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Write $z_i=H-h$ for the height of the warm-layer interface above the heat source. A negligible-volume point heat source supplies a pure plume with conserved buoyancy flux $B_0$. The given plume buoyancy implies its volume flux at the interface isAt steady displacement ventilation, the plume feeds the warm layer at the same rate it empties through the upper vent: $q_v=Q_p(z_i)$. The upper-layer buoyancy balance gives $q_v{g}^{\prime }=B_0$, soEliminate $q_v$ and $g^{\prime }$ to getThe height equation has one solution with $0<h<H$. Its height is independent of source strength in this ideal pure-plume model, while its warm-layer density contrast increases as $B_0^{2/3}$. The sketch shows cool replacement air below, an entraining plume through that region, and the upper mixed layer feeding the ceiling vent.

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.

Assume both openings have positive area, no imposed wind pressure, a light buoyant plume, and a Boussinesq approximation. Write $S=\pi Q_s$ and $B_s=\pi F_s$ for physical source volume and buoyancy flux. Include discharge coefficients $C_H,C_0$; if ideal openings are intended, set these to one. In a steady two-layer displacement ventilation state with interface height $z_{\ast }$, let the floor inflow be $q_0$ and roof outflow $q_H$. Volume and buoyancy balances giveThe hydrostatic pressure head across the warm upper layer must equal the two opening losses:A positive unprocessed lower layer needs $q_0>0$. At the boundary where this layer disappears, $z_{\ast }=0$, the plume can entrain no lower ambient fluid, so $q_0=0$, $q_H=S$ and $g_u^{\prime }=B_s/S$. The critical condition is therefore $S^3=2C_H^2{A}_H^2{B}_{s}H$.

To determine the inequality, consider a fully processed room. Its steady reduced gravity is $g^{\prime }=B_s/S$. If $p_0$ is the interior-to-exterior pressure difference at the floor, source-volume blocking of fresh floor inflow requires $p_0\ge 0$. Both openings then exhaust, withThe right-hand side is increasing in $p_0$ and has its minimum at zero. Consequently a fully processed state exists exactly whenThis is source-volume blocking of displacement ventilation. At equality floor inflow vanishes; above it the floor can also exhaust and no fresh lower-layer supply is needed. Below it, the floor pressure must be negative, admitting fresh fluid and supporting a positive lower layer. In fact the steady interface equation has a unique positive root: its loss side increases with $z_{\ast }$ because $Q(z_{\ast })$ increases, while its buoyancy-head side decreases.

The floor-opening area affects the positive-interface state and its approach, but drops out of the blocking threshold because floor flux is zero at the threshold. Equal splitting of pressure head, or using a source-free effective opening area at this transition, would miss this distinction. If the floor opening is sealed, the fresh-inflow argument no longer applies: a finite source with only roof exhaust instead cycles the room asymptotically as in the ceiling-opening case of part (c).

Write $\theta =T-T_e$ and use the Boussinesq approximation for air, ${\rho }_e-\rho \simeq \rho \theta /T_e$. A uniformly heated, well-mixed room has the buoyancy pressure difference $\Delta p_b=\rho gH\theta /T_e$ between its openings. Let $C_o$ denote the combined hydraulic coefficient of the two equal openings, so the effective opening area for pressure-driven ventilation givesFor example, if each orifice has conventional discharge coefficient $C_d$ in $V=C_{d}A\sqrt{2\Delta p/\rho }$, the two pressure losses add and $C_o=C_d$. The question's coefficient is the prefactor in the final heat-load law; define $c=(2C_o^2{)}^{1/3}$. This explicitly states the numerical loss-coefficient convention. Steady heat balance is $Q=\rho C_{p}V\theta$. Eliminating $\theta$ givesHere $Q$ is total heating power and $C_p$ is specific heat capacity at constant pressure. Writing $c=C_d$ without this conversion would incorrectly change the power of the discharge coefficient.

After the reduction, cooler newly heated air occupies the bottom and displaces the old warmer air upward. Use the ideal two-layer displacement ventilation model: each layer is well mixed, the upper layer remains at ${\theta }_0$, walls have negligible heat capacity, and the sharp interface is advected without exchange between the layers. Let the lower layer have depth $h$ and excess temperature $\theta$. Integrating the hydrostatic buoyancy over the full height givesThe room's excess thermal energy is $E_T=\rho C_p{A}_r[h\theta +(H-h){\theta }_0]$. The floor supplies $\lambda Q$, incoming air has zero excess temperature, and the ceiling removes $\rho C_{p}V{\theta }_0$. Thus the required heat balance isThis moving-interface energy term must include the loss of old upper-layer volume; balancing only the lower-layer temperature would miss it.

