Normal Stress Balance at a Fluid-fluid Interface

[tse8682]

I am trying to determine a normal stress balance at an axisymmetric and dynamic fluid-fluid interface, $z(r,t)$. For a static, free surface, this simplifies to the Young-Laplace equation: $$ \Delta p=\rho gz-\sigma2H=\rho gz-\frac{\sigma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right) $$ where $H$ is the mean curvature of the interface. Assuming the overhead pressure is simply atmospheric pressure, and the pressure below the interface can also be written axisymmetrically, $p(r,t)$, this simplifies to: $$ \frac{\sigma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)=\rho gz-p $$ I am trying to determine a more general form of this for a fluid-fluid interface that is moving, as opposed to a liquid-gas interface that is static. If the interface is moving, i.e. it is dynamic, the viscous stress needs to be included. A general form I found is given here as: $$ n\cdot \hat{T}\cdot n-n\cdot T\cdot n=\sigma\left(\nabla \cdot n\right)$$ where $n$ is the normal vector, $T=-p+\mu\left[\nabla u+\left(\nabla u\right)^T\right]$, and $\hat{T}=-\hat{p}+\hat{\mu}\left[\nabla \hat{u}+\left(\nabla \hat{u}\right)^T\right]$, $u$ is the velocity, and the hatted symbols are for the upper fluid and the unhatted symbols are for the lower fluid. I know $\nabla \cdot n$ is the mean curvature so the right hand side translates to the left hand side of the previous equation. I think I need to add the hydrostatic pressure term as well. My best guess right now is that it looks something like this: $$ \frac{\sigma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)=(\rho-\hat{\rho})gz-p+\hat{p}+2\mu\frac{\partial u_n}{\partial n}-2\hat{\mu}\frac{\partial \hat{u_n}}{\partial n} $$ and that the normal velocity gradient can be estimated as $\frac{\partial u_n}{\partial n}=-2\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t}$. I get this by first considering an interface of constant curvature. $R$. which is moving radially outwards. By conservation of mass, the radial velocity can be given as $u(r)=\frac{R^2}{r^2}\frac{dR}{dt}$ where $\frac{dR}{dt}$ is the interface velocity. From this, the velocity gradient at the interface can be written as $\frac{du}{dr}(R)=-\frac{2}{R}\frac{dR}{dt}$. The inverse of radius of curvature, $\frac{1}{R}$, is the interface curvature, $\kappa$, which can be estimate based on the interface shape as $\frac{\partial^2 z}{\partial r^2}$ for nonconstant values assuming that $\left(\frac{\partial z}{\partial r}\right)^2\ll 1$. With this assumption, I also assume that $\frac{dR}{dt}\approx\frac{\partial z}{\partial t}$. So the final equation would look something like this:$$ \frac{\sigma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)=(\rho-\hat{\rho})gz-p+\hat{p}-4\mu\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t}-4\hat{\mu}\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t} $$ This is what I have come up with so far but am hoping someone could confirm this or possible give their own insights. Any advice or suggestions would be greatly appreciated, thanks!

 

[Chestermiller]

In my judgment, the $\rho g z$ term does not belong in the original equation (or in any of the equations for that matter). How did it get in there in the first place? I haven't checked out you math in rendering the normal components of the stresses. Also understand that the tangential components of the stresses must match at the interface.

 

[tse8682]

In my judgment, the $\rho g z$ term does not belong in the original equation (or in any of the equations for that matter). How did it get in there in the first place? I haven't checked out you math in rendering the normal components of the stresses. Also understand that the tangential components of the stresses must match at the interface.

The $\rho gz$ term is for the hydrostatic pressure. Technically it could be grouped with the pressure acting downward on the interface, I'm separating it away from some external pressure acting on the interface. And yes, thank you. I know the tangential shear stress must be continuous across the interface, but for the moment I am only concerned with the normal stress.

 

[Chestermiller]

The $\rho gz$ term is for the hydrostatic pressure. Technically it could be grouped with the pressure acting downward on the interface, I'm separating it away from some external pressure acting on the interface. And yes, thank you. I know the tangential shear stress must be continuous across the interface, but for the moment I am only concerned with the normal stress.

Like I said, that term does not belong in the equation. The Young-Laplace equation is just a force balance on the (massless) interface between two fluids with a curved interface. Since the interface is massless, there is no gravitational force on it. The pressure on each side of the interface already includes hydrostatics.

