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In a bidisperse particle suspension transported by a steady depositing gravity current, species with concentrations $c_i$ and downward settling velocity magnitudes $v_i$ satisfy $c_i(x)=c_{i0}{e}^{-v_{i}x/q}$. Their local particle deposition fluxes are $v_i{c}_i$. The local deposited particle-volume fraction belonging to species one isFor $v_1>v_2>0$ and both species present, $f_1$ decreases strictly downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q$. If $g_i^{\prime }=g_{0i}{c}_i$, use $c_i=g_i^{\prime }/g_{0i}$ rather than buoyancy ratios unless $g_{01}=g_{02}$. A number fraction additionally divides each species volume flux by its single-particle volume. Integrating deposition over $[0,x]$ instead gives cumulative weights $q{c}_{i0}(1-e^{-v_{i}x/q})$.

Under the same box-model assumptions, $M_0={\rho }_p{\varphi }_{0}W$ is initial particle mass and $m_d$ is final deposited mass per downstream length. Integrate the vertical particle deposition flux across the instantaneous triangular-valley width $2\lambda h$ from arrival of the front at $x$ until settling is complete. This gives the displayed profile on $0\le x\le R$ and zero beyond the runout. Its integral is $M_0$, verifying particle mass conservation. Areal deposit density also depends on the lateral coordinate because the wetted valley width changes as the current thins.

// Target: geophysics.bigb

A dilute particle suspension can supply both excess mass density and heat. To first order its equation of state is $\rho /{\rho }_0=1+\gamma \varphi -\theta$, where $\varphi$ is the particle volume fraction and $\theta$ is fractional thermal expansion. Particle deposition flux reduces $\varphi$, while particle heating increases $\theta$. The competition can reverse the reduced gravity and create unstable density stratification.

A well-mixed particle-laden gravity current in a prismatic channel with $b(z)=C\sqrt{z}$ has volume $V=2C{h}^{3/2}L/3$. Negligible entrainment gives fixed $V$. Vertical particle deposition flux is $W_s\varphi b(h)L$, so $\dot{\varphi }=-3W_s\varphi /(2h)$. Close the front by $\dot{L}=F\sqrt{G\varphi h}$, where $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Eliminating time givesThe concentration tends to zero at a finite limiting runout length of a gravity current, although reaching that limit takes infinite time in this model.

Use an inertial gravity-current box model in a horizontal prismatic V-shaped valley. Let its section have area $A(h)=\lambda h^2$ and width $2\lambda h$, where the dimensionless constant $\lambda >0$ is half-width divided by depth. Let $X(t)$ be front position and $h(t)$ the representative uniform current depth. Assume negligible entrainment and volume loss, hydrostatic cross-sections, large Reynolds number, and negligible bed friction and axial bed slope. These assumptions describe the bulk spreading after the short dam-break adjustment; they do not give a universal law for a steep or friction-dominated valley.

For water of fixed volume $W$, conservation gives $W=\lambda h^{2}X$. Close the front speed by the gravity-current front condition $\dot{X}=\mathrm{Fr}\sqrt{gh}$, where $\mathrm{Fr}$ is a constant order-one Froude number for the chosen inertial front model. ThereforeIntegration gives $X^{5/4}=X_0^{5/4}+5K_{w}t/4$. Once the initial reservoir length $X_0$ is negligible, the inertial dam-break current in a triangular valley satisfiesThe time exponent differs from a flat-bottomed channel because a V-shaped section stores volume proportional to $h^2$. A viscous V-shaped channel gravity current obeys a different spreading law, so retaining the inertial assumption is essential.

For the short-lived eruption, interpret $W$ as the bulk mixture volume released during the event. Let $\varphi (t)$ be the well-mixed particle volume fraction, initially ${\varphi }_0$, let ${\rho }_p$ and ${\rho }_a$ be particle and ambient carrier densities, and assume a dilute Boussinesq approximation for the mixture. The particles provide reduced gravityAssume thermal buoyancy, compressibility, erosion and continued injection are negligible, and that the carrier volume remains $W$. Turbulent mixing keeps the particles suspended uniformly while they settle vertically at a constant terminal settling velocity $w_s$. For small spherical particles of radius $a_p$ in carrier viscosity ${\mu }_a$, the Stokes settling velocity is $w_s=2({\rho }_p-{\rho }_a)g{a}_p^2/(9{\mu }_a)$ when the particle Reynolds number is small. If an eruption volume is quoted as solid particle volume rather than mixture volume, replace the $W$ used here by that volume divided by ${\varphi }_0$.

