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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.

Matching a slowly varying grounded Newtonian sheet to a freely extending Newtonian shelf gives, in a standard two-dimensional model,where $h_G$ is the flotation thickness and $g^{\prime }=g({\rho }_w-\rho )/{\rho }_w$ is the reduced gravity.

A particle-laden gravity current is driven by the excess density of suspended particles. Deposition decreases its reduced gravity, so a finite-volume current can approach a finite runout length.

Assume a sharp interface, uniformly mixed warm upper air with fixed reduced gravity $g^{\prime }=g\Delta \rho /{\rho }_0$, cool ambient air entering below, no wind, no wall heat exchange, negligible source volume, and quasi-steady Boussinesq flow through the small vents. Neglect the interior approach velocity. The hydrostatic buoyancy head between the two vents is ${\rho }_0{g}^{\prime }h$.

Both openings carry the same volume flux $q_v$. The discharge coefficient law gives a pressure loss ${\rho }_0{q}_v^2/(2C_d^2{A}^2)$ at each opening. Adding the two losses gives $q_v=C_{d}A\sqrt{g^{\prime }h}$, not the single-opening value with an extra factor $\sqrt{2}$. Cool replacement air displaces the warm layer, so $S\dot{h}=-q_v$. With $h(0)=H$, integrate to obtainThe ideal model sets $h=0$ thereafter. The formula applies only while the parenthesis is nonnegative; squaring a negative value would invent renewed filling. A finite aperture or a diffuse final interface invalidates the idealization near exhaustion.

Assume a steady inviscid light layer of constant reduced gravity $g^{\prime }$ under a roof at $z=H(x)$, with depth $h$, interface $z_i=H-h$, and slowly varying width $W(x)$. The ambient is deep and approximately motionless; neglect entrainment, friction and approach variations across the section. Let $p^{\prime }$ be pressure relative to the ambient hydrostatic pressure. Then $p_z^{\prime }={\rho }_0{g}^{\prime }$ in the light layer and $p^{\prime }={\rho }_0{g}^{\prime }(z-z_i)$ when the ambient pressure offset is chosen zero. The steady Bernoulli equation along a light-fluid streamline is $u^2/2+p^{\prime }/{\rho }_0-g^{\prime }z=\mathrm{constant}$. With $Q=Whu$, this becomes the hydraulic flow in an inverted channel energyAt fixed $x$ and fixed flux,The energy curve has a minimum at $h_c=[Q^2/(g^{\prime }{W}^2){]}^{1/3}$, with $E_c=3g^{\prime }{h}_c/2-g^{\prime }H$. Above that minimum there are a thin supercritical branch and a thick subcritical branch. A lowered roof increases $E_c$; it is the inverted counterpart of raising a bed crest.

For a single smooth obstruction in a constant-width channel, energy exceeding its largest $E_c$ permits entirely subcritical passage or entirely supercritical passage, depending on the boundary data. At the limiting head a subcritical upstream current can pass through critical flow at the roof minimum and emerge supercritical; this is the usual freely discharging control. A reverse transcritical branch would need suitable imposed downstream data and is not the usual free discharge. If the supplied head is below the required minimum, no smooth steady flow with that flux exists: upstream adjustment changes the discharge or head. A supercritical current encountering deep downstream warm fluid can undergo a hydraulic jump to the subcritical branch, with an energy loss rather than a second conservative branch switch.

For a rectangular section the jump conserves $h_1{u}_1=h_2{u}_2=q$ and $q^2/h+g^{\prime }{h}^2/2$. Hence $h_2/h_1=[\sqrt{1+8F_1^2}-1]/2$ for $F_1>1$, and the lost specific energy is $g^{\prime }(h_2-h_1{)}^3/(4h_1{h}_2)$. This remains positive after vertical reflection. The energy and profile sketches label the roof and the lower boundary of the light layer explicitly.

