Motion of Particles in a Fluid

CHAPTER 6

Motion of Particles in a Fluid 6.1 Introduction Processes for the separation of particles of various sizes and shapes often depend on the variation in the behaviour of the particles when they are subjected to the action of a moving fluid. Further, many of the methods for the determination of the sizes of particles in the subsieve ranges involve relative motion between the particles and a fluid. The flow problems considered in Volume 1A are unidirectional, with the fluid flowing along a pipe or channel, and the effect of an obstruction is discussed only in so far as it causes an alteration in the forward velocity of the fluid. In this chapter, the force exerted on a body as a result of the flow of fluid past it is considered, and, as the fluid is generally diverted all round it, the resulting three-dimensional flow is more complex. Conversely, it is frequently needed to estimate the force required to move an object (or a particle) in a fluid medium at a given velocity. The flow of fluid relative to an infinitely long cylinder, a spherical particle, and a nonspherical particle is considered, followed by a discussion of the motion of particles in both gravitational and centrifugal fields.

6.2 Flow Past a Cylinder and a Sphere The crossflow of fluid past an infinitely long cylinder, in a direction perpendicular to its axis, is considered in the first instance because this involves only two-directional flow, with no flow parallel to the axis. For a nonviscous fluid flowing past a cylinder, as shown in Fig. 6.1, the velocity and direction of flow varies around the circumference. Thus, at points A and D, the fluid is brought to rest, and at B and C, the velocity is at a maximum. Because the fluid is nonviscous, there is no drag, and an infinite velocity gradient exists at the surface of the cylinder. If the fluid is incompressible, and the cylinder is small (so that the changes in the potential energy are negligible), the sum of the kinetic energy and the pressure energy is constant at all points on the surface. The kinetic energy is a maximum at B and C and zero at A and D, so that the pressure falls from A to B and from A to C and rises again from B to D and from C to D, the pressure at A and D being the same. No net force is, therefore, exerted by the fluid on the cylinder. It is found that, although the predicted pressure variation for a nonviscous fluid agrees well with the results obtained with a viscous fluid over the front face of the cylinder, very considerable differences occur at the rear face. Coulson and Richardson’s Chemical Engineering. https://doi.org/10.1016/B978-0-08-101098-3.00007-X # 2019 Elsevier Ltd. All rights reserved.

281

282 Chapter 6

Fig. 6.1 Flow round a cylinder.

It is shown in Volume 1B, Chapter 3 that, when a viscous fluid flows over a surface, the fluid is retarded in the boundary layer which is formed near the surface, and that the boundary layer increases in thickness with increase in distance from the leading edge. If the pressure is falling in the direction of flow, the retardation of the fluid is less, and the boundary layer is thinner in consequence. If the pressure is rising, however, there will be a greater retardation, and the thickness of the boundary layer increases more rapidly. The force acting on the fluid at some point in the boundary layer may then be sufficient to bring it to rest or to cause flow in the reverse direction with the result that an eddy current is set up. A region of reverse flow then exists near the surface where the boundary layer has separated, as shown in Fig. 6.2. The velocity rises from zero at the surface to a maximum negative value and falls again to zero. It then increases in the positive direction until it reaches the main stream velocity at the edge of the boundary layer, as shown in Fig. 6.2. At PQ the velocity in the X-direction is zero and the direction of flow in the eddies must be in the Y-direction. For the flow of a viscous fluid past the cylinder, the pressure decreases from A to B and from A to C so that the boundary layer is thin and the flow is similar to that obtained with a nonviscous fluid. From B to D and from C to D, the pressure is rising and therefore the boundary layer rapidly thickens with the result that it tends to separate from the surface. If separation occurs, eddies are formed in the wake of the cylinder and energy is thereby dissipated and an additional force, known

Fig. 6.2 Flow of fluid over a surface against a pressure gradient.

Motion of Particles in a Fluid 283 as form drag, is set up. In this way, on the forward surface of the cylinder, the pressure distribution is similar to that obtained with the ideal fluid of zero viscosity, although on the rear surface, the boundary layer is thickening rapidly and pressure variations are very different in the two cases. All bodies immersed in a fluid are subject to a buoyancy force. In a flowing fluid, there is an additional force which is made up of two components: the skin friction (or viscous drag) and the form drag (due to the pressure distribution). At low rates of flow, no separation of the boundary layer takes place, although as the velocity is increased, separation occurs, and the skin friction forms a gradually decreasing proportion of the total drag. If the velocity of the fluid is very high, however, or if turbulence is artificially induced, the flow within the boundary layer will change from streamline to turbulent before separation takes place. Because the rate of transfer of momentum through a fluid in turbulent motion is much greater than that in a fluid flowing under streamline conditions, separation is less likely to occur, because the fast-moving fluid outside the boundary layer is able to keep the fluid within the boundary layer moving in the forward direction. If separation does occur, this takes place nearer to point D in Fig. 6.1, the resulting eddies are smaller, and the total drag will be reduced. Turbulence may arise either from an increased fluid velocity or from artificial roughening of the forward face of the immersed body. Prandtl roughened the forward face of a sphere by fixing a hoop to it, with the result that the drag was considerably reduced. Further experiments have been carried out in which sand particles have been stuck to the front face, as shown in Fig. 6.3. The tendency for separation, and hence, the magnitude of the form drag, are also dependent on the shape of the body.

Fig. 6.3 Effect of roughening front face of a sphere (A) 216 mm diameter ball entering water at 298 K (B) as above, except for 100 mm diameter patch of sand on nose.

284 Chapter 6 Conditions of flow relative to a spherical particle are similar to those relative to a cylinder, except that the flow pattern is three-directional. The flow is characterised by the Reynolds number Re0 (¼ udρ/μ), in which ρ is the density of the fluid, μ is the viscosity of the fluid, d is the diameter of the sphere, and u is the velocity of the fluid relative to the particle. For the case of creeping flow, that is, flow at very low velocities relative to the sphere, the drag force F on the particle was obtained in 1851 by Stokes1 who solved the hydrodynamic equations of motion, the Navier–Stokes equations, to give: F ¼ 3πμdu

(6.1)

Eq. (6.1), which is known as Stokes’ law, is applicable only at very low values of the particle Reynolds number, and deviations become progressively greater as Re0 increases. Skin friction constitutes two-thirds of the total drag on the particle, as given by Eq. (6.1). Thus, the total force F is made up of two components:
ðiÞ skin friction : 2π du total 3π μdu ðiiÞ form drag : π μdu As Re0 increases, skin friction becomes proportionately less, and, at values greater than about 20, flow separation occurs with the formation of vortices in the wake of the sphere. At high Reynolds numbers, the size of the vortices progressively increases until, at values of between 100 and 200, instabilities in the flow give rise to vortex shedding. The effect of these changes in the nature of the flow on the force exerted on the particle is now considered.

6.3 The Drag Force on a Spherical Particle 6.3.1 Drag Coefficients The most satisfactory way of representing the relation between drag force and velocity involves the use of two dimensionless groups, similar to those used for correlating information on the pressure drop for flow of fluids in pipes. The first group is the particle Reynolds number Re0 (¼udρ/μ). The second is the group R0 /ρu2, in which R0 is the force per unit projected area of particle in a plane perpendicular to the direction of motion. For a sphere, the projected area is that of a circle of the same diameter as the sphere. Thus : R0 ¼ and

F ðπd 2 =4Þ

R0 4F ¼ 2 2 2 ρu πd ρu

(6.2) (6.3)

Motion of Particles in a Fluid 285 R0 /ρu2 is a form of drag coefficient, often denoted by the symbol CD0 . Frequently, a drag coefficient CD is defined as the ratio of R0 to 12 ρu2 . Thus : CD ¼ 2C0D ¼

2R0 ρu2

(6.4)

It is seen that C0D is analogous to the friction factor ϕ(¼R/ρu2) for pipe flow, and CD is analogous to the Fanning friction factor f. When the force F is given by Stokes’ law (Eq. 6.1), then: R0 μ 1 ¼ 12 ¼ 12Re0 2 ρu udρ

(6.5)

Eqs (6.1) and (6.5) are applicable only at very low values of the Reynolds number Re0 . Goldstein2 has shown that, for values of Re0 up to about 2, the relation between R0 /ρu2 and Re0 is given by an infinite series of which Eq. (6.5) is just the first term. Thus:


R0 12 3 0 19 71 30,179 122,519 02 03 04 05 ¼ 1 + Re  Re + Re  Re + Re  ⋯ ρu2 Re0 16 1280 20, 480 34, 406,400 560,742,400 (6.6)

Oseen3 employs just the first two terms of Eq. (6.6) to give:   R0 3 0 0 1 ¼ 12Re 1 + Re : ρu2 16

(6.7)

The correction factors for Stokes’ law from both Eq. (6.6) and Eq. (6.7) are given in Table 6.1. It is seen that the correction becomes progressively greater as Re0 increases. Several researchers have used numerical methods for solving the equations of motion for flow at higher Reynolds numbers relative to spherical and cylindrical particles. These include, Jenson,4 le Clair et al.,5 Fornberg,6, 7 and Johnson and Patel.8 The relation between R0 /ρu2 and Re0 is conveniently given in graphical form by means of a logarithmic plot as shown in Fig. 6.4. The graph may be divided into four regions as shown. The four regions are now considered in turn. Region (a) (104 < Re0 < 0.2)

R0 In this region, the relationship between 2 and Re0 is a straight line of slope  1 represented by ρu Eq. (6.5): R0 1 ¼ 12Re0 ðEq:6:5Þ 2 ρu

286 Chapter 6 Table 6.1 Correction factors for Stokes’ law

Re0 0.01 0.03 0.1 0.2 0.3 0.6 1 2

Goldstein (Eq. 6.6) 1.002 1.006 1.019 1.037 1.055 1.108 1.18 1.40

Oseen (Eq. 6.7)

Schiller and Naumann (Eq. 6.9)

Wadell (Eq. 6.12)

Khan and Richardson (Eq. 6.13)

1.002 1.006 1.019 1.037 1.056 1.113 1.19 1.38

1.007 1.013 1.03 1.05 1.07 1.11 1.15 1.24

0.983 1.00 1.042 1.067 1.115 1.346 1.675 1.917

1.038 1.009 1.006 1.021 1.038 1.085 1.137 1.240

Fig. 6.4 R0 /ρu2 versus Re0 for spherical particles.

The limit of 104 is imposed because reliable experimental measurements have not been made at lower values of Re0 , although the equation could be applicable down to very low values of Re0 , provided that the dimensions of the particle are large compared with the mean free path of the fluid molecules so that the fluid behaves as a continuum.

Motion of Particles in a Fluid 287 The upper limit of Re0 ¼ 0.2 corresponds to the condition where the error arising from the application of Stokes’ law is about 4%. This limit should be reduced if a greater accuracy is required, and it may be raised if a lower level of accuracy is acceptable. Region (b) (0.2 < Re0 < 500–1000) In this region, the slope of the curve changes progressively from 1 to 0 as Re0 increases. Several researchers have suggested approximate equations for flow in this intermediate region. Dallavelle9 proposed that R0 /ρu2 may be regarded as being composed of two component parts, one due to Stokes’ law and the other, a constant, due to additional nonviscous effects. Thus :

R0 1 ¼ 12Re0 + 0:22 ρu2

(6.8)

Schiller and Naumann10 gave the following simple equation which gives a reasonable approximation for values of Re0 up to about 1000:   R0 0 1 0 0:687 (6.9) ¼ 12Re 1 + 0:15Re ρu2 Region (c) (500–1000 < Re0 < c.2  105) In this region, Newton’s law is applicable, and the value of R0 /ρu2 is approximately constant giving: R0 ¼ 0:22 ρu2

(6.10)

Region (d) (Re0 > 2  105) When Re0 exceeds about 2  105, the flow in the boundary layer changes from streamline to turbulent and the separation takes place nearer to the rear of the sphere. The drag force is decreased considerably and: R0 ¼ 0:05 ρu2

(6.11)

A recent study11 sheds some light on the underlying physics of the drag behaviour in this regime. Values of R0 /ρu2 using Eqs (6.5), (6.9), (6.10), and (6.11) are given in Table 6.2 and plotted in Fig. 6.4. The curve shown in Fig. 6.4 is really continuous, and its division into four regions is merely a convenient means by which a series of simple equations can be assigned to limited ranges of values of Re0 .