The particle deposition flux per horizontal projected bed area is $w_s\varphi$. The current's projected settling area is $2\lambda hX$, soThe factor two reflects the triangular section's mean depth $h/2$. Put $s=\sqrt{\varphi /{\varphi }_0}$ and $K=\mathrm{Fr}\sqrt{g_0^{\prime }}(W/\lambda {)}^{1/4}$. The front and concentration balances becomeTaking the initial release length negligible compared with the final runout, integration yields the constant-settling runout in a triangular valley:In particular $R\propto W^{3/7}(g_0^{\prime }{)}^{2/7}{w}_s^{-4/7}$ at fixed valley shape and front coefficient. With a finite initial length, replace $R^{7/4}$ by $X_0^{7/4}+7K\sqrt{W/\lambda }/(4w_s)$ in the corresponding integrated relation.

The whole time dependence is determined analytically by the quadratureEarly spreading has $X\simeq (5Kt/4{)}^{4/5}$, the water law with $g$ replaced by $g_0^{\prime }$. At late times the concentration loss slows the flow, and $R-X$ decays exponentially with rate $7K/(4R^{5/4})=w_s/h_R$, where $h_R=\sqrt{W/(\lambda R)}$. The final length $R$ is a limiting runout reached as $t\to \infty$, not a finite-time stopping event in this ideal model.

To calculate the deposit profile of a finite triangular-valley particle current, let $m_d(x)$ be final deposited particle mass per unit downstream length, and $M_0={\rho }_p{\varphi }_{0}W$ the initial particle mass. Deposition at a fixed section starts at arrival time $t_a(x)$, when the front first reaches $x$. With the instantaneous front coordinate denoted $\xi$, use $dt={\xi }^{1/4}d\xi /[K(1-(\xi /R{)}^{7/4})]$ to obtainWriting $\eta =x/R$, this evaluates toand $m_d=0$ beyond $R$. The profile decreases downstream, vanishes quadratically at the runout, and obeys ${\int }_0^R{m}_d(x)dx=M_0$. This integral check ensures that all initially suspended particle mass is accounted for.

For an areal rather than a linear deposit density, the changing valley width must also be specified. At lateral coordinate $y_{\perp }$, the current covers the bed only while $|y_{\perp }|\le \lambda h$. Set $Z=\min (R,\lambda W/y_{\perp }^2)$, with $Z=R$ on the valley axis. The mass per horizontal projected bed area is zero if $Z<x$, and otherwise isOn the axis this reduces to ${\rho }_p{\varphi }_0{h}_R[\frac{7}{5}(1-{\eta }^{5/4})-\frac{7}{12}(1-{\eta }^3)]$. Divide projected areal density by $\sqrt{1+{\lambda }^{-2}}$ to obtain mass per actual sloping bed area. A specified packed deposit density then converts this areal mass to local deposit thickness; the spreading model itself predicts deposited mass, not a packing fraction.

For dilute sediment volume concentration $\varphi$, define $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Neglecting particle volume in the fluid-volume balance,This applies the Boussinesq approximation to the bulk mixture, requiring $G\varphi /g\ll 1$, and uses a deep ambient layer $h\ll H$. A gravity-current front condition closes the unresolved nose. Write it as $\dot{L}=F\sqrt{g^{\prime }h}$, with a prescribed order-one front coefficient $F$. If the chosen Froude number is instead $\mathrm{Fr}=\dot{L}/c$, then $F=\sqrt{2/3}\mathrm{Fr}$.

Use a parabolic-channel sediment-current box model for one side of the dam: initially its length is $L_0$, depth $h_0$, and uniform concentration ${\varphi }_0$. Its conserved fluid volume isFor vertical settling onto the channel sides, the projected deposition width is $b(h)=\sqrt{h}$. Thus the total particle deposition flux is $W_s\varphi \sqrt{h}L$ andThe geometric factor $3/2$ follows from $b(h)/A(h)$; using a rectangular deposition formula would give a different result.

