21-Mat-B1 Hydrometallurgy and Electrometallurgy · December 2016
Question 2 of 6: Two-Stage Grinding-Circuit Mass Balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Professional Examinations, December 2016 — 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 here, since this set is a study resource rather than an exam script.
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.
Note on recycled data
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), thickening (Ch. 15), tailings management.
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).
Question 2 — Two-Stage Grinding-Circuit Mass Balance (20 marks)
Ore specific gravity 3.0; rod mill runs open-circuit (no recycle); ball mill runs closed-circuit with the rake classifier.
Find. (a) overflow solids (t/h); (b) sands solids (t/h); (c) dilution water added at the classifier (t/h); (d) SG of the ball mill discharge slurry.
Solved grinding-circuit mass balance: 30 t/h new feed enters the open-circuit rod mill unchanged in tonnage; the rake classifier splits 90 t/h of combined solids (30 t/h rod-mill discharge + 60 t/h recycled ball-mill discharge) into 30 t/h overflow (final product) and 60 t/h sands (closed-loop ball-mill feed).
Approach. Because the rod mill is open-circuit, its discharge tonnage equals its feed tonnage; because the ball mill is closed-circuit with no other in/out streams, its discharge tonnage equals its feed (the classifier sands). At steady state the only stream that leaves the whole circuit is the classifier overflow, so an overall-circuit solids balance fixes it immediately; the size-fraction (%−100 µm) data is then used as a "tracer" to balance the classifier itself and back out the internal circulating load (sands = ball-mill discharge). A separate water balance on each stream (from %solids) then gives the dilution-water addition, and a solids/water weighted-average gives the discharge slurry SG.
Part (a) — Classifier overflow tonnage. The rod mill is open-circuit, so its solids discharge equals its solids feed: $R=30\ \text{t/h}$. The ball mill is closed-circuit, so its discharge solids equal its feed solids: sands $S=$ ball-mill discharge $X$. Around the WHOLE circuit (rod mill + classifier + ball mill as one box), the only solids entering are the 30 t/h new feed and the only solids leaving are the classifier overflow $O$ (every other stream is internal): $$O=30\ \text{t/h}$$ (this also falls straight out of the classifier's own solids balance, $R+X=S+O$, once $S=X$ is substituted: $30+X=X+O\Rightarrow O=30$). \(\boxed{O=30\ \text{t/h solids}}\)
Part (b) — Classifier sands (recirculating load). With $S=X$, balance the −100 µm size fraction across the classifier (in: rod-mill discharge $R$ at 20% and recycled ball-mill discharge $X$ at 47.5%; out: sands $S$ at 20% and overflow $O$ at 75%): $$R(0.20)+S(0.475)=S(0.20)+O(0.75)$$ $$30(0.20)+0.475S=0.20S+30(0.75)$$ $$6+0.475S=0.20S+22.5$$ $$0.275S=16.5\ \Rightarrow\ S=60\ \text{t/h}$$ Check: total solids into the classifier $=R+X=30+60=90=S+O=60+30=90$. \(\boxed{S=X=60\ \text{t/h solids}}\) (a 200% circulating load relative to new feed — typical of a fine, closed-circuit ball-mill stage).
Part (c) — Dilution water added. Convert each stream's %solids to a water/solids mass ratio, $\text{water}=\text{solids}\times(1-f_s)/f_s$: rod-mill discharge water $=30\times(0.20/0.80)=7.5\ \text{t/h}$; ball-mill discharge (recycle) water $=60\times(0.25/0.75)=20.0\ \text{t/h}$; sands water $=60\times(0.25/0.75)=20.0\ \text{t/h}$ (identical, since the sands and the ball-mill discharge share the same 60 t/h solids and 75% solids content); overflow water, using $33.3\%\approx1/3$, $=30\times\dfrac{1-1/3}{1/3}=30\times2=60.0\ \text{t/h}$. Water balance around the classifier (in: rod-mill-discharge water + recycled-discharge water + dilution water $D_w$; out: sands water + overflow water): $$7.5+20.0+D_w=20.0+60.0$$ the two 20.0 t/h terms cancel (same tonnage and %solids in the sands as in the ball-mill discharge), leaving $$D_w=60.0-7.5=52.5\ \text{t/h}$$ \(\boxed{D_w=52.5\ \text{t/h}}\)
Part (d) — SG of the ball mill discharge slurry. Ball-mill discharge is 75% solids by weight (SG 3.0) and 25% water by weight (SG 1.0), so the slurry SG is independent of tonnage. For a slurry, $1/\text{SG}_{slurry}$ is the weighted average of $1/\text{SG}$ by weight fraction: $$\dfrac{1}{\text{SG}_{slurry}}=\dfrac{0.75}{3.0}+\dfrac{0.25}{1.0}=0.25+0.25=0.50$$ $$\text{SG}_{slurry}=\dfrac{1}{0.50}=2.0$$ \(\boxed{\text{SG}_{slurry}=2.0}\)
