Lesson 15. Empirical laws of Size Reduction

15.1 Rittinger’s law

Rittinger’s assumption was based on the fact that particles do not deform before breaking, therefore being infinitely brittle. Rittinger state that the energy required in size reduction is proportional to new surface created and gave the power n a value of 2 thus by integrating equation (1.9) obtained the so called Rittinger’s law:

${{dE} \over {dx}}$ =  $- C\,x^{ - 2}$

$\int\limits_0^E {dE}$ =  $- C\,\int\limits_{x_f }^{x_p } {x^{ - 2} } \,dx$
this can be evaluated to give

E = $C_R \left( {{1 \over {x_p }} - {1 \over {x_f }}\,} \right)$     ........................... (4.10)

where, E is the energy per unit mass required for the production of a new surface by reduction, CR is called Rittinger’s constant  which has a unit of  J m/kg and is determined for a particular equipment and material, xf is the average initial feed size, and xp is the average final product size. Rittinger’s law has been found to hold better for fine grinding, where a large increase in surface results. It suggests that the energy required is proportional to the increase in surface area per unit mass.

15.2 Kick’s law

Kick state that the energy required for a given size reduction was proportional to the sizen reduction ratio, and gave the power n a value of 1. Thus by integration of Eq. (13.9), the following relationship, known as Kick’s law is obtained:

$\int\limits_0^E {dE}$ =  $- C\,\int\limits_{x_f }^{x_p } {x^{ - 1} } \,dx$

this gives

E =$C_K \,\ln \,\left( {{{x_f } \over {x_p }}} \right)$ .....................   (4.11)

where xf/xp is the size reduction ratio and CK is called Kick’s constant which unit of J/kg . Kick’s law has been found to hold more accurately for coarser crushing, where most of the energy is used in causing fracture along existing cracks. In Kick’s law the energy input is proportional to the size reduction ratio. In other words the same energy input is required for a reduction in size from 1 cm to 1 mm as for a reduction from 100 to 10 μm. In using either of the models due to Rittinger and Kick the relevant constant must be obtained by experiment using both the same equipment and the same material.

15.3 Bond’s law and Work Index

Bond state that the energy required for size reduction is proportional to the square root of the surface- volume ratio of the product. In Bond’s consideration n takes the value of 3/2, yielding the following version (Bond’s law), also by integrating Eq. (13.9):

$\int\limits_0^E {dE}$ =$- C\,\int\limits_{x_f }^{x_p } {x^{ - {3 \over 2}} } \,dx$

E = $2\,C_B \left[ {x^{ - {1 \over 2}} } \right]_{x_f }^{x_p }$
E= $2\,C_B \left( {{1 \over {\sqrt {x_p } }} - {1 \over {\sqrt {x_f } }}\,} \right)$   ............................... (4.12a)
where xf and xp are measured in micrometers and E in kWh/ton, Bond put the constant CB = 5Wi, where Wi is the Bond Work Index, defined as the gross energy requirement in kWh/ton of feed to reduce a unit mass of material from an infinite particle size to a size such that 80% (i.e., Φ = 2) of the product passing through a 100 µm sieve (i.e, xp = 100 µm) . The Bond Work Index is material specific and is obtained from laboratory crushing tests on the feed material. Hard and brittle materials have a work index in the range 4 × 104 − 8 × 104J/kg. Bond’s law holds reasonably well for a variety of materials undergoing coarse, medium and fine size reduction. (Eq. 14.3a) can also be reduced to

E= $0.3162\,\,W_i \left( {{1 \over {\sqrt {x_p } }} - {1 \over {\sqrt {x_f } }}\,} \right)$..............................(4.12b)

where xf and xp are measured in mm and E in kWh/ton, and Wi is the Bond Work Index,

Worked Example 4.1

A material consisting originally of 20 mm particle crushed to an average size of 5 mm and required 30 kJ/kg for this size reduction. Determine the energy required to crush the material from 20 mm to 3 mm assuming (a) Rittinger’s law, (b) Kick’s law and (c) Bond’s Law.

Solution:

(a)    Applying Rittinger’s law as expressed by equation (4.10)

20 = $C_R \left( {{1 \over 5} - {1 \over {20}}\,} \right)$

Hence CR = 133.33 and so  with xp= 3 mm

E = $133.33\left( {{1 \over 3} - {1 \over {20}}\,} \right)$

Hence E = 37.78 kJ/kg

(b)  Applying Kick’s law as expressed by equation (4.11)

20 = $C_K \,\ln \left( {{{20} \over 5}} \right)$
Hence CR = 14.43 and so  with xp= 3 mm

E =$14.43\,\ln \left( {{{20} \over 3}} \right)$

Hence E = 27.36 kJ/kg

(c)  Applying Bond’s law as expressed by equation (4.12b)

20 =$0.3162\,W_{i\,} \left( {{1 \over {\sqrt 5 }} - {1 \over {\sqrt {20} }}\,} \right)$

Hence Wi = 282.87 and so with xp = 3 mm
E = $0.3162\,\, \times \,282.87\,\,\left( {{1 \over {\sqrt 3 }} - {1 \over {\sqrt {20} }}\,} \right)$

Hence E = 31.64 kJ/kg

Worked Example 4.2

Grain is milled at a rate of 10 t/h and the power required for this operation is 67.5 kW. Assuming that Bond’s law best describes the relationship between energy required and change in particle size, determine the work index for the grain and thus find the total power requirement to mill down to a distribution where 80% passes 100 μm.

 Initial distribution Final distribution Sieve size (μm) Mass fraction Sieve size (μm) Mass fraction 6730 0.00 605 0.00 4760 0.05 425 0.08 3360 0.15 300 0.12 2380 0.70 212 0.65 1680 0.10 150 0.11 100 0.04

Solution:

From the tabulated data, 80% of the grain in the initial distribution passes a 3360 µm sieve and therefore xf = 3360 µm = 3.36 mm. Similarly final distribution gives xp = 300 µm = 0.3mm.

The power requirement to achieve this degree of size reduction is 67.5 kW. Dividing this figure by the mass flow rate of grain in kg/s gives the energy input per unit mass. Thus

E = ${{67.5\, \times \,3600} \over {10\, \times \,10^3 }}$
E = 24.3 kJ/kg

The work index can be found from Eq. (4.12b)

24.3 = $0.3162\,\,W_i \left( {{1 \over {\sqrt {0.3} }} - {1 \over {\sqrt {3.36} }}\,} \right)$

From which, Wi  = 60 kJ/kg

For a final grain size distribution where xp = 100 μm = 0.1, Substituting this figure and the work index into Eq. (14.3b) gives the required energy input as

E =$0.3162\,\, \times \,60\,\left( {{1 \over {\sqrt {0.1} }} - {1 \over {\sqrt {3.36} }}\,} \right)\, \times \,{{10^4 } \over {3600}}\,$ kW
Thus E = 137.9 kW

The power requirement for the problem in example 1.2 calculated by using Rittinger’s law gave 215.6 kW and Kick’s law give a value of 98.2 kW. Thus Rittinger’s law tends to overestimate the energy input required for size reduction whereas Kick’s law usually gives an underestimate.