Pressure and the Young-Laplace Equation

[member 428835]

I solved a fluids problem recently and had a question about the answer. The question was, given a fluid in a rotating cylinder with gravity acting axially, what is the profile of the meniscus. I found the answer, which matched the book's, to be a parabolic meniscus. However, when solving I did not use the Young-Laplace equation. So when should you use Young-Laplace equation when dealing with pressure, and when should you consider surface tension?

 

[Chestermiller]

There has to be a dimensionless group that tells you that, in your particular system of interest, surface tension forces are negligible.
https://en.wikipedia.org/wiki/List_of_dimensionless_quantities

 

[member 428835]

The two numbers that come to mind are the Bond number (gravity / surface tension $\Delta \rho g L^2/ \sigma$) and the Capillary number (viscous forces / surface tension $\mu V / \sigma$). In the above-described problem I reduced Navier Stokes to $$-\rho\frac{v_\theta^2}{r}=-p_x\\0 = \mu d_r(r^{-1}d_r(rv_\theta))\\0=-p_z-\rho g$$ I then solved for $v_\theta$ using the $\theta$ equation and then solved for the pressure distribution by integrating the other two. However, I did not apply Young-Laplace: $\Delta P = \sigma(R_1^{-1}+R_2^{-1})$, where I believe I could approximate $R_1^{-1} \approx r^{-1}$ and $R_2^{-1} \approx z''(r)$. In order to know whether to apply Young-Laplace to the pressure, would I have to check both of the above listed non-dimensional numbers and ensure they are both small?

 

[Chestermiller]

Please tell us your solution.

 

[member 428835]

Please tell us your solution.

Sorry, I realized a typo: my first equation should read $-p_r$ on the rhs. At any rate, the solution I obtain is $z=z_0+(\omega^2/2g)r^2$ where $z$ is the meniscus height, $z_0$ is the given lowest height of the flow, $\omega$ is the angular velocity, $g$ is acceleration of gravity, and $r$ is the radius.

My issue is the non-constant curvature seems to suggest, via Young-Laplace, that pressure is higher at the center; both principal radii of curvature at $r=0$ are smaller than at $r=R$. However, clearly $r=0$ has the least pressure since the fluid is lowest there. Your thoughts?

 

[Chestermiller]

Let's work this out without using the Young-Laplace equation. Our goal will be to determine an equation the pressure on the fluid side of the free surface as a function of radial location and the shape of the free surface (and surface tension).

Let s represent the tangential distance along the free surface, at a constant value of $\theta$, measured from the axis of the cylinder. The shape of the free surface can be expressed parametrically in terms of s by writing R = R(s) and Z = Z(s), where R and Z are coordinates on the free surface. We are going to do a differential force balance on the portion of the free surface located between s and $s+\Delta s$. How does that grab you so far?

After we have derived the differential force balance, we are going to reduce all the equations to dimensionless form, so that they can tells us what the dimensionless groups for the system are.

To get started, in terms of R, Z, s, and the unit vectors in the r and z directions, what is the equation for a unit tangent vector to the free surface? What is the equation for a unit outwardly directed normal vector from the fluid side of the free surface to air side of the free surface?

 

[Chestermiller]

I carried out the formulation of this problem, including surface tension, and arrived at the following dimensionless group:
$$M=\frac{\sigma \omega^4}{\rho g^3}$$where $\sigma$ is the surface tension and M is what I call the McCraney number. When this group is small, surface tension effects are not important.

 

[member 428835]

To get started, in terms of R, Z, s, and the unit vectors in the r and z directions, what is the equation for a unit tangent vector to the free surface? What is the equation for a unit outwardly directed normal vector from the fluid side of the free surface to air side of the free surface?

If I'm understanding your parameterization the outward-oriented normal and tangent vectors to the surface $s$ would respectively be $$\hat{n} = \frac{dr\,\hat{k}-dz\,\hat{r}}{\sqrt{dz^2+dr^2}}\\
\hat{t} = \frac{dr\,\hat{r}+dz\,\hat{k}}{\sqrt{dz^2+dr^2}}.$$ How does this look to you?

 

[member 428835]

I carried out the formulation of this problem, including surface tension, and arrived at the following dimensionless group:
$$M=\frac{\sigma \omega^4}{\rho g^3}$$where $\sigma$ is the surface tension and M is what I call the McCraney number. When this group is small, surface tension effects are not important.

Hahhahaha this actually made me laugh! Ok, so how did you get there? Also, is there a reason you did not use Navier-Stokes directly, but rather took a control balance?

 

[Chestermiller]

Hahhahaha this actually made me laugh! Ok, so how did you get there?

