= Solution
Continue with downward coordinate $y=H-z$ and $W=W_s(a)>0$. By <Stokes drag law>, the small-particle isolated <settling velocity> is $W/4$. Normalize concentrations by $\phi_{\max}$: write $q_i=\phi_i/\phi_{\max}$. In suspension the specified independent <hindered settling> laws give fluxes $j_1/\phi_{\max}=Wq_1(1-q_1)$ and $j_2/\phi_{\max}=(W/4)q_2(1-q_2)$. The initial state is $(q_1,q_2)=(1/8,1/8)$.
Two upper <sedimentation shocks> clear the species separately. The faster front removes large particles, leaving $(0,1/8)$ behind it; the small <concentration> is unchanged across this front because the species fluxes are independent. The slower front separates that small-only suspension from clear fluid. Their downward velocities are
$$
\boxed{\widehat U_1=\frac78W,\qquad\widehat U_2=\frac7{32}W.}
$$
There is also a growing mixed deposit at the bottom. Let its normalized solid concentrations be $(p_1,p_2)$, with $p_1+p_2=1$, and let its upward speed magnitude be $s=-\widehat U_3>0$. Applying the <Rankine-Hugoniot condition> to each species gives
$$
s(p_1-1/8)=\frac7{64}W,\qquad s(p_2-1/8)=\frac7{256}W.
$$
Adding these relations proves
$$
\boxed{s=\frac{35}{192}W,\qquad \widehat U_3=-\frac{35}{192}W,\qquad
(p_1,p_2)=\left(\frac{29}{40},\frac{11}{40}\right).}
$$
The deposited flux is zero even though each species occupies only part of the packing: the settling law applies to the suspension, while the deposit is a separate stationary state. These jump relations determine the <deposit composition from sedimentation jump conditions>.
The large-particle clearing front has height $H-7Wt/8$ and meets the mixed-deposit front at height $35Wt/192$. Thus
$$
\boxed{t_1=\frac{192}{203}\frac HW,\qquad h_1=\frac{35}{203}H=\frac5{29}H.}
$$
At this merger all large particles have entered the deposit. Only the small particles remain suspended, at normalized <concentration> $1/8$. The new material deposited above the mixed layer is pure small-particle solid, at total <concentration> $\phi_{\max}$. Its <sedimentation shock> therefore has downward speed
$$
\boxed{\widehat U_4=\frac{0-(W/4)(1/8)(7/8)}{1-1/8}=-\frac W{32}.}
$$
For $t>t_1$, the deposit top and the remaining clearing front have heights
$$
z_{\rm bed}=h_1+\frac W{32}(t-t_1),\qquad z_{\rm clear}=H-\frac7{32}Wt.
$$
Their meeting, or the total solid-volume balance, gives
$$
\boxed{t_2=\frac{24}{7}\frac HW,\qquad h_2=\frac H4.}
$$
This is <two-stage bidisperse batch sedimentation>: the fast species finishes first, followed by deposition of the remaining slow species.
The final solid composition by <volume>, and also by <mass> because the particle densities are identical, is
$$
\boxed{\begin{cases}
72.5\%\ \text{large},\ 27.5\%\ \text{small},&0<z<5H/29,\\
0\%\ \text{large},\ 100\%\ \text{small},&5H/29<z<H/4.
\end{cases}}
$$
The upper pure-small layer has thickness $9H/116$. As checks, $(29/40)h_1=H/8$ accounts for all large-particle solid <volume>, and $(11/40)h_1+(h_2-h_1)=H/8$ accounts for all small-particle solid <volume>, in units of $\phi_{\max}$. If “percentage of particles” means number rather than <volume>, a large sphere occupies eight times the <volume> of a small sphere: the mixed bottom layer contains $29/117\simeq24.8\%$ large particles by number and $88/117\simeq75.2\%$ small. The upper layer remains entirely small under either convention.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-74-bidisperse-shocks.png]
{title=Bidisperse sedimentation: three initial shocks, merger at t1 and h1, final settling at t2 and h2, and solid-volume composition of the deposit}
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