14	Flow in Porous Media

“If your experiment needs statistics, you ought to have done a better experiment.”

—Ernest Rutherford, 1871–1937, Physicist

I.    DESCRIPTION OF POROUS MEDIA

By a “porous medium” is meant a solid, or a collection of solid particles, with sufficient open space in or around the particles to enable a fluid to pass through or around them. There are various conceptual ways of describing a porous medium.

One concept is a continuous solid body with pores in it, such as a brick or a block of sandstone. Such a medium is referred to as consolidated, and the pores may be unconnected (“closed cell” or impermeable) or connected (“open cell” or permeable). Another concept is a collection (or “pile”) of solid particles in a packed bed, where the fluid can pass through the voids in between the particles. This is referred to as unconsolidated media. A schematic representation is shown in Figure 14.1. Either of these concepts may be valid, depending upon the specific medium under consideration, and both have been used as the basis for developing the equations that describe fluid flow behavior within such a medium. In practice, porous media may range from a “tight” oil-bearing rock formation to a packed column containing relatively large packing elements and large void spaces.

The “pile of solid particles” concept is useful for either consolidated or unconsolidated media as a basis for analyzing the flow phenomena, since many consolidated media are actually made up of individual particles that are just stuck together (e.g., sandstone). One of the key properties of a porous medium is its porosity, ε, or void fraction, which is defined by

$\text{ε}=\frac{\text{Total}\text{Volume}-\text{Volume}\text{of}\text{solids}}{\text{Total}\text{volume}}$

(14.1)

For an isotropic and homogeneous medium, the porosity can be written in terms of the corresponding areas as

$\text{ε}=1-\frac{A_{solid}}{A}=\frac{A_{void}}{A}$

where A_solid is the area of the solid phase in a total cross section of area A. For instance, certain rocks and sandstone may have very low values of porosity (~0.15–0.2), whereas fibrous beds, glass wool, metallic or plastic foams, structured packing, etc., may have high values (~≥0.9) of porosity.

We also distinguish between the velocity of approach, or the “superficial” velocity of the fluid,

$V_s=\frac{Q}{A}$

(14.2)

and the “interstitial” velocity, which is the actual velocity within the pores or voids:

$V_i=\frac{Q}{\text{ε}A}=\frac{V_s}{\text{ε}}$

(14.3)

FIGURE 14.1 Schematic representation of porous media (a) consolidated and (b) unconsolidated.

A.    HYDRAULIC DIAMETER

Because the fluid in a porous medium follows a tortuous path through channels of varying size and shape, one method of describing the flow behavior in the pores is to consider the flow path as a “noncircular conduit.” This requires an appropriate definition of the hydraulic diameter, that is,

$\begin{array}{ll}{D}_h\hfill & =4\frac{A_i}{W_p}=4\frac{A{i}_L}{W_{p}L}=4\frac{\text{Flow}\text{volume}}{\text{Internal}\text{wetted}\text{surfaearea}}\hfill \\ \hfill & =4\frac{\text{ε}\times \text{Bed}\text{volume}}{\left(\text{Number}\text{of}\text{particles}\right)\left(\text{Surface}\text{area/Particle}\right)}\hfill \end{array}$

(14.4)

The medium, with overall dimensions of height L and cross-sectional area A, is assumed to be made up of a collection of individual particles that may be either consolidated or unconsolidated. The number of particles in the medium can be expressed as

$\begin{array}{ll}\text{Number}\text{of}\text{particles}\hfill & =\frac{\left(\text{Bed}\text{volume}\right)\left(\text{Fraction}\text{of}\text{solids}\text{ln}\text{bed}\right)}{\text{Volume/Particle}}\hfill \\ \hfill & =\frac{\left(\text{Bed}\text{volume}\right)\left(1-\text{ε}\right)}{\text{Volume/Particle}}\hfill \end{array}$

(14.5)

Substitution of this into Equation (14.4) leads to

$D_h=4\left(\frac{\text{ε}}{1-\text{ε}}\right)\frac{1}{a_s}$

(14.6)

where a_s is the specific surface area of each particle, that is, the particle surface area per unit volume of the particle. If the particles are spherical with diameter d, then a_s = 6/d. Thus, for a medium composed of uniform spherical particles, the hydraulic diameter is

$D_h=\frac{2d\text{ε}}{3\left(1-\text{ε}\right)}$

(14.7)

For non-spherical particles, the parameter d may be replaced by the equivalent diameter defined as

$d=\text{ψ}{d}_s=\frac{6}{a_s}$

(14.8)

where ψ is the sphericity factor, which is defined by

$\text{ψ}=\frac{\text{Surface}\text{area}\text{of}\text{a}\text{sphere}\text{with}\text{same}\text{volume}\text{as}\text{the}\text{particle}}{\text{Surface}\text{area}\text{of}\text{the}\text{particle}}$

(14.9)

and d_s is the diameter of the sphere with the same volume as the particle. This definition of ψ is identical to that given by Equation 12.25 or 12.26.

