21-Mat-A3 Structure and Characterization of Materials · May 2015
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 2015 — 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, pyrometallurgy, iron and steelmaking, and magnesium and zinc production — 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.
F. Habashi, Textbook of Hydrometallurgy, 2nd ed. — leaching, purification, electrowinning.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — refining, molten-salt electrolysis.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — comminution, classification, flotation, thickening, pulp density.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities and reaction/transformation enthalpies.
ASM Handbook, Vol. 2 (Nonferrous Alloys) — magnesium and zinc production practice.
Find. (a) Tailings grade; (b) % Cu recovery to concentrate; (c) % Cu loss to tailings; (d) enrichment ratio $c/f$; (e) specific gravity of a 20 % solids pulp.
Figure 2.1 — Two-product split on a 100 t feed basis: 9.5 t concentrate at 21 % Cu, 90.5 t tailings at the back-calculated grade.
Approach. Take a 100 t feed basis; close the copper mass balance around the two-product split ($F=C+T$, $Ff=Cc+Tt$) to get the tailings grade, then read recovery, loss and enrichment ratio directly off the same balance; treat the pulp as solids plus water by volume for part (e).
Set the basis and mass-balance the solids. Take $F=100$ t feed. Then $C=9.5$ t concentrate and $T=F-C=90.5$ t tailings.
Copper balance to get the tailings grade (a). Copper in equals copper out: $Ff=Cc+Tt$.
$$100(2.1)=9.5(21)+90.5\,t \;\Rightarrow\; t=\frac{210-199.5}{90.5}$$
$$\boxed{t = 0.116\%\text{ Cu in tailings}}$$
Recovery (b). Recovery is the fraction of feed copper reporting to the concentrate:
$$R=\frac{Cc}{Ff}\times100=\frac{9.5\times21}{100\times2.1}\times100=\frac{199.5}{210}\times100$$
$$\boxed{R = 95.0\%}$$
Loss (c). The complement of recovery, since all feed copper reports to one stream or the other:
$$\text{Loss}=100-R=100-95.0$$
$$\boxed{\text{Loss} = 5.0\%}$$
(Cross-check: $Tt/Ff\times100 = 90.5(0.116)/210\times100 = 5.0\%$ — consistent.)
Enrichment ratio (d). The ratio of concentrate grade to feed grade:
$$\text{ER}=\frac{c}{f}=\frac{21}{2.1}$$
$$\boxed{\text{ER} = 10.0}$$
Pulp specific gravity (e). Take a 100 kg pulp basis at 20 % solids: 20 kg ore (SG 2.8) plus 80 kg water (SG 1.0). Volumes add:
$$V=\frac{20}{2.8}+\frac{80}{1.0}=7.14+80.0=87.14\ \text{L}$$
$$\text{SG}_{\text{pulp}}=\frac{100\ \text{kg}}{87.14\ \text{L}}$$
$$\boxed{\text{SG}_{\text{pulp}} = 1.15}$$
