12

External Flows

You have to learn the rules of the game. And then you have to play better than anyone else.”

—Albert Einstein, 1879–1955, Physicist

I.    THE DRAG COEFFICIENT

When a fluid flows past a solid body, or the body moves through the fluid (e.g., Figure 12.1), the force (FD) exerted on the body by the fluid is proportional to the relative rate of momentum (mass flow rate × velocity = ρV2A) transported by the fluid. This can be expressed in terms of a dimensionless drag coefficient (CD), which is defined by the following equation:

FDA=CD2ρV2

(12.1)

Here

ρ is the density of the fluid

V is the relative velocity between the fluid and the solid body

A is the cross-sectional area of the body normal to the velocity vector V. For example, for a sphere of diameter d it is πd2/4

Note that the definition of the drag coefficient from Equation 12.1 is analogous to that of the friction factor for flow in a conduit, that is,

τw=f2ρV2

(12.2)

where τw is the force exerted by the moving fluid on the wall of the pipe per unit area. In the case of τw, however, the area is the total contact area between the fluid and the conduit wall, as opposed to the cross-sectional area normal to the flow direction (or projected area) in the case of CD. One reason for this is that the fluid interaction with the tube wall is uniform over the entire surface for fully developed one-dimensional flow, whereas for a body immersed in a moving fluid, the nature and degree of interaction varies with the position around the body, which is two- or three-dimensional flow.

A.    STOKES FLOW

If the relative velocity is sufficiently low, the fluid streamlines can follow the contour of the body almost completely all the way around without incurring any loss of momentum (this is called creeping flow). For this case, the microscopic momentum balance equations in spherical coordinates for the two-dimensional flow [vr(r, θ), vθ(r, θ)] of a Newtonian fluid were solved by Stokes for the distribution of the pressure, velocity, and local stress components. These equations can then be integrated over the surface of the sphere to determine the total drag FD acting on the sphere in the direction of flow, 2/3 of which results from viscous drag and 1/3 from the nonuniform pressure distribution (form drag). That is:

FD=2πμdV(viscousdrag)+πμdV(formdrag)

(12.3)

Image

FIGURE 12.1 Drag on a sphere.

This result can be expressed in dimensionless form using Equation 12.1, as a theoretical expression for the drag coefficient:

CD=24NRe

(12.4)

where

NRe=dVρμ

(12.5)

This is known as Stokes flow, and Equation 12.4 has been found to be accurate for flow over a sphere in an unconfined expanse of fluid for NRe < 0.1 (or to within about 5% for NRe < 1). Note the similarity between Equation 12.4 and the dimensionless Hagen–Poiseuille equation for laminar tube flow, that is, f = 16/NRe. However, note that the limiting value of the Reynolds number for the validity of these equations varies from 2100 (for a pipe) to ~0.1 for a sphere. This is due to the difference in the nature of the transition to turbulent flow, as explained below.

B.    FORM DRAG

As the fluid flows over the forward part of the sphere, the velocity increases because the available flow area decreases, and the pressure must decrease in order to satisfy the conservation of energy. Conversely, as the fluid flows around the back side of the body, the velocity decreases due to the expanding flow area, and the pressure increases. This is not unlike the flow in a diffuser or a converging-diverging duct. The flow behind the sphere thus moves from a low-pressure point near the equator to a higher pressure, that is, into a region of “adverse pressure gradient” behind the sphere. This is inherently unstable so that, as the relative velocity (and NRe) increases, it becomes more difficult for the streamlines to follow the contour of the body and they eventually break away from the surface. This condition is called separation, although it is the smooth streamline that is separating from the surface, not the fluid itself. When separation occurs, eddies or vortices form behind the body, as illustrated in Figure 12.1, which form a “wake” behind the sphere.

As the velocity (and NRe) increases, the point of streamline separation from the surface moves further upstream and the wake gets larger, both in the lateral and axial directions. The wake region contains circulating eddies of a three-dimensional turbulent nature, so it is a region of relatively high velocity and hence low pressure. Thus, the pressure in the wake is lower than that on the front of the sphere, and the product of this pressure difference and the projected area of the wake results in a net force acting on the sphere in the direction of the flow, that is, in the same direction as the drag force. This additional force resulting from the low pressure in the wake is called form drag (i.e., that component of the drag due to the pressure distribution, in excess of the viscous drag). The total drag is thus a combination of Stokes drag and wake drag, and consequently the drag coefficient is greater than that given by Equation 12.4 for NRe > 0.1. This is illustrated in Figure 12.2, which shows CD versus NRe for spheres (as well as for cylinders and disks oriented normal to the flow direction). For 1000 < NRe < 1 × 105, the sphere drag coefficient is approximately CD = 0.45. In this region, the wake, and consequently the wake drag, is a maximum, and the streamlines actually separate slightly ahead of the equator of the sphere. The drag at this point is completely dominated by the wake, which is actually larger in diameter than the sphere itself (see Figure 12.4).

Image

FIGURE 12.2 Drag coefficients for spheres, disks, and cylinders. (From Perry, J.H., Ed., Chemical Engineers’ Handbook, 6th edn., McGraw-Hill, New York, 1984.)

C.    ALL REYNOLDS NUMBERS

For NRe > 0.1 (or >1, within ~5%), a variety of expressions for CD versus NRe (mostly empirical) have been proposed in the literature (Clift et al., 1978; Chhabra, 2006). However, a simple and very useful equation, which represents the entire range of CD versus NRe reasonably well (within experimental error) up to about NRe = 2 × 105, has been given by Dallavalle (1948):

CD=(0.632+4.8NRe)2

(12.6)

(Actually, according to Coulson et al. [1991], this equation was first presented by Wadell in 1934.) A comparison of Equation 12.6 with the measured values is shown in Figure 12.2. A somewhat more accurate equation, although more complex, has been proposed by Khan and Richardson (1987):

CD=(2.25NRe0.31+0.35NRe0.06)3.45

(12.7)

Although Equation 12.7 is more accurate than Equation 12.6 at intermediate values of NRe, Equation 12.6 provides a sufficiently accurate prediction for most applications. Also it is simpler to manipulate, so we will prefer it as the analytical expression describing the drag coefficient for a sphere.

D.    CYLINDER DRAG

For flow past a circular cylinder normal to the cylinder axis, with L/d ≫ 1, the flow is similar to that for the sphere. An equation that adequately represents the cylinder drag coefficient over the entire range of NRe (up to about 2 × 105), which is analogous to the Dallavalle equation, is

CD=(1.05+1.9NRe)2

(12.8)

A comparison of this equation with measured values is also shown in Figure 12.2. It may be noted here that for the case of a long circular cylinder, no analytical expression akin to Equation 12.3 is possible in the limit of low Reynolds numbers (see, e.g., Lamb, 1945).

E.    BOUNDARY LAYER EFFECTS

As seen in Figure 12.2, the drag coefficient for the sphere exhibits a sudden drop from 0.45 to about 0.15 (almost 70%) at a Reynolds number of 2–5 × 105. For the cylinder, the drop is from about 1.1 to about 0.35. This drop is due to the transition of the flow inside the boundary layer from laminar to turbulent flow and can be explained as follows.