Let $W_1$ be the first doorway width and $W_2$ the conservatory's outer doorway width. For a hydraulic comparison assume the same reduced gravity and conserved volume flux through both, negligible losses except at any hydraulic jump, and roofs at the same doorway height $D$. Their critical depths and limiting energies areThe narrower doorway has the larger minimum required head. This gives the possible two successive hydraulic controls rather than a universal jump location.

If $W_2>W_1$, the first doorway can remain the discharge control. Its supercritical outflow may pass through the wider second doorway without being controlled there. Alternatively, if warm air accumulates in the conservatory, the first jet can undergo an inverted hydraulic jump, feed a subcritical intermediate reservoir, and accelerate to a second critical section at the outer doorway. The wider second throat has a lower critical energy, so the difference can accommodate the positive head loss at the jump. Whether this second configuration occurs, and where its jump stands, depends on conservatory dimensions and downstream head.

If $W_2<W_1$, an outflow set solely by critical flow at the first doorway lacks the energy to pass the narrower second doorway: $E_{c2}>E_{c1}$, and a jump cannot increase energy. Warm air backs up in the conservatory. The downstream doorway becomes the dominant control, and the first doorway is subcritical or submerged by the intermediate layer. During adjustment an upstream-moving bore or jump can propagate towards or through the first doorway. A permanently freely controlled first doorway followed by a narrower same-height control is impossible under the conserved-buoyancy assumptions.

If $W_2=W_1$, the equal-energy idealization is marginal. Both throats can reach critical conditions in a lossless configuration, but any positive interposed jump loss prevents two independent critical sections at the same flux and buoyancy. Small losses or differences in effective discharge geometry select the downstream restriction and can drown the first. A constriction should not be assumed controlled simply because it exists.

The sketches show these admissible regimes. Mixing with cool conservatory air, heat loss, finite return-flow inertia, different roof elevations or aperture losses change the comparison and need additional data. In particular relative widths alone do not determine the actual conservative-room temperature or the position of a jump.

Let $\theta =T-T_0>0$ and ${\theta }_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming volume flux, equal to the outgoing volume flux, isOnly one of these equal exchange fluxes multiplies the temperature difference in the heat budget: incoming fluid brings $q{T}_0$, outgoing fluid removes $qT$. With constant mass density and specific heat capacity, those factors cancel and the well-mixed ventilation temperature balance isIntegrating from ${\theta }_1$ givesThe cooling is algebraic rather than exponential because the buoyancy-driven single-opening exchange flow weakens as its driving reduced gravity decreases. The formula assumes negligible wall heat capacity, no heat input, the specified aperture geometry, and a uniform interior temperature maintained by mixing.

With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving stable density stratification. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of buoyancy, that changes the physical outcome.

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.

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.

With $W=0$ and $V>0$, $Q=Q_0$ and $\varphi =e^{-kx}$, where $k=V/Q_0$. No carrier momentum is withdrawn; hence the integrated momentum flux is exactly constant:On an initially small-Froude number branch, the pressure term dominates. At leading orderA slightly more accurate large-distance amplitude is $\sqrt{2K/g_0^{\prime }}$, so $h\sim \sqrt{2K/g_0^{\prime }}{e}^{kx/2}$. The Froude number satisfies ${\mathrm{Fr}}^2\sim {\mathrm{Fr}}_0^2{e}^{-kx/2}$ and remains small. The growing depth can eventually violate confinement or the shallow water assumption; this does not make the asymptotic momentum calculation inconsistent within its intended regime.

On an initially large-Froude number branch, horizontal momentum flux dominates instead. The leading depth is constant, $\boxed{h\sim h_0}$, while the decaying reduced gravity makes the Froude number still larger. To resolve the small change, put $\delta ={\mathrm{Fr}}_0^{-2}\ll 1$ and $y=h/h_0$:soThe depth decreases slightly and tends exactly to $h_{\infty }=Q_0^2/K=h_0/(1+\delta /2)$. Its Froude number grows like ${\mathrm{Fr}}_0{e}^{kx/2}$. Both approximations agree with the exact depth equation $h^{\prime }=kh/[2(1-{\mathrm{Fr}}^2)]$: the subcritical flow deepens, while the supercritical flow thins.

