Stokes' Law: Equal-Settling-Velocity Particle Diameters: 21-Mat-B1 Hydrometallurgy and Electrometallurgy, December 2013, Q4

Question 4 — Stokes' Law: Equal-Settling-Velocity Particle Diameters (5 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. Quartz: $d_q=30\ \mu\text{m}$, $\text{SG}_q=2.65$; Coal: $\text{SG}_c=1.4$; medium 1 = water ($\text{SG}=1.0$); medium 2 = air ($\text{SG}\approx0.0012$); both particles obey Stokes' law in each medium; same $g$, same $\mu$ within a given medium.

Find. The coal particle diameter $d_c$ that settles at the same velocity as the 30 µm quartz particle, first in water, then in air.

Approach. For two particles settling at the SAME velocity in the SAME medium, $g$ and $\mu$ (and $V$ itself) cancel between the two Stokes'-law expressions, leaving a simple ratio of $d^2(\text{SG}_{particle}-\text{SG}_{fluid})$ terms — the classical "equal-settling-ratio" concept used to size gravity/classification equipment for mixed-density feeds.

Part (a) — Settling in water. Equal velocity, same medium: $$d_q^2(\text{SG}_q-\text{SG}_w)=d_c^2(\text{SG}_c-\text{SG}_w)$$ $$d_c=d_q\sqrt{\dfrac{\text{SG}_q-\text{SG}_w}{\text{SG}_c-\text{SG}_w}}=30\sqrt{\dfrac{2.65-1.0}{1.4-1.0}}=30\sqrt{\dfrac{1.65}{0.40}}=30\sqrt{4.125}=30(2.031)$$ $$d_c=60.9\ \mu\text{m}$$ $\boxed{d_c\approx60.9\ \mu\text{m (in water)}}$ — more than double the quartz diameter, because water's density (SG 1.0) subtracts a much larger FRACTION from coal's low density (1.4) than from quartz's (2.65), so the lighter coal particle must be substantially coarser to sink at the same rate.
Part (b) — Settling in air. Repeating with $\text{SG}_{air}\approx0.0012$ (negligible next to either solid, but carried through for completeness): $$d_c=30\sqrt{\dfrac{2.65-0.0012}{1.4-0.0012}}=30\sqrt{\dfrac{2.6488}{1.3988}}=30\sqrt{1.8935}=30(1.376)$$ $$d_c=41.3\ \mu\text{m}$$ $\boxed{d_c\approx41.3\ \mu\text{m (in air)}}$ — a noticeably SMALLER equal-settling coal diameter than in water, because with the fluid density essentially negligible in both terms the ratio collapses toward $\sqrt{\text{SG}_q/\text{SG}_c}=\sqrt{2.65/1.4}=1.376$, whereas water's SG of 1.0 (comparable in size to both particle SGs) inflates the ratio by preferentially depressing the coal term.

Medium	Equal-settling coal diameter
(a) Water	60.9 µm
(b) Air	41.3 µm

Check: both results assume Stokes' (laminar, creeping-flow) drag applies to both particles in both media at these sizes — a reasonable assumption for tens-of-micron mineral/coal particles in water or air, since particle Reynolds numbers stay well below the Re ≈ 1 upper bound of the Stokes regime at these sizes and typical settling velocities.

Topic: Stokes' law and the equal-settling-ratio concept used in gravity concentration and classification.

Key relations: Stokes' law $V=gd^2(\rho_s-\rho_f)/18\mu$; equal-settling ratio $d_1/d_2=\sqrt{(\rho_2-\rho_f)/(\rho_1-\rho_f)}$ for two minerals settling at the same velocity in the same fluid.

Why this works: Because $V$, $g$ and $\mu$ are common to both particles in a single medium, equal-velocity settling reduces to matching $d^2(\rho_s-\rho_f)$ between the two particles; the fluid density $\rho_f$ enters only as a subtractive term, so its relative importance — and hence the equal-settling ratio — changes with the medium, which is exactly why the same two minerals separate differently by gravity concentration in water than they would in an air-classifier.

Common pitfall: Treating the equal-settling ratio as a constant that transfers unchanged between water and air (it does not, because $\rho_f$ is not negligible relative to the particle densities in water but is nearly negligible in air), or forgetting that a LOWER-density particle must always be LARGER, never smaller, than a higher-density particle to settle at the same rate.

Source: Wills & Finch, Mineral Processing Technology, 8th ed., Ch. 10 (gravity concentration, equal-settling particles).

Question 4 of 6: Stokes' Law: Equal-Settling-Velocity Particle Diameters

Question 4 — Stokes' Law: Equal-Settling-Velocity Particle Diameters (5 marks)