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For the pressure-neglecting confined plume with compensating return flow, define specific plume momentum flux $M=b^2{w}^2$ and area fraction $s=Q^2/(R^{2}M)$. Then $J=M/(1-s)$ andThe branch issuing from a pure plume reaches a fold at $s=1/2$, where plume and ambient have equal areas, opposite velocities and equal specific momentum fluxes. The numerator is nonzero on this branch, so $M^{\prime }$ diverges although $M,Q,w$ remain finite. This signals failure of the steady uniform-profile description, not infinite physical velocity.

For an axisymmetric top-hat plume model in uniform ambient density, use full kinematic fluxes and $c=2{\alpha }_e\sqrt{\pi }$. The equations $Q^{\prime }=c\sqrt{M}$, $M^{\prime }=F_{0}Q/M$ conserve $M^{5/2}-5F_0{Q}^2/(4c)$. Solve this invariant for $M(Q)$, then integrate $dz=dQ/(c\sqrt{M})$ to obtain the displayed exact quadrature. It determines the transition from the near-source jet limit of the top-hat plume model to the distant pure plume. The two regimes cannot be added by linear superposition because the flux equations are nonlinear.

A forced plume has both imposed momentum flux and positive buoyancy flux. Its near-source region behaves like a jet, while its far field approaches a pure plume. The crossover is measured by a jet length.

If two equal pure plumes merge while keeping their common velocity and reduced gravity, conservation of cross-sectional area gives radius $\sqrt{2}$ times the individual radius. All three integral fluxes double. The plume balance parameter increases by $2^3/2^{5/2}=\sqrt{2}$, so the immediate combined plume is a lazy plume. Similarity attraction may restore pure balance farther downstream, but instantaneous pure balance does not follow from adding two pure plumes.

A constant surviving buoyancy flux sets the amplitudes of a pure plume: $Q\propto F^{1/3}$ and $M\propto F^{2/3}$. Relative-flux perturbations have the neutral ratio $(\delta Q/Q,\delta M/M,\delta F/F)\propto (1,2,3)$. This is the zero-eigenvalue limit of growing mode of power-law plume similarity; nonzero endpoint stratification can still create cumulative logarithmic flux loss.

For an ideal-gas heated pure plume with entrainment proportional to $w\sqrt{\rho /{\rho }_0}$, the excess heat flux implies conserved kinematic buoyancy flux $B$. Put $q=Q/{\rho }_0$, $m=M/{\rho }_0$. Their balances are $q^{\prime }=2\alpha \sqrt{m}$ and $m^{\prime }=Bq/m$, identical to the pi-suppressed Boussinesq pure-plume flux equations. With $d=6\alpha /5$ and $s=[25B/(48{\alpha }^2){]}^{1/3}$, $q=d^{2}s{z}^{5/3}$, $m=d^2{s}^2{z}^{4/3}$ and $z_b^{5/3}=B/(g{d}^{2}s)$. The radius is $b=dz\sqrt{1+(z_b/z{)}^{5/3}}$ and reduced gravity is $g/[1+(z/z_b{)}^{5/3}]$. At $z=z_b$ density is half ambient; far above this height the Boussinesq approximation becomes accurate.

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.

For a rising light plume use $g^{\prime }=g(\rho -{\rho }_p)/{\rho }_0>0$ in the Boussinesq approximation, with ambient density as the inertial reference. The full cross-sectional kinematic volume flux, momentum flux and buoyancy flux of a top-hat plume areHere momentum flux per unit reference density is meant, not momentum per unit volume. The Batchelor entrainment hypothesis says ambient fluid crosses the plume edge at mean inward speed $\alpha w$, where $\alpha$ is the dimensionless entrainment coefficient. Volume added per unit height is $2\pi b\alpha w$. With $a_e=2\alpha \sqrt{\pi }$, the integrated plume balances in uniform ambient density areSeek a pure plume with no persistent source volume or momentum scale. Equating powers of $z$ gives $Q\propto z^{5/3}$ and $M\propto z^{4/3}$; constant buoyancy gives $B=B_0$. Equating coefficients yields, relative to the ideal source or a fitted plume virtual origin,Thus $Q\sim B_0^{1/3}{z}^{5/3}$, $M\sim B_0^{2/3}{z}^{4/3}$ and $b=6\alpha z/5$. A finite nozzle or nonzero source momentum requires a near-source transition rather than this singular point-source solution at arbitrarily small $z$.

