UNIT OPERATIONS IN FOOD PROCESSING
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CHAPTER 10
MECHANICAL SEPARATIONS (cont'd)
CENTRIFUGAL SEPARATIONS


Liquid Separation
Centrifuge Equipment


The separation by sedimentation of two immiscible liquids, or of a liquid and a solid, depends on the effects of gravity on the components. Sometimes this separation may be very slow because the specific gravities of the components may not be very different, or because of forces holding the com-ponents in association, for example as occur in emulsions. Also, under circumstances when sedimentation does occur there may not be a clear demarcation between the components but rather a merging of the layers.

For example, if whole milk is allowed to stand, the cream will rise to the top and there is eventually a clean separation between the cream and the skim milk. However, this takes a long time, of the order of one day, and so it is suitable, perhaps, for the farm kitchen but not for the factory.

Much greater forces can be obtained by introducing centrifugal action, in a centrifuge. Gravity still acts and the net force is a combination of the centrifugal force with gravity as in the cyclone. Because in most industrial centrifuges, the centrifugal forces imposed are so much greater than gravity, the effects of gravity can usually be neglected in the analysis of the separation.


The centrifugal force on a particle that is constrained to rotate in a circular path is given by

      Fc = mrw2                                                                                                        (10.5)

where Fc is the centrifugal force acting on the particle to maintain it in the circular path, r is the radius of the path, m is the mass of the particle, and w(omega) is the angular velocity of the particle.

Or, since w = v/r, where v is the tangential velocity of the particle

      Fc = (mv2)/r                                                                                                      (10.6)

Rotational speeds are normally expressed in revolutions per minute, so that eqn. (10.6) can also be written, as w = 2pN/60 (as it has to be in s-1, divide by 60)

      Fc = mr( 2pN/60)2 = 0.011 mrN2                                                                        (10.7)

where N is the rotational speed in revolutions per minute.

If this is compared with the force of gravity (Fg) on the particle, which is Fg = mg , it can be seen that the centrifugal acceleration, equal to 0.011 rN2, has replaced the gravitational acceleration, equal to g. The centrifugal force is often expressed for comparative purposes as so many "g".


EXAMPLE 10.3. Centrifugal force in a centrifuge.
How many "g" can be obtained in a centrifuge which can spin a liquid at 2000 rev/min at a maximum radius of 10 cm?

      Fc = 0.011 mrN2
      Fg = mg

Fc /Fg = (0.011 rN2) / g
          = (0.011 x 0.1 x 20002)/9.81
          = 450

The centrifugal force depends upon the radius and speed of rotation and upon the mass of the particle. If the radius and the speed of rotation are fixed, then the controlling factor is the weight of the particle so that the heavier the particle the greater is the centrifugal force acting on it. Consequently, if two liquids, one of which is twice as dense as the other, are placed in a bowl and the bowl is rotated about a vertical axis at high speed, the centrifugal force per unit volume will be twice as great for the heavier liquid as for the lighter. The heavy liquid will therefore move to occupy the annulus at the periphery of the bowl and it will displace the lighter liquid towards the centre. This is the principle of the centrifugal liquid separator, illustrated diagrammatically in Fig. 10.3.

FIG. 10.3 Liquid separation in a centrifuge
Figure 10.3 Liquid separation in a centrifuge


Rate of separation


The steady-state velocity of particles moving in a streamline flow under the action of an accelerating force is, from eqn. (10.1),

     vm = D2a(rp - rf) /18m

If a streamline flow occurs in a centrifuge we can write, from eqns. (10.6) and (10.7) as a is the tangential acceleration;:

      Fc = ma
  Fc/m = a = r(2pN/60)2

so that

     vm = D2r(2pN/60)2(rp - rf) /18m

          = D2N2r(rp - rf)/1640m                                                                                  (10.8)


EXAMPLE 10.4. Centrifugal separation of oil in water
A dispersion of oil in water is to be separated using a centrifuge. Assume that the oil is dispersed in the form of spherical globules 5.1 x 10-5 m diameter and that its density is 894 kg m-3. If the centrifuge rotates at 1500 rev/min and the effective radius at which the separation occurs is 3.8 cm, calculate the velocity of the oil through the water. Take the density of water to be 1000 kg m-3 and its viscosity to be
 0.7 x 10-3 N s m-2. (The separation in this problem is the same as that in Example 10.2, in which the rate of settling under gravity was calculated.)

