21-Mat-B1 Hydrometallurgy and Electrometallurgy · December 2013
Question 3 of 6: Flotation Kinetics
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Professional Examinations, December 2013 — 10-Met-B1, Mineral Processing. Three hours, closed book, approved Casio/Sharp calculator only. Six numbered Problems plus a two-mark Bonus Question; the rubric requires all problems except Problem 5, which is answered as any SIX of ten short sketch-and-describe topics. All ten topics of Problem 5 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, sampling theory, classification, gravity concentration and froth flotation — i.e. 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), particle size analysis and sampling (Ch. 3–4), classification and hydrocyclones (Ch. 9), gravity concentration (Ch. 10), froth flotation (Ch. 12), dense-medium separation (Ch. 11).
P. M. Gy, Sampling of Particulate Materials: Theory and Practice — the Gy sampling-constant equation used in Question 1(e).
F. F. Aplan (ed.), SME Mineral Processing Handbook — flotation reagents (collectors, frothers, activators, depressants), flotation kinetics.
A. Taggart, Handbook of Mineral Dressing — classical terminology (probable error, imperfection, ratio of concentration, jigging, crusher types).
Find. (a) $R_\infty$ and $k$; (b) time $t$ for $R=80\%$.
Approach. The recovery is identical (90%) at both $t=12$ and $t=15$ min, so the test has already reached its ultimate (plateau) recovery by 12 minutes — that plateau value IS $R_\infty$. With $R_\infty$ fixed, the single early data point ($t=1$ min, $R=60\%$) is enough to solve for $k$; the model is then checked against the two plateau points before being used to answer part (b).
Part (a) — Determine $R_\infty$ and $k$. Since $R(12)=R(15)=90\%$, the recovery has stopped changing with time, i.e. the process has reached its asymptote: $$R_\infty=90\%$$ Substituting the $t=1$ min point: $$60=90[1-\exp(-k\cdot1)]\ \Rightarrow\ 1-e^{-k}=\dfrac{60}{90}=0.6667\ \Rightarrow\ e^{-k}=0.3333$$ $$k=-\ln(0.3333)=\ln 3=1.099\ \text{min}^{-1}$$ Checking the model reproduces the plateau: at $t=12$, $R=90[1-\exp(-1.099\times12)]=90(1-1.9\times10^{-6})\approx90.0\%$; at $t=15$, $R\approx90.00\%$ to four figures — both match the data essentially exactly, confirming the fit. \(\boxed{R_\infty=90\%,\ \ k=\ln 3\approx1.10\ \text{min}^{-1}}\)
Part (b) — Time for 80% recovery. $$80=90[1-\exp(-kt)]\ \Rightarrow\ 1-e^{-kt}=\dfrac{80}{90}=0.8889\ \Rightarrow\ e^{-kt}=0.1111=\dfrac{1}{9}$$ $$-kt=\ln\!\left(\dfrac19\right)=-\ln 9\ \Rightarrow\ t=\dfrac{\ln 9}{k}=\dfrac{\ln 9}{\ln 3}=\dfrac{2\ln 3}{\ln 3}=2.00\ \text{min}$$ \(\boxed{t=2.00\ \text{min}}\) (an exact result here, because $\ln 9=2\ln 3$ — the clean numbers in the test data were evidently chosen so this falls out to two minutes exactly).