Don't worry, I'll get you there. But, it will take a little work.

Also, is there a reason you did not use Navier-Stokes directly, but rather took a control balance?

My initial focus is on the free surface, and you can't use Navier-Stokes on a free surface. For the rest of the system, the fluid is rotating as a rigid body, so why bother with Navier-Stokes?

 

[Chestermiller]

If I'm understanding your parameterization the outward-oriented normal and tangent vectors to the surface $s$ would respectively be $$\hat{n} = \frac{dr\,\hat{k}-dz\,\hat{r}}{\sqrt{dz^2+dr^2}}\\
\hat{t} = \frac{dr\,\hat{r}+dz\,\hat{k}}{\sqrt{dz^2+dr^2}}.$$ How does this look to you?

Good, but I asked for it in terms of $ds=\sqrt{(dr)^2+(dz)^2}$. So, $$\hat{t}=\frac{dr}{ds}\hat{r}+\frac{dz}{ds}\hat{z}$$
$$\hat{n}=-\frac{dz}{ds}\hat{r}+\frac{dr}{ds}\hat{z}$$If we define the angle $\phi$ as the "contour angle" that the tangent vector to the free surface makes with the horizontal, then$$\frac{dr}{ds}=\cos{\phi}$$and $$\frac{dz}{ds}=\sin{\phi}$$So, $$\hat{t}=\cos{\phi}\ \hat{r}+\sin{\phi}\ \hat{z}$$
$$\hat{n}=-\sin{\phi}\ \hat{r}+\cos{\phi}\ \hat{z}$$

Now, we are going to do a force balance on the portion of the free surface between z and $z+\Delta z$ (more precisely, between contour locations s and $s+\Delta s$). There are surface tension forces acting on the cuts at s and at $s+\Delta s$, and there is a pressure force acting normal to conical segment of length $\Delta s$. If $\sigma$ is the surface tension, what is the net surface tension force acting on the cut at $s+\Delta s$?

 

[member 428835]

If $\sigma$ is the surface tension, what is the net surface tension force acting on the cut at $s+\Delta s$?

Would the surface tension force be $r \sigma \Delta \theta \hat{t}$ and the pressure force $-P r \Delta \theta \Delta s \hat{n}$? Then the net forces would be integrating each of these, so that net surface tension force would be $2\pi r \sigma \hat{t}$, though I'm unsure how to integrate the area integral of the pressure force ($\theta\in[0,2\pi]$ but how about $s$)?

 

[Chestermiller]

Would the surface tension force be $r \sigma \Delta \theta \hat{t}$ and the pressure force $-P r \Delta \theta \Delta s \hat{n}$? Then the net forces would be integrating each of these, so that net surface tension force would be $2\pi r \sigma \hat{t}$

This result is close to correct, but not quite. When you integrated with respect to $\theta$, you forgot to take into account the fact that $\hat{r}$ is a function of $\theta$. As a result, the net force on the cut at $s+\Delta s$ should be $$[2\pi r\sigma \hat{t}_z]_{s+\Delta s}=[2\pi r\sigma \sin{\phi}]_{s+\Delta s}\ \hat{z}$$where $\hat{t}_z=(\hat{t}\centerdot \hat{z})\hat{z}$ is the vertical component if $\hat{t}$. Similarly, the net force on the cut at s should be $$-[2\pi r\sigma \sin{\phi}]_{s}\ \hat{z}$$

though I'm unsure how to integrate the area integral of the pressure force ($\theta\in[0,2\pi]$ but how about $s$)?

The pressure force can be obtained similarly. It is the integral with respect to $\theta$ of $+Pr(\Delta s)\hat{n}d\theta$, where P is the gauge pressure on the fluid side of the interface. So, what result do you get for this integral?

 

[member 428835]

When you integrated with respect to $\theta$, you forgot to take into account the fact that $\hat{r}$ is a function of $\theta$.

How is $r$ a function of $\theta$; the body is symmetric in $\theta$ (cylindrical container).

As a result, the net force on the cut at $s+\Delta s$ should be $$[2\pi r\sigma \hat{t}_z]_{s+\Delta s}=[2\pi r\sigma \sin{\phi}]_{s+\Delta s}\ \hat{z}$$

Isn't this just the vertical component of force, not the net force?

The pressure force can be obtained similarly. It is the integral with respect to $\theta$ of $+Pr(\Delta s)\hat{n}d\theta$, where P is the gauge pressure on the fluid side of the interface. So, what result do you get for this integral?

