3.1. FLOW PAST IMMERSED OBJECTS AND PACKED AND FLUIDIZED BEDS

3.1A. Definition of Drag Coefficient for Flow Past Immersed Objects

1. Introduction and types of drag

In Chapter 2 we were concerned primarily with the momentum transfer and frictional losses for flow of fluids inside conduits or pipes. In this section we consider in some detail the flow of fluids around solid, immersed objects.

The flow of fluids outside immersed bodies appears in many chemical engineering applications and other processing applications. These occur, for example, in flow past spheres in settling, flow through packed beds in drying and filtration, flow past tubes in heat exchangers, and so on. It is useful to be able to predict the frictional losses and/or the force on the submerged objects in these various applications.

In the examples of fluid friction inside conduits that we considered in Chapter 2, the transfer of momentum perpendicular to the surface resulted in a tangential shear stress or drag on the smooth surface parallel to the direction of flow. This force exerted by the fluid on the solid in the direction of flow is called skin or wall drag. For any surface in contact with a flowing fluid, skin friction will exist. In addition to skin friction, if the fluid is not flowing parallel to the surface but must change direction to pass around a solid body such as a sphere, significant additional frictional losses will occur; this is called form drag.

In Fig. 3.1-1a the flow of fluid is parallel to the smooth surface of the flat, solid plate, and the force F in newtons on an element of area dA m2 of the plate is the wall shear stress τw times the area dA, or τw dA. The total force is the sum of the integrals of these quantities evaluated over the entire area of the plate. Here the transfer of momentum to the surface results in a tangential stress or skin drag on the surface.

Figure 3.1-1. Flow past immersed objects: (a) flat plate, (b) sphere, (c) streamlined object.


In many cases, however, the immersed body is a blunt-shaped solid which presents various angles to the direction of the fluid flow. As shown in Fig. 3.1-1b, the free-stream velocity is ν0 and is uniform on approaching the blunt-shaped body suspended in a very large duct. Lines called streamlines represent the path of fluid elements around the suspended body. The thin boundary layer adjacent to the solid surface is shown as a dashed line; at the edge of this layer the velocity is essentially the same as the bulk fluid velocity adjacent to it. At the front center of the body, called the stagnation point, the fluid velocity will be zero; boundary-layer growth begins at this point and continues over the surface until the layer separates. The tangential stress on the body because of the velocity gradient in the boundary layer is the skin friction. Outside the boundary layer the fluid changes direction to pass around the solid and also accelerates near the front and then decelerates. Because of these effects, an additional force is exerted by the fluid on the body. This phenomenon, called form drag, is in addition to the skin drag in the boundary layer.

In Fig. 3.1-1b, as shown, separation of the boundary layer occurs and a wake, covering the entire rear of the object, occurs where large eddies are present and contribute to the form drag. The point of separation depends on the shape of the particle, Reynolds number, and so on, and is discussed in detail elsewhere (S3).

Form drag for bluff bodies can be minimized by streamlining the body (Fig. 3.1-1c), which forces the separation point toward the rear of the body, greatly reducing the size of the wake. Additional discussion of turbulence and boundary layers is given in Section 3.10.

2. Drag coefficient

From the preceding discussion it is evident that the geometry of the immersed solid is a major factor in determining the amount of total drag force exerted on the body. Correlations of the geometry and flow characteristics for solid objects suspended or held in a free stream (immersed objects) are similar in concept and form to the friction factor-Reynolds number correlation given for flow inside conduits. In flow through conduits, the friction factor was defined as the ratio of the drag force per unit area (shear stress) to the product of density times velocity head, as given in Eq. (2.10-4).

