When a uniform particle is crushed, after the first crushing the size of the particles produced will vary a great deal from relatively coarse to fine and even to dust. As the grinding continues, the coarser particles will be further reduced but there will be less change in the size of the fine particles. Careful analysis has shown that there tends to be a certain size that increases in its relative proportions in the mixture and which soon becomes the predominant size fraction. For example, wheat after first crushing gives a wide range of particle sizes in the coarse flour, but after further grinding the predominant fraction soon becomes that passing a 250 mm sieve and being retained on a 125 mm sieve. This fraction tends to build up, however long the grinding continues, so long as the same type of machinery, rolls in this case, is employed.
The surface area of a fine particulate material is large and can be important. Most reactions are related to the surface area available, so the surface area can have a considerable bearing on the properties of the material. For example, wheat in the form of grains is relatively stable so long as it is kept dry, but if ground to a fine flour has such a large surface per unit mass that it becomes liable to explosive oxidation, as is all too well known in the milling industry. The surface area per unit mass is called the specific surface. To calculate this in a known mass of material it is necessary to know the particle-size distribution and, also the shape factor of the particles. The particle size gives one dimension that can be called the typical dimension, Dp, of a particle. This has now to be related to the surface area.
We can write, arbitrarily:Vp = pDp3and
Ap= 6qDp2.
where Vp is the volume of the particle, Ap is the area of the particle surface, Dp is the typical dimension of the particle and p, q are factors which connect the particle geometries.(Note subscript p and factor p)
For example, for a cube, the volume is Dp3 and the surface area is 6Dp2; for a sphere the volume is (p/6)Dp3 and the surface area is pDp2 In each case the ratio of surface area to volume is 6/Dp.
A shape factor is now defined as q/p = l (lambda), so that for a cube or a sphere l = 1. It has been found, experimentally, that for many materials when ground, the shape factor of the resulting particles is approximately 1.75, which means that their surface area to volume ratio is nearly twice that for a cube or a sphere.
The ratio of surface area to volume is:Ap/Vp =( 6q/p)Dp = 6l/Dp (11.5)
and so Ap= 6q Vp/pDp = 6l(VP/DP)
If there is a mass m of particles of density rp, the number of particles is m/rpVP each of area Ap.
So total area At = (m/rpVP) x ( 6qVp/pDP) = 6qm/rp pDp
= 6lm/rDp (11.6)
where At is the total area of the mass of particles. Equation (11.6) can be combined with the results of sieve analysis to estimate the total surface area of a powder.
EXAMPLE 11.2. Surface area of salt crystals
In an analysis of ground salt using Tyler sieves, it was found that 38% of the total salt passed through a 7 mesh sieve and was caught on a 9 mesh sieve. For one of the finer fractions, 5% passed an 80 mesh sieve but was retained on a 115 mesh sieve. Estimate the surface areas of these two fractions in a 5 kg sample of the salt, if the density of salt is 1050 kg m-3 and the shape factor (l) is 1.75.
Aperture of Tyler sieves, 7 mesh = 2.83 mm, 9 mesh = 2.00 mm, 80 mesh = 0.177 mm, 115 mesh = 0.125 mm.
Mean aperture 7 and 9 mesh = 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Now from Eqn. (11.6)
Mean aperture 7 and 9 mesh = 2.41 mm = 2.4 x 10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x 10-3m
Now from Eqn. (11.6)
A1 = (6 x 1.75 x 0.38 x 5)/(1050 x 2.41 x 10-3)
= 7.88 m2
A2 = (6 x 1.75 x 0.05 x 5)/(1050 x 0.151 x 10-3)
= 16.6 m2.
http://www.nzifst.org.nz/unitoperations/sizereduction1.htm
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