Module 4 · Products of size reduction

Size distribution and closed-circuit grinding

Turn a sieve analysis into differential and cumulative plots, fit the Rosin–Rammler–Sperling and Gates–Gaudin–Schuhmann laws, and then close the circuit: a mill feeding a classifier that returns its oversize.

Screen analysis:
data (cumulative passing at each aperture) RRS fit GGS fit D₅₀, D₈₀

What you're seeing

A differential analysis gives the mass fraction \(x_i\) held between two screens; it is plotted at the mean size \(\bar D_i\) of the pair. The cumulative analysis adds the fractions to give the percentage passing each aperture. Two mean sizes matter for the energy laws:

$$\bar D_s = \frac{1}{\sum x_i/\bar D_i}\quad\text{(volume–surface mean, Rittinger and Kick)},\qquad D_{80}\quad\text{(80\% passing, Bond)}$$

Rosin–Rammler–Sperling describes the oversize \(R\) of crushed products: \(R = \exp[-(x/x')^n]\), so \(\ln\ln(1/R) = n\ln x - n\ln x'\) is a straight line; \(x'\) is the size that 63.2% passes and \(n\) measures the spread. Gates–Gaudin–Schuhmann fits the fine end as a power law of the undersize: \(F = (x/k)^m\), a straight line on log–log axes, with \(k\) the size modulus. Both are fitted by least squares on the linearised forms; switch the plot to see the straight lines.

Try this

  1. Finer setting. Load MSH's second grind (crusher set closer). \(\bar D_s\) falls from 0.843 to 0.441 mm and \(D_{80}\) from 2.10 to 1.27 mm.

  2. Is it really RRS? On the linearised RRS plot the points bend away from the line at both ends (R² = 0.93): crushed products from a narrow feed are only roughly RRS.

  3. Edit the data. Move 5 g from the 8-mesh screen to the pan and watch \(\bar D_s\) drop from 0.843 to 0.606 mm while \(D_{80}\) only moves from 2.10 to 1.92 mm: the surface mean is sensitive to fines.

Closed-circuit grinding

Fresh feed joins the recycled oversize, the mill grinds it, and a classifier (here a hydrocyclone; a screen works the same way) splits the discharge. At steady state the product leaving equals the fresh feed, so whatever the classifier rejects builds up as the circulating load.

Mill and classifier flowsheet
fresh feed and mill feed mill discharge product (overflow) recycle (underflow) Pipe width ∝ mass flow; dot size ∝ particle size (log).
mill discharge product recycle partition: fraction of each size sent to the recycle

What you're seeing

Mass balances, with \(F\) fresh feed, \(M\) mill throughput, \(R\) recycle and \(P = F\) product:

$$M = F + R,\qquad M\,m = P\,p + R\,r \;\Rightarrow\; \frac{R}{F} = \frac{p - m}{m - r}$$

where \(m\), \(p\), \(r\) are the fractions finer than a chosen size in the mill discharge, product and recycle. If the classifier returns all the oversize and recovers a fraction \(E\) of the fines (fraction \(a\) of the discharge):

$$\frac{C}{F} = \frac{1}{E\,a} - 1$$

The classifier is described by a partition curve \(c(x) = B_p + (1 - B_p)\left[1 - e^{-0.693(x/d_{50})^s}\right]\): fines bypass to the underflow with the water (\(B_p\)), and the cut is not sharp (\(s\)). MSH's screen effectiveness is the product of fines recovery and coarse rejection.

Simplification: the mill discharge distribution is set directly (RRS with \(x'\), \(n\)) rather than computed from a breakage model, so it does not change with the circulating load. Real mills give a coarser discharge as the load rises and the residence time falls.

Closed-circuit operation with free crushing removes fines as soon as they are made: less over-grinding, a narrower product and lower energy per tonne than open-circuit choke feeding (slides).

Try this

  1. Finer product. Lower the cut to 45 μm: the product becomes finer (P₈₀ ≈ 40 μm) and C/F rises from 1.45 to about 2.6.

  2. Better classifier. A sharp cut with no bypass returns less finished product, so C/F falls to 1.2 and the effectiveness rises to 0.88.

