Question 4 of 4: Short-Answer Items

Question 4 — Short-Answer Items (40 marks; all 10 of 10 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.

Given / Find. Ten independent short-answer items, each self-contained; given data and the requested quantity are restated at the start of each item below.

1. (1) Fire-assay gold grade. Given: gold bead $=0.5\ \text{mg}$; 1 assay-ton $=29.17\ \text{g}$ represents the sample. Approach: the bead mass recovered from one assay-ton sample scales directly to a full tonne. $$\text{Grade}=\dfrac{0.5\ \text{mg}}{29.17\ \text{g}}\times1{,}000{,}000\ \dfrac{\text{g}}{\text{t}} = \dfrac{0.0005\ \text{g}}{29.17\ \text{g}}\times10^6$$ $$\boxed{\text{Grade}\approx 17.1\ \text{g/t Au}}$$
2. (2) Bond Grindability work index. Given: closing screen $P_1=105\ \mu\text{m}$ (the −105 µm test screen); net grindability $G_{rp}=1.6\ \text{g/rev}$; $P_{80}=89\ \mu\text{m}$; $F_{80}=2350\ \mu\text{m}$. Approach: substitute directly into the standard Bond grindability formula. $$W_i=\dfrac{44.5}{(105)^{0.23}(1.6)^{0.82}\left(\dfrac{10}{\sqrt{89}}-\dfrac{10}{\sqrt{2350}}\right)}=\dfrac{44.5}{(2.917)(1.470)(1.060-0.206)}$$ $$\boxed{W_i\approx 12.2\ \text{kWh/short ton}}$$ (the classical Bond grindability formula is calibrated in kWh per short ton, not metric tonne, since it was developed on US test data).
3. (3) Copper tailings grade. Given: feed 1.2% Cu; concentrate 32% Cu; recovery 95%. Approach: two-product balance, basis 100 units feed. Cu in feed $=1.2$; Cu to concentrate $=1.2\times0.95=1.14$; concentrate mass $=1.14/0.32=3.5625$; tailing mass $=100-3.5625=96.4375$; Cu to tailing $=1.2-1.14=0.06$. $$\boxed{\text{Tailings grade}=\dfrac{0.06}{96.4375}\times100\approx 0.0622\%\ \text{Cu}}$$
4. (4) Maximum operating speed of a tumbling mill. Given: mill radius $R=2.6\ \text{m}$, so diameter $D=5.2\ \text{m}$. Approach: beyond the critical speed $N_c$, centrifugal force pins the charge to the mill's inner wall and grinding stops, so $N_c$ is the theoretical ceiling on proper operation (in practice mills run at 65–80% of $N_c$, but the question asks for the maximum). $$N_c=\dfrac{42.3}{\sqrt{D}}=\dfrac{42.3}{\sqrt{5.2}}$$ $$\boxed{N_c\approx 18.5\ \text{rpm}}$$
5. (5) Gravity separability of chalcopyrite and pyrite. Given: $SG_{\text{pyrite}}=5.0$ (heavy), $SG_{\text{chalcopyrite}}=4.3$ (light), water $SG_f=1.0$. Approach: evaluate the concentration criterion $CC=\dfrac{SG_h-SG_f}{SG_l-SG_f}$ and compare against Taggart's feasibility bands. $$CC=\dfrac{5.0-1.0}{4.3-1.0}=\dfrac{4.0}{3.3}\approx1.21$$ $$\boxed{CC\approx1.21<1.25\ \Rightarrow\ \text{NOT separable by gravity, at any practical particle size}}$$ The two minerals are too close in density; a criterion below 1.25 is Taggart's "impossible at any size" band.
6. (6) Hydrocyclone underflow wt% solids. Given: slurry volume $670\ \text{mL}$; slurry mass $830\ \text{g}$; solids $SG=3.0$. Approach: let $M_s$ be the solids mass; solids and water volumes must sum to the measured slurry volume. $$\dfrac{M_s}{3.0}+(830-M_s)=670\ \Rightarrow\ M_s\left(\dfrac{1}{3}-1\right)=670-830\ \Rightarrow\ M_s=240\ \text{g}$$ $$\boxed{\text{wt\% solids}=\dfrac{240}{830}\times100\approx28.9\%}$$