For a genuine reduction assume $0<\lambda <1$. SetThe integrated buoyancy factor is $H{\theta }_0(1-x)$, so $V/V_0=\sqrt{1-x}$. Since $Q=\rho C_p{V}_0{\theta }_0$, the thermal balance becomes the ventilation after a heating reduction equationThe material interface also moves with the upward displacement velocity, $A_r\dot{h}=V$. Therefore ${\eta }_s=\sqrt{1-x}$, and subtracting the two equations givesThe second relation is also the lower-layer energy balance: its total excess energy increases at rate $\lambda Q$. It does not assume its temperature stays constant. Since $x_s>0$ until $x=1-{\lambda }^2$, the solution remains $0\le x<1-{\lambda }^2$, and $V/V_0\ge \lambda$. Hence ${\eta }_s\ge \lambda$, so the lower layer reaches $\eta =1$ in finite time $s_H\le 1/\lambda$. Also ${\eta }_s\le 1$, so $s_H\ge 1$.

An explicit implicit solution is useful. Put $y=\sqrt{1-x}$, which decreases from $1$ toward $\lambda$. Separation givesFilling occurs when $1=1-y_H^2+\lambda s_H$, so $\lambda s_H=y_H^2$ and ${\theta }_H/{\theta }_0=y_H^2$. These relations determine the exit time and the continuous temperature at that time. For complete switch-off $\lambda =0$, the limiting solution is $y=1-s/2$, $x=\eta =s-s^2/4$ for $0\le s\le 2$, so the cold lower layer fills the room at $s_H=2$ even though its final ventilation rate is zero.

After the interface exits, the whole room is well mixed. For $z=\theta /{\theta }_0$, the natural ventilation rate is $V=V_0\sqrt{z}$, and the well-mixed ventilation temperature balance becomesEquivalently, $\rho C_p{A}_{r}H\dot{\theta }=\lambda Q-\rho C_p{C}_{o}A\sqrt{gH/T_e}{\theta }^{3/2}$. The heat-removal term is strictly increasing in $\theta \ge 0$, so every nonnegative solution tends to the unique stable equilibriumFor $\lambda =0$ the final temperature is $T_e$. The unchanged-load case $\lambda =1$ has no distinct new temperature layer and is outside the reduction scenario.

Let $q_b$ be floor inflow and $q_t$ roof outflow. Use effective-area discharge $A\sqrt{\delta p/\rho }$, matching the factors in the supplied relations. With negligible source volume, both equal $q$. The lower region has exterior mass density while the warm upper layer has reduced gravity $g^{\prime }$ and depth $H-h$. hydrostatic pressure therefore supplies a total opening pressure head $g^{\prime }(H-h)$. Equal openings sharing equal discharge take half each, givingSteady lower-layer volume conservation sets this ventilation rate equal to the entrained turbulent plume flux at the interface, $q=\lambda B^{1/3}{h}^{5/3}$. Upper-layer buoyancy conservation gives $B=q{g}^{\prime }$. These are the required balances for displacement ventilation:Eliminating $g^{\prime }$ and $q$ gives the displacement-ventilation interface height equationThe left side is strictly increasing from zero to infinity, so there is one interface height. It depends on opening geometry and plume entrainment, not on $B$; the resulting $g^{\prime }=B^{2/3}/(\lambda h^{5/3})$ and throughflow do depend on $B$.

For finite source volume $Q$, conservation instead gives $q_t=q_b+Q$. The two pressure drops are no longer equal:Let $q_p(h;Q)$ be a physically entraining source plume: $q_p(0;Q)=Q$ and $q_p(h;Q)>Q$ for $h>0$. Its interface balance is $q_p=q_t=Q+q_b$. Therefore, as $q_b$ reaches zero, a positive-depth lower layer has no replenishment and cannot remain steady: this is no steady displacement layer without ambient supply. The limiting interface is $h=0$. At the threshold $q_t=Q$ and $g^{\prime }=B/Q$, and the entire hydrostatic head drives the roof discharge. HenceThis is source-volume blocking of displacement ventilation. The half-head factor of the negligible-source state must not be kept after the floor inflow vanishes. In the usual physical-area convention $q=A_{\mathrm{phys}}{C}_d\sqrt{2\delta p/\rho }$, the equivalent formula is $Q_c=(2C_d^2{A}_{\mathrm{phys}}^{2}BH{)}^{1/3}$.

The quoted pure-plume law cannot be used unmodified down to the origin for a source with nonzero volume: it would incorrectly give zero flux there. A compatible finite-source example uses a virtual origin,It recovers the supplied pure-plume limit when $Q\to 0$, and gives the same blocking threshold. The complete finite-$Q$ interface trajectory depends on that plume model, but the threshold needs only source-volume conservation and positive entrainment. For $Q>Q_c$, the assumed steady regime with a lower inflow no longer exists.