  3. Fines short-circuit. 35% bypass sends fines back to the mill: C/F climbs to 2.4 and they are over-ground.

  4. Overloaded mill. A coarser mill discharge (x′ = 250 μm) pushes C/F to 3.3: the circuit is limited by the mill.

Circulating-load calculators

From screen analyses of the three streams

From classifier efficiency

Worked examples

Example 1Mean sizes and fitted laws from a gyratory-crusher productMcCabe–Smith–Harriott, 5th ed., Problems 29.1–29.2 (first grind)

Trap rock (uniform 2-in. feed) is crushed in a gyratory crusher; the product screen analysis is the "Grind 1" table. Find the volume–surface mean diameter, the 80%-passing size and the Bond energy (Wi = 19.32 kWh/ton). Fit RRS and GGS.

  1. Mean sizes of each fraction: arithmetic mean of the two screens, e.g. 6/8 mesh: \((3.327 + 2.362)/2 = 2.845\) mm; pan: \(0.208/2 = 0.104\) mm.

  2. Volume–surface mean: \(\sum x_i/\bar D_i = 1.187\ \text{mm}^{-1}\) → \(\bar D_s = 0.843\) mm.

  3. D₈₀: 13.4% is retained on 8 mesh (2.362 mm) and 33.4% on 10 mesh (1.651 mm); interpolating log-linearly to 20% retained:

    $$D_{80} = 2.362\left(\frac{1.651}{2.362}\right)^{(20 - 13.4)/(33.4 - 13.4)} = 2.10\ \text{mm}$$

  4. Bond: \(0.3162 \times 19.32\,(1/\sqrt{2.10} - 1/\sqrt{50.8}) = 3.36\) kWh/ton, i.e. 370 kW at 110 ton/h.

  5. Fits: RRS \(n = 2.06\), \(x' = 1.52\) mm (R² = 0.93); GGS \(m = 1.86\), \(k = 1.85\) mm on the points up to 90% passing.

Answer\(\bar D_s = 0.843\) mm, \(D_{80} = 2.10\) mm, 3.36 kWh/ton by Bond.
Example 2Circulating load from screen analysesExam-style (closed-circuit grinding, BUET 2013-14 flow diagram)

A ball mill in closed circuit with a classifier receives 50 t/h of fresh feed. Screen analyses show 40% finer than 200 mesh in the mill discharge, 85% in the classifier overflow and 15% in the underflow. Find the circulating load and the mill throughput.

  1. Balances: \(M = F + R\) and \(0.40\,M = 0.85\,F + 0.15\,R\) (product = fresh feed at steady state).

  2. Circulating load: \(\dfrac{R}{F} = \dfrac{p - m}{m - r} = \dfrac{0.85 - 0.40}{0.40 - 0.15} = 1.8\)

  3. Flows: \(R = 1.8 \times 50 = 90\) t/h; \(M = 50 + 90 = 140\) t/h.

AnswerCirculating load 180% (R/F = 1.8); the mill handles 140 t/h.
Example 3Circulating load from classifier efficiencyExam-style

A mill grinds 40 t/h of fresh feed in closed circuit. 30% of the mill discharge is finer than the product size, and the classifier sends 80% of those fines to the product and returns all the oversize. Find C/F and the mill throughput.

  1. Product balance: fines leaving \(= E\,a\,M = F\) → \(M = \dfrac{F}{E\,a} = \dfrac{40}{0.8 \times 0.3} = 166.7\) t/h.

  2. Circulating load: \(C/F = \dfrac{1}{E\,a} - 1 = \dfrac{1}{0.24} - 1 = 3.17\), so \(R = 126.7\) t/h.

  3. Comment: raising the classifier efficiency to 0.95 would cut C/F to 2.5; a finer mill discharge (larger \(a\)) also lowers it.

AnswerC/F = 3.17 (317%); the mill passes about 167 t/h.