Question 1 of 8: Crushing circuit size distribution and circulating load

Question 1: Crushing circuit size distribution and circulating load (1/6)

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. Circuit-product cumulative % passing at four log-log stations: (2 mm, 46.3%), (6 mm, 67.1%), (9 mm, 85.3%), (15 mm, 99.2%). Screen aperture 15 mm; screen feed (crusher discharge) is 51.0% passing 15 mm; circuit product (screen undersize) is 99.2% passing 15 mm; screen efficiency $E=90\%$ (fraction of the -15 mm material in the feed that reports to the undersize).

Find. (a) The Gates-Gaudin-Schuhmann (GGS) distribution modulus $m$ and size modulus $K$. (b) The circulating load ratio of stream T (oversize returned to the crusher) as a percentage of new/product feed.

Approach. (a) Fit $Y=100(x/K)^m$ by least-squares linear regression of $\ln Y$ against $\ln x$ over the four stations; (b) close a two-product mass balance around the screen using the feed/product -15 mm fractions and the stated efficiency.

1. Linearize the GGS model. $Y=100(x/K)^m \Rightarrow \ln(Y/100) = m\ln x - m\ln K$, a straight line in $\ln x$ vs. $\ln(Y/100)$. $$\begin{aligned} x_i &= (2,\,6,\,9,\,15)\ \text{mm}, \quad \ln x_i = (0.693,\,1.792,\,2.197,\,2.708) \\ Y_i &= (46.3,\,67.1,\,85.3,\,99.2)\%, \quad \ln(Y_i/100) = (-0.770,\,-0.399,\,-0.159,\,-0.008) \end{aligned}$$
2. Least-squares slope and intercept. With $n=4$, $\sum\ln x = 7.390$, $\sum\ln(Y/100) = -1.336$, $\sum(\ln x)^2 = 14.514$, $\sum \ln x\cdot\ln(Y/100) = -1.936$: $$m=\frac{n\sum xy - \sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{4(-1.936)-(7.390)(-1.336)}{4(14.514)-(7.390)^2}=0.386$$ The intercept gives the size modulus (100% passing size, extrapolated): $\boxed{K = 15.1\ \text{mm}}$, with distribution modulus $\boxed{m = 0.386}$.
3. Screen mass balance (basis: screen feed $F=1$). Let $f=0.510$ (fraction -15 mm in the screen feed = crusher discharge), $u=0.992$ (fraction -15 mm in the undersize = circuit product), and efficiency $E=(Uu)/(Ff)=0.90$ as defined in the question. $$U = \frac{Ef}{u}F = \frac{0.90\times0.510}{0.992}(1) = 0.4627$$
4. Oversize (stream T) and circulating load ratio. By overall mass balance $O=F-U$, and the product leaving the circuit equals the undersize ($P=U$), so the circulating load ratio is referenced to the new/product feed: $$O = 1-0.4627 = 0.5373 \qquad \text{C.L.}=\frac{O}{U}\times100\% = \frac{0.5373}{0.4627}\times100\%=\boxed{116.1\%}$$