Dynamic boundary condition

Question

I need to compute the dynamic boundary condition for a small drop slowly spreading on a completely wetting, solid substrate . We are using cylindrical coordinates (s,z), and there is no flow in the $\phi$ direction. The shape of the drop is given by h(s,t). The velocity field in s is u(s,z), and the velocity field in the z direction is w(s,z).

The stress tensor $\sigma_1$ is in cylindrical coordinates given by:

$\sigma_{ss}=2\eta \nabla_{s}u-p$ $\\$

$\sigma_{zz}=2\eta \nabla_{z}w-p$ $\\$

$\sigma_{sz}=\eta( \nabla_{z}u+\nabla_{s}w)$ $\\$

The dynamic boundary condition is the continuity of the stress vector across an interface i.e. $ \ \sigma_{1}n_{1}-\sigma_2n_2=0$=> $\sigma_{1}n_{1}=\sigma_2n_2$

I guess a normal vector could be $n_1=n_2=(1,0,0)$.

I am a bit unsure if that works.

The other thing I am not sure about is the stress tensor for air $\sigma_2$.

So I am not sure how to compute this dynamic BC.

The second thing I need to show is that in lowest order of $h'=\nabla_sh$ the conditions can be written as:

$\nabla_{z}u(s,z=h) \approx 0$ and $p(s,z=h) \approx P.$ where P is the laplace pressure

Any ideas ?

Answer 1

If you apply the Cauchy stress relationship at the free surface and take the traction in the air to be zero (i.e., zero gauge pressure (p=0) and zero tangential stress), then the stress boundary conditions at the free surface are: $$\sigma_{zz}=\sigma_{sz}\frac{\partial h}{\partial s}$$and$$\sigma_{sz}=\sigma_{ss}\frac{\partial h}{\partial s}$$ This neglects surface tension.

Answer 2

The shape of the drop is given by the equation $q(s,z,t)=z-h(s,t)=0$, in which $s$ is the radial coordinate, $z$ is the axial coordinate referred to a cylindrical coordinate system whose axis coincides with the symmetry axis of the spreading drop, $h$ and $q$ are functions, and $t$ is time. The outward normal to this surface (i.e. normal pointing into the air) is given by: $$\mathbf{n}=\frac{\nabla q}{|\nabla q|}$$

Now $\nabla q=(-\frac{\partial h}{\partial s},0,1)$ in cylindrical coordinates. Therefore: $$\frac{1}{|\nabla q|}=\left(1+\left(\frac{\partial h}{\partial s}\right)^2\right)^{-1/2}=1-\frac{1}{2}\left(\frac{\partial h}{\partial s}\right)^2+O\left(\left(\frac{\partial h}{\partial s}\right)^4\right)$$

Therefore if $\frac{\partial h}{\partial s}\ll 1$ then correct to linear order in $\frac{\partial h}{\partial s}$ we have: $\mathbf{n}=\nabla q$.

If $\sigma_{0}, \sigma,$ are stress tensors inside air and the drop respectively, then at the drop surface the corresponding stress vectors are $\mathbf{f}_0=\sigma_{0}\cdot\mathbf{n}$ and $\mathbf{f}=\sigma\cdot(-\mathbf{n})$. There is a jump in the normal stress due to surface tension of the drop-air interface. If $P(t)$ be the associated Laplace pressure then $-P\mathbf{n}$ is the force exerted by surface tension. Then force balance gives: $$\mathbf{f}_0+\mathbf{f}-P\mathbf{n}=\mathbf{0}\\\Rightarrow\sigma_{0}\cdot\mathbf{n}-\sigma\cdot\mathbf{n}=P\mathbf{n}$$ which is the dynamic boundary condition you need.

The stress tensor for air is a required input for further calculations. Usually the shear stress exerted by air on the drop is assumed to be zero. If you make this assumption (which must be correct to at least linear order in $\frac{\partial h}{\partial s}$), then the stress tensor for air becomes isotropic: $\sigma_0=-p_0\mathbf{I}$, in which $\mathbf{I}$ is the identity tensor and $p_0$ is the constant hydrostatic pressure due to air. If the shear stress exerted by air is zero, then from a force balance in the tangential direction at the drop surface, it follows that shear stress on the liquid side (evaluated at the drop surface) must also be zero. Therefore $\sigma=-p\mathbf{I}$ (when evaluated at the drop surface). Substituting into the dynamic boundary condition, and dotting the resulting expression with $\mathbf{n}$, we get: $$p(t)-p_0=P(t)$$ If you are considering only gauge pressures then you may set $p_0=0$ in the equation above.

P.S. If $\mathbf{t}$ is a tangent to the drop surface then we must have $\mathbf{t}\cdot\mathbf{n}=0$. Since $\mathbf{n}=(-\frac{\partial h}{\partial s},0,1)$, a tangent would be $\mathbf{t}=(1,0,\frac{\partial h}{\partial s})$. Another orthogonal tangent would then be $\tilde{\mathbf{t}}=\mathbf{n}\times\mathbf{t}=(0,1,0)$. Now dotting the dynamic boundary condition with $\mathbf{t}$ and $\tilde{\mathbf{t}}$ in turn, and using the last result, you can perhaps derive the other identities you need correct to linear order in $\frac{\partial h}{\partial s}$. I didn't try because it involves tedious algebra.