In a similar manner, for flow past immersed objects the drag coefficient CD is defined as the ratio of the total drag force per unit area to /2:

Equation 3.1-1


where FD is the total drag force in N, Ap is an area in m2, CD is dimensionless, ν0 is free-stream velocity in m/s, and ρ is density of fluid in kg/m3. In English units, FD is in 1bf, ν0 is in ft/s, ρ is in 1bm/ft3, and Ap is in ft2. The area Ap used is the area obtained by projecting the body on a plane perpendicular to the line of flow. For a sphere, Ap = , where Dp is sphere diameter; for a cylinder whose axis is perpendicular to the flow direction, Ap = LDP, where L = cylinder length. Solving Eq. (3.1-1) for the total drag force,

Equation 3.1-2


The Reynolds number for a given solid immersed in a flowing liquid is

Equation 3.1-3


where G0 = ν0ρ.

3.1B. Flow Past Sphere, Long Cylinder, and Disk

For each particular shape of object and orientation of the object with respect to the direction of flow, a different relation of CD versus NRe exists. Correlations of drag coefficient versus Reynolds number are shown in Fig. 3.1-2 for spheres, long cylinders, and disks. The face of the disk and the axis of the cylinder are perpendicular to the direction of flow. These curves have been determined experimentally. However, in the laminar region for low Reynolds numbers, less than about 1.0, the experimental drag force for a sphere is the same as the theoretical Stokes' law equation as follows:

Equation 3.1-4


Figure 3.1-2. Drag coefficients for flow past immersed spheres, long cylinders, and disks. (Reprinted with permission from C. E. Lapple and C. B. Shepherd, Ind. Eng. Chem., 32, 606 (1940). Copyright by the American Chemical Society.)


Combining Eqs. (3.1-2) and (3.1-4) and solving for CD, the drag coefficient predicted by Stokes' law is

Equation 3.1-5


The variation of CD with NRe (Fig. 3.1-2) is quite complicated because of the interaction of the factors that control skin drag and form drag. For a sphere, as the Reynolds number is increased beyond the Stokes' law range, separation occurs and a wake is formed. Further increases in NRe cause shifts in the separation point. At about NRe = 3 × 105 the sudden drop in CD is the result of the boundary layer becoming completely turbulent and the point of separation moving downstream. In the region of NRe about 1 × 103 to 2 × 105, the drag coefficient is approximately constant for each shape and CD = 0.44 for a sphere. Above NRe about 5 × 105, the drag coefficients are again approximately constant, with CD being 0.13 for a sphere, 0.33 for a cylinder, and 1.12 for a disk. Additional discussions and theory concerning flow past spheres are given in Section 3.9E.

For derivations of theory and detailed discussions of the drag force for flow parallel to a flat plate, Section 3.10 on boundary-layer flow and turbulence should be consulted. The flow of fluids normal to banks of cylinders or tubes occurs in heat exchangers and other processing applications. The banks of tubes can be arranged in a number of different geometries. Because of the many possible geometric tube configurations and spacings, it is not possible to have one correlation for the data on pressure drop and friction factors. Details of the many correlations available are given elsewhere (P1).

EXAMPLE 3.1-1. Force on a Submerged Sphere

Air at 37.8°C and 101.3 kPa absolute pressure flows at a velocity of 23 m/s past a sphere having a diameter of 42 mm. What are the drag coefficient CD and the force on the sphere?

Solution: From Appendix A.3, for air at 37.8°C, ρ = 1.137 kg/m3 and μ = 1.90 × 10-5 Pa · s. Also, Dp = 0.042 m and ν0 = 23.0 m/s. Using Eq. (3.1-3),


From Fig. 3.1-2 for a sphere, CD = 0.47. Substituting into Eq. (3.1-2), where Ap = for a sphere,



EXAMPLE 3.1-2. Force on a Cylinder in a Tunnel

Water at 24°C is flowing past a long cylinder at a velocity of 1.0 m/s in a large tunnel. The axis of the cylinder is perpendicular to the direction of flow. The diameter of the cylinder is 0.090 m. What is the force per meter length on the cylinder?

