Normal Stress Balance at a Fluid-fluid Interface

[Chestermiller]

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 $p$.

Sure, if the interface is moving very slowly, then it's quasi-hydrostatic, and the fluid viscosity doesn't come in at all. Otherwise, you need to include the hydrodynamics (involving viscosity). You would obtain closure on the pressure by using the boundary condition that $p=p_{\infty}$ at z = 0 and r ---> $\infty$.

 

[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 $p_{\infty}-\rho g z$).

Would it be valid to consider $p$ as the sum of the hydrostatic and hydrodynamic pressures so $p=p_{\infty}-\rho gz+p_{dynamic}$? That seems to be what they did in that paper to derive their equation 1, I’m just wondering what assumptions are made to allow that. It seems like they use that assumption which then gives: $$n \cdot \sigma \cdot n=-p_{\infty}-\rho gz+p_{dynamic}+\tau_{nn}$$ That is then applied to a free surface in which the upper fluid density and viscosity are considered negligible which gives: $$[n \cdot \sigma^L \cdot n] - [n \cdot \sigma^H \cdot n]=\gamma (\nabla \cdot n)$$ $$[-p_{\infty}]-[-p_{\infty}-\rho^H gz+p_{dynamic}^H+\tau_{nn}^H]=\gamma \left[\frac{\frac{\partial z}{\partial r}}{r\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{1/2}}+\frac{\frac{\partial^2 z}{\partial r^2}}{\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{3/2}}\right]$$ Assuming $\left(\frac{\partial z}{\partial r}\right)^2 \ll 1$, this gives an almost identical equation to what they have: $$\rho^H gz-p_{dynamic}^H-\tau_{nn}^H=\gamma \left[\frac{1}{r}\frac{\partial z}{\partial r}+\frac{\partial^2 z}{\partial r^2}\right]=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)$$ where the superscripted $H$ and $L$ are for the high and low density fluids and $\gamma$ is the surface tension. They do not have the viscous stress term but that may be due to the presence of a thin film below the interface as opposed to infinite liquid. I know you said the hydrostatic term doesn’t belong, I’m just trying to reconcile that with what they have given.

 

[Chestermiller]

Yes. You can always define (and work with) as scalar $P=p+\rho g z$. But you have to be very careful with the mathematics, and with the discontinuities at the interface. Also, in the equations you have written, the "dynamic contribution" must still come from the coupling to the Navier Stokes equations (even for negligible intertia).

Also, in your equations above, I don't see why the dynamic term is not included on the L side of the interface. Also, is the viscosity the same on both sides of the interface?

 

[tse8682]

Yes. You can always define (and work with) as scalar $P=p+\rho g z$. But you have to be very careful with the mathematics, and with the discontinuities at the interface. Also, in the equations you have written, the "dynamic contribution" must still come from the coupling to the Navier Stokes equations (even for negligible intertia).

Also, in your equations above, I don't see why the dynamic term is not included on the L side of the interface. Also, is the viscosity the same on both sides of the interface?

No, the viscosity is different on either side as is density. In the above equation, I am only considering a free surface (such as a water-air interface) and saying the viscosity and density on the L (low density or fluid on top) side are very small. I assumed that would reduce the dynamic term to be negligibly small on that side?

 

[Chestermiller]

No, the viscosity is different on either side as is density. In the above equation, I am only considering a free surface (such as a water-air interface) and saying the viscosity and density on the L (low density or fluid on top) side are very small. I assumed that would reduce the dynamic term to be negligibly small on that side?

In that case, correct.

 

[tse8682]

