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Rate of Extraction
Stage-equilibrium Extraction
Extraction and Washing Equipment

It is often convenient to use a liquid in order to carry out a separation process. The liquid is thoroughly mixed with the solids or other liquid from which the component is to be removed and then the two streams are separated.

In the case of solids, the separation of the two streams is generally by simple gravity settling. Sometimes it is the solution in the introduced liquid that is the product required, such as in the extraction of coffee from coffee beans with water. In other cases, the washed solid may be the product as in the washing of butter. The term washing is generally used where an unwanted constituent is removed in a stream of water. Extraction is also an essential stage in the sugar industry when soluble sucrose is removed by water extraction from sugar cane or beet. Washing occurs so frequently as to need no specific examples.

To separate liquid streams, the liquids must be immiscible, such as oil and water. Liquid-liquid extraction is the name used when both streams in the extraction are liquid. Examples of extraction are found in the edible oil industry in which oil is extracted from natural products such as peanuts, soya beans, rapeseeds, sunflower seeds and so on. Liquid-liquid extraction is used in the extraction of fatty acids.

Factors controlling the operation are:
area of contact between the streams,
time of contact,
properties of the materials so far as the equilibrium distribution of the transferred component is concerned,
number of contact stages employed.

In extraction from a solid, the solid matrix may hinder diffusion and so control the rate of extraction.

Rate of Extraction

The solution process can be considered in terms of the usual rate equation

        rate of solution = driving force/resistance.

In this case, the driving force is the difference between the concentration of the component being transferred, the solute, at the solid interface and in the bulk of the solvent stream. For liquid-liquid extraction, a double film must be considered, at the interface and in the bulk of the other stream.

For solution from a solid component, the equation can be written
                             dw/dt = KlA(ys - y)                                                                             (9.8)

where dw/dt is the rate of solution, Kl is the mass-transfer coefficient, A is the interfacial area, and ys and y, are the concentrations of the soluble component in the bulk of the liquid and at the interface. It is usually assumed that a saturated solution is formed at the interface and ys is the concentration of a saturated solution at the temperature of the system.

Examination of eqn. (9.8) shows the effects of some of the factors, which can be used to speed up rates of solution. Fine divisions of the solid component increases the interfacial area A. Good mixing ensures that the local concentration is equal to the mean bulk concentration. In other words, it means that there are no local higher concentrations arising from bad stirring increasing the value of y and so cutting down the rate of solution. An increase in the temperature of the system will, in general, increase rates of solution by not only increasing Kl, which is related to diffusion, but also by increasing the solubility of the solute and so increasing ys.

In the simple case of extraction from a solid in a contact stage, a mass balance on the solute gives the equation:
                                 dw = Vdy                                                                                       (9.9)

where V is the quantity of liquid in the liquid stream.

Substituting for dw in eqn. (9.8) we then have:

                    Vdy/dt = Kl A(ys - y)

which can then be integrated over time t during which time the concentration goes from an initial value of y0 to a concentration y, giving

loge [(ys - y0)/ (ys - y)] = tKlA/V.                                                                                 (9.10)

Equation (9.10) shows, as might be expected, that the approach to equilibrium is exponential with time. The equation cannot often be applied because of the difficulty of knowing or measuring the interfacial area A. In practice, suitable extraction times are generally arrived at by experimentation under the particular conditions that are anticipated for the plant.

Stage-equilibrium Extraction

Analysis of an extraction operation depends upon establishing the equilibrium and operating conditions. The equilibrium conditions are, in general, simple. Considering the extraction of a solute from a solid matrix, it is assumed that the whole of the solute is dissolved in the liquid in one stage, which in effect accomplishes the desired separation. However, it is not possible then to separate all of the liquid from the solid because some solution is retained with the solid matrix and this solution contains solute. As the solid retains solution with it, the content of solute in this retained solution must be then progressively reduced by stage contacts. For example, in the extraction of oil from soya bean seeds using hexane or other hydrocarbon solvents, the solid beans matrix may retain its own weight, or more, of the solution after settling. This retained solution may therefore contain a substantial proportion of the oil. The equilibrium conditions are simple because the concentration of the oil is the same in the external solution that can be separated as it is in the solution that remains with the seed matrix. Consequently, y, the concentration of oil in the "light" liquid stream, is equal to x, the concentration of oil in the solution in the "heavy" stream accompanying the seed matrix. The equilibrium line is, therefore, plotted from the relation y = x.

