= Hindered-settling runout in a rectangular channel
{title2=$L_\infty^{5/2}=L_0^{5/2}+\frac{5FM^{3/2}\sqrt{g_0'}}{V_s}\operatorname{artanh}\sqrt{\phi_0}$}
For a constant-volume unit-width <particle-laden gravity current> with $M=Lh$, define the particle-fluid reference <reduced gravity> $g_0'=g(\rho_p-\rho_0)/\rho_0$, so the current has $g'=g_0'\phi$. A constant front <Froude number> and absorbing-bed <hindered settling> give $\dot L=F\sqrt{g_0'\phi M/L}$ and $\dot\phi=-(V_sL/M)\phi(1-\phi)$. Eliminating time and integrating gives the displayed <runout length of a gravity current>. For large runout and dilute initial concentration, $L_\infty^5\sim25F^2M^3g_0'(\phi_0+2\phi_0^2/3+\cdots)/V_s^2$. This $g_0'$ is buoyancy per unit particle fraction, not the initial current reduced gravity $g_0'\phi_0$.
Back to article page