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.

The shallow-water approximation requires the depth to be small compared with the horizontal scale of variation, small vertical accelerations and nearly hydrostatic pressure. Take cross-sectionally uniform lower-layer velocity and a sufficiently deep, effectively stationary upper layer. Under the Boussinesq approximation, $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0$ is the reduced gravity. Conservation of lower-layer volume gives constant discharge $Q=bhu$.

Dividing the Bernoulli equation by $g^{\prime }$ gives the shallow-water specific energyAt fixed $Q,b$, the energy curve has its minimum at the critical depth $h_c=(Q^2/(g^{\prime }{b}^2){)}^{1/3}$, where $E_c=3h_c/2$. Above this minimum there are two depth branches: deeper subcritical flow has $F<1$, and shallower supercritical flow has $F>1$.

The two long-wave speeds relative to the bed are $u\pm \sqrt{g^{\prime }h}$. In subcritical flow one family can propagate upstream and one downstream; in supercritical flow both travel downstream for $u>0$. At hydraulic control, $u-\sqrt{g^{\prime }h}=0$, and a smooth transcritical solution must also satisfy the geometric regularity condition. The flow may remain entirely on either branch, cross smoothly through the control from subcritical to supercritical, or take the reverse smooth branch under appropriate boundary conditions. A downstream hydraulic jump can return supercritical flow to the subcritical branch while losing energy.

The sketch distinguishes smooth branch changes at a critical section from a jump: a jump conserves mass and momentum, not the original Bernoulli head.

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.

Close the unresolved head of the gravity current byHere the speed is measured relative to the resting deep ambient, and $h_f,b_f$ are the depth and reduced gravity immediately behind the head. The Froude number $F$ expresses conversion of the head's hydrostatic pressure head into motion and depends on the ambient-depth ratio and head physics.

A front condition can be motivated by a steady control volume in the front frame: balance mass conservation, momentum conservation and the upstream/downstream Bernoulli equation, with a stagnation region and the appropriate ambient return flow. Assumptions include small density contrast, approximately hydrostatic far fields, negligible imposed ambient current, specified far-field depths, and a prescription for any mixing or energy loss. Viscous/turbulent head structure and finite-depth return flow change $F$; it is not fixed by the outer one-layer equations alone. A deep-ambient ideal head often motivates a constant order-one value, but no numerical choice is needed here.

The shallow-water approximation fails locally in the head because its vertical and horizontal dimensions are comparable and vertical accelerations are appreciable. The Froude number closure is a boundary condition for the outer shallow water equations, not a claim that they resolve the front itself.

Let the two populations have settling speeds $w_1\gg w_2>0$ and particle volume fractions $c_1,c_2$. In the same mixed box their scalar equations are ${\dot{c}}_j=-w_j{c}_j/h$, while the common reduced gravity is $g^{\prime }=b_1{c}_1+b_2{c}_2$ after neglecting thermal buoyancy. Both populations are carried by the same radius and depth:The coarse ash deposits rapidly near the source; the fine ash remains aloft longer, sustains the current's buoyancy and is carried farther. The composition of the deposit therefore fines outward. Since the remaining fine ash drives the flow transporting both classes, the runout is not obtained by superposing two independent monodisperse currents.

Define the settling exposure for a particle size distribution $J={\int }_0^{t}ds/h(s)$. Then $c_j=c_{j0}{e}^{-w_{j}J}$ andThe integral is finite when both settling speeds are positive. Each deposited mass is ${\Sigma }_j(r)={\rho }_{pj}{w}_j{\int }_{t_a(r)}^{\infty }{c}_j(t)dt$; their sum is the total deposit, with a coarse proximal component and a finer distal tail. If the actual hot-gas density deficit is retained, early loss of coarse ash can make $g^{\prime }$ vanish and loft the remaining fine-rich cloud, shortening its ground-hugging reach relative to this particle-only calculation.