The operating conditions can be analysed by writing mass balances round the stages to give the eqn. (9.5). The plant is generally arranged in the form of a series of mixers, followed by settlers in which the two streams are separated prior to passing to the next stage of mixers. For most purposes of analysis, the solid matrix need not be considered; the solids can be thought of as just the means by which the two solution streams are separated after each stage. So long as the same quantity of solid material passes from stage to stage, and also the solids retain the same quantity of liquid after each settling operation, the analysis is straightforward. In eqn. (9.5), V refers to the liquid overflow stream from the settlers, and L to the mixture of solid and solution that is settled out and passes on with the underflow.

If the underflow retains the same quantity of solution as it passes from stage to stage, eqn. (9.5) simplifies to eqn. (9.6). The extraction operation can then be analysed by application of step-by-step solution of the equations for each stage, or by the use of the McCabe-Thiele graphical method.

EXAMPLE 9.6. Counter current extraction of oil from soya beans with hexane
Oil is to be extracted from soya beans in a counter current stage-contact extraction apparatus, using hexane. If the initial oil content of the beans is 18%, the final extract solution is to contain 40% of oil, and if 90% of the total oil is to be extracted, calculate the number of contact stages that are necessary. Assume that the oil is extracted from the beans in the first mixer, that equilibrium is reached in each stage, and that the crushed bean solids in the underflow retain in addition half their weight of solution after each settling stage.
The extraction plant is illustrated diagrammatically in Fig. 9.5.

FIG. 9.5 Extraction stages
FIG. 9.5 Hexane extraction of oil from soya beans in stages

Each box represents a mixing-settling stage and the stages are numbered from the stage at which the crushed beans enter.

The underflow will be constant from stage to stage (a constant proportion of solution is retained by the crushed beans) except for the first stage in which the entering crushed beans (bean matrix and oil) are accompanied by no solvent. After the first stage, the underflow is constant and so all stages but the first can be treated by the use of eqn. (9.6).

To illustrate the principles involved, the problem will be worked out from stage-by-stage mass balances, and using the McCabe-Thiele graphical method.

Basis for calculation: 100 kg raw material (bean solids and their associated oil) entering stage 1. Concentrations of oil will be expressed as weight fractions.

Overall mass balance

In 100 kg raw material there will be 18% oil, that is 82 kg bean solids and 18 kg oil.
In the final underflow, 82 kg beans will retain 41 kg of solution, the solution will contain 10% of the initial oil in the beans, that is, 1.8 kg so that there will be (18 - 1.8) = 16.2 kg of oil in the final overflow,

Extract contains (16.2 x 60/40) = 24.3 kg of solvent
Total volume of final overflow = 16.2 + 24.3 = 40.5 kg
Total solvent entering = (39.2 + 24.3) = 63.5 kg

Note that the solution passing as overflow between the stages is the same weight as the solvent entering the whole system, i.e. 63.5 kg.

Basis: 100 kg beans

Mass in (kg) Mass out ( kg)
Underflow      Underflow  
Raw beans         100.0   Extracted beans + solution         123.0
    Bean solids 82.0       Bean solids 82.0
    Oil 18.0       Oil   1.8
          Solvent 39.2
Overflow     Overflow  
Solvent           63.5   Total extract           40.5
          Solvent 24.3
          Oil 16.2
Total          163.5   Total          163.5

Analysis of stage 1

Oil concentration in underflow = product concentration = 0.4.
It is an equilibrium stage, so oil concentration in underflow equals oil concentration in overflow. Let y2 represent the concentration of oil in the overflow from stage 2 passing in to stage 1. Then oil entering stage 1 equals oil leaving stage 1.
Therefore balance on oil:

            63.5y2 + 18 = 41 x 0.4 + 40.5 x 0.4
                          y2 = 0.23.

Analysis of stage 2

                                  x2 = y2 = 0.23
Therefore balance on oil:
            41 x 0.4 + 63.5y3 = 63.5 x 0.23 + 41 x 0.23,
                                  y3 = 0.12.

Analysis of stage 3

                                  x3 = y3= 0.12
Therefore balance on oil:
          41 x 0.23 + 63.5y4 = 63.5 x 0.12 + 41 x 0.12
                                   y4= 0.049

Analysis of stage 4

                                  x4 = y4 = 0.049,

Therefore balance on oil

         41 x 0.12 + 63.5 y5 = 63.5 x 0.049 + 41 x 0.049,
                                  y5 = 0.00315.

