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24-MMP-A3 Mineral Processing · December 2018

Question 3 of 4: Mineral Processing Short-Answer Bank

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

EGBC National Exam — Mining and Mineral Processing Engineering, 09-MMP-A3 Mineral Processing, 2018-Dec. 3 hours duration, closed book; only an approved Casio or Sharp calculator permitted. Four questions constitute a complete exam paper (100 marks total).

Reference texts: Wills & Finch, Wills' Mineral Processing Technology, 8th ed. (comminution, crushers and mills – Ch. 6; classification, hydrocyclones and partition curves – Ch. 9; gravity concentration – Ch. 10; froth flotation, cells, reagents and flotation columns – Ch. 12; metallurgical balances, recovery/enrichment ratio – Ch. 1 & 12; solid-liquid separation, thickening and filtration – Ch. 15); SME Mining Engineering Handbook, 3rd ed. (porphyry copper mill flowsheets); BC Health, Safety and Reclamation Code for Mines, and the MEND/GARD Guide (Global Acid Rock Drainage Guide) for acid mine drainage prediction and control in the Canadian regulatory context.

Question 3: Mineral Processing Short-Answer Bank (40 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.

Check – the paper allows choosing any 8 of these 12 items (5 marks each, 40 marks total). All twelve are answered in full below so this solution also serves as a complete study reference for the item bank for choose-N-of-M exams.

3(1) — Hydrocyclone underflow slurry density (5 marks)

Given. Underflow solids content 70 wt%; solids SG $=2.6$ (so $\rho_s=2600\ \text{kg/m}^3$); water $\rho_w=1000\ \text{kg/m}^3$. Find. Pulp (slurry) density.

  1. Reciprocal (weighted-volume) pulp density formula. $$\frac{1}{\rho_{\text{pulp}}}=\frac{x_s}{\rho_s}+\frac{x_w}{\rho_w}=\frac{0.70}{2600}+\frac{0.30}{1000}=0.0002692+0.0003000=0.0005692$$ $$\rho_{\text{pulp}}=\frac{1}{0.0005692}=\boxed{1757\ \text{kg/m}^3}$$

3(2) — Xanthate solution dosage (5 marks)

Given. Sample $=2\ \text{kg}=0.002\ \text{t}$; dosage $=20\ \text{g/t}$; xanthate solution strength $=1.0\ \text{wt}\%$. Find. Grams of solution to add.

  1. Mass of pure xanthate required. $$m_{\text{xanthate}}=20\ \text{g/t}\times0.002\ \text{t}=0.040\ \text{g}$$
  2. Mass of 1.0 wt% solution carrying that much xanthate. $$m_{\text{solution}}=\frac{m_{\text{xanthate}}}{0.01}=\frac{0.040}{0.01}=\boxed{4.0\ \text{g of solution}}$$

3(3) — Gravity separation of pyrite and chalcopyrite (5 marks)

Given. $SG_{\text{pyrite}}=5.0$ (heavy mineral), $SG_{\text{chalcopyrite}}=4.3$ (light mineral), medium $=$ water, $SG_{\text{medium}}=1.0$. Find. Whether gravity separation is feasible.

  1. Concentration criterion. $$CC=\frac{SG_{\text{heavy}}-SG_{\text{medium}}}{SG_{\text{light}}-SG_{\text{medium}}}=\frac{5.0-1.0}{4.3-1.0}=\frac{4.0}{3.3}=\boxed{1.21}$$
  2. Interpretation. $CC<1.25$ (below about 1.25 gravity separation is not commercially feasible at any particle size; 1.25–1.75 permits only very coarse separations), so pyrite and chalcopyrite cannot be separated by gravity concentration – their densities are simply too close. Froth flotation (as used throughout the rest of this paper) is the correct route for this mineral pair.

3(4) — Silver grade in the theoretical galena concentrate (5 marks)

Given. Ore $=3.5$ wt% galena (PbS); Ag $=150\ \text{g/t}$ ore, hosted entirely in the galena; 100% galena recovery to a pure PbS concentrate. Find. Ag grade of that concentrate.

  1. Silver is conserved; only the host mass shrinks to the galena fraction. Per tonne of ore, all 150 g of Ag reports to the $0.035$ t of galena concentrate produced: $$\text{Ag grade}=\frac{150\ \text{g/t ore}}{0.035\ \text{t concentrate / t ore}}=\boxed{4286\ \text{g/t}}$$

3(5) — Gaudin–Schuhmann size distribution (5 marks)

Given. Distribution modulus $m=0.56$; size modulus $k=220\ \mu\text{m}$. Find. Mass % in the range $-150+74\ \mu\text{m}$.

