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Question : (a) Calculate the terminal falling velocities in water of gl...

(a) Calculate the terminal falling velocities in water of glass particles of diameter 12 mm and density 2500 kg/m^{3}, and of metal particles of diameter 1.5 mm and density 7500 kg/m^{3}.
It may be assumed that the particles are spherical and that, in both cases, the friction factor, {R}'/\rho u^{2} is constant at 0.22, where R’ is the force on the particle per unit of projected area of the particle, \rho is the fluid density and u the velocity of the particle relative to the fluid.

(b) Why is the sedimentation velocity lower when the particle concentration in the suspension is high? Compare the behaviour of the concentrated suspension of particles settling under gravity in a liquid with that of a fluidised bed of the same particles.

(c) At what water velocity will fluidised beds of the glass and metal particles have the same densities? The relation between the fluidisation velocity u_{c} terminal velocity u_{0} and bed voidage e is given for both particles by:

u_{c}/u_{0}=e^{2.30}

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For spheres, a take balance gives:

R'(\pi /4)d^{2}=(\pi /6)d^{3}(\rho _{s}-\rho )g

or: R'=(2d/3)(\rho _{s}-\rho )g=(R'/\rho u_{0}^{2})\rho u_{0}^{2}=0.22\rho u_{0}^{2}\approx (2/g)\rho u_{0}^{2}

Thus:u_{0}=[3dg(\rho_{s} -\rho )/\rho ]^{0.5}

For the metal particles:

u_{0}=[(3\times 1.5\times 10^{-3}\times 6500\times 9.81)/1000 ]^{0.5}

= 0.536 m/s

For the glass particles:

u_{0}=[(3\times 1.5\times 10^{-3}\times 1500\times 9.81)/1000 ]^{0.5}

= 0.727 m/s.

For the fluidised bed:

The density of the suspension = e\rho +(1-e)\rho _{s}

For the metal particles:

(u_{c}/0.536)=e_{1}^{2.30} (i)

For the glass particles:

(u_{c}/0.727)=e_{2}^{2.30} (ii)

and from equations (i) and (ii):

(e_{1}/e_{2})=(0.727/0.536)^{1/2.30}=1.142

For equal bed densities:

e_{1}\rho (1-e_{1})\rho _{s1}=e_{2}\rho+(1-e_{2})\rho _{s2} (iii)

Thus: (1.142e_{1})1000+(1-1.142e_{2})7500=1000e_{2}+(1-e_{2})2500

from which: e_{3}=0.844

and: u_{c}=(0.727\times 0.844^{2.30})=0.492m/s

Substituting in equation (iii):

e_{1}=0.964

and: u_{c}=(0.536\times 0.964^{2.30})=0.493 m/s