Starting from the top, I'm going to change some notations to make everything clear. The top, less dense, fluid will be designated with a superscript $T$ and the bottom, more dense, fluid will be designated using the superscript $B$. The interfacial tension will be designated with $\gamma$. The stress tensor will be designated with $\sigma=-p\mathbf{I}+\tau$ where $\mathbf{I}$ is the Kronecker delta and $\tau=\mu[\nabla u +(\nabla u)^\top]$ is the viscous stress in which $\mu$ is the fluid viscosity and $u$ the fluid velocity. The pressure, $p$, will be considered as the sum of the hydrostatic and hydrodynamic pressure such that $p=p_\infty-\rho gz+p_D$. Starting in the general form for a normal stress balance at a fluid-fluid interface: $$\mathbf{n} \cdot \mathbf{\sigma^T} \cdot \mathbf{n}-\mathbf{n} \cdot \mathbf{\sigma^B} \cdot \mathbf{n}=\gamma(\nabla \cdot \mathbf{n})\tag{1}$$ $$[-p_{\infty}+\rho^T gz-p_D^T+\tau_{nn}^T]-[-p_{\infty}+\rho^B gz-p_D^B+\tau_{nn}^B]=-\gamma \left[\frac{\frac{\partial z}{\partial r}}{r\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{1/2}}+\frac{\frac{\partial^2 z}{\partial r^2}}{\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{3/2}}\right]\tag{2}$$ Assuming $\left(\frac{\partial z}{\partial r}\right)^2 \ll 1$ and noting the negative sign on the right side of equation 2 reduces it to: $$(\rho^B-\rho^T) gz-p_D^B+p_D^T+\tau_{nn}^B-\tau_{nn}^T=\gamma \left[\frac{1}{r}\frac{\partial z}{\partial r}+\frac{\partial^2 z}{\partial r^2}\right]=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\tag{3}$$ The normal viscous stress can be expressed as: $$\tau_{nn}=2\mu \frac{\partial u_n}{\partial n}\tag{4}$$ The normal velocity gradient is estimated based on the interfacial shape and velocity assuming the tangential velocity at the interface is zero. This isn't necessarily true but the assumption is made to simplify the problem. $$\frac{\partial u_n}{\partial n}=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\frac{\partial z}{\partial t}\tag{5}$$ Assuming this can be used for the normal velocity gradient in both the top and bottom fluid, equation 3 can be reduced to:$$(\rho^B-\rho^T) gz-p_D^B+p_D^T+2(\mu^B-\mu^T)\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\frac{\partial z}{\partial t}=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\tag{6}$$ The last piece of the puzzle is how to determine the dynamic pressures $p_D^B$ and $p_D^T$. Somehow, these must come from the continuity and Navier-Stokes equations. In cylindrical coordinates for axisymmetric conditions, these simplify to: $$\frac{1}{r}\frac{\partial}{\partial r}(ru_r)+\frac{\partial u_z}{\partial z}=0\tag{7}$$ $$\rho \left(\frac{\partial u_r}{\partial t}+u_r\frac{\partial u_r}{\partial r}+u_z\frac{\partial u_r}{\partial z}\right)=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial r}(ru_r)\right)+\frac{\partial^2u_r}{\partial z^2}\right]\tag{8}$$ $$\rho \left(\frac{\partial u_z}{\partial t}+u_r\frac{\partial u_z}{\partial r}+u_z\frac{\partial u_z}{\partial z}\right)=-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u_z}{\partial r}\right)+\frac{\partial^2u_z}{\partial z^2}\right]+\rho g\tag{9}$$ These might be able to be simplified further so I could get an actual solution for the dynamic pressure but I'm unsure of what more could be done.

 

[Chestermiller]

I disagree with much of this.

Starting from the top, I'm going to change some notations to make everything clear. The top, less dense, fluid will be designated with a superscript $T$ and the bottom, more dense, fluid will be designated using the superscript $B$. The interfacial tension will be designated with $\gamma$. The stress tensor will be designated with $\sigma=-p\mathbf{I}+\tau$ where $\mathbf{I}$ is the Kronecker delta and $\tau=\mu[\nabla u +(\nabla u)^\top]$ is the viscous stress in which $\mu$ is the fluid viscosity and $u$ the fluid velocity. The pressure, $p$, will be considered as the sum of the hydrostatic and hydrodynamic pressure such that $p=p_\infty-\rho gz+p_D$. Starting in the general form for a normal stress balance at a fluid-fluid interface: $$\mathbf{n} \cdot \mathbf{\sigma^T} \cdot \mathbf{n}-\mathbf{n} \cdot \mathbf{\sigma^B} \cdot \mathbf{n}=\gamma(\nabla \cdot \mathbf{n})\tag{1}$$ $$[-p_{\infty}+\rho^T gz-p_D^T+\tau_{nn}^T]-[-p_{\infty}+\rho^B gz-p_D^B+\tau_{nn}^B]=-\gamma \left[\frac{\frac{\partial z}{\partial r}}{r\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{1/2}}+\frac{\frac{\partial^2 z}{\partial r^2}}{\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{3/2}}\right]\tag{2}$$