 

[tse8682]

Like I said, that term does not belong in the equation. The Young-Laplace equation is just a force balance on the (massless) interface between two fluids with a curved interface. Since the interface is massless, there is no gravitational force on it. The pressure on each side of the interface already includes hydrostatics.

I think I have misrepresented what I am trying to ask. Instead of a normal stress balance, I think I'm looking for the pressure drop across a dynamic fluid-fluid interface which is experiencing an applied pressure. For a static, free surface, would this then be a balance between the applied pressure, hydrostatic pressure, and surface tension as represented by the second equation shown here?

 

[Chestermiller]

I think I have misrepresented what I am trying to ask. Instead of a normal stress balance, I think I'm looking for the pressure drop across a dynamic fluid-fluid interface which is experiencing an applied pressure. For a static, free surface, would this then be a balance between the applied pressure, hydrostatic pressure, and surface tension as represented by the second equation shown here?

Like I said. The hydrostatic term does not belong in the equation.

 

[tse8682]

Like I said. The hydrostatic term does not belong in the equation.

Here is the Soft Matter paper I am basing this from in which this exact equation is used to describe a free surface, here. Is this paper incorrect in how it is describing the shape of an axisymmetric free surface?

 

[tse8682]

Like I said. The hydrostatic term does not belong in the equation.

Also, on page 17 of the lecture notes given here as part of an MIT course on interfacial phenomena, again the $\rho gz$ term appears in an example on the shape of a static meniscus. It states "The normal stress balance thus requires that the interface satisfy the Young-Laplace Equation: $\rho gz=\sigma \nabla \cdot n$."

 

[Chestermiller]

Here is the Soft Matter paper I am basing this from in which this exact equation is used to describe a free surface, here. Is this paper incorrect in how it is describing the shape of an axisymmetric free surface?

In the specific hydrostatic problem that they solve, they have replaced $\Delta P$ by $\rho g z$, but both are not present simultaneously in the Young-Laplace equation. More generally, this replacement cannot be done and, in any case, they are never present together in the Young-Laplace equation.

 

[Chestermiller]

Here is the Soft Matter paper I am basing this from in which this exact equation is used to describe a free surface, here. Is this paper incorrect in how it is describing the shape of an axisymmetric free surface?

I'm too lazy to go through this entire paper, but it seems to me that the pressure inside the bubble is p and the pressure throughout most of the region outside the bubble is hydrostatic, and thus has been replaced by the $\rho g z$ term. There also seems like there is an fluid viscous deformation (lubrication) term outside the bubble represented by the $\Pi$. So the total normal stress outside the bubble is $\rho g z +\Pi$.

 

[tse8682]

I'm too lazy to go through this entire paper, but it seems to me that the pressure inside the bubble is p and the pressure throughout most of the region outside the bubble is hydrostatic, and thus has been replaced by the $\rho g z$ term. There also seems like there is an fluid viscous deformation (lubrication) term outside the bubble represented by the $\Pi$. So the total normal stress outside the bubble is $\rho g z +\Pi$.

Yeah, no need to. The rest isn't really relevant to my question, just that first equation. The pressure outside of the bubble, present in the film between the bubble and interface, is given by $p$, $\Pi$ is the disjoining pressure but is later ignored since the film thickness is relatively large.

 

[tse8682]

In the specific hydrostatic problem that they solve, they have replaced $\Delta P$ by $\rho g z$, but both are not present simultaneously in the Young-Laplace equation. More generally, this replacement cannot be done and, in any case, they are never present together in the Young-Laplace equation.

In the paper where both are present, then perhaps they have done the same thing? Replaced $\Delta p$ by $\rho gz$ and then they state the $p$ in their equation is due to fluid motion so maybe that is just the film pressure as opposed to the pressure acting on the interface.

 

[Chestermiller]

All I can say is that I am in agreement with everything presented in the MIT document, and it is totally consistent with what I have been saying. So, I stand by what I've said.

I feel like you an I are on different wavelengths in discussing this matter. So, I'm going to withdraw from the discussion and give others a chance to weight in.

Chet

 

[tse8682]

All I can say is that I am in agreement with everything presented in the MIT document, and it is totally consistent with what I have been saying. So, I stand by what I've said.

