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In an impermeable cylinder, an axisymmetric turbulent plume with negligible source volume flux requires a downward return flow. With uniform upward plume velocity $w$ and ambient velocity $v$, volume conservation gives $b^{2}w+(R^2-b^2)v=0$. An entrainment coefficient $\alpha$ based on the relative speed gives $Q^{\prime }=2\alpha b(w-v)$. A specified pressure closure is essential for momentum conservation. If the departure from ambient hydrostatic pressure and wall force are neglected in the whole-section integral, the total specific momentum flux $J=b^2{w}^2+(R^2-b^2)v^2$ obeys $J^{\prime }=b^2{g}^{\prime }$. Homogeneous unmodified ambient and no buoyancy loss give $b^{2}w{g}^{\prime }=B_s$. These are leading integral-model closures; restoring the ambient dynamic pressure can change the singularity criterion.

For a prismatic channel of width $b(z)=z$, a Boussinesq current of depth $h$ occupies area $A=h^2/2$. Ambient fluid at rest enters across the interface at speed $w_e$, supplying volume but no horizontal momentum or excess density. The volume conservation, momentum and excess-density balances areHere $g^{\prime }$ is the reduced gravity. With $D={\partial }_t+u{\partial }_x$, these give $Dh+(h/2)u_x=w_e$, $Du+g^{\prime }{h}_x+(h/3)g_x^{\prime }=-2u{w}_e/h$, and $D{g}^{\prime }=-2g^{\prime }{w}_e/h$. The momentum source in primitive form describes acceleration of newly entrained fluid, even when bed drag vanishes.

In a filling box model, the first front separates fluid already modified by plume discharge from the initially unmodified fluid below. If the plume occupies a negligible fraction of horizontal area $A$, its volume flux across the front is $P(h)$, and the ceiling exhaust is $F$, volume conservation gives $A\dot{h}=F-P(h)$. For a closed box with a negligible-volume buoyancy source, $F=0$. For a finite injected source flux $Q$ with ceiling overflow $Q$, $F=Q$. The first front need not delimit a uniformly mixed upper layer throughout the unventilated transient.

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.

A steady well-mixed dilute particle-laden gravity current above an absorbing porous bed loses carrier fluid at speed $W$ and particles at total downward speed $W+V$, where $V$ is their relative settling velocity magnitude. Volume conservation and the particle budget give $Q=Q_0-Wx$ and $\varphi =(Q/Q_0{)}^{V/W}$ for $W>0$. The momentum conservation equation is $(Q^2/h+g_0^{\prime }\varphi h^2/2{)}^{\prime }=-Wu$ if the withdrawn fluid carries its local horizontal velocity. Consequently $u{u}^{\prime }+g_0^{\prime }\varphi h^{\prime }+g_0^{\prime }h{\varphi }^{\prime }/2=0$. At $W=0$, $\varphi =e^{-Vx/Q_0}$ and the momentum flux is constant.

Take a dense lower layer and write $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0>0$. In the Boussinesq approximation the fractional density difference is small; use the reference density in inertia and retain the difference in buoyancy. The one-layer model additionally treats the ambient as much deeper, with negligible distributed horizontal inertia away from the head. The shallow water approximation requires depth small relative to the horizontal variation length, small vertical accelerations, nearly hydrostatic pressure, and an approximately depth-uniform horizontal velocity. Neglect mixing, viscosity, bed friction and rotation. A slowly varying rectangular channel allows cross-sectional averages without large lateral separation.

A slice of width $b(x)$ contains volume $bhdx$. Its outgoing volume flux is $bhu$, so volume conservation gives $(bh{)}_t+(bhu{)}_x=0$. The integrated hydrostatic excess pressure is ${\rho }_0{g}^{\prime }b{h}^2/2$. Pressure on the sloping side walls contributes ${\rho }_0{g}^{\prime }{h}^2{b}^{\prime }/2$ to the longitudinal force. The conservative momentum equation is thereforeUse the volume equation to simplify it. The resulting shallow water equations areOmitting the wall force would incorrectly introduce a width term into the material acceleration equation.