B.    POROUS MEDIUM FRICTION FACTOR

The expressions for the hydraulic diameter and the superficial velocity can be incorporated into the definition of the friction factor (i.e., Equation 7.5) to give an equivalent expression for the porous medium friction factor, as follows:

$f\equiv \frac{e_f}{\left(4L/D_h\right)\left(V_i^2/2\right)}=\frac{e_{f}d\text{ε}}{3L\left(1-\text{ε}\right)V_i^2}=\frac{e_{f}d{\text{ε}}^3}{3L\left(1-\text{ε}\right)V_s^2}$

(14.10)

Most references use Equation 14.10 without the numerical factor of 3 in the denominator as the definition of the porous medium friction factor, i.e.

$f_{PM}\equiv \frac{e_{f}d{\text{ε}}^3}{L\left(1-\text{ε}\right)V_s^2}$

(14.11)

C.    POROUS MEDIUM REYNOLDS NUMBER

In a like fashion, the hydraulic diameter and the superficial velocity can be introduced into the definition of the Reynolds number to give

$N_{Re}=\frac{D_h{V}_i\rho }{\mu }=\frac{2d\text{ε}{V}_i\rho }{3\left(1-\text{ε}\right)\mu }=\frac{2d{V}_s\rho }{3\left(1-\text{ε}\right)\mu }$

(14.12)

Here again, the usual porous medium Reynolds number is defined by Equation 14.12 without the numerical factor (2/3):

$N_{RePM}=\frac{d{V}_s\rho }{\left(1-\text{ε}\right)\mu }$

(14.13)

II.    FRICTION LOSS IN POROUS MEDIA

A.    LAMINAR FLOW

By analogy with laminar flow in a tube, the friction factor in laminar flow would be

$f=\frac{16}{N_{Re}}\text{or}{f}_{PM}=\frac{72}{N_{RePM}}$

(14.14)

However, this expression assumes that the total resistance to flow is due to the shear deformation of the fluid, as in a uniform pipe. In reality, the resistance is a result of both shear and stretching (extensional) deformation as the fluid moves through the nonuniform converging–diverging flow cross section within the pores. This “stretching resistance” is the product of the extension (stretch) rate and the extensional viscosity. The extension rate in porous media is of the same order as the shear rate, and the extensional viscosity for a Newtonian fluid is three times the shear viscosity. Thus, in practice, a value of 150–180 instead of 72 is in closer agreement with experimental observations at low Reynolds numbers, i.e.:

$f_{PM}=\frac{180}{N_{RePM}}\text{for}{N}_{RePM}<10$

(14.15)

This is known as the Blake–Kozeny equation and, as noted, applies for N_RePM < 10. Alternately, the use of the numerical value of 150 (or 180) as opposed to the theoretical value of 72 in Equation 14.14 can be partially attributed to the tortuosity of a porous medium, that is, an actual fluid particle travels a longer distance than the length of the bed L. For example, for a fluid particle to travel around a spherical particle, the tortuosity is π/2 and 72 × π/2 ≈ 115.

B.    TURBULENT FLOW

At high Reynolds numbers (i.e., high turbulence levels), the flow is dominated by inertial forces and “wall roughness,” as in pipe flow. The porous medium can be considered an “extremely rough” conduit, with ε/d ~1. Thus, the flow at a sufficiently high Reynolds numbers should be fully turbulent and the friction factor should be constant. This has been confirmed by observations, with the value of the constant equal to approximately 1.75:

$f_{PM}=1.75\text{for}{N}_{RePM}>1000$

(14.16)

This is known as the Burke–Plummer equation and, as noted, applies for N_RePM > 1000.

C.    ALL REYNOLDS NUMBERS

An expression that adequately represents the porous medium friction factor over all values of Reynolds number is

$f_{PM}=1.75+\frac{180}{N_{RePM}}$

(14.17)

This equation (with a value of 150 instead of 180) is called the Ergun equation and is simply the sum of Equations 14.15 and 14.16. Obviously, for N_RePM < 10, the first term is small relative to the second, and the Ergun equation reduces to the Blake–Kozeny equation. Likewise, for N_RePM > 1000, the first term is much larger than the second, and the equation reduces to the Burke–Plummer equation.