As the fluid encounters the solid boundary and proceeds along the surface, a boundary layer forms as illustrated in Figure 12.3. The boundary layer is the region of the fluid near a solid boundary in which viscous forces dominate and the velocity varies with the distance, from zero at the wall to a maximum value (V) at the edge of the boundary layer. Outside the boundary layer, the fluid velocity is that of the free stream that is unaffected by the presence of the wall. Adjacent to the wall in the boundary layer, the velocity is small and and the flow is laminar and stable. However, the boundary layer thickness (δ) grows along the plate (in the x direction), in proportion to NRex1/2 (where NRex = xVρ/μ). As the boundary layer grows, inertial forces increase and the boundary layer becomes less stable until it reaches a point (at NRex ≈ 2 × 105) where it becomes unstable, that is, turbulent. Within the turbulent boundary layer, the flow streamlines are no longer parallel to the boundary but break up into three-dimensional eddy structures.

Image

FIGURE 12.3 Boundary layer over a flat plate.

Image

FIGURE 12.4 Two bowling balls falling in still water at 25 ft/s. The ball on the left is smooth and the one on the right has a patch of sand on the nose. (From Coulson, J.M. et al., Chemical Engineering, Vol. 2, 4th edn., Pergamon Press, 1991.)

With regard to the flow over an immersed body (e.g., a sphere), the boundary layer grows from the impact (stagnation) point along the front of the body and remains laminar until NRex ≈ 2 × 105, where x is the distance traveled along the surface of the sphere, at which point it becomes turbulent. If the boundary layer is laminar at the point where streamline separation occurs, the separation point can be ahead of the equator of the sphere, resulting in a wake diameter which is larger than that of the sphere. However, if the boundary layer becomes turbulent before separation occurs, the three-dimensional eddy structure in the turbulent boundary layer carries momentum components inward toward the surface, which delays the separation of the streamline and tends to stabilize the wake. This delayed separation results in a smaller wake and a corresponding reduction in the form drag, which is the cause of the sudden drop in CD at NRe ≈ 2 × 105.

This shift in the size of the wake can be rather dramatic, as illustrated in Figure 12.4, which shows two pictures of a bowling ball falling in water with the wake clearly visible. The ball on the left shows a large wake since the boundary layer is laminar and the separation point is ahead of the equator. The ball on the right has a rougher surface that promotes turbulence, and the boundary layer has become turbulent before separation occurs resulting in a much smaller wake due to the delayed separation. The primary effect of surface roughness on the flow around immersed objects is to promote transition to a turbulent boundary layer and to delay the separation of the streamlines, and thus to slightly lower the value of NRe at which the sudden drop (or “kink”) in the CDNRe curve occurs. This apparent paradox—wherein the promotion of turbulence actually results in lower drag—has been exploited in various ways, such as the dimples on golf balls and the “boundary layer spoilers” on airplane wings and automobiles.

II.    FALLING PARTICLES

Many engineering operations involve the separation of solid particles from fluids, in which the motion of the particles is the result of a gravitational force (or other potential such as centrifugal force in a centrifuge or electric field in electrostatic precipitators). To illustrate this, consider a spherical solid particle with diameter d and density ρs, surrounded by a fluid of density ρ and viscosity μ, which is released and begins to fall (in the x = −z direction) under the influence of gravity. A momentum balance on the particle is simply ΣFx = max, where the forces include gravity acting on the solid particle (Fg), the buoyant force due to the fluid displaced by the solid particle (Fb), and the drag exerted by the fluid (FD). The inertial term involves the product of the acceleration (ax = dVx/dt) and the effective mass (m) of the particle. The mass that is accelerated includes that of the solid (ms) as well as the “virtual mass” (mf) of the fluid, which is displaced by the body as it accelerates. It can be shown that the latter is equal to one-half of the total mass of the displaced fluid (Clift et al., 1978), that is, mf = ½ ms (ρ/ρs). Thus, the x component of the momentum balance becomes

g(ρsρ)πd36CDρπd2V28=πd3(ρs+ρ/2)6dVdt

(12.9)

At t = 0, V = 0 and the drag force is zero. As the particle accelerates, the drag force increases, which decreases the acceleration. This process continues until the acceleration drops to zero, at which time the particle falls at a constant velocity resulting from the balance of forces due to drag and gravity (buoyancy). This steady-state terminal velocity of the body is given by the solution to Equation 12.9 with the acceleration equal to zero:

Vt=(4gΔρd3ρCD)1/2

(12.10)

where Δρ = (ρs – ρ). It is evident that the velocity cannot be determined until the drag coefficient is known, which in turn depends on the velocity. If the Stokes flow regime prevails, then CD = 24/NRe, and Equation 12.10 becomes

Vt=gΔρd218μ

(12.11)

However, the criterion for Stokes flow (i.e., NRe < 1) cannot be tested until Vt is known, and if it is not valid, then Equation 12.11 will not apply. This will be addressed shortly. It is, however, useful to note here that under these conditions, the terminal falling velocity of a spherical particle varies as Vtd2. Indeed, this forms the basis of classifying mixtures of particles of different sizes in settling ponds, centrifuges, thickeners, etc.

There are several types of problems that we may encounter with falling particles, depending upon what is known and what is to be found. All of these problems involve the two primary dimensionless variables CD and NRe or combinations thereof. The former is determined, for gravity-driven motion, by Equation 12.10, that is,

CD=4gΔρd3ρVt2

(12.12)

and CD can be related to NRe by the Dallavalle equation (Equation 12.6) over the entire practical range of NRe. The following procedures for the various types of problems apply to Newtonian fluids under all flow conditions.

A.    UNKNOWN VELOCITY

In this case, the unknown velocity (Vt) appears in the definitions of both the drag coefficient CD (Equation 12.12) and the Reynolds number NRe (Equation 12.5). Hence, a suitable dimensionless group that does not contain the unknown Vt can be formulated as follows:

CDNRe2=4d3ρgΔρ3μ2=43NAr

(12.13)

where NAr is known as the Archimedes number (also sometimes called the Galileo number). The most appropriate set of dimensionless variables to use for this problem is therefore NAr and NRe. An equation for NAr can be obtained by multiplying both sides of the Dallavalle equation (Equation 12.6) by NRe, and the result can then be rearranged for NRe to give

NRe=[(14.42+1.827NAr)1/23.798]2

(12.14)

The procedure for determining the unknown velocity is, therefore, as follows:

Given: d, ρ, ρs, μ Find: Vt

1.  Calculate the Archimedes number from Equation 12.13:

NAr=d3ρgΔρμ2

(12.15)

2.  Insert this value into Equation 12.14 and calculate NRe.

3.  Determine Vt from NRe, that is, Vt = NReμ/dρ.

If NAr < 15, then the system is within the Stokes law range and the terminal velocity is given by Equation 12.11.

B.    UNKNOWN DIAMETER

It often happens that we know or can measure the particle velocity and wish to know the size of the falling particle. In this case, we may form a dimensionless group that does not contain d as follows:

CDNRe=4μΔρg3ρ2Vt3

(12.16)

This group can be related to the Reynolds number by dividing Equation 12.6 by NRe and then solving the resulting equation for 1/(NRe)1/2 to give

1NRe=(0.00433+0.208CDNRe)1/20.0658

(12.17)

The two appropriate dimensionless variables are now CD/NRe and NRe. The procedure is as follows:

Given: Vt, ρs, ρ, μ Find: d

1.  Calculate CD/NRe from Equation 12.16.

2.  Insert the result into Equation 12.17 and calculate 1/(NRe)1/2, and hence NRe.

3.  Calculate d = μNRe/Vtρ.

If CD/NRe > 30, the flow is within the Stokes law range, and the diameter can be calculated directly from Equation 12.11:

d=(18μVtgΔρ)1/2

(12.18)

This is often used to estimate the size of small particles by measuring their terminal velocity in a container much larger than the particles, with a fluid of known properties.

