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; screening 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 $m$, size modulus $K$, and $d_{50}$, $d_{80}$, $d_{20}$ for the circuit product. (b) The circulating load ratio of stream T (oversize returned to the crusher via the bin) as a percentage of the 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 stipulated stations, then invert the fit for $d_{50}$, $d_{80}$, $d_{20}$; (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. $\ln(Y/100)$. $$x_i=(2,6,9,15)\ \text{mm},\quad Y_i=(46.3,67.1,85.3,99.2)\%$$
2. Least-squares slope and intercept. With $n=4$, regressing $\ln Y_i$ on $\ln x_i$: $$m=\frac{\sum(\ln x_i-\overline{\ln x})(\ln Y_i-\overline{\ln Y})}{\sum(\ln x_i-\overline{\ln x})^2}=0.386$$ $$K=\exp\!\left(\frac{\ln100-b}{m}\right)=\boxed{15.1\ \text{mm}}\qquad m=\boxed{0.386}$$ ($K$ is the extrapolated 100%-passing size — the size modulus — not a value actually tabulated.)
3. Invert the fit for the requested size fractions. $d_P = K(P/100)^{1/m}$: $$d_{50}=15.1(0.50)^{1/0.386}=\boxed{2.50\ \text{mm}}\qquad d_{80}=15.1(0.80)^{1/0.386}=\boxed{8.45\ \text{mm}}$$ $$d_{20}=15.1(0.20)^{1/0.386}=\boxed{0.233\ \text{mm}}\qquad \frac{d_{80}}{d_{20}}=\frac{8.45}{0.233}=\boxed{36.3}$$ ($d_{50}$, $d_{80}$ cross-check within a few percent against direct linear interpolation of the full table — 2.58 mm and 8.28 mm respectively — since the fit uses only the four stipulated stations; $d_{20}$ lies below the smallest tabulated size, 0.5 mm, so it is necessarily a GGS extrapolation.)
4. 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), efficiency $E=(Uu)/(Ff)=0.90$: $$U=\frac{Ef}{u}F=\frac{0.90\times0.510}{0.992}=0.4627$$
5. Oversize (stream T) and circulating load ratio. $O=F-U$, and the product leaving the circuit equals the undersize ($P=U$), so: $$O=1-0.4627=0.5373\qquad \text{C.L.}=\frac{O}{U}\times100\%=\frac{0.5373}{0.4627}\times100\%=\boxed{116.1\%}$$