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If a buoyant thermal receives volume inflow $J$ with reduced gravity $g_p$ and receives no other fluid entrainment, its mass and volume balances imply $\dot{V}=J$ and $d(g_{t}V)/dt=g_{p}J$. If $V\propto t^{9/4}$ and $g_t\propto t^{-5/4}$, then $g_t=9g_p/4$.

A dilute bubbly thermal is a rising, entraining fluid volume whose gas bubbles provide part of its buoyancy. With negligible bubble slip and gas loss, Boyle's law gives gas volume proportional to $(1-z/H_p{)}^{-1}$, where $H_p=p_0/({\rho }_{0}g)$. Spherical fluid entrainment with normal entrainment velocity $\alpha W$ gives $da/dz=\alpha$. Bubble concentration follows by dividing the gas volume by the growing thermal volume. A diameter-based Froude number closure is $W=F\sqrt{2a{g}^{\prime }}$.

An eddy diffusivity is a transport closure representing unresolved turbulence by a mean scalar flux proportional to a negative mean concentration gradient. A local estimate is $K\sim U\ell$, where $U$ and $\ell$ are turbulent speed and length scales. In a line plume with constant characteristic speed and width proportional to $z$, this gives $K\propto z$. Its coefficient is an independent closure parameter; fluid entrainment conservation alone does not determine it.

A filling box model describes a turbulent plume rising through a confined fluid and spreading beneath its ceiling. Fluid entrainment transfers initially unmodified ambient fluid into the plume; the discharged fluid forms a descending filling-box first front. A maintained heat source can have negligible source volume flux while maintaining nonzero buoyancy flux. A finite injected fluid flux requires an equal exhaust if the container volume is fixed.

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 $\ell$ and $\mathcal{T}$ denote longitudinal length and time scales, and let $U$ be a typical streamwise velocity. The shallow water approximation requires $h/\ell \ll 1$, slow changes in the channel geometry $h|B^{\prime }/B|\ll 1$, and small longitudinal slopes. The vertical velocity is then of order $Uh/\ell$; its acceleration must be small compared with the restoring acceleration, for example $Uh/(\mathcal{T}\ell )+U^{2}h/{\ell }^2\ll g^{\prime }$ in a reduced-gravity layer. The leading pressure is hydrostatic pressure, and cross-sectional horizontal velocity is approximated by $u(x,t)$. High Reynolds number permits negligible interior viscous terms but does not itself guarantee the geometric or hydrostatic assumptions. We also require $B>0$ and $h>0$ in the wetted interior. The dimensionless $B$ need not be small: it measures the cross-section's opening angle, while the smallness condition concerns streamwise variation.

The cross-sectional area and volume transport areFor an impermeable bed and no fluid entrainment, volume conservation givesor, equivalently,The geometrical source term expresses the depth change needed when a moving layer encounters a wider or narrower triangular section.

Use a gravity-current box model: approximate the current by uniform depth $h(t)$ and length $L(t)$, with constant reduced gravity, no deposition and no fluid entrainment. The triangular cross-sectional area is $h^2/2$, so the fixed released volume isThe Benjamin deep-ambient front condition closes the integral model:For a finite initial length $L_0$, direct integration givesThe ideal point-release limit takes $L_0=0$ and has $L\propto t^{4/5}$, $h\propto t^{-2/5}$. Its divergent initial depth is outside the shallow water approximation, so a real finite reservoir regularizes the early stage. This is the finite-volume triangular-channel current integral law; it does not assert a spatially uniform exact solution of the local equations.

Use $G=g\beta (T-T_0(z))$ for the plume reduced gravity, and take $z$ upwards. For a symmetric line plume define fluxes per unit source length,The integrals span the plume; its edges can move horizontally with $z$. The fluid entrainment rate at one moving edge is the inward normal transport relative to that edge. For an edge $x=b(z)$ it is $E=w{b}^{\prime }-u$; symmetry supplies equal transport through the other edge. The Batchelor entrainment hypothesis closes its mean value as $E=\alpha w_c$, where $w_c$ is a representative vertical velocity and $\alpha$ is the entrainment coefficient. The integrated volume conservation equation is therefore $q^{\prime }=2\alpha w_c$.

