We have a small drop spreading on a completely wetting solid substrate. The drop shape is h(s,t). The coordinate system is cylindrical (s,z). The velocity fields are: u(s,z) in the s direction and w(s,z) in the z direction,$\vec{U}=(u(s,z, w(s,z))$. There is no flow in the $\phi$ direction. I need to show that combining kinematic boundary condition with the continuity equation gives the mass conservation in the following form: $\nabla_th+\frac{1}{s}\nabla_s(sq)$ where $q=\int_{0}^{h}u(s,z)dz$.

I assume that the flow is incompressible, so the continuity equation is given by $\nabla \cdot U=\frac{1}{s}\frac{\partial}{\partial s}(s u(s,z))+\frac{\partial}{\partial z}(w(s,z))=0$.

I am a bit unsure about the kinematic boundary condition. I guess I can use the no slip condition for the kinematic boundary condition. I guess then the velocity is zero at the bottom u(s,z=0)=0. Again, I am not sure what this gives me. Another kinematic boundary condition I was thinking about is that the normal component of the velocity field should be zero at the boundary since no flux can be there. I was thinking about expressing it as $\vec{U} \cdot \vec{n}=0$, but do not know how to get the normal vector. Do you have a hint ?

  • $\begingroup$ Integrate the continuity equation with respect to z between 0 and h. $\endgroup$ – Chet Miller Apr 27 '18 at 23:55

Continuity equation: $$\frac{1}{s}\frac{\partial (us)}{\partial s}+\frac{\partial w}{\partial z}=0$$Integrating with respect to z: $$\frac{1}{s}\int_0^h{\frac{\partial (us)}{\partial s}dz}+w|_h=0\tag{1}$$ Applying the Leibnitz rule:$$\int_0^h{\frac{\partial (us)}{\partial s}dz}=\frac{\partial}{\partial s}\left(s\int_0^h{udz}\right)-su|_h\frac{\partial h}{\partial s}\tag{2}$$Combining Eqns. 1 and 2:
$$\frac{1}{s}\frac{\partial (sq)}{\partial s}+w|_h-u|_h\frac{\partial h}{\partial s}=0\tag{3}$$But, kinemtaically, $$w|_h=\frac{\partial h}{\partial t}+u|_h\frac{\partial h}{\partial s}\tag{4}$$Therefore, combining Eqns. 3 and 4, we obtain: $$\frac{1}{s}\frac{\partial (sq)}{\partial s}+\frac{\partial h}{\partial t}=0\tag{5}$$


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