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If a finite-volume gravity current loses or dilutes the scalar that supplies its reduced gravity, its front can approach a finite runout length of a gravity current. A box model couples the scalar balance to a gravity-current front condition; eliminating time then gives the front position directly as a function of the remaining scalar concentration.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.

Because $\hat{W}<0$, species 1 now settles more slowly and species 2 more rapidly than in the equal-speed case. The supplied inequality says that species 2 provides the larger initial particle-density contribution to the reduced gravity; with exact Boussinesq density contrasts, the corresponding condition is ${\tilde{\varphi }}_1({\rho }_1-{\rho }_a)<{\tilde{\varphi }}_2({\rho }_2-{\rho }_a)$. The dominant buoyancy contribution is therefore removed earlier, so $g^{\prime }(t)$ and the gravity-current front condition fall below their equal-settling values at first order in $ϵ$. The runout length decreases. The slower loss of the weaker species-1 contribution only partly compensates for this effect.

Let the fixed current volume per unit channel width beA suitable high-Reynolds number deep-ambient gravity-current front condition iswhere the order-one Froude number $\mathrm{Fr}$ records the selected front closure. In the dilute limit the mixture's reduced gravity iswith ambient and carrier-fluid density ${\rho }_a$.

The well-mixed particle volume of species $i$ is $V_0{\varphi }_i$. Its deposition rate through the base of length $L$ is $L{\hat{W}}_i{\varphi }_i$, where ${\hat{W}}_i<0$. The resulting gravity-current box model is thereforeIt conserves fluid volume while suspended particle volume, and therefore the driving buoyancy, decreases by deposition.