This equation should read:
$$[-p^T+\tau_{nn}^T]-[-p^B+\tau_{nn}^B]=-\gamma \left[\frac{\frac{\partial z}{\partial r}}{r\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{1/2}}+\frac{\frac{\partial^2 z}{\partial r^2}}{\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{3/2}}\right]\tag{2}$$
As I said before, the $\rho g's$ should not be in the boundary condition. So the next equation should read:

$$(-p^B+p^T+\tau_{nn}^B-\tau_{nn}^T=\gamma \left[\frac{1}{r}\frac{\partial z}{\partial r}+\frac{\partial^2 z}{\partial r^2}\right]=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\tag{3}$$

The normal viscous stress can be expressed as: $$\tau_{nn}=2\mu \frac{\partial u_n}{\partial n}\tag{4}$$ The normal velocity gradient is estimated based on the interfacial shape and velocity assuming the tangential velocity at the interface is zero. This isn't necessarily true but the assumption is made to simplify the problem. $$\frac{\partial u_n}{\partial n}=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\frac{\partial z}{\partial t}\tag{5}$$

These equations for the normal stresses are not correct. I have no idea where you came up with Eqn. 5. Even though the velocity components are continuous at the interface, the velocity gradients are not. The correct equations for the viscous normal stresses at the interface (within the framework of your approximation are given by

$$\tau_{nn}^B=2\eta^B\left[\frac{\partial u_z^B}{\partial z}-\frac{dz}{dr}\left(\frac{\partial u_r^B}{\partial z}-\frac{\partial u_z^B}{\partial r}\right)\right]$$
$$\tau_{nn}^T=2\eta^T\left[\frac{\partial u_z^T}{\partial z}-\frac{dz}{dr}\left(\frac{\partial u_r^T}{\partial z}-\frac{\partial u_z^T}{\partial r}\right)\right]$$

Assuming this can be used for the normal velocity gradient in both the top and bottom fluid

In my judgment, your version can't be used.

Navier-Stokes equations. In cylindrical coordinates for axisymmetric conditions, these simplify to: $$\frac{1}{r}\frac{\partial}{\partial r}(ru_r)+\frac{\partial u_z}{\partial z}=0\tag{7}$$ $$\rho \left(\frac{\partial u_r}{\partial t}+u_r\frac{\partial u_r}{\partial r}+u_z\frac{\partial u_r}{\partial z}\right)=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial r}(ru_r)\right)+\frac{\partial^2u_r}{\partial z^2}\right]\tag{8}$$ $$\rho \left(\frac{\partial u_z}{\partial t}+u_r\frac{\partial u_z}{\partial r}+u_z\frac{\partial u_z}{\partial z}\right)=-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u_z}{\partial r}\right)+\frac{\partial^2u_z}{\partial z^2}\right]+\rho g\tag{9}$$ These might be able to be simplified further so I could get an actual solution for the dynamic pressure but I'm unsure of what more could be done.

This set of equations must be applied separately to the top fluid and the bottom fluid because the densities, the viscosities, and the velocity gradients in the two fluids are different throughout the separate regions and across the interface. Only the velocities match at the interface, not the velocity gradients. One thing you can do is neglect the inertial terms from the lhs's of these equations, since they are probably going to be negligible, except at very high frequencies and large displacements of the cylinder.

 

[tse8682]

This equation should read:
$$[-p^T+\tau_{nn}^T]-[-p^B+\tau_{nn}^B]=-\gamma \left[\frac{\frac{\partial z}{\partial r}}{r\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{1/2}}+\frac{\frac{\partial^2 z}{\partial r^2}}{\left(1+\left(\frac{\partial z}{\partial r}\right)^2\right)^{3/2}}\right]\tag{2}$$
As I said before, the $\rho g's$ should not be in the boundary condition. So the next equation should read:
$$(-p^B+p^T+\tau_{nn}^B-\tau_{nn}^T=\gamma \left[\frac{1}{r}\frac{\partial z}{\partial r}+\frac{\partial^2 z}{\partial r^2}\right]=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)\tag{3}$$

So $p^B$ and $p^T$ can't be considered as the sum of their respective hydrostatic and hydrodynamic pressures?