For $(h,u)$ the principal matrix is $\left(\begin{array}{cc}u& h\\ g^{\prime }& u\end{array}\right)$. Its eigenvalues are $u\pm c$, where $c=\sqrt{g^{\prime }h}$. For $h>0$ they are real and distinct, so the system is strictly hyperbolic. It degenerates at a dry front. From the volume equation, $c_t+u{c}_x+(c/2)u_x=-(cu/2)b^{\prime }/b$. Combine this with the momentum equation to obtain the variable-width shallow-water characteristics:In a constant-width channel $u\pm 2c$ are Riemann invariants along the corresponding characteristic curves; with varying width they satisfy compatibility equations with a source term.

For the unit-width channel put $\eta =x/L(t)$, $h=h_{f}H(\eta )$ and $u=\dot{L}U(\eta )$. Fixed volume gives $V=h_{f}L{\int }_0^{1}Hd\eta$, soThe closed rear wall has $U(0)=0$. Substituting into volume conservation gives $-H-\eta H^{\prime }+(HU{)}^{\prime }=0$, or $[H(U-\eta ){]}^{\prime }=0$. The wall sets this constant to zero. Wherever $H>0$, $U(\eta )=\eta$. Thus the velocity is $u=(\dot{L}/L)x$ and its material acceleration is $\ddot{L}\eta$.

Use a constant-Froude gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }{h}_f}$, with the front depth normalized by $H(1)=1$. Differentiating its square and using the volume relation gives $\ddot{L}=-{\dot{L}}^2/(2L)$. The Euler equation now reduces toIntegrate from $\eta =1$ to obtainA filled rear-wall solution needs $0<F<2$; $F=2$ is the limiting zero rear-depth profile. Integrating the front equation gives the complete similarity solutionHere $u$ is the horizontal velocity; the virtual origin $t_v$ accounts for matching to the finite initial release. With the Benjamin deep-ambient front condition, $F=\sqrt{2}$ and $A_F=2/3$, this simplifies toThe volume, wall condition and finite front depth can all be checked directly. Choosing another physically appropriate nose Froude number changes the coefficient and profile; the interior equations alone cannot choose it.

The room's nearly motionless upper reservoir has interface $z_i=H-h$. A doorway roof at height $D$ has available light-layer head $\delta =D-z_i=h-(H-D)>0$. Evaluate the Bernoulli equation in the room and at a freely controlled doorway: $E=-g^{\prime }{z}_i$ upstream and $E=3g^{\prime }{h}_c/2-g^{\prime }D$ at its critical section. Therefore $h_c=2\delta /3$ andThis is the ideal broad-crested weir law, not a small-orifice law with $C_d$. A nonideal doorway would require its own measured loss correction.

When the floor heat source is on, volume conservation and buoyancy flux conservation still give $Q=B_0^{1/3}{z}_i^{5/3}/\gamma$ and $g^{\prime }=\gamma B_0^{2/3}{z}_i^{-5/3}$. Combine these with the doorway relation to obtain the plume-fed doorway ventilation equilibrium:There is one interface-height root in the indicated interval. The result is physically appropriate only if its upper outflow is thin enough for the one-layer doorway model to remain valid.

If $D<H$, air in the volume above the doorway is part of the upper thermal reservoir and can store heat and alter transients. But its extra depth $H-D$ does not add available head at the doorway: only $D-z_i$ enters the critical condition. In the steady ideal well-mixed model it shares the same density as the outflow. Treating the whole $h$ as the doorway head would overestimate discharge.

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.

Constant density in each layer and negligible ambient acceleration far from the front giveThis primitive momentum conservation equation remains valid with slowly varying $B$. In conservative form it iswhere the right side is the longitudinal component of the hydrostatic pressure force on the sloping sidewalls. Omitting that term while retaining variable $B$ would give the wrong acceleration.