But isn't pressure force $\vec{F}_p = -\iint_S P \hat{n} \, dS$ where $P$ is pressure, $S$ is the surface, and $\hat{n}$ is the outward oriented normal vector, which is $-\sin\phi\hat{r}+\cos\phi\hat{z}$, which implies we integrate with respect to $\theta$ of $-Pr(\Delta s)\hat{n}d\theta$, giving $-2\pi Pr \Delta s \hat{n}$. This is the pressure force on a thin ring of meniscus with area $\Delta S r 2 \pi$.

I have a feeling you disagree; can you say why?

 

[Chestermiller]

How is $r$ a function of $\theta$; the body is symmetric in $\theta$ (cylindrical container).

I didn't say that r is a function of $\theta$. I said that the unit vector in the radial direction $\hat{r}$ is a function of $\theta$. In particular, $$\hat{r}=\hat{i}\cos{\theta}+\hat{j}\sin{\theta}$$ What do you get when you integrate that between $\theta = 0$ and $\theta = 2\pi$? The radial component has to cancel out because of the axial symmetry (as you yourself pointed out).

Isn't this just the vertical component of force, not the net force?

It is both the vertical component of force and the net force.

But isn't pressure force $\vec{F}_p = -\iint_S P \hat{n} \, dS$ where $P$ is pressure, $S$ is the surface, and $\hat{n}$ is the outward oriented normal vector, which is $-\sin\phi\hat{r}+\cos\phi\hat{z}$, which implies we integrate with respect to $\theta$ of $-Pr(\Delta s)\hat{n}d\theta$, giving $-2\pi Pr \Delta s \hat{n}$. This is the pressure force on a thin ring of meniscus with area $\Delta S r 2 \pi$.

I don't know where you are getting the minus sign in front of the pressure. The compressive force per unit area normal to the free surface on the fluid side of the free surface is $+P\hat{n}$ (just like you would get with a rigid boundary). Again, in obtaining the pressure force on our area element of free surface, the radial component cancels out. We are then left with a pressure force exerted by the fluid on the surface area element as:

$$F=2\pi r (\Delta s)\cos{\phi}P\ \hat{z}$$

 

[member 428835]

I didn't say that r is a function of $\theta$. I said that the unit vector in the radial direction $\hat{r}$ is a function of $\theta$. In particular, $$\hat{r}=\hat{i}\cos{\theta}+\hat{j}\sin{\theta}$$ What do you get when you integrate that between $\theta = 0$ and $\theta = 2\pi$? The radial component has to cancel out because of the axial symmetry (as you yourself pointed out).

Shoot, don't know how I missed this. Okay, yea sine and cosine are zero when integrate. Now I understand what you meant when you said "It is both the vertical component of force and the net force."

I don't know where you are getting the minus sign in front of the pressure. The compressive force per unit area normal to the free surface on the fluid side of the free surface is $+P\hat{n}$ (just like you would get with a rigid boundary).

I was thinking about a control volume approach, and since the control volume is the fluid the pressure force exerted on the fluid is the inward oriented normal vector, so $-\hat{n}$. Can you explain why it is instead $+\hat{n}$, which to me seems like the pressure the fluid exerts on the air.

We are then left with a pressure force exerted by the fluid on the surface area element as:
$$F=2\pi r (\Delta s)\cos{\phi}P\ \hat{z}$$

Awesome, and this makes perfect sense. So are we then going to put these forces into Newton's second law, where the centrifugal force is the acceleration?

 

[Chestermiller]

Shoot, don't know how I missed this. Okay, yea sine and cosine are zero when integrate. Now I understand what you meant when you said "It is both the vertical component of force and the net force."

Excellent. It looks like we're making progress.

I was thinking about a control volume approach, and since the control volume is the fluid the pressure force exerted on the fluid is the inward oriented normal vector, so $-\hat{n}$. Can you explain why it is instead $+\hat{n}$, which to me seems like the pressure the fluid exerts on the air.

It's not the pressure the fluid exerts on the air. Remember, we are treating the interface as a membrane (like a balloon). It is the gauge pressure that the fluid exerts on one side of the interface membrane. The air is exerting a force in the opposite direction on the other side of the interface membrane, but, since we are using gauge pressure, the net force per unit area from both the fluid on one side of the interface membrane and the air on the opposite side of the interface membrane is just $+P\hat{n}$.

Awesome, and this makes perfect sense. So are we then going to put these forces into Newton's second law, where the centrifugal force is the acceleration?

No. Our initial focus is to determine the pressure on the fluid side of the membrane. The membrane itself has no mass, so the "ma" term in its force balance is zero. So can you please now sum up the forces acting on the portion of the interface membrane between s and $s+\Delta s$ and complete the force balance? Then, divide the resulting equation by $\Delta s$, and let $\Delta s$ go to zero. What do you obtain?