Model the post-eruption flow as an axisymmetric, shallow, hydrostatic, turbulently mixed layer on a horizontal bed. Neglect entrainment, erosion, heat exchange, rotation and large-scale viscous drag. Particles are dilute, with density ${\rho }_p$, uniform-in-depth particle volume fraction $c(r,t)$ and constant downward settling speed $w_s>0$ relative to the carrier gas. Deposited particles are not resuspended. Their volume is small enough that settling does not significantly change the bulk carrier volume. Let ${\rho }_a$ be ambient air density and ${\rho }_g$ carrier-gas density at its local temperature, and use a leading Boussinesq approximation for the mixture motion. DefineThe hot gas reduces the density excess; the ground-hugging current model requires $g^{\prime }>0$. If density contrasts are too large for Boussinesq dynamics, density-weighted balances would be needed instead. The equations below are a stated shallow-layer model, not a description of the initial explosive vent jet.

For layer height $h$ and radial depth-averaged speed $u$, the appropriate shallow-water balances areThe hydrostatic force gives the nonconservative momentum equation $u_t+u{u}_r=-g^{\prime }{h}_r-(h/2)g_r^{\prime }$. In particular a spatially varying particle fraction must not be treated as constant reduced gravity when differentiating pressure. With no heat exchange or entrainment, a carrier-temperature anomaly $\Theta$ is advected:At this order $b_T$ is proportional to $-\Theta$. At the bed the carrier has no normal penetration and particles have absorbing deposition flux $w_{s}c$; at the top the pressure equals ambient pressure. Axis regularity requires $u(0,t)=0$ and smooth scalar fields, with no further source after the initial pulse. The initial height, particle inventory and temperature anomaly specify the released volume and ash content. Outside the front $r=R(t)$, the current layer is absent. At a resolved advancing head, use the material-front condition and the gravity-current front condition $\dot{R}=\mathsf{F}\sqrt{g_f^{\prime }{h}_f}$, where $\mathsf{F}$ is an order-one front Froude number and $h_f,g_f^{\prime }$ refer to the fluid feeding the head. The front closure replaces unresolved head dynamics; simply prescribing both a zero-depth mathematical edge and a finite-depth box-head speed would be inconsistent.

To obtain a gravity-current box model, assume horizontal as well as vertical mixing, uniform $h(t),c(t),\Theta (t)$ in a disk of radius $R(t)$, and a negligible-volume head whose feeding depth is the box depth. The conserved bulk volume $\mathcal{V}$ givesA bulk velocity $u=(\dot{R}/R)r$ satisfies interior continuity for this box. Rather than impose a uniform pressure on the full shallow-water equations and thereby lose the head force, replace their momentum dynamics by the front closure. Total suspended particle volume is $\mathcal{V}c$, so bed deposition givesThese ordinary differential equations, together with initial $R_0,c_0,\mathcal{V},{\Theta }_0$, are the axisymmetric settling box model. The maintained negative thermal term may eventually cancel particle buoyancy and cause lift-off; beyond $g^{\prime }=0$ this horizontal gravity-current model does not apply.

For the requested particle-only calculation set $b_T=0$ and define the initial reduced gravity $g_0^{\prime }=b_p{c}_0$. The equations reduce toEliminating time gives $d\sqrt{c}/dR=-w_s{R}^3/[2\mathsf{F}\sqrt{b_p}{A}^{3/2}]$. Integrating from the release state yieldsThus the monodisperse circular ash runout has limiting radiusThe maximum outward distance beyond the initial release edge is $R_{\infty }-R_0$. When the vent scale is negligible, set $R_0=0$ in this late-time source idealization. The resulting radius scales as ${\mathcal{V}}^{3/8}(g_0^{\prime }{)}^{1/8}{w}_s^{-1/4}$; it is not obtained by conserving reduced gravity after particles settle out.

