SETTLINGANDSEDIMENTATION

Filtration versus settling and sedimentation: Filtration The solid particles are removed from the slurry by forcing the fluid through a filter medium, which blocks the passage of the solid particles and allows the filtrate to pass through. Settling and sedimentation The particles are separated from the fluid by forces acting on the particles.Introduction (1/4)

Applications of settling and sedimentation:*Removal of solids from liquid sewage wastes

*Settling of solid food particles from a liquid foodIntroduction (2/4)

Free settling versus hindered settling:Free settlingA particle is at a sufficient distance from the walls of the container and from other particles so that the fall is not affected.Interference is less than 1% if the ratio of the particle diameter to the container diameter is less than 1:200 or if the particle concentration is less than 0.2 vol% in the solution.Hindered settlingOccurred when the particles are crowded so that they settle at a lower rate.

What is sedimentation? The separation of a dilute slurry or suspension by gravity settling into a clear fluid and a slurry of higher solid content.

For a rigid particle of mass m moving in a fluid, there are three forces acting on the body:Gravity force, Fg, acting downward

Buoyant force, Fb, acting upwardwherer = density of the fluidrs = density of the solid particleVs = volume of the particleTHEORY OF PARTICLE MOVEMENT THROUGH A FLUID

For a rigid particle of mass m moving in a fluid, there are three forces acting on the body:THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (3) Drag force, FD, acting in opposite direction to the particle motion whereCD = the drag coefficientA = the projected area of the particleThe resultant force equals the force due to acceleration.

The falling of the body consists of two periods:(1) The period of accelerated fall The initial acceleration period is usually very short, of the order of a tenth of a second or so.(2) The period of constant velocity fallSetand solve the above equation for v. * vg is called the free settling velocity or terminal velocity.THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

For spherical particles of diameter d, THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

The drag coefficient for rigid spheres has been shown to be a function of the Reynolds number.THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

In the Stokes' law region (NRe < 1), THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

Brownian motion: the random motion imparted to the particle by collisions between the molecules of the fluid surrounding the particle and the particle.*If the particles are quite small, Brownian motion is present.This movement of the particles in random directions tends to suppress the effect of gravity. Settling of the particles may occur more slowly or not at all.THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

Brownian motion (continued)*At particle sizes of a few micrometers, the Brownian effect becomes appreciable and at sizes of less than 0.1 mm, the effect predominates.In very small particles, application of centrifugal force helps reduce the effect of Brownian motion.THEORY OF PARTICLE MOVEMENT THROUGH A FLUID

[Example] Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden beads or microcarriers have a density of 1.02 g/cm3 and a diameter of 150 mm. A 50-liter stirred tank is used to cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the stirring is stopped and the microcarriers are allowed to settle. The microcarrier-free fluid is then withdrawn to isolate the vaccine. The tank has a liquid height to diameter ratio of 1.5; the carrier-free fluid has a density of 1.00 g/cm3 and a viscosity of 1.1 cP. (a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity. (b) Estimate the time to reach this velocity. Hint:(To be continued)

Example: settling of dextran beadsData: d = 150 mm = 0.015 cm; m = 1.1 cP = 0.011 g/cm-s; rs = 1.02 g/cm3; r = 1.00 g/cm3; g = 980 cm/s2 Solution:(a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity. vg = 0.022 cm/s Check:Liquid volume, V h = 52.3 cm Settling time (To be continued)

(b) Estimate the time to reach the terminal velocity. Solution (contd):Force balance: ;(I.C.: t = 0, v = 0) Example: settling of dextran beads(To be continued)

At steady state (t ), When When t >> 1.16 10-3 s, v = vg For v = 0.99vg, t = 5.34 10-3 s#(b) Estimate the time to reach the terminal velocity. Solution (contd):Example: settling of dextran beads

ISOPYCNIC (SAME-DENSITY) SEDIMENTATION To capture particles in a solution having density gradient. Application: determining the density of the solute or suspended particle.*There are three methods for establishing conditions for isopycnic sedimentation:(1)Layer solutions of decreasing density, starting at the bottom of the tube.(2)Centrifuge the solution containing a density-forming solute (such as CsCl) at extremely high speed.(3)Use the gradient mixing method.

[Example] You wish to capture 3 mm particles in a linear density gradient having a density of 1.12 g/cm3 at the bottom and 1.00 g/cm3 at the top. You layer a thin particle suspension on the top of the 6 cm column of fluid with a viscosity of 1.0 cp and allow particles to settle at 1 g. How long must you wait for the particles you want (density = 1.07 g/cm3) to sediment to within 0.1 cm of their isopycnic level? Is it possible to determine the time required for particles to sediment to exactly their isopycnic level?Solution:(a)(To be continued)

The dependence of liquid density r on the distance x is:The isopycnic level of r = 1.07 g/cm3 is: The time needed for the particle to sediment to 3.4 cm can be obtained from:(To be continued)

(b) It is not possible to determine the time required for particles to sediment to exactly their isopycnic level (3.5 cm).#

DIFFERENTIAL SETTLING (or CLASSIFICATION)Separation of solid particles into several size fractions based upon the settling velocities in a medium.

If the light and heavy materials both have a range of particle sizes, the smaller, heavy particles settle at the same terminal velocity as the larger, light particles.The terminal settling velocities of components A and B are:For particles of equal settling velocities, vgA = vgB.

In the turbulent Newton's law region, CD is constant.For laminar Stokes law settling,

In the turbulent Newton's law region, CD is constant,For laminar Stokes law settling, For transition flow between laminar and turbulent flow,

Settling a mixture of particles of materials A (the heavier) and B (the lighter) with a size range of d1 to d4 for both types of material:*Size range dA3 to dA4: pure fraction of A No B particles settle as fast as the A particles in this size range.*Size range dB1 to dB2: pure fraction of B No particles of A settle as slowly.

*Size range of A particles from dA1 to dA3 and size range of B particles from dB2 to dB4: form a mixed fraction of A and B *Increasing the density r of the medium. The spread between dA and dB is increased.

[Example] A mixture of silica and galena (; PbS) solid particles having a size range of 5.21 10-6 m to 2.50 10-5 m is to be separated by hydraulic classification using free settling conditions in water at 20C. The specific gravity of silica is 2.65 and that of galena is 7.5. Calculate the size range of the various fractions obtained in the settling. The water viscosity at 20C is 1.005 10-3 Pa-s. Solution: A particles: galena; B particles: silica Assuming Stokes law settling, Check the validity of the Stokes law region.(To be continued)

Solution (contd):For the largest particle and the biggest density, dA = 2.50 10-5 m and rsA = 7.5 g/cm3 = 7500 kg/m3Check:= 0.0547 < 1 O.K. with the Stokes law region.(To be continued)

dA3 = 1.260 10-5 mThe size range of pure A (galena) is: dA3 = 1.260 10-5 m to dA4 = 2.50 10-5 m For particles of equal settling velocities, Solution (contd):(To be continued)

dB2 = 1.033 10-5 mThe size range of pure B (silica) is: dB1 = 5.21 10-6 m to dB2 = 1.033 10-5 m The mixed-fraction size range is:dA1 = 5.21 10-6 m to dA3 = 1.260 10-5 mdB2 = 1.033 10-5 m to dB4 = 2.50 10-5 mSolution (contd):#