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The assessment of mixed small volumes, which can be taken or sampled, is what mixing measurement is all about. Sample compositions move from the initial state to the mixed state, and measurements of mixing must reflect this.

The problem at once arises, what size of sample should be chosen? To take extreme cases, if the sample is so large that it includes the whole mixture, then the sample composition is at once the mean composition and there remains no mixing to be done. At the other end of the scale, if it were possible to take samples of molecular size, then every sample would contain only one or other of the components in the pure state and no amount of mixing would make any difference. Between these lie all of the practical sample sizes, but the important point is that the results will depend upon sample size.

In many practical mixing applications, process conditions or product requirements prescribe suitable sample sizes. For example, if table salt is to contain 1% magnesium carbonate, the addition of 10 kg of magnesium carbonate to 990 kg of salt ensures, overall, that this requirement has been met.

However, if the salt is to be sold in 2 kg packets, the practical requirement might well be that each packet contains 20 g of magnesium carbonate with some specified tolerance, and adequate mixing would have to be provided to achieve this.

A realistic sample size to take from this mixture, containing 1000 kg of mixture, would be 2 kg. As mixing proceeds, greater numbers of samples containing both components appear and their composition tends towards 99% salt and 1% magnesium carbonate.

It can be seen from this discussion that the deviation of the sample compositions from the mean composition of the overall mixture represents a measure of the mixing process. This deviation decreases as mixing progresses. A satisfactory way of measuring the deviation is to use the statistical term called the standard deviation. This is the mean of the sum of the squares of the deviations from the mean, and so it gives equal value to negative and positive deviation and increasingly greater weight to larger deviations because of the squaring. It is given by:

s2 = 1/n [(x1 - )2 + (x2 - )2 + …. + (xn - )2]                                                        (12.1)

where s is the standard deviation, n is the number of samples taken, x1, x2, ... xn , are the fractional compositions of component X in the 1, 2, ... n samples and is the mean fractional composition of component X in the whole mixture.

Using eqn. (12.1) values of s can be calculated from the measured sample compositions, taking the n samples at some stage of the mixing operation. Often it is convenient to use s2 rather than s, and s2 is known as the variance of the fractional sample compositions from the mean composition.

EXAMPLE 12.1. Mixing salt and magnesium carbonate
After a mixer mixing 99 kg of salt with 1 kg of magnesium carbonate had been working for some time, ten samples, each weighing 20 g, were taken and analysed for magnesium carbonate. The weights of magnesium carbonate in the samples were: 0.230, 0.172, 0.163, 0.173, 0.210, 0.182, 0.232, 0.220, 0.210,  0.213g. Calculate the standard deviation of the sample compositions from the mean composition.

Fractional compositions of samples, that is the fraction of magnesium carbonate in the sample, are respectively:

0.0115, 0.0086, 0.0082, 0.0087, 0.0105, 0.0091, 0.0116, 0.0110, 0.0105, 0.0107 (x)

Mean composition of samples, overall = 1/100 = 0.01 ()

Deviation of samples from mean, (0.0115 - 0.01), (0.0086 - 0.01), etc.

s2 = 1/10[(0.0115 - 0.01)2 + (0.0086 - 0.01)2 + ...]
    = 2.250 x 10-6

 s = 1.5 x 10-3.

At some later time samples were found to be of composition: 0.0113, 0.0092, 0.0097, 0.0108, 0.0104, 0.0098, 0.0104, 0.0101, 0.0094, 0.0098, giving:
         s = 3.7 x 10-7.
and showing the reducing standard deviation. With continued mixing the standard deviation diminishes further.

The process of working out the differences can be laborious, and often the standard deviation can be obtained more quickly by making use of the mathematical relationship, proof of which will be found in any textbook on statistics:

s2 = 1/n[S(x12) - S()2]
    = 1/n[
S(x12) - n)2]
    = 1/n[
S(x12)] - ()2.


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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology