CHE4162 – TUTORIAL 2 - SOLUTIONS
PARTICLE SIZE ANALYSIS (Chapter 1)
EXERCISE 1.1:
For a regular cuboid particle of dimensions 1.00 x 2.00 x 6.00 mm, calculate the
following diameters:
(a) equivalent volume sphere diameter
(b) equivalent surface sphere diameter
(c) surface-volume diameter (the diameter of a sphere having the same external
surface to volume ratio as the particle).
(d) sieve diameter (the width of the minimum aperture through which the particle will
pass)
(e) projected area diameters (the diameter of a circle having the same area as the
projected area of the particle resting in a stable position).
SOLUTION TO EXERCISE 1.1:
(a) volume of cuboid = 12 mm3
If xv is the equivalent volume sphere diameter, then π
6
x
3v
= 12
Hence, xv = 2.840 mm.
(b) surface area of cuboid = 6 ( ) × 2 × 2 +(6 ×1 × 2)+ (1 × 2 × 2)= 40 mm2
If xs is the equivalent surface sphere diameter, then πxs 2 = 40 . Hence, xs = 3.568mm.
(c) Surface to volume ratio of the cuboid = 40
12
= 3.333 mm2 / mm3
If xsv is the surface-volume sphere diameter, then 6
x
sv
= 3.333 (surface-volume ratio
for a sphere of diameter x is 6/x). Hence, xsv = 1.8 mm.
(d) Sieve diameter is the second largest dimension, i.e. 2 mm.
(e) The cuboid has three stable resting positions and so has three projected areas:
projected area 1 = 6 mm2
projected area 2 = 2 mm2
projected area 3 = 12 mm2
If x
p is the projected area diameter, then