If the definitions of f_PM and N_RePM are inserted into the Ergun equation, the resulting expression for the frictional energy loss (dissipation) per unit mass of fluid in the medium is

$e_f=1.75\frac{V_s^2}{d}\frac{\left(1-\text{ε}\right)}{{\text{ε}}^{\text{3}}}L+180\frac{V_s\mu {\left(1-\text{ε}\right)}^{2}L}{d^2{\text{ε}}^3\rho }$

(14.18)

III.    PERMEABILITY

The “permeability” of a porous medium (K) is defined as the proportionality constant that relates the flow rate through the medium to the pressure drop, cross-sectional area, fluid viscosity, and net flow length through the medium, that is,

$Q=K\frac{\left(-\Delta P\right)A}{\mu L}$

(14.19)

This equation defines permeability (K) and is known as Darcy’s law. The most common unit for permeability is the “darcy,” which is defined as the flow rate in cm³/s that results when a pressure drop of one atmosphere is applied to a porous medium that is 1 cm² in cross-sectional area and one cm long, for a fluid with a viscosity of one centipoise. It should be evident that the dimensions of the darcy are L² and the conversion factors are (approximately) 10⁻⁸ cm²/darcy = 10⁻¹¹ ft²/darcy. The flow properties of tight crude oil–bearing rock formations are often described in permeability units of millidarcies. Typical values of permeability range from ~10⁻¹⁴ m² for silica powder to 10^-11 m² for fiberglass.

If the Blake–Kozeny equation for laminar flow is used to describe the friction loss, which is then equated to ΔP/ρ from the Bernoulli equation, the resulting expression for the flow rate is

$Q=\frac{-\Delta PA}{\mu L}\left(\frac{d^2{\text{ε}}^3}{180{\left(1-\text{ε}\right)}^2}\right)$

(14.20)

By comparison of Equations 14.19 and 14.20, it is evident that permeability is identical to the term in brackets in Equation 14.20. This shows how the permeability is related to the equivalent particle size and porosity of the medium. Since Equation 14.20 applies only to laminar flow, it is evident that permeability has no meaning under turbulent flow conditions. It should be noted that although permeability increases with increasing porosity, one cannot be predicted from the other due to the presence of both “’open pores” and “closed pores” (Dullien, 1992).

IV.    MULTIDIMENSIONAL FLOW

Flow in a porous medium in two or three dimensions is important in situations such as the production of crude oil from reservoir formations, for example. Thus, it is of interest to consider this situation briefly and to point out some characteristics of the governing equations.

Consider the flow of an incompressible fluid through a 2D porous medium, as illustrated in Figure 14.2. Assuming that the kinetic energy change is negligible and that the flow is laminar as characterized by Darcy’s law, the Bernoulli equation becomes

FIGURE 14.2 Schematic of two-dimensional flow in a porous medium.

$-\left(\frac{\Delta P}{\rho }+g\Delta z\right)=e_f=\frac{\mu V_{s}L}{K\rho }$

(14.21)

or

$\Delta \left(\frac{\Phi }{\rho }\right)=-\frac{\mu V_{s}L}{K\rho }$

(14.22)

where the density cancels out if the fluid is incompressible. Equation 14.22 can be applied in both the x and y directions, by taking L = Δx for the x direction and L = Δy for the y direction:

$\frac{\Delta \Phi }{\Delta x}=-\frac{\mu V_x}{K}=\frac{\partial \Phi }{\partial x}$

(14.23)

and

$\frac{\Delta \Phi }{\Delta y}=-\frac{\mu V_y}{K}=\frac{\partial \Phi }{\partial y}$

(14.24)

If Equation 14.23 is differentiated with respect to x and Equation 14.24 is differentiated with respect to y and the results are added, assuming μ and K to be constant, we get

$\frac{{\partial }^2\Phi }{\partial x^2}+\frac{{\partial }^2\Phi }{\partial y^2}=-\frac{\mu }{K}\left(\frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}\right)=0$

(14.25)

For an incompressible fluid, the term in brackets is zero as a result of the conservation of mass (e.g., the microscopic continuity equation). In this case, Equation 14.25 can be generalized to three dimensions as

${▽}^2\Phi =0$

(14.26)

This is called the Laplace equation. The solution of this equation, along with appropriate boundary conditions, determines the potential (e.g., pressure) distribution within the medium. The derivatives of this potential then determine the velocity distribution in the medium (e.g., Equations 14.23 and 14.24). The Laplace equation also governs the 3D (potential) flow of an inviscid fluid. Note that the Laplace equation follows from Equation (14.25) for either an incompressible viscous fluid, by virtue of the continuity equation, or for any flow with negligible viscosity effects (e.g., compressible flow outside the boundary layer near a solid boundary). It is interesting that the same equation governs both of these extreme cases.

