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Let $\theta =T-T_0>0$ and ${\theta }_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming volume flux, equal to the outgoing volume flux, isOnly one of these equal exchange fluxes multiplies the temperature difference in the heat budget: incoming fluid brings $q{T}_0$, outgoing fluid removes $qT$. With constant mass density and specific heat capacity, those factors cancel and the well-mixed ventilation temperature balance isIntegrating from ${\theta }_1$ givesThe cooling is algebraic rather than exponential because the buoyancy-driven single-opening exchange flow weakens as its driving reduced gravity decreases. The formula assumes negligible wall heat capacity, no heat input, the specified aperture geometry, and a uniform interior temperature maintained by mixing.

With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving stable density stratification. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of buoyancy, that changes the physical outcome.

LetThe global steady heat, or buoyancy flux, balance equates the floor-source input to the buoyancy carried out by the one-way upper-layer exhaust:The corresponding upper-to-lower density jump is fixed by the warm plume crossing the interface,The first balance also shows that the descending cold plume has buoyancy-flux magnitude per unit wall length

At a steady interface, the upward axisymmetric-plume volume flux equals the total downward wall-plume volume flux. Using parts (c) and (d),After cancellation of $B^{1/3}$, the required implicit geometric relation isThus the ideal steady interface fraction is independent of the source strength: increasing $B$ multiplies both opposing plume volume fluxes by $B^{1/3}$. The floor area $A$ and room height $H$ affect the transient filling time and admissibility of the assumed ordering, but not this steady integral balance, provided $0<h<z_V$ and the plumes remain separated.

SetCombining the single-opening exchange flow relation $Q_V=K_V\sqrt{g_H^{\prime }}$ with $B=Q_V{g}_H^{\prime }$ givesIt also gives $g_H^{\prime }=(B/K_V{)}^{2/3}$, after which the second density balance determines $g_L^{\prime }$.

Assume the two identical openings and their wall plumes behave symmetrically and do not interact before reaching the interface. If $Q_V^{(1)}$ is the one-way rate through either vent, the global buoyancy balance isEach wall plume therefore has buoyancy flux per unit span $\mathcal{B}=B/(2L)$. The steady volume balance now includes two descending plumes:Hence the interface height is determined implicitly byFor completeness, if each opening retains the same single-opening exchange flow coefficient $K_V$, then