1. TUTORIAL 1 PARTICULATE SOLIDS

2. Example 1
The screen analysis shown in Table 28.1 applies to a sample of crushed quartz. The density of the particles is 2,650 kg/m3 ( g/mm3), and the shape factors are a = 0.8 and Фs = For the material between 4-mesh and 200-mesh in particle size, calculate:
Aw in square millimeters per gram and Nw in particles per gram Dv Ds
Dw and
Ni for the 150/200-mesh increment
What fraction of the total number of particles is in the 150/200-mesh increment

3. Solution 1
To find Aw and Nw,

5. For the 4/6-mesh increment Dpi is the arithmetic mean of the mesh openings of the defining screens; or, from the Table 28.1,
( ) / 2 = mm
For this increment xi =
hence xi /Dpi = /4.013 =
xi /Dpi3 =
Corresponding quantities are calculated for the other 11 increment to give:
Σ xi /Dpi = and
Σ xi /Dpi3 =

6. Table 28.1 Screen analysis Mesh Dpi xi Avg Dpi 4 4.699 6 3.327 0.02516
3.327
0.0251
4.013 8
2.362
0.1250 10
1.651
0.3207 14
1.168
0.2570 20
0.833
0.1590 28
0.589
0.0538 35
0.417
0.0210 48
0.295
0.0102 65
0.208
0.0077
100
0.147
0.0058
150
0.104
0.0041
200
0.074
0.0031
Pan -
0.0075

7. Mesh
Dpi xi
Avg Dpi
xi/Avg. Dpi
xi/(Avg. Dpi)^3 4
4.699 6
3.327
0.0251
4.013
0.0063
0.0004 8
2.362
0.1250
2.845
0.0439
0.0054 10
1.651
0.3207
2.007
0.1598
0.0397 14
1.168
0.2570
1.409
0.1824
0.0919 20
0.833
0.1590
1.001
0.1588
0.1585 28
0.589
0.0538
0.711
0.0757
0.1497 35
0.417
0.0210
0.503
0.0417
0.1650 48
0.295
0.0102
0.356
0.0287
0.2261 65
0.208
0.0077
0.252
0.0306
0.4812
100
0.147
0.0058
0.178
0.0326
1.0284
150
0.104
0.0041
0.126
0.0325
2.0496
200
0.074
0.0031
0.089
0.0348
4.3974
sum
0.8278
8.7932

8. For the material between 4-mesh and 200-mesh in particle size, the pan fraction is excluded.
Therefore, the fraction of the material between 4-mesh and 200-mesh = 1 – =
Therefore, Aw = 3307 mm2/g and Nw = 4179 particles/g.

9. (b) (c) The volume-surface mean diameter is calculated from:

10. (d) Mass mean diameter Dw is obtained from this equation:xi
Avg Dpi
xi*Avg. Dpi
0.0000
0.0251
4.013
0.1007
0.125
2.845
0.3556
0.3207
2.007
0.6436
0.257
1.409
0.3621
0.159
1.001
0.1592
0.0538
0.711
0.0383
0.021
0.503
0.0106
0.0102
0.356
0.0036
0.0077
0.252
0.0019
0.0058
0.178
0.0010
0.0041
0.126
0.0005
0.0031
0.089
0.0003
Sum
1.6772

11. (e) The number of particles in the 150/200-mesh increment is found from this equation: This is 2074/4179 = 0.496, or 49.6 percent of the particles in the top 12 increments. For the materials in the pan fraction, the number of particles and specific area are enormously greater than for the coarser materials, but they cannot be accurately estimated from the data in Table 28.1.

12. TUTORIAL 2 SIZE REDUCTION

13. Exercise 1
What is the power required to crush 100 ton/h of limestone if 80% of the feed passes a 2-in. screen and 80% of the product a 1/8-in. screen?

15. Solution 1
From Table 28.2,the work index for limestone is Other quantities for substitution into
are:
The power required is:

16. Exercise 2
It is desired to crush 20 ton/h of iron ore (hematite). The size of the feed is such that 80% passes a 3 inch screen and 80% of the product is pass to 1/8 inch screen. Calculate the gross power required in SI unit.

17. Solution 2
From Table 28.2,the work index for iron ore (hematite) is

18. Exercise 3
Calculate the power ratio to crush the coffee extract from 38 mm to 17 mm and from 10 mm to 6 mm. Use Rittinger’s law.

19. Solution 3
From Rittinger’s Law:

20. THANK YOU