Write $h(t)$ for the height of the filling-box first front above the floor and $P(h)$ for the upward plume volume flux crossing it. Neglect the plume's horizontal area compared with $A$, use the Boussinesq approximation, and measure buoyancy relative to the initially unmodified ambient. The first front separates fluid already modified by plume discharge from the unmodified fluid below; the entire upper region need not be uniformly mixed during an unventilated filling box model transient.

A finite injected volume flux $Q$ cannot enter a hermetically sealed, fixed-volume room under the incompressible flow assumption. If $Q>0$, an equal overflow or exhaust is needed. Taking that exhaust at the ceiling, volume conservation of the region above the front givesThe source volume flux is subtracted because only entrained room fluid is removed from the initially unmodified region. The printed pure plume law has $P(0)=0$, so it cannot be an exact finite-$Q$ source law down to the physical floor. In the far field where $Q\ll P(h)$, or for a heat source whose injected volume flux is negligible, it givesHere $t=0$ is when the plume discharge first reaches the ceiling. Retaining $Q$ while using the far-field law gives $-\dot{h}=[\lambda B^{1/3}{h}^{5/3}-Q]/A$ only while that law remains a valid approximation; it must not be extrapolated to predict negative plume entrainment. A plume virtual origin below the floor is one way to match a nonzero source volume flux.

For displacement ventilation, let $F$ be the total ceiling exhaust, $S$ the total buoyant-source volume flux, and $L=F-S$ the separate ambient inflow into the lower layer. Volume conservation and the upper-layer buoyancy flux balance giveThe ambient inflow has zero reference buoyancy. In the ideal two-layer closure the upper layer is uniformly mixed with reduced gravity $b>0$, and the lower layer has zero reference buoyancy. The plume supplies the upper layer while lighter upper-layer air is removed at the ceiling. An interface is driven downward when $P(h)>F$ and upward when $P(h)<F$; since $P^{\prime }(h)>0$, its equilibrium is stable.

In the negligible-source-volume pure plume limit, identifying $F=Q_v$ gives the intended one-source answerThe prescribed inequality makes $h_1/H\ll 1$, so both layers fit in the room. It does not, by itself, justify neglecting a finite source $Q$: the pure plume source region must also be small compared with $h_1$, normally requiring $Q\ll Q_v$. If $Q_v$ denotes additional background ventilation rather than total exhaust, replace $Q_v$ in both formulas by $Q+Q_v$. If it is total exhaust, the separate ambient inlet carries $Q_v-Q$. These conventions must not be mixed.

For $n$ independent equal axisymmetric pure plumes, adding their volume fluxes givesThus the conventional multiple-plume displacement ventilation calculation with negligible source volume flux and separate ambient ventilation givesSplitting the buoyancy flux increases total fluid entrainment, so the same ventilation balances the plumes at a lower interface. The global buoyancy flux is unchanged, which leaves the upper-layer reduced gravity unchanged. Independence requires sufficiently separated sources and plumes that do not merge below the interface.

There is, however, a substantive problem with the literal source-volume specification in this paragraph of the PDF. It puts all $Q_v$ through the buoyant sources, so $S=F=Q_v$ and $L=0$. For every positive interface height, entraining plumes haveThere is no fresh ambient inflow to replace the lower-layer fluid being entrained. Consequently the stated positive steady two-layer configuration does not follow from those literal hypotheses. This is no steady displacement layer without ambient supply, not a small correction to the $n^{-2/5}$ scaling. For example, a matched plume virtual origin givesWhen $S=F$, this yields $h_{\ast }=0$. The unshifted answer $h_n$ equals the omitted $z_0$ in this case, so neglecting the source region cannot be justified. Also each source has reference buoyancy $(B/n)/(Q_v/n)=B/Q_v$, already equal to the proposed upper-layer value; a steady positive plume cannot entrain zero-buoyancy lower fluid and still deliver that same value. The conventional answer is valid for a corrected arrangement with negligible-volume buoyancy sources and a separate ambient inlet, or with source flux $S<Q_v$ and the remaining $Q_v-S$ supplied directly to the lower layer. A real room may have other circulation or continuous stratification, but that requires a different model.

For the intended displacement ventilation model, reducing the exhaust to $F=Q_v/2$ makes the new equilibriumThe interface initially descends since its old plume volume flux exceeds the new exhaust. In ventilated filling-box relaxation, writing $V_{\ast }=A(H-h_{\ast })$ and taking the linearization of the two balances above givesAs $P_n^{\prime }(h_{\ast })=5F/(3h_{\ast })$, the two decay times areA dimensional analysis estimate of the full adjustment time is thereforeThe interface adjusts on the shorter lower-layer scale; replacing the buoyancy stored throughout the upper layer is slower. Exact equilibrium is approached asymptotically, and a specified small relative tolerance adds a logarithmic factor. For the literal all-inflow-through-sources arrangement there is no positive two-layer equilibrium to return to; $2AH/Q_v$ is only a flushing-scale estimate for a different, whole-room adjustment.

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 $0<h<H$.