Fluid pressure in Neumann's triangle for triple junctions

Question

Consider a point of contact where three fluid phases meet. The Neumann's triangle rule for this case is usually written as, \begin{equation} \frac{\sigma_{\alpha\beta}}{\sin \gamma} = \frac{\sigma_{\beta\gamma}}{\sin \alpha} = \frac{\sigma_{\gamma\alpha}}{\sin \beta} \end{equation}

where $\sigma_{\alpha\beta}$ is the surface tension between two fluids and the angle $\gamma$ shows the angle created by phase $\gamma$ at the triple contact point. The angles naturally satisfy $\alpha + \beta +\gamma = 2\pi$.

In the case of diffuse interface modelling, the surface tension effects on the fluid momentum balance (Navier-Stokes equations) are expressed as, \begin{equation} \frac{\partial \rho\mathbf{u}}{\partial t} + \nabla\cdot\nabla\cdot(\rho\mathbf{u}\mathbf{u}) = -\nabla p + \nabla\cdot S -\sum_{\alpha<\beta}\sigma_{\alpha\beta}\kappa_{\alpha\beta}\nabla\varphi_{\alpha\beta} \label{Eq:NV} \end{equation}

where $S$ is the shear stress tensor, $\kappa_{\alpha\beta}$ is curvature, $\nabla\varphi_{\alpha\beta}$ is gradient of the phase indicator function of the interface. In equilibrium, when the fluid velocity is zero $\mathbf{u} = 0$, we can write, \begin{equation} -\nabla p -\sum_{\alpha<\beta}\sigma_{\alpha\beta}\kappa_{\alpha\beta}\nabla\varphi_{\alpha\beta} = 0 \end{equation} I have some questions regarding this:

1) I understand the Lami's law for force balance. But, in order to extend the Lami's law we must specify how the surface tension is converted to an equivalent force. Specifically, we should mention which length scale is divided with surface tension $\sigma_{\alpha\beta}$ etc so that a force term is generated. Also why this length scale is same for all the three interfaces?

2) From the two phase fluid case, I know that the pressure difference across the interface is related to the surface tension as $\Delta p = \sigma\kappa$, where $\kappa$ is the curvature. Why we do not consider the pressure drop across different interfaces while considering the equilibrium situation of three phase contact? Surely there will a component of the pressure drop $\Delta p_{\alpha\beta}$ in a direction tangent to the interface for $\alpha\gamma$ and $\beta\gamma$ interfaces. What am I missing here?

3) The Neumann's triangle relation is equivalent, \begin{equation} \sigma_{\alpha\beta}\hat{\tau}_{\alpha\beta} + \sigma_{\beta\gamma}\hat{\tau}_{\beta\gamma} + \sigma_{\gamma\alpha}\hat{\tau}_{\gamma\alpha} = 0 \end{equation} where $\hat{\tau}_{\gamma\alpha}$ is a unit vector along tangential direction to $\gamma\alpha$ interface. How do the tangential components come into play when in the Navier-Stokes equation (see Eq.~(\ref{Eq:NV})) we include surface tension forces in a direction normal to the interface?

Any hint or justification will be highly appreciated.

Answer 1

some hints here:

1. The Lami law is two-dimensional, while the point of contact of three fluids is actually a line. The "unit length" is this third dimension, the interface line, that rarely appears in the drawings. All drawings i found represent only a 2D slice. Strictly speaking, the tension at the point (line) is a force per unit length.
2. At the interface point we disregard the curvature, since its contribution is small; the dominant contribution is the tension along the surface. Note also that difference in pressure has negligible contribution at the point of contact (arbitrary small surface). Locally, the surface tension is much greater than the force due to pressure difference.
3. I do not know how to write Navier-Stokes... I will say as before, at the point you have forces only tangential, nothing normal to the interface