C.    UNKNOWN VISCOSITY

The viscosity of a Newtonian fluid can be determined by measuring the terminal velocity of a sphere of known diameter and density if the fluid density is known. If the Reynolds number is low enough for Stokes flow to apply (NRe < 0.1), then the viscosity can be determined directly by rearrangement of Equation 12.11 as follows:

μ=d2gΔρ18Vt

(12.19)

This equation forms the basis of commercially available falling ball viscometers (Gupta, 2014). The Stokes flow criterion is rather stringent e.g., a 1 mm diameter sphere would have to fall at a rate of 1 mm/s or slower in a fluid with a viscosity of 10 cP and SG = 1 to be in the Stokes range, which means that the density of the solid would have to be within 2% of the density of the fluid!

However, with only a slight loss in accuracy, the Dallavalle equation can be used to extend the useful range of this measurement to a much higher Reynolds number, as follows. From the known quantities, CD can be calculated from Equation 12.12. The Dallavalle equation (Equation 11.6) can be rearranged to give NRe:

NRe=(4.8CD1/20.632)2

(12.20)

The viscosity is then determined from the known value of NRe:

μ=dVtρNRe

(12.21)

Note that when NRe > 1000, CD ≈ 0.45 (constant), which, from Equation 12.20, gives μ = 0! Although this may seem strange, it is consistent because in this range the drag is dominated by form (wake) drag and viscous forces are negligible. It should be evident that one cannot determine the viscosity from measurements made under conditions that are insensitive to viscosity, which means that the utility of Equation 12.20 is limited in practice to approximately NRe < 50–100.

III.    CORRECTION FACTORS

A.    WALL EFFECTS

All expressions presented so far have assumed that the particles are surrounded by an infinite sea of fluid, that is, that the boundaries of the fluid container are far enough from the particle that their influence is negligible. For a falling particle, this might seem to be a reasonable assumption if d/D < 0.01 (say), where D is the container diameter. However, the presence of the wall is felt by the particle over a much greater distance than one might expect. This is because as the particle falls it must displace an equal volume of fluid, which must then flow upward around the particle to fill the space just vacated by the particle. Thus, the relative velocity between the particle and the adjacent fluid is much larger in a confined space than it would be in an infinite fluid, that is, the effective “free stream” (relative) velocity is no longer zero, as it would be for an infinite stagnant fluid. A variety of analyses of this problem have been performed, as reviewed by Clift et al. (1978) and Chhabra (2002, 2006). These can be represented by a wall correction factor (Kw) that is a multiplier for the “infinite fluid” terminal velocity and corrects for the wall effect. (This is also equivalent to correcting the Stokes law drag force by a factor of Kw.) The following equation due to Francis (1933) is found to be valid for d/D < 0.97 and NRe < 1:

Kwo=(1d/D10.475d/D)4

(12.22)

For larger Reynolds numbers, the following expression is found to be satisfactory for d/D < 0.8 and NRe > 1000:

Kw=1(dD)1.5

(12.23)

Although these wall correction factors appear to be independent of the Reynolds number for small (Stokes) and large values of NRe (> 1000), the value of Kw is a function of both NRe and (d/D) for intermediate Reynolds numbers (Chhabra, 2002, 2006). Further corrections to the measured falling velocities are required for off-center settling and/or in short tubes due to bottom effects (Wham et al., 2002; Chhabra, 2006).

B.    EFFECT OF PARTICLE SHAPE

It is readily conceded that nonspherical particles like cylinders, disks, thin wires, cones, prisms, oblates and prolates, or of irregular shape, etc. are encountered much more frequently than the idealized case of spherical particles. Naturally, the drag and terminal settling velocity of such particles in fluids are strongly influenced by their shape and orientation. For instance, the settling velocity of a given cylinder (known length and diameter) can vary by a large factor depending upon its orientation during its sedimentation. Therefore, the need to estimate the terminal settling velocity (or drag force) often arises in process design calculations for slurry pipe lines, fluidized beds, hydrocyclones, settling chambers, etc. Most of the results available in the literature (Chhabra et al., 1999; Chhabra, 2006) fall into two distinct categories: in the first approach, the particle shape and orientation are maintained constant and drag coefficient is correlated with the Reynolds number by numerical solution of the Navier–Stokes equations or experiments. Naturally, this approach is accurate but has limited appeal due to its applicability to a fixed shape and orientation. In the second approach, the results for variously shaped particles are consolidated via a single correlation. The following discussion is based mainly on the latter approach.

At least two attributes of a nonspherical particle are needed to account for the settling behavior, namely, size and shape. A sphere is unique in so far that one only needs to specify its radius or diameter as it exhibits perfect symmetry. This is not so for a nonspherical particle, for example, diameter and length of a cylinder, minor and major axes of spheroids, etc. are needed to describe their “size.” One of the simplest definitions of the effective diameter of a nonspherical particle is the so-called Stokes diameter, which is the diameter of the sphere that has the same terminal velocity as the actual nonspherical particle. This can be determined from a direct measurement of the settling rate of the particles and provides the best value of equivalent diameter for use in applications involving fluid drag on particles. Another widely used measure of the size of a nonspherical particle is the diameter of the sphere with a volume equal to that of the particle, ds. For a nonspherical particle of volume Vp, the equivalent diameter, ds, is

ds=(6Vpπ)1/3

(12.24)

Next, it is conceivable that a cylinder and a spheroid can have the same ds value, yet their terminal falling velocity (and drag force experienced) differs from each other significantly due to intrinsic difference in their shapes. The simplest measure of the shape (or deviation from a sphere) is the so-called sphericity of a particle, ψ. The sphericity is defined as

ψ=surfacearea of an equal volume spheresurface area of the actual particle

(12.25)

Equation 12.25 can also be expressed in terms of the volume (Vp) and surface area (Ap) of the particle as

ψ=πds2Ap=π(6Vp/π)2/3Ap=(62π)1/3Vp2/3Ap=4.84Vp2/3Ap

(12.26)

Obviously for a sphere, ψ = 1 and, since for a given volume, the sphere has the minimum surface area/volume, ψ < 1 for all other shapes. It is also possible to define an effective particle diameter by d = ψ ds = 6/as, where as = particle surface area/particle volume. If the particles are spherical with diameter d, then as = 6/d. This definition is commonly used in the context of packed and fluidized beds, as will be seen in Chapters 15 and 16. These two definitions of ψ are equivalent.

Representative values of ψ for some familiar shapes are given in Table 12.1. The smaller the value of ψ, the greater is the departure from a spherical shape.

