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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.

Consider entraining plumes supplied with total source volume flux $S$, a separate lower-layer ambient inflow $L$, and ceiling exhaust $F=S+L$. Write $P(h)=S+E(h)$, where $E(h)$ is total fluid entrainment from the lower layer. Volume conservation gives $A\dot{h}=L-E(h)$. A steady positive lower layer therefore requires $L=E(h)>0$. If $L=0$ and $E(h)>0$ for every $h>0$, no positive steady displacement-ventilation interface height exists. In a shifted pure plume approximation $P_i(h)=\lambda B_i^{1/3}(h+z_i{)}^{5/3}$ matched to source flux $S_i$ at $h=0$, this follows immediately from $P_i(h)>S_i$ for $h>0$. This is a statement about the ideal two-layer model, not a prohibition of all possible stratification in a real ventilated room.

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.

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 $0<h<H$.