problem.

 

13 minutes ago, studiot said:

The weight of the drop is derivable from the geometry and in particular the surface area.

 

Thank you, but why is my expression of weight force not exact? Is there something hidden in it that I am unaware of? Any clarification is appreciated.

 

14 minutes ago, studiot said:

This is then an extremal (maximisation) problem which could be attacked by good old trial and error or by the calculus of variations if you are seeking an analytical solution.

How to approach these two methods? Any suggestions? Thanks again.

[Supercazzola]

On 11/9/2023 at 1:30 PM, studiot said:

This is not a simple standard school problem so it would be helpful to tell us the context in which it set.

As said the contact angle at the ceiling must be zero.

So you then need a 'hump' shaped line that flattens out to parallel to the horizontal at both sides. As in Fig A.

Rotating this line will produce a domed hat like an inverted ww1 or ww2 tommie's hat with a dome and a wide flat brim.

This is modelled as in Fig B.

You can reduce this to a one dimensional solution by assuming radial symmetry. This would be expected by the action of surface tension.

This is shown in Fig C.

You have identified the important features of this problem, including the fact that the brim plays a crucial role in maintaining the adhesion since it contains little or no water  (weight) but presents an large surface area to the atmousphere to press against.

As a first estimate I would suggest that surface tension holds the droplet together but does not hold the droplet against gravity.

A more sophisticated model would include the change in surface energy cause by the droplet breaking away, which could be incorporated in the virtual work equation below.

Your force balance can be conveniently solved by the theorem of virtual work, either using a differential virtual vertical displacement or a differential virtual increase in the weight of the drop.

I don't think Young's formula will help you here, but it was a good idea.

The weight of the drop is derivable from the geometry and in particular the surface area.

This is then an extremal (maximisation) problem which could be attacked by good old trial and error or by the calculus of variations if you are seeking an analytical solution.

 

 

 

Okay, reviewing the problem I came to these conclusions.

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 cross-section of your droplet. These completely determine the droplet’s shape due to rotational symmetry. 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.

 

Could you help me formalize this in terms of physical relations?

[Theodore Lewis]

So, besides spherical drops and oval drops, there are no others.

[Supercazzola]

On 11/12/2023 at 10:58 AM, Theodore Lewis said:

So, besides spherical drops and oval drops, there are no others.

Forgive me, I don't think I understand.

[Genady]

The OP is not genuine. It is plagiarized from https://www.physicsforums.com/threads/height-of-a-stable-droplet-on-a-perfectly-wetting-surface.1057037/post-6964327.

[studiot]

3 hours ago, Genady said:

The OP is not genuine. It is plagiarized from https://www.physicsforums.com/threads/height-of-a-stable-droplet-on-a-perfectly-wetting-surface.1057037/post-6964327.

Thanks I lost my last response and further thoughts to mathjax the mangler when I was collecting the OP's further questions to reply to.

That is on top of the fact that we had had something of a crisis here, chez moi,  and a hectic couple of weeks.

 

I see from your link that PhysicsForums and Stack Exchange are also discussing this problem.