Stokes' law particle settling: 21-Mat-B1 Hydrometallurgy and Electrometallurgy, December 2014, Q3

Problem 3 — Stokes' law particle settling (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Stokes' law $V=gd^2(D_s-D_f)/18\mu$; a $d_1=100$-micron particle settles 1 m in $t_1=1.5$ s (same fluid, same solid and fluid densities, so $g,\mu,D_s,D_f$ are all identical between the two particles); target particle $d_2=10$ microns.

Find. (a) the settling time $t_2$ for the 10-micron particle; (b) two limitations of Stokes' law.

Approach. With every term except $d$ held constant, Stokes' law reduces to $V\propto d^2$; since both particles settle the same 1 m distance, $t=1/V\propto1/d^2$, so the ratio of settling times can be found without computing $g,\mu,D_s,D_f$ individually.

Scale settling time with particle size. $V_1/V_2=(d_1/d_2)^2$ and $t=1/V$ (fixed 1 m drop) gives $$\dfrac{t_2}{t_1}=\left(\dfrac{d_1}{d_2}\right)^2=\left(\dfrac{100}{10}\right)^2=100.$$
Solve for $t_2$. $$t_2=100\times t_1=100\times1.5=\boxed{150\ \text{s}}\ (2.5\ \text{minutes}).$$ A ten-fold reduction in diameter increases the settling time by the square, i.e. a hundred-fold — fine particles settle very slowly, which is the practical reason classification and thickening equipment must be sized generously for the fines fraction.
(b) Limitations of Stokes' law. It is valid only in the laminar (viscous) settling regime, $Re_p=\rho_f v d/\mu\lesssim1$ — larger or denser particles settle fast enough to enter the transitional/turbulent regime, where drag no longer scales linearly with velocity and Stokes over-predicts the settling velocity. It also assumes rigid, smooth spheres settling individually in an infinite, quiescent Newtonian fluid: real ore particles are angular (higher drag than an equal-volume sphere), and at the solids concentrations found in an actual slurry, particles interfere with each other's flow field (hindered settling), which further slows the true settling velocity below the Stokes prediction.

Quantity	Result
Settling time, 10-micron particle	$\boxed{150\ \text{s} = 2.5\ \text{min}}$
Limitation 1	Laminar-flow assumption ($Re_p\lesssim1$) fails for coarser/faster particles
Limitation 2	Assumes ideal spheres in dilute, quiescent suspension — angularity and hindered settling both violate this

Check. The $d^2$ scaling used here holds only because the problem keeps particle and fluid density fixed between the two cases (an "ore particle" of unspecified but presumably identical mineralogy in both parts) — if the two particles were different minerals, $(D_s-D_f)$ would also differ and the simple ratio would not apply.

Concept & Theory

Topic. Stokes' law of hindered-free settling and its validity range.

Key relations. $V=gd^2(\rho_s-\rho_f)/18\mu$; particle Reynolds number $Re_p=\rho_f Vd/\mu$; Stokes' regime $Re_p<1$; hindered-settling correction factors (e.g. Richardson-Zaki) for concentrated slurries.

Why this works. Stokes balances gravity (net of buoyancy) against viscous drag on a sphere at low Reynolds number, giving a velocity proportional to the square of the diameter — this is exactly why classifiers and thickeners are so sensitive to the finest size fraction in the feed, and why size analysis by sedimentation (e.g. the Andreasen pipette) is itself built on this law.

Common pitfall. Applying the $d^2$ scaling across particles of different density (forgetting $(D_s-D_f)$ is size-independent only for a fixed mineral) or applying Stokes' law at all to a coarse (millimetre-scale) particle, where the true settling regime is turbulent (Newton's law, drag $\propto V^2$) and Stokes badly over-predicts velocity.

Source. Wills & Finch, Mineral Processing Technology, 8th ed., Ch. 9 (gravity concentration and classification, Stokes' and Newton's settling laws).

Question 3 of 7: Stokes' law particle settling

Problem 3 — Stokes' law particle settling (10 marks)