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Mixing of Widely Different Quantities
Rates of Mixing
Energy Input in Mixing

If particles are to be mixed, starting out from segregated groups and ending up with the components randomly distributed, the expected variances (s2) of the sample compositions from the mean sample composition can be calculated.

Consider a two-component mixture consisting of a fraction p of component P and a fraction q of component Q. In the unmixed state virtually all small samples taken will consist either of pure P or of pure Q. From the overall proportions, if a large number of samples are taken, it would be expected that a proportion p of the samples would contain pure component P. That is their deviation from the mean composition would be (1 - p), as the sample containing pure P has a fractional composition 1 of component P. Similarly, a proportion q of the samples would contain pure Q, that is, a fractional composition 0 in terms of component P and a deviation (0 - p) from the mean. Summing these in terms of fractional composition of component P and remembering that p + q = 1.

so2 = 1/n [pn(1 - p)2 + (1 - p)n(0 - p)2] (for n samples)
      = p(1 - p)                                                          (12.2)

When the mixture has been thoroughly dispersed, it is assumed that the components are distributed through the volume in accordance with their overall proportions. The probability that any particle picked at random will be component Q will be q, and (1 - q) that it is not Q. Extending this to samples containing N particles, it can be shown, using probability theory, that:

sr2 = p(1 - p)/N = so2/N.                                                                                          (12.3)

This assumes that all the particles are equally sized and that each particle is either pure P or pure Q. For example, this might be the mixing of equal-sized particles of sugar and milk powder. The subscripts o and r have been used to denote the initial and the random values of s2, and inspection of the formulae, eqn. (12.2) and eqn. (12.3), shows that in the mixing process the value of s2 has decreased from p(1 - p) to 1/Nth of this value. It has been suggested that intermediate values between so2 and sr2 could be used to show the progress of mixing. Suggestions have been made for a mixing index, based on this, for example:

(M) = (so2 - s2)/(so2 - sr2)                                                                                         (12.4)

which is so designed that (M) goes from 0 to 1 during the course of the mixing process. This measure can be used for mixtures of particles and also for the mixing of heavy pastes.

EXAMPLE 12.2. Mixing of yeast into dough
For a particular bakery operation, it was desired to mix dough in 95 kg batches and then at a later time to blend in 5 kg of yeast. For product uniformity it is important that the yeast be well distributed and so an experiment was set up to follow the course of the mixing. It was desired to calculate the mixing index after 5 and 10 min mixing.
Sample yeast compositions, expressed as the percentage of yeast in 100 g samples were found to be:
After 5 min
      (%)       0.0  16.5  3.2  2.2   12.6  9.6  0.2  4.6  0.5  8.5
Fractional    0.0  0.165  0.032.......
After 10 min
       (%)      3.4  8.3    7.2  6.0   4.3    5.2  6.7  2.6  4.3  2.0
Fractional   0.034  0.083 ..........
Using the formula 1/n[
S(x12)] - ()2

Calculating   s52 = 3.0 x 10-3
                 s102 = 3.8 x 10-4

The value of   so2 = 0.05 x 0.95 = 4.8 x 10-2

and sr2 0 as the number of "particles" in a sample is very large,

         (M)5 = (4.8 - 0.3)/(4.8 - 0)
                = 0.93
        (M)10 = (4.8 - 0.04)/4.8 - 0)
                = 0.99.

Mixing of Widely Different Quantities

The mixing of particles varying substantially in size or in density presents special problems, as there will be gravitational forces acting in the mixer which will tend to segregate the particles into size and density ranges. In such a case, initial mixing in a mixer may then be followed by a measure of (slow gravitational) un-mixing and so the time of mixing may be quite critical.

Mixing is simplest when the quantities that are to be mixed are roughly in the same proportions. In cases where very small quantities of one component have to be blended uniformly into much larger quantities of other components, the mixing is best split into stages, keeping the proportions not too far different in each stage. For example, if it were required to add a component such that its final proportions in relatively small fractions of the product are 50 parts per million, it would be almost hopeless to attempt to mix this in a single stage. A possible method might be to use four mixing stages, starting with the added component in the first of these at about 30:1. In planning the mixing process it would be wise to take analyses through each stage of mixing, but once mixing times had been established it should only be necessary to make check analyses on the final product.

EXAMPLE 12.3. Mixing vitamin addition into powdered cereal
It is desired to mix vitamin powder at the level of 10-3 % by weight into a 1 tonne batch of powdered cereal. Two double-cone blenders are available, one (L) with a capacity of 100 to 500 kg powder and another (S) with a capacity of 1 to 10 kg. Both will mix adequately in 10 min so long as the minor constituent constitutes not less than 10%. Suggest a procedure for the mixing.