288 Chapter 6 Table 6.2 R0 /ρu2, (R0 /ρu2)Re0 2 and (R0 /ρu2)Re0 21 as a function of Re0 Re0

Re0 /ρu2

103 2  103 5  103 102 2  102 5  102 101 2  101 5  101 100 2  100 5  100 10 2  10 5  10 102 2  102 5  102 103 2  103 5  103 104 2  104 5  104 105 2  105 5  105 106 2  106 5  106 107

12,000 6000 2400 1200 600 240 124 63 26.3 13.8 7.45 3.49 2.08 1.30 0.768 0.547 0.404 0.283 0.221 0.22 0.22 0.22 0.22 0.22 0.22 0.05 0.05 0.05 0.05 0.05 0.05

(R0 /ρu2)Re0 2

(R0 /ρu2)Re0 21

1.20  101 2.40  101 6.00  101 1.24 2.52 6.4 1.38  10 2.98  10 8.73  10 2.08  102 5.20  102 1.92  103 5.47  103 1.62  104 7.08  104 2.21  105 8.8  105 5.5  106 2.2  107

1.20  105 3.00  104 4.80  103 1.24  103 3.15  102 5.26  10 1.38  10 3.73 7.00  101 2.08  101 6.50  102 1.54  102 5.47  103 2.02  103 5.70  104 2.21  104 1.1  104 4.4  105 2.2  105

A comprehensive review of the various equations proposed to relate drag coefficient to particle Reynolds number has been carried out by Clift et al.12 and Michaelides.13 One of the earliest equations applicable over a wide range of values of Re0 is that due to Wadell,14 which may be written as: !2 R0 3:39 ¼ 0:445 + (6.12) ρu2 √Re0 Subsequently, Khan and Richardson15 have examined the experimental data and suggest that a very good correlation between R0 /ρu2 and Re0 , for values of Re0 up to 105, is given by: i3:45 R0 h 0 0:31 0 0:06 ¼ 1:84Re + 0:293Re (6.13) ρu2

Motion of Particles in a Fluid 289 In Table 6.3, values of R0 /ρu2, calculated from Eqs (6.12) and (6.13), together with values from the Schiller and Naumann Eq. (6.9), are given as a function of Re0 over the range 102 < Re0 < 105. Values are plotted in Fig. 6.5, from which it will be noted that Eq. (6.13) gives a shallow minimum at Re0 of about 104, with values rising to 0.21 at Re0 ¼ 105. This agrees with the limited experimental data which are available in this range.

Table 6.3 Values of drag coefficient R0 /ρu2 as a function of Re0 Schiller and Naumann10 (Eq. 6.9)

Re0 0.01 0.1 1 10 100 500 1000 3000 10,000 30,000 100,000

1208 124 13.8 2.07 0.55 0.281 0.219 0.151 0.10 – –

Drag coefficient (R⬘/(ru2))

Wadell

(Eq. 6.12)

Khan and Richardson15 (Eq. 6.13)

1179 125 14.7 2.3 0.62 0.356 0.305 0.257 0.229 0.216 0.208

50 100

20,000 10,000

14

1246 121 13.7 2.09 0.52 0.283 0.234 0.200 0.187 0.191 0.210

1000

10,000

100,000 1

1000

100

10

0.1 Legend Wadell

1

Khan and Richardson Schiller and Naumann 0.1 0.001

0.01

0.1

1 10 100 Reynolds number (Re⬘)

1000

10,000 100,000

Fig. 6.5 R0 /ρu2 versus Re0 for spherical particles from Eqs (6.9) (Schiller and Naumann10), (6.12) (Wadell14) and (6.13) (Khan and Richardson15). The enlarged section covers the range of Re0 from 50 to 105.

290 Chapter 6 For values of Re0 < 2, correction factors for Stokes’ law have been calculated from Eqs (6.9), (6.12), and (6.13), and these are included in Table 6.1.

6.3.2 Total Force on a Particle The force on a spherical particle may be expressed using Eqs (6.5), (6.9), (6.10), and (6.11) for each of the regions a, b, c, and d as follows.   μ 12uμ ¼ (6.14) In regionðaÞ : R0 ¼ 12ρu2 udρ d The projected area of the particle is πd2/4. Thus, the total force on the particle is given by: 12uμ 1 2 πd ¼ 3πμdu d 4 This is the expression originally obtained by Stokes1 already given as Eq. (6.1). F¼

(6.15)

In region (b), from Eq. (6.9):  12uμ  0:687 1 + 0:15Re0 d   0:687 and therefore : F ¼ 3πμdu 1 + 0:15Re0

(6.17)

In regionðcÞ : R0 ¼ 0:22ρu2

(6.18)

R0 ¼

1 and : F ¼ 0:22ρu2 πd2 ¼ 0:055πd2 ρu2 4 This relation is often known as Newton’s law.

(6.16)

(6.19)

In region (d): R0 ¼ 0:05ρu2

(6.20)

F ¼ 0:0125πd2 ρu2

(6.21)

Alternatively, using Eq. (6.13), which is applicable over the first three regions (a), (b), and (c) gives:   π 0:31 0:06 3:45 + 0:293Re0 (6.22) F ¼ d2 ρu2 1:84Re0 4

6.3.3 Terminal Falling Velocities If a spherical particle is allowed to settle in a fluid under gravity, its velocity will increase until the accelerating force is exactly balanced by the resistance force. Although this state is approached exponentially, the effective acceleration period is generally of short duration for

Motion of Particles in a Fluid 291 very small particles. If this terminal falling velocity is such that the corresponding value of Re0 is <0.2, the drag force on the particle is given by Eq. (6.15). If the corresponding value of Re0 lies between 0.2 and 500, the drag force is given approximately by Schiller and Naumann in Eq. (6.17). It may be noted, however, that if the particle has started from rest, the drag force is given by Eq. (6.15) until Re0 exceeds 0.2. Again, if the terminal falling velocity corresponds to a value of Re0 greater than about 500, the drag on the particle is given by Eq. (6.19). Under terminal falling conditions, velocities are rarely high enough for Re0 to approach 105, with the small particles generally used in industry. The accelerating force due to gravity is given by:   1 3 ¼ πd ðρs  ρÞg 6

(6.23)

where ρs is the density of the solid. The terminal falling velocity u0 corresponding to region (a) is given by:   1 3 πd ðρs  ρÞg ¼ 3πμdu0 6 and : u0 ¼

d2 g ðρ  ρÞ 18μ s

(6.24)

The terminal falling velocity corresponding to region (c) is given by:   1 3 πd ðρs  ρÞg ¼ 0:055πd 2 ρu20 6 or : u20 ¼ 3dg

ðρs  ρÞ ρ

(6.25)

In the expressions given for the drag force and the terminal falling velocity, the following assumptions have been made: (a) That the settling is not affected by the presence of other particles in the fluid. This condition is known as ‘free settling’. When the interference of other particles is appreciable, the process is known as ‘hindered settling’. (b) That the walls of the containing vessel do not exert an appreciable retarding effect. (c) That the fluid can be considered as a continuous medium, that is the particle is large compared with the mean free path of the molecules of the fluid, otherwise the particles may occasionally ‘slip’ between the molecules and thus attain a velocity higher than that calculated using the aforementioned expressions. These factors are considered further in Sections 6.3.4 and 6.3.5 and in Chapter 8.

292 Chapter 6 From Eqs (6.24) and (6.25), it is seen that terminal falling velocity of a particle in a given fluid becomes greater as both particle size and density are increased. If for a particle of material A of diameter dA and density ρA, Stokes’ law is applicable, then the terminal falling velocity u0A is given by Eq. (6.24) as: u0A ¼

dA2 g ðρ  ρÞ 18μ A

(6.26)

dB2 g ðρ  ρÞ 18μ B

(6.27)

Similarly, for a particle of material B: u0B ¼

The condition for the two terminal velocities to be equal is then:   dB ρ  ρ 1=2 ¼ A dA ρB  ρ

(6.28)

If Newton’s law is applicable, Eq. (6.25) holds, and: u20A ¼

3dA gðρA  ρÞ ρ

and u20B ¼ For equal settling velocities:

3dB gðρB  ρÞ ρ

  dB ρA  ρ ¼ dA ρB  ρ

In general, the relationship for equal settling velocities is:   dB ρA  ρ S ¼ dA ρB  ρ where S ¼

(6.29) (6.30)

(6.31)

(6.32)

1 for the Stokes’ law region, S ¼ 1 for Newton’s law, and, as an approximation, 2

1 < S < 1 for the intermediate region. 2 This method of calculating the terminal falling velocity is satisfactory provided that it is known a priori which equation should be used for the calculation of drag force or drag coefficient. It has already been seen that the equations give the drag coefficient in terms of the particle Reynolds number Re00 (¼u0dρ/μ), which is itself a function of the terminal falling velocity u0 which is to be determined. The problem is analogous to that discussed in Volume 1A, where the calculation of the velocity of flow in a pipe in terms of a known pressure difference

Motion of Particles in a Fluid 293 presents difficulties, because the unknown velocity appears in both the friction factor and the Reynolds number. The problem is most effectively solved by introducing a new dimensionless group which is independent of the particle velocity. The resistance force per unit projected area of the particle under terminal falling conditions R00 is given by: 1 1 R00 πd2 ¼ πd 3 ðρs  ρÞg 4 6 2 or : R00 ¼ dðρs  ρÞg 3 Thus :

(6.33)

R00 2dg ð ρ  ρÞ ¼ 2 ρu0 3ρu20 s

(6.34)

The dimensionless group (R00 /ρu20)Re0 20 does not involve u0 because: R00 u20 d2 ρ2 2dgðρs  ρÞ u20 d2 ρ2 ¼ μ2 3ρu20 ρu20 μ2

(6.35)

2d 3 ðρs  ρÞρg ¼ 3μ2

d 3 ρðρs  ρÞg is known as the Galileo number Ga or sometimes the Archimedes The group μ2 number Ar. Thus :

R00 0 2 2 Re ¼ Ga ρu20 0 3

(6.36)

Using Eqs (6.5), (6.9), and (6.10) to express R0 /ρu2 in terms of Re0 over the appropriate range of Re0 , then: Ga ¼ 18Re00 ðGa < 3:6Þ

(6.37)

1:687 

Ga ¼ 18Re00 + 2:7Re0 0

3:6 < Ga < 105



1 2  Ga ¼ Re0 0 Ga > 105 3 (R00 /ρu20)Re0 20 can be evaluated if the properties of the fluid and the particle are known.

(6.38) (6.39)

294 Chapter 6 Table 6.4 Values of log Re0 as a function of log{(R0 /ρu2)Re0 2} for spherical particles log{(R0 /ρu2) Re0 2} 2 1 0 1 2 3 4 5

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

3:919 2.908 1.874 0.738 1.491 2.171 2.783

2:018 1.007 1.967 0.817 1.562 2.236 2.841

2:117 1.105 0.008 0.895 1.632 2.300 2.899

2:216 1.203 0.148 0.972 1.702 2.363 2.956

2.315 1.301 0.236 1.048 1.771 2.425 3.013

2.414 1.398 0.324 1.124 1.839 2.487 3.070

2.513 1.495 0.410 1.199 1.907 2.548 3.127

3.620 2.612 1.591 0.495 1.273 1.974 2.608 3.183

3.720 2.711 1.686 0.577 1.346 2.040 2.667 3.239

3.819 2.810 1.781 0.659 1.419 2.106 2.725 3.295

In Table 6.4, values of log Re0 are given as a function of log{(R0 /ρu2)Re0 2} and the data taken from tables given by Heywood,16 are represented in graphical form in Fig. 6.6. In order to determine the terminal falling velocity of a particle, (R00 /ρu20)Re0 20 is evaluated, and the corresponding value of Re00 , and hence, of the terminal velocity, is found either from Table 6.4 or from Fig. 6.6.