Solution: From Appendix A.2, for water at 24°C, ρ = 997.2 kg/m3 and μ = 0.9142 × 103 Pa · s. Also, Dp = 0.090 m, L = 1.0 m, and v0 = 1.0 m/s. Using Eq. (3.1-3),


From Fig. 3.1-2 for a long cylinder, CD = 1.4. Substituting into Eq. (3.1-2), where Ap = LDp = 1.0(0.090) = 0.090 m2,



3.1C. Flow in Packed Beds

1. Introduction

A system of considerable importance in chemical and other process engineering fields is the packed bed or packed column, which is used for a fixed-bed catalytic reactor, adsorption of a solute, absorption, filter bed, and so on. The packing material in the bed may be spheres, irregular particles, cylinders, or various kinds of commercial packings. In the discussion to follow it is assumed that the packing is everywhere uniform and that little or no channeling occurs. The ratio of diameter of the tower to packing diameter should be a minimum of 8:1 to 10:1 for wall effects to be small. In the theoretical approach used, the packed column is regarded as a bundle of crooked tubes of varying cross-sectional area. The theory developed in Chapter 2 for single straight tubes is used to develop the results for the bundle of crooked tubes.

2. Laminar flow in packed beds

Certain geometric relations for particles in packed beds are used in the derivations for flow. The void fraction ε in a packed bed is defined as

Equation 3.1-6


The specific surface of a particle av in m−1 is defined as

Equation 3.1-7


where Sp is the surface area of a particle in m2 and vp the volume of a particle in m3. For a spherical particle,

Equation 3.1-8


where Dp is diameter in m. For a packed bed of nonspherical particles, the effective particle diameter Dp is defined as

Equation 3.1-9


Since (1 - ε) is the volume fraction of particles in the bed,

Equation 3.1-10


where a is the ratio of total surface area in the bed to total volume of bed (void volume plus particle volume) in m−1.

EXAMPLE 3.1-3. Surface Area in Packed Bed of Cylinders

A packed bed is composed of cylinders having a diameter D = 0.02 m and a length h = D. The bulk density of the overall packed bed is 962 kg/m3 and the density of the solid cylinders is 1600 kg/m3.

  1. Calculate the void fraction ε.

  2. Calculate the effective diameter Dp of the particles.

  3. Calculate the value of a in Eq. (3.1-10).

Solution: For part (a), taking 1.00 m3 of packed bed as a basis, the total mass of the bed is (962 kg/m3)(1.00 m3) = 962 kg. This mass of 962 kg is also the mass of the solid cylinders. Hence, volume of cylinders = 962 kg/(1600 kg/m3) = 0.601 m3. Using Eq. (3.1-6),


For the effective particle diameter Dp in part (b), for a cylinder where h = D, the surface area of a particle is


The volume vp of a particle is


Substituting into Eq. (3.1-7),


Finally, substituting into Eq. (3.1-9),


Hence, the effective diameter to use is Dp = D = 0.02 m. For part (c), using Eq. (3.1-10),



The average interstitial velocity in the bed is v m/s and is related to the superficial velocity v' based on the cross section of the empty container by

Equation 3.1-11


The hydraulic radius rH for flow defined in Eq. (2.10-21) is modified as follows (B2):

Equation 3.1-12


Combining Eqs. (3.1-10) and (3.1-12),

Equation 3.1-13


Since the equivalent diameter D for a channel is D = 4rH, the Reynolds number for a packed bed is as follows, using Eq. (3.1-13) and v' = εv:

Equation 3.1-14


For packed beds Ergun (E1) defined the Reynolds number as above but without the 4/6 term:

Equation 3.1-15


where G' = ν'ρ.

For laminar flow, the Hagen-Poiseuille equation (2.10-2) can be combined with Eq. (3.1-13) for rH and Eq. (3.1-11) to give

Equation 3.1-16


The true ΔL is larger because of the tortuous path, and use of the hydraulic radius predicts too large a v'. Experimental data show that the constant should be 150, which gives the Blake-Kozeny equation for laminar flow, void fractions less than 0.5, effective particle diameter Dp, and NRe,p < 10:

Equation 3.1-17


3. Turbulent flow in packed beds

For turbulent flow we use the same procedure by starting with Eq. (2.10-5) and substituting Eqs. (3.1-11) and (3.1-13) into this equation to obtain

Equation 3.1-18


For highly turbulent flow the friction factor should approach a constant value. Also, it is assumed that all packed beds should have the same relative roughness. Experimental data indicate that 3f = 1.75. Hence, the final equation for turbulent flow for NRe,p > 1000, which is called the Burke-Plummer equation, becomes

Equation 3.1-19


Adding Eq. (3.1-17) for laminar flow and Eq. (3.1-19) for turbulent flow, Ergun (E1) proposed the following general equation for low, intermediate, and high Reynolds numbers, which has been tested experimentally:

Equation 3.1-20


Rewriting Eq. (3.1-20) in terms of dimensionless groups,

Equation 3.1-21


See also Eq. (3.1-33) for another form of Eq. (3.1-21). The Ergun equation (3.1-21) can be used for gases by taking the density ρ of the gas as the arithmetic average of the inlet and outlet pressures. The velocity v' changes throughout the bed for a compressible fluid, but G' is a constant. At high values of NRe,p, Eqs. (3.1-20) and (3.1-21) reduce to Eq. (3.1-19), and to Eq. (3.1-17) for low values. For large pressure drops with gases, Eq. (3.1-20) can be written in differential form (P1).

EXAMPLE 3.1-4. Pressure Drop and Flow of Gases in Packed Bed

Air at 311 K is flowing through a packed bed of spheres having a diameter of 12.7 mm. The void fraction ε of the bed is 0.38 and the bed has a diameter of 0.61 m and a height of 2.44 m. The air enters the bed at 1.10 atm abs at the rate of 0.358 kg/s. Calculate the pressure drop of the air in the packed bed. The average molecular weight of air is 28.97.

Solution: From Appendix A.3, for air at 311 K, μ = 1.90 × 10−5 Pa · s. The cross-sectional area of the bed is A = (π/4)D2 = (π/4)(0.61)2 = 0.2922 m2. Hence, G' = 0.358/0.2922 = 1.225 kg/m2 · s (based on empty cross section of container or bed). Dp = 0.0127 m, ΔL = 2.44 m, and inlet pressure p1 = 1.1(1.01325 × 105) = 1.115 × 105 Pa.

From Eq. (3.1-15),


To use Eq. (3.1-21) for gases, the density ρ to use is the average at the inlet p1 and outlet p2 pressures, or at (pl + p2)/2. This is trial and error since p2 is unknown. Assuming that Δp = 0.05 × 105 Pa, p2 = 1.115 × 105 - 0.05 × 105 = 1.065 × 105 Pa. The average pressure is pav = (1.115 × 105 + 1.065 × 105)/2 = 1.090 × 105 Pa. The average density to use is

Equation 3.1-22


Substituting into Eq. (3.1-21) and solving for Δp,


Solving, Δp = 0.0497 × 105 Pa. This is close enough to the assumed value, so a second trial is not needed.


4. Shape factors

Many particles in packed beds are often irregular in shape. The equivalent diameter of a particle is defined as the diameter of a sphere having the same volume as this particle. The sphericity shape factor φS of a particle is the ratio of the surface area of this sphere having the same volume as the particle to the actual surface area of the particle. For a sphere, the surface area and the volume Hence, for any particle, , where Sp is the actual surface area of the particle and Dp is the diameter (equivalent diameter) of the sphere having the same volume as the particle. Then

Equation 3.1-23


From Eq. (3.1-7),

Equation 3.1-24


Then Eq. (3.1-10) becomes

Equation 3.1-25


For a sphere, φS = 1.0. For a cylinder where diameter = length, φS is calculated to be 0.874, and for a cube, φS is calculated to be 0.806. For granular materials it is difficult to measure the actual volume and surface area to obtain the equivalent diameter. Hence, Dp is usually taken to be the nominal size from a screen analysis or visual length measurements. The surface area is determined by adsorption measurements or measurement of the pressure drop in a bed of particles. Then Eq. (3.1-23) is used to calculate φS (Table 3.1-1). Typical values for many crushed materials are between 0.6 and 0.7. For convenience, for the cylinder and the cube, the nominal diameter is sometimes used (instead of the equivalent diameter), which then gives a shape factor of 1.0.

