Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 3 / e / Solution

The final state is displacement ventilation: fresh air of density ${\rho }_0$ forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height $h$. Steady volume conservation requires the plume volume flux there to equal the imposed ventilation flux,

Below the interface the ambient is uniform, so $B_0=F_0/{\rho }_0$ is constant. Put $J=M/{\rho }_0$. The triangular-profile wall line plume equations reduce toThe pure plume conditions $V,J\to 0$ at the radiator select the similarity solutionConsequently the displacement-ventilation interface height isThe upper-layer reduced gravity follows from its steady buoyancy balance as $g_u^{\prime }=B_0/A=F_0/({\rho }_{0}A)$. The two-layer solution applies when the calculated interface satisfies $0<h<H$.