Using the two parameters ds and ψ, Haider and Levenspiel (1989) collated much of the literature data for drag behavior of nonspherical particles, as shown in Figure 12.5. While the general trend for a fixed value of ψ is similar to that seen for a sphere in Figure 12.2 (except for the sudden drop in the value of the drag coefficient at NRe ~ 2 × 105), the drag experienced by a nonspherical particle is always higher than that of a sphere at the same Reynolds number. Further inspection of Figure 12.5 shows that the onset of fully turbulent regime (constant value of drag coefficient) occurs at lower Reynolds numbers for smaller values of ψ. For thin disks oriented normal to the flow, this limiting behavior is approached at about NRe ~ 10 as opposed to that for a sphere, that is, NRe ~ 103.

Another interesting observation is that the drag is seen to be influenced by shape much more at high Reynolds numbers than that at low Reynolds numbers. Generally, the size of the particle (surface area) matters more at small Reynolds numbers, whereas the shape of the particle (ψ) is more relevant at high Reynolds numbers. Based on the literature data, Haider and Levenspiel (1989) put forth the following correlation for calculating the drag of a nonspherical particle:

CD=24NRe[1+8.172e4.066ψNRe0.0964+0.557ψ]+73.7e5.075ψNReNRe+5.38e6.21ψ

(12.27)

Note that in both Figure 12.5 and Equation 12.27, d = ds is used in the definitions of the Reynolds number and drag coefficient. Equation 12.27 purports to reproduce the literature data with a root mean square (RMS) error of 5.8%.

TABLE 12.1

Values of Sphericity

Shape

ψ

Sphere

1

Cube

(π/6)1/3

Circular cylindera (length L, diameter d)

{(3/2)(L/d)(L/d)+(1/2)}1/2

a For large values of (L/d), the factor (1/2) in the denominator arising from the two ends can be neglected.

Image

FIGUR 12.5 Drag coefficients for nonspherical particles. (Redrawn from A. Haider and O. Levenspiel, Powder Technol., 58, 63, 1989.)

In order to facilitate the estimation of the unknown velocity or the unknown diameter (Sections II.A and II.B), Haider and Levenspiel (1989) introduced a dimensionless falling velocity V* and a dimensionless diameter d* defined as follows:

V*=(43NReCD)1/3=(ρ2Vt3μ(Δρ)g)1/3

(12.28)

which is similar to Equation 12.16, and

d*=(34CDNRe2)1/3=(NAr)1/3

(12.29)

Using the same experimental data, Haider and Levenspiel (1989) rearranged these results in terms of d* and V* as follows:

V*=[18(d*)2+2.3351.74ψd*]1

(12.30)

Figure 12.6 shows a graphical representation of Equation 12.30. It is clearly seen that for a given particle volume (i.e., constant value of ds), nonspherical particles settle slower than the same size sphere under identical conditions, that is, a nonspherical particle experiences more drag than a sphere.

It should be noted here that although both ds and ψ are measures of the particle size and shape, respectively, this approach does not account for the orientation of the particle. Thus, this approach will not distinguish whether a cylinder falls with its long axis aligned with or transverse to the direction of gravity. This issue has been discussed by Ganser (1993) and Rajitha et al. (2006), among others. In essence, this necessitates introducing another ratio (dn/ds), where dn is the diameter of the circle with the same area as the projected area of the particle during its fall. Obviously, dn/ds = 1 for a sphere. However, even this measure will not distinguish the cases of a cone falling with its apex upward or downward. Some correlations are available that tend to be far more complex than Equation 12.27, which account for the orientation of the particle during settling. Another weakness of the discussion presented here is that it is applicable only for regular shaped particles for which the surface area can be calculated or measured. Some ideas for treating irregular shaped particles are also available in the literature (Zakhem et al., 1992; Tran-Cong et al., 2004).

Image

FIGURE 12.6 Graphical representation of Equation 12.30. (Based on A. Haider and O. Levelspiel, Drag coefficient and terminal velocity of spherical and non-spherical particles, Powder Technology, 58, 63, 1989.)

Example 12.1

Calculate the terminal falling velocity of a 1 mm diameter sphere made of plastic (ρs = 1100 kg/m3), glass (ρs = 2500 kg/m3), and steel (ρs = 7870 kg/m3) in air, water, and oil (ρ = 875 kg/m3, μ = 27 mPa s) at 25°C.

Solution:

First, we look up density and viscosity values for air and water in Appendix A:

At 25°C

Air

Water

Oil

μ (mPa s)

0.01849

0.8904

27

ρ (kg/m3)

1.184

1000

875

For the 1 mm diameter plastic sphere (ρs = 1100 kg/m3) falling in air, from Equation 12.13 NAr = d3ρgΔρ/μ2

NAr=(1×103m)3(1.184kg/m3)(9.81m/s2)(11001.84)kg/m3(0.01849×103kg/ms)2=37,331

From Equation 12.14, NRe=[(14.42+1.827NAr)1/23.798]2=236.2

Now, NRe = ρVd

V=NReμρd=(236.2)(0.01849×103kg/ms)(1.184kg/m3)(0.001m)=3.333m/s

Similarly, calculations for the other spheres yield the following results:

Plastic

Glass

Steel

Air

NAr

3.73 × 104

8.66 × 104

2.67 × 105

NRe

236

388

738

V (m/s)

3.69

6.06

11.5

Water

NAr

1.24 × 103

1.92 × 104

9.74 × 104

NRe

25.7

158

415

V (m/s)

0.0229

0.1404

0.3698

Oil

NAr

2.65

19.7

82.4

NRe

0.139

0.901

3.13

V (m/s)

0.00428

0.02779

0.09666

Example 12.2

Calculate the terminal falling velocity of the following particles of different shapes but equal volume made of plastic (ρs = 1100 kg/m3) in air and water at 25°C.

(a)  5 mm diameter sphere

(b)  Cube

(c)  Cylinder with L/d = 1, 2, 5

(d)  Spheroids of aspect ratios 0.5 and 2

Physical properties of air and water from Appendix A:

At 25°C

Air

Water

μ (mPa s)

0.01849

0.8904

ρ (kg/m3)

1.184

1000

Solution:

For the particle of plastic (ρs = 1100 kg/m3) falling in air from Equation 12.29,

d*=(34CDNRe2)1/3=(NAr)1/3

For equal volume particles, the value of ds will be the same for each shape, though the value of ψ will vary with shape. Thus, d = ds = 5 mm.

d*=(d3ρg(ρsρ)1/3μ2)=((0.005m)3(1.184kg/m3)(9.81m/s2)(11001.184)kg/m3(0.01849×103kg/ms)2)1/3=167.12

(a)  For the 5 mm diameter sphere, from Equation 12.30,

V*=[18(d*)2+2.3351.74ψd*]1

For the sphere, ψ = 1

V*=21.43

The terminal falling velocity of the particle, from Equation 12.28, is

V*=[ρ2Vt3μ(ρsρ)g]1/3Vt=V*[μ(ρsρ)gρ2]1/3=(21.43)×[(0.01849×103kg/ms)(11001.184)(kg/m3)(9.81m/s2)(1.184kg/m3)2]1/3Vt=11.18m/s

The corresponding value of the Reynolds number is

NRe=ρVdsμ=(1.184kg/m3)(11.18m/s)(0.005m)(0.01849×103kg/ms)=3580

which is below the critical Reynolds number.