The Laplace equation also applies to the distribution of electrical potential and current flow in an electrically conducting medium, as well as the temperature distribution and heat flow in a thermally conducting medium. For example, if Φ ⇒ E, V ⇒ i, and μ/K ⇒ r_e, where r_e is the electrical resistivity (r_e = RA/Δx), Equation 14.22 becomes Ohm’s law:

$\frac{\partial E}{\partial x}=-r_e{i}_x,{\nabla }^{2}E=0,\text{and}\left(\frac{\partial i_x}{\partial x}+\frac{\partial i_y}{\partial y}\right)=0$

(14.27)

Furthermore, if Φ ⇒ T, V ⇒ q, and K/μ ⇒ k, where q is the heat flux and k is the thermal conductivity, the same equations govern the flow of heat in a thermally conducting medium (i.e., Fourier’s law):

$\frac{\partial T}{\partial x}=-\frac{1}{k}{q}_x,{\nabla }^{2}T=0,\text{and}\left(\frac{\partial q_x}{\partial x}+\frac{\partial q_y}{\partial y}\right)=0$

(14.28)

By making use of these analogies, electrical analog models can be constructed that can be used, for example, to determine the pressure and flow distribution in a porous medium from measurements of the voltage and current distribution in a conducting medium. The process becomes more complex, however, when the local permeability varies with position within the medium (i.e., in anisotropic and/or nonhomogeneous porous media), which is often the case in practice.

V.    PACKED COLUMNS

At the other end of the spectrum from a “porous rock” is the unconsolidated medium composed of beds of relatively large packing elements. These elements may include a variety of shapes, such as rings, saddles, grids, and meshes, which are generally used to provide a large gas–liquid interface for promoting mass transfer in operations such as distillation, absorption, or liquid–liquid extraction. A typical application might be the removal of an impurity from a gas stream by selective absorption with a solvent in an absorption column filled with packing. The gas (or lighter liquid, in the case of liquid–liquid extraction) typically enters the bottom of the column, and the heavier liquid enters the top and drains by gravity, the flow being countercurrent as illustrated in Figure 14.3.

FIGURE 14.3 Schematic of a packed column.

FIGURE 14.4 Effect of liquid rate on pressure drop.

For single-phase flow through packed beds, the pressure drop can generally be predicted adequately by the Ergun equation, Equation 14.17. However, because the flow in packed columns is normally countercurrent two-phase flow, this situation is more complex. The effect of increasing the liquid mass flow rate (L) on the pressure drop through the column for a given gas mass flow rate (G), starting with dry packing, is illustrated in Figure 14.4. The pressure drop for wet drained packing is higher than for dry packing, since the liquid occupies some of the void space between packing elements even in the “drained” condition. As the liquid flow rate increases, the liquid occupies an increasing portion of the void space, so that the area available to the gas is reduced and therefore the total pressure drop increases. As the liquid flow rate increases, the curve of ΔP versus G becomes increasingly steeper. The points labeled “l” in Figure 14.4 are referred to as the “loading” points and indicate the point where there is a marked increase in the interaction between the liquid and the gas, which is the desired operating point for the column. The points labeled “f” on Figure 14.4 are the “flooding” points. At these points, the pressure drop through the column is equal to the static head of liquid in the column. When this occurs, the pressure drop due to the gas flow balances the static head of liquid so that the liquid can no longer drain through the packing by gravity and the column is said to be “flooded.” It is obviously undesirable to operate a packed column at or near the flooding point, since a slight increase in gas flow at this point will carry the liquid out of the top of the column.

The pressure drop through packed columns, and the flooding conditions, can be estimated from the generalized correlation of Leva (1992), as shown in Figure 14.5. The pressure gradient in (mm of water) per (m of packed height) is the parameter on the curves, and interpolation is usually necessary to determine the pressure drop (note that the pressure drop is not linearly proportional to the spacing between the curves). Correction factors for liquid density and viscosity, which are applied to the y axis of this correlation, are also shown. The parameter F in this correlation is called the packing factor. Values of F are given in Table 14.1, which shows the dimensions and physical properties of a variety of packings. Note that in Table 14.1, the term S_B is equal to a_s (1 – ε), where a_s is the specific surface area (i.e., the surface area per unit volume of the packing element). The packing factor F is comparable to the term S_B/ε³ in the definition of f_PM but is an empirical factor that characterizes the packing somewhat better than S_B/ε³.

FIGURE 14.5 Generalized correlation for pressure drop in packed columns. L, liquid mass flux [lb_m/(s ft²)], kg/(s m²); G, gas mass flux [lb_m/(sft²)]; ρ_L, liquid density (lb_m/ft³, kg/m³); F, packing factor (Table 14.1); μ_L, liquid viscosity (mN s/m²); g, (9.81 m/s², 32.2 ft/s²); w, water at same T and P as column. (From Coulson, J.M. et al., Chemical Engineering, Vol. 2, 5th edn., Butterworth-Heinemann, Oxford, U.K., 2002.)