I feel like you an I are on different wavelengths in discussing this matter. So, I'm going to withdraw from the discussion and give others a chance to weight in.

Chet

Thank you for your insights, Chet. I think I'm starting to get a better handle on it.

 

[tse8682]

Taking another shot at this with the above in mind. To give a physical sense to the problem, I will say I am considering two problems: (1) a static cylindrical rod of radius $R$ placed at a fluid-fluid interface, and (2) starting from equilibrium, the rod is then moved up and down sinusoidally. Assuming axisymmetric conditions, I am trying to determine a normal stress balance at the fluid-fluid interface to later be used to analytically determine the interface shape. Starting with the static meniscus shape, page 17 here gives the normal stress balance for a static meniscus as:$$\Delta \rho gz=\sigma (\nabla \cdot n)$$ The interface curvature in the case of an axisymmetric interface (given at Appendix B of same source) is: $$ \nabla \cdot n=\frac{r \frac{dz}{dr}+r^2 \frac{d^2z}{dr^2}}{r^2\left(1+\frac{dz}{dr}^2\right)^{3/2}} \approx \frac{1}{r} \frac{dz}{dr}+\frac{d^2z}{dr^2}=\frac{1}{r} \frac{d}{dr}\left(r\frac{dz}{dr}\right)$$ Thus, the normal stress balance is written: $$ \Delta \rho gz=\frac{\sigma}{r} \frac{d}{dr}\left(r\frac{dz}{dr}\right)$$ This can be rearranged to look like the modified Bessel equation which has an analytical solution in terms of the modified Bessel function of the second kind of order zero, $K_0$. Since $z(\infty)=0$, the modified Bessel function of the first kind is ignored as a possible solution since it is unbounded at larger argument values. You need to apply some boundary conditions to determine the coefficient of the function, but I'm not concerned about that for the moment.
Moving onto the second problem, at some time, $t=t_0$, the rod begins to move up and down sinusoidally. My main question is this: what would the normal stress balance then look like? I believe it would be similar to the stress balance above, only it would be necessary to include the viscous stress at the interface and could look something like this: $$\Delta \rho gz+2\mu\frac{\partial u_n}{\partial n}-2\hat{\mu}\frac{\partial \hat{u}_n}{\partial n}=\frac{\sigma}{r} \frac{d}{dr}\left(r\frac{dz}{dr}\right)$$ Using the analysis given in the original post for the normal velocity gradient, this could then be written in terms of the interface shape and velocity as: $$\Delta \rho gz-4\mu\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t}+4\hat{\mu}\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t}=\Delta \rho gz-4\Delta\mu\frac{\partial^2 z}{\partial r^2}\frac{\partial z}{\partial t}=\frac{\sigma}{r} \frac{d}{dr}\left(r\frac{dz}{dr}\right)$$ I think this looks a bit better than what I had originally, but I'm still unsure if it is correct. I'd like to make sure it's right before trying to solve this.

 

[Chestermiller]

I don't get the geometry. The interface is horizontal, right? Is the rod horizontal or vertical?

What about contact angle?

 

[tse8682]

I don't get the geometry. The interface is horizontal, right? Is the rod horizontal or vertical?

What about contact angle?

Link to schematic. For the time being, I am considering the height of the meniscus at the rod edge, $h$, as a known value just for the sake of simplicity. In case (1) $h(t)$ is constant, in case (2) $h(t)=A\sin{\omega t}$ where amplitude $A$ and frequency $\omega$ are known. Using this instead of contact angle in hopes that applying this boundary condition will be simpler.

 

[Chestermiller]

Taking another shot at this with the above in mind. To give a physical sense to the problem, I will say I am considering two problems: (1) a static cylindrical rod of radius $R$ placed at a fluid-fluid interface, and (2) starting from equilibrium, the rod is then moved up and down sinusoidally. Assuming axisymmetric conditions, I am trying to determine a normal stress balance at the fluid-fluid interface to later be used to analytically determine the interface shape. Starting with the static meniscus shape, page 17 here gives the normal stress balance for a static meniscus as:$$\Delta \rho gz=\sigma (\nabla \cdot n)$$ The interface curvature in the case of an axisymmetric interface (given at Appendix B of same source) is: $$ \nabla \cdot n=\frac{r \frac{dz}{dr}+r^2 \frac{d^2z}{dr^2}}{r^2\left(1+\frac{dz}{dr}^2\right)^{3/2}} \approx \frac{1}{r} \frac{dz}{dr}+\frac{d^2z}{dr^2}=\frac{1}{r} \frac{d}{dr}\left(r\frac{dz}{dr}\right)$$

I'm going to take a shot at this on my own and see what I come up with. But meanwhile, please answer this question: What happened to the curvature in the other direction (circumferential)? That also contributes to the pressure differential across the free surface.