From eqn. (10.8)

     vm = (5.1 x 10-5)2 x (1500)2 x 0.038 x (1000 - 894)/(1.64 x 103 x 0.7 x 10-3)
          = 0.02 m s-1.

Checking that it is reasonable to assume Stokes' Law

     Re = (Dvr/m)
          = (5.1 x 10-5 x 0.02 x 1000)/(7.0 x 10-4)
          = 1.5
so that the flow is streamline and it should obey Stokes' Law.


Liquid Separation

The separation of one component of a liquid-liquid mixture, where the liquids are immiscible but finely dispersed, as in an emulsion, is a common operation in the food industry. It is particularly common in the dairy industry in which the emulsion, milk, is separated by a centrifuge into skim milk and cream. It seems worthwhile, on this account, to examine the position of the two phases in the centrifuge as it operates. The milk is fed continuously into the machine, which is generally a bowl rotating about a vertical axis, and cream and skim milk come from the respective discharges. At some point within the bowl there must be a surface of separation between cream and the skim milk.

FIG. 10.4 Liquid centrifuge (a) pressure difference (b) neutral zone
Figure 10.4 Liquid centrifuge (a) pressure difference (b) neutral zone


Consider a thin differential cylinder, of thickness dr and height b as shown in Fig. 10.4(a): the differential centrifugal force across the thickness dr is given by equation (10.5):

         dFc = (dm)rw2

where dFc is the differential force across the cylinder wall, dm is the mass of the differential cylinder, w is the angular velocity of the cylinder and r is the radius of the cylinder. But,
                 dm = 2prrbdr
where r is the density of the liquid and b is the height of the cylinder. The area over which the force dFc acts is 2prb, so that:

dFc /2prb = dP =rw2rdr

where dP is the differential pressure across the wall of the differential cylinder.

To find the differential pressure in a centrifuge, between radius r1 and r2, the equation for dP can be integrated, letting the pressure at radius r1 be P1 and that at r2 be P2, and so

    P2 - P1 = rw2 (r22 - r12)/2                                                                                    (10.9)

Equation (10.9) shows the radial variation in pressure across the centrifuge.

Consider now Fig. 10.4(b), which represents the bowl of a vertical continuous liquid centrifuge. The feed enters the centrifuge near to the axis, the heavier liquid A discharges through the top opening 1 and the lighter liquid B through the opening 2. Let r1 be the radius at the discharge pipe for the heavier liquid and r2 that for the lighter liquid. At some other radius rn there will be a separation between the two phases, the heavier and the lighter. For the system to be in hydrostatic balance, the pressures of each component at radius rn must be equal, so that applying eqn. (10.9) to find the pressures of each component at radius rn, and equating these we have:

rAw2 (rn2 - r12)/2 = rBw2(rn2– r22)/2

                     rn2 = (rAr12 - rBr22) / (rA - rB)                                                           (10.10)

where rA is the density of the heavier liquid and rB is the density of the lighter liquid.

Equation (10.10) shows that as the discharge radius for the heavier liquid is made smaller, then the radius of the neutral zone must also decrease. When the neutral zone is nearer to the central axis, the lighter component is exposed only to a relatively small centrifugal force compared with the heavier liquid. This is applied where, as in the separation of cream from milk, as much cream as possible is to be removed and the neutral radius is therefore kept small. The feed to a centrifuge of this type should be as nearly as possible into the neutral zone so that it will enter with the least disturbance of the system. This relationship can, therefore, be used to place the feed inlet and the product outlets in the centrifuge to get maximum separation.


