Question 2 of 6: A2 — Spin-coating film-thinning law

Question 2: A2 — Spin-coating film-thinning law (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A liquid film of thickness $h(r,t)$ on a wafer spinning at constant angular velocity $\omega$; liquid density $\rho$, viscosity $\mu$, kinematic viscosity $\nu=\mu/\rho$; the film is thin so lubrication (thin-film) scaling applies — radial pressure gradients are negligible and the free surface is shear-free.

Find. (a) $h(t)$ from an initial thickness $h_0$; (b) the long-time (very-thin-film) limiting form.

Approach. Balance the centrifugal body force against viscous shear to get the radial velocity profile, integrate it for the outward flux per unit circumference, then close a thin-film continuity equation to obtain an ODE for $h(t)$; integrate and take the long-time limit.

1. Radial momentum (lubrication) balance. With the free surface shear-free and pressure gradients negligible, the centrifugal body force is balanced by viscous shear: $\mu\dfrac{\partial^2 u_r}{\partial z^2}=-\rho\omega^2 r,$ where $z$ is measured up from the wafer.
2. Velocity profile. Integrating with no-slip at the wafer $u_r(0)=0$ and zero shear at the free surface $\partial u_r/\partial z|_{z=h}=0$: $$u_r(r,z)=\frac{\rho\omega^2 r}{\mu}\left(hz-\frac{z^2}{2}\right).$$
3. Outward flux per unit circumference. $q=\displaystyle\int_0^h u_r\,dz=\frac{\rho\omega^2 r}{\mu}\left(\frac{h^3}{2}-\frac{h^3}{6}\right)=\frac{\rho\omega^2 r\,h^3}{3\mu}=\frac{\omega^2 r\,h^3}{3\nu}.$
4. Thin-film continuity. Conservation of liquid on the axisymmetric film gives $\dfrac{\partial h}{\partial t}+\dfrac1r\dfrac{\partial (r\,q)}{\partial r}=0.$ Substituting $q$ and noting that a spatially-uniform film has $\partial h/\partial r=0$ so $\tfrac1r\partial_r(r\,q)=\tfrac{2\omega^2h^3}{3\nu}$: $$\boxed{\dfrac{\partial h}{\partial t}=-\frac{2\omega^2}{3\nu}\,h^3}.$$
5. Integrate the thinning ODE. Separating $-h^{-3}dh=\tfrac{2\omega^2}{3\nu}dt$ from $h_0$ at $t=0$: $\dfrac{1}{2h^2}-\dfrac{1}{2h_0^2}=\dfrac{2\omega^2}{3\nu}t$, i.e. $$\boxed{h(t)=\frac{h_0}{\sqrt{\,1+\dfrac{4\omega^2 h_0^2}{3\nu}\,t\,}}}\qquad\text{(part a)}.$$
6. Very-thin-film (long-time) limit. When $\dfrac{4\omega^2 h_0^2}{3\nu}t\gg1$ the "1" and hence $h_0$ drop out: $$\boxed{h(t)\to\sqrt{\frac{3\nu}{4\omega^2 t}}=\frac12\sqrt{\frac{3\nu}{\omega^2 t}}}\qquad\text{(part b)}.$$ The thickness becomes independent of the initial amount deposited.