Total weight vitamin needed = 1000 x 10-5 kg = 10 g
Divide into two, 2 x 5 g as two final 500 kg batches will be needed. Then:

1. Hand blend 5 g and 50 g cereal mixture (1), and then (1) and 945 g cereal mixture (2). This may need analytical checking to set up a suitable hand-mixing procedure.
2. Take (2) + 9 kg cereal and mix in mixer (S)
mixture (3).
3. Take (3) + 90 kg cereal and mix in (L)
mixture (4).
4. Take (4) + 400kg cereal and mix in (L)
mixture (5) - product.
5. Repeat, with other 5 g vitamin and 500 kg cereal, steps (1) to (4).

Rates of Mixing

Once a suitable measure of mixing has been found, it becomes possible to discuss rates of accomplishing mixing. It has been assumed that the mixing index ought to be such that the rate of mixing at any time, under constant working conditions such as in a well-designed mixer working at constant speed, ought to be proportional to the extent of mixing remaining to be done at that time. That is,

         dM/dt = K[(1 - (M))]                                                                                         (12.5)

where (M) is the mixing index and K is a constant, and on integrating from t = 0 to t = t during which (M) goes from 0 to (M),

   [(1 - (M))] = e-Kt
or         (M) = 1 - e-Kt                                                                                              (12.6)

This exponential relationship, using (M) as the mixing index, has been found to apply in many experimental investigations at least over two or three orders of magnitude of (M). In such cases, the constant K can be related to the mixing machine and to the conditions and it can be used to predict, for example, the times required to attain a given degree of mixing.

EXAMPLE 12.4. Blending starch and dried vegetables for a soup mix
In a batch mixer, blending starch and dried-powdered vegetables for a soup mixture, the initial proportions of dried vegetable to starch were 40:60. If the variance of the sample compositions measured in terms of fractional compositions of starch was found to be 0.0823 after 300 s of mixing, for how much longer should the mixing continue to reach the specified maximum sample composition variance for a 24 particle sample of 0.02?

Assume that the starch and the vegetable particles are of approximately the same physical size.

Then taking the fractional content of dried vegetables to be p = 0.4,

        (1 - p) = (1 - 0.4) = 0.6
            so2 = 0.4 x 0.6 = 0.24
             sr2 = s02/N = 0.24/24 from eqn. (12.3)
                  = 0.01
Substituting in eqn. (12.4) we have:

            (M) = (so2 - s2)/(so2 - sr2)
                  = (0.24 - 0.0823)/(0.24 - 0.01)
                  = 0.685

Substituting in eqn.12.6

        e-300K = 1 - 0.685 = 0.315
         300K = 1.155,
              K = 3.85 x 10-3
For         s2 = 0.02,
             (M) = (0.24 - 0.02)/(0.24 - 0.01)
                  = 0.957
         0.957 = 1 - e-0.00385t
       - 0.043 = - e-0.00385t
         3.147 = 0.00385t

                        t = 817s, say 820 s,

the additional mixing time would be 820s - 300 s = 520 s.

Energy Input in Mixing

Quite substantial quantities of energy can be consumed in some types of mixing, such as in the mixing of plastic solids. There is no necessary connection between energy consumed and the progress of mixing: to take an extreme example there could be shearing along one plane in a sticky material, then recombining to restore the original arrangement, then repeating which would consume energy but accomplish no mixing at all. However, in well-designed mixers energy input does relate to mixing progress, though the actual relationship has normally to be determined experimentally. In the mixing of flour doughs using high-speed mixers, the energy consumed, or the power input at any particular time, can be used to determine the necessary mixing time. This is a combination of mixing with chemical reaction as flour components oxidize during mixing in air which leads to increasing resistance to shearing and so to increased power being required to operate the mixer.

EXAMPLE 12.5. Mixing time for bread dough
In a particular mixer used for mixing flour dough for breadmaking, it has been found that mixing can be characterized by the total energy consumed in the mixing process and that sufficient mixing has been accomplished when 8 watt hours of energy have been consumed by each kg of dough.
Such a mixer, handling 2000 kg of dough, is observed just after starting to be consuming 80 amperes per phase, which rises steadily over 10s to 400 amperes at which level it then remains effectively constant. The mixer is driven by a 440-volt, 3-phase electric motor with a power factor (cos
f)) of 0.89, and the overall mechanical efficiency between motor and mixing blades is 75%. Estimate the necessary mixing time.

Power to the motor = 3 EI cosf  where I is the current per phase and E and cosf in this case are 440 volts and 0.89, respectively.
In first 10 s, Iaverage = ½(80 + 400) = 240 amperes.

Energy consumed in first 10 s
                      = power x time =
3 EI cosf
3 x 440 x 0.75 x 240 x 0.89 x 10/3600
                      = 339 watt h
                      = 339/2000
                      = 0.17 W h kg-1.

Energy still needed after first 10 s

                      = (8 - 0.17)
                      = 7.83 W h kg-1.

Additional time needed, at steady 400 ampere current

                      = (7.83 x 2000 x 3600)/(1.73 x 440 x 0.75 x 400 x 0.89)
                      = 277 s.

       Total time = 277 + 10 s = 4.8 min.

In actual mixing of doughs the power consumed would decrease slowly and more or less uniformly, but only by around 10-15% over the mixing process.


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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