Using volume conservation, the vertical acceleration equation has conservative form ${\partial }_x(uw)+{\partial }_z(w^2)=G$. Incoming ambient has zero vertical momentum; the integrated momentum conservation equation is consequently $m^{\prime }=\int Gdx$. For the temperature anomaly, $T_0$ depends on height andAmbient fluid enters with $G=0$, giving the buoyancy flux balance. The integral model isHere $N$ is the buoyancy frequency when $N^2>0$; the same algebra permits unstable temperature gradients with $N^2<0$.

These are mean integral balances. The advective equations printed in the question omit explicit turbulent stresses and heat fluxes; a turbulent plume interpretation requires the usual entrainment closure for that unresolved transverse transport. Assuming vanishing edge stresses and heat fluxes, and negligible additional axial turbulent-flux contributions, gives the displayed bulk model. It would be inconsistent to treat a discontinuous top-hat profile as a smooth exact pointwise solution of the inviscid temperature-advection equation.

In the top-hat plume model, vertical velocity $w(z)$ and reduced gravity $G(z)$ are uniform over $-b<x<b$. The ambient has zero vertical velocity, negligible ambient turbulence and a prescribed $T_0(z)$. Thenand the equations becomeEquivalently, the ordinary differential equations for the three unknowns areThe last equation displays dilution by fluid entrainment as well as loss of relative buoyancy while rising through a stably stratified ambient. The model assumes a steady slender two-dimensional turbulent plume, large Reynolds number, the Boussinesq approximation, constant $\beta$ and $\alpha$, negligible heat loss, negligible imposed crossflow, and negligible pressure departure from ambient hydrostatic pressure. The factors of two count both entraining edges; $b$ is a half-width. A finite source also requires its volume, momentum and buoyancy fluxes as boundary data.

Let the closed box have ceiling height $H$ above the source. The printed width $2R$ alone does not specify this height, so it must appear as an additional geometrical parameter. Take $t=0$ here when the plume has first reached the ceiling and its thin outflow has spread across the box. The leading filling box model assumes a negligible source volume, insulated boundaries, fast horizontal spreading, slow evolution compared with the plume transit time, negligible diffusion, and $b\ll R$. Below the interface the ambient still has its original temperature, so the local plume flux remains $q(z)=2\alpha w_{\ast }z$ from part (c).

The ceiling outflow creates a warm upper region. Fluid entrainment draws the original lower fluid into the plume, making that upper region deepen. The filling-box first front is the lower material boundary of fluid already modified by plume discharge, separating it from the initially unmodified ambient; it need not bound a uniformly mixed upper layer. With negligible occupied plume area, volume conservation at its height $h(t)$ givesThereforeIf source switch-on is the time origin instead, replace $t$ by elapsed time after the initial transit and spreading stage; the quasi-steady model does not resolve that stage.

The buoyancy jump requires distinguishing ambient fluid at the front from the plume crossing it. Just after the first ceiling outflow, the ambient buoyancy immediately above the nascent front isOutside the narrow plume, ambient temperature is transported without diffusion. If $G_a$ is its buoyancy relative to the original fluid, its equation isThe first front follows precisely these material trajectories, so the temperature of the fluid immediately above it is conserved. The initially unmodified fluid below has $G_a=0$. Thus the ambient jump isBy contrast, the contemporaneous plume immediately below the front has $G_p(h,t)=w_{\ast }^2/h(t)$, which grows as the front descends. That plume value is not the ambient jump across the material first front. Equating the two would inadvertently impose a uniformly mixed upper layer and discard the stable density stratification of the standard filling box model. Diffusion or extensive interfacial mixing can smooth and alter the ideal jump.

The filling box model requires the first outflow to spread across the enclosure and remain above unmodified fluid, while a narrow nearly vertical plume rises below it. Several mechanisms can defeat those assumptions. A tall narrow box makes the plume a substantial fraction of the width, so its compensating return velocity is no longer small and modifies both fluid entrainment and momentum flux. A fast ceiling current can descend the sidewalls and overturn the initially stable upper region. In both cases a horizontally uniform slowly descending filling-box first front is inappropriate. Strong plume displacement from the centre, imposed crossflow, large source volume, vents or substantial wall cooling likewise alter the volume and heat budgets.

A useful diagnostic is whether the ceiling outflow's inertial acceleration, of order $U_c^2/{\ell }_c$, overwhelms the restoring buoyancy of the nascent layer. If $U_c^2/(G_H{d}_c)$ is not small for its depth $d_c$, the outflow can penetrate or overturn the stratification rather than remain trapped above it. This is a diagnostic Froude number, not a universal threshold without specifying the wall-current geometry and mixing. Low Reynolds number invalidates a constant turbulent entrainment coefficient, and thermal diffusion or wall heat loss can erase the sharp first-front jump. Thus large aspect ratio, strong return circulation, overturning, ventilation or diffusion require a model beyond the standard filling-box approximation.

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.

Put $G=g({\rho }_p-{\rho }_0)/{\rho }_0$ and $b=G\varphi >0$, the reduced gravity. Use the Boussinesq approximation, hydrostatic pressure, uniform vertical velocity and concentration profiles, negligible fluid entrainment and drag, and a flat bed. Integrating volume conservation, momentum conservation and the particle budget over the layer givesThe hydrostatic pressure resultant is ${\rho }_{0}b{h}^2/2$; the last right-hand side is the deposition flux. With $D={\partial }_t+u{\partial }_x$, these areThe quasilinear coefficient matrix in variables $(h,u,b)$ isIts eigenvalues are $u,u\pm c$, where $c=\sqrt{bh}$. Therefore the characteristic curves and their compatibility ordinary differential equations areFor the last equation the left eigenvectors are $(\pm c/h,1,\pm h/(2c))$, and multiplying the quasilinear equations by them gives the displayed source terms. This is the sedimenting rectangular-channel characteristic compatibility relation; it supplies one differential relation on each wave family, rather than three independently closed scalar equations there. The contact eigenvector has $du=0$ and $d(b{h}^2)=0$. For reference, an equivalent form isOnly for constant $b$ without settling do $u\pm 2c$ become genuine Riemann invariants. With settling, the spatial buoyancy gradient cannot be silently omitted.

Let $V=4\pi b^3/3$ and $B=Vg({\rho }_0-\rho )/{\rho }_{\mathrm{ref}}$. The four balances describe, respectively, fluid entrainment through the spherical surface at normal speed $\alpha u$, the increase of vertical momentum due to buoyancy, conservation of total buoyancy in a uniform ambient, and motion of the thermal centre. Entrained ambient fluid initially has no vertical momentum, so the momentum equation includes the inertia needed to accelerate it. This model neglects added mass and drag; these effects would alter the numerical similarity coefficients.

With $|\rho -{\rho }_0|\ll {\rho }_0$, the Boussinesq approximation replaces inertial density by ${\rho }_0$ while retaining the density difference in buoyancy. Take ${\rho }_{\mathrm{ref}}={\rho }_0$ to this accuracy. ThenFor a point source, $b(0)=z(0)=0$ and the initial integrated impulse is zero. Therefore $b=\alpha z$ and $Vu=B_{0}t$. Integrating $b^3\dot{b}=3\alpha B_{0}t/(4\pi )$ yieldsThe mass density follows from conserved buoyancy:Thus the radius and rise height grow as $t^{1/2}$, the rise speed decays as $t^{-1/2}$, and the density deficit decays as $t^{-3/2}$. In height coordinates, put $K=4\pi {\alpha }^3/3$ and $C=\sqrt{B_0/(2K)}$; then $V=K{z}^3$ and $u=C/z$.

If a reference density different from the inertial ambient density is retained, $d(Vu)/dt=({\rho }_{\mathrm{ref}}/{\rho }_0)B_0$: replace $B_0$ in the kinematic formulas by $({\rho }_{\mathrm{ref}}/{\rho }_0)B_0$, while retaining the original $B_0$ in the density-deficit formula. This keeps reference-density normalization separate from the physical acceleration.

The point-source spherical thermal similarity is singular at $t=0$ and cannot satisfy the small-density-difference condition arbitrarily close to its formal origin. A finite-radius, zero-impulse regularization has $b=b_s+\alpha z$ and $b^4=b_s^4+3\alpha B_0{t}^2/(2\pi )$, with $u=B_{0}t/V$. This displays the required initial data rather than attributing the singularity to a physical infinite initial speed.

In stable density stratification, ambient density decreases with height. A rising buoyant thermal encounters progressively lighter surroundings and entrains denser fluid from below, so its net buoyancy decreases. It reaches neutral buoyancy, overshoots through its remaining momentum, and eventually stops rising, spreads as a buoyant intrusion or excites internal gravity waves. The finite rise is a consequence of this restoring effect, not just drag.

Total buoyancy has dimensions $[B_0]=L^4{T}^{-2}$ and buoyancy frequency has $[N]=T^{-1}$. For an ideal point release with no other dimensional source length or time, dimensional analysis therefore givesThe dimensionless coefficient depends on entrainment coefficient, shape and the momentum model. A finite source radius or impulse supplies additional dimensionless parameters; dimensional analysis cannot determine the coefficient by itself.

Within precisely the spherical model, $b=\alpha z$ and the evolving thermal buoyancy satisfies $dB/dz=-N^{2}V=-K{N}^2{z}^3$, so $B(z)=B_0-K{N}^2{z}^4/4$. Neutrality occurs at $z_n^4=4B_0/(K{N}^2)$. For $P=Vu$, the momentum equation and $\dot{z}=P/V$ giveThe first nonzero-height stopping point is $z_{\max }^4=8B_0/(K{N}^2)$, so $z_{\max }=2^{1/4}{z}_n$. This illustrates both the dimensional scale and the distinction between neutral height and maximum rise; the precise coefficients rely on the specified ideal fluid entrainment and impulse model.

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.

The Boussinesq approximation retains small density differences only in the gravitational buoyancy force and uses a constant reference density in inertia and volume conservation. The entrainment coefficient $\alpha$ specifies the inward velocity of ambient fluid relative to the thermal boundary, $u_e=\alpha W$. Turbulent interface motions engulf ambient water and mix it into the coherent rising volume.

For a spherical buoyant thermal, $V=4\pi a^3/3$. To leading order in the dilute gas fraction, expansion of the gas does not dominate the total-volume balance, so fluid entrainment givesBut $dV/dt=4\pi a^{2}da/dt$ and $dz/dt=W$. Therefore, wherever the thermal is rising,This growth law for a dilute bubbly thermal is independent of stratification under the constant-$\alpha$ model. Stratification changes the speed and eventual stopping height; it does not enter this volume-per-distance relation. In real flows the entrainment coefficient itself can vary with stratification and flow structure.

For constant buoyancy frequency in the Boussinesq approximation, the ambient density isConservation of liquid mass during fluid entrainment gives $d(V\rho )/dz={\rho }_a(z)V^{\prime }(z)$, with $\rho (0)={\rho }_0$. Integrating by parts,Since $a=a_0+\alpha z$ and $V\propto a^3$,Hence the water density within the thermal isThe entrained water is denser than its new surroundings because it retains a mixture of fluid drawn from lower levels.

The bulk density is $\rho (1-\varphi )$. Dropping products of the small density contrast with $\varphi$, its net reduced gravity isThe first term is the expanding-bubble contribution and the second is the restoring effect of stratification. In the quasi-steady Froude model, ascent stops when $g^{\prime }=0$. The neutral height of a bubbly thermal in a stratified fluid therefore satisfiesThis is a fifth-degree polynomial, with leading coefficient $-{\alpha }^4/H_p$ at finite release pressure. The relevant height is the first positive zero reached while the thermal is still below the surface, dilute and within the Boussinesq range. Depending on parameters there may be no such zero before the surface, or more than one mathematical zero; a later root beyond a first stopping point is not another continuous ascent solution. The model's stopping height is a neutral-buoyancy estimate, not a claim that inertial overshoot is absent.

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.