Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 76 / 6

Assume Boussinesq approximation, no wind, a uniform exterior, negligible source volume, sharp well-mixed internal layers, point-source pure plume entrainment and thin openings represented by discharge coefficients $C_b,C_t$. Let $g_1^{\prime }=g\Delta {\rho }_1/\rho$ be the upper-layer reduced gravity. Steady volume conservation makes the magnitudes of inlet and outlet flow equal to $Q$.

If the opening pressure drops are $\Delta p_b$ and $\Delta p_t$, the orifice laws and hydrostatic pressure difference giveDefine the reduced area byThen the throughflow in this displacement ventilation model isLower-layer conservation requires $Q=Q_p(h_1)=C{B}_1^{1/3}{h}_1^{5/3}$. Global buoyancy conservation gives $g_1^{\prime }Q=B_1$. Eliminating $Q,g_1^{\prime }$ yieldsThis uniquely determines an interface height between zero and $H$ for a given positive area parameter. Opening area controls the nondimensional height; source strength controls the density contrast and flow magnitude. Factors of two depend on the stated effective-area convention, which has been fixed explicitly here.

Let $g_i^{\prime }=g\Delta {\rho }_i/\rho$ denote the density deficit of each warm layer relative to outside. The integrated hydrostatic driving head isThe same pair of orifice laws therefore givesBoth plumes remove fluid from the lower layer. If $q_1=C{B}_1^{1/3}{h}_1^{5/3}$ and $q_{21}=C{B}_2^{1/3}{h}_1^{5/3}$ are their fluxes at its upper interface, lower-layer conservation also givesIn steady state the weak plume supplies the middle layer, while the strong plume entrains that supply above $h_1$ and carries the total volume into the upper layer. Consequently $Q_2(h_2)=Q$ and $g_2^{\prime }Q=B_1+B_2$. These balances assume the plumes remain spatially separate and wall heat loss is negligible.

Put $\psi =B_1/B_2$. The middle layer matches the density of plume 1 at $h_1$, so $g_1^{\prime }=B_1/q_1$. The upper layer exports both sources, so $g_2^{\prime }=(B_1+B_2)/Q$. With $Q=q_1+q_{21}$ and $q_{21}/q_1={\psi }^{-1/3}$, the two-plume three-layer displacement ventilation density ratio isThe driving head can therefore be rewritten as $\mathcal{H}=g_2^{\prime }\mathcal{L}$, whereCombining $Q^2=(A^{\ast }{)}^2{g}_2^{\prime }\mathcal{L}$ with $g_2^{\prime }Q=B_2(1+\psi )$ gives $Q^3=(A^{\ast }{)}^2{B}_2(1+\psi )\mathcal{L}$. But $Q^3=C^3{B}_2(1+{\psi }^{1/3}{)}^3{h}_1^5$, and henceThe square root requires a positive remaining buoyancy head.

As $\psi \to 0$, the weak source vanishes and $g_1^{\prime }/g_2^{\prime }\to 0$: the middle and lower layers have the same density, leaving a single physical interface at $h_2$. The old $h_1$ no longer marks an independent density interface. The single-plume two-layer model is the meaningful limit; the additional closure in part (e) also gives $h_2/h_1\to 1$.

As $\psi \to 1$, the density ratio tends to one, so the middle and upper layers merge. There is then a single physical lower interface $h_1$, fed by two equal, separate plumes. The area prefactor tends to two and the head reduces to $g^{\prime }(H-h_1)$. The mathematical label $h_2$ ceases to identify a distinct density interface, so a finite limiting value assigned to it by an extra closure is not evidence for three distinct layers. Two separate equal plumes entrain more than one coincident plume of combined source strength; the distinction survives after the layers merge.

Above $h_1$, plume 2 entrains middle-layer fluid rather than the original ambient. That fluid is already warm and light, so the plume's buoyancy relative to its surroundings is reduced. Its pre-interface density contrast and zero-source pure plume formula cannot simply be continued unchanged through the interface.

A consistent model treats the middle layer as a uniform new ambient, assumes a sharp interface and weak overall density differences, and carries the incident volume and momentum fluxes continuously across it. The relative buoyancy flux immediately above the interface isContinue an entraining plume model from these nonzero incident fluxes, using buoyancy measured relative to the middle layer. In that uniform layer the relative buoyancy flux is conserved, while entrainment raises the absolute buoyancy carried relative to the outside environment. This is generally a finite-flux or forced plume continuation, not a newly reset point-source similarity solution. A virtual-origin approximation can be used only after matching its fluxes appropriately.

At its matching level $h_2$, the plume density reaches the upper-layer density; it spreads and detrains, with possible finite-momentum overshoot. The model neglects mutual plume entrainment, wall heat transfer, interface thickness and detailed intrusion dynamics, and assumes each ambient layer is well mixed. In the equal-source limit $B_2^{+}=0$, the continuation is a neutrally buoyant momentum-driven flow, emphasizing why the upper interface loses its independent density meaning.

The strong plume's outgoing flux equals the total lower-layer supply, $Q_2(h_2)=Q$. Apply the additional closure in this part and the lower-layer balance from part (b):Dividing by $C{B}_2^{1/3}{h}_1^{5/3}$ givessoFor $0<\psi <1$ this puts the second matching level above the first. It tends to one as $\psi \to 0$ and to $2^{2/5}$ as $\psi \to 1$. At the latter endpoint the two warm layers already have identical density, so $h_2$ is only a formal matching label, not a distinct observable interface. The expression follows from the stipulated closure; modeling the actual plume in the changed ambient requires the matched continuation discussed in part (d).