 

[member 428835]

It's not the pressure the fluid exerts on the air. Remember, we are treating the interface as a membrane (like a balloon). It is the gauge pressure that the fluid exerts on one side of the interface membrane. The air is exerting a force in the opposite direction on the other side of the interface membrane, but, since we are using gauge pressure, the net force per unit area from both the fluid on one side of the interface membrane and the air on the opposite side of the interface membrane is just $+P\hat{n}$.

Thanks for clearing this up!

No. Our initial focus is to determine the pressure on the fluid side of the membrane. The membrane itself has no mass, so the "ma" term in its force balance is zero. So can you please now sum up the forces acting on the portion of the interface membrane between s and $s+\Delta s$ and complete the force balance? Then, divide the resulting equation by $\Delta s$, and let $\Delta s$ go to zero. What do you obtain?

Ahhhhhh this makes a lot of sense! You did all the hard work, let's hope my algebra can do the rest hahaha. Ok, so we'd have $$\sigma\left(\sin\phi+r\frac{d\phi}{ds}\right)+r P = 0$$ How does this look?

 

[Chestermiller]

Thanks for clearing this up!

Ahhhhhh this makes a lot of sense! You did all the hard work, let's hope my algebra can do the rest hahaha. Ok, so we'd have $$\sigma\left(\sin\phi+r\frac{d\phi}{ds}\right)+r P = 0$$ How does this look?

Josh! You're the man!

So, the fluid pressure just below the membrane interface is sub-atmospheric at:$$P=-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)$$The two terms in parenthesis are the principal curvatures.

So, what do you do next?

 

[member 428835]

Josh! You're the man!

Hahahaha yea because calculus and algebra is soooo hard after you do all the physics, but thanks!

So, the fluid pressure just below the membrane interface is sub-atmospheric at:$$P=-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)$$The two terms in parenthesis are the principal curvatures.

So, what do you do next?

I'm not sure. Any ideas or hints?

 

[Chestermiller]

Hahahaha yea because calculus and algebra is soooo hard after you do all the physics, but thanks!

I'm not sure. Any ideas or hints?

Let the bottom of the free surface be the datum for elevation, z = 0, and let P(s) be the pressure on the fluid side of the membrane interface at contour location s. So, at radial location r(s) and depth z*, the pressure is $$p(r(s),-z*)=P(s)+\rho g [z(s)+z^*]$$ Similarly, at radial location r = 0 and depth z*, the pressure is $$p(0,-z*)=P(0)+\rho g z^*$$So, $$p(r(s),-z^*)-p(0,-z*)=P(s)+\rho g z(s)-P(0)$$But, from the force balance down below at z*, $$p(r(s),-z^*)-p(0,-z*)=\frac{\rho \omega^2r^2(s)}{2}$$So, combining:$$P(s)+\rho g z(s)-P(0)=\frac{\rho \omega^2r^2(s)}{2}$$
Any thoughts what to do after this?

 

[member 428835]

But, from the force balance down below at z*, $$p(r(s),-z^*)-p(0,-z*)=\frac{\rho \omega^2r^2(s)}{2}$$

I lost you here. Assuming steady flow and only pressure and body forces, the force balance would look like $$\iint_S\rho \vec{V}\vec{V}\cdot\vec{dS}=-\iint_SP\vec{dS}+\iint_S\rho g z\, \vec{dS}$$ But what surface $S$ would yield your balance? I'm thinking some sort of ring, but I'm unsure of this.

So, combining:$$P(s)+\rho g z(s)-P(0)=\frac{\rho \omega^2r^2(s)}{2}$$
Any thoughts what to do after this?

At this point would we write $P(s)$ and $P(0)$ using Young-Laplace, for example, $P(s) = \sigma(r(s)^{-1}+r_1(s)^{-1})$, where $r_1(s)$ is the radius of curvature of the free surface that is orthogonal to $r(s)$?

 

[Chestermiller]

I lost you here. Assuming steady flow and only pressure and body forces, the force balance would look like $$\iint_S\rho \vec{V}\vec{V}\cdot\vec{dS}=-\iint_SP\vec{dS}+\iint_S\rho g z\, \vec{dS}$$ But what surface $S$ would yield your balance? I'm thinking some sort of ring, but I'm unsure of this.

You are aware that $$\left(\frac{\partial P}{\partial r}\right)_z=\rho \omega^2 r$$
Correct?

At this point would we write $P(s)$ and $P(0)$ using Young-Laplace, for example, $P(s) = \sigma(r(s)^{-1}+r_1(s)^{-1})$, where $r_1(s)$ is the radius of curvature of the free surface that is orthogonal to $r(s)$?

