Normal Stress Balance at a Fluid-fluid Interface

[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.