Example 6.1 What is the terminal velocity of a spherical steel particle, 0.40 mm in diameter, settling in an oil of density 820 kg/m3 and viscosity 10 mN s/m2? The density of steel is 7870 kg/m3. Solution For a sphere:

¼

R00 0 2 2d 3 ðρs  ρÞρg Re ¼ ðEq:6:35Þ ρu20 0 3μ2 2  0:00043  820ð7870  820Þ9:81 2

3ð10  103 Þ ¼ 24:2 log 10 24:2 ¼ 1:384 From Table 6:4 : log 10 Re00 ¼ 0:222 Thus : Re00 ¼ 1:667 1:667  10  103 and : u0 ¼ 820  0:004 ¼ 0:051m=s or 51mm=s

Motion of Particles in a Fluid 295

Fig. 6.6 (R0 /ρu2)Re0 2 and (R0 /ρu2)Re0 1 versus Re0 for spherical particles.

296 Chapter 6 Example 6.2 A finely ground mixture of galena and limestone in the proportion of 1–4 by mass is subjected to elutriation by an upward-flowing stream of water flowing at a velocity of 5 mm/s. Assuming that the size distribution for each material is the same, and is as shown in the following table, estimate the percentage of galena in the material carried away and in the material left behind. The viscosity of water is 1 mN s/m2 and Stokes’ Eq. (6.1) may be used. Diameter (μm) Undersize (percent by mass)

20 15

30 28

40 48

50 54

60 64

70 72

80 78

100 88

The densities of galena and limestone are 7500 and 2700 kg/m3, respectively. Solution The first step is to determine the size of a particle which has a settling velocity equal to that of the upward flow of fluid, that is 5 mm/s. Taking the largest particle, d ¼ (100  106) ¼ 0.0001 m   and : Re0 ¼ 5  103  0:0001  1000 = 1  103 ¼ 0:5 Thus, for the bulk of particles, the flow will be within region (a) in Fig. 6.4, and the settling velocity is given by Stokes’ equation:  u0 ¼ d 2 g=18μ ðρs  ρÞ ðEq:6:24Þ For a particle of galena settling in water at 5 mm/s: 

  5  103 ¼ d 2  9:81 = 18  103 ð7500  1000Þ ¼ 3:54  106 d 2 and : d ¼ 3:76  105 mor37:6 μm

For a particle of limestone settling at 5 mm/s: 

  5  103 ¼ d 2  9:81 = 18  103 ð2700  1000Þ ¼ 9:27  105 d 2 and : d ¼ 7:35  105 mor 73:5 μm

Thus, particles of galena of <37.6 μm and particles of limestone of <73.5 μm will be removed in the water stream. Interpolation of the data given shows that 43% of the galena and 74% of the limestone will be removed in this way. In 100 kg feed, there is 20 kg galena and 80 kg limestone. Therefore, galena removed ¼ (20  0.43) ¼ 8.6 kg, leaving 11.4 kg, and limestone removed ¼ (80  0.74) ¼ 59.2 kg, leaving 20.8 kg. Hence, in the material removed: Concentration of galena ¼ ð8:6  100Þ=ð8:6 + 59:2Þ ¼ 12:7% by mass and in the material remaining: Concentration of galena ¼ ð11:4  100Þ=ð11:4 + 20:8Þ ¼ 35:4% by mass

Motion of Particles in a Fluid 297 As an alternative, the data used for the generation of Eq. (6.13) for the relation between drag coefficient and particle Reynolds number may be expressed as an explicit relation between Re00 (the value of Re0 at the terminal falling condition of the particle) and the Galileo number Ga. The equation takes the form15:  13:3 Re00 ¼ 2:33Ga0:018  1:53Ga0:016 (6.40) The Galileo number is readily calculated from the properties of the particle and the fluid, and the corresponding value of Re00 , from which u0 can be found, is evaluated from Eq. (6.40). A similar difficulty is encountered in calculating the size of a sphere having a given terminal falling velocity, because Re00 and R00 /ρu2 are both functions of the diameter d of the particle. This calculation is similarly facilitated by the use of another combination, (R00 /ρu20)Re0 1 0 , which is independent of diameter. This is given by: R00 0 1 2μg Re 0 ¼ 2 3 ðρs  ρÞ 2 ρu0 3ρ u0

(6.41)

Log Re0 is given as a function of log[(R0 /ρu2)Re0 1] in Table 6.5 and the functions are plotted in Fig. 6.6. The diameter of a sphere of known terminal falling velocity may be calculated by 0 evaluating (R00 /ρu20)Re0 1 0 , and then finding the corresponding value of Re0 , from which the diameter may be calculated. As an alternative to this procedure, the data used for the generation of Eq. (6.13) may be expressed to give Re00 as an explicit function of {(R00 /ρu20)Re0 1 0 }, which from Eq. (6.40) is equal to 2/3[(μg/ρ2u30)(ρs  ρ)]. Then, writing KD ¼ (μg/ρ2u30)(ρs  ρ)], Re00 may be obtained from:  3:56 Re00 ¼ 1:47KD0:14 + 0:11KD0:4

(6.42)

d may then be evaluated because it is the only unknown quantity involved in the Reynolds number. Table 6.5 Values of log Re0 as a function of log{(R0 /ρu2)Re0 21} for spherical particles log{(R0 /ρu2) Re0 21} 5 4 3 2 1 0 1 2 3 4

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

3.316 2.517 1.829 1.200 0.616 0.072 1.554 1.047 2.544

3.231 2.443 1.763 1.140 0.560 0.019 1.503 2.996 2.493

3.148 2.372 1.699 1.081 0.505 1.969 1.452 2.946 2.443

3.065 2.300 1.634 1.022 0.449 1.919 1.401 2.895 2.393

2.984 2.231 1.571 0.963 0.394 1.865 1.350 2.845 2.343

2.903 2.162 1.508 0.904 0.339 1.811 1.299 2.794 2.292

2.824 2.095 1.496 0.846 0.286 1.760 1.249 2.744

2.745 2.027 1.383 0.788 0.232 1.708 1.198 2.694

2.668 1.961 1.322 0.730 0.178 1.656 1.148 2.644

3.401 2.591 1.894 1.260 0.672 0.125 1.605 1.097 2.594

298 Chapter 6

6.3.4 Rising Velocities of Light Particles Although there appears to be no problem in using the standard relations between drag coefficient and particle Reynolds number for the calculation of terminal falling velocities of particles denser than the liquid, Karamanev et al.17 have shown experimentally that, for light particles rising in a denser liquid, an overestimate of the terminal rising velocity may result. This can occur in the Newton’s law region and may be associated with an increase in the drag coefficient CD0 from the customary value of 0.22 for a spherical particle up to a value as high as 0.48. Vortex shedding behind the rising particle may cause it to take a longer spiral path, thus, reducing its vertical component of velocity. A similar effect is not observed with a falling dense particle because its inertia is too high for vortex-shedding to have a significant effect. Further experimental work by Dewsbury et al.18 with shear-thinning power-law solutions of CMC (carboxymethylcellulose) has shown similar effects.

6.3.5 Effect of Boundaries The discussion so far relates to the motion of a single spherical particle in an effectively infinite expanse of fluid. If other particles are present in the neighborhood of the sphere, the sedimentation velocity will be decreased, and the effect will become progressively more marked as the concentration is increased. There are three contributory factors. First, as the particles settle, they will displace an equal volume of fluid, and this gives rise to an upward flow of liquid. Second, the buoyancy force is influenced because the suspension has a higher density than the fluid. Finally, the flow pattern of the liquid relative to the particle is altered, and velocity gradients are affected. The settling of concentrated suspensions is discussed in detail in Chapter 8. The boundaries of the vessel containing the fluid in which the particle is settling will also affect its settling velocity. If the ratio of diameter of the particle (d) to that of the tube (dt) is significant, the motion of the particle is retarded. Two effects arise. First, as the particle moves downwards, it displaces an equal volume of liquid which must rise through the annular region between the particle and the wall. Second, the velocity profile in the fluid is affected by the presence of the tube boundary. There have been several studies19–25 of the influence of the walls, most of them in connection with the use of the ‘falling sphere’ method of determining viscosity, in which the viscosity is calculated from the settling velocity of the sphere. The resulting correction factors have been tabulated by Clift et al.12 and Chhabra.25 The effect is difficult to quantify accurately because the particle will not normally follow a precisely uniform vertical path through the fluid. It is, therefore, useful also to take into account work on the sedimentation of suspensions of uniform spherical particles at various concentrations, and to extrapolate the results to zero concentration to obtain the free falling velocity for different values of the ratio d/dt. The correction factor for the influence of the walls of the tube on the

Motion of Particles in a Fluid 299 settling velocity of a particle situated at the axis of the tube was calculated by Ladenburg,19 who has given the equation:   u0t d 1 ¼ 1 + 2:4 ðd=dt < 0:1Þ (6.43) u0 dt where u0t is the settling velocity in the tube, and u0 is the free falling velocity in an infinite expanse of fluid. Eq. (6.43) was obtained for the Stokes’ law regime. It overestimates the wall effect, however, at higher particle Reynolds number (Re0 > 0.2). Similar effects are obtained with noncylindrical vessels, although, in the absence of adequate data, it is best to use the correlations for cylinders, basing the vessel size on its hydraulic mean diameter, which is four times the ratio of the cross-sectional area to the wetted perimeter. The particles also suffer a retardation as they approach the bottom of the containing vessel because the lower boundary then influences the flow pattern of the fluid relative to the particle. This problem has been studied by Ladenburg,19 Tanner,26 and Sutterby.27 Ladenburg19 gives the following equation:   u0t d 1 ¼ 1 + 1:65 0 (6.44) u0 L where L0 is the distance between the centre of the particle and the lower boundary, for the Stokes’ law regime.

6.3.6 Behaviour of Very Fine Particles Very fine particles, particularly in the submicron range (d < 1 μm), are very readily affected by natural convection currents in the fluid, and great care must be taken in making measurements to ensure that temperature gradients are eliminated. The behaviour is also affected by Brownian motion. The molecules of the fluid bombard each particle in a random manner. If the particle is large, the net resultant force acting at any instant may be large enough to cause a change in its direction of motion. This effect has been studied by Davies,28 who has developed an expression for the combined effects of gravitation and Brownian motion on particles suspended in a fluid. In the preceding treatment, it has been assumed that the fluid constitutes a continuum, and that the size of the particles is small compared with the mean free path λ of the molecules. Particles of diameter d < 0.1 μm in gases at atmospheric pressure (and for larger particles in gases at low pressures) can “slip” between the molecules, and, therefore, attain higher than predicted settling velocities. According to Cunningham,29 the slip factor is given by:

300 Chapter 6 1+β

λ d

(6.45)

Davies30 gives the following expression for β: β ¼ 1:764 + 0:562e0:785ðd=λÞ

(6.46)

More details can be found in a book by Reist.31

6.3.7 Effect of Turbulence in the Fluid If a particle is moving in a fluid which is in laminar flow, the drag coefficient is approximately equal to that in a still fluid, provided that the local relative velocity at the particular location of the particle is used in the calculation of the drag force. When the velocity gradient is sufficiently large to give a significant variation of velocity across the diameter of the particle; however, the estimated force may be somewhat in error. When the fluid is in turbulent flow, or where turbulence is generated by some external agent such as an agitator, the drag coefficient may be substantially increased. Brucato et al.32 have shown that the increase in drag coefficient may be expressed in terms of the Kolmogoroff scale of the eddies (λE) given by: h i1=4 (6.47) λE ¼ ðμ=ρÞ3 =ε where ε is the mechanical power generated per unit mass of fluid by an agitator, for example. The increase in the drag coefficient CD over that in the absence of turbulence CD0 is given by: η ¼ ðCD  CD0 Þ=CD0 ¼ 8:76  104 ðd=λE Þ3

(6.48)

Values of η of up to about 30, have been reported.