 

[tse8682]

I'm going to take a shot at this on my own and see what I come up with. But meanwhile, please answer this question: What happened to the curvature in the other direction (circumferential)? That also contributes to the pressure differential across the free surface.

I also had that exact question. My current line of thinking is this. Consider some function $y(x)$. The curvature of that function is given by $\kappa=\frac{y”}{\left(1+y’^2\right)^{3/2}}$. Now separating out the numerator of the equation for mean curvature gives this: $$\nabla \cdot n=\frac{z’}{r\left(1+z’^2\right)^{3/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}$$ The second term on the right side in the above equation is the same as just the curvature for some function $y(x)$. So my thought is the curvature in the radial direction is given by that term and the curvature in the circumferential direction is given by the first term.

 

[Chestermiller]

I also had that exact question. My current line of thinking is this. Consider some function $y(x)$. The curvature of that function is given by $\kappa=\frac{y”}{\left(1+y’^2\right)^{3/2}}$. Now separating out the numerator of the equation for mean curvature gives this: $$\nabla \cdot n=\frac{z’}{r\left(1+z’^2\right)^{3/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}$$ The second term on the right side in the above equation is the same as just the curvature for some function $y(x)$. So my thought is the curvature in the radial direction is given by that term and the curvature in the circumferential direction is given by the first term.

Thanks. I get close to the same result, but not quite. I get:
$$\nabla \cdot n=-\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]$$ Note the exponent of 1/2 in the first term rather than the exponent 3/2. And I get it with a minus sign. Is the $\Delta \rho$ in your equation $(\rho_H-\rho_L)$ or $(\rho_L-\rho_H)$, where H refers to the higher density fluid and the L refers to the lower density fluid.

In the end, I get: $$-\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]=(\rho_H-\rho_L)gz$$

I hope I got the signs right.

 

[tse8682]

Thanks. I get close to the same result, but not quite. I get:
$$\nabla \cdot n=-\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]$$ Note the exponent of 1/2 in the first term rather than the exponent 3/2. And I get it with a minus sign. Is the $\Delta \rho$ in your equation $(\rho_H-\rho_L)$ or $(\rho_L-\rho_H)$, where H refers to the higher density fluid and the L refers to the lower density fluid.

In the end, I get: $$-\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]=(\rho_H-\rho_L)gz$$

I hope I got the signs right.

The $\Delta \rho$ is low minus high on mine so we are getting the same thing. You are correct with the power of 1/2. That’s my error. Glad we’re finally on the same page. Unfortunately, keeping track of signs only gets more confusing after bringing the viscous stresses into the problem.

 

[Chestermiller]

The $\Delta \rho$ is low minus high on mine so we are getting the same thing. You are correct with the power of 1/2. That’s my error. Glad we’re finally on the same page. Unfortunately, keeping track of signs only gets more confusing after bringing the viscous stresses into the problem.

I dom’t Think that should be a problem. I’ll be back tomorrow to help. What

 

[Chestermiller]

I did find a sign error in my analysis, so, with the sign error corrected, the equation without viscous terms should read:
$$\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]=(\rho_H-\rho_L)gz\tag{1}$$
Now for the viscous terms. The unit normal vector from the high-density side of the interface to the low density side of the interface is given by:
$$\mathbf{n}=\frac{-\mathbf{i_r}\frac{dz}{dr}+\mathbf{i_z}}{\sqrt{1+\left(\frac{dz}{dr}\right)^2}}$$From this, it follows that
$$\mathbf{n}\centerdot \boldsymbol{\sigma}\centerdot \mathbf{n}=-p_{\infty}+\rho gz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}$$where $$\tau_{rr}=2\eta \frac{\partial u}{\partial r}$$
$$\tau_{zz}=2\eta\frac{\partial w}{\partial z}$$
and $$\tau_{rz}=\eta\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r}\right)$$So, to include viscous stresses, we replace $\rho_Hgz$ in Eqn 1 by $$\rho_Hgz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}^H+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}^H-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}^H$$and similarly for $\rho_Lgz$.

