Question 3 of 3: Mineral Processing Short-Answer Bank

Question 3: Mineral Processing Short-Answer Bank (44 marks – 11 of 13 required; all 13 answered)

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.

1. (1) Control actions for a finer cyclone overflow (re: Question 2). With reference to the Question 2 circuit, two standard operator actions to make the cyclone overflow (product) finer are: (a) increase the dilution water added at the sump, which lowers the %-solids (and hence the effective viscosity/mass loading) of the cyclone feed and shifts the classifier's cut size $d_{50c}$ finer; and (b) increase the cyclone operating (inlet) pressure – e.g. by increasing pump speed or throttling less – which raises the centrifugal force in the cyclone and likewise lowers $d_{50c}$. (A smaller vortex-finder diameter or a smaller apex/spigot orifice would have the same finishing effect, but changing dilution water or feed pressure are the two actions available to an operator in real time without a shutdown.)
2. (2) Particle size analysis techniques other than sieving. (a) Laser diffraction (e.g. Malvern/Mastersizer), which infers a size distribution from the angular scattering pattern of a dispersed suspension; and (b) gravity/centrifugal sedimentation (e.g. the Andreasen pipette or a Sedigraph), which infers size from measured settling velocity via Stokes' law. (Electrical sensing-zone counting, the Coulter principle, is a further common alternative.)
3. (3) Why heap leaching of gold is usually more economical than stirred-tank leaching. Heap leaching needs only coarse crushing (no fine grinding), no mechanical agitation energy, and minimal capital – no leach tanks, agitators or dedicated reactor vessels – so it can profitably treat very large tonnages of LOW-grade ore that could never carry the capital and operating cost of agitated tank leaching. The trade-off is much slower kinetics (weeks to months of percolation versus hours to days of agitation) and typically lower ultimate recovery, which is acceptable because the ore grade is too low to justify the faster, more expensive route in the first place.
4. (4) Two limitations of Stokes' equation. $v=g(\rho_s-\rho_f)d^2/18\mu$. (a) It is valid only in the laminar (Stokes') settling regime, Reynolds number $\lesssim1$ – for coarser or denser particles settling faster, inertial/turbulent drag dominates and the equation under-predicts the drag (over-predicts $v$). (b) It assumes perfectly smooth, rigid, SPHERICAL particles settling alone (unhindered, dilute suspension) in a Newtonian fluid – real ore particles are angular/irregular (different, shape-dependent drag coefficient) and industrial slurries are often concentrated enough that particle–particle hindrance measurably slows settling relative to the Stokes prediction.
5. (5) Canadian commodities routinely processed by Dense (Heavy) Medium Separation. (a) Diamonds – Canadian diamond mines (e.g. in the Northwest Territories) use ferrosilicon dense-medium cyclones to reject barren kimberlite ahead of the final recovery stages; and (b) coal – Canadian coal preparation plants routinely use DMS vessels/cyclones to reject high-density rock and shale from run-of-mine coal.
6. (6) Definition. "The percentage of the mineral occurring as free particles in the ore in relation to the total content of the mineral" is the definition of the degree of liberation (liberation) of that mineral.
7. (7) $E_p=\dfrac{d_{75}-d_{25}}{2}$. $E_p$ is the probable error of the classification (or separation), read off the partition (Tromp) curve at the 75% and 25% partition-to-underflow sizes. It measures the SHARPNESS of the classification/separation: a small $E_p$ means the cut between the two products is sharp (a narrow size range straddles the 50% partition point), while a large $E_p$ means the separation is diffuse. It is often reported non-dimensionally as the imperfection $I=E_p/d_{50c}$.
8. (8) Structural formula of sodium ethyl xanthate. Sodium ethyl xanthate is the sodium salt of ethyl dithiocarbonic acid: an ethyl group bonded through oxygen to a thiocarbonyl carbon that also carries a thiolate sulphur bonded to sodium, $\text{CH}_3\text{CH}_2\text{-O-C(=S)-S}^-\text{Na}^+$.
   Sodium ethyl xanthate: ethyl–O–C(=S)–S–Na (skeletal formula; vertices and chain end are CH₂/CH₃ carbons).
9. (9) Specific surface diameter from BET area.

   Given. Specific surface area $S_w=1.4\ \text{m}^2/\text{g}=1.4\times10^4\ \text{cm}^2/\text{g}$; solids density $\rho_s=2.6\ \text{g/cm}^3$. Find. Specific surface (surface-volume mean) diameter $d_{sv}$.

   $$d_{sv}=\frac{6}{\rho_s\,S_w}=\frac{6}{2.6\times1.4\times10^4}=1.648\times10^{-4}\ \text{cm}=\boxed{1.65\ \mu\text{m}}$$
10. (10) wt% solids of a hydrocyclone underflow sample.

    Given. Sample mass $1{,}142$ g, volume $650$ mL, solids density $2.6\ \text{g/cm}^3$ (water $1.0\ \text{g/cm}^3$). Find. wt% solids.

    Let $M_s=$ solids mass; volumes are additive: $M_s/2.6+(1142-M_s)/1.0=650$, so $$M_s\left(\frac{1}{2.6}-1\right)=650-1142\ \Rightarrow\ M_s=\frac{650-1142}{0.3846-1}=\frac{-492}{-0.6154}=799.5\ \text{g}$$ $$\text{wt\% solids}=\frac{799.5}{1142}\times100=\boxed{70.0\%}$$
11. (11) vol% solids (same sample as Q10). Solids volume $V_s=799.5/2.6=307.5$ mL: $$\text{vol\% solids}=\frac{307.5}{650}\times100=\boxed{47.3\%}$$
12. (12) Fire-assay gold grade.

    Given. Bead mass $0.5$ mg; 1 assay-ton $=29.17$ g. Find. Gold grade of the ore, g/t.

    The assay-ton is sized so that milligrams of gold per assay-ton read directly as troy ounces per short ton (0.5 mg/AT $=0.5$ oz/st); in metric units the grade is simply bead mass over sample mass: $$\text{Grade}=\frac{0.5\times10^{-3}\ \text{g}}{29.17\ \text{g}}\times10^6\ \text{g/t}=\boxed{17.1\ \text{g/t}}$$
13. (13) Bond Work Index from a locked-cycle Grindability test.

    Given. $-6$ mesh feed, $F_{80}=2350\ \mu\text{m}$; steady state at 250% circulating load; net production $G=1.6$ g/rev of $-105\ \mu\text{m}$ material; closing screen $P_1=105\ \mu\text{m}$; product $P_{80}=89\ \mu\text{m}$. Find. $W_i$.

    $$W_i=\frac{44.5}{P_1^{0.23}\,G^{0.82}\left(\dfrac{10}{\sqrt{P_{80}}}-\dfrac{10}{\sqrt{F_{80}}}\right)}=\frac{44.5}{105^{0.23}(1.6)^{0.82}\left(\dfrac{10}{\sqrt{89}}-\dfrac{10}{\sqrt{2350}}\right)}$$ $$=\frac{44.5}{2.917\times1.470\times0.8537}=\boxed{12.16\ \text{kWh/t}}$$ The 44.5 constant printed on the paper is Bond's original, strictly giving kWh per short ton; the metric equivalent (constant 49.1, i.e. $\times1.1$) is about 13.4 kWh per tonne. The paper's own formula is used for the boxed answer.