The required terminal condition is that the underflow from the final nth stage will have less than 1.8 kg of oil, that is, that xn is less than 1.8/41 = 0.044.
Since xn = yn and we have calculated that y5 is 0.00315 which is less than 0.044 (whereas y4 = 0.049 was not), four stages of contact will be sufficient for the requirements of the process.

Using the graphical method, the general eqn. (9.6) can be applied to all stages after the first.

From the calculations above for the first stage, we have x1 = 0.4, y2 = 0.23 and these can be considered as the entry conditions xa and ya for the series of subsequent stages.

Applying eqn. (9.6) the operating line equation:
                               yn+1 = xnL/V + ya - xaL/V

Now L = 41, V = 63.5, ya = 0.23 xa = 0.4
                               yn+1 = 0.646 xn - 0.028

The equilibrium line is:
                                  yn = xn

The operating line and the equilibrium line have been plotted on Fig. 9.6.

FIG. 9.6 Extraction: McCabe-Thiele plot
FIG. 9.6 Hexane extraction of oil from soyabeans: McCabe-Thiele plot

The McCabe-Thiele construction has been applied, starting with the entry conditions to stage 2 (stage 1 being the initial mixing stage) on the operating line, and it can be seen that three steps are not sufficient, but that four steps give more than the minimum separation required. Since the initial stage is included but not shown on the diagram, four stages are necessary, which is the same result as was obtained from the step-by-step calculations.

The step construction on the McCabe-Thiele diagram can also be started from the nth stage, since we know that yn+1 which is the entering fresh solvent, equals 0. This will also give the same number of stages, but it will apparently show slightly different stage concentrations. The apparent discrepancy arises from the fact that in the overall balance, a final (given) concentration of oil in the overflow stream of 0.044 was used, and both the step-by-step equations and the McCabe-Thiele operating line depend upon this. In fact, this concentration can never be reached using a whole number of steps under the conditions of the problem and to refine the calculation it would be necessary to use trial-and-error methods. However, the above method is a sufficiently close approximation for most purposes.

In some practical extraction applications, the solids may retain different quantities of the solvent in some stages of the plant. For example, this might be due to rising concentrations of extract having higher viscosities. In this case, the operating line is not straight, but step-by-step methods can still be used. For some of the more complex situations other graphical methods using triangular diagrams can be employed and a discussion of their use may be found in Charm (1970), Coulson and Richardson (1978) or Treybal (1970).

It should be noted that in the chemical engineering literature what has here been called extraction is more often called leaching, the term extraction being reserved for liquid-liquid contacting using immiscible liquids. "Extraction" is, however, in quite general use in the food industry to describe processes such as the one in the above example, whereas the term "leaching" would probably only cause confusion.


Washing is almost identical to extraction, the main distinction being one of the emphasis in that in washing the inert material is the required product, and the solvent used is water which is cheap and readily available. Various washing situations are encountered and can be analysed. That to be considered is one in which a solid precipitate, the product, retains water which also contains residues of the mother liquor so that on drying without washing these residues will remain with the product. The washing is designed to remove them, and examples are butter and casein and cheese washing in the dairy industry.

Calculations on counter current washing can be carried out using the same methods as discussed under extraction, working from the operating and equilibrium conditions. In washing, fresh water is often used for each stage and the calculations for this are also straightforward.

In multiple washing, the water content of the material is xw (weight fraction) and a fraction of this, x, is impurity, and to this is added yxw of wash water, and after washing thoroughly, the material is allowed to drain. After draining it retains the same quantity, approximately, of water as before, xw. The residual yxw of wash liquid, now at equilibrium containing the same concentration of impurities as in the liquid remaining with the solid, runs to waste. Of course in situations in which water is scarce counter current washing may be worthwhile.

The impurity which was formerly contained in xw of water is now in a mass (xw + yxw): its concentration x has fallen by the ratio of these volumes, that is to x [xw/(xw + yxw)].

So the concentration remaining with the solid after one washing, x1, is given by:

x1 = x[xw /xw(1 + y)] = x[1/(1 + y)]
             after two washings:
x2 = x1[1/(1 + y)] = x[1/(1 + y)]2
             and so after n washings:
xn = x[1/(1 + y)]n                                                                                                     (9.11)

If, on the other hand, the material is washed with the same total quantity of water as in the n washing stages, that is nyxw, but all in one stage, the impurity content will be:

x'n = x[1/(ny + 1)]                                                                                                    (9.12)

and it is clear that the multiple contact washing is very much more efficient in reducing the impurity content that is single contact washing, both using the same total quantity of water.