Eliminate time and use $h^{3/2}=h_0^{3/2}{L}_0/L$:The runout length of a gravity current is the limit where suspended concentration vanishes:The model approaches this length asymptotically in time. It assumes negligible entrainment, erosion and water loss, and does not resolve the initial release or nose. If the initially disturbed region supplies both sides, apply the formula to each side using its share of the initial volume and length.

The Boussinesq approximation requires $\delta =(\rho -{\rho }_0)/{\rho }_0=\varphi ({\rho }_p/{\rho }_0-1)\ll 1$. The reference mass density ${\rho }_0$ can then be used in inertia, while the small density excess is retained in buoyancy. Dilute particle volume fraction $\varphi \ll 1$ alone does not suffice if ${\rho }_p/{\rho }_0$ is exceptionally large. Define $G=g({\rho }_p-{\rho }_0)/{\rho }_0$, so the reduced gravity is $g^{\prime }=G\varphi$.

The shallow water equations require depth small compared with the horizontal evolution scale, weak vertical acceleration and approximately hydrostatic pressure. Here the ambient is deep and quiescent, so its leading contribution is a hydrostatic reference pressure; the excess pressure in the layer is ${\rho }_{0}G\varphi (h-z)$. The model also needs nearly uniform streamwise velocity across the section and the stated absence of secondary circulation. The advancing gravity current nose is a separate region where these assumptions can fail.

A nearly uniform particle volume fraction can be maintained by turbulent mixing with eddy diffusivity $K_z$, provided $h^2/K_z\ll h/w$ and the mixing time is short compared with horizontal advection and bulk evolution. Equivalently, $wh/K_z\ll 1$ keeps the bulk settling-induced gradient small. The particles should respond rapidly enough to follow the mixing motions. A thin boundary layer adjacent to the absorbing bed or walls need not be well mixed.

Uniform bulk concentration does not imply zero sedimentation. The relative downward particle velocity still supplies a particle deposition flux of approximately $w\varphi$ to an absorbing boundary. Mixing redistributes the remaining particles and maintains the bulk concentration profile while its overall level decreases; it need not suspend every particle indefinitely. Deposition requires negligible resuspension in the model.

For identical suspended particles, reduced gravity is proportional to particle volume fraction. If particles settle through the bed at speed $v_s$ and are not resuspended, the downward particle deposition flux removes a buoyancy-equivalent amount $v_s{g}^{\prime }$ per unit bed area. The depth-integrated scalar mass conservation equation isFor a steady particle-laden gravity current, $B_x=-v_{s}B/(uh)$. If $v_s/u\ll 1$, with fixed positive $E$ and drag/slope coefficients, the locally adjusted leading solution of the momentum and volume equations is $u\sim {\beta }_{\ast }{B}^{1/3}$ and $h\sim h_0+E(x-x_0)$. The weak-settling attenuation of a sloping gravity current then givesand henceFor an inlet at $x_0=0$ this has the requested logarithmic form. If the inlet is genuinely at $x_0\ne 0$, the argument must contain $x-x_0$ in order to satisfy the specified inlet condition.

The speed factor must be the ${\beta }_{\ast }$ derived in the previous part. The printed $\beta$ omits the factor one-half multiplying $E\cos \theta$ and uses an undefined uppercase $C$ instead of the governing equation's $c$. Taking $C=c$ still leaves a nonzero leading momentum residual $-\frac{1}{2}E(B/u)\cos \theta$ for the printed speed; this discrepancy cannot be repaired by a correction of order $v_s/u$. With the given hydrostatic momentum equation, the logarithmic law is valid with ${\beta }_{\ast }$, not the printed speed factor.

