Question 4 of 5: Sieve Analysis — Cumulative Size Distribution
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Mining and Mineral Processing Engineering, 09-MMP-A3 Mineral Processing, 2013-May. 3 hours duration, closed book; only an approved Casio or Sharp calculator permitted. Question 3 allows a choice of any six (6) of nine sub-terms. Question 5 and its bonus (page 5) were to be handed in with the exam booklet; Question 4's log-log plot (page 6) is reproduced here as a computed inline figure.
Reference texts: Wills & Finch, Wills' Mineral Processing Technology, 8th ed. (flotation circuit design and metallurgical balances – Ch. 12; comminution, Bond's law and circulating load – Ch. 3 & 6; particle size analysis – Ch. 4; sampling theory, Gy's equation – Ch. 3; gravity concentration, dense medium separation, magnetic/electrostatic separation – Ch. 10, 11 & 13); Taggart, Handbook of Mineral Dressing (classical economic-recovery/economic-efficiency formula used in Question 1(vi)).
Question 4: Sieve Analysis — Cumulative Size Distribution (10 marks)
Given. The sieve mass distribution in the table above; total sample mass $160\ \text{g}$.
Find. The cumulative % passing curve; the 50%-passing (mass median) size; the mass % between 75 and 37 μm.
Approach. Convert the retained-on-each-screen masses to cumulative % passing at each screen boundary, plot on log-log axes (Gates–Gaudin–Schuhmann distributions plot as straight lines on log-log paper), then read/interpolate $d_{50}$ and extrapolate the trend of the finest measured interval to 37 μm.
Cumulative % passing at each screen. Total mass $=17+31+24+19+15+12+42=160\ \text{g}$; cumulative retained above each screen, then % passing $=100\times(160-\text{cum.\ retained})/160$:
$$425\,\mu\text{m}:89.4\%,\ \ 300\,\mu\text{m}:70.0\%,\ \ 212\,\mu\text{m}:55.0\%,\ \ 150\,\mu\text{m}:43.1\%,\ \ 106\,\mu\text{m}:33.8\%,\ \ 75\,\mu\text{m}:26.3\%$$
(the $-75\,\mu\text{m}$ pan fraction, $42/160=26.3\%$ of the sample, matches the cumulative % passing 75 μm exactly, confirming the arithmetic).
Plot on log-log paper.
Figure 4.1 — Cumulative % passing vs. particle size, log-log axes. The 50% line is interpolated between the 150 μm and 212 μm points (d50); the curve's last measured segment (106→75 μm) is extended to 37 μm to estimate the −75+37 μm fraction.
(i) Mass median size, $d_{50}$. 50% passing lies between the 150 μm (43.1%) and 212 μm (55.0%) points; interpolating log-log,
$$\log d_{50}=\log150+\frac{\log50-\log43.1}{\log55.0-\log43.1}(\log212-\log150)\ \Rightarrow\ \boxed{d_{50}\approx185\ \mu\text{m}}$$
(a straight-line interpolation on the plotted data gives 186 μm – consistent to within plotting accuracy).
(ii) −75+37 μm fraction. 37 μm is below the finest sieved screen (75 μm), so % passing 37 μm is estimated by extending the slope of the last measured segment (106→75 μm) on the log-log plot:
$$\%\text{passing }37\,\mu\text{m}\approx15.7\%$$
$$\%(-75+37\,\mu\text{m}) = \%\text{passing }75-\%\text{passing }37 = 26.3-15.7=\boxed{10.5\%}$$
Check. 37 μm falls outside the sieved range (finest screen is 75 μm), so the −75+37 μm estimate in step 4 extrapolates the measured log-log trend one interval below the data – standard graphical practice, but less certain than an interpolated value; a direct 37 μm (400-mesh) sieve or a laser-diffraction check would firm this figure up.