(b)  Cube

For a cube of equal volume to a sphere,

 Volume of cube = Volume of sphere

l3=(π6)ds3l=(π6)1/3(0.005m)=0.004m

For a cube,

ψ=πds26l2=0.82V*=[18(d*)2+2.3351.74ψd*]1=[18(167.12)2+2.3351.74(0.82)167.12]1=14.05Vt=V*[μ(ρsρ)gρ2]1/3=(14.05)×[(0.01849×103kg/ms)(11001.184)(kg/m3)(9.81m/s2)(1.184kg/m3)2]1/3Vt=7.34m/s

(c)  For cylinders with L/d = 1, 2, 5

L/d=1L=d

For a cylinder of volume equal to the sphere,

 Volume of cylinder = Volume of sphere

π4d2L=π6ds3d=(23)1/3ds=(23)1/3(0.005m)=0.0044mL=0.0044m

For the cylinder,

ψ=πds2πdL+2((π/4)d2)=0.87V*=[18(d*)2+2.3351.74ψd*]1=[18(167.12)2+2.3351.74(0.87)167.12]1=15.69Vt=V*[μ(ρsρ)gρ2]1/3=(15.69)×[(0.01849×103kg/ms)(11001.184)(kg/m3)(9.81m/s2)(1.184kg/m3)2]1/3Vt=8.19m/s

Similarly, for a cylinder with L/d = 2,

ψ=0.85,V*=14.58,Vt=7.58m/sL/d=5:ψ=0.74,V*=12.23,Vt=6.38m/s

(d)  For a spheroid with an aspect ratio of 0.5,

Aspectratio=b/a=0.5,b=0.5a

For a spheroid of equal volume to a sphere,

 Volume of spheroid = Volume of sphere

43πab2=π6ds3a=(12)1/3ds=(12)1/3(0.005m)=0.004mb=0.002m

For the spheroid,

ψ=πds22πb2+(2πab/e)sin1e=0.92

where

e=[1(ba)2]1/2=0.87V*=[18(d*)2+2.3351.74ψd*]1=[18(167.12)2+2.3351.74(0.92)167.12]1=17.28Vt=V*[μ(ρsρ)gρ2]1/3=(17.28)×[(0.01849×103kg/ms)(11001.184)(kg/m3)(9.81m/s2)(1.184kg/m3)2]1/3Vt=9.02m/s

For the spheroid with an aspect ratio of 2,

Aspectratio=a/b=2,a=2b

For the spheroid of equal volume to a sphere,

 Volume of spheroid = Volume of sphere

43πa2b=π6ds3b=(132)1/3ds=(132)1/3(0.005m)=0.0016ma=0.0032m

For the spheroid,

ψ=πds22πb2+(πa2/e)ln((1+e)/(1e))=0.71

where

e=[1(ba)2]1/2=0.87V*=[18(d*)2+2.3351.74ψd*]1=[18(167.12)2+2.3351.74(0.71)167.12]1=11.68Vt=V*[μ(ρsρ)gρ2]1/3=(11.63)×[(0.01849×103kg/ms)(11001.184)(kg/m3)(9.81m/s2)(1.184kg/m3)2]1/3Vt=6.10m/s

Summary of results

Vt (m/s)

ψ

Air

Water

Sphere

1

11.18

0.109

Cube

0.82

 7.32

0.073

Cylinder

L/d = 1

0.87

 8.19

0.081

L/d = 2

0.84

 7.58

0.076

L/d = 5

0.74

 6.38

0.064

Spheroid

b/a = 0.5

0.92

 9.02

0.089

a/b = 2

0.71

 6.10

0.061

As the value of ψ increases, that is, the deviation from a spherical shape decreases, the terminal falling velocity of nonspherical particles is seen to approach that of the sphere, as expected.

C.    DROPS AND BUBBLES

Because of the surface tension forces, very small drops and bubbles are nearly rigid spheres and behave in a manner similar to rigid particles. However, larger fluid drops or bubbles may experience a considerably different settling behavior because the shear stress on the drop surface can be transmitted to the fluid inside the drop, which, in turn, results in circulation of the internal fluid. This internal circulation dissipates energy, which is extracted from the energy of the bubble motion and is equivalent to an additional drag force. For example, in Stokes flow of spherical drops or bubbles (e.g., NRe < 1), it has been shown by Hadamard and Rybczynski (see, e.g., Clift et al., 1978; Grace, 1983) that the drag coefficient can be corrected for this effect as follows:

Cd=24NRe(κ+2/3κ+1)

(12.31)

where κ = μio, μo is the viscosity of dispersed (“inside”) fluid and μo is the viscosity of the continuous (“outside”) fluid.

For larger Reynolds numbers (1 < NRe < 500), Rivkind and Ryskind (Grace, 1983) have proposed the following equation for the drag coefficient for spherical drops and bubbles:

CD=1κ+1[κ(24NRe+4NRe1/3)+14.9NRe0.78]

(12.32)

As the size of the drop or bubble increases, however, it will become distorted due to the unbalanced forces around it. The viscous shear stresses tend to elongate the shape, whereas the pressure distribution tends to flatten it out in the direction normal to the flow. Thus, the shape tends to progress from spherical, to ellipsoidal, to a “spherical cap” form as the size increases. Above a certain size, the deformation is so great that the drag force is approximately proportional to the volume, and the terminal velocity becomes nearly independent of size. More detailed discussion can be found in the books of Clift et al. (1978) and Michaelides (2006).

IV.    NON-NEWTONIAN FLUIDS

The motion of solid particles, drops, or bubbles through non-Newtonian fluid media is encountered frequently and has been the subject of considerable research (e.g., Chhabra, 2006). We will present some relations here that are applicable to purely viscous non-Newtonian fluids, although there is also much interest and activity in viscoelastic fluids as well. Despite the relative large amount of work that has been done in this area, there is still no general agreement as to the “right,” or even the “best,” description of the drag on a sphere in non-Newtonian fluids. This is due not only to the complexity of the momentum equations that must be solved for the various models but also the difficulty in obtaining good, reliable, representative data for fluids with well-characterized unambiguous rheological properties.

A.    POWER LAW FLUIDS

The usual approach for non-Newtonian fluids is to start with the known results for Newtonian fluids and modify them to account for the non-Newtonian properties. For example, the definition of the Reynolds number for a power law fluid can be obtained by replacing the viscosity in the Newtonian definition by an appropriate shear rate–dependent viscosity function. If the characteristic shear rate for flow over a sphere is taken to be (V/d), for example, the power law viscosity function becomes

μη(γ˙)m(Vd)n1

(12.33)

and the corresponding expression for the Reynolds number is

NRepl=ρVdm(V/d)n1=ρV2ndnm

(12.34)

The corresponding (creeping flow) drag coefficient can be characterized by a correction factor (X) applied to the Stokes law drag coefficient:

CD=24NRe,plX

(12.35)

A variety of theoretical expressions, as well as experimental values, for the correction factor X as a function of the power law flow index (n) have been summarized by Chhabra (2006). In a series of papers, Chhabra (1995, 2006) and Tripathi et al. (1994, 1995) presented the results of numerical calculations for the drag on spheroidal particles in a power law fluid in terms of CD = fn(NRe, n). Darby (1996) analyzed these results and has shown that this function can be expressed in a form equivalent to the Dallavalle equation, which applies over the entire range of n and NRe. This equation is

CD=(C1+4.8NRe,pl/X)2

(12.36)

where both X and C1 are functions of the flow index, n. These functions were determined by Darby (1996) by empirically fitting the following equations to the values given by Chhabra (1995):

1C1=[(1.82n)8+34]1/8

(12.37)

X=1.33+0.37n1+0.7n3.7

(12.38)

Image

FIGURE 12.7 Plot of 1/C1 versus n for power law fluid.