TABLE 14.1

Design Data for Various Column Packings

Note: The packing factor F replaces the term S_B/ε³. Use of the given value of F in Fig. 14.5 permits more predictable performance of designs incorporating packed beds, because the values quoted are derived from operating characteristics of the packings rather than from their physical dimensions.

Source: Coulson et al. (2002).

VI.    FILTRATION

For fine suspended solids that are too small to separate from the liquid by gravitational or centrifugal methods, a “barrier” method such as a filter may be used. The liquid is passed through a filter medium (usually a cloth or screen), which provides a support for a cake formed from the solid particles removed from the slurry. In reality, the pores in the filter medium are frequently larger than the particles, which may penetrate some distance into the medium before being trapped. The layer of solids that builds up on the surface of the medium is called the cake, and it is the cake that provides the actual filtration. The pressure–flow characteristics of the porous cake primarily determine the performance of the filter. A recent article by Gabelman (2015) provides a good overview of filtration and Branan (1994) and Cheremisinoff (1998) treat filtration in greater detail.

A.    GOVERNING EQUATIONS

A schematic of the flow through the cake and the filter medium is shown in Figure 14.6. The slurry flow rate is Q, and the total volume of the filtrate that passes through the filter is $\tilde{V}$. The flow through the cake and the filter medium is invariably laminar, so the resistance can be described by Darcy’s law, Equation 14.19, and the permeability of the medium (K):

$\frac{-\Delta P}{L}=\frac{Q\mu }{KA}$

(14.29)

Applying this relation across both the cake and the filter medium in series gives

$P_1-P_2={\left(\frac{L}{K}\right)}_{cake}\frac{Q\text{μ}}{A}$

(14.30)

$P_2-P_3={\left(\frac{L}{K}\right)}_{FM}\frac{Q\mu }{A}$

(14.31)

The total pressure drop across the filter is the sum of these:

$P_1-P_3=\frac{Q\mu }{A}\left[{\left(\frac{L}{K}\right)}_{cake}+{\left(\frac{L}{K}\right)}_{FM}\right]$

(14.32)

where

(L/K)_cake is the resistance of the cake

(L/K)_FM is the resistance of the filter medium

FIGURE 14.6 Schematic of flow through filter cake and medium.

The latter is higher for a “dirty” filter medium than for a clean one, but once the initial particles become imbedded in the medium and the cake starts to build up, it remains relatively constant. The cake resistance, on the other hand, continues to increase with time as the cake thickness increases. The cake thickness is directly proportional to the volume of solids that has been deposited from the slurry and inversely proportional to the area:

$L_{cake}=\frac{{\tilde{V}}_{cake}}{A}=\frac{{\tilde{V}}_{solids}}{A\left(1-\text{ε}\right)}=\frac{M_{solids}}{A{\rho }_s\left(1-\text{ε}\right)}$

(14.33)

Now $\left(M_{solids}/\tilde{V}\right)$ is the mass of solids per unit volume of liquid in the slurry feed (e.g., the “solids loading” of the slurry). $\tilde{V}$ is the volume of liquid (filtrate) that has passed to deposit M_solids on the cake

Thus, the cake thickness can be expressed as

$L_{cake}=\left(\frac{M_{solids}}{\tilde{V}}\right)\left(\frac{\tilde{V}}{A}\right)\frac{1}{{\rho }_s\left(1-\text{ε}\right)}=W\frac{\tilde{V}}{A}$

(14.34)

where $W=\left(M_{solids}/\tilde{V}\right)/{\rho }_s\left(1-\text{ε}\right)$ is a property of the specific slurry and cake. The density of the cake is given by

${\rho }_c=\left(1-\text{ε}\right){\rho }_s+\epsilon {\rho }_{liq}$

(14.35)

Substituting Equation 14.34 into Equation 14.32 and rearranging the result in the basic equation governing the filter performance

$\begin{array}{|c|}\hline \frac{Q}{A}=\frac{1}{A}\frac{d\tilde{V}}{dt}=\frac{P_1-P_3}{\mu \left[\left(\tilde{V}W/AK\right)+a\right]}\\ \hline\end{array}$

(14.36)

where a is the filter medium resistance, that is, a = (L/K)_FM.

It should be recognized that the operation of a filter is an unsteady cyclic process. As the cake builds up, its resistance increases with time, and either the flow rate (Q) will drop or the pressure drop (ΔP) will increase with time. The specific behavior depends upon how the filter is operated, that is, constant pressure or constant flow operation, as follows.

