21-Mat-B1 Hydrometallurgy and Electrometallurgy · December 2014
Question 4 of 7: Flotation kinetics of an oil sands sample
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Professional Examinations, December 2014 — 10-Met-B1, Mineral Processing. Three hours, closed book, approved Casio/Sharp calculator only. Six numbered Problems (all compulsory except Problem 5) plus a two-mark Bonus Question. Problem 5's rubric asks for any SIX of eleven sketch-and-describe topics; all eleven topics are answered below.
Note on the exam title
Nothing on the paper is a hydrometallurgy (leaching, solvent extraction, electrowinning) or electrometallurgy question; the syllabus actually examined is comminution and grinding-circuit mass balance, particle settling, flotation kinetics, and mineral-processing equipment/terminology — the physical/mechanical beneficiation stage that precedes hydro- or pyro-metallurgical extraction.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
B. A. Wills and J. A. Finch, Wills' Mineral Processing Technology, 8th ed. — comminution and Bond's law (Ch. 6), classification and hydrocyclones (Ch. 9), gravity concentration and jigs (Ch. 10), froth flotation and flotation kinetics (Ch. 12), dense-medium separation and thickening (Ch. 11, 17).
F. F. Aplan (ed.), SME Mineral Processing Handbook — flotation reagents (collectors, frothers, activators), plant flowsheets, comminution circuit design.
A. Taggart, Handbook of Mineral Dressing — classical terminology (probable error, ratio of concentration/enrichment, jig hutch, crusher types).
Given. First-order flotation model $R=R_I[1-\exp(-kt)]$; data $(t,R)$ = (1 min, 50 %), (12 min, 90 %), (15 min, 90 %).
Find. (a) the ultimate recovery $R_I$ and rate constant $k$; (b) the time to reach 80 % recovery.
Approach. Recovery is identical (90 %) at both 12 and 15 minutes, so the test has already reached its plateau by 12 minutes — that plateau value is the ultimate recovery $R_I$. With $R_I$ fixed, the single early data point (1 min, 50 %) is enough to solve for $k$ algebraically.
(a) Ultimate recovery from the plateau. Because $R(12)=R(15)=90\ \%$, the curve has flattened well before 12 minutes, so $$R_I=\boxed{90\ \%}.$$ Substituting the $t=1$ min point: $$0.50=0.90[1-\exp(-k\cdot1)]\ \Rightarrow\ \exp(-k)=1-\dfrac{0.50}{0.90}=0.4444\ \Rightarrow\ k=-\ln(0.4444)=\boxed{0.811\ \text{min}^{-1}}.$$ Check: at $t=12$ min, $R=90[1-\exp(-0.811\times12)]=90(1-0.00006)=89.99\ \%\approx90\ \%$ — matches both remaining data points, confirming the fit.
(b) Time for 80 % recovery. $$0.80=0.90[1-\exp(-kt)]\ \Rightarrow\ \exp(-kt)=1-\dfrac{0.80}{0.90}=0.1111\ \Rightarrow\ t=\dfrac{-\ln(0.1111)}{0.811}=\dfrac{2.197}{0.811}=\boxed{2.71\ \text{minutes}}\ (\approx2\ \text{min}\ 43\ \text{s}).$$
Quantity
Result
Ultimate recovery, $R_I$
$\boxed{90\ \%}$
Rate constant, $k$
$\boxed{0.811\ \text{min}^{-1}}$
Time to 80 % recovery
$\boxed{2.71\ \text{min}}$
Check. Reading $R_I$ directly off a plateau (rather than fitting all three points by least squares) is only valid because two of the three points already agree to the reported precision; a noisier data set would need a proper nonlinear regression for both parameters simultaneously.