Image

FIGURE 12.8 Plot of X versus n for power law fluid.

The agreement between these values of C1 and X and the values given by Chhabra is shown in Figures 12.7 and 12.8 up to NRe,pl ≤ 100. Equations 12.36 through 12.38 are equivalent to the Dallavalle equation for a sphere in a power law fluid. A comparison of the values of CD predicted by Equation 12.36 with the values given by Chhabra is shown in Figure 12.9. The deviation is the greatest for highly dilatant fluids, in the Reynolds number range of about 5–50, although the agreement is quite reasonable above and below this range, and for pseudoplastic fluids over the entire range of Reynolds number. We will illustrate the application of these equations by outlining the procedure for solving the “unknown velocity” and the “unknown diameter” problems.

1.  Unknown Velocity

The expressions for CD and NRe,pl can be combined to give a group that is independent of V, as follows:

CD2n(NReplX)2=(ρXm)2(4gΔρ3ρ)2ndn+2=Nd

(12.39)

Image

FIGURE 12.9 Comparison of Equation 12.36 with results of Tripathi (1994) and Chhabra (1995, 2006).

which is similar to Chhabra’s D+ parameter. Using Equation 12.36 to eliminate CD gives

Nd=[C1(NRe,plX)1/(2n)+4.8(NRe,plX)n/2(2n)]2(2n)

(12.40)

Although this equation cannot be solved analytically for NRe,pl, it can be solved by iteration (or using the “solve” command on a calculator or spreadsheet), since all other parameters are known. The unknown velocity is then given by

V=(mNRe,plρdn)1/(2n)

(12.41)

2.  Unknown Diameter

The diameter can be eliminated from the expressions for CD and NRe,pl as follows:

CDnXNRe,pl=(4gΔρ3ρ)n(Xmρ)V(n+2)=NV

(12.42)

Again using Equation 12.36 to eliminate CD gives

NV=[C1(XNRe,pl)1/2n+4.8(XNRe,pl)(1+n)/2n]2n

(12.43)

As before, everything in this equation is known except for NRe,pl, which can be determined by iteration (or the “solve” spreadsheet or calculator command). When this is found, the unknown diameter is given by

d=(mNRe,plρV2n)1/n

(12.44)

Example 12.3:  Unknown Velocity and Unknown Diameter of a Sphere Settling in a Power Law Fluid

Table E12.1 summarizes the procedure and shows the results of a spreadsheet calculation for an application of this method to the same three examples given by Chhabra (1995). Examples 1 and 2 are “unknown velocity” problems, and Example 3 is an “unknown diameter” problem (Table E12.2). The line labeled “Equation” refers to Equation 12.40 for the unknown velocity cases and Equation 12.43 for the unknown diameter case. The “Stokes” value is from Equation 12.10, which only applies for NRe,pl < 1 (e.g., Example 1). It is seen that the solutions for Examples 1 and 2 are virtually identical to Chhabra’s, and that for Example 3 is within 5% of Chhabra’s. The values labeled “Using Data” were obtained by iteration using the data from Figure 4 of Tripathi et al. (1994). These values are only approximate, since they were obtained by interpolating from the (very compressed) log scale of the plot. The method shown here has several advantages over that reported by Chhabra (1995), namely:

TABLE E12.1

Procedure for Determining Unknown Velocity or Unknown Diameter for Particles Settling in a Power Law Fluid

Problem

Unknown Velocity

Unknown Diameter

Given:

Particle diameter (d) and fluid properties (m, n, and ρ)

Particle settling velocity (V) and fluid properties (m, n, and ρ)

Step 1

Using value of n, calculate C1 and X from Equations 12.37 and 12.38

Using value of n, calculate C1 and X from Equations 12.37 and 12.38

Step 2

Calculate Nd from Equation 12.39

Calculate Nv from Equation 12.42

Step 3

Solve Equation 12.40 for NRe,pl by iteration (or using “solve” function)

Solve Equation 12.43 for NRe,pl by iteration (or using “solve” function)

Step 4

Get V from Equation 12.41

Get d from Equation 12.44

TABLE E12.2

Comparison of Calculated Settling Properties Using Equation 12.35 with Literature Values, Chhabra (1995)

Example 1

Example 2

Example 3

Given data:

d = 0.002 m

d = 0.002 m

V = 0.2 m/s

m = 1.3 Pa sn

m = 0.015 Pa sn

m = 0.08 Pa sn

n = 0.6

n = 0.8

n = 0.5

ρ = 1002 kg/m3

ρ = 1050 kg/m3

ρ = 1005 kg/m3

ρs = 7780 kg/m3

ρs = 2500 kg/m3

ρs = 8714 kg/m3

Calculated values:

X = 1.403

X = 1.244

X = 1.438

C1 = 0.329

C1 = 0.437

C1 = 0.275

Nd = 15.4

Nd = 2830

Nd = 0.064

NRe,pl = 0.082

NRe,pl = 55.7

NRe,pl = 29.8

Darby (1996)

V = 0.0208 m/s

V = 0.165 m/s

d = 7.05 × 10-4 m

Chhabra (1995)

V = 0.020 m/s

V = 0.167 m/s

d = 6.67× 10-4 m

Stokes’ law

V = 0.0206 m/s

V = 0.514 m/s

d = 5.31 × 10-4 m

Tripathi et al. (1994)

V = 0.167 m/s

d = 7.18 × 10-4 m

NRe,pl = 57

NRe,pl = 30

CD = 1.3

CD = 1.8

(1)  All expressions are given in equation form, and it is not necessary to read any plots to solve the problems (i.e., the empirical data are represented analytically by curve-fit equations).

(2)  The method is more general, in that it is a direct extension of the technique of solving similar problems for Newtonian fluids and applies over all values of the Reynolds number.

(3)  Only one calculation procedure is required, regardless of the value of the Reynolds number for the specific problem.

(4)  The calculation procedure is simple and straightforward and can be done quickly using a spreadsheet.

B.    WALL EFFECTS

The wall effect for particles settling in power law non-Newtonian fluids appears to be significantly smaller than for Newtonian fluids. For power law fluids, the wall correction factor in creeping flow, as well as for very high Reynolds numbers, appears to be independent of the Reynolds number, similar to the behavior seen in Newtonian fluids. For creeping flow, the wall correction factor given by Chhabra (2002, 2006) is

Kwo=11.6(dD)

(12.45)

whereas for high Reynolds numbers, he gives

Kw=13(dD)3.5

(12.46)

For intermediate Reynolds numbers, the wall factor depends upon the Reynolds number as well as d/D. Over a range of 10−2 < NRe,pl < 103, 0 < d/D < 0.5, and 0.53 < n < 0.95, the following equation describes the Reynolds number dependence of the wall factor quite well:

(1/Kw)(1/Kw)(1/Kwo)(1/Kw)=(1+1.3NRe,pl2)0.33

(12.47)

It may be concluded that the predictions of Equation 12.47 are in good agreement with the numerical predictions of Missirlis et al. (2001) in the creeping flow regime, whereas the correspondence between the predictions and experiments deteriorates with the increasing Reynolds number and/or decreasing value of the power law index (Song et al., 2009). Also, note that Equations 12.45 through 12.47 implicitly include the dependence on the power law index, whereas the numerical results (Song et al., 2009) suggest a stronger influence of the power law index (n) than these equations.