We already went over this in post # 19:
$$P(s)=-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)$$

 

[member 428835]

You are aware that $$\left(\frac{\partial P}{\partial r}\right)_z=\rho \omega^2 r$$
Correct?

Yes of course, this comes from Bernoulli's, right?

We already went over this in post # 19:
$$P(s)=-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)$$

Shoot, sorry, not sure how I missed this.

 

[Chestermiller]

Yes of course, this comes from Bernoulli's, right?

Not really. It just comes from a force balance on a differential volume $\Delta r (r\Delta\theta)\Delta z$

Shoot, sorry, not sure how I missed this.

So, are we in agreement now?

 

[member 428835]

So, are we in agreement now?

We are!

 

[Chestermiller]

We are!

OK. So, if you combine the final equations of posts #21 and 23, what do you get? This is the equation that one would have to solve to get the shape of the free surface. It would be solved for $\phi$ as a function of s.

 

[member 428835]

OK. So, if you combine the final equations of posts #21 and 23, what do you get?

A big mess!

$$-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)+\rho g z(s)-P(0)=\frac{\rho \omega^2r^2(s)}{2}$$

How did you get the "McCraney" number $M=\sigma \omega^4/ \rho g^3$ from this? Also, how do we interpret $P(s=0)$ since $ds^2=r^2+z^2$?

 

[Chestermiller]

$$-\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)+\rho g z(s)+\sigma\left(\frac{\sin\phi}{r}+\frac{d\phi}{ds}\right)_{s=0}=\frac{\rho \omega^2r^2(s)}{2}$$
I don't want to spend any time discussing how you would solve this equation because it would involve a lot of work. But instead, I will focus on how to reduce the equation to dimensionless form.

If we were to solve this problem, we would be solving for $\phi$ as a function of s. The boundary condition at s = 0 is $\phi=0.$. There is another boundary condition required on phi now that we are including surface tension. That boundary condition is $\phi=\phi_{contact}$ at r = a, where $\phi_{contact}$ is the contact angle at the outer solid surface r =a.

Now to begin to reduce this equation and the boundary conditions to dimensionless form, please make the following substitutions into the equation and boundary conditions:
$$S=s/s_0$$
$$R=r/s_0$$
$$Z=z/s_0$$
where $s_0$ is a characteristic length scale for the problem.

 

[member 428835]

Now to begin to reduce this equation and the boundary conditions to dimensionless form, please make the following substitutions into the equation and boundary conditions:
$$S=s/s_0$$
$$R=r/s_0$$
$$Z=z/s_0$$
where $s_0$ is a characteristic length scale for the problem.

$$-\sigma\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)+s_0^2\rho g Z(S)+\sigma\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}=s_0^3\frac{\rho \omega^2R^2(S)}{2}$$
Is this what you had in mind?

 

[Chestermiller]

$$-\sigma\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)+s_0^2\rho g Z(S)+\sigma\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}=s_0^3\frac{\rho \omega^2R^2(S)}{2}$$
Is this what you had in mind?

Yes. Now divide the equation by $\rho \omega^2 s^3_0$. What do you get?

 

[member 428835]

Yes. Now divide the equation by $\rho \omega^2 s^3_0$. What do you get?

$$\frac{\sigma}{\rho \omega^2 s^3_0}\left[\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}-\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)\right]+\frac{g }{\omega^2 s_0}Z(S)=\frac{R^2(S)}{2}$$ Starting to look like the "McCraney number".

 

[Chestermiller]

$$\frac{\sigma}{\rho \omega^2 s^3_0}\left[\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}-\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)\right]+\frac{g }{\omega^2 s_0}Z(S)=\frac{R^2(S)}{2}$$ Starting to look like the "McCraney number".

Good. Now, there are a couple of ways to proceed further with the dimensional analysis. One way is to now set $s_0=g/\omega^2$, so that the coefficient of Z is unity. What does this give you?

 

[member 428835]

Good. Now, there are a couple of ways to proceed further with the dimensional analysis. One way is to now set $s_0=g/\omega^2$, so that the coefficient of Z is unity. What does this give you?

Awesome, this is perfect. Thanks so much!
$$M\left[\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}-\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)\right]+Z(S)=\frac{R^2(S)}{2}$$

 

[Chestermiller]

Awesome, this is perfect. Thanks so much!
$$M\left[\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)_{S=0}-\left(\frac{\sin\phi}{R}+\frac{d\phi}{dS}\right)\right]+Z(S)=\frac{R^2(S)}{2}$$

The important thing is to assimilate this kind of methodology for reducing the equations for a system to dimensionless form.