7. (7) Xanthate dosing. Given: $4\ \text{kg}$ ore; dosage $50\ \text{g/tonne}$; xanthate solution $5\ \text{wt\%}$; pulp $40\ \text{wt\%}$ solids. Approach: scale the per-tonne dosage to the actual sample mass, back out the solution volume from its concentration, then build a liquid-phase balance from the pulp density. Xanthate needed $=50\times(4/1000)=0.20\ \text{g}$. Mass of 5% solution needed $=0.20/0.05=4.0\ \text{g}\approx4.0\ \text{mL}$ (dilute aqueous solution, $\rho\approx1\ \text{g/mL}$). $$\boxed{\text{Add}\approx4\ \text{mL of xanthate solution}}$$ At 40 wt% solids, total pulp mass $=4000/0.40=10{,}000\ \text{g}$, so liquid (water) phase $=10{,}000-4000=6000\ \text{g}=6.0\ \text{kg}$. $$\boxed{\text{Xanthate in liquid phase}=\dfrac{200\ \text{mg}}{6.0\ \text{kg}}\approx33.3\ \text{mg/kg}}$$
8. (8) Why quartz is hydrophilic and graphite is hydrophobic. The difference is in what kind of chemical bond gets broken when each mineral fractures. Quartz (SiO₂) is a three-dimensional network of strong, polar Si–O covalent/ionic bonds; any fracture surface, in any direction, must break these bonds, exposing broken, charge-unsatisfied Si and O sites that immediately react with water to form polar silanol (Si–OH) surface groups. These polar groups hydrogen-bond strongly with water molecules, so the surface is wetted — hydrophilic. Graphite has the opposite, anisotropic structure: strong covalent bonds hold carbon atoms together WITHIN each hexagonal sheet, but adjacent sheets are held together only by weak van der Waals forces. Graphite cleaves preferentially along these weak interlayer planes, exposing basal-plane carbon surfaces that have no broken bonds and no polar sites at all — there is nothing for a water molecule to hydrogen-bond to, so the surface is non-wetting — hydrophobic. This is exactly why graphite (like talc and molybdenite, which share the same sheet structure) floats naturally with little or no collector, while quartz needs a strongly adsorbing collector (or depressant strategy) to be floated at all.
9. (9) Meaning of $E_p=(d_{75}-d_{25})/2$. $E_p$ is the probable error (in French mineral-processing literature, "Écart Probable") of a classification or screening operation, read directly off its own partition (Tromp) curve: $d_{75}$ and $d_{25}$ are the particle sizes at which 75% and 25% (respectively) of that size fraction reports to the oversize/coarse product. $$\boxed{E_p\ \text{measures the SHARPNESS of separation}}$$ A small $E_p$ means the partition curve rises steeply through its 25–75% band (close to an ideal step function at $d_{50}$, i.e. a sharp, efficient separation); a large $E_p$ means the curve is shallow and diffuse (a poor separation, with substantial misplaced coarse-to-fine and fine-to-coarse material). Applying this to Question 2's own partition curve, $d_{75}\approx13.8\ \text{mm}$ and $d_{25}\approx8.8\ \text{mm}$, giving $E_p\approx2.5\ \text{mm}$ — a useful cross-reference showing the same curve answers both questions.
10. (10) Specific surface (Sauter) diameter from BET area. Given: $S_v=1.4\ \text{m}^2/\text{g}=14{,}000\ \text{cm}^2/\text{g}$; solids density $\rho=2.6\ \text{g/cm}^3$. Approach: the surface-volume mean (Sauter) diameter of an equivalent sphere satisfies $S_v=6/(\rho\, d_{sv})$. $$d_{sv}=\dfrac{6}{\rho\, S_v}=\dfrac{6}{2.6\times14{,}000}\ \text{cm}=1.648\times10^{-4}\ \text{cm}$$ $$\boxed{d_{sv}\approx1.65\ \mu\text{m}}$$