Question 4 of 4: Niobium Flotation Plant Audit

Question 4: Niobium Flotation Plant Audit (25 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. Six size fractions (+208 down to −37 µm) of rougher Feed, Concentrate and Tail, each with its own weight% and $\text{Nb}_2\text{O}_5$ assay (table above; each stream's weight% is on its own 100% basis). Find. Size-fraction enrichment ratios and recoveries for the +208 and −37 µm fractions, plus the overall rougher recovery, with conclusions.

Approach. First collapse each stream to a single overall grade by weight-averaging its six size fractions; that gives the classical two-product feed/concentrate/tail grades needed for the overall mass yield and recovery. Size-specific recoveries then need the ABSOLUTE tonnage of each fraction in the concentrate (mass yield $\times$ that fraction's own weight%), compared against the same fraction's tonnage in the feed – not simply the two streams' weight% figures, which are each on their own, different, 100% basis.

1. Enrichment ratios, +208 and −37 µm fractions (Part 1). Enrichment ratio is simply concentrate assay over feed assay for that same size fraction: $$ER_{+208}=\frac{4.09}{0.52}=\boxed{7.87}$$ $$ER_{-37}=\frac{2.12}{1.46}=\boxed{1.45}$$
2. Overall stream grades (needed for Parts 3–4). Weight-average each stream's six fractions: $$f=\sum w_i f_i/100=\frac{22.1(0.52)+21.4(0.65)+21.2(0.78)+12.0(0.85)+12.9(1.14)+10.4(1.46)}{100}=\boxed{0.8203\%}$$ $$c=\frac{8.6(4.09)+10.8(4.98)+14.4(4.68)+11.0(4.00)+23.0(2.86)+32.2(2.12)}{100}=\boxed{3.344\%}$$ The tail is weight-averaged the same way from its own six fractions: $t=\boxed{0.1147\%}$.
3. Overall rougher recovery (Part 3). Two-product metal balance, exactly as in Q2's magnetic-separator balance but applied to grade (not tonnage) data: $$R=\frac{c(f-t)}{f(c-t)}\times100=\frac{3.344(0.8203-0.1147)}{0.8203(3.344-0.1147)}\times100=\frac{3.344\times0.7056}{0.8203\times3.229}\times100=\boxed{89.1\%}$$ The same balance gives the overall mass yield to concentrate, needed for step 4: $Y=(f-t)/(c-t)=0.7056/3.229=0.2185$, i.e. 21.85% of the feed mass reports to rougher concentrate.
4. Size-specific recovery, +208 and −37 µm fractions (Part 4). On a common "per 100 units of total feed" basis, the tonnage of a size fraction in the feed is just its own feed weight%; the tonnage of that same fraction in the concentrate is $Y\times(\text{its concentrate weight}\%)$. Size recovery is the ratio of contained $\text{Nb}_2\text{O}_5$: $$R_i=100\times Y\times\frac{w_{i,c}\,c_i}{w_{i,f}\,f_i}$$ $$R_{+208}=100\times0.2185\times\frac{8.6\times4.09}{22.1\times0.52}=100\times0.2185\times\frac{35.17}{11.49}=\boxed{66.9\%}$$ $$R_{-37}=100\times0.2185\times\frac{32.2\times2.12}{10.4\times1.46}=100\times0.2185\times\frac{68.26}{15.18}=\boxed{98.2\%}$$

Part 2 – conclusions from the enrichment ratios. The coarse (+208 µm) fraction enriches far more strongly (7.87×) than the fine (−37 µm) fraction (1.45×). This is the classic size-by-size flotation signature of an under-liberated coarse fraction combined with fine-particle entrainment: the coarse niobium-bearing particles that DO report to concentrate are well liberated and float on true hydrophobicity (hence the high grade jump), while the fine fraction's concentrate grade sits close to its feed grade because a large share of the fine material reporting to concentrate is unselective – carried up mechanically in the water film around rising bubbles rather than genuinely floated.

Part 5 – conclusions from the recoveries. The pattern inverts: the coarse fraction recovers poorly (66.9%) despite its high grade, while the fine fraction recovers almost completely (98.2%) despite its poor grade. Read together with Part 2, this points at two distinct, size-specific loss mechanisms rather than one plant-wide problem: (i) a substantial share of the +208 µm feed is composite, gangue-locked niobium that is simply too coarse/poorly liberated to attach to a bubble at all, so it reports to tailings (low recovery) – the fix is finer primary grind or a regrind stage ahead of roughing, not more reagent; (ii) the −37 µm fraction floats (and entrains) almost completely regardless of true hydrophobicity, so its recovery is high but its grade is diluted by entrained gangue – the fix there is wash water / froth depth or a cleaning stage, not more collector. A plant-wide reagent increase would not address either root cause.