Critical Particle Size in a Settling Lagoon: 23-Chem-A2 Unit Operations and Separation Processes, December 2017, Q3

Question A3: Critical Particle Size in a Settling Lagoon (25 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.

Quantity	Value
Lagoon (W × L × depth)	10 × 50 × 2 m
Wash-water flow $Q$	360 m³/hr = 0.10 m³/s
Ore / water density	2800 / 1000 kg/m³
Water viscosity $\mu$	$1\times10^{-3}$ Pa·s ($Re_p<0.4$)

Find. The smallest ore-particle diameter the lagoon can capture (and a check that Stokes' law applies).

Figure A3 — A particle is captured if it settles from surface to floor within its residence time; the cut is set by the overflow rate $Q/A_\text{plan}$, independent of depth.

Approach. A continuous basin captures a particle whenever its terminal settling velocity exceeds the overflow rate $Q/A_\text{plan}$; set the two equal, invert Stokes' law for the diameter, and confirm $Re_p<0.4$.

Overflow (surface-loading) rate. The basin is an ideal clarifier, so the cut velocity is the flow divided by the plan area (depth cancels): $$v_t^{\,\text{crit}}=\frac{Q}{A_\text{surface}}=\frac{0.10\ \text{m}^3/\text{s}}{(10)(50)\ \text{m}^2}=\boxed{2.0\times10^{-4}\ \text{m/s}}.$$ Any particle settling faster than this reaches the floor before the outlet.
Invert Stokes' law for the critical diameter. The smallest retained particle settles at exactly $v_t^{\,\text{crit}}$. From $v_t=\dfrac{g\,d^2(\rho_p-\rho_f)}{18\mu}$, $$d=\sqrt{\frac{18\mu\,v_t}{g(\rho_p-\rho_f)}}=\sqrt{\frac{18(1\times10^{-3})(2.0\times10^{-4})}{9.81(2800-1000)}}=\boxed{1.43\times10^{-5}\ \text{m}=14.3\ \mu\text{m}}.$$
Check the Stokes-law validity. The particle Reynolds number at this size is $$Re_p=\frac{\rho_f\,v_t\,d}{\mu}=\frac{1000(2.0\times10^{-4})(1.43\times10^{-5})}{1\times10^{-3}}=2.9\times10^{-3}\ \ll0.4.\ \checkmark$$ Deeply viscous — Stokes' law is fully valid.
Model bound. The largest size for which Stokes still holds ($Re_p=0.4$) is $d_\text{max}=\left[\dfrac{0.4(18)\mu^2}{\rho_f g(\rho_p-\rho_f)}\right]^{1/3}\approx74\ \mu\text{m}$. The 14.3 µm cut lies well inside the valid range, so the answer is self-consistent: the lagoon retains ore $\gtrsim14.3\ \mu\text{m}$ and finer particles overflow.

Quantity	Value
Overflow rate $v_t^{\,\text{crit}}$	2.0×10⁻⁴ m/s
Critical (smallest retained) diameter	14.3 µm
Particle Reynolds number $Re_p$	2.9×10⁻³ (<0.4 ✓)
Stokes-valid upper size ($Re_p=0.4$)	≈74 µm

Concept & Theory

Topic. Overflow-rate (surface-loading) design of an ideal settling basin.

Key relations. Cut criterion $v_t\ge Q/A_\text{plan}$; Stokes' law $v_t=g d^2(\rho_p-\rho_f)/18\mu$; validity $Re_p=\rho_f v_t d/\mu<0.4$–1.

Why this works. In the Hazen model a particle enters at the surface and drifts horizontally while settling; whether it reaches the floor before the outlet depends only on the ratio of settling velocity to the overflow velocity $Q/A_\text{plan}$. Depth cancels because a deeper tank lengthens both the settling path and the residence time in equal proportion — so to capture finer particles you add area, not depth.

Common pitfall. Using the tank cross-section or volume instead of the plan area, or applying Stokes' law outside its range — above $Re_p\approx1$ the drag deviates and one must iterate on the intermediate/Newton regimes.

Source. McCabe, Smith & Harriott ch. 7 (motion of particles through fluids); Geankoplis ch. 14 (settling & sedimentation); Coulson & Richardson Vol. 2 ch. 3.

Question 3 of 6: Critical Particle Size in a Settling Lagoon

Part A — Unit Operations

Question A3: Critical Particle Size in a Settling Lagoon (25 marks)