Mass Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2014, Q2

Question 2 — Mass Balance (20 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 ball mill runs in closed circuit with a cyclone. Fresh ore (6 % moisture) joins the cyclone underflow recycle ahead of the mill; the mill discharge is diluted with water before the cyclone. The cyclone overflow, at 100 t/h dry solids, is the final product fed to flotation.

Given data
Quantity	Symbol	Value
Dry ore fed to flotation (cyclone overflow, dry)	O_s	100 t/h
Moisture in ore-bin feed	—	6 %
Cyclone feed, % solids	f	30 %
Cyclone underflow, % solids	u	75 %
Cyclone overflow, % solids	o	15 %

Find. (a) The circulating load — the dry-solids tonnage in the underflow recycled to the mill; (b) the water that must be added between the mill discharge and the cyclone to dilute the pulp to 30 % solids.

Figure 1 — Closed-circuit grinding: fresh feed and cyclone underflow both enter the ball mill; the discharge is diluted with water before the cyclone splits it into product overflow and recycled underflow.

Approach. Use a solids balance around the whole circuit to fix the fresh-feed rate, then a coupled solids/water balance around the cyclone alone (its three %solids values) to get the circulating underflow tonnage; finally balance water from the ore bin through the mill to the cyclone feed to isolate the dilution water.

Fix the fresh-feed dry-solids rate. At steady state the underflow recycle enters and leaves the mill loop unchanged, so a solids envelope around the whole circuit (ore bin in, cyclone overflow out) gives fresh feed = product: $$O_s = 100\ \text{t/h dry} \;\Rightarrow\; \text{fresh feed dry solids} = 100\ \text{t/h}$$
Balance total mass and solids across the cyclone. Let $U_s$ and $O_s$ be the dry-solids tonnages in the underflow and overflow. Total (wet) cyclone feed equals underflow plus overflow, and dividing each stream's solids by its own %solids gives its wet tonnage: $$\frac{U_s+O_s}{f}=\frac{U_s}{u}+\frac{O_s}{o}$$ with $f=0.30$, $u=0.75$, $o=0.15$. This is the standard two-product balance around a classifier.
Solve for the circulating load $U_s$. Substituting the %solids and clearing fractions, $$0.30\left(\frac{U_s}{0.75}+\frac{O_s}{0.15}\right)=U_s+O_s \;\Rightarrow\; 0.4\,U_s+2\,O_s = U_s+O_s \;\Rightarrow\; O_s = 0.6\,U_s$$ $$U_s=\frac{O_s}{0.6}=\frac{100}{0.6}$$ $$\boxed{U_s = 166.7\ \text{t/h dry solids (circulating load)}}$$ As a ratio to fresh feed this is a circulating-load ratio of 167 %, well inside the 100–400 % range typical of a ball-mill–cyclone circuit.
Cross-check with the classic %solids ratio formula. The textbook relation for the wet-tonnage ratio underflow/overflow around a cyclone is $U_{\text{wet}}/O_{\text{wet}}=(f-o)/(u-f)=(30-15)/(75-30)=0.333$. Computing $U_{\text{wet}}=U_s/u=222.2$ t/h and $O_{\text{wet}}=O_s/o=666.7$ t/h gives the same ratio, $222.2/666.7=0.333$, confirming step 3.
Build up the water at each point, starting with the fresh ore-bin feed. At 6 % moisture, the wet ore-bin tonnage is $$\dot m_{NF}=\frac{100}{1-0.06}=106.4\ \text{t/h},\qquad \text{water}_{NF}=106.4-100=6.4\ \text{t/h}$$
Add the water carried by the recycled underflow. At 75 % solids, $U_{\text{wet}}=166.7/0.75=222.2$ t/h, so its water content is $222.2-166.7=55.6$ t/h. The mill neither adds nor removes water (dilution water is added downstream, before the cyclone), so the mill discharge carries the sum of what entered: $$\text{water}_{\text{disc}} = 6.4+55.6 = 61.9\ \text{t/h},\qquad \dot m_{\text{disc}} = 106.4+222.2 = 328.6\ \text{t/h}\ (81.2\ \%\ \text{solids})$$
Set the cyclone-feed water demand at 30 % solids. The cyclone feed carries the same $266.7$ t/h of solids ($=U_s+O_s$) but must arrive at 30 % solids: $$\dot m_{CF}=\frac{266.7}{0.30}=888.9\ \text{t/h},\qquad \text{water}_{CF}=888.9-266.7=622.2\ \text{t/h}$$
Take the difference to get the added dilution water. $$\text{water added} = \text{water}_{CF}-\text{water}_{\text{disc}} = 622.2-61.9$$ $$\boxed{\text{water added} \approx 560.3\ \text{t/h}}$$

Final results — Question 2
Part	Quantity	Result
(a)	Circulating load (underflow, dry)	166.7 t/h (167 % of fresh feed)
(b)	Dilution water added before the cyclone	560.3 t/h
—	Supporting: mill discharge (before dilution)	328.6 t/h, 81.2 % solids
—	Supporting: cyclone feed (after dilution)	888.9 t/h, 30 % solids

Topic. Circulating load and water balance in closed-circuit ball-mill grinding.

Key relations. Two-product cyclone balance $(U_s+O_s)/f=U_s/u+O_s/o$; equivalently $U_{\text{wet}}/O_{\text{wet}}=(f-o)/(u-f)$; water balance closes independently of the solids balance at every junction.

Why this works. A hydrocyclone does not add or remove solids — it only redistributes the same dry tonnage between a coarse, dense underflow and a fine, dilute overflow. Because %solids ties total mass to solids mass at each of the three streams, three %solids numbers plus one known solids flow are exactly enough equations to fix everything else. The circulating load is large (167 % of fresh feed) precisely because the underflow must run dense (75 % solids) while the overflow must run dilute (15 % solids) to achieve the required separation, and that constraint is what the (f−o)/(u−f) ratio expresses.

Common pitfall. The most common error is applying the %solids ratio to get a water answer directly, skipping the intermediate mill-discharge water; the dilution water is a difference between the cyclone feed's required water and whatever water already arrived via the ore bin and the recycle, not the cyclone feed's water alone. A second common slip is forgetting the ore bin's own 6 % moisture, which is small next to the recycle's water but not negligible.

Source. Wills' Mineral Processing Technology, 8th ed., Ch. 9 (classification, circulating load); Habashi, Textbook of Pyrometallurgy, Ch. 2 (mass balances).

Question 2 of 7: Mass Balance (20 marks)

Question 2 — Mass Balance (20 marks)