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![[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 µm. 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fluid is then withdrawn to isolate the vaccine.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.
g
d
v sg )(
18
2
ρρ
µ
−=Hint:
(To be continued)](https://image.slidesharecdn.com/bs-4-sedimentation-150104042708-conversion-gate02/85/Bs-4-sedimentation-14-320.jpg)
![[Example] You wish to capture 3 µm 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)
g
d
dt
dx
v s )(
18
2
ρρ
µ
−==
ρρ
µ
−
=
s
dx
gd
dt
18
2
(To be continued)](https://image.slidesharecdn.com/bs-4-sedimentation-150104042708-conversion-gate02/85/Bs-4-sedimentation-19-320.jpg)
![[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, g
d
v sg )(
18
2
ρρ
µ
−=
Check the validity of the Stokes’ law region.
(To be continued)](https://image.slidesharecdn.com/bs-4-sedimentation-150104042708-conversion-gate02/85/Bs-4-sedimentation-28-320.jpg)
Uploaded byR.K. JAIN
Bs 4-sedimentation
The document discusses settling and sedimentation processes. It defines filtration versus settling, where settling involves particles separating from fluid due to gravitational forces. It also discusses free versus hindered settling, sedimentation, and the theory of particle movement through fluids. Key factors like particle size, density, fluid properties, and drag forces determine particle settling rates and behaviors. Differential settling can separate particles into fractions based on size-dependent settling velocities.
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Bs 4-sedimentation
- 1.
- 2. • Filtration versussettling 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)
- 3. • Applications ofsettling and sedimentation: * Removal of solids from liquid sewage wastes * Settling of solid food particles from a liquid food Introduction (2/4)
- 4. • Free settlingversus 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 issedimentation? The separation of a dilute slurry or suspension by gravity settling into a clear fluid and a slurry of higher solid content.
- 6. For a rigidparticle of mass m moving in a fluid, there are three forces acting on the body: (1)Gravity force, Fg, acting downward (2) Buoyant force, Fb, acting upward s sb gm gVF ρ ρ ρ == where ρ = density of the fluid ρs = density of the solid particle Vs = volume of the particle THEORY OF PARTICLE MOVEMENT THROUGH A FLUID mgFg =
- 7. For a rigidparticle 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 2 2 DD v ACF ρ = where CD = the drag coefficient A = the projected area of the particle The resultant force equals the force due to acceleration. 2 2 D v AC gm mg dt dv m s ρ ρ ρ −−=
- 8. The falling ofthe 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 Set 0= dt dv and solve the above equation for v. AC mg vv Ds s g ρρ ρρ )(2 − == * vg is called the free settling velocity or terminal velocity. 0 2 2 =−−= v AC gm mg dt dv m D s ρ ρ ρ THEORY OF PARTICLE MOVEMENT THROUGH A FLUID
- 9. AC mg vv Ds s g ρρ ρρ )(2 − == Forspherical particles of diameter d, 4 and 6 23 d A d m s ππρ == D s g C dg v ρ ρρ 3 )(4 − = THEORY OF PARTICLE MOVEMENT THROUGH A FLUID
- 10. The drag coefficientfor rigid spheres has been shown to be a function of the Reynolds number. THEORY OF PARTICLE MOVEMENT THROUGH A FLUID
- 11. In the Stokes'law region (NRe < 1), µρ / 2424 Re D vdN C == gs gss g gv ddvdg C dg v )( 18243 )(4 3 )(4 2 D ρρ µµ ρ ρ ρρ ρ ρρ −=× − = − = g d v sg )( 18 2 ρρ µ −= THEORY OF PARTICLE MOVEMENT THROUGH A FLUID
- 12. • 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
- 13. • Brownian motion(continued) * At particle sizes of a few micrometers, the Brownian effect becomes appreciable and at sizes of less than 0.1 µm, 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
- 14. [Example] Many animalcells 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 µm. 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. g d v sg )( 18 2 ρρ µ −=Hint: (To be continued)