In a stably stratified ambient, entrained fluid is progressively lighter with increasing height. Relative to that ambient, the plume's buoyancy flux satisfies $B^{\prime }=-N^{2}Q<0$. It therefore becomes neutrally buoyant before its upward momentum vanishes. It overshoots, becomes negatively buoyant, and eventually turns and spreads laterally near its neutral-buoyancy level. The top-hat ascent equations cease to describe the overturning cap and lateral intrusion.

For constant $N$, the balances give a useful exact relation:for an ideal zero-momentum source. The rising branch begins at $B=B_0$, attains its largest momentum at $B=0$, and reaches its terminal zero-momentum limit at $B=-B_0$. Entrainment and decreasing buoyancy make the associated rise interval finite.

An estimate follows already from the unstratified pure plume flux: $N^2{\int }_0^{H}Qdz\sim B_0$, so $N^2{B}_0^{1/3}{H}^{8/3}\sim B_0$. HenceThe coefficient depends on the entrainment coefficient and the distinction between neutral height, maximum rise and final spreading level. The scaling can also be obtained from the dimensions $[B_0]=L^4{T}^{-3}$ and $[N]=T^{-1}$. The sketch distinguishes the neutral level from the maximum rise.

The ratio $\lambda =H/z_s$ compares the reference constant-stratification rise length with the height at which the actual buoyancy frequency equals $N_s$. Equivalently, the actual gradient sampled around that reference height is $N^2(H)/N_s^2={\lambda }^{\beta }$. Positive $\beta$ means stratification strengthens upwards; negative $\beta$ means it weakens upwards.

For a finite-rise branch at fixed $\beta >0$, increasing $\lambda$ makes the gradient stronger at each fixed normalized height $z/H$, so $H^{\prime }/H$ decreases. A small $\lambda$ means a larger region of weaker-than-reference stratification and permits relatively greater rise. For $\beta <0$, the same comparison reverses: increasing $\lambda$ weakens the sampled stratification, increases $H^{\prime }/H$, and can permit escape when the upper stratification decays sufficiently fast. At $\beta =0$ the ratio is one. Comparison at $\lambda =1$ is not an exact equality for nonzero $\beta$, because the plume samples a whole range of heights, not only $z=H$.

Using the pure plume as a scaling estimate before appreciable buoyancy loss,when $\beta >-8/3$ and the lower integral is regular. This gives the preceding monotonic trends, but not a universal numerical height. In particular, sufficiently negative powers are singular at a literal point source at $z=0$. A finite source elevation or a regularized near-source stratification must then be specified, as in part (d); no finite positive source buoyancy can absorb an infinite near-source buoyancy loss. Once such a finite source is used, $N^{2}Q\sim z^{\beta +5/3}$ has an integrable far-field tail for $\beta <-8/3$. A sufficiently buoyant plume can then retain positive buoyancy and rise without bound. The exact decaying-buoyancy infinite-rise branch in part (d) is another, specially matched possibility.

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.

Normalize all fluxes by $\pi$, rather than including that factor in only some of them. The specific plume volume flux, momentum flux and buoyancy flux areThe ambient transport is $-Q$ and its specific momentum flux isPut $s=b^2/R^2=Q^2/(R^{2}M)$, with $0<s<1$. The total specific momentum flux isThe first differential equation follows immediately from relative-speed fluid entrainment:For the second, differentiate $J=M/(1-s)$ and use $s^{\prime }=2s{Q}^{\prime }/Q-s{M}^{\prime }/M$. Since $J^{\prime }=b^2{g}^{\prime }=B_{s}Q/M$, this yieldsThis is the requested two-equation system, with $B=B_s$ and zero total volume flux as the additional conservation conditions. Near the pure source, $Q,M\to 0$ and $s\to 0$. Eliminating height in that limit givesThe zero integration constant selects a pure plume; a nonzero imposed source momentum would select a different branch and generally a different turning height. The leading radius is $b\sim 6\alpha z/5$, which recovers the unconfined pure-plume limit near the source.

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.

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 Batchelor entrainment hypothesis assumes that the mean ambient inflow normal to the edge of a turbulent turbulent plume is proportional to its representative axial velocity, $v_e=\alpha w$. With top-hat plume model fields and circumference $2\pi b$, the added volume per height is $2\pi b\alpha w=2\pi \alpha \sqrt{M}$. It is an integral turbulent mixing closure; it does not assert that every instantaneous eddy has the same inward speed. Use constant positive $\alpha$, uniform resting ambient, Boussinesq approximation, negligible drag and no buoyancy loss.

Since $B^{\prime }=0$, $B=B_0$. Eliminate $z$ between the volume and momentum balances:The zero-source-flux condition gives $M^{5/2}=5B_0{Q}^2/(8\alpha )$, the pure plume balance. Substitution in $Q^{\prime }=2\alpha \sqrt{M}$ and integration givesRecovering the top-hat fields from $b=Q/\sqrt{M}$, $w=M/Q$ and $g^{\prime }=B/Q$ yieldsThe paper's ratio is thereforeThis is the constant inverse-normalized ratio used here, not the catalog's normalized plume balance parameter, which is one for a pure plume. The zero-flux point source is a limiting similarity origin: the speed and density anomaly diverge there, so a physically finite source must regularize the near field. The expressions are valid in the region where the Boussinesq approximation and turbulent closure apply.

Differentiate $b=Q/\sqrt{M}$ and use both balances:Substituting $Q^2=8\alpha M^{5/2}/(5B_0)+c{Q}_0^2$ givesFor $c<0$, the radius grows faster than the pure-plume slope $6\alpha /5$ and its growth rate decreases toward that slope as $M$ increases. For $c=0$, it grows at that slope exactly. For $0<c<1$, it grows more slowly, possibly initially contracting, and its slope increases toward $6\alpha /5$. Since $M^{\prime }=B_{0}Q/M>0$ and $M\to \infty$, all branches satisfy $db/dz\to 6\alpha /5$ as $z\to \infty$.

Initial necking requiresThus a sufficiently lazy plume initially narrows as buoyancy accelerates its fluid faster than fluid entrainment supplies additional volume. It still has increasing volume and momentum fluxes; shrinking radius does not imply volume loss. At equality the initial slope is zero, and for $c>3/5$ the slope later becomes positive once $M^{5/2}=5c{B}_0{Q}_0^2/(12\alpha )$. This gives a finite minimum radius before the eventual pure plume regime.

Use $F=F_s>0$ and constant entrainment parameter $\alpha >0$. The printed integrals are normalized by $\pi$, so a top-hat plume model has $Q=b^{2}w$, $M=b^2{w}^2$ and $F=b^{2}w{g}^{\prime }$; the physical area is still $\pi b^2$. Keeping this normalization avoids spurious factors of $\pi$ in the flux equations.

Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both $M$ and $Q$ increase. In particular $Q^{\prime }\ge 2\alpha \sqrt{M_s}>0$, so $Q\to \infty$ as $z\to \infty$. Therefore $C/(\lambda Q^2)\to 0$, whatever the finite source imbalance, and $M\sim {\lambda }^{2/5}{Q}^{4/5}$. Integrating $Q^{\prime }\sim 2\alpha {\lambda }^{1/5}{Q}^{2/5}$ gives the attracting similarity solutionThis proves attraction to pure plume similarity for physical positive source fluxes. Sources with a vanishing initial volume or momentum flux enter the positive regime immediately and can be obtained by the corresponding limiting solutions; a point buoyancy source has the same similarity form with zero virtual-origin offset. The statement assumes positive buoyancy flux: a jet with $F=0$ is a different asymptotic problem.

