Grinding and cutting reduce the size of solid materials by mechanical action, dividing them into smaller particles. Perhaps the most extensive application of grinding in the food industry is in the milling of grains to make flour, but it is used in many other processes, such as in the grinding of corn for manufacture of corn starch, the grinding of sugar and the milling of dried foods, such as vegetables.
Cutting is used to break down large pieces of food into smaller pieces suitable for further processing, such as in the preparation of meat for retail sales and in the preparation of processed meats and processed vegetables.
In the grinding process, materials are reduced in size by fracturing them. The mechanism of fracture is not fully understood, but in the process, the material is stressed by the action of mechanical moving parts in the grinding machine and initially the stress is absorbed internally by the material as strain energy. When the local strain energy exceeds a critical level, which is a function of the material, fracture occurs along lines of weakness and the stored energy is released. Some of the energy is taken up in the creation of new surface, but the greater part of it is dissipated as heat. Time also plays a part in the fracturing process and it appears that material will fracture at lower stress concentrations if these can be maintained for longer periods. Grinding is, therefore, achieved by mechanical stress followed by rupture and the energy required depends upon the hardness of the material and also upon the tendency of the material to crack - its friability.
The force applied may be compression, impact, or shear, and both the magnitude of the force and the time of application affect the extent of grinding achieved. For efficient grinding, the energy applied to the material should exceed, by as small a margin as possible, the minimum energy needed to rupture the material . Excess energy is lost as heat and this loss should be kept as low as practicable.The important factors to be studied in the grinding process are the amount of energy used and the amount of new surface formed by grinding.
Grinding is a very inefficient process and it is important to use energy as efficiently as possible. Unfortunately, it is not easy to calculate the minimum energy required for a given reduction process, but some theories have been advanced which are useful.
These theories depend upon the basic assumption that the energy required to produce a change dL in a particle of a typical size dimension L is a simple power function of L:
dE/dL = KLn (11.1)
where dE is the differential energy required, dL is the change in a typical dimension, L is the magnitude of a typical length dimension and K, n, are constants.
Kick assumed that the energy required to reduce a material in size was directly proportional to the size reduction ratio dL/L. This implies that n in eqn. (11.1) is equal to -1. If
K = KKfcwhere KK is called Kick's constant and fc is called the crushing strength of the material, we have:
dE/dL = KKfcL-1which, on integration gives:
E = KKfc loge(L1/L2) (11.2)
Rittinger, on the other hand, assumed that the energy required for size reduction is directly proportional, not to the change in length dimensions, but to the change in surface area. This leads to a value of -2 for n in eqn. (11.1) as area is proportional to length squared. If we put:
K = KRfcwhere KR is called Rittinger's constant, and integrate the resulting form of eqn. (11.1), we obtain:
and so
dE/dL = KRfcL-2
E = KRfc(1/L2– 1/L1) (11.3)
Equation (11.3) is known as Rittinger's Law. As the specific surface of a particle, the surface area per unit mass, is proportional to 1/L, eqn. (11.3) postulates that the energy required to reduce L for a mass of particles from 10 cm to 5 cm would be the same as that required to reduce, for example, the same mass of 5 mm particles down to 4.7 mm. This is a very much smaller reduction, in terms of energy per unit mass for the smaller particles, than that predicted by Kick's Law.
It has been found, experimentally, that for the grinding of coarse particles in which the increase in surface area per unit mass is relatively small, Kick's Law is a reasonable approximation. For the size reduction of fine powders, on the other hand, in which large areas of new surface are being created, Rittinger's Law fits the experimental data better.
Bond has suggested an intermediate course, in which he postulates that n is -3/2 and this leads to:E = Ei (100/L2)1/2[1 - (1/q1/2)] (11.4)
Bond defines the quantity Ei by this equation: L is measured in microns in eqn. (11.4) and so Ei is the amount of energy required to reduce unit mass of the material from an infinitely large particle size down to a particle size of 100 mm. It is expressed in terms of q, the reduction ratio where q = L1/L2.
Note that all of these equations [eqns. (11.2), (11.3), and (11.4)] are dimensional equations and so if quoted values are to be used for the various constants, the dimensions must be expressed in appropriate units. In Bond's equation, if L is expressed in microns, this defines Ei and Bond calls this the Work Index.
The greatest use of these equations is in making comparisons between power requirements for various degrees of reduction.
EXAMPLE 11.1. Grinding of sugar
Sugar is ground from crystals of which it is acceptable that 80% pass a 500 mm sieve(US Standard Sieve No.35), down to a size in which it is acceptable that 80% passes a 88 mm (No.170) sieve, and a 5-horsepower motor is found just sufficient for the required throughput. If the requirements are changed such that the grinding is only down to 80% through a 125 mm (No.120) sieve but the throughput is to be increased by 50% would the existing motor have sufficient power to operate the grinder? Assume Bond's equation.
Using the subscripts 1 for the first condition and 2 for the second, and letting m kg h-1 be the initial throughput, then if x is the required power
Now 88mm = 8.8 x 10-6m , 125 mm = 125 x 10-6m , 500mm = 500 x 10-6m
Now 88mm = 8.8 x 10-6m , 125 mm = 125 x 10-6m , 500mm = 500 x 10-6m
E1 = 5/m = Ei(100/88 x 10-6)1/2 [1 – (88/500)1/2]
E2 = x/1.5m = Ei(100/125 x 10-6)1/2 [1 - (125/500)1/2]
E2/E1 = x/(1.5 x 5) = (88 x 10-6)1/2[1 - (125/500)1/2]
(125 x 10-6)1/2[1 – (88/500)1/2]
x/(7.5) = 0.84 x (0.500/0.58)
= 0.72
x = 5.4 horsepower.
So the motor would be expected to have insufficient power to pass the 50% increased throughput, though it should be able to handle an increase of 40%.
http://www.nzifst.org.nz/unitoperations/sizereduction1.htm
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