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Suppose $n$ independent axisymmetric pure plumes share total buoyancy flux $B$ equally, have negligible source volume flux, and are supplied with a separate ambient ventilation flux $F$ at floor level. Their combined flux is $P(h)=\lambda n^{2/3}{B}^{1/3}{h}^{5/3}$. Volume conservation at a steady displacement-ventilation interface height and buoyancy flux conservation giveA positive two-layer state needs $0<h_{\ast }<H$, sources small relative to $h_{\ast }$, and plumes sufficiently separated to remain independent. Splitting the sources increases total fluid entrainment and reduces the interface height by $n^{-2/5}$. A finite source flux cannot simultaneously be neglected and supply all the replacement air; no steady displacement layer without ambient supply explains this restriction.

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.