To determine the time, put $s=R^2/R_{\infty }^2$, $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. The concentration relation becomes $\sqrt{c/c_0}=(1-s^2)/(1-s_0^2)$, and the front equation reduces toConsequentlyThe furthest reach is a limiting radius, approached asymptotically rather than attained at a finite stopping time. As the residual particle concentration vanishes, the front slows exponentially. A finite observational threshold for remaining ash would give a finite practical stopping time, but no such threshold is part of these balances.

The local ash deposit density is most naturally its mass per horizontal area, $\Sigma (r)$. Once the front has arrived at $r$, its local deposition flux is ${\rho }_p{w}_{s}c(t)$. For $r<R_{\infty }$, define $s_a=\max (s_0,r^2/R_{\infty }^2)$; the arrival time is zero inside the initial footprint and otherwise $t_a=\tau [\mathrm{artanh}{s}_a-\mathrm{artanh}{s}_0]$. Sincewe can integrate the flux exactly. Using $\int {\mathrm{sech}}^4\xi d\xi =\tanh \xi -({\tanh }^3\xi )/3$ gives the circular ash deposit profileIt is constant under the initial disk, then decreases to zero at the limiting edge. Particle solid volume per area is $\Sigma /{\rho }_p$; deposit thickness is $\Sigma /{\rho }_d$ if a bulk deposit density ${\rho }_d$ including voids is supplied. The problem does not determine that packing density.

For a negligible initial radius this simplifies toIts integral $2\pi {\int }_0^{R_{\infty }}r\Sigma (r)dr={\rho }_p{c}_0\mathcal{V}$ confirms that the complete initial ash inventory is deposited. The finite-$R_0$ formula gives the same budget including its central flat portion.

The diagram uses the negligible-vent-radius idealization and time normalized by $\tau$. It illustrates both the asymptotic approach to $R_{\infty }$ and the smoothly vanishing outer deposit. All these predictions depend on the mixing and constant-Froude head closure; they do not claim to resolve the initial blast or the ultimate inertia-dominated motion of almost particle-free gas.

Interpret ${\rho }_2$ as the ambient density: the PDF's phrase assigning it a “depth” is a typographical error. Take $0<{\rho }_1-{\rho }_2\ll {\rho }_2$, and use a common reference density in inertia while retaining the density difference in buoyancy. This is the Boussinesq approximation, with reduced gravity $g^{\prime }=g({\rho }_1-{\rho }_2)/{\rho }_{\mathrm{ref}}>0$. For the shallow water approximation, the lower-layer depth must be small compared with longitudinal variation scales, vertical accelerations must be negligible and pressure hydrostatic. The deep ambient is approximated as a hydrostatic pressure reservoir away from the nose; assume a uniform velocity over each lower-layer cross-section and neglect drag and entrainment.

Write the prismatic triangular area as $A(h)=k{h}^2$ with fixed $k>0$, so the width at height $z$ is $2kz$. The excess hydrostatic pressure force divided by reference density isThe prismatic-channel shallow water equations are thereforeFor positive depth, divide the first equation by $2kh$ and use it in the second to obtainThe coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/2\\ g^{\prime }& u\end{array}\right)$. Its two distinct real characteristic speeds areThus the system is strictly hyperbolic for $h>0$ and $g^{\prime }>0$, though it degenerates at zero depth. Since $dc/dh=c/(2h)$, direct substitution in the two equations givesHence $R_{\pm }=u\pm 4c$ are the Riemann invariants conserved along their respective characteristic curves. The factor four, rather than the rectangular-channel factor two, comes from the triangular relation $A\propto h^2$.