 

[tse8682]

I did find a sign error in my analysis, so, with the sign error corrected, the equation without viscous terms should read:
$$\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]=(\rho_H-\rho_L)gz\tag{1}$$
Now for the viscous terms. The unit normal vector from the high-density side of the interface to the low density side of the interface is given by:
$$\mathbf{n}=\frac{-\mathbf{i_r}\frac{dz}{dr}+\mathbf{i_z}}{\sqrt{1+\left(\frac{dz}{dr}\right)^2}}$$From this, it follows that
$$\mathbf{n}\centerdot \boldsymbol{\sigma}\centerdot \mathbf{n}=-p_{\infty}+\rho gz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}$$where $$\tau_{rr}=2\eta \frac{\partial u}{\partial r}$$
$$\tau_{zz}=2\eta\frac{\partial w}{\partial z}$$
and $$\tau_{rz}=\eta\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r}\right)$$So, to include viscous stresses, we replace $\rho_Hgz$ in Eqn 1 by $$\rho_Hgz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}^H+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}^H-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}^H$$and similarly for $\rho_Lgz$.

Just to be clear, high is referring to the higher density liquid located at at the bottom?

 

[Chestermiller]

Just to be clear, high is referring to the higher density liquid located at at the bottom?

Yes.

 

[Chestermiller]

If I were doing this, for the case in which there is fluid flow, I would not have $(\Delta \rho) gz$ in the boundary condition. I would instead have $p_H$ and $p_L$ in the boundary condition at the free surface. Instead, the $\rho g's$ should go in the Navier Stokes equation. For the case without flow, on the other hand, it is valid to write $\frac{\partial p}{\partial z}+\rho g=0$

 

[tse8682]

If I were doing this, for the case in which there is fluid flow, I would not have $(\Delta \rho) gz$ in the boundary condition. I would instead have $p_H$ and $p_L$ in the boundary condition at the free surface. Instead, the $\rho g's$ should go in the Navier Stokes equation. For the case without flow, on the other hand, it is valid to write $\frac{\partial p}{\partial z}+\rho g=0$

My ultimate goal with this is to determine an analytical differential equation for $z(r,t)$ that is only dependent on $z$ and it’s derivatives, $r$, and fluid properties and then use the $z(\infty,t)=0$ and $z(R,t)=h(t)$ as boundary conditions with $z(r,0)=0$ as the initial condition. If an analytical solution to that equation isn’t possible, then just a basic ode solver in Matlab like ode15s could be used by discretizing in the spatial domain. My concern with using Navier Stokes in the dynamic case is that I don’t have access to the fluid velocity field around the interface, just the interface velocity itself. Even with the method we’ve been working on, I still need a way to estimate the partial derivatives in $u$ and $w$ from the interface velocity and shape. With Navier-Stokes, I think I would need boundary conditions for the velocity to integrate it and come up with that differential equation for $z(r,t)$?

 

[Chestermiller]

My ultimate goal with this is to determine an analytical differential equation for $z(r,t)$ that is only dependent on $z$ and it’s derivatives, $r$, and fluid properties and then use the $z(\infty,t)=0$ and $z(R,t)=h(t)$ as boundary conditions with $z(r,0)=0$ as the initial condition. If an analytical solution to that equation isn’t possible, then just a basic ode solver in Matlab like ode15s could be used by discretizing in the spatial domain. My concern with using Navier Stokes in the dynamic case is that I don’t have access to the fluid velocity field around the interface, just the interface velocity itself. Even with the method we’ve been working on, I still need a way to estimate the partial derivatives in $u$ and $w$ from the interface velocity and shape. With Navier-Stokes, I think I would need boundary conditions for the velocity to integrate it and come up with that differential equation for $z(r,t)$?

Well the first thing I would do would be to drop the inertial terms from the Navier Stokes equations. Then I would start examining using a stream function for determining the velocities so that the continuity equation is automatically satisfied and the number of dependent variables is reduced. This would involve careful examination of the kinematics. This is not a simple problem.