6.3.8 Effect of Motion of the Fluid If the fluid is moving relative to some surface other than that of the particle, there will be a superimposed velocity distribution, and the drag on the particle may be altered. Thus, if the particle is situated at the axis of a vertical tube up which fluid is flowing in streamline motion, the velocity near the particle will be twice the mean velocity because of the parabolic velocity profile in the fluid. The drag force is then determined by the difference in the velocities of the fluid and the particle at the axis. The effect of turbulence in the fluid stream has been studied by Richardson and Meikle,33 who suspended a particle on a thread at the centre of a vertical pipe up which water was passed under conditions of turbulent flow. The upper end of the thread was attached to a lever

Motion of Particles in a Fluid 301 fixed on a coil free to rotate in the field of an electromagnet. By passing a current through the coil, it was possible to bring the level back to a null position. After calibration, the current required could be related to the force acting on the sphere. The results were expressed as the friction factor (R0 /ρu2), which was found to have a constant value of 0.40 for particle Reynolds numbers (Re0 ) over the range from 3000 to 9000, and for tube Reynolds numbers (Re) from 12,000 to 26,000. Thus, the value of R0 /ρu2 has been approximately doubled as a result of turbulence in the fluid. By surrounding the particle with a fixed array of similar particles on a hexagonal spacing, the effect of neighboring particles was measured. The results are discussed in Chapter 8. Rowe and Henwood34 made similar studies by supporting a spherical particle 12.7 mm diameter, in water, at the end of a 100 mm length of fine nichrome wire. The force exerted by the water when flowing in a 150 mm square duct was calculated from the measured deflection of the wire. The experiments were carried out at low Reynolds numbers with respect to the duct (<1200), corresponding to between 32 and 96 relative to the particle. The experimental values of the drag force were about 10% higher than those calculated from the Schiller and Naumann equation, Eq. (6.9). The work was then extended to cover the measurement of the force on a particle surrounded by an assemblage of particles, as described in Chapter 8. If Re0 is of the order of 105, the drag on the sphere may be reduced if the fluid stream is turbulent. The flow in the boundary layer changes from streamline to turbulent and the size of the eddies in the wake of the particle is reduced. The higher the turbulence of the fluid, the lower is the value of Re0 at which the transition from region (c) to region (d) occurs. The value of Re0 at which R0 /ρu2 is 0.15 is known as the turbulence number and is taken as an indication of the degree of turbulence in the fluid.

6.4 Nonspherical Particles 6.4.1 Effect of Particle Shape and Orientation on Drag There are two difficulties which soon become apparent when attempting to assess the very large amount of experimental data which are available on drag coefficients and terminal falling velocities for nonspherical particles. The first is that an infinite number of nonspherical shapes exists, and the second is that each of these shapes is associated with an infinite number of orientations which the particle is free to take up in the fluid, and the orientation may oscillate during the course of settling. In a recent comprehensive study, Chhabra et al. 35 have found that the most satisfactory characteristic linear dimension to use is the diameter of the sphere of equal volume, and that the most relevant characteristic of particle shape is the sphericity, (surface area of particle/surface

302 Chapter 6 area of sphere of equal volume). The limitation of this whole approach is that mean errors are often as high as about 16%, and maximum errors may be of the order of 100%. The extent of the errors may be reduced, however, by using separate shape factors in the Stokes’ and Newton’s law regions. Another problem is that, when settling, a nonspherical particle will not travel vertically in a fixed orientation unless it has a plane of symmetry which is horizontal. In general, the resistance force to movement in the gravitational field will not act vertically, and the particle will tend to spiral, to rotate, and to wobble. A spherical particle is unique in that it presents the same area to the oncoming fluid whatever its orientation. For nonspherical particles, the orientation must be specified before the drag force can be calculated. The experimental data for the drag can be correlated in the same way as for the sphere, by plotting the dimensionless group R0 /ρu2 against the Reynolds number, Re0 ¼ ud0 ρ/μ, using logarithmic coordinates, and a separate curve is obtained for each shape of particle and for each orientation. In these groups, R0 is taken, as before, as the resistance force per unit area of particle, projected on to a plane perpendicular to the direction of flow. d0 is defined as the diameter of the circle having the same area as the projected area of the particle and is, therefore, a function of the orientation, as well as the shape, of the particle. The curve for R0 /ρu2 against Re0 may be divided into four regions, (a), (b), (c), and (d), as before. In region (a) the flow is entirely streamline, and, although no theoretical expressions have been developed for the drag on the particle, the practical data suggest that a law of the form: R0 1 ¼ K Re0 ρu2

(6.49)

is applicable. The constant K varies somewhat according to the shape and orientation of the particle, although it always has a value of about 12. In this region, a particle falling freely in the fluid under the action of gravity will normally move with its longest surface nearly parallel to the direction of motion. At higher values of Re0 , the linear relation between R0 /ρu2 and Re0 1 no longer holds and the slope of the curve gradually changes until R0 /ρu2 becomes independent of Re0 in region (c). Region (b) represents transition conditions and commences at a lower value of Re0 , and a correspondingly higher value of R0 /ρu2, than in the case of the sphere. A freely falling particle will tend to change its orientation as the value of Re0 changes, and some instability may be apparent. In region (c), the particle tends to fall so that it is presenting the maximum possible surface to the oncoming fluid. Typical values of R0 /ρu2 for nonspherical particles in region (c) are given in Table 6.6. It may be noted that all these values of R0 /ρu2 are higher than the value of 0.22 for a sphere. Clift et al.12 and Darby and Chhabra36 have critically reviewed the information available on the drag of nonspherical particles.

Motion of Particles in a Fluid 303 Table 6.6 Drag coefficients for nonspherical particles Configuration Thin rectangular plates with their planes perpendicular to the direction of motion Cylinders with axes parallel to the direction of motion Cylinders with axes perpendicular to the direction of motion

Length/Breadth

R0 /ρu2

1–5 20 ∞ 1

0.6 0.75 0.95 0.45

1 5 20 ∞

0.3 0.35 0.45 0.6

6.4.2 Terminal Falling Velocities Heywood16 has developed an approximate method for calculating the terminal falling velocity of a nonspherical particle, or for calculating its size from its terminal falling velocity. The method is an adaptation of his method for spheres. A mean projected diameter of the particle dp is defined as the diameter of a circle having the same area as the particle when viewed from above and lying in its most stable position. Heywood selected this particular dimension because it is easily measured by microscopic examination. If dp is the mean projected diameter, the mean projected area is πd2p/4 and the volume is k0 d3p, where k0 is a constant whose value depends on the shape of the particle. For a spherical particle, k0 is equal to π/6. For rounded isometric particles, that is, particles in which the dimension in three mutually perpendicular directions is approximately the same, k0 is about 0.5, and for angular particles, k0 is about 0.4. For most minerals, the value of k0 lies between 0.2 and 0.5. The method of calculating the terminal falling velocity consists in evaluating (R00 /ρu2)Re0 20, using dp as the characteristic linear dimension of the particle and πd2p/4 as the projected area in a plane perpendicular to the direction of motion. The corresponding value of Re00 is then found from Table 6.4 or from Fig. 6.6, which both refer to spherical particles, and a correction is then applied to the value of log Re00 to account for the deviation from spherical shape. Values of this correction factor, which is a function both of k0 and of (R0 /ρu2)Re0 2, are given in Table 6.7. A similar procedure is adopted for calculating the size of a particle of given terminal velocity, using Tables 6.5 and 6.8. The method is only approximate because it is assumed that k0 completely defines the shape of the particle, whereas there are many different shapes of particle for which the value of k0 is the same. Further, it assumes that the diameter dp is the same as the mean projected diameter d0 . This is very nearly so in regions (b) and (c), although in region (a), the particle tends to settle so

304 Chapter 6 Table 6.7 Corrections to log Re0 as a function of log{(R0 /ρu2)Re0 2} for nonspherical particles log{(R0 /ρu2)Re0 2} 2 1 0 1 2 2.5 3 3.5 4 4.5 5 5.5 6

k0 5 0.4

k0 5 0.3

k0 5 0.2

k0 5 0.1

0.022 0.023 0.025 0.027 0.031 0.033 0.038 0.051 0.068 0.083 0.097 0.109 0.120

0.002 0.003 0.005 0.010 0.016 0.020 0.032 0.052 0.074 0.093 0.110 0.125 0.134

+0.032 +0.030 +0.026 +0.021 +0.012 0.000 0.022 0.056 0.089 0.114 0.135 0.154 0.172

+0.131 +0.131 +0.129 +0.122 +0.111 +0.080 +0.025 0.040 0.098 0.146 0.186 0.224 0.255

Table 6.8 Corrections to log Re0 as a function of {log(R0 /ρu2)Re0 21} for nonspherical particles log{(R0 /ρu2)Re0 21}

k0 5 0.4

k0 5 0.3

k0 5 0.2

k0 5 0.1

4 4.5 3 3.5 2 2.5 1 1.5 0 1 2 3 4 5

+0.185 +0.149 +0.114 +0.082 +0.056 +0.038 +0.028 +0.024 +0.022 +0.019 +0.017 +0.015 +0.013 +0.012

+0.217 +0.175 +0.133 +0.095 +0.061 +0.034 +0.018 +0.013 +0.011 +0.009 +0.007 +0.005 +0.003 +0.002

+0.289 +0.231 +0.173 +0.119 +0.072 +0.033 +0.007 0.003 0.007 0.008 0.010 0.012 0.013 0.014

+0.282 +0.170 +0.062 0.018 0.053 0.061 0.062 0.063 0.064 0.065 0.066 0.066

that the longest face is parallel to the direction of motion and some error may, therefore, be introduced in the calculation, as indicated by Heiss and Coull.37 For a nonspherical particle settling under the influence of gravity: 1 Total drag force, F ¼ R00 πdp2 ¼ ðρs  ρÞgk0 dp3 4 Thus :

R00 4k0 dp g ¼ ðρ  ρÞ ρu20 πρu20 s

(6.50) (6.51)

Motion of Particles in a Fluid 305 R00 0 2 4k0 ρdp3 g Re ¼ ðρs  ρÞ μ2 π ρu20 0 and :

R00 0 1 4k0 μg Re ¼ 2 3 ðρs  ρÞ ρu20 0 πρ u0

(6.52) (6.53)

Provided k0 is known, the appropriate dimensionless group may be evaluated and the terminal falling velocity, or diameter, calculated. Numerous other methods38–40 have been proposed in the literature for estimating the terminal falling velocities of nonspherical particles, but none of these has proved to be entirely satisfactory. Depending upon the information available for the particle size and shape, one can use these methods to establish upper and lower bounds on the settling velocity of a particle in a given situation. Broadly, size is important in the viscous regime (low Reynolds numbers), and shape and orientation matter more in the inertial regime (high Reynolds numbers). Most of the methods mentioned so far are applicable for regular-shaped particles for which it is possible to calculate/measure the volume and surface area. In many situations, one encounters particles of irregular shape with edges and corners, or agglomerates of fine particles such as flocs, composite particles, etc. It is, thus, not very practical to calculate or measure their surface area. Some ideas based on fractal dimensions, circularity, Corey shape factors, etc., are available in the literature for such situations.41–45 Example 6.3 What will be the terminal velocities of mica plates, 1 mm thick and ranging in area from 6 to 600 mm2, settling in an oil of density 820 kg/m3 and viscosity 10 mN s/m2? The density of mica is 3000 kg/m3. Solution A0 dp d3p Volume k0

Smallest Particles

Largest Particles

6  106 m2 √(4  6  106/π) ¼ 2.76  103 m 2.103  108 m3 6  109 m3 0.285

6  104 m2 √(4  6  104/π) ¼ 2.76  102 m 2.103  105 m3 6  107 m3 0.0285



 R’0 4k’ ’2 Re ¼ ðρ  ρÞρdp3 g ðEq:6:52Þ: 0 ρu2 μ2 π s   ¼ 4  0:285=π  0:012 ð3000  820Þ 820  2:103  108  9:81 ¼ 1340 f or the smallest particles and, similarly, 134, 000 f or the largest particles:

306 Chapter 6 Thus: 

R00 0 2 log Re ρu20 0 logRe00



Correction from Table 6.6 Corrected Re00 Re00 u0

Smallest Particles

Largest Particles

3.127

5.127

1.581 0.038 1.543 34.9 0.154 m/s

2.857 (from Table 6.4) 0.300 (estimated) 2.557 361 0.159 m/s

Thus, it is seen that all the mica particles settle at approximately the same velocity.