B.    CONSTANT PRESSURE OPERATION

If the slurry is fed to the filter using a centrifugal pump, which delivers (approximately) a constant head, or if the filter is operated by a controlled pressure or vacuum, the pressure drop across the filter will remain essentially constant during filtering operation and the flow rate will drop as the cake thickness (resistance) increases. In this case, Equation 14.36 can be integrated at constant pressure (i.e., ΔP = P₁ − P₃) to give

$C_1{\left(\frac{\tilde{V}}{A}\right)}^2+C_2\left(\frac{\tilde{V}}{A}\right)=\left(-\Delta P\right)t$

(14.37)

where

C₁ = μW/2K

C₂ = μa

Both of these are assumed to be independent of pressure (we will consider compressible cakes later). The constant of integration has been evaluated using the condition that at t = 0, $\tilde{V}=0$. In Equation 14.37, t is the time required to pass volume $\tilde{V}$ of the filtrate through the filter.

As C₁ and C₂ are unique properties of a specific slurry–cake system, it is usually appropriate to determine their values from laboratory tests using samples of the specific slurry and the filter medium that are to be used in the plant. For this purpose, it is more convenient to rearrange Equation 14.37 in the following form:

$\left(\frac{-\Delta Pt}{\tilde{V}/A}\right)=C_1\left(\frac{\tilde{V}}{A}\right)+C_2$

(14.38)

If $\tilde{V}$ is measured as a function of t in a lab experiment for given values of ΔP and A, the data can be arranged in the form of Equation 14.38. When the left hand side is plotted versus $\tilde{V}/A$, the result should be a straight line with slope C₁ and intercept C₂ (which are easily determined by linear regression).

C.    CONSTANT FLOW OPERATION

If the slurry is fed to the filter by a positive displacement pump, the flow rate will be constant regardless of the pressure drop, which will increase with time. In this case, noting that $\tilde{V}=Qt$, Equation 14.36 can be rearranged to give

$-\Delta P=2C_1{\left(\frac{Q}{A}\right)}^{2}t+C_2\left(\frac{Q}{A}\right)$

(14.39)

This shows that for given Q and A, the plot of ΔP versus t should be a straight line and the system constants C₁ and C₂ can be determined from the slope, 2C₁(Q/A)², and intercept, C₂(Q/A).

It is evident that the filter performance is governed by the system constants C₁ and C₂, regardless of whether the operation is at constant pressure or constant flow rate. These constants can be evaluated from laboratory data taken under either type of operation and used to analyze the performance of the full-scale filter for either type of operation.

D.    CYCLE TIME

As noted earlier, the operation of a filter is cyclic. As the filtration proceeds, and the pressure either increases or the flow rate drops either a cake will eventually build up to fill the space available for it or the pressure drop will reach the operational limit. At that point, the filtration must cease and the cake removed. There is often a wash cycle prior to removal of the cake in order to remove the slurry carrier liquid from the pores of the cake, using a clean liquid. The pressure–flow behavior during the wash period is a steady-state operation, controlled by the maximum cake and filter medium resistance, since no solids are deposited during this period. The cake can be removed by physically disassembling the filter, removing the filter medium and the cake (as for a plate-and-frame filter), and then reassembling the filter and starting the cycle over. Or in the case of a rotary drum filter, the wash period and cake removal are part of the rotating drum cycle. The drum rotates continuously, although the filtration operation is still cyclic (as discussed in Section F below).

The variable t in the equations given earlier is the actual time (t_filter) that is required to pass a volume $\tilde{V}$ of the filtrate through the medium and is only part of the total time of the cycle (t_cycle). The rest of the cycle time, which may include wash time, disassembly and assembly time, and cleaning time, we shall call “dead” time (t_dead):

$t_{cycle}=t_{filter}+t_{dead}$

(14.40)

The net (average) filter capacity is determined by the amount of slurry processed during the total cycle time, not just the “filter” time, and represents the average flow rate $\left(\overline{Q}\right)$:

$\overline{Q}={\left(\frac{{\tilde{V}}_{cycle}}{t_{cycle}}\right)}_{const\Delta P}={\left(\frac{Q{t}_{filter}}{t_{cycle}}\right)}_{constQ}$

(14.41)

E.    PLATE-AND-FRAME FILTER

A plate-and-frame filter press consists of alternate solid plates and hollow frames in a “sandwich” arrangement. The open frames are covered by a filter medium (e.g., the filter cloth), and the slurry enters through the frames and deposits a cake on the filter medium. The operation is “batch,” in that the filter must be disassembled when the cake fills the available frame space and then cleaned and reassembled, after which the entire process is repeated. A schematic of a plate-and-frame press is shown in Figure 14.7. In the arrangement shown, all of the frames are in parallel, and the total filter area (which appears in the equations) is

$A=2n{A}_f$

(14.42)

where

n is the number of frames

A_f is the filter area of (one side) of the frame

The flow rate Q in the equations is the total flow rate, and Q/A = Q/(2nA_f) is the total flow per unit total filtering area, or the flow rate per filter side per unit area of the filter side.