EXAMPLE 10.5. Centrifugal separation of milk and cream
If a cream separator has discharge radii of 5 cm and 7.5 cm and if the density of skim milk is 1032 kg m-3 and that of cream is 915 kg m-3, calculate the radius of the neutral zone so that the feed inlet can be designed.
For skim milk, r1 = 0.075m, rA = 1032 kg m-3, cream r2 = 0.05 m, rB= 915 kg m-3

From eqn. (10.10)

                     rn2 = [1032 x (0.075)2 - 915 x (0.05)2] / (1032 - 915)
                          = 0.03 m2
                      rn = 0.17 m
                          = 17 cm


Centrifuge Equipment

The simplest form of centrifuge consists of a bowl spinning about a vertical axis, as shown in Fig. 10.4(a). Liquids, or liquids and solids, are introduced into this and under centrifugal force the heavier liquid or particles pass to the outermost regions of the bowl, whilst the lighter components move towards the centre.

If the feed is all liquid, then suitable collection pipes can be arranged to allow separation of the heavier and the lighter components. Various arrangements are used to accomplish this collection effectively and with a minimum of disturbance to the flow pattern in the machine. To understand the function of these collection arrangements, it is very often helpful to think of the centrifuge action as analogous to gravity settling, with the various weirs and overflows acting in just the same way as in a settling tank even though the centrifugal forces are very much greater than gravity.

In liquid/liquid separation centrifuges, conical plates are arranged as illustrated in Fig. 10.5(a) and these give smoother flow and better separation.

FIG. 10.5 Liquid centrifuges: (a) conical bowl, (b) nozzle
FIG. 10.5 Liquid centrifuges: (a) conical bowl, (b) nozzle


Whereas liquid phases can easily be removed from a centrifuge, solids present much more of a problem.

In liquid/solid separation, stationary ploughs cannot be used as these create too much disturbance of the flow pattern on which the centrifuge depends for its separation. One method of handling solids is to provide nozzles on the circumference of the centrifuge bowl as illustrated in Fig. 10.5(b). These nozzles may be opened at intervals to discharge accumulated solids together with some of the heavy liquid. Alternatively, the nozzles may be open continuously relying on their size and position to discharge the solids with as little as possible of the heavier liquid. These machines thus separate the feed into three streams, light liquid, heavy liquid and solids, the solids carrying with them some of the heavy liquid as well. Another method of handling solids from continuous feed is to employ telescoping action in the bowl, sections of the bowl moving over one another and conveying the solids that have accumulated towards the outlet, as illustrated in Fig. 10.6(a).

FIG. 10.6 Liquid/solid centrifuges (a) telescoping bowl, (b) horizontal bowl-scroll discharge
FIG. 10.6 Liquid/solid centrifuges (a) telescoping bowl, (b) horizontal bowl, scroll discharge


The horizontal bowl with scroll discharge, centrifuge, as illustrated in Fig.10.6(b) can discharge continuously. In this machine, the horizontal collection scroll (or screw) rotates inside the conical-ended bowl of the machine and conveys the solids with it, whilst the liquid discharges over an overflow towards the centre of the machine and at the opposite end to the solid discharge. The essential feature of these machines is that the speed of the scroll, relative to the bowl, must not be great. For example, if the bowl speed is 2000 rev/min, a suitable speed for the scroll might be 25 rev/min relative to the bowl which would mean a scroll speed of 2025 or 1975 rev/min. The differential speeds are maintained by gearing between the driving shafts for the bowl and the scroll. These machines can continuously handle feeds with solid contents of up to 30%.

A discussion of the action of centrifuges is given by Trowbridge (1962) and they are also treated in McCabe and Smith (1975) and Coulson and Richardson (1977).


Mechanical separations > FILTRATION


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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