Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 75 / 2

Write $\beta$ for the thermal expansion coefficient, $\nu$ for kinematic viscosity, $\kappa$ for thermal diffusivity, and $k_T=\rho c_p\kappa$ for thermal conductivity. For a temperature difference $\Delta >0$ across depth $h$,The four-thirds convective heat-transfer law posits $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ at high Rayleigh number, and thereforeThe layer depth cancels. A useful physical justification is a mixed interior with thin conductive boundary layers of thickness $\delta$, each maintained near a fixed critical local Rayleigh number: $g\beta {\Delta }_b{\delta }^3/(\nu \kappa )\sim {\mathrm{Ra}}_{\mathrm{crit}}$ and $q\sim k_T{\Delta }_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one boundary layer changes the coefficient, not the exponent.

Assume Boussinesq approximation, constant material properties, a laterally extensive statistically turbulent Rayleigh-Bénard convection layer, negligible side-wall heat transfer, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified Prandtl number regime when choosing $C$. The closure is empirical and regime dependent; high Rayleigh number alone does not make it exact or extend it to the impulsive conductive initial transient.

Let $\overline{T}=T_0+\vartheta (t)$, with $0\le \vartheta \le \Delta T$. Apply the same one-boundary four-thirds convective heat-transfer law coefficient $K$ to both boundary layers. The lower inward heat flux is $K(\Delta T-\vartheta {)}^{4/3}$ and the upper outward heat flux is $K{\vartheta }^{4/3}$. The layer's conservation of energy per unit area givesHere $\rho c_p$ is the volumetric heat capacity. This uses a uniform bulk temperature, negligible boundary layer heat storage, constant depth and material properties, prescribed boundary temperatures, no lateral losses or internal heat generation, and identical upper and lower closure coefficients. It deliberately extrapolates the turbulent closure to $t=0$, as the model requires; an actual impulsive start first has a conductive transient. This is the requested evolution equation, without solving it.

A linear, or “three-thirds”, boundary heat transfer closure needs its own dimensional conductance $L>0$, so $q=L\Delta$. One cannot simply keep the dimensions of the previous nonlinear coefficient $K$. With equal conductances at both boundaries, the mean-temperature equation becomesIntegrating this linear ordinary differential equation givesThe relaxation time is $\rho c_{p}h/(2L)$. If the conductances differed, conservation of energy would instead select their weighted boundary temperature; the arithmetic mean follows from the assumed upper/lower symmetry.