# Fabrizio Sarghini » 18.Separation Equipments – Part II

### Separation

Centrifugal separation

We can define centrifugation as a separation process that uses the action of centrifugal force to promote accelerated settling of particles in a solid-liquid or liquid-liquid mixture:

$F_c=mr\omega^2$

where Fc is the centrifugal force acting on the particle, r is the radius of the trajectory, m is the mass of the particle and ω is the angular velocity of the particle. If we define ω = v / r where v is the velocity of the particle, it results:

$F_c=m\frac {v^2}r$

### Separation (cont’d)

Centrifugal separation

The speed of rotation is usually expressed in revolutions per minute (rpm) N, and the equation then can be written as:

$F_c=mr(2\pi N/60)^2=0.011 mrN^2$

Since ω=2 π N/60 (s-1)and where N is the speed of rotation in rpm.

Compared with the force of gravity Fg on the particle, with Fg = mg, we can see that the centrifugal acceleration, equal to 0.011rN2, replaced gravitational acceleration g.

For this reason the ability to separate in a centrifuge is often expressed in terms of n x g, i.e. corresponding to n-times g.

### Separation (cont’d)

Centrifugal separation

The centrifugal force depends on the radius, the speed of rotation and the mass of a particle.

If the radius and the speed of rotation a are fixed, the most important factor is the weight of the particles, so that the more the weight of the particle, the more the centrifugal force acting on it.

Therefore, if two fluids, one of them twice as thick as the other, are placed in a cup and the cup is placed in fast rotation over a vertical axis, then the centrifugal force per unit volume will be two times greater for the heavy liquid than for the lighter one.

The heavy liquid will tend to move and occupy the outer ring at the periphery of the cup, while the lighter liquid will shift toward the centre.

This is the principle of centrifugal separator tube, shown schematically in the next figure.

### Separation (cont’d)

Centrifugal separation. Liquid distribution in a vertical centrifuge

### Separation (cont’d)

Centrifugal separation

We saw earlier that the separation speed limit of a particle moving in the fluid under acceleration due to a generic force a is given by:

$v_m=\frac{D^2a(\rho_p-\rho_f)}{18 \mu}~~~~~~~~(12)$

If we consider the centrifugal force, we can write:

$F_c=ma~~~~~~~~(13)$

### Separation (cont’d)

$a=F_c/m=r\biggl(\frac{2\pi N}{60}\biggr)^2 ~~~~~~~~~~~~~~~~~(14)$

obtaining

$v_m=\frac{D^2r(2\piN/60)^2(\rho_p-\rho_f)}{18\mu}=\frac{D^2rN^2(\rho_p-\rho_f)}{1640\mu}~~~~~~~(15)$

### Separation (cont’d)

Centrifugal separation

The separation of a component of a liquid-liquid mixture, where the liquids are immiscible but finely dispersed, as in an emulsion, is a common operation for the food industry.

It is particularly common the example in the dairy sector in which the emulsion, the milk, is separated by a centrifuge into skimmed milk and cream, or in the process of clarification andseparation in the production of oil.

In the case of milk we can examine the position of the two phases in a centrifuge to see how it operates.

Milk is introduced continuously into the machine, and generally constituted by a cup rotating on a vertical axis, resulting inside the cup in a surface separating the cream and skimmed milk.

### Separation (cont’d)

Centrifugal separation. Pressure balance in a vertical centrifuge

### Separation (cont’d)

Centrifugal separation

Let us consider a circular cylindrical element inside the centrifuge cup, with thickness dr and height b as described in Figure 3. The force acting on the element of thickness dr is given by:

$dF_c=dm\omega^2r$

where dFc is the differential of the force through the cylindrical ring, dm is the mass of the cylinder, ω is the angular speed and r is the the generic radius of cylinder.
Since:

$dm=2\pi\rho rbdr$

where ρ is the density of the liquid, the area on which the force acts is equal to 2πrb and therefore the pressure:

$\frac{dF_c}{2\pi rb}=dP=\rho \omega^2 rdr$

where dP is the pressure differential through the cylinder’ surface.

### Separation (cont’d)

Centrifugal separation

To determine the difference in pressure in a centrifuge of the radius r1 and r2, the equation of the differential pressure dP must be integrated andwe can call P1 the pressure corresponding to the surface of radius r1 and P2 that in correspondence of R2, resulting:

$P_2-P_1=\rho \omega^2(r_2^2-r_1^2)/2$

### Separation (cont’d)

Centrifugal separation

The equation 19 defines the radial variation of pressure through thecentrifuge .

Let us consider Figure 3b, which represents the cup of a separatorcontinuous centrifugal liquid-liquid.

The mixture enters the centrifuge near the vertical axis, while the heavier liquid exits from the opening 1 and the lighter liquid from opening 2.

r1 is the radius at which distance the liquid 1 output of liquid ant r2 that of the lighter one.

At a generic distance rn the separation will appear between the two phases.

### Separation (cont’d)

Centrifugal separation

If the system is in hydrostatic equilibrium,the pressures of the two components must be equal, and therefore we get:

$\rho_{A^{\omega}^2} (r_n^2-r_1^2)/2=\rho_{B^{\omega}^2}(r_n^2-r_2^2)/2$

where ρA is the density of the heavier liquid and ρB the density of the lighter liquid:

$r_n^2=(\rho_A r^2_1 - \rho_B r_2^2)/(\rho_A-\rho_B)$

### Separation (cont’d)

Centrifugal separation: milk-cream plate centrifuge From discoverarmfield

### Separation (cont’d)

Centrifugal separation From flottweg ag

### Separation (cont’d)

Centrifugal separation From flottweg ag

### Separation (cont’d)

Centrifugal separation From flottweg ag

### Separation (cont’d)

Centrifugal separation From flottweg ag

### Separation (cont’d)

Centrifugal separation: 2 way decanter From flottweg ag

### Separation (cont’d)

Centrifugal separation: 2 way decanter From flottweg ag

### Separation (cont’d)

Centrifugal separation: 3 way decanterFrom flottweg ag

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