 

[Chestermiller]

So $p^B$ and $p^T$ can't be considered as the sum of their respective hydrostatic and hydrodynamic pressures?

Only if there is no flow or if you can figure out a way of manipulating the variables mathematically in my equations.

 

[tse8682]

Doing some more reading, I found a book and journal paper by the same author on interfacial hydrodynamics which gives a balance of normal forces for a quasi-flat interface in cartesian coordinates. It gives the following description in the book:

"The second boundary condition, the so-called dynamic condition, expresses the continuity of forces at the interface. Following the decision to resolve these into normal and tangential components, we write first, for the balance of normal forces: $$p^B-p^T+(\rho^T-\rho^B) gz+\tau_{zz}^B-\tau_{zz}^T+\gamma\nabla _{II}^2z=0$$ The pressures $p^B$ and $p^T$ may have dynamic (inertial) as well as hydrostatic components, which are referenced back to the datum plane. The next term is the hydrostatic adjustment to any elevation difference between the actual interface and the datum plane. The terms $\tau_{zz}^B$ and $\tau_{zz}^T$ are the normal viscous stresses, which for Newtonian incompressible fluids are given by: $$\tau_{zz}^B=-2\mu^B\left(\frac{\partial u_z^B}{\partial z}\right), etc.$$ The final term on the left hand side is the Young-Laplace curvature term..."

I'm not sure how they have hydrostatic components in $p$ and also as the third term in the equation. Seems like that'd be double counting or perhaps they cancel each other out.

 

[Chestermiller]

I'm as puzzled by this as you are. All I can say is that I am totally confident about what I've been saying. If you can figure out a way of manipulating my equations to a form consistent with their expressions, I'd be interested in seeing it.

 

[tse8682]

I'm as puzzled by this as you are. All I can say is that I am totally confident about what I've been saying. If you can figure out a way of manipulating my equations to a form consistent with their expressions, I'd be interested in seeing it.

My only guess is they substituted the sum of the hydrostatic and dynamic pressures in. So in your equation, letting $p=p_H+p_D$ where $p_H=p_0-\rho gz$ and $p_0$ is the hydrostatic pressure at the undisturbed interface. Dropping the $D$ subscript from the dynamic terms gives their equation. That's the only manipulation I could think of but it doesn't quite match their text description of the equation.

 

[Chestermiller]

My only guess is they substituted the sum of the hydrostatic and dynamic pressures in. So in your equation, letting $p=p_H+p_D$ where $p_H=p_0-\rho gz$ and $p_0$ is the hydrostatic pressure at the undisturbed interface. Dropping the $D$ subscript from the dynamic terms gives their equation. That's the only manipulation I could think of but it doesn't quite match their text description of the equation.

I don't know what to tell you. If I were doing this, I would be playing with the mathematics, starting with the equations I presented. I would be manipulating the equations and variables until I arrived at a form that I would be comfortable working with (trying many different possible options). But this would have to be done very carefully because of the complexity associated with the discontinuities at the free surface. I'm confident that I could do this if I wanted to and come up with a formulation that would accurately deliver the correct answer. But I'm not going to do it for you. I just don't want to put in the time on it, and, in addition, it's really your problem to run with. One thing I will say is that I don't have confidence in the analysis in your reference, and I don't think you do either. To be able to accept that reference, I would have to be able to start with my equations, and carefully and mathematically arrive at theirs.

 

[tse8682]

