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tse8682
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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!
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