C.    CARREAU FLUIDS

As discussed in Chapter 3, the Carreau viscosity model is one of the most general and useful and reduces to many of the common two-parameter models (power law, Ellis, Sisko, Bingham, etc.) as special cases. This model can be written as

η=η+(ηoη)[1+(λγ˙)2](n1)/2

(12.48)

where n = (1 – 2p) is the flow index for the power law region (p is the shear thinning parameter in the form of this model given in Equation 3.27). Since the shear conditions surrounding particles virtually never reaches the levels corresponding to the high shear viscosity (η), this parameter can be neglected and the parameters reduced to three: ηo, λ, and n. Chhabra and Uhlherr (1980) and Bush and Phan-Thien (1984) have determined the Stokes flow correction factor, X, for this model, which is a function of the dimensionless parameters n and Nλ = λV/d. The following equation represents their results for the CD correction factor over a wide range of data to ±10%, for (0.4 < n < 1) and (0 < Nλ < 400):

X=1[1+(0.275Nλ)2](1n)/2

(12.49)

where the Stokes equation uses ηo for the viscosity in the definition of the Reynolds number.

D.    BINGHAM PLASTICS

A particle will not fall through a fluid with a yield stress unless the weight of the particle is sufficient to overcome the yield stress. Because the stress is not uniform around the particle and the distribution is very difficult to determine, it is not possible to determine the critical “yield” criterion exactly. However, it should be possible to characterize this state by a dimensionless “gravity yield” parameter, at least for a sphere falling under its own weight:

YG=τogdΔρ

(12.50)

By equating the vertical component of the yield stress acting over the surface of the sphere to the weight of the particle, a critical value of YG = 0.17 results (Chhabra, 2006; Chhabra and Richardson, 2008). Experimentally, however, the results appear to fall into groups: one, for which YG ≈ 0.2 and the other for which YG ≈ 0.04–0.08. There seems to be no consensus as to the correct value, and the difference may well be due to the fact that the yield stress is not an unambiguous parameter, in as much as values determined from “static” measurements can differ significantly from the values determined from “dynamic” measurements, that is, whether the yield point is approached from above or below.

With regard to the drag on a sphere that is moving in a Bingham plastic medium, the drag coefficient (CD) must be a function of the Reynolds number as well as either the Hedstrom number or the Bingham number (NBi = NHe/NRe = τodV). One approach is to reconsider the Reynolds number from the perspective of the ratio of inertial to viscous momentum flux. For a Newtonian fluid in a tube, this is equivalent to

NRe=DVρμ=8ρV2μ(8V/D)=8ρV2τw

(12.51)

which follows from the Hagen–Poiseuille equation, since τw = μ(8V/D) is the drag per unit area of pipe wall, and the shear rate at the wall of the pipe is γ˙w=8V/D. By analogy, the drag force per unit area on a sphere is F/A = CDρV2/2, which, for Stokes flow (i.e., CD = 24/NRe), becomes F/A = 12μV/d. If F/A for the sphere is considered to be analogous to the “wall stress” (τ) on the sphere, the corresponding “effective wall shear rate” is 12V/d. Thus, the sphere Reynolds number could be written as

NRe=dVρμ=12ρV2μ(12V/d)=12ρV2τ

(12.52)

For a Bingham plastic, the corresponding expression would be

12ρV2τ=12ρV2μ(12V/d)+τo=NRe1+NBi/12

(12.53)

Equation 12.53 is thus the effective Reynolds number for correlating the drag coefficient.

Another approach is to consider the effective shear rate over the sphere to be V/d, as was done in Equation 12.33 for the power law fluid. If this approach is applied to a sphere in a Bingham plastic, the result is

NReBP=NRe1+NBi

(12.54)

This is similar to the analysis obtained by Ansley and Smith (1967) using the slip line theory from soil mechanics, which results in a dimensionless group called the plasticity number:

NPl=NRe1+2πNBi/24

(12.55)

A finite element analysis (Blackery and Mitsoulis, 1997) resulted in an equivalent Stokes law correction factor X (= CD NRe/24) that is a function of NBi for NBi < 1000, as follows:

X=1+aNBib

(12.56)

where a = 2.93 and b = 0.83 for a sphere in an unbounded fluid, and 2.93 > a > 1.63 and 0.83 < b < 0.95 for 0 < d/D < 0.5. Also, based upon available data, Chhabra and Uhlherr (1988) found that the “Stokes flow” relation (CD = 24X/NRe) applies up to NRe100NBi0.4 for Bingham plastics. Equation 12.56 is equivalent to a Bingham plastic Reynolds number (NReBP) of

NReBP=NRe1+2.93NBi0.83

(12.57)

Unfortunately, there are insufficient experimental data reported in the literature to verify or confirm any of these expressions. Thus, for the lack of any other information, Equation 12.57 is recommended, because it is based on the most detailed analysis.

This can be extended beyond the Stokes flow region by incorporating Equation 12.57 into the equivalent Dallavalle equation:

CD=(0.632+4.8NReBP)2

(12.58)

which can be used to solve the “unknown velocity” and “unknown diameter” problems as previously discussed. However, in this case, rearrangement of the dimensionless variables CD and NReBP into an alternate set of dimensionless groups in which the unknown appears in only one group is not possible due to the form of NReBP. Thus, the procedure would be to equate Equations 12.58 and 12.12 and solve the resulting equation directly by iteration for the unknown V or d (as the case requires).

More detailed discussions concerning the combined effects of shear thinning and yield stress on drag and wake phenomena and on wall effects for a sphere can be found in the literature (Nirmalkar et al., 2013; Das et al., 2015). The corresponding results for a cylinder are also available in the literature (Nirmalkar and Chhabra, 2014).

SUMMARY

The key concepts that should be retained from this chapter include

•  Similarities and differences between fluid-particle drag and fluid-wall drag in a pipe

•  Cause and effect of “form drag”

•  Using a falling particle in a fluid to determine (a) particle velocity, (b) particle diameter, and (c) fluid viscosity

•  Influence of the container walls on the velocity of a falling particle

•  Effect of geometry of nonspherical particles on the terminal falling velocity

•  The flow behavior of drops and bubbles

•  The falling velocity of particles in non-Newtonian fluids

PROBLEMS

1.  By careful streamlining, it is possible to reduce the drag coefficient of an automobile from 0.4 to 0.25. How much power would this save at 40 mph and 60 mph, assuming the effective projected area of the car is 25 ft2?

2.  If your pickup truck has a drag coefficient equivalent to a 5 ft diameter disk, and the same projected frontal area, how much horsepower is required to overcome wind drag at 40 mph? What horsepower is required at 70 mph?

3.  You take a tumble while water skiing. The handle attached to the tow rope falls beneath the water and remains perpendicular to the direction of the boat’s heading. If the handle is one inch in diameter and 1 ft long, and the boat is moving at 20 mph, how much horsepower is required to pull the handle through the water?

