MOTION OF PARTICLES TROUGH FLUIDS

PART 2

By: Dr Akmal Hadi Bin Ma’ Radzi

Particle settling

Free settling When a particle is at sufficient distance from the wall of the container and

from other particle, so that its fall is not affected by them, the process is called free settling.

Terminal velocity is also known as free settling velocity.

Hindered settling When the particles are crowded, they settle at a lower rate and the process

is called hindered settling. The particles will interfere with the motion of individual particles The velocity gradient of each particle are affected by the close presence of

other particles.

Hindered settling

The velocity for hindered setting can be computed by this equation:

Where, ε is volume fraction of the slurry mixture and Ψp is empirical correction factor.

Bulk density of mixture –

Empirical correction factor -

Correction factorStokes Law

Hindered settling

The Reynolds number is then based on the velocity relative to the fluid is

Where the viscosity of the mixture µm is given by;

Exercise

Calculate the settling velocity of glass spheres having a diameter of 1.554 x10-4 m in water at 20ºC. The slurry contains 60 wt % solids. The density of the glass sphere and water is 2467kg/m3 and 998 kg/m3 respectively.

Calculate the Reynold number for this settling.

Solution

Density of the water = 998 kg/m3, and viscosity of water μ = 1.005 x 10-3 Pa.s. To calculate the volume fraction of the liquid,

The bulk density of the slurry m is

Then,

The terminal velocity is:

The Reynolds number can be calculated as follows:

NRe

Fluidization

Fluidization is a process whereby a granular material is converted from a static solid-like state to a dynamic fluid-like state.

This process occurs when a fluid (liquid or gas) is passed up through the granular material.

The most common reason for fluidizing a bed is to obtain vigorous agitation of the solids in contact with the fluid, leading to an enhanced transport mechanism (diffusion, convection, and mass/energy transfer). 

Fluidized Bed Reactor

When a gas flow is introduced through the bottom of a bed of solid particles, it will move upwards through the bed via the empty spaces between the particles.

At low gas velocities, aerodynamic drag on each particle is also low, and thus the bed remains in a fixed state.

Increasing the velocity, the aerodynamic drag forces will begin to counteract the gravitational forces, causing the bed to expand in volume as the particles move away from each other

Further increasing the velocity, it will reach a critical value at which the upward drag forces will exactly equal the downward gravitational forces, causing the particles to become suspended within the fluid. At this critical value, the bed is said to be fluidized and will exhibit fluidic behavior.

By further increasing gas velocity, the bulk density of the bed will continue to decrease, and its fluidization becomes more violent, until the particles no longer form a bed and are “conveyed” upwards by the gas flow.

When fluidized, a bed of solid particles will behave as a fluid, like a liquid or gas.

Objects with a lower density than the bed density will float on its surface, bobbing up and down if pushed downwards, while objects with a higher density sink to the bottom of the bed

The fluidic behavior allows the particles to be transported like a fluid, channeled through pipes, not requiring mechanical transport

Applications of Fluidization

Conditions for Fluidization

Based on the Figure:

If the superficial velocity, VO is gradually increased, the pressure drop will increases, but the particles do not move and the height (L) remains the same.

At a certain velocity, the pressure drop across the bed counterbalances the forces of gravity on the particles or the weight of the bed

At point A = Any further increase in velocity, causes the particles to move

At point B = Further increase in velocity, the particles become separate enough to move about in the bed and true fluidization begins.

From point B to point C = Once bed is fluidized, the pressure drop across the bed stays constant , but the bed heights continues to increase with increasing velocity.

From point C to B = If the velocity is gradually reduced, the pressure drop remains constant and the bed height decreases.

*The pressure drop required for the liquid or the gas to flow through the column at a specific flow rate

Minimum Fluidization Velocity

Minimum velocity of fluidization took place at incipient (beginning) fluidization.

During this stage, the ratio of pressure drop to the vessel height (L) is given by;

Where is the minimum porosity

The minimum fluidization velocity can be obtained by this equation;

For roughly spherical particles, is generally between 0.4 and 0.45 (commonly taken as 0.45) which increasing slightly with decreasing particle diameter.

If the Reynolds number is used, then,

The equation becomes:

Exercise 2

A bed of ion exchange beads 8 ft deep is to be backwashed with water to remove dirt. The particles have a density of 1.24 g/cm3 and an average size of 1.1 mm. The beads are assumed to be spherical (Φ =1) and is taken as 0.4. What is the minimum fluidization velocity using water at 20ºC with density of 980 kg/m3?

Solution 2

𝟏𝟓𝟎(𝟏 .𝟎𝟎𝟓𝐱𝟏𝟎−𝟑)𝑽 𝑶𝑴

𝟏𝟐 (𝟏 .𝟏𝐱𝟏𝟎−𝟑 )𝟐(𝟏−𝟎 .𝟒 )𝟎 .𝟒𝟑 +

𝟏 .𝟕𝟓(𝟗𝟖𝟎)𝑽 𝟐𝑶𝑴

𝟏 (𝟏.𝟏𝐱𝟏𝟎−𝟑)𝟏

𝟎 .𝟒𝟑=𝟗 .𝟖𝟏 (𝟏𝟐𝟒𝟎−𝟗𝟖𝟎 )

𝑽 𝑶𝑴=𝟎 .𝟎𝟎𝟐𝟏𝒎/𝒔

Exercise 3

Solid particles having a size of 0.12 mm, a shape factor Фs of 0.88, and a density of 1000 kg/m3 are to be fluidized using air at 2.0 atm abs and 25 oC. The voidage at minimum fluidizing condition is 0.42.(a) If the cross section of the empty bed is 0.30 m2

and the bed contains 300 kg of solid, calculate the minimum height of the fluidized bed.

(b) Calculate the pressure drop at minimum fluidizing conditions

(c) Calculate the minimum velocity of fluidization

For part (a), the volume of solids = 300 kg/(1000 kg/m3) = 0.300 m3. The height of the solids would occupy in the bed if ɛ1 = 0 is L1 = 0.300m3/(0.30m2 cross section) = 1.00 m.

Solving Lmf = 1.724 m

The physical properties of air at 2.0 atm and 25 oC are μ = 1.845 x 10-5 Pa.s, = 1.187 x 2 = 2.374 kg/m3, and p = 2.0265 x 105 Pa. For the particle, Dp = 0.00012 m, p= 1000 kg/m3, s = 0.88 and ɛmf = 0.42.

For part (b),

To calculate Vmf’ for part (c),

Solving,