6.5 Motion of Bubbles and Drops The drag force acting on a gas bubble or a liquid droplet will not, in general, be the same as that acting on a rigid particle of the same shape and size because circulating currents are set up inside the bubble. The velocity gradient at the surface is thereby reduced, and the drag force is, therefore, less than that for the rigid particle. Hadamard46 showed that, if the effects of surface tension are neglected, the terminal falling velocity of a drop, as calculated from Stokes’ law, must be multiplied by a factor Q, to account for the internal circulation, where: Q¼

3μ + 3μl 2μ + 3μl

(6.54)

In this equation, μ is the viscosity of the continuous fluid, and μl is the viscosity of the fluid forming the drop or bubble. This expression applies only in the range for which Stokes’ law is valid. If μl/μ is large, Q approaches unity. If μl/μ is small, Q approaches a value of 1.5. Thus, the effect of circulation is small when a liquid drop falls in a gas, and it is large when a gas bubble rises in a liquid. If the fluid within the drop is very viscous, the amount of energy which has to be transferred in order to induce circulation is large, and circulation effects are, therefore, small. Hadamard’s work was later substantiated by Bond47 and by Bond and Newton,48 who showed that Eq. (6.54) is valid provided that surface tension forces do not play a large role. With very small droplets, the surface tension forces tend to nullify the tendency for circulation, and the droplet falls at a velocity close to that of a solid sphere. In addition, drops and bubbles are subject to deformation because of the differences in the pressures acting on various parts of the surface. Thus, when a drop is settling in a still fluid, both the hydrostatic and the impact pressures will be greater on the forward face than on the rear face and will tend to flatten the drop, whereas the viscous drag will tend to elongate it.

Motion of Particles in a Fluid 307 Deformation of the drop is opposed by the surface tension forces, so that very small drops retain their spherical shape, whereas large drops may be considerably deformed and the resistance to their motion thereby increased. For drops above a certain size, the deformation is so great that the drag force increases at the same rate as the volume, and the terminal falling velocity therefore becomes independent of size. Garner and Skelland49, 50 have shown the importance of circulation within a drop in determining the coefficient of mass transfer between the drop and the surrounding medium. The critical Reynolds number at which circulation commences has been shown49 to increase at a rate proportional to the logarithm of viscosity of the liquid constituting the drop and to increase with interfacial tension. The circulation rate may be influenced by mass transfer because of the effect of concentration of diffusing material on both the interfacial tension and on the viscosity of the surface layers. As a result of circulation, the falling velocity may be up to 50% greater than for a rigid sphere, whereas oscillation of the drop between oblate and prolate forms will reduce the velocity of fall.50 Terminal falling velocities of droplets have also been calculated by Hamielec and Johnson51 and others52 from approximate velocity profiles at the interface, and the values so obtained compare well with experimental values for droplet Reynolds numbers up to 80.

6.6 Drag Forces and Settling Velocities for Particles in Non-Newtonian Fluids Only a very limited amount of data is available on the motion of particles in non-Newtonian fluids, and the following discussion is restricted to their behaviour in shear-thinning power-law fluids and in fluids exhibiting a yield stress, both of which are discussed in Volume 1A, Chapter 3.

6.6.1 Power-Law Fluids Because most shear-thinning fluids, particularly polymer solutions and flocculated suspensions, have high apparent viscosities, even relatively coarse particles may have velocities in the creeping-flow or the Stokes’ law regime. Chhabra53, 54 has proposed that both theoretical and experimental results for the drag force F on an isolated spherical particle of diameter d moving at a velocity u may be expressed as a modified form of Stokes’ law: F ¼ 3πμc duY

(6.55)

where the apparent viscosity μc is evaluated at a characteristic shear rate u/d, and Y is a correction factor which is a function of the rheological properties of the fluid. The best available theoretical estimates of Y for power-law fluids are given in Table 6.9.

308 Chapter 6 Table 6.9 Values of Y for power-law fluids53 n

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Y

1

1.14

1.24

1.32

1.38

1.42

1.44

1.46

1.41

1.35

Several expressions of varying forms and complexity have been proposed36, 53, 54 for the prediction of the drag on a sphere moving through a power-law fluid. These are based on a combination of numerical solutions of the equations of motion and extensive experimental results. In the absence of wall effects, dimensional analysis yields the following functional relationship between the variables for the interaction between a single isolated particle and a fluid:  (6.56) 2C0D ¼ CD ¼ f Re0n , n where CD and CD0 are drag coefficients defined by Eq. (6.4), n is the power-law index, and Ren0 is the particle Reynolds number given by:  (6.57) Re0n ¼ u2n dn ρ =k where k is the consistency coefficient in the power-law relation. Combining Eqs (6.55) and (6.56):   1 (6.58) CD ¼ 24Re0 n Y   1 C0D ¼ 12Re0 n Y

(6.59)

From Table 6.9 it is seen that, depending on the value of n, the drag on a sphere in a power-law fluid may be up to 46% higher than that in a Newtonian fluid at the same particle Reynolds number. Practical measurements lie in the range 1 < Y < 1.8, with considerable divergences between the results of the various researchers. In view of the general uncertainty concerning the value of Y, it may be noted that the unmodified Stokes’ law expression gives a acceptable first approximation. The terminal settling velocity u0 of a particle in the gravitational field is then given by equating the buoyant weight of the particle to the drag force to give:  u0 ¼


1=n gdn + 1 ðρs  ρÞ 18kY

(6.60)

where (ρs  ρ) is the density difference between the particle and the fluid. From Eq. (6.60), it is readily seen that in a shear-thinning fluid (n < 1), the terminal velocity is more strongly dependent on d, g and ρs  ρ than in a Newtonian fluid, and a small change in any of these variables produces a larger change in u0.

Motion of Particles in a Fluid 309 Outside the creeping flow regime, experimental results for drag on spheres in power-law fluids have been presented by Tripathi et al.,55 Graham and Jones56 and Song et al.57 for values of Ren0 up to 100, and these are reasonably well correlated by the following expressions with an average error of about 10%:

 

  CD ¼ 35:2Re0n 1:03 2n + n 1  20:9Re0n 1:11 2n   (6.61a) 0:2 < 2n Re0n < 24

  CD ¼ 37Re0n 1:1 2n + ½0:36n + 0:25   24 < 2n Re0n < 100 (6.61b) h  i h   i 1:03 1:11 C0D ¼ 17:6Re0 n 2n + n 0:5  10:5Re0 n 2n (6.62a)

  0:2 < 2n Re0n < 24 h  i 1:1 2n + ½0:18n + 0:125 C0D ¼ 18:5Re0 n (6.62b)

  24 < 2n Re0n < 100 It may be noted that these two equations do not reduce exactly to the relation for a Newtonian fluid (n ¼ 1). Extensive comparisons of predictions and experimental results for drag on spheres suggest that the influence of non-Newtonian characteristics progressively diminishes as the value of the Reynolds number increases, with inertial effects then becoming dominant, and the standard curve for Newtonian fluids may be used with little error. Experimentally determined values of the drag coefficient for power-law fluids (1 < Ren0 < 1000; 0.4 < n < 1) are within 30% of those given by the standard drag curve.55, 56 Suffice it to add here that the other methods36, 53, 58 available in the literature do not offer any significant improvement over Eqs (6.61) and (6.62). While Eqs (6.62a) and (6.62b) are convenient for estimating the value of the drag coefficient, they need to be rearranged in order to enable the settling velocity u0 of a sphere of given diameter and density to be calculated, because both CD(CD0 ) and Ren0 are functions of the unknown settling velocity. By analogy with the procedure used for Newtonian fluids (Eq. 6.36), the dimensionless Galileo number Gan which is independent of u0 may be defined by: 2 ½2=ð2nÞ ¼ gd ½ðn + 2Þ=ð2nÞ ðsr  1Þðρ=kÞ½2=ð2nÞ Gan ¼ C0D Re0 3

(6.63)

where sr is the ratio of the densities of the particle and of the fluid (ρs/ρ). Eq. (6.56) may be written as: Re0n ¼ f ðGan , nÞ

(6.64)

310 Chapter 6 Experimental results comprising about 1000 data points from a large number of sources cover the following range of variables: 1 < d < 20ðmmÞ;1190 < ρs < 16, 600ðkg=m3 Þ; 990 < ρ < 1190ðkg=m3 Þ; 0:4 < n < 1; and 1 < Re0n < 104 : These are satisfactorily correlated by: Re0n

 ¼ r1

2 Gan 3

r2 (6.65)

where r1 ¼ 0.1{exp[(0.5/n)  0.73n] and r2 ¼ (0.954/n)  0.16. Thus, in Eq. (6.65) only Ren0 includes the terminal falling velocity, which may then be calculated for a spherical particle in a power-law fluid.

6.6.2 Fluids With a Yield Stress Much less is known about the settling of particles in fluids exhibiting a yield stress. Barnes59 suggests that this is partly due to the fact that considerable confusion exists in the literature as to whether or not the fluids used in the experiments do have a true yield stress.59 Irrespective of this uncertainty, which usually arises from the inappropriateness of the rheological techniques used for their characterisation, many industrially important materials, notably particulate suspensions, have rheological properties closely approximating to viscoplastic behavior. By virtue of its yield stress, an unsheared viscoplastic material is capable of supporting the immersed weight of a particle for an indefinite period of time, provided that the immersed weight of the particle does not exceed the maximum upward force which can be exerted by virtue of the yield stress of the fluid. The conditions for the static equilibrium of a sphere are now discussed. Static equilibrium Many investigators53 have reported experimental results on the necessary conditions for the static equilibrium of a sphere. The results of all such studies may be represented by a factor Z, which is proportional to the ratio of the forces due to the yield stress τY and those due to gravity. Thus : Z ¼

τY dgðρs  ρÞ

(6.66)

The critical value of Z which indicates the point at which the particle starts to settle from rest appears to lie in the range 0.04 < Z < 0.2.

Motion of Particles in a Fluid 311 Drag force Under conditions where a spherical particle is not completely supported by the forces attributable to the yield stress, it will settle at a velocity such that the total force exerted by the fluid on the particle balances its weight. For a fluid whose rheological properties may be represented by the Herschel-Bulkley model discussed in Volume 1A, Chapter 3, the shear stress τ is a function of the shear rate γ_ or: 0 γ_ nHB τ ¼ τY + kHB

(6.67)

From dimensional considerations, the drag coefficient is a function of the Reynolds number for the flow relative to the particle, the exponent, nHB, and the so-called Bingham number Bi, which is proportional to the ratio of the yield stress to the viscous stress attributable to the settling of the sphere. Thus: C0D ¼ R0 =ρu2 f ðRe0 mHB , nHB , BiÞ 0 ðu=dÞnHB Bi ¼ τY =kHB

(6.68) (6.69)

Using the scant data in the literature and their own experimental results, Atapattu et al.60 suggest the following expression for the drag on a sphere moving through a Herschel-Bulkley fluid in the creeping flow regime: C0D ¼ 12Re0HB ð1 + BiÞ1

(6.70)

It may be noted that an iterative solution to Eq. (6.70) is required for the calculation of the unknown settling velocity u0, because this term appears in all three dimensionless groups, ReHB0 , Bi, and CD0 , for a given combination of properties of sphere and fluid. It is useful to mention here that the predictions of Eq. (6.70) are also in line with the recent numerical predictions61 of drag on a sphere up to ReHB0 ¼ 100. The effect of particle shape on the forces acting when the particle is moving in a shear-thinning fluid has been investigated by Tripathi et al.,55 and by Venumadhav and Chhabra.62 In addition, some information is available on the effects of viscoelasticity of the fluid.53 In an extensive study dealing with the free settling of nonspherical particles in power-law fluids, Rajitha et al.63 collated much of the literature data and modified a previous drag formula developed for spheres falling in power-law fluids.64 Their approach hinges on evaluating three parameters to characterise the shape and orientation of the particle: equal volume sphere diameter (ds), sphericity, ψ, defined as the ratio of the surface areas of the equal volume sphere and of the actual particle and the diameter, dn, of a circle of area equal to the projected area of the falling particle normal to the direction of sedimentation. Furthermore, Rajitha et al.63 introduced a composite shape parameter χ defined as:

312 Chapter 6 χ¼

  4 ds 2 ψ dn

(6.71)

Evidently, χ ¼ 4 for a sphere (ψ ¼ 1; ds ¼ dn) and χ can be greater or smaller than 4 depending upon the shape and orientation of the falling particle. The drag expression due to Rajitha et al.63 is given as follows: 8 > > <