- 15. Example: settling ofdextran beads Data: d = 150 µm = 0.015 cm; µ = 1.1 cP = 0.011 g/cm-s; ρs = 1.02 g/cm3 ; ρ = 1.00 g/cm3 ; g = 980 cm/s2 Solution: g d v sg )( 18 2 ρρ µ −= (a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity. vg = 0.022 cm/s Check: 103.0 011.0 015.0022.01 Re <= ×× == µ ρvd N Liquid volume, V L50 5.144 22 = =× = h h h dt ππ h = 52.3 cm Settling time s2379 cm/s0.22 cm3.52 === gv h (To be continued)
- 16. (b) Estimate thetime to reach the terminal velocity. Solution (cont’d): Force balance: 2 2 v AC gm mg dt dv m D s ρ ρ ρ −−= 6 3 sd m ρπ = vd vd vd v ACD µπ ρπ ρ µρ 3 24 24 2 222 = =; ss d vg g dt dv ρ µ ρ ρ 2 18 −−= gv ddt dv ss −=+ ρ ρ ρ µ 1 18 2 (I.C.: t = 0, v = 0) − −−= t d g d v s s ρ µ ρρ µ 2 2 18 exp1)( 18 Example: settling of dextran beads (To be continued)
- 17. − −−= t d g d v s s ρ µ ρρ µ2 2 18 exp1)( 18 At steady state (t → ∞), g d vv sg )( 18 2 ρρ µ −== When When t >> 1.16 × 10−3 s, v = vg g2 ,1 18 vv d t s =>> ρ µ For v = 0.99vg, t = 5.34 × 10−3 s # (b) Estimate the time to reach the terminal velocity. Solution (cont’d): Example: settling of dextran beads
- 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 wishto capture 3 µm 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) g d dt dx v s )( 18 2 ρρ µ −== ρρ µ − = s dx gd dt 18 2 (To be continued)
- 20. ρρ µ − = s dx gd dt 18 2 The dependence ofliquid density ρ on the distance x is: xx 02.000.1 6 00.112.1 00.1 += − +=ρ The isopycnic level of ρ = 1.07 g/cm3 is: cm5.3 02.0 00.107.1 = − =x The time needed for the particle to sediment to 3.4 cm can be obtained from: ∫∫ − = 3.4 0 2 0 18 ρρ µ s t dx gd dt (To be continued)
- 21. ∫∫ − = 3.4 0 2 0 18 ρρ µ s t dx gd dt ( )∫ +−×× × = − 3.4 0 24 )02.000.1(07.1980103 01.018 x dx t ∫ − − − = 4.3 0 02.007.0 )02.007.0( 02.0 2041 x xd h100.8s823,362 07.0 002.0 ln 02.0 2041 ==−= (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 (orCLASSIFICATION) Separation of solid particles into several size fractions based upon the settling velocities in a medium.
- 23. If the lightand 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: DB BsB gB DA AsA gA C gd v C gd v ρ ρρ ρ ρρ 3 )(4 and 3 )(4 − = − = For particles of equal settling velocities, vgA = vgB. DB DA sA sB B A DB BsB DA AsA C C d d C d C d × − − = − = − ρρ ρρρρρρ or )()(
- 24. DB DA sA sB B A C C d d × − − = ρρ ρρ In the turbulentNewton's law region, CD is constant. ρρ ρρ − − = sA sB B A d d BgB DB AgA DA dv C dv C ρ µ ρ µ 24 and 24 == For laminar Stokes’ law settling, 5.0 − − = ρρ ρρ sA sB B A d d DB DA sA sB B A C C d d × − − = ρρ ρρ A B sA sB DB DA sA sB B A d d C C d d × − − =× − − = ρρ ρρ ρρ ρρ
- 25. DB DA sA sB B A C C d d × − − = ρρ ρρ In the turbulentNewton's law region, CD is constant, ρρ ρρ − − = sA sB B A d d For laminar Stokes’ law settling, 5.0 − − = ρρ ρρ sA sB B A d d For transition flow between laminar and turbulent flow, 1 2 1 where << − − = n d d n sA sB B A ρρ ρρ
- 26. • Settling amixture 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 rangeof 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 ρ of the medium. The spread between dA and dB is increased.
- 28. [Example] A mixtureof 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, g d v sg )( 18 2 ρρ µ −= Check the validity of the Stokes’ law region. (To be continued)
- 29. Solution (cont’d): For thelargest particle and the biggest density, dA = 2.50 × 10-5 m and ρsA = 7.5 g/cm3 = 7500 kg/m3 g d v sg )( 18 2 ρρ µ −= m/s1020.2)8.9)(10007500( )10005.1(18 )1050.2( 3 3 25 gA − − − ×=− × × =v Check : 3 53 Re 10005.1 )1050.2)(1020.2(1000 − −− × ×× == µ ρvd N = 0.0547 < 1 O.K. with the Stokes’ law region. (To be continued)
- 30. 5.0 4 3 − − = ρρ ρρ sA sB B A d d 5.0 5 3 15.7 165.2 1050.2 − − = × − Ad 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 For particles of equal settling velocities, Solution (cont’d): 5.0 − − = ρρ ρρ sA sB B A d d (To be continued)
- 31. 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 5.0 2 1 − − = ρρ ρρ sA sB B A d d 5.0 2 6 15.7 165.21021.5 − − = × − Bd 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 Solution (cont’d): #