Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 332 / 2 / vi / Solution

The approximation in part v requires the dimensionless stagnant depthto remain small. Since $0<\theta \le 1$ makes the numerator order one, this condition isUsing the solution from part v, the equivalent time range isor, at the level of asymptotic equivalence,The cooling becomes appreciable on the shorter scale $\tau =O({\mathcal{F}}^{-1})$, so these ranges overlap widely when $\mathcal{F}\gg 1$.

The Rayleigh number based on the convecting depth isThe definition of $\mathcal{F}$ implicit in part iii givesIt follows thatIf $1-\eta =O(1)$ and $\theta \gg {\mathcal{F}}^{-3/5}$, then $\mathrm{Ra}\gg O({\mathcal{F}}^{9/5})$. The thermal convection therefore remains strongly supercritical throughout the range in which the thin stagnant-layer approximation is valid.