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Question : (c) A liquid fluidised bed consists of equal volumes of sphe...

(c) A liquid fluidised bed consists of equal volumes of spherical particles 0.5 mm and 1.0 mm in diameter. The bed is fluidised and complete segregation of the two species occurs. When the liquid flow is stopped the particles settle to form a segregated two-layer bed. The liquid flow is then started again. When the velocity is such that the larger particles are at their incipient fluidisation point what will be the approximate voidage of the fluidised bed composed of the smaller particles?
It may be assumed that the drag force F of the fluid on the particles under the free falling conditions is given by Stokes’ law and that the relation between the fluidisation velocity u_{c} and voidage, e, for particles of terminal velocity, u_{0} , is given by:

u_{c}/u_{0}=e^{4.8}

For Stokes’ law, the force F on the particles is given by F=3\pi \mu du_{0}, where d is the particle diameter and µ is the viscosity of the liquid.

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A force balance gives:

3\pi \mu du_{c}=(\pi /6)d^{3}(\rho _{s}-\rho )g

Thus: u_{c}=(d^{2}g/18\mu )(\rho_{s} -\rho )g

That is: u_{c}=k_{i}d^{2} (i)

For larger particles and assuming that the voidage at u_{mf}=0.45,

then: u_{c}/u_{0L}=0.45^{n} (ii)

For smaller particles:

u_{c}/u_{0S}=e^{n}

u_{c} has the same k_{i} both large and small particles and from equation (i):

u_{0S}=u_{0L}/4

Thus: 4u_{c}/u_{0L}=e^{n} (iii)

Dividing equation (iii) by equation (ii) gives:

4=(e/0.45)^{n}=(e/0.45)^{4.8}

Thus: 1.334 = e/0.45

and: e = 0.60