9δo > > =

8 > > <

911 12 > >  2δo = 24 24Y 6Ybo 6Ybo     δ1 + CD∞ (6.72) CD ¼ 0 Y + χCD∞ > > > 24Y 24Y > Ren Re0n > > > > :6Ybo + :6Ybo + 128 ; ; Re0n Re0n The three constants δo, δ1 and bo are evaluated using the numerical results of Tripathi et al.55 as follows: bo ¼ exp f3ðα  ln6Þg   
αo  α1 αo  α1 ln3 δ1 ¼ exp 3 2αo α1 2αo α1 ( " 2 pffiffiffi #) 11 pffiffiffi αo  α1 61 pffiffiffi δo ¼ ln 6 1  exp 2ðαo  1Þα1 48 6

(6.73a) (6.73b)

(6.73c)

Here, αo ¼ 3 and it corresponds to the value of α1 for n ¼ 0, see Eq. (6.73e). The two functions Y and α1 related to each other as follows:
pffiffiffin1  3 Y¼ 6 n2 + n + 1

(6.73d)

This is based on the numerical results in the creeping flow summarised in Table 6.9. Finally, α1 is given as: n o1=ðn + 1Þ (6.73e) α1 ¼ 6ð1nÞ=2 Y   It is easily seen that for n ¼ 0, Y ¼ p3ffiffi6 , which, in turn, yields the value of α1 ¼ αo ¼ 3. Some comments about such a complex form of Eq. (6.72) are in order. First, in the limit of n ¼ 1, Eq. (6.72) reproduces the standard drag curves for a sphere (CD∞ ¼ 0.44) and for a long cylinder (CD∞ ¼ 1) rather well.65 Subsequently, it has been shown that in the limit of χ ¼ 4, Eq. (6.72) correlates much of the experimental data for spheres falling in power-law fluids with acceptable levels of accuracy.64 Because with the increasing Reynolds number, the effects of fluid viscosity progressively diminish, Rajitha et al.63 recommended the value of CD∞ ¼ 0.44 for spheres in power-law fluids also. Eq. (6.72) reproduced nearly 1500 data points relating to

Motion of Particles in a Fluid 313 cylinders, cones, oblates, prolates, prisms, disks, etc., spanning the ranges of conditions as 0.33  ψ  0.98; 0.3  n  1 and 105  Ren 0  270 with a mean error of 30%. It is seen, therefore, that, in the absence of any entirely satisfactory theoretical approach or reliable experimental data, it is necessary to adopt a highly pragmatic approach to the estimation of the drag force on a particle in a non-Newtonian fluid.

6.7 Accelerating Motion of a Particle in the Gravitational Field 6.7.1 General Equations of Motion The motion of a particle through a fluid may be traced because the value of the drag factor R0 /ρu2 for a given value of the Reynolds number is fixed. The behaviour of a particle undergoing acceleration or retardation has been the subject of a very large number of investigations, which have been critically reviewed by Torobin and Gauvin66 and others.13 The results of different researchers are not consistent, although it is shown that the drag factor is often related, not only to the Reynolds number, but also to the number of particle diameters traversed by the particle since the initiation of the motion. A relatively simple approach to the problem, which gives results closely in accord with practical measurements, is to consider the mass of fluid which is effectively given the same acceleration as the particle, as discussed by Mironer.67 This is only an approximation because elements of fluid at different distances from the particle will not all be subject to the same acceleration. For a spherical particle, this added or hydrodynamic mass is equal to the mass of fluid whose volume is equal to one half of that of the sphere. This can give rise to a very significant effect in the case of movement through a liquid, and can result in accelerations substantially less than those predicted when the added mass is neglected. For movement through gases, the contribution of the added mass term is generally negligible. Added mass is most important for the motion of gas bubbles in a liquid because, in that case, the surrounding liquid has a much greater density than the bubble. The total mass of particle and associated fluid is sometimes referred to as the virtual mass (m0 ). Thus : Virtual mass ¼ Mass of particle + Added mass: π π For a sphere : m0 ¼ d3 ρs + d3 ρ 6 12     π 3 ρ ρ 0 ¼m 1+ or : m ¼ d ρs 1 + 6 2ρs 2ρs

(6.74)

Considering the motion of a particle of mass m in the earth’s gravitational field, at some time t, the particle will be moving at an angle α to the horizontal with a velocity u as shown in Fig. 6.7. _ in the horizontal and vertical The velocity u may then be resolved into two components, x_ and y,

314 Chapter 6 x

a

X

u

y

Y

Fig. 6.7 Two-dimensional motion of a particle.

directions. x_ and x€ will be taken to denote the first and second derivatives of the displacement x in the X-direction with respect to time, and y_ and y€ the corresponding derivatives of y. Thus : cosα ¼

x_ u

y_ u pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and : u ¼ ðx_2 + y_ 2 Þ sin α ¼

(6.75) (6.76) (6.77)

There are two forces acting on the body: (a) In the vertical direction, the apparent weight of the particle,   ρ 0 W ¼ mg 1  ρs

(6.78)

(b) The drag force which is equal to R0 A0 and acts in such a direction as to oppose the motion of the particle. Its direction, therefore, changes as α changes. Here, A0 is the projected area of the particle on a plane at right angles to the direction of motion, and its value varies with the orientation of the particle in the fluid. The drag force can be expressed by: F¼

R0 2 0 ρu A ρu2

This has a component in the X-direction of: R0 2 0 R0 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρu A cos α ¼ A ρx_ ðx_ 2 + y_ 2 Þ ρu2 ρu2 and in the Y -direction of:

(6.79)

Motion of Particles in a Fluid 315 R0 2 0 R0 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρu A sin α ¼ A ρy_ ðx_ 2 + y_ 2 Þ ρu2 ρu2 The equations of motion in the X- and Y -directions are, therefore: mx€¼ 

 2 2 R0 0 _ x√ x_ + y_ ρA ρu2

  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R0 ρ 0 2 2 and : my€¼  2 ρA y_ ðx_ + y_ Þ + mg 1  ρu ρs

(6.80) (6.81)

If allowance is now made for the added mass, m0 ¼ m[1 + (ρ/2ρs)] must be substituted for m. Eqs (6.80) and (6.81) refer to conditions where the particle is moving in the positive sense in the X-direction and in the positive (downward) sense in the Y -direction. If the particle is moving in the negative X-direction, the form of solution is unchanged, except that all increments of x will be negative. If, however, the particle is initially moving upwards, the sign of only the frictional term in Eq. (6.81) is changed, and the form of solution will, in general, be different from that for downward movement. Care must therefore be exercised in the application of the equation, particularly if a change of sense may occur during the motion of the particle. It may be noted that the equations of motion for the two directions (X and Y) are coupled, with x_ and y_ appearing in each of the equations. General solutions are, therefore, not possible, except as will be seen later for motion in the Stokes’ law region. π π Putting m ¼ d 3 ρs and A0 ¼ d 2 for a spherical particle, then: 6 4  R0 3ρ _ x_ 2 + y_ 2 (6.82) x√ x€¼  2 ρu 2dρs If allowance is made for the added mass and m0 is substituted for m in Eq. (6.81), then: x€¼ 

 2 2 R0 3 ρ _ x√ x_ + y_ ρu2 d ð2ρs + ρÞ

  R0 3 ρ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ρ 2 2 Similarly : y€¼  2 y_ ðx_ + y_ Þ + g 1  ρu 2d ρs ρs

(6.83)

(6.84)

and allowing for the added mass: y€¼ 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2gðρs  ρÞ R0 3 ρ _ ðx_ 2 + y_ 2 Þ + y ρu2 d ð2ρs + ρÞ ð2ρs + ρÞ

(6.85)

where the minus sign in Eqs (6.84) and (6.85) is applicable for downward motion and the positive sign for upward motion (with downwards taken as the positive sense).

316 Chapter 6

6.7.2 Motion of a Sphere in the Stokes’ Law Region Under these conditions, from Eq. (6.5): R0 12μ 1 ¼ 12Re0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ρu dρ ðx_ 2 + y_ 2 Þ

(6.86)

Substituting in Eqs (6.82)–(6.86) gives: x€¼  or : x€¼ 

18μ x_ ¼ ax_ d2 ρs 36μ x_ ¼ a0 x_ s + ρÞ

d 2 ð2ρ

(6.87) (6.88)

if allowance is made for the added mass.

  18μ ρ ¼ ay_ + b y€¼  2 y_ + g 1  d ρs ρs

or : y€¼ 

36μ 2gðρs  ρÞ y_ + ¼ a0 y0 + b0 ð2ρs + ρÞ s + ρÞ

d 2 ð2ρ

(6.89) (6.90)

if allowance is made for the added mass. In this particular case, the equations of motion for the X and Y directions are mutually independent and therefore can be integrated separately. Furthermore, because the frictional term is now a linear function of velocity, the sign will automatically adjust to take account of whether motion is downwards or upwards. The equations are now integrated, ignoring the effects of added mass which can be accounted for by replacing a by a0 and b by b0 . For the Y -direction, integrating Eq. (6.89) with respect to t: y_ ¼ ay + bt + constant: The axes are chosen so that the particle is at the origin at time t ¼ 0. If the initial component of the velocity of the particle in the Y -direction is v, then, when t ¼ 0, y ¼ 0 and y_ ¼ v, and the constant ¼ v, or y_ + ay ¼ bt + v Thus : eat y_ + eat ay ¼ ðbt + vÞeat Z eat eat at Thus : e y ¼ ðbt + vÞ  b dt a a eat b at ¼ ðbt + vÞ  2 e + constant a a

Motion of Particles in a Fluid 317 b v  . a2 a   b v b b v Thus : y ¼ t +  2 + 2  eat a a a a a

When t ¼ 0, y ¼ 0, and the constant ¼

where : a ¼ 18

μ d2 ρ

(6.91) (6.92)

s

  ρ g: and : b ¼ 1  ρs

(6.93)

It may be noted that b/a ¼ u0, the terminal falling velocity of the particle. This equation enables the displacement of the particle in the Y -direction to be calculated at any time t. For the X-direction, Eq. (6.87) is of the same form as Eq. (6.89) with b ¼ 0. Substituting b ¼ 0 and writing w as the initial velocity in the X-direction, Eq. (6.91) becomes: w x ¼ ð1  eat Þ a

(6.94)

Thus, the displacement in the X-direction may also be calculated for any time t. By eliminating t between Eqs (6.92), (6.93) and (6.94), a relation between the displacements in the X- and Y -directions is obtained. Equations of this form are useful for calculating the trajectories of particles in size-separation equipment. From Eq. (6.94): ax w 1  ax and : t ¼  ln 1  a w eat ¼ 1 

Substituting in Eq. (6.91) gives: 
  b 1  ax v b b v  ax  ln 1  +  2 + 2 1 y¼ a a w a a a a w   b ax bx vx ¼  2 ln 1   + a w aw w The values of a and b can now be substituted and the final relation is:     
gρs ðρs  ρÞd4 18μx 18μx 18vμ + ln 1  1 2 y¼ 324μ2 wρs d2 wρs d2 d ðρs  ρÞg

(6.95)

(6.96)

(6.97)

If allowance is made for added mass, a0 and b0 are substituted for a and b, respectively.