There are a variety of arrangements that operate in the same manner as the plate-and-frame filter, such as the “leaf filter,” which may consist of one or more “frames” covered by the filter medium, which are immersed in the slurry. They are often operated by a vacuum that draws the filtrate through the filter, and the cake collects on the filter medium on the outside of the frame.

FIGURE 14.7 Schematic of a plate-and-frame filter.

F.    ROTARY DRUM FILTER

The rotary drum filter is a “continuous” process, since it does not have to be shut down during the cycle, although the operation is still cyclic. A schematic is shown in Figure 14.8. The drum rotates at a rate N (rpm), and the filter area is the total drum surface, that is, A = πDL. However, if the fraction of the drum that is in contact with the slurry is f, then the total time of the cycle during which any one point on the surface is actually filtering is f/N:

FIGURE 14.8 Schematic of a rotary drum filter.

$t_{cycle}=\frac{1}{N},t_{filt}=\frac{f}{N}$

(14.43)

The rotary drum filter is commonly operated under constant pressure conditions (i.e. under a vacuum).

G.    COMPRESSIBLE CAKE

The equations presented so far all assume that the cake is incompressible, that is, the porosity, permeability, and density of the cake are constant. For many cakes, this is not so, since the cake properties may vary with pressure due to consolidation (e.g., flocs, gels, fibers, pulp). For such cases, the basic filter equation (Equation 14.36) can be expressed in the following form:

$\frac{Q}{A}=\frac{{\left(-\Delta P\right)}^{1-s}}{\mu \left[\alpha \left(\tilde{V}/A\right)\left(M_s/\tilde{V}\right)+a\right]}=\frac{1}{A}\frac{d\tilde{V}}{dt}$

(14.44)

where the pressure dependence is characterized by the parameter s, and α and a are the pressure-independent properties of the cake. There are several modes of performance of the filter, depending upon the value of s:

1.  If s = 0, then Q proportional (∝) ΔP (the cake is incompressible).

2.  If s < 1, then Q increases as ΔP increases (slightly compressible).

3.  If s = 1, then Q is independent of ΔP (compressible).

4.  If s > 1, then Q decreases as ΔP increases (highly compressible).

In case 4, the increasing pressure compresses the cake to such an extent that it actually “squeezes off” the flow so that as the pressure increases the flow rate decreases. This situation can be compensated for by adding a “filter aid” to the slurry, which is a rigid dispersed solid that forms an incompressible cake (e.g., diatomaceous earth, sand, etc.). This provides “rigidity” to the cake and enhances its permeability, thus increasing the filter capacity (it may seem like a paradox that adding more solids to the slurry feed actually increases the filter performance, but it works).

The equations that apply for a compressible cake are as follows:

Constant pressure drop:

$C_1{\left(\frac{\tilde{V}}{A}\right)}^2+C_2\left(\frac{\tilde{V}}{A}\right)=t{\left(-\Delta P\right)}^{1-s}$

(14.45)

Constant flow rate:

$2C_1{\left(\frac{Q}{A}\right)}^{2}t+C_2\left(\frac{Q}{A}\right)={\left(-\Delta P\right)}^{1-s}$

(14.46)

where

$C_1=\left(\mu \alpha /2\right)\left(M_s/\tilde{V}\right)$

C₂ = μa

There are now three parameters that must be determined empirically from laboratory measurements: C₁, C₂, and s. The easiest way to do this would be to utilize the constant pressure mode in the laboratory (e.g., a Buchner funnel, with a set vacuum pressure difference) and obtain several sets of data for $\tilde{V}$ as a function of t, with each set at a different value of ΔP. For each data set, the plot of $t/\tilde{V}$ versus $\tilde{V}$ should yield a straight line, with a slope of C₁/[A²(−ΔP)^1–s] and intercept of C₂/[A(–ΔP)^1–s]. Thus, a log–log plot of either the slope or the intercept versus ΔP should have a slope of (s – 1), which determines s.

SUMMARY

The following are the major points that should be retained from this chapter:

•  The difference between consolidated and unconsolidated porous media

•  The definitions of porosity, hydraulic diameter, sphericity factor, and permeability for porous media

•  Laminar and turbulent friction loss in porous media

•  The pressure drop behavior and the concept of flooding in packed columns

•  Filtration under constant pressure or constant flow conditions

•  Operation of plate-and-frame and rotary drum filters

PROBLEMS

POROUS MEDIA

1.    A packed bed is composed of crushed rock with a density of 175 lb_m/ft³ having a size and shape such that the average ratio of surface area to volume for the particles is 50 in.²/in.³ The bed is 6 ft deep, has a porosity of 0.3, and is covered by a 2 ft deep layer of water that drains by gravity through the bed. Calculate the flow rate of water through the bed in gpm/ft², assuming it exits at one atmosphere pressure.

