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SETTLING AND SEDIMENTATION

SETTLING AND SEDIMENTATION. Introduction (1/4).  Filtration versus settling and sedimentation: Filtration

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SETTLING AND SEDIMENTATION

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  1. SETTLING AND SEDIMENTATION

  2. Introduction (1/4) 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.

  3. Introduction (2/4) Applications of settling and sedimentation: * Removal of solids from liquid sewage wastes * Settling of solid food particles from a liquid food

  4. Free settling versus hindered settling: Free settling  A 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 settling Occurred when the particles are crowded so that they settle at a lower rate.

  5. 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.

  6. THEORY 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: • Gravity force, Fg, acting downward • Buoyant force, Fb, acting upward where r = density of the fluid rs = density of the solid particle Vs = volume of the particle

  7. THEORY 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: (3) Drag force, FD, acting in opposite direction to the particle motion where CD = the drag coefficient A = the projected area of the particle The resultant force equals the force due to acceleration.

  8. THEORY OF PARTICLE MOVEMENT THROUGH A FLUID 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 fall and solve the above equation for v. Set  *vgis called the free settling velocity orterminal velocity.

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

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

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

  12. 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.

  13. 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 helpsreduce the effect of Brownian motion.

  14. [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)

  15. Example: settling of dextran beads Data: 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 (a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity. Solution: vg = 0.022 cm/s Check: Liquid volume, V h = 52.3 cm Settling time (To be continued)

  16. Example: settling of dextran beads (b) Estimate the time to reach the terminal velocity. Solution (cont’d): Force balance: ;   (I.C.: t = 0, v = 0)  (To be continued)

  17. Example: settling of dextran beads (b) Estimate the time to reach the terminal velocity. Solution (cont’d): At steady state (t), When  When t >> 1.16  10-3 s, v = vg For v = 0.99vg, t = 5.34  10-3 s #

  18. 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.

  19. [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)

  20. 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)

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

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

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

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

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

  26. 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.

  27. * 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.

  28. [Example] A mixture of silica and galena (方鉛礦; 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 20C. 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 20C 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)

  29. Solution (cont’d): For the largest particle and the biggest density, dA = 2.50  10-5 m and rsA = 7.5 g/cm3 = 7500 kg/m3 Check: = 0.0547 < 1  O.K. with the Stokes’ law region. (To be continued)

  30. Solution (cont’d): For particles of equal settling velocities,  dA3 = 1.260  10-5 m The size range of pure A (galena) is: dA3 = 1.260  10-5 m to dA4 = 2.50  10-5 m (To be continued)

  31. Solution (cont’d):  dB2 = 1.033  10-5 m The 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 m dB2 = 1.033  10-5 m to dB4 = 2.50  10-5 m #

  32. SETTLING AND SEDIMENTATION

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