318 Chapter 6 Then: gð2ρs + ρÞðρs  ρÞd4 y¼ 648μ2





  
36μx 36μx 18vμ + ln 1  1 2 wð2ρs + ρÞd 2 wð2ρs + ρÞd 2 d ðρs  ρÞg (6.98)

6.7.3 Vertical Motion (General Case) For the Stokes’ law regime Eqs (6.91)–(6.93) are applicable. For the Newton’s law regime, R0 /ρu2 is a constant and equal to 0.22 for a spherical particle. Therefore, substituting in Eq. (6.84) and putting x_ ¼ 0 for vertical motion, and using the negative sign for downward motion (and neglecting the effect of added mass):   1 ρ 2 ρ (6.99) y_ + g 1  y€¼  3d ρs ρs y€¼ cy_ 2 + b Thus : and :

d y_ ¼ dt b  cy_ 2 dy_ f 2  y_ 2

¼ cdt

1 ρ 3d ρs s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi ffi s ffi b 3dðρs  ρÞg ¼ and : f ¼ c ρ where : c ¼

Integrating Eq. (6.101) gives:

  1 f + y_ ln ¼ ct + constant 2f f  y_   1 f +v When t ¼ 0, y_ ¼ v, say, and, therefore, the constant ¼ ln . 2f f v    1 f + y_ f v ln ¼ ct 2f f  y_ f +v    f + y_ f v Thus : ¼ e2f ct f  y_ f +v

Thus :

(6.100)

(6.101) (6.102)

(6.103)

Motion of Particles in a Fluid 319 2f  f + v 2f ct 1+ e f v Z dt   Thus : y_ ¼ ft  2f f + v 2f ct 1+ e f v Z dt ¼ ft  2f ðsayÞ 1 + jept and : ðf  y_ Þ ¼



Putting : s ¼ 1 + jept then : ds ¼ pjept dt ¼ pðs  1Þdt Z ds Thus : y ¼ ft  2f psðs  1Þ 
1 s1 ln + constant ¼ ft  2f p s ¼ ft 

1 ln c

1  + constant f  v 2f ct 1+ e f +v 

When t ¼ 0, y ¼ 0 and: 1 1 f +v  ¼ ln f v c 2f 1+ f +v 
1 f +v f  v 2f ct 1+ e Thus : y ¼ ft + ln c 2f f +v constant ¼

1 ln c



Thus, for downward motion: y ¼ ft +

 1 1 ln f + v + ðf  vÞe2f ct c 2f

(6.104)

If the added mass is taken into account, f remains unchanged, but c must be replaced by c0 in Eqs (6.100) and (6.104), where: c0 ¼

2ρ : 3d ð2ρs + ρÞ

(6.105)

For vertical upwards motion in the Newton’s law regime; the positive sign in Eq. (6.84) applies and, thus, by analogy with Eq. (6.101):

320 Chapter 6 dy_ f 2 + y_ 2 Integrating :

¼ cdt

(6.106)

1 y_ tan 1 ¼ ct + constant f f

When t ¼ 0, y_ ¼ v, say, and the constant ¼ (1/f ) tan1(v/f ), v is a negative quantity. Thus: 1 1 v y_ tan 1 ¼ ct + tan 1 f f f f   y_ 1 v ¼ tan f ct + tan f f   f 1 v and : y ¼ ln cos f ct + tan + constant fc f When t ¼ 0, y ¼ 0, and: 1 v ln cos tan 1 , c f   v cos f ct + tan 1 f ln v 1 cos tan f     1 v 1 v cosf ct cos tan  sinf ct sin tan f f   ln v cos tan 1 f   v ln cosf ct  sin f ct f constant ¼

then : y ¼ 

1 c

¼

1 c

or : y ¼ 

1 c

(6.107)

The relation between y and t may also be obtained graphically, though the process is more tedious than that of using the analytical solution appropriate to the particular case in question. When Re0 lies between 0.2 and 500, there is no analytical solution to the problem, and a numerical or graphical method must be used. When the spherical particle is moving downwards, that is when its velocity is positive:   3 ρ R0 2 ρ g ðfrom Eq:6:84Þ y_ + 1  y€¼  2d ρs ρu2 ρs   μ dRe0 3 ρ R0 0 2 μ2 ρ g Re 2 2 + 1  ¼ d ρ ρd dt 2d ρs ρu2 ρs

Motion of Particles in a Fluid 321 Z and : t ¼

Re02

Re01

dRe0 dρðρs  ρÞg 3μ R0 0 2  2 Re μρs 2d ρs ρu2

If the particle is moving upwards, the corresponding expression for t is: Z Re0 2 d Re0 t¼ 3μ R0 0 2 Re01 dρðρs  ρÞg + 2 Re μρs 2d ρs ρu2

(6.108)

(6.109)

Eqs (6.108) and (6.109) do not allow for added mass. If this is taken into account, then: Z Re0 2 dRe0 t¼ (6.110) 3μ R0 0 2 Re01 2d ðρs  ρÞg   2 Re d ð2ρs + ρÞ ρu2 μ 2ρρ + ρ where the positive sign applies to upward motion and the negative sign to downward motion. From these equations, Re0 may be obtained as a function of t. The velocity y_ may then be calculated. By means of a second graphical integration, the displacement y may be found at any time t. In using the various relations which have been obtained, it must be noted that the law of motion of the particle will change as the relative velocity between the particle and the fluid changes. If, for example, a particle is initially moving upwards with a velocity v, so that the corresponding value of Re0 is greater than about 500, the relation between y and t will be given by Eq. (6.107). The velocity of the particle will progressively decrease, and, when Re0 is <500, the motion is obtained by application of Eq. (6.109). The upward velocity will then fall still further until Re0 falls below 0.2. While the particle is moving under these conditions, its velocity will fall to zero and will then gradually increase in the downward direction. The same Eq. (6.91) may be applied for the whole of the time the Reynolds group is <0.2, irrespective of sense. Then, for higher downward velocities, the particle motion is given by Eqs (6.108) and (6.104). Unidimensional motion in the vertical direction, under the action of gravity, occurs frequently in elutriation and other size separation equipment, as described in Chapter 1. Example 6.4 A material of density 2500 kg/m3 is fed to a size separation plant where the separating fluid is water rising with a velocity of 1.2 m/s. The upward vertical component of the velocity of the particles is 6 m/s. How far will an approximately spherical particle, 6 mm diameter, rise relative to the walls of the plant before it comes to rest relative to the fluid? Solution Initial velocity of particle relative to fluid, v ¼ (6.0–1.2) ¼ 4.8 m/s.

322 Chapter 6   Thus : Re0 ¼ 6  103  4:8  1000 = 1  103 ¼ 28, 800 When the particle has been retarded to a velocity such that Re0 ¼ 500, the minimum value for which Eq. (6.107) is applicable: y_ ¼ ð4:8  500Þ=28, 800 ¼ 0:083m=s In this solution, the effect of added mass is not taken into account. Allowance may be made by adjustment of the values of the constants in the equations as indicated in Section 6.7.3. When Re0 is >500, the relation between the displacement of the particle y and the time t is:   1 v ðEq:6:107Þ y ¼  ln cos f ct  sin f ct c f 1 ρ  where : c ¼ ¼ 0:33=6  103 ð1000=2500Þ ¼ 22:0 ðEq:6:102Þ 3d ρs s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3d ðρs  ρÞg ¼ ½ð6  103  1500  9:81Þ=ð0:33  1000Þ f¼ ρ ¼ 0:517 ðEq:6:103Þ 

Thus : y ¼ 

and : v ¼ 4:8m=s

1 4:8 ln cos 0:517  22t + sin 0:517  22t 22:0 0:517



¼ 0:0455 ln ð cos 11:37t + 9:28 sin 11:37tÞ 0:0455ð11:37 sin 11:37t + 9:28  11:37 cos 11:37t Þ cos11:37t + 9:28 sin 11:37t 0:517ð9:28 cos 11:37t  sin 11:37tÞ ¼ cos 11:37t + 9:28 sin11:37t The time taken for the velocity of the particle relative to the fluid to fall from 4.8 m/s to 0.083 m/s is given by: y_ ¼ 

0:517ð9:28 cos 11:37t  sin 11:37t Þ cos11:37t + 9:28 sin 11:37t or : cos 11:37t + 9:28 sin 11:37t ¼ 6:23 sin 11:37t + 57:8cos11:37t

0:083 ¼ 

or : 56:8 cos 11:37 ¼ 15:51 sin 11:37t ; sin 11:37t ¼ 3:66 cos 11:37t squaring : 1  cos 2 11:37t ¼ 13:4 cos 2 11:37t ; cos 11:37t ¼ 0:264  and : sin 11:37t ¼ √ 1  0:2642 ¼ 0:965

(i)

The distance moved by the particle relative to the fluid during this period is, therefore, given by: y ¼ 0:0455 ln ð0:264 + 9:28  0:965Þ ¼ 0:101m

Motion of Particles in a Fluid 323 If Eq. (6.107) is applied for a relative velocity down to zero, the time taken for the particle to come to rest is given by: 9:28 cos 11:37t ¼ sin 11:37t squaring : 1  cos 2 11:37t ¼ 86:1 cos 2 11:37t and : cos 11:37t ¼ 0:107 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and : sin 11:37t ¼ ð1  0:1072 Þ ¼ 0:994 The corresponding distance the particle moves relative to the fluid is then given by: y ¼ 0:0455 ln ð0:107 + 9:28  0:994Þ ¼ 0:102m that is, the particle moves only a very small distance with a velocity of <0.083 m/s. If form drag were neglected for all velocities <0.083 m/s, the distance moved by the particle would be given by:   b v b b v y ¼ t +  2 + 2  2 eat ðEq:6:91Þ a a a a a   b b and : y_ ¼   v eat a a   μ 18  0:001 where : a ¼ 18 2 ¼ ¼ 0:20 ðEq:6:92Þ d ρs 0:0062  2500   ρ b¼ 1 g ¼ ½1  ð1000=2500Þ9:81 ¼ 5:89 ðEq:6:93Þ ρs Thus : b=a ¼ 29:43 and : v ¼ 0:083m=s   0:083 29:43  Thus : y ¼ 29:43t  + 1  e0:20t 0:20 0:20 29:51  1  e0:20t ¼ 29:43t  0:20 and : y_ ¼ 29:43  29:51e0:20t When the particle comes to rest in the fluid, y_ ¼ 0, and: e0:20t ¼ 29:43=29:51 and : t ¼ 0:0141s The corresponding distance moved by the particle is given by:  y ¼ 29:43  0:0141  ð29:51=0:20Þ 1  e0:200:0141 ¼ 0:41442  0:41550 ¼ 0:00108 m

324 Chapter 6 Thus, whether the resistance force is calculated by Eq. (6.15) or Eq. (6.19), the particle moves a negligible distance with a velocity relative to the fluid of <0.083 m/s. Further, the time is also negligible, and thus, the fluid also has moved through only a very small distance. It may, therefore, be taken that the particle moves through 0.102 m before it comes to rest in the fluid. The time taken for the particle to move this distance, on the assumption that the drag force corresponds to that given by Eq. (6.19), is given by: cos 11:37t ¼ 0:264 ðfrom Eq:ðiÞ aboveÞ ; 11:37t ¼ 1:304 and : t ¼ 0:115 s The distance travelled by the fluid in this time ¼ (1.2  0.115) ¼ 0.138 m. Thus, the total distance moved by the particle relative to the walls of the plant ¼ 0:102 + 0:138 ¼ 0:240m or 240mm

Example 6.5 Salt of density 2350 kg/m3 is charged to the top of a reactor containing a 3 m depth of aqueous liquid of density 1100 kg/m3 and viscosity 2 mN s/m2, and the crystals must dissolve completely before reaching the bottom. If the rate of dissolution of the crystals is given by:   dd=dt ¼ 3  106 + 2  104 u where d is the size of the crystal (m) at time t (s), and u its velocity in the fluid (m/s), calculate the maximum size of crystal which should be charged. The inertia of the particles may be neglected, and the resistance force may be taken as that given by Stokes’ law (3πμdu), where d is taken as the equivalent spherical diameter of the particle. Solution Assuming that the salt always travels at its terminal velocity, then for the Stokes’ law region, this is given by Eq. (6.24). u0 ¼ (d2/g/18μ)(ρ8  ρ) or, in this case, u0 ¼ (d2  9.81)/(18  2  103) (2350–1100) ¼ 3.406  105d2 m/s   The rate of dissolution : dd=dt ¼ 3  106 + 2  104 u m=s and substituting : dd=dt ¼ ð3  106 Þ  ð2  104  3:406  105 d 2 Þ ¼ 3  106  68:1d 2 The velocity at any point h from the top of the reactor is u ¼ dh/dt and:  dh dh dt ¼ ¼ 3:406  105 d 2 = 3  106  68:1d 2 dd dt dd Z

3

Thus : 0

Z dh ¼  d

0

ð3:406  105 d 2 dd Þ ð3  106 + 68:1d 2 Þ

Motion of Particles in a Fluid 325 Z or : 3 ¼ 3:406  10

d

5 0

dd C1  C2 C22

Z

d

0

dd ðC1 =C2 Þ + d 2



where C1 ¼ 3  106 and C2 ¼ 68.1. 8 " !#d 9 < d  d =  C1 1 d Thus : 3 ¼ 3:406  105  2 tan 1 0:5 0:5 : C2 0 ; C2 ðC1 =C2 Þ ðC1 =C2 Þ 0 

 and : 3 ¼ 3:406  105 =C2 d  ðC1 =C2 Þ0:5 tan 1 d ðC1 =C2 Þ0:5 Substituting for C1 and C2: d ¼ (6  104) + (2.1  104) tan1(4.76  103d), and solving by trial and error: d ¼ 8.8  104 m or 0.88 mm. The integration may also be carried out numerically with the following results:  d (m) 0 1  104 2  104 3  104 4  104 5  104 6  104 7  104 8  104 9  104

d2 (m2) 0 1  108 4  108 9  108 1.6  107 2.5  107 3.6  107 4.9  107 6.4  107 8.1  107

3:406  105 d 2 3  106 + 68:1d 2

0 9.25  102 2.38  103 3.358  103 3.922  103 4.25  103 4.46  103 4.589  103 4.679  103 4.74  103

 Interval of d (m)

Mean Value of Function in Interval

Integral Over Interval

Total Integral

1  104 1  104 1  104 1  104 1  104 1  104 1  104 1  104 1  104

4.63  102 1.65  103 2.86  103 3.64  103 4.09  103 4.35  103 4.52  103 4.634  103 4.709  103

0.0463 0.1653 0.2869 0.364 0.409 0.435 0.452 0.463 0.471

0.0463 0.2116 0.4985 0.8625 1.2715 1.706 2.158 2.621 3.09

From which d ¼ 0.9 mm.