Table 3.1-1. Shape Factors (Sphericity) of Some Materials
MaterialShape Factor, φSReference
Spheres1.0 
Cubes0.81 
Cylinders, Dp = h (length)0.87 
Berl saddles0.3(B4)
Raschig rings0.3(C2)
Coal dust, pulverized0.73(C2)
Sand, average0.75(C2)
Crushed glass0.65(C2)

5. Mixtures of particles

For mixtures of particles of various sizes we can define a mean specific surface avm as

Equation 3.1-26


where xi is volume fraction. Combining Eqs. (3.1-24) and (3.1-26),

Equation 3.1-27


where Dpm is the effective mean diameter for the mixture.

EXAMPLE 3.1-5. Mean Diameter for a Particle Mixture

A mixture contains three sizes of particles: 25% by volume of 25 mm size, 40% of 50 mm, and 35% of 75 mm. The sphericity is 0.68. Calculate the effective mean diameter.

Solution: The following data are given: x1 = 0.25, Dp1 = 25 mm; x2 = 0.40, Dp2 = 50; x3 = 0.35, Dp3 = 75; φS = 0.68. Substituting into Eq. (3.1-27),



6. Darcy's empirical law for laminar flow

Equation (3.1-17) for laminar flow in packed beds shows that the flow rate is proportional to Δp and inversely proportional to the viscosity μ and length ΔL. This is the basis for Darcy's law, as follows, for purely viscous flow in consolidated porous media:

Equation 3.1-28


where v' is superficial velocity based on the empty cross section in cm/s, q' is flow rate in cm3/s, A is empty cross section in cm2, μ is viscosity in cp, Δp is pressure drop in atm, ΔL is length in cm, and k is permeability in (cm3 flow/s) · (cp) · (cm length)/(cm2 area) · (atm pressure drop). The units used for k of cm2 · cp/s · atm are often given in darcy or in millidarcy (1/1000 darcy). Hence, if a porous medium has a permeability of 1 darcy, a fluid of 1 cp viscosity will flow at 1 cm3/s per 1 cm2 cross section with a Δp of 1 atm per cm length. This equation is often used in measuring permeabilities of underground oil reservoirs.

3.1D. Flow in Fluidized Beds

1. Types of fluidization in beds

In a packed bed of small particles, when a fluid enters at sufficient velocity from the bottom and passes up through the particles, the particles are pushed upward and the bed expands and becomes fluidized. Two general types of fluidization, particulate fluidization and bubbling fluidization, can occur.

In particulate fluidization, as the fluid velocity is increased the bed continues to expand and remains homogeneous for a time. The particles move farther apart and their motion becomes more rapid. The average bed density at a given velocity is the same in all regions of the bed. An example is catalytic cracking catalysts fluidized by gases. This type of fluidization is very desirable in promoting intimate contact between the gas and solids. Liquids often give particulate fluidization.

In bubbling fluidization, the gas passes through the bed as voids or bubbles which contain few particles, and only a small percentage of the gas passes in the spaces between individual particles. The expansion of the bed is small as gas velocity is increased. Sand and glass beads provide examples of this behavior. Since most of the gas is in bubbles, little contact occurs between the individual particles and the bubbles.

Particles which behave as above have been classified by Geldart (G2) into classes. Those called Type A, which exhibit particulate fluidization in gases, fall into the following approximate ranges: For Δρ = (ρpρ) = 2000 kg/m3, Dp = 20–125 μm; Δρ = 1000, Dp = 25−250; Δρ = 500, Dp = 40–450; Δρ = 200, Dp = 100–1000. For those called Type B, which exhibit bubbling fluidization, approximate ranges are as follows: Δρ = 2000, Dp = 125−700 μm; Δρ = 1000, Dp = 250−1000; Δρ = 500, Dp = 450−1500; Δρ = 200, Dp = 1000−2000.

Another type of behavior, called slugging, can occur in bubbling since the bubbles tend to coalesce and grow as they rise in the bed. If the column is small in diameter with a deep bed, bubbles can become large and fill the entire cross section and travel up the tower separated by slugs of solids.