Together with volume conservation, the principal coefficient matrix in $(h,u)$ isIts real eigenvalues and characteristic curves areThese are downstream and upstream gravity waves relative to the moving fluid. The factor $1/2$ is geometric: the hydraulic depth is $A/b(h)=h/2$. Both waves move downstream in a sufficiently fast current, whereas one can carry information upstream in a slower current.

For $B=1$, differentiate $c^2=g^{\prime }h/2$ and combine the two equations. The Riemann invariants areFor variable $B$ they instead obeywhich also checks that the invariant claim depends on a prismatic channel.

Let $W\ge 0$ be the downward fluid suction speed and $V\ge 0$ the downward particle settling velocity relative to that fluid. Set $Q=uh$ and $g^{\prime }=g_0^{\prime }\varphi$. The carrier volume conservation and particle-volume balances per unit channel width areThe particle sink includes suction as well as relative settling: the particles follow the well-mixed withdrawn fluid, in addition to settling through it. Subtracting $\varphi Q^{\prime }$ yields $Q{\varphi }^{\prime }=-V\varphi$. Thus for $W>0$, as long as $Q>0$,For $W=0$ the continuous limit is $Q=Q_0$ and $\varphi =e^{-Vx/Q_0}$. In particular suction alone does not change the well-mixed particle fraction: $V=0$ implies $\varphi =1$.

Withdrawn fluid carries its local horizontal momentum, so the momentum flux and hydrostatic pressure force obeyExpand this and use $Q^{\prime }=-W$. The suction momentum terms cancel in the primitive acceleration equation, leavingThese are the governing equations for a particle-laden current with suction. Substituting $u=Q/h$ and the concentration equation gives the convenient depth equationThe steady carrier flux is exhausted at the formal length $\boxed{x_{\ast }=Q_0/W}$ when $W>0$. Positive $Q$ cannot continue beyond it. Near that limit a chosen depth branch, mixing or shallow-layer assumptions may fail, so this is the mass-budget extent, not a guarantee that every arbitrary inlet state has a regular shallow continuation to it. When $W=0$, settling makes the driving reduced gravity exponentially small but does not remove carrier fluid; this ideal steady model has no finite mass-budget runout.

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.

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.

Use the explicit printed width law $b(z)=\alpha z^2$. Its wetted area and hydrostatic pressure integral areAssuming hydrostatic pressure, a horizontal prismatic bed and cross-sectionally uniform velocity, volume conservation and momentum conservation give the prismatic-channel shallow water equationsEquivalently,The coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/3\\ g& u\end{array}\right)$. Its eigenvalues and the corresponding gravity-wave speed areFor $h>0$ these eigenvalues are distinct and real, establishing strict hyperbolicity in the hyperbolic partial differential equation sense. Since $dc/dh=c/(2h)$, substitution in the two equations showsFor example $({\partial }_t+(u+c){\partial }_x)(u+6c)=0$. This provides both the slopes and the conserved characteristic quantities. At a dry bed $h=0$, strict hyperbolicity degenerates and the front is interpreted as the limit from positive depth.

The descriptive word “parabolic” alone would normally suggest $b\propto \sqrt{z}$ and different coefficients. The explicit $z^2$ width in the PDF determines the calculation here; it must not be silently replaced by that other geometry.

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.

The Batchelor entrainment hypothesis makes the inward entrainment velocity proportional to the local plume speed:where $\alpha >0$ is the dimensionless entrainment coefficient. For an upward plume $w>0$, this is $u_e=\alpha w$. The hypothesis represents turbulent ingestion of ambient fluid; it is a closure rather than a consequence of mass conservation.

Let $A=b^2$, suppressing the common factor $\pi$ from cross-sectional integrals. In the Boussinesq approximation, use a constant reference density ${\rho }_r$ in the leading-order mass conservation equation. Dividing that equation by ${\rho }_r$ gives volume conservation:The factor two is the ratio of the perimeter $2\pi b$ to the suppressed area factor $\pi$. Density differences are retained in buoyancy, even though they are neglected in the leading inertial and volume terms.