  1. Cumulative % passing at each size. $Y(x)=100(x/k)^m$: $$Y(150)=100\left(\frac{150}{220}\right)^{0.56}=100(0.6818)^{0.56}=\boxed{80.7\%}$$ $$Y(74)=100\left(\frac{74}{220}\right)^{0.56}=100(0.3364)^{0.56}=\boxed{54.3\%}$$
  2. Size fraction by difference. $$\text{mass\%}_{-150+74}=Y(150)-Y(74)=80.7-54.3=\boxed{26.4\%}$$

3(6) — Copper activation of ZnS flotation (5 marks)

Sphalerite (ZnS)surfaceCu-activated surface(CuS-like layer)Hydrophobic Cu-xanthatecoating -- floatsCu2+ ion exchangexanthate chemisorbs
Fig. 2 – Copper-ion activation mechanism for sphalerite (ZnS) flotation.

Sphalerite (ZnS) is only weakly floatable with a xanthate collector on its own, because Zn–xanthate is too soluble/unstable to give a robust hydrophobic coating. Copper sulfate is added as an activator: $\text{Cu}^{2+}$ ions adsorb onto the ZnS surface and undergo an ion-exchange reaction, $\text{ZnS}+\text{Cu}^{2+}\rightarrow \text{CuS(surface)}+\text{Zn}^{2+}$, leaving a thin CuS-like layer on the sphalerite surface. Isopropyl xanthate then chemisorbs strongly onto this copper-rich surface (Cu–xanthate is far less soluble and more stable than Zn–xanthate), producing a robust hydrophobic coating that floats readily. At pH 9 (mildly alkaline, typical Zn-circuit pH with lime present) the activation reaction is efficient and the copper-xanthate surface complex is stable, which is why pH 9 is specified rather than a strongly acidic or strongly alkaline pH.

3(7) — $d_{50}$ and $d_{50c}$ in hydrocyclone operation (5 marks)

particle size (log scale)mass recovery to underflow050100d50d50cactual partition curvecorrected (water-split removed)
Fig. 3 – Actual vs. corrected partition curve, showing $d_{50}$ and $d_{50c}$.

$d_{50}$ is the particle size that has a 50% probability of reporting to the underflow, read directly off the actual partition (efficiency) curve – the curve built straight from the measured mass split of each size fraction between overflow and underflow. Because some fine slurry always reports to the underflow purely by being carried along in the water that leaves with it (rather than by true centrifugal classification) – the "water-split" or short-circuit / bypass effect – the actual curve never starts at 0% recovery even at very fine sizes. $d_{50c}$ is the corrected $d_{50}$: it is read off the corrected partition curve, obtained by mathematically removing the bypass fraction (subtracting the water-split recovery and rescaling), so that it represents the size at which true, size-dependent classification (not simple water entrainment) is 50:50. $d_{50c}$ is always coarser than the raw $d_{50}$ (removing the bypass lowers the recovery at every size, so the corrected curve reaches 50% at a larger size, as Fig. 3 shows) and is the figure that should be used to compare a cyclone's genuine sharpness of cut between operating conditions.

3(8) — How a flotation column works (5 marks)

froth zonecollection zonewash waterconcentratefeedsparger (air/N2 bubbles)tailings
Fig. 4 – Schematic flotation column.

A flotation column is a tall (often 8–15 m), unagitated vessel with two zones stacked vertically instead of the several agitated cells used in conventional flotation. Feed slurry enters part-way up the column into the collection zone, where fine bubbles generated by spargers near the base rise counter-currently through the settling slurry; hydrophobic particles collide with and attach to these bubbles as they rise. The loaded bubbles pass up through a deep froth zone at the top, onto which clean wash water is sprayed downward – this displaces entrained gangue-laden water back down out of the froth, so only genuinely attached (truly hydrophobic) particles survive to report to the concentrate launder that overflows the top. Barren tailings slurry leaves the bottom. Because there is no mechanical agitation and the froth is continuously washed, columns achieve much higher selectivity (grade) than conventional cells for a given recovery, and are typically used as a final cleaning stage (e.g. after 2nd stage cleaner cells) rather than as roughers.

3(9) — Acid mine drainage (AMD) (5 marks)

Sulfide minerals(e.g. FeS2) + O2 + H2OFe2+, SO4(2-), H+releasedFe(OH)3 ppt + further H+-> low-pH, metal-laden AMDoxidationbacterial catalysis
Fig. 5 – Acid mine drainage generation mechanism.