 

[tse8682]

I did find a sign error in my analysis, so, with the sign error corrected, the equation without viscous terms should read:
$$\left[\frac{z’}{r\left(1+z’^2\right)^{1/2}}+\frac{z”}{\left(1+z’^2\right)^{3/2}}\right]=(\rho_H-\rho_L)gz\tag{1}$$
Now for the viscous terms. The unit normal vector from the high-density side of the interface to the low density side of the interface is given by:
$$\mathbf{n}=\frac{-\mathbf{i_r}\frac{dz}{dr}+\mathbf{i_z}}{\sqrt{1+\left(\frac{dz}{dr}\right)^2}}$$From this, it follows that
$$\mathbf{n}\centerdot \boldsymbol{\sigma}\centerdot \mathbf{n}=-p_{\infty}+\rho gz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}$$where $$\tau_{rr}=2\eta \frac{\partial u}{\partial r}$$
$$\tau_{zz}=2\eta\frac{\partial w}{\partial z}$$
and $$\tau_{rz}=\eta\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r}\right)$$So, to include viscous stresses, we replace $\rho_Hgz$ in Eqn 1 by $$\rho_Hgz+\frac{\left(\frac{dz}{dr}\right)^2}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rr}^H+\frac{1}{1+\left(\frac{dz}{dr}\right)^2}\tau_{zz}^H-\frac{2\frac{dz}{dr}}{1+\left(\frac{dz}{dr}\right)^2}\tau_{rz}^H$$and similarly for $\rho_Lgz$.

Instead of representing the viscous stress tensor in $r$ and $z$, would it still be valid to use the normal, $n$, and tangent, $t$? So instead, the following could be written: $$ n \cdot \sigma \cdot n=-p_\infty+\rho gz+\tau_{nn}$$ where $$\tau_{nn}=2\eta \frac{\partial u_n}{\partial n}$$ in which $u_n$ is the velocity in the normal direction.

 

[Chestermiller]

Instead of representing the viscous stress tensor in $r$ and $z$, would it still be valid to use the normal, $n$, and tangent, $t$? So instead, the following could be written: $$ n \cdot \sigma \cdot n=-p_\infty+\rho gz+\tau_{nn}$$ where $$\tau_{nn}=2\eta \frac{\partial u_n}{\partial n}$$ in which $u_n$ is the velocity in the normal direction.

That would be correct. However, as I said in an earlier post, when the fluid is deforming, it is not valid to include the hydrostatic gravitational term in the equation for the normal stress. Your equation should really read $$ n \cdot \sigma \cdot n=-p+\tau_{nn}$$where p is the fluid pressure at the interface (and it is not equal to $p_{\infty}-\rho g z$).

 

[tse8682]

That would be correct. However, as I said in an earlier post, when the fluid is deforming, it is not valid to include the hydrostatic gravitational term in the equation for the normal stress. Your equation should really read $$ n \cdot \sigma \cdot n=-p+\tau_{nn}$$where p is the fluid pressure at the interface (and it is not equal to $p_{\infty}-\rho g z$).

In your equation, $p$ is referring to $p_{nn}$ I believe?

 

[tse8682]

That would be correct. However, as I said in an earlier post, when the fluid is deforming, it is not valid to include the hydrostatic gravitational term in the equation for the normal stress. Your equation should really read $$ n \cdot \sigma \cdot n=-p+\tau_{nn}$$where p is the fluid pressure at the interface (and it is not equal to $p_{\infty}-\rho g z$).

And is that also true for your derivation of $n \cdot \sigma \cdot n$ that the hydrostatic term is not valid for a deforming fluid? Or is it still there in the $r$ and $z$ coordinate system?

 

[Chestermiller]

In your equation, $p$ is referring to $p_{nn}$ I believe?

The pressure p is the isotropic part of the stress tensor, so it is the same in all directions.

 

[Chestermiller]

And is that also true for your derivation of $n \cdot \sigma \cdot n$ that the hydrostatic term is not valid for a deforming fluid? Or is it still there in the $r$ and $z$ coordinate system?

Yes. I really shouldn't have included the hydrostatic term in my derivation of the boundary condition at the interface for a deforming fluid.

 

[tse8682]

Yes. I really shouldn't have included the hydrostatic term in my derivation of the boundary condition at the interface for a deforming fluid.

Could an approximation or assumption be made that if the interface is not moving too quickly, the problem can be treated as quasi-static and the hydrostatic term is dominant and can thus be representative of the pressure $p$ at the interface? Otherwise, I'll need some other way to determine $p$.