Particle Size Analysis

First the diameter of the particles to be analysed is measured. The measurement is most easily accomplished using a digital image processor which determines the chord length of the particles in 4 to 16 directions then averages thoses lengths to determine the average diameter.

Alternatively the area occupied by the particle is determined by counting the number of pixels that represent the particle and then calculating the diamter of the particle assuming a circular shape. The example above shows how a digtal image anlyser would sample the particle.

After collecting the size data for a sufficiently large number of particles, a histogram of particles sizes is created by dividing the maximum diameter by an integer k (7<kČ15) to define a size interval D .

It is then assumed that the actual diameter of particles in the first group is D, in the second group 2D and in the k’th group kD. The number of particles per unit volume in group 1 is N_V(1), in group 2 is N_V(2) and in group k is N_V(k).

Those particles in group 1 can only produce sections with diameters between 0 and D and so all sections from particles in group 1 will fall in the histogram bin#1. Those particles in group 2 can produce sections in both histogram bin#1 and bin#2. Continuing this arguement up through particles of size kD ie. particles in group k, we can see that the measured number of particles per unit area in any group N_A(i) is made up of the contribution of particles with diameters in that group and higher. Thus

Simple so far.....Now for the tricky bit...

We now need to work out the number of sections observed in unit area in the plane of polish that derive from a given number of particles per unit volume.

Take a particle from group j whose size is jD and consider a section such that the section falls in group i. This section will have diameters from iD to (i-1)D. The centre of the particle must lie a distance d from the plane of sections such that . For unit area on the plane of polish, the volume containing the centre of the particle is of width h_i-h_i-1 lying above and below the plane of polish

If in 1 mm³ of the sample there are N_V(j) particles with diameters equal to jD then the number of sections with diameters between (i-1)D and iD is given by

We can use pythagarus’ theorum to rewrite h_i and h_i-1 in terms of i,j and D

Having define dthe contribution of each particle size group to the number of observed sections in each group smaller than the particles actual size we can write out ...

Just for the sake of clarity we shall assume only 3 size groups but the principle works for any number of groups. We now substitute for each of the N_a(i,j) terms.

Note that in any value of N_A(i) the magnitude of the coefficients decrease rapidly with increasing values of j. i.e. the contribution of large particles to the smaller observed sections becomes insignificant very rapidly - just look at the micrograph below for confirmation (The particles are a monodispersion of 1/8” ball bearings in epoxy - note the paucity of small section sizes!). It should now be apparent why the maximum number of groupings is 15! as the values of the coefficients are less than 0.1%.

In the above set of 3 equations there are 3 measured quantities N_A(i) and 3 unknown quantities N_V(j). If we extende dthe analysis to the full 15 groups then there would be 15 equations, and 15 unknowns. The problem can therefore, be solved! The solution is carried out by backward substitution since the last equation has only 1 unknown..

The constants do not change regardless of the number of groups. They are summarised below

Particle Size Distributions Recitation Problem 1

The following data was recorded for the measurements of particle sizes. The size interval was 0.4mm and the area measured was 3000 mm².