EXAMPLE 9.7. Washing of casein curd
After precipitation and draining procedures, it is found that 100 kg of fresh casein curd has a liquid content of 66% and this liquid contains 4.5% of lactose. The curd is washed three times with 194 kg of fresh water each time. Calculate the residual lactose in the casein after drying. Also calculate the quantity of water that would have to be used in a single wash to attain the same lactose content in the curd as obtained after three washings. Assume perfect washing, and draining of curd to 66% of moisture each time.

100 kg of curd contain 66 kg solution. The 66 kg of solution contain 4.5% that is 3 kg of lactose.

In the first wash (194 + 66) = 260 kg of solution contain 3 kg lactose.
In 66 kg of solution remaining there will be (66/260) x 3 = 0.76 kg of lactose.

After the second wash the lactose remaining will be (66/260) x 0.76 = 0.19kg

After the third wash the lactose remaining will be (66/260) x 0.19     = 0.048 kg

Or, after three washings lactose remaining will be 3 x (66/260)3       = 0.048 kg as above

So, after washing and drying 0.048 kg of lactose will remain with 34 kg dry casein so that

lactose content of the product = 0.048/34.05
                                            = 0.14%
and total wash water              = 3 x 194   =   582 kg

To reduce the impurity content to 0.048 kg in one wash would require x kg of water, where
                    (3 x 66)/(x + 66) = 0.048 kg
                                          x = 4060 kg
and so the total wash water                     = 4060 kg

Alternatively using eqns. (9.11) and (9.12)
                                        xn = x[1/(1 + y)]n
                                            = 3[1/(1 + 194/66)]3
                                            = 0.049
                                       xn' = x[1/(ny + 1)]
                                   0.049 = 3[1/(ny+ 1)]
                                       ny = 61.5.

Total wash water                    = nyxw        = 61.5 x 66
                                                              = 4060 kg.

Extraction and Washing Equipment

The first stage in an extraction process is generally mechanical grinding, in which the raw material is shredded, ground or pressed into suitably small pieces or flakes to give a large contact area for the extraction. In some instances, for example in sugar-cane processing and in the extraction of vegetable oils, a substantial proportion of the desired products can be removed directly by expression at this stage and then the remaining solids are passed to the extraction plant. Fluid solvents are easy to pump and so overflows are often easier to handle than underflows and sometimes the solids may be left and solvent from successive stages brought to them.

This is the case in the conventional extraction battery. In this a number of tanks, each suitable both for mixing and for settling, are arranged in a row or a ring. The solids remain in the one mixer-settler and the solvent is moved progressively round the ring of tanks, the number, n, often being about 12. At any time, two of the tanks are out of operation, one being emptied and the other being filled. In the remaining (n - 2), tanks extraction is proceeding with the extracting liquid solvent, usually water, being passed through the tanks in sequence, the "oldest" (most highly extracted) tank receiving the fresh liquid and the "youngest" (newly filled with fresh raw material) tank receiving the most concentrated liquid. After leaving this "youngest" tank, the concentrated liquid passes from the extraction battery to the next stage of the process. After a suitable interval, the connections are altered so that the tank which has just been filled becomes the new “youngest" tank. The former "oldest" tank comes out of the sequence and is emptied, the one that was being emptied is filled and the remaining tanks retain their sequence but with each becoming one stage "older". This procedure which is illustrated in Fig. 9.7 in effect accomplishes counter current extraction, but with only the liquid physically having to be moved apart from the emptying and filling in the terminal tanks.

FIG. 9.7 Extraction battery
FIG. 9.7 Extraction battery

In the same way and for the same reasons as with counter flow heat exchangers, this counter current (or counter flow) extraction system provides the maximum mean driving force, the log mean concentration difference in this case, contrasting with the log mean temperature difference in the heat exchanger. This ensures that the equipment is used efficiently.

In some other extractors, the solids are placed in a vertical bucket conveyor and moved up through a tower down which a stream of solvent flows. Other forms of conveyor may also be used, such as screws or metal bands, to move the solids against the solvent flow. Sometimes centrifugal forces are used for conveying, or for separating after contacting.

Washing is generally carried out in equipment that allows flushing of fresh water over the material to be washed. In some cases, the washing is carried out in a series of stages. Although water is cheap, in many cases very large quantities are used for washing so that attention paid to more efficient washing methods may well be worthwhile. Much mechanical ingenuity has been expended upon equipment for washing and many types of washers are described in the literature.

Contact-Equilibrium Processes - APPLICATIONS > CRYSTALLIZATION

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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