Three-Product Laboratory Flotation Test on a Lead-Zinc Ore: 24-MMP-A3 Mineral Processing, May 2015, Q2

Question 2: Three-Product Laboratory Flotation Test on a Lead-Zinc Ore (7 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.

Given. A selective lead-zinc flotation test in which the whole charge has been weighed and assayed as three products. The feed itself was not assayed, which is the point of the question.

Given data, Problem 2: laboratory test products
Product	Weight, g	Assay, % Pb	Assay, % Zn
Lead concentrate	40	65	7
Zinc concentrate	30	4	50
Tailing	430	0.5	1
Total charge	500	—	—

Find. (a) the back-calculated head assay in lead and in zinc, and (b) the percentage of each metal reporting to each of the three products.

Approach. Convert every product into metal units, the product of weight and assay; units are conserved, so they can be summed to give the head and divided to give the distribution.

Figure 2. The three-product test as a mass split. Metal units - weight times assay - are conserved across the split, so summing them recovers the feed assay that was never measured.

Form the metal units for each product. A metal unit is the weight of a product multiplied by its assay; it is proportional to the mass of contained metal and is the natural currency for this kind of accounting: $$u_i = W_i \times a_i$$
Tabulate the units for lead and for zinc. Lead: $40 \times 65 = 2\,600$, $30 \times 4 = 120$ and $430 \times 0.5 = 215$, giving 2 935 units in total. Zinc: $40 \times 7 = 280$, $30 \times 50 = 1\,500$ and $430 \times 1 = 430$, giving 2 210 units in total.
Divide the total units by the total weight to get the head assay. Because units are conserved, the feed must carry the sum of the products' units in the sum of their weights: $$\begin{aligned} f_{Pb} &= \frac{2\,935}{500} = 5.87\ \%\ \text{Pb} \\ f_{Zn} &= \frac{2\,210}{500} = 4.42\ \%\ \text{Zn} \end{aligned}$$ $$\boxed{\text{calculated head} = 5.87\ \%\ \text{Pb and } 4.42\ \%\ \text{Zn}}$$
Form each distribution as a share of the total units. The distribution of a metal to a product is that product's units divided by the total units: $$D_i = \frac{W_i a_i}{\sum W_j a_j} \times 100$$ For lead to the lead concentrate this is $2\,600/2\,935 = 88.59$ per cent; the remaining figures follow the same way and are collected below.
Check that each metal's distributions add to 100 per cent. Lead: $88.59 + 4.09 + 7.33 = 100.01$, and zinc: $12.67 + 67.87 + 19.46 = 100.00$, the small residue being rounding only. A set that does not close to 100 means an arithmetic slip, not a real metallurgical loss. $$\boxed{\text{Pb: } 88.59\ /\ 4.09\ /\ 7.33\ \%\ \text{ and Zn: } 12.67\ /\ 67.87\ /\ 19.46\ \%}$$

Read metallurgically, the test is a good separation on lead and a mediocre one on zinc. Lead recovery to the lead concentrate is 88.6 per cent at a grade of 65 per cent lead, close to a saleable galena concentrate. Zinc, however, recovers only 67.9 per cent into its own concentrate, with 12.7 per cent misplaced into the lead concentrate and, more seriously, 19.5 per cent lost to the tailing. The 7 per cent zinc in the lead concentrate points to inadequate depression of sphalerite in the lead circuit — more zinc sulphate and cyanide, or a finer primary grind to liberate the two sulphides from one another, would be the first things a mill metallurgist would try.

Results, Problem 2: calculated head assay and metal distribution
Product	Weight, %	% Pb	% Zn	Pb distribution, %	Zn distribution, %
Lead concentrate	8.0	65	7	88.59	12.67
Zinc concentrate	6.0	4	50	4.09	67.87
Tailing	86.0	0.5	1	7.33	19.46
Calculated feed	100.0	5.87	4.42	100.00	100.00

Topic. Metallurgical accounting for a three-product separation, and the back-calculated head assay.

Key relations. Metal units $u = W a$; calculated head $f = \sum W_i a_i / \sum W_i$; distribution $D_i = W_i a_i / \sum W_j a_j$; the two-product formula of Problem 1 is the special case of the same algebra with one concentrate.

Why this works. Conservation of metal is the only physics in the question: whatever lead entered the cell must be in one of the three products, so the sum of the products' lead units equals the feed's lead units. Dividing that sum by the total weight recovers the assay the feed would have shown had it been sampled, and dividing each product's units by the total gives the fraction of the metal it captured. The calculated head is in fact preferred over a directly assayed head in plant practice, because a single grab sample of feed is far less reliable than the weighted average of three carefully split products; where both exist, the difference between them is a standard measure of sampling quality.

Common pitfall. Confusing grade with distribution: the zinc concentrate assays 50 per cent zinc but accounts for only 67.9 per cent of the zinc, and those two numbers answer different questions. The other frequent error is averaging the assays directly — $(65 + 4 + 0.5)/3 = 23.2$ per cent lead — which ignores the weights entirely and is out by a factor of four.

Source. Wills and Finch, Wills' Mineral Processing Technology, 8th ed., ch. 3, "Metallurgical accounting, control and simulation"; Taggart, Handbook of Mineral Dressing, section 19, for the units convention.

Question 2 of 6: Three-Product Laboratory Flotation Test on a Lead-Zinc Ore

Question 2: Three-Product Laboratory Flotation Test on a Lead-Zinc Ore (7 marks)