The acceleration of the particle to its terminal velocity has been neglected, and, in practice, the time taken to reach the bottom of the reactor will be slightly larger, allowing a somewhat larger crystal to dissolve completely.

6.8 Motion of Particles in a Centrifugal Field In most practical cases where a particle is moving in a fluid under the action of a centrifugal field, gravitational effects are comparatively small and may be neglected. The equation of motion for the particles is similar to that for motion in the gravitational field, except that the gravitational acceleration g must be replaced by the centrifugal acceleration rω2, where r is the radius of rotation, and ω is the angular velocity. It may be noted, however, that in this case, the acceleration is a function of the position r of the particle. For a spherical particle in a fluid, the equation of motion for the Stokes’ law region is:

326 Chapter 6 π 3 dr π d2 r d ðρs  ρÞrω2  3πμd ¼ d 3 ρs 2 6 dt 6 dt

(6.111)

As the particle moves outwards, the accelerating force increases, and, therefore, it never acquires an equilibrium velocity in the fluid. If the inertial terms on the right-hand side of Eq. (6.111) are neglected, then: dr d2 ðρs  ρÞrω2 ¼ 18μ dt 2 d ðρs  ρÞg rω2 ¼ g 18μ  2 rω ¼ u0 g

(6.112)

(6.113)

Thus, the instantaneous velocity (dr/dt) is equal to the terminal velocity u0 in the gravitational field, increased by a factor of rω2/g. Returning to the exact form (Eq. 6.111), this may be rearranged to give: d2 r 18μ dr ρs  ρ 2 + +  ω r¼0 ρs dt2 d2 ρs dt or :

(6.114)

d2 r dr + a  qr ¼ 0 2 dt dt

(6.115)

The solution of Eq. (6.115) takes the form: r ¼ B1 e½a=2 + √ða =4 + qÞt + B2 e½a=2√ða =4 + qÞt

 ¼ eat=2 B1 ekt + B2 ekt   pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 18μ ρ where : a ¼ 2 , q ¼ 1  ω2 and k ¼ ½ða2 =4Þ + q d ρs ρs 2

2

(6.116) (6.117) (6.118)

The effects of added mass, which have not been taken into account in these equations, require the replacement of a by a0 and q by q0 , where: a0 ¼

36μ 2ðρs  ρÞ 2 and q0 ¼ ω + ρÞ ð2ρs + ρÞ

d 2 ð2ρs

Eq. (6.117) requires the specification of two boundary conditions so that the constants B1 and B2 may be evaluated.

Motion of Particles in a Fluid 327 If the particle starts (t ¼ 0) at a radius r1 with zero velocity (dr/dt) ¼ 0, then from Eq. (6.117):

 a

 dr ¼ eat=2 kB1 ekt + kB2 ekt  eat=2 B1 ekt + B2 ekt dt n o   a2 a ¼ eat=2 k  B2 ekt  k + B1 ekt 2 2

(6.119)

Substituting the boundary conditions into Eqs (6.117) and (6.119): r1 ¼ B1 + B2   a a 0 ¼ B2 k   B1 k + 2 2 k  a=2 k + a=2 r1 and B2 ¼ r1 2k 2k 
at=2 k  a=2 kt k + a=2 kt Thus : r ¼ e r1 e + r1 e 2k 2k n o r a and : ¼ eat=2 cosh kt + sinh kt r1 2k Hence : B1 ¼

(6.120)

(6.121)

Hence, r/r1 may be directly calculated at any value of t, although a numerical solution is required to determine t for any particular value of r/r1. If the effects of particle acceleration may be neglected, Eq. (6.115) simplifies to: a

dr  qr ¼ 0 dt

Direct integration gives: ln

r q d2 ðρs  ρÞω2 ¼ t¼ t 18μ r1 a

(6.122)

Thus, the time taken for a particle to move to a radius r from an initial radius r1 is given by: t¼

18μ r ln d2 ω2 ðρ2  ρÞ r1

(6.123)

For a suspension fed to a centrifuge, the time taken for a particle initially situated in the liquid surface (r1 ¼ r0) to reach the wall of the bowl (r ¼ R) is given by: t¼

18μ d 2 ω2 ðρs  ρÞ

ln

R r0

If h is the thickness of the liquid layer at the walls, then: h ¼ R  r0

(6.124)

328 Chapter 6

  R R h Then : ln ¼ ln ¼  ln 1  r0 Rh R  2 h 1 h +⋯ ¼ + R 2 R

If h is small compared with R, then: ln

R h r0 R

Eq. (6.123) then becomes: t¼

18μh s  ρÞR

(6.125)

d 2 ω2 ðρ

For the Newton’s law region, the equation of motion is:   π 3 π 2 dr 2 π 3 d 2 r 2 ¼ d ρs 2 d ðρs  ρÞrω  0:22 d ρ 6 4 dt 6 dt This equation can only be solved numerically. If the acceleration term may be neglected, then:  2   dr 2 ρs  ρ ¼ 3dω r dt ρ   
1=2 1=2 dr 2 ρs  ρ (6.126) ¼ 3dω Thus : r dt ρ Integration gives:
1=2    1=2 1=2 2 ρs  ρ ¼ 3dω t 2 r  r1 ρ  1=2   ρ 1=2 1=2 2 r  r or : t ¼ 1 3dω2 ðρs  ρÞ

(6.127)

6.9 Nomenclature

A0 a

Projected area of particle in plane perpendicular to direction of motion 18 μ/d2ρs

Units in SI System

Dimensions in M, L, T

m2

L2

s1

T1

Motion of Particles in a Fluid 329 a0 B1, B2 b b0 b0 c CD CD0 CD0 CD∞ d dp dt d0 dn ds F f g h j K KD k kHB k0 L0 m m0 n nHB q q0 p

36 μ/d2(2ρs + ρ) Coefficients in Eq. (6.114) [1  (ρ/ρs)]g 2 g(ρs  ρ)/(2ρs + ρ) Constant, Eq. (6.72) ρ/3dρs Drag coefficient 2R0 /ρu2 Drag coefficient R0 /ρu2 Drag coefficient in the absence of turbulence Limiting value of drag for a sphere Diameter of sphere or characteristic dimension of particle Mean projected diameter of particle Diameter of tube or vessel Linear dimension of particle Diameter of a circle of area equal to the projected area of a particle Equal volume sphere diameter Total force on particle √(b/c) Acceleration due to gravity Thickness of liquid layer (f + v)/(f  v) Constant for given shape and orientation of particle  0 2 0 2 3 ðR =ρu Þ Re0 1 —see Eq. (6.42) Consistency coefficient for power-law fluid Consistency coefficient for Herschel–Bulkley fluid (Eq. 6.67) Constant for calculating volume of particle Distance of particle from bottom of container Mass of particle Virtual mass (mass + added mass) Power-law index for non-Newtonian fluid Power-law index for Herschel–Bulkley fluid (Eq. 6.67) [1  (ρ/ρs)]ω2 (Eqs 6.114 and 6.110) 2(ρs  ρ)ω2/(2ρs + ρ) 2 fc

s1 m m/s2 m/s2 – m1 – – – – m

T1 L LT2 LT2 – L1 – – – – L

m m m m

L L L L

m N m/s m/s2 m – –

L MLT2 LT1 LT2 L – –

– Nsn/m2 Nsn/m2

– ML1Tn2 ML1Tn2

– m kg kg – –

– L M M – –

s2 s2 s1

T2 T2 T1

330 Chapter 6 Q R R0 R00 r r0 r1 r2 S s sr t u u0 u0t v W0 w x x_ x€ Y y y_ y€ Z α α1 β γ_ δ1 δ0 η

Correction factor for velocity of bubble Radius of basket, or Shear stress at wall of pipe Resistance per unit projected area of particle Resistance per unit projected area of particle at free falling condition Radius of rotation Radius of inner surface of liquid Coefficient in Eq. (6.65) Exponent in Eq. (6.65) Index in Eq. (6.32) 1 + j ept Density ratio (ρs/ρ) Time Velocity of fluid relative to particle Terminal falling velocity of particle Terminal falling velocity of particle in vessel Initial component of velocity of particle in Y-direction Apparent (buoyant) weight of particle Initial component of velocity of particle in X-direction Displacement of particle in X-direction at time t Velocity of particle in X-direction at time t Acceleration of particle in X-direction at time t Correction factor in Stokes’ law for power-law fluid Displacement of particle in Y-direction at time t Velocity of particle in Y direction at time t Acceleration of particle in Y-direction at time t Ratio of forces due to yield stress and to gravity (Eq. 6.66) Angle between direction of motion of particle and horizontal Constant, Eq. (6.73e) Coefficient in Eqs (6.45) and (6.46) Shear rate Constant, Eq. (6.72) Constant, Eq. (6.72) (CD  CD0)/CD0

– m N/m2 N/m2 N/m2

– L ML1 T2 ML1 T2 ML1 T2

m m – – – – – s m/s m/s m/s m/s

L L – – – – – T LT1 LT1 LT1 LT1

N m/s

MLT2 LT1

m m/s m/s2 –

L LT1 LT2 –

m m/s m/s2 –

L LT1 LT2 –

–

–

– – s1 – – –

– – T1 – – –

Motion of Particles in a Fluid 331 λ λE μ μ1 μc ϕ χ τY ρ ρs ψ ω Bi Ga Gan Re0 Re00 0 ReHB Ren0 Suffixes A, B

Mean free path Kolmogoroff scale of turbulence (Eq. 6.46) Viscosity of fluid Viscosity of fluid in drop or bubble Viscosity at shear rate (u/d) Pipe friction factor R/ρu2 Composite shape factor, Eq. (6.71) Yield stress Density of fluid Density of solid Sphericity Angular velocity Bingham number (Eq. 6.69) Galileo number d3(ρs  ρ)ρg/μ2 Galileo number for power-law fluid (Eq. 6.63) Reynolds number udρ/μ or ud0 ρ/μ Reynolds number u0dρ/μ Reynolds number for spherical particle in a Herschel–Bulkley fluid Reynolds number for spherical particle in power-law fluid

m m N s/m2 N s/m2 Ns/m2 – – N/m2 kg/m3 kg/m3 – rad/s – – – – – –

L L ML1 T1 ML1 T1 ML1 T1 – – ML1 T2 ML3 ML3 – T1 – – – – – –

–

–

Particle A, B

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334 Chapter 6

Further Reading Curle N, Davies HJ. Modern fluid dynamics. Volume 1. Incompressible flow. Van Nostrand; 1968. Orr C. Particulate technology. Macmillan; 1966. Ortega-Rivas E. Unit operations of particulate solids. Boca Raton, FL: CRC Press; 2012.