2. Minimum velocity and porosity for particulate fluidization

When a fluid flows upward through a packed bed of particles at low velocities, the particles remain stationary. As the fluid velocity is increased, the pressure drop increases according to the Ergun equation (3.1-20). Upon further increases in velocity, conditions finally occur where the force of the pressure drop times the cross-sectional area equals the gravitational force on the mass of particles. Then the particles begin to move, and this is the onset of fluidization or minimum fluidization. The fluid velocity at which fluidization begins is the minimum fluidization velocity in m/s based on the empty cross section of the tower (superficial velocity).

The porosity of the bed when true fluidization occurs is the minimum porosity for fluidization, εmf. Some typical values of εmf for various materials are given in Table 3.1-2. The bed expands to this voidage or porosity before particle motion appears. This minimum voidage can be determined experimentally by subjecting the bed to a rising gas stream and measuring the height of the bed Lmf in m. Generally, it appears best to use gas as the fluid rather than a liquid, since liquids give somewhat higher values of εmf.

Table 3.1-2. Void Fraction, εmf, at Minimum Fluidization Conditions (L2)
Type of ParticlesParticle Size, DP (mm)
0.060.100.200.40
 Void fraction, εmf
Sharp sand (φS = 0.67)0.600.580.530.49
Round sand (φS = 0.86)0.530.480.43(0.42)
Anthracite coal (φS = 0.63)0.610.600.560.52
Absorption carbon0.710.69  
Fischer Tropsch catalyst (φS = 0.58) 0.580.56(0.54)

As stated earlier, the pressure drop increases as the gas velocity is increased until the onset of minimum fluidization. Then, as the velocity is further increased, the pressure drop decreases very slightly, and then it remains practically unchanged as the bed continues to expand or increase in porosity with increases in velocity. The bed resembles a boiling liquid. As the bed expands with increase in velocity, it continues to retain its top horizontal surface. Eventually, as the velocity is increased much further, entrainment of particles from the actual fluidized bed becomes appreciable.

The relation between bed height L and porosity ε is as follows for a bed having a uniform cross-sectional area A. Since the volume LA(1 − ε) is equal to the total volume of solids if they existed as one piece,

Equation 3.1-29


Equation 3.1-30


where L1 is height of bed with porosity ε1 and L2 is height with porosity ε2.

3. Pressure drop and minimum fluidizing velocity

As a first approximation, the pressure drop at the start of fluidization can be determined as follows. The force obtained from the pressure drop times the cross-sectional area must equal the gravitational force exerted by the mass of the particles minus the buoyant force of the displaced fluid:

Equation 3.1-31


Hence,

Equation 3.1-32


Often we have irregular-shaped particles in the bed, and it is more convenient to use the particle size and shape factor in the equations. First we substitute for the effective mean diameter Dp the term φSDP, where DP now represents the particle size of a sphere having the same volume as the particle and φS the shape factor. Often, the value of DP is approximated by using the nominal size from a sieve analysis. Then Eq. (3.1-20) for pressure drop in a packed bed becomes

Equation 3.1-33


where ΔL = L, bed length in m.

Equation (3.1-33) can now be used by a small extrapolation for packed beds to calculate the minimum fluid velocity at which fluidization begins by substituting for v', εmf for ε, and Lmf for L, and combining the result with Eq. (3.1-32) to give

Equation 3.1-34


Defining a Reynolds number as

Equation 3.1-35


Eq. (3.1-34) becomes

Equation 3.1-36


When NRe,mf < 20 (small particles), the first term of Eq. (3.1-36) can be dropped, and when NRe,mf > 1000 (large particles), the second term drops out.

If the terms εmf and/or φS are not known, Wen and Yu (W4) found for a variety of systems that

Equation 3.1-37


Substituting into Eq. (3.1-36), the following simplified equation is obtained:

Equation 3.1-38


This equation holds for a Reynolds-number range of 0.001 to 4000, with an average deviation of ± 25%. Alternative equations are available in the literature (K1, W4).

