Question 6 of 6: Fill in the Blanks

Question 6: Fill in the Blanks (8 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.

6(1) — Sieve / sub-sieve cutoff (2 marks)

Answer: $\boxed{38\ \mu\text{m}}$ (approximately – corresponding to a 400-mesh Tyler/ASTM screen).

Dry sieving becomes unreliable below roughly this size because fine particles agglomerate, blind the screen apertures with electrostatic/surface-tension effects, and take impractically long to pass; below the cutoff, sub-sieve techniques (laser diffraction, sedimentation/Andreasen pipette, the Coulter counter) are used instead.

6(2) — Contained value of the deposit (2 marks)

Given. Copper price $= \text{USD}\ 6.83/\text{kg}$; grade $=1.2\%\ \text{Cu}$, no other value metal.

1. Contained Cu per tonne of ore, then its value. One tonne of ore at 1.2% Cu contains $1000\times0.012=12\ \text{kg Cu}$, so: $$\text{Contained value}=12\ \text{kg}\times\text{USD}\,6.83/\text{kg}=\boxed{\text{USD}\ 81.96/\text{t}}$$

6(3) — Circulating load of the grinding circuit (2 marks)

Given. Fresh feed $=34\ \text{t/h}$; mill (fresh feed + recycle) treats $85\ \text{t/h}$ total.

1. Recirculating tonnage and load ratio. The mill treats fresh feed plus the recycled underflow, so the recirculating load is the difference: $$\text{Recirculating load}=85-34=51\ \text{t/h}$$ $$\text{Circulating load}=\frac{51}{34}\times100=\boxed{150\%}$$

6(4) — Identity of the surface solid (2 marks)

[Figure not reproduced: Figure 6.4 — Redrawn from the paper: a flat solid at the surface of the aqueous solution with an air bubble touching its underside at a single point; the bubble does not spread over the solid (zero contact angle). See the official exam paper.]

Answer: $\boxed{\text{hydrophilic (water-wetting)}}$.

The figure shows the bubble pressed against the solid yet still perfectly spherical, touching it only at one point: water has not been displaced from the solid surface, so no three-phase contact line has formed and the contact angle is $\theta\approx0^\circ$. A surface that water wets completely in this way is hydrophilic – the work of adhesion between bubble and particle, $W_{sa}=\gamma_{wa}(1-\cos\theta)$, is zero at $\theta=0$, so the bubble cannot hold the particle and it will not report to a froth product. A hydrophobic solid would instead let the bubble spread across its surface with a finite contact angle (the larger $\theta$, the stronger the attachment), which is what a collector is dosed to achieve.