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

Write $\hat{\rho }(z)$ for the ambient density far from the wall and $\rho (z)$ for the density at the wall. Across the plume, let $y=x/b$ and use the prescribed triangular profilesDirect integration gives the volume flux, mass flux, momentum flux, and density-weighted buoyancy flux, all per unit radiator length:These coefficients distinguish the triangular-profile wall line plume from a top-hat plume model.

The Batchelor entrainment hypothesis takes the inflow speed through the plume's exposed outer edge to be $\alpha W$, where $\alpha$ is the entrainment coefficient. A wall plume has only one such edge, so $dV/dz=\alpha W$.

The Boussinesq approximation replaces density by a constant reference value ${\rho }_0$ in inertia and mass flux while retaining the small density deficit in buoyancy. It requires $|\hat{\rho }-\rho |/{\rho }_0\ll 1$. A sufficiently hot radiator can violate this near the source, where thermal expansion is large and the developed-plume description may also fail.

Put $B_0=F_0/{\rho }_0$ for the kinematic buoyancy flux per unit length. Dimensional analysis for a line plume givesThe plume rise time is therefore $t_{\mathrm{rise}}\sim H/B_0^{1/3}$. Changing the room stratification requires a plume volume comparable with $XH$, so $t_{\mathrm{room}}\sim XH/V(H)\sim X/B_0^{1/3}$. HenceThe plume consequently follows the slowly changing ambient through a quasi-steady approximation when $X\gg H$.

Choose ${\rho }_0$ as a representative room density, for example the fresh-air density, and neglect relative density variations everywhere except in buoyancy. Withthe triangular profiles giveSolving these algebraic relations,

The Batchelor entrainment hypothesis, vertical momentum conservation, and mass conservation giveAn ascending parcel entrains ambient fluid from progressively lower ambient density. With the buoyancy frequencythe change of ambient reference density subtracts $Q{N}^2$ from its density-weighted buoyancy flux, so

The source supplies the kinematic buoyancy flux $B_0=F_0/{\rho }_0$. The unstratified line plume velocity scale is $B_0^{1/3}$. Comparing this with the stratification time scale $N_0^{-1}$ gives the stratified line-plume height scaleIf $z_s\gg H$, the initial plume is only weakly affected by the stable density stratification and reaches the ceiling much like an unstratified wall plume. If $z_s\sim H$, its buoyancy flux falls substantially during the rise; it approaches neutral buoyancy near the upper room, overshoots because of its momentum flux, and spreads as a horizontal buoyant intrusion. If $z_s\ll H$, the plume reaches neutral buoyancy low in the room and forms a low intrusion after a modest overshoot. In each sketch the plume widens by entrainment, while increasing $N_0$ lowers the neutral-buoyancy and overshoot heights.

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 $0<h<H$.