Well I have multiple sources now which give the following equation for the normal force balance as well as my own analysis that gives me that equation so I'm just going to run with it. If need be, I'll discount the hydrostatic terms later if they're already involved in the pressure calculations. $$(\rho^B-\rho^T) gz-p_D^B+p_D^T+2\mu^B \frac{\partial u^B_n}{\partial n}-2\mu^T \frac{\partial u^T_n}{\partial n}=\frac{\gamma}{r}\frac{\partial}{\partial r}\left(r\frac{\partial z}{\partial r}\right)$$ Moving forward, I'm going to try to apply some wave theory here. If I could determine the velocity potential, $\phi$, in each liquid, that might give me something usable in the stress balance. Assuming the flow in each is irrotational and incompressible, its valid to write:$$\nabla^2 \phi(r,z,t)=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial \phi}{\partial r}\right)+\frac{\partial^2\phi}{\partial z^2}=\frac{\partial^2\phi}{\partial r^2}+\frac{1}{r}\frac{\partial\phi}{\partial r}+\frac{\partial^2\phi}{\partial z^2}=0$$ Using separation of variables and letting $\phi(r,z,t)=R(r)Z(z)T(t)$, this simplifies to: $$Z\frac{\partial^2R}{\partial r^2}+\frac{Z}{r}\frac{\partial R}{\partial r}+R\frac{\partial^2Z}{\partial z^2}=0$$ Separating this into two functions gives: $$\left(\frac{\partial^2R}{\partial r^2}+\frac{1}{r}\frac{\partial R}{\partial r}\right)R^{-1}=-Z^{-1}\frac{\partial^2Z}{\partial z^2}=E^2$$ where $E^2$ is some constant. Solving each of these gives solutions of:$$R(r)=C_1I_0(Er)+C_2K_0(Er)$$ $$Z(z)=-\frac{z^3}{6}E^2+C_3z+C_4$$ Boundary conditions must then be defined to determine the value of the constants. I'm changing the problem slightly, so defining the interface as $z=\eta(r,t)$ and instead of a rod I'll say just the interface at $r=r_i$ is moving such that $\eta(r_i,t)=A\sin(\omega t)$. I will assume that no disturbance exists far away so $\frac{\partial \phi}{\partial t}=\nabla \phi=0$ and $p=p_0-\rho gz$ at $r=r_f$ and/or at $z=H$. Since $I_0$ is unbounded at large argument, this leaves me to assume $C_1=0$. At the inner boundary, I can apply $R(r_i)=A\sin(\omega t)=C_2K_0(Er_i)$ to say $C_2=\frac{1}{K_0(Er_i)}A\sin(\omega t)$. I think the unsteady Bernoulli equation is used too which is:$$p+\rho\left[\frac{\partial \phi}{\partial t}+\frac{1}{2}\mid\nabla\phi\mid^2+gz\right]=c(t)$$ Thus, $p$ is written as:$$p=-\rho\left[\frac{\partial \phi}{\partial t}+\frac{1}{2}\mid\nabla\phi\mid^2+gz\right]+c(t)$$In free surface wave theory, they say $p=p_{atm}$ at $z=\eta$ and let $c(t)=p_{atm}$ so for $z=\eta$ it can be written that $-\rho\left[\frac{\partial \phi}{\partial t}+\frac{1}{2}\mid\nabla\phi\mid^2+g\eta\right]=0$. This isn't true for this problem so still working on that part. I can also say that $\frac{\partial \phi}{\partial z}=\frac{\partial \eta}{\partial t}$ at $z=\eta$. To make it a little simpler to solve, it can also be shown that these conditions can be taken at $z=0$ instead of $z=\eta$ by taking the Taylor's series expansion about $z=0$. The next step will be applying these BCs to determine and equation for $\phi$. Linearizing the equations may also help.

 

[Chestermiller]

I think I may have figured out how they might have arrived at the approximation that they used for the normal stress boundary condition. The "divergence form" of the z component of the axisymmetric equation of motion for an incompressible fluid in cylindrical coordinates is given by:
$$\frac{\partial (\rho u_z)}{\partial t}+\frac{1}{r}\frac{\partial (\rho ru_ru_z)}{\partial r}+\frac{\partial (\rho u_z^2)}{\partial z}=-\frac{\partial p}{\partial z}-\rho g+\frac{1}{r}\frac{\partial \tau_{rz}}{\partial r}+\frac{\partial \tau_{zz}}{\partial z}$$where we use the convention that the $\tau's$ represent tensile stresses.