4.  Your new car is reported to have a drag coefficient of 0.3. If the cross-sectional area of the car is 20 ft2, how much horsepower is used to overcome wind resistance at 40 mph, 55 mph, 70 mph, and 100 mph? (T = 70°F)

5.  The supports for a tall chimney must be designed to withstand a 120 mph wind. If the chimney is 10 ft in diameter and 40 ft high, what is the wind force on the chimney at this speed? T = 50°F.

6.  A speedboat is propelled by a water jet motor that takes water in at the bow through a 10 cm diameter duct and discharges it through a 50 mm diameter nozzle at a rate of 80 kg/s. Neglecting friction in the motor and internal ducts and assuming that the drag coefficient for the boat hull is the same as for a 1 m diameter sphere, determine

(a)  The static thrust developed by the motor when it is stationary

(b)  The maximum velocity attainable by the boat

(c)  The power (kW) required to drive the motor

(Assume seawater density to be 1030 kg/m3, viscosity to be 1.2 cP.)

7.  After blowing up a balloon, you release it without tying off the opening, and it flies out of your hand. If the diameter of the balloon is 6 in., the pressure inside it is 1 psig, and the opening is 1/2 in. in diameter, what is the balloon velocity? You may neglect friction in the escaping air and the weight of the balloon and assume that an instantaneous steady state (i.e., a “pseudo” steady state) applies.

8.  A mixture of titanium (SG = 4.5) and silica (SG = 2.65) particles, with diameters ranging from 50 to 300 μm, is dropped into a tank in which water is flowing upward. What is the velocity of the water if all of the silica particles are carried out with the water?

9.  A small sample of ground coal is introduced at the top of a column of water 30 cm high, and the time required for the particles to settle out is measured. If it takes 26 s for the first particle to reach the bottom and 18 h for all particles to settle, what is the range of particle sizes in the sample? (T = 60°F, SGcoal = 1.4)

10.  You want to determine the viscosity of an oil which has an SG of 0.9. To do this, you drop a spherical glass bead (SG = 2.7) with a diameter of 0.5 mm into a large vertical column of the oil and measure its settling velocity. If the measured velocity is 3.5 cm/s, what is the viscosity of the oil?

11.  A solid particle with a diameter of 5 mm and an SG of 1.5 is immersed in a liquid with a viscosity of 10 P and an SG of 1. How long will it take for the particle to reach 99% of its terminal velocity after it is released?

12.  A hot air popcorn popper operates by blowing air through the popping chamber, which carries the popped corn up through a duct and out of the popper leaving the un-popped grains behind. The un-popped grains weigh 0.15 g, half of which is water, and have an equivalent spherical diameter of 4 mm. The popped corn loses half of the water to steam and has an equivalent diameter of 12 mm. What are the upper and lower limits of the air volumetric flow rate at 200°F over which the popper will operate properly, for a duct diameter of 8 cm?

13.  You have a granular solid with SG = 4, which has particle sizes of 300 μm and smaller. You want to separate out all of the particles with a diameter of 20 μm and smaller by pumping water upward through a slurry of the particles in a column with a diameter of 10 cm. What flow rate is required to ensure that all particles less than 20 μm are swept out of the top of the column? If the slurry is fed to the bottom of the column through a vertical tube, what should the diameter of this tube be to ensure that none of the particles settle out in it?

14.  You want to reproduce the experiment shown in the text which illustrates the wake behind a sphere falling in water at the point where the boundary layer undergoes transition from laminar to turbulent. If the sphere is made of steel with a density of 500 lbm/ft3, what should the diameter be?

15.  You have a sample of crushed coal containing a range of particle sizes from 1 to 1000 μm in diameter. You wish to separate the particles according to size by entrainment, in which they are dropped into a vertical column of water that is flowing upward. If the water velocity in the column is 3 cm/s, which particles will be swept out of the top of the column and which will settle at the bottom? (SG of the solid is 2.5)

16.  A gravity settling chamber consists of a horizontal rectangular duct that is 6 m long, 3.6 m wide, and 3 m high. The chamber is used to trap sulfuric acid mist droplets entrained in an air stream. The droplets settle out as the air passes through the duct and may be assumed to behave as rigid spheres. If the air stream has a flow rate of 6.5 m3/s, what is the diameter of the largest particle that will not be trapped in the duct?

acid = 1.74 g/cm3; ρair = 0.01 g/cm3; μair = 0.02 cP; μacid = 2 cP)

17.  A small sample of a coal slurry containing particles with equivalent spherical diameters from 1 to 500 μm is introduced at the top of a water column 30 cm high. The particles that fall to the bottom are continuously collected and weighed to determine the particle size distribution in the slurry. If the solid SG is 1.4 and the water viscosity is 1 cP, over what time range must the data be obtained in order to collect and weigh all of the particles in the sample?

18.  Construct a plot of CD versus NReBP for a sphere falling in a Bingham plastic fluid over the range of 1 < NRe < 100 and 10 < NBi < 1000 using Equation 12.58. Compare the curves for this relation based upon Equations 12.54, 12.55, and 12.57.

19.  The viscosity of applesauce at 80°F was measured to be 24.2 P at a shear rate of 10 1/s and 1.45 P at 500 1/s. The density of the applesauce is 1.5 g/cm3. Determine the terminal velocity of a solid sphere 1 cm in diameter with a density of 3.0 g/cm3 falling in the applesauce, if the fluid is described by (a) the power law model and (b) the Bingham plastic model.

20.  Determine the size of the smallest sphere of SG = 3 that will settle in the applesauce with properties given in Problem 19, assuming it is best described by the Bingham plastic model. Find the terminal velocity of the sphere that has a diameter twice this size.

NOTATION

A

Cross-sectional area of particle normal to flow direction, [L2]

CD

Particle drag coefficient, [—]

d

Particle diameter, [L]

FD

Drag force on particle, [F = M L/t2]

g

Acceleration due to gravity, [L/t2]

Kwo

Low Reynolds number wall correction factor, [—]

Kw

High Reynolds number wall correction factor, [—]

m

Power law consistency coefficient, [M/(L t2–n)]

n

Power law flow index, [—]

NAr

Archimedes number, Equation (12.13), [—]

NReBP

Bingham plastic Reynolds number, Equation (12.54), [—]

NBi

Bingham number = NRe/NHe = (dτoV), [—]

NRe

Reynolds number, [—]

NRe,pl

Power law Reynolds number, [—]

Nλ

Dimensionless time constant, λV/d, [—]

V

Relative velocity between fluid and particle, [L/t]

X

Correction factor to Stokes law to account for non-Newtonian properties, [—]

YG

“Gravity yield” parameter for Bingham plastics, defined by Equation 12.50, [—]

GREEK

κ

μio, [—]

ηo

Low shear limiting viscosity, [M/(L t)]

η

High shear limiting viscosity, [M/(L t)]

Δρ

s – ρ), [M/L3]

ρ

Density, [M/L3]

λ

Carreau fluid time constant parameter, [t]

μ

Viscosity (constant), [M/(L t)]

μ

Bingham plastic limiting viscosity, [M/(L t)]

τo

Bingham plastic yield stress, [F/L2 = M/(L t2)]

SUBSCRIPTS

i

Distributed (“inside”) liquid phase

o

Continuous (“outside”) liquid phase

s

Solid

t

Terminal velocity condition

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