Question 7 of 9: Particle-Size Averages from a Screen Analysis, and the Sphericity of a Cube

Question 7: Particle-Size Averages from a Screen Analysis, and the Sphericity of a Cube (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.

(a) Given. The particle-count size distribution \((D_{pi},N_i)\) tabulated above; five size classes from 0.2 to 0.6 cm.

Find. The length mean diameter \(\bar D_L\), surface mean diameter \(\bar D_S\), and volume-surface (Sauter) mean diameter \(\bar D_{vs}\).

Approach. Each mean diameter is a different moment ratio of the same count distribution: the length mean is the first moment, the surface mean is the square root of the second moment, and the volume-surface mean is the ratio of the third to the second moment.

1. Length mean diameter. $$\bar D_L = \frac{\sum N_iD_{pi}}{\sum N_i} = \frac{17.40}{45} = \boxed{0.3867\ \text{cm}}.$$
2. Surface mean diameter. $$\bar D_S = \left(\frac{\sum N_iD_{pi}^2}{\sum N_i}\right)^{1/2} = \left(\frac{7.12}{45}\right)^{1/2} = \boxed{0.3978\ \text{cm}}.$$
3. Volume-surface (Sauter) mean diameter. $$\bar D_{vs} = \frac{\sum N_iD_{pi}^3}{\sum N_iD_{pi}^2} = \frac{3.06}{7.12} = \boxed{0.4298\ \text{cm}}.$$ Each mean is built from a progressively higher moment of the distribution, so \(\bar D_L < \bar D_S < \bar D_{vs}\) here, exactly as expected — the Sauter mean weights the larger (0.4–0.6 cm) particles most heavily because it is dominated by the volume/surface ratio those particles contribute.

(b) Given. A cube of side length \(W\). Find. Its sphericity \(\phi\) (ratio of the surface area of a volume-equivalent sphere to the cube's own surface area). Approach. Find the diameter \(D_p\) of a sphere with the same volume as the cube, then compare surface areas.

1. Equivalent sphere diameter. Equating volumes, $$\frac{\pi}{6}D_p^3 = W^3 \;\Rightarrow\; D_p = W\left(\frac{6}{\pi}\right)^{1/3}.$$
2. Form the sphericity ratio. Sphericity is the equivalent sphere's surface area divided by the actual (cube) surface area: $$\phi = \frac{\pi D_p^2}{6W^2} = \frac{\pi\, W^2(6/\pi)^{2/3}}{6W^2} = \left(\frac{\pi}{6}\right)^{1/3} = \boxed{0.806}.$$ The cube's side length \(W\) cancels completely — sphericity is a pure shape factor, independent of size, exactly as it must be to compare particles of different sizes on their shape alone.