Define the plume balance parameterPure plume balance means $\Gamma =1$ at each height, or equivalently $C=0$: momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has $\Gamma <1$, meaning excess momentum relative to this balance; a lazy plume has $\Gamma >1$, meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while $\Gamma \to 1$. In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with $z$ replaced by $Z=z+z_0$. Writing $a=(9\alpha F_s/10{)}^{1/3}$,Thus the radius growth of an axisymmetric pure plume is $b^{\prime }=6\alpha /5$ and the plume virtual origin lies at $-z_0$, with $z_0=5b_s/(6\alpha )$. Equivalently, in terms of top-hat source properties, pure plume balance is $g_s^{\prime }{b}_s/w_s^2=8\alpha /5$.

For the stated source-to-source geometry, the relevant horizontal distance is $L_s$. Consequently the edge criterion is $b_m=L_s$, rather than the radius at which two expanding edges first touch. Since $b(z)=b_s+6\alpha z/5$,Here $b_s=Q_s/\sqrt{M_s}={\lambda }^{-1/5}{Q}_s^{3/5}$. A positive merging height requires $L_s>b_s$; equality puts it at the source, while an already wider source satisfies the geometric condition from the start. The no-interaction premise extrapolates independent plumes even though their edges would first touch earlier, when each radius is $L_s/2$.

Use $M^{5/2}=\lambda Q^2$ with $Q=b^{2}w$ and $M=b^2{w}^2$ to get $F_s=(8\alpha /5)b{w}^3$. The individual plume properties at the specified height are thereforeThe source volume flux affects the height through the virtual origin, but it does not affect these properties at a prescribed radius in an exactly pure plume.

For the stipulated merger of two equal pure plumes, the new radius is $b_c=\sqrt{2}{b}_m$. This factor is outside $b_m$, as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are $Q_c=2Q_m$, $M_c=2M_m$ and $F_c=2F_s$. Their balance parameter isThe immediate combined plume is lazy. The same result follows from $\Gamma =5g^{\prime }b/(8\alpha w^2)$: enlarging the radius at fixed $g^{\prime },w$ increases this parameter.

At fixed $w_m,g_m^{\prime }$, the radius required for pure plume balance isThat is half the stipulated combined area $2\pi L_s^2$. This fixed-speed pure-plume merger and flux conservation comparison exposes the limitation of demanding immediate pure balance with unchanged speed and reduced gravity: reducing the area to this value would also halve all three merged fluxes. A conservative merger must instead adjust its properties and begin out of pure balance. Under the given constant-entrainment equations, the resulting lazy plume subsequently approaches the attracting similarity solution with total buoyancy flux $2F_s$.

Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains $-2m{M}^{\prime }$, since the reciprocal of $\hat{M}$ contributes another $-M^{\prime }$.

For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval $q,m>0$ and $f<0$, $P(0)<0$ while $P(\lambda )\to +\infty$ as $\lambda \to +\infty$. There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as $e^{\lambda \xi }\propto z^{\lambda }$; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.

At the upper endpoint,The limiting polynomial is $\lambda (\lambda +1)(\lambda +10/3)$, reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with $\lambda \sim -2f/3$. The two negative modes decay as $z^{-1}$ and $z^{-10/3}$; the remaining mode is neutral at the limiting linear system. Its eigenvector is $(1,2,3)$ in relative-flux coordinates.

The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with $Q\propto F^{1/3}{z}^{5/3}$ and $M\propto F^{2/3}{z}^{4/3}$. A small fractional change in $F$ produces changes in $Q,M$ in the ratio $1:2:3$. It is retained rather than damped out.

There is a qualification at the exact endpoint. For fixed nonzero $K$ and $\beta =-8/3$, taking $Q\propto z^{5/3}$ gives $F_z\propto -z^{-1}$, not zero. The flux coefficients above also diverge as $f\to 0$. Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.

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.