A controlled local approximation also needs $v_s/(Eu)\ll 1$ if $E$ is allowed to become small. Indeed the exact volume balance is $h_x=E-h{u}_x/u$; under $u\sim {\beta }_{\ast }{B}^{1/3}$ its second term is $v_s/(3u)$. The neglected depth and acceleration corrections become significant as $B$ approaches zero.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

The exact mixture equation of state is $\rho =(1-\varphi ){\rho }_f+\varphi {\rho }_p$. Keeping first-order terms in the small particle volume fraction and thermal expansion givesThe omitted term is ${\rho }_0\varphi \theta$. The downward solid-volume particle deposition flux is $J_v=W_s\varphi$; the corresponding particle mass flux is $J_m={\rho }_p{W}_s\varphi$. Uniform vertical mixing and constant layer depth implyFor $W_s>0$, set $r=W_s/H$. Thenand henceThe heated particle-laden layer loses its stable density stratification when this contrast vanishes. For $\beta >0$, solving for the neutral-buoyancy time givesThe layer is neutral at $T_s$ and statically unstable for $t>T_s$. The subsequent uniformly mixed lower-layer solution cannot represent the resulting overturning.

If $W_s=0$, then $\varphi ={\varphi }_0$, $\theta =\beta {\varphi }_{0}t$, and the continuous limit is $T_s=\gamma /\beta$. If $\beta =0$, the layer stays denser than its surroundings at every finite time and becomes neutral only asymptotically when $W_s>0$; thus $T_s=\infty$. These limiting cases must replace the printed formula when its denominator vanishes.

For a dilute particle suspension, take $W_s>0$ as the downward speed relative to the carrier fluid. The particle deposition flux through a horizontal absorbing bed is $W_s{\varphi }_b$, where ${\varphi }_b$ is the near-bed particle volume fraction. Without stirring, settling can create a particle-free upper layer and a descending concentration front, so uniform concentration throughout the remaining fluid is not automatic. Weak turbulence can maintain a nearly uniform suspended concentration while leaving the mean settling velocity approximately equal to the isolated-particle value; this justifies using $W_s\varphi$ in a mixed-layer balance. Strong stirring can produce resuspension and requires a separate erosion model.

For a bed inclined at angle $\vartheta$ to the horizontal, the normal settling speed is $W_s\cos \vartheta$, giving deposition flux $W_s{\varphi }_b\cos \vartheta$ per unit actual bed area. There is also downslope particle motion, and near-wall effects or resuspension can alter this simplest absorbing-boundary model. Integrating the normal projection over a sloping bed is equivalent to integrating $W_s{\varphi }_b$ over its horizontal projection.

The mixture mass density is $\rho ={\rho }_0+({\rho }_p-{\rho }_0)\varphi$, so its reduced gravity isThe Boussinesq approximation requires $({\rho }_p-{\rho }_0)\varphi /{\rho }_0\ll 1$, in addition to the dilution assumption.

A uniform-depth particle-laden gravity current in fixed triangular cross section has conserved volume proportional to $h^{2}X$. A gravity-current front condition and absorbing-boundary particle deposition flux give $\dot{X}=\mathrm{Fr}\sqrt{G\varphi h}$ and $\dot{\varphi }=-2w(\varphi )\varphi /h$. The box describes late-time bulk evolution; it does not resolve the nose or immediate dam-break flow.

For fixed channel width $b(z)=\beta z$, cross-sectional area is $A=\beta h^2/2$. With reduced gravity $G\varphi$, uniform cross-sectional fields and downward settling velocity magnitude $w$, volume conservation, momentum and particle deposition flux giveThe hydrostatic pressure force contains ${\int }_0^h(h-z)\beta zdz=\beta h^3/6$. No streamwise widening term occurs in a prismatic channel. Deposition uses the projected boundary width $\beta h$ under the vertical-settling model.

A finite-volume gravity-current box model in a widening triangular channel has volume proportional to $h^2{L}^2$, so $hL=K$ is constant. A gravity-current front condition and absorbing-bed particle deposition flux giveEliminating time makes $\sqrt{\varphi }$ affine in $L^{5/2}$, producing a finite limiting runout length of a gravity current despite an infinite idealized stopping time.

In a channel with width $2\lambda xz$ at height $z$, a well-mixed particle-laden gravity current of depth $h$ occupies area $\lambda x{h}^2$. Its volume conservation, momentum and particle deposition flux balances givewhere the reduced gravity is $g^{\prime }=d\varphi$. The $h/3$ coefficient is the cross-section-weighted mean of $h-z$ in the hydrostatic pressure force. The characteristic curves have speeds $u$ and $u\pm \sqrt{g^{\prime }h/2}$.