Acid mine drainage is acidic, metal- and sulfate-laden water produced when sulfide minerals (most commonly pyrite, $\text{FeS}_2$) in mine tailings, waste rock or exposed pit walls are oxidised on contact with oxygen and water. The reaction releases ferrous iron, sulfate and acidity ($\text{H}^+$); acidophilic bacteria (e.g. Acidithiobacillus ferrooxidans) catalyse the further oxidation of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$, which precipitates as $\text{Fe(OH)}_3$ and releases still more acid, and $\text{Fe}^{3+}$ itself becomes an additional oxidant that keeps attacking fresh sulfide surfaces – a self-sustaining, autocatalytic cycle once started. The result, if unmanaged, is a long-term seepage of low-pH water carrying dissolved heavy metals into the receiving environment. AMD potential is predicted before a tailings facility is built (and is a routine Canadian regulatory requirement under provincial mine permitting and the federal Metal and Diamond Mining Effluent Regulations) using acid–base accounting (ABA): measure the tailings' Acid Potential (AP, from total or sulfide sulfur content) and Neutralization Potential (NP, from carbonate/alkaline mineral content) and compute the Net Neutralization Potential $NNP=NP-AP$ (or the ratio $NPR=NP/AP$); a strongly negative NNP ($NPR\lesssim 1$–2, per the MEND/GARD Guide screening criteria) flags the material as acid-generating. A complementary static test is the Net Acid Generation (NAG) test, which reacts a sample with hydrogen peroxide to force complete sulfide oxidation and directly titrates the resulting acidity; samples classed as uncertain are then confirmed with kinetic tests (humidity cells or leach columns run for weeks to months) that measure actual oxidation, neutralization-depletion and metal-release rates.

3(10) — Polymer flocculants and thickener settling rate (5 marks)

Dispersed fine particles(mutually repulsive)Polymer chain adsorbs onmultiple particle surfacesBridged floc -- larger d,settles faster (Stokes)flocculant addedbridging
Fig. 6 – Bridging flocculation mechanism.

Fine mineral particles in a thickener feed are typically mutually repulsive (like surface charge) and settle individually, each obeying Stokes' Law where settling velocity scales with the square of particle diameter – so very fine particles settle extremely slowly on their own. A high-molecular-weight polymer flocculant, dosed at parts-per-million levels, has long chains with charged or polar functional groups that simultaneously adsorb onto several different particles' surfaces at once. This "bridges" many individual fine particles together into a much larger, loosely bound floc. Because the effective settling diameter of the floc is many times that of a single particle, its Stokes settling velocity increases dramatically (roughly with the square of the floc-to-particle size ratio), which raises the achievable thickener throughput (or, equivalently, produces a clearer overflow and a denser underflow for the same unit area).

3(11) — Rod mill vs. SAG mill length-to-diameter ratio (5 marks)

Rod millL/D > 1.25(rods must stay parallel --a short mill lets them tangle/skew)SAG millL/D much less than 1(large-diameter drum lifts ore/ballsthrough a long fall -- power scales with D)
Fig. 7 – Rod mill vs. SAG mill proportions.

A rod mill needs $L/D>1.25$ because the grinding media are long steel rods that must stay aligned parallel to the mill axis to grind selectively (coarse particles preferentially, by a rolling/crushing action along the rod's length) and to avoid "coarse breakage" by tangled or crossed rods; a mill that is too short relative to its diameter lets the rods skew and tangle at the ends, damaging both the rods and the mill liners and destroying the selective grinding action rods are chosen for. A SAG (semi-autogenous) mill instead needs $L/D$ much less than 1 (a short, large-diameter drum) because its grinding action relies on ore itself (plus a modest steel ball charge) being lifted by the rotating drum most of the way around a large diameter and then cascading/cataracting down through a long fall onto the toe of the charge – the impact energy of that fall, and therefore the mill's power draw and breakage rate, scale strongly with diameter. A large diameter also lets big run-of-mine lumps be accepted directly without a fine crushing stage; making the mill long as well as wide would only add dead retention time and support cost without adding useful breakage energy.

3(12) — Carrier flotation (5 marks)

Ultra-fine target mineral(too fine to float alone)Coarse carrier particles+ collector addedFines heterocoagulate ontocarrier -- both float togethermixed togetherattachment
Fig. 8 – Carrier flotation mechanism.

Carrier flotation is used to recover ultra-fine (typically sub-10 µm) target mineral particles that are individually too small to float efficiently on their own – fine particles have very low momentum, so their collision-and-attachment efficiency with rising air bubbles is poor, and they are also prone to non-selective entrainment in either direction. In carrier flotation, coarser "carrier" particles (which may be a different, deliberately added mineral, or coarser particles of a related gangue/host mineral) are conditioned together with the fine target particles and a suitable collector. The fine particles heterocoagulate – attach by surface (often electrostatic or hydrophobic) forces – onto the much larger carrier particle surfaces, forming composite aggregates. These aggregates behave, for flotation purposes, like a single coarse hydrophobic particle: they collide with and attach to bubbles far more efficiently than the original fines could alone, and float together with the carrier, recovering the ultra-fine target mineral into the concentrate as a "passenger" on the carrier.