EXAMPLE 3.1-6. Minimum Velocity for Fluidization

Solid particles having a size of 0.12 mm, a shape factor φS of 0.88, and a density of 1000 kg/m3 are to be fluidized using air at 2.0 atm abs and 25°C. The voidage at minimum fluidizing conditions is 0.42.

  1. If the cross section of the empty bed is 0.30 m2 and the bed contains 300 kg of solid, calculate the minimum height of the fluidized bed.

  2. Calculate the pressure drop at minimum fluidizing conditions.

  3. Calculate the minimum velocity for fluidization.

  4. Use Eq. (3.1-38) to calculate assuming that data for φS and εmf are not available.

Solution: For part (a), the volume of solids = 300 kg/(1000 kg/m3) = 0.300 m3. The height the solids would occupy in the bed if ε1 = 0 is L1 = 0.300 m3/(0.30 m2 cross section) = 1.00 m. Using Eq. (3.1-30) and calling Lmf = L2 and εmf = ε2,


Solving, Lmf = 1.724 m.

The physical properties of air at 2.0 atm and 25°C (Appendix A.3) are μ = 1.845 × 10-5 Pa · s, ρ = 1.187 × 2 = 2.374 kg/m3, and p = 2.0265 × 105 Pa. For the particle, DP = 0.00012 m, ρp = 1000 kg/m3, φS = 0.88, and εmf = 0.42.

For part (b), using Eq. (3.1-32) to calculate Δp,


To calculate for part (c), Eq. (3.1-36) is used:


Solving,


Using the simplified Eq. (3.1-38) for part (d),


Solving, = 0.004618 m/s.


4. Expansion of fluidized beds

For the case of small particles and where NRe,f = Dpν'ρ/μ < 20, we can estimate the variation of porosity or bed height L as follows. We assume that Eq. (3.1-36) applies over the whole range of fluid velocities, with the first term being neglected. Then, solving for ν',

Equation 3.1-39


We find that all terms except ε are constant for the particular system, and ε depends upon ν'. This equation can be used with liquids to estimate ε with ε < 0.80. However, because of clumping and other factors, errors can occur when used for gases.

The flow rate in a fluidized bed is limited on the one hand by the minimum and on the other by entrainment of solids from the bed proper. This maximum allowable velocity is approximated as the terminal settling velocity of the particles. (See Section 13.3 for methods for calculating this settling velocity.) Approximate equations for calculating the operating range are as follows (P2). For fine solids and NRe,f < 0.4,

Equation 3.1-40


For large solids and NRe,f > 1000,

Equation 3.1-41


EXAMPLE 3.1-7. Expansion of Fluidized Bed

Using the data from Example 3.1-6, estimate the maximum allowable velocity . Using an operating velocity of 3.0 times the minimum, estimate the voidage of the bed.

Solution: From Example 3.1-6, NRe,mf = 0.07764, = 0.005029 m/s, and εmf = 0.42. Using Eq. (3.1-40), the maximum allowable velocity is


Using an operating velocity ν' of 2.0 times the minimum,


To determine the voidage at this new velocity, we substitute into Eq. (3.1-39) using the known values at minimum fluidizing conditions to determine K1:


Solving K1 = 0.03938. Then using the operating velocity in Eq. (3.1-39),


Solving, the voidage of the bed ε = 0.503 at the operating velocity.


5. Minimum bubbling velocity

The fluidization velocity at which bubbles are first observed is called the minimum bubbling velocity, . For Group B particles, which exhibit bubbling fluidization, is reasonably close to . For Group A particles, is substantially greater than . The following equation (G3) can be used to calculate :

Equation 3.1-42


where is minimum bubbling velocity in m/s, μ is viscosity in Pa · s, and ρ is gas density in kg/m3. If the calculated is greater than the calculated by Eq. (3.1-42), then the should be used as the minimum bubbling velocity (G4).

EXAMPLE 3.1-8. Minimum Bubbling Velocity

Using data from Example 3.1-6, calculate the .

Solution: Substituting into Eq. (3.1-42),


This is substantially greater than the of 0.005029 m/s.


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