21-Mat-A3 Structure and Characterization of Materials · May 2018
Question 2 of 7: Mass Balance (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2018 — 10-Met-A3, Metal Extraction Processes. Three hours, closed book, one approved calculator (Casio or Sharp). Seven problems of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All seven are solved here, because this set is a study resource rather than an exam script.
Note on the exam title. The printed exam header reads 10-Met-A3, Metal Extraction Processes. The content is extractive metallurgy — mineral processing, mass and heat balances, pyrometallurgical roasting, zinc production, ironmaking and electrometallurgy — and is answered as such.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, smelting, mass and heat balances.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — blast-furnace ironmaking, thermal and electrolytic reduction, electrometallurgy.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — pulp density, comminution, froth flotation.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, transformation enthalpies, electrochemistry.
ASM Handbook, Vol. 15 (Casting), and CIM (Canadian Institute of Mining) practice literature for Canadian smelter/refinery practice (e.g. the historic Cominco zinc-lead smelter at Trail, BC).
Given. A single flotation test on a copper ore, reported as grades and a mass split; part (e) is an independent pulp-density question sharing only the ore's own specific gravity.
Given data
Quantity
Symbol
Value
Feed grade
$f$
2.0 % Cu
Concentrate grade
$c$
20 % Cu
Concentrate mass, as % of feed
$C/F$
9.09 %
Ore specific gravity
SG$_s$
2.8
Pulp solids by weight (part e)
$x$
22.5 %
Find. The tailings grade, the copper recovery and loss, the enrichment ratio, and (independently) the specific gravity of a 22.5 % solids pulp of this ore.
Figure 2.1 — Two-product flotation balance on a 100 kg feed basis. Copper is conserved across the circuit: whatever does not report to the 9.09 kg concentrate leaves in the 90.91 kg tailings.
Approach. Work on a convenient 100 kg feed basis, close a copper mass balance around the circuit to get the tailings grade, then read recovery, loss and enrichment ratio directly off the same balance. Part (e) is solved separately with the volume-additive pulp-density mixing rule.
Set the basis and the copper in each stream. Take $F=100$ kg feed. The concentrate is $C=9.09$ kg (9.09 % of $F$), so the tailings mass is $T=F-C=100-9.09=90.91$ kg. Copper entering with the feed is
$$m_{Cu,F}=F\,f=100\times0.020=2.000\ \text{kg}$$
and copper leaving in the concentrate is
$$m_{Cu,C}=C\,c=9.09\times0.20=1.818\ \text{kg}$$
Part (a) — close the balance to get the tailings grade. Copper not in the concentrate is in the tailings:
$$m_{Cu,T}=m_{Cu,F}-m_{Cu,C}=2.000-1.818=0.182\ \text{kg}$$
$$t=\frac{m_{Cu,T}}{T}\times100=\frac{0.182}{90.91}\times100$$
$$\boxed{t\approx0.20\ \%\ \text{Cu in the tailings}}$$
This is exactly the kind of number a flotation metallurgist watches day to day — it is what the plant is failing to recover.
Part (b) — percentage copper recovery. Recovery is the fraction of the feed's copper reporting to the concentrate:
$$R=\frac{m_{Cu,C}}{m_{Cu,F}}\times100=\frac{1.818}{2.000}\times100$$
$$\boxed{R=90.9\ \%}$$
Part (c) — percentage copper loss. By the same balance, whatever is not recovered is lost to tailings:
$$L=\frac{m_{Cu,T}}{m_{Cu,F}}\times100=\frac{0.182}{2.000}\times100$$
$$\boxed{L=9.1\ \%}$$
As a check, $R+L=90.9+9.1=100.0\ \%$ — recovery and loss are complementary by definition, since every kilogram of feed copper must leave in one stream or the other.
Part (d) — enrichment ratio. The enrichment ratio compares concentrate grade to feed grade:
$$ER=\frac{c}{f}=\frac{20}{2.0}$$
$$\boxed{ER=10.0}$$
The concentrate is ten times richer in copper than the ore that was fed to the cell — a compact single number for how much "upgrading" the circuit achieved.
Part (e) — specific gravity of the pulp. This part shares only the ore's SG with parts (a)–(d); it uses the same volume-additive mixing rule that relates pulp density to solids fraction, written here in specific-gravity form (water's own density cancels out of the ratio):
$$\frac{1}{SG_p}=\frac{x}{SG_s}+\frac{1-x}{SG_w}$$
With $x=0.225$, $SG_s=2.8$ and $SG_w=1.0$ (water):
$$\frac{1}{SG_p}=\frac{0.225}{2.8}+\frac{0.775}{1.0}=0.08036+0.775=0.8554$$
$$\boxed{SG_p=1.169}$$
A pulp that is less than a quarter solids by weight still has a specific gravity well above water's, because the solid fraction, though a minority by weight here, is nearly three times denser than the water carrying it.