Now at large values of r, the flow disturbance caused by the cylinder movement is going to be negligible. So we are going to integrate the equations of motion over the following path (within the top fluid T):

1. Start at z = 0 and r = r* (where r* is large and the pressure is $p_{\infty}$) and integrate vertically to large z = z*

2. Integrate radially inward from at constant z = z* from r = r* to a value of r within the radial region where the interface is curved

3. Start at z = z* and r, and integrate downward at constant r down to the curved interface.

For Step 1, we obtain $$p^T(r^*,z^*)=p_{\infty}-\rho^Tg z^*$$

For Step 2, since at large z, $u_r\rightarrow 0$ at large z, we obtain $$p^T(r,z^*)=p_{\infty}-\rho^Tg z^*$$

For Step 3, since we are assuming that $\frac{dz}{dr}<<1$, we are going to neglect the radial derivatives in the z equation of motion (for purposes only of the boundary condition development). This leads to
$$\frac{\partial (\rho u_z^2)}{\partial z}=-\frac{\partial p}{\partial z}-\rho g+\frac{\partial \tau_{zz}}{\partial z}$$where we have also neglected the time derivative. If we integrate this in Step 3, we obtain at the interface: $$p^T(r,z)=p_{\infty}-\rho^T g z-\rho^T(u_z^T)^2+\tau^T_{zz}$$

Thoughts?

 

[tse8682]

I think I may have figured out how they might have arrived at the approximation that they used for the normal stress boundary condition. The "divergence form" of the z component of the axisymmetric equation of motion for an incompressible fluid in cylindrical coordinates is given by:
$$\frac{\partial (\rho u_z)}{\partial t}+\frac{1}{r}\frac{\partial (\rho ru_ru_z)}{\partial r}+\frac{\partial (\rho u_z^2)}{\partial z}=-\frac{\partial p}{\partial z}-\rho g+\frac{1}{r}\frac{\partial \tau_{rz}}{\partial r}+\frac{\partial \tau_{zz}}{\partial z}$$where we use the convention that the $\tau's$ represent tensile stresses.

Now at large values of r, the flow disturbance caused by the cylinder movement is going to be negligible. So we are going to integrate the equations of motion over the following path (within the top fluid T):

1. Start at z = 0 and r = r* (where r* is large and the pressure is $p_{\infty}$) and integrate vertically to large z = z*

2. Integrate radially inward from at constant z = z* from r = r* to a value of r within the radial region where the interface is curved

3. Start at z = z* and r, and integrate downward at constant r down to the curved interface.

For Step 1, we obtain $$p^T(r^*,z^*)=p_{\infty}-\rho^Tg z^*$$

For Step 2, since at large z, $u_r\rightarrow 0$ at large z, we obtain $$p^T(r,z^*)=p_{\infty}-\rho^Tg z^*$$

For Step 3, since we are assuming that $\frac{dz}{dr}<<1$, we are going to neglect the radial derivatives in the z equation of motion (for purposes only of the boundary condition development). This leads to
$$\frac{\partial (\rho u_z^2)}{\partial z}=-\frac{\partial p}{\partial z}-\rho g+\frac{\partial \tau_{zz}}{\partial z}$$where we have also neglected the time derivative. If we integrate this in Step 3, we obtain at the interface: $$p^T(r,z)=p_{\infty}-\rho^T g z-\rho^T(u_z^T)^2+\tau^T_{zz}$$

Thoughts?

This all looks sound to me. I think the second to last term in the first equation should read $\frac{1}{r}\frac{\partial(r \tau_{rz})}{\partial r}$ if I'm doing my tensor divergence correctly, but it's being neglected later anyway. For step two, would it also be necessary to assume $u_z=0$? I can work with that too, I'm just thinking if it were an infinitely long rod, there would still be $u_z$ in at a value $r$ where the interface is curved. If its just something like a thin puck moving up and down at the interface though, step two should be valid. I guess my only other thought is what assumption would be needed to neglect the time derivative, maybe just that that the interface velocity is not changing too quickly.

 

[Chestermiller]

This all looks sound to me. I think the second to last term in the first equation should read $\frac{1}{r}\frac{\partial(r \tau_{rz})}{\partial r}$

Yes. I mis-typed that term.

if I'm doing my tensor divergence correctly, but it's being neglected later anyway. For step two, would it also be necessary to assume $u_z=0$?

No. Just that, at large z, $u_z$ is a function only of r and t and $u_r$ is zero.

I guess my only other thought is what assumption would be needed to neglect the time derivative, maybe just that that the interface velocity is not changing too quickly.

I guess.