Three-Product Metallurgical Balance: 21-Mat-B1 Hydrometallurgy and Electrometallurgy, Undated paper, Q1

Question 1 — Three-Product Metallurgical Balance (15 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. All values above, basis 100 mass units of feed. Three products leave the circuit (Pb concentrate, Zn concentrate, final tailing); every cell shown as "—" is to be found by mass balance.

Find. The Final Tailing's weight %, Pb and Zn assays, and the Pb/Zn distribution % of all four products.

Fig. Q1 — three-product split of the Buick Concentrator circuit, basis 100 units feed.

Approach. Take a basis of 100 mass units of feed; the Final Tailing's mass is the remainder after the two concentrates, and its Pb/Zn content is the remainder after subtracting what reports to each concentrate, from which the tailing assays and every distribution percentage follow directly.

Final tailing mass. Only three products leave the circuit, so $$T = 100 - C_{Pb} - C_{Zn} = 100 - 6.2 - 3.3 = 90.5\ \text{units}$$
Lead balance. Total Pb in feed $=100\times5.05=505.0$ units. Pb reporting to the two concentrates $=6.2\times76.33+3.3\times1.76=473.246+5.808=479.054$ units. By difference, Pb in the tailing $=505.0-479.054=25.946$ units, so $$\boxed{\text{Pb assay of tailing} = \dfrac{25.946}{90.5} = 0.287\%\ \text{Pb}}$$
Zinc balance. Total Zn in feed $=100\times2.26=226.0$ units. Zn reporting to the two concentrates $=6.2\times3.03+3.3\times55.45=18.786+182.985=201.771$ units. By difference, Zn in the tailing $=226.0-201.771=24.229$ units, so $$\boxed{\text{Zn assay of tailing} = \dfrac{24.229}{90.5} = 0.268\%\ \text{Zn}}$$
Distribution percentages. Each product's distribution is its own contained metal divided by the total metal in the feed. For Pb: $$\%\text{Pb to Pb Conc.}=\dfrac{473.246}{505.0}\times100=93.71\%,\quad \%\text{Pb to Zn Conc.}=\dfrac{5.808}{505.0}\times100=1.15\%,\quad \%\text{Pb to tailing}=\dfrac{25.946}{505.0}\times100=5.14\%$$ these sum to 100.0%, closing the Pb balance. For Zn: $$\%\text{Zn to Pb Conc.}=\dfrac{18.786}{226.0}\times100=8.31\%,\quad \%\text{Zn to Zn Conc.}=\dfrac{182.985}{226.0}\times100=80.97\%,\quad \%\text{Zn to tailing}=\dfrac{24.229}{226.0}\times100=10.72\%$$ which likewise sum to 100.0%. $$\boxed{\text{Distributions above close both balances exactly}}$$

Product	Weight, %	Pb Assay, %	Zn Assay, %	Pb Dist., %	Zn Dist., %
Feed	100.0	5.05	2.26	100.0	100.0
Pb Concentrate	6.2	76.33	3.03	93.71	8.31
Zn Concentrate	3.3	1.76	55.45	1.15	80.97
Final tailing	90.5	0.287	0.268	5.14	10.72

Check: this is the standard two-metal generalisation of the two-product formula — each product's assay/distribution is found by simple mass-and-metal bookkeeping (no recovery or grade assumptions needed) because the problem gives every product's own assay directly except the tailing's, which closes by difference. No selectivity index or separation-efficiency assumption is required for this question.

Topic: Multi-product metallurgical accounting (mass and metal balance closure).

Key relations: Total mass balance $F=\sum{\text{products}}$; metal balance $F\cdot a_F=\sum(\text{product mass}\times\text{product assay})$; distribution $\%=\dfrac{\text{product mass}\times\text{product assay}}{F\cdot a_F}\times100$.

Why this works: Mass and each individual metal are separately conserved through a physical separation process (no chemical reaction), so writing one mass balance and one metal balance per element gives exactly as many independent equations as unknowns when only one product's data is missing. Distribution percentages are simply each metal balance restated as a fraction of the feed's own metal content, and by construction they must sum to 100% for every element — a free internal check.

Common pitfall: Forgetting that with THREE products (not the usual two), the tailing must be found from a metal balance across BOTH concentrates simultaneously, not by treating either concentrate in isolation; students sometimes also confuse "weight %" (mass fraction) with "distribution %" (fraction of a specific metal), which are numerically different for every product except the feed.

Source: Wills & Finch, Mineral Processing Technology, 8th ed., §3.3–3.4 (two- and three-product metallurgical balances).

Question 1 of 4: Three-Product Metallurgical Balance

Question 1 — Three-Product Metallurgical Balance (15 marks)