 Solution

ID: past-exam-of-the-fluids-course-of-the-university-of-oxford/2021/3/b/solution

Integrating the radial velocity over the thickness gives the radial volume flux per unit width:

q_{r} = \int_{0}^{h} v_{r} d z = \frac{ρ g h _{r} + γ κ _{r}}{2 μ} [\frac{h ^{3}}{3} - h^{3}] = - \frac{h ^{3}}{3 μ} (ρ g h_{r} + γ κ_{r}) .

Q (r, t) = 2 π r q_{r} = - \frac{2 π r h ^{3}}{3 μ} (ρ g h_{r} + γ κ_{r}) .

An annulus of radial thickness

d r

contains volume

2 π r h d r

to leading order. Its rate of gain equals incoming minus outgoing volume flux, so

2 π r h_{t} d r = Q (r, t) - Q (r + d r, t) = - Q_{r} d r + o (d r) .

Dividing and taking

d r \to 0

yields the axisymmetric thin-film equation:

h_{t} = \frac{1}{r} \partial_{r} [\frac{r h ^{3}}{3 μ} (ρ g h_{r} + γ κ_{r})] .

The equation assumes no evaporation or injection through the boundaries. Integrating it over the whole drop also verifies conservation of mass when the edge carries zero volume flux.

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