Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 4

The Boussinesq approximation uses a constant reference density in inertia and continuity while retaining the small temperature-dependent density deficit in buoyancy. Here $({\rho }_0-\rho )/{\rho }_0=(T-T_0)/T_0$, so the plume remains incompressible but rises under $g(T-T_0)/T_0$.

The Batchelor entrainment hypothesis sets the ambient inflow speed at the exposed plume edge to $EW$, where $E$ is a dimensionless entrainment coefficient. For this one-sided heated wall plume, it gives $dQ/dz=EW$.

With thermal diffusivity ${\kappa }_T$, temperature obeysScale $z$ with $h_0$, $x$ with $b_0=ϵh_0$, and $w$ with $w_0$. Continuity gives the cross-plume velocity scale $u_0\sim (b_0/h_0)w_0=ϵw_0$, so both advective terms scale as $w_0\Delta T/h_0$. Horizontal and vertical diffusion scale as ${\kappa }_T\Delta T/b_0^2$ and ${\kappa }_T\Delta T/h_0^2$, respectively. Their ratio is

For top-hat profiles, $Q=bW$, $M=b{W}^2$, and $\mathcal{B}=Qg(T-T_0)/T_0$. One-sided entrainment, vertical momentum, and the wall heat input giveIntegrating the temperature equation across the plume givesCombining this with the definition of buoyancy flux yields

Far above the source, $\mathcal{B}\sim \beta z$. Suppose $Q\sim z^p$ and $M\sim z^r$. The first integral equation gives $p-1=r-p$, while the second gives $r-1=1+p-r$. Solving,Hence

Introduce the virtual-origin coordinateso that $\mathcal{B}=\beta \zeta$. Seek $Q=A{\zeta }^{4/3}$ and $M=C{\zeta }^{5/3}$. Substitution givesTherefore the exact similarity solution isand

At the source, $\zeta =z_0={\mathcal{B}}_0/\beta$. The initial width required to lie exactly on the similarity solution is thereforeThe corresponding $Q_0=A{z}_0^{4/3}$ and $M_0=C{z}_0^{5/3}$ are the compatible initial fluxes stipulated in the question.