Question

Q. Spherical glass particles (12 mm diameter and 2500 kg/m 3 density) and spherical metal particles (1.5 mm diameter and 7500 kg/m3) are falling in water (density= 1000 kg/m3) .

(1) Calculate the terminal falling velocities of glass and metal particles in water for a constant friction factor of 0.22.

(2) At what water velocity will fluidized beds of glass particles and metal particles have the same bed densities?

The relation between fluidization velocity (u_c), terminal velocity (u_t) and bed voidage (e) for a spherical particle is given by the equation (uclu_t) = e^2·3

Answer 1

Step 1

Given that:

Spherical glass particle has a diameter(D1) =12mm and density() = 2500 kg/m3,

Spherical metal particle has diameter(D2) = 1.5mm and density() = 7500 kg/m3

Falling water density() = 1000 kg/m3

Step 2

Equation for the sphere balance,

where,

That is equal to the

For metal particles,

Put all the values in the above equation to get the velocity of the terminal falling velocity of the metal particle,

on solving the above equation we get velocity,

u0 = 0.5356 m/s

For the glass particle,

u0 = 0.7278 m/s

Step 3

We have an equation for the fluidized bed,

The density of the suspension =

uc is the fluidized velocity, ut is the terminal velocity and e is the bed voidage.

For glass particles:

(uc/0.7278) = e12.30

For the metal particles:

(uc/0.536) = e22.30

By dividing equation metal particles to the glass particles we get,

On cancellation of uc

0.7278/0.536 = (e2/e1)2.30

(e2/e1) = (0.7278/0.536)1/2.30 = 1.142

e2 = 1.142e1

b) For equal bed densities:

Put all the values in the above equation to get the value of bed voidage(e)

e1 * 1000kg/m3 + (1 - e1) * 2500kg/m3 = 1.142e1 * 1000kg/m3 + (1 - 1.142e1) * 7500kg/m3

1000e1 + 2500kg/m3 - 2500(kg/m3)e1 = 1142(kg/m3)e1 + 7500kg/m3 - 8565(kg/m3)e1

Taking e1 common on both side we get,

e1(1000kg/m3 -2500kg/m3 -1142kg/m3 +8565kg/m3 ) = 7500kg/m3 -2500kg/m3

e1(5926kg/m3) = 5000kg/m3

dividing both side by 5926 we get,

e1 = 0.844

and e2 = 1.142* e1 = 1.142*0.844 => 0.96355

For glass particles

(uc/0.7278) = e12.30

uc = (0.844)2.30 * 0.7278 => 0.492 m/s

By putting the value of e1 we have uc

uc = 0.492 m/s

For the metal particles:

(uc/0.536) = e22.30

uc = (0.96355)2.30 * 0.536 => 0.492 m/s

By putting the value of e2 we have uc

uc = 0.492 m/s

At 0.492 m/s water velocity would be same bed densities.