2.    An impurity in a water stream at a very small concentration is to be removed in a charcoal trickle bed filter. The filter is in a cylindrical column that is 2 ft in diameter, and the bed is 4 ft deep. The water is stored at a level that is 2 ft above the top of the bed, and it trickles through by gravity flow. If the charcoal particles have a geometric surface area to volume ratio of 48 in.⁻¹ and they pack with a porosity of 0.45, what is the flow rate of water through the column, in gpm?

3.    A trickle bed filter is composed of a packed bed of broken rock that has a shape such that the average ratio of the surface area to volume for the rock particles is 30 in.⁻¹ The bed is 2 ft deep, has a porosity of 0.3, and is covered by a layer of water that is 2 ft. deep and drains by gravity through the bed.

(a)  Determine the volume flow rate of the water through the bed per unit bed area (in gpm/ft²).

(b)  If the water is pumped upward through the bed (e.g., to flush it out), calculate the flow rate (in gpm/ft² of bed area) that would be required to fluidize the bed.

(c)  Calculate the corresponding flow rate that would sweep the rock particles away with the water. The rock density is 120 lb_m/ft³.

PACKED COLUMNS

4.    A packed column that is 3 ft in diameter with a packing height of 25 ft is used to absorb an impurity from a methane gas stream using an amine solution absorbent. The gas flow rate is 2000 scfm, and the liquid has a density of 1.2 g/cm³ and a viscosity of 2 cP. If the column operates at 1 atm and 80°F, determine the liquid flow rate at which flooding would occur in the column and the pressure drop at 50% of the flooding liquid rate for the following packings:

(a)  2 in. ceramic Raschig rings

(b)  2 in. plastic Pall rings

5.    A packed column is used to scrub SO₂ from air using water. The gas flow rate is 500 scfm/ft² and the column operates at 90°F and 1 atm. If the column contains No. 1 plastic Intalox packing, what is the maximum liquid flow rate (per unit cross section of column) that could be used without flooding?

6.    A stripping column packed with 2 in. metal Pall rings uses air at 5 psig, 80°C, to strip an impurity from an absorber oil (SG = 0.9, viscosity = 5 cP, T = 20°C). If the flow rate of the oil is 500 lb_m/min and that of the air is 20 lb_m/min, determine the following:

(a)  What is the minimum column diameter that can be used without flooding?

(b)  If the column diameter is 50% greater than the minimum size, what is the pressure drop per ft of column height?

7.    A packed column, which is 0.6 m in diameter and 4 m high containing 25 mm Raschig rings, is used in a gas absorption process to remove an impurity from the gas stream by absorbing it in a liquid solvent. The liquid, which has a viscosity of 5 cP and SG = 1.1, enters the top of the column at a rate of 2.5 kg/s m², and the gas, which may be assumed to have the same properties as air, enters the bottom of the column at a rate of 0.6 kg/s m². The column operates at atmospheric pressure and 25°C. Determine

(a)  The pressure drop through the column, in inches of water

(b)  How high the liquid rate could be increased before the column would flood

8.    A packed column is used to absorb SO₂ from flue gas using an ethanol amine solution. The column is 4 ft in diameter and has a packed height of 20 ft and is packed with 2 in. plastic Pall rings. The flue gas is at a temperature of 180°F and has an average molecular weight of 31. The amine solution has a specific gravity of 1.02 and a viscosity at the operating temperature of 1.5 cP. If the gas must leave the column at 25 psig and a flow rate of 10,000 scfm, determine

(a)  The flow rate of the liquid (in gpm) that is 50% of that at which flooding would occur

(b)  The horsepower that would be required for the blower to move the gas through the column, if the blower is 80% efficient

9.    A packed absorption tower is used to remove SO₂ from an air stream by absorption in a solvent. The tower is 5 ft. in diameter and 60 ft. high and contains 1.5 in. plastic Pall rings. The temperature and pressure in the tower are 90°F and 30 psig. The gas stream flow rate is 6500 scfm, and the liquid SG is 1.25, with a viscosity of 25 cP.

(a)  What is the liquid flow rate (in gpm) at which the column will flood?

(b)  If the column operates at a liquid rate that is 75% of the flooding value, what is the total pressure drop through the tower in psi?

10.  A packed absorption column removes an impurity from a gas stream by contact with a liquid solvent. The column is 3 ft in diameter and contains 25 ft of No. 2 plastic Super Intalox packing. The gas has a MW of 28 and enters the column at 120°F and leaves at 10 psig at