To derive the density-deficit balance, first leave the small density differences explicit in the supplied mass equation. Because the ambient density ${\rho }_0(z)$ is time independent, the product rule givesThus the ambient mass entrained through the perimeter cancels exactly against its contribution to the volume budget, yielding the required buoyancy balance. It is understood to the accuracy of the underlying Boussinesq approximation; the volume equation is not an additional exact constraint on a fully variable-density fluid.

For later use define reduced gravity $g^{\prime }=g({\rho }_0-\rho )/{\rho }_r$ and the squared buoyancy frequency $N^2=-g{\rho }_{0z}/{\rho }_r$. With $B=A{g}^{\prime }$ and $Q=Aw$, the balance isA stably stratified ambient has $N^2>0$, so upward motion consumes the plume's positive density deficit; an unstable ambient reverses this sign.

Use the conventional kinematic plume fluxesThen $\pi Q$ is the volume flux, $\pi {\rho }_{r}Q$ the dimensional mass flux, $\pi {\rho }_{r}M$ the dimensional momentum flux, and $\pi F$ the kinematic buoyancy flux. Calling $\pi Q$ a mass flux uses a density-normalized convention: the reference density has been suppressed. If density-weighted momentum and buoyancy fluxes are desired, multiply $M$ and $F$ by the same constant ${\rho }_r$ throughout. Here the conventional kinematic definitions make the Boussinesq approximation explicit.

For an upward plume with $Q,M>0$, the inverse relations areThe volume conservation, vertical momentum conservation, and buoyancy balances from part a becomeHere the vertical momentum conservation equation uses ${\rho }_r$ in the inertia and retains $({\rho }_0-\rho )gA={\rho }_{r}B$ in the buoyancy force. The ambient has no imposed vertical momentum at entrainment.

To classify the resulting partial differential equations, work in the nonsingular primitive variables $(A,w,g^{\prime })$ with $A>0$. Subtract $w$ times the volume equation from the momentum equation and $g^{\prime }$ times the volume equation from the buoyancy equation. This givesThe principal matrix is thereforeAll three characteristic speeds are real and equal to $w$, so the model is a hyperbolic system if that term means only reality of the characteristic roots. However, for $A>0$, $(K-wI)v=0$ requires $v_w=0$, leaving only two independent eigenvectors. HenceThis weak hyperbolicity of the unsteady top-hat plume prevents a complete set of independent characteristic fields. For example, freezing $A,w$ in the homogeneous principal part gives $D_t\delta w=0$ and $D_t\delta A=-A{\partial }_z\delta w$. A Fourier perturbation of wave number $k$ therefore produces a term proportional to $kt$ in $\delta A$. Real characteristic roots alone do not provide uniform control of high-frequency perturbations. The flux-variable transformation above is invertible in the stated region, so changing to $(Q,M,F)$ does not remove this degeneracy.

Assume the dense current lies on the bottom, $\Delta \rho >0$, and its height is small compared with the confined aquifer depth $w$. This permits the hydrostatic approximation and neglect of leading-order resistance from the return flow of the original fluid. The dense-layer pressure anomaly on the basal seal is $\Delta \rho gh$. The along-layer Darcy flux, integrated over the current thickness, and the downward leakage flux are respectivelyVolume conservation requires $\varphi h_t+{\partial }_x{q}_x=-q_{\ell }$. Thus the porous gravity current with hydrostatic leakage through a basal seal obeysHere permeability of a porous medium has units of area, and $\Omega$ has units of inverse time. The original PDF instead prints the sink coefficient $\lambda k\Delta \rho g/(\mu b)$. With $\lambda$ defined there as a permeability, that coefficient has units of area per time and cannot multiply $h$ to give $h_t$. It also omits the division of leakage by porosity. This is a genuine source error, not a Darcy law convention. The remaining parts solve the stated linear-drainage equation for a constant $\Omega >0$; the physically consistent coefficient is the one above. Formally treating the printed coefficient as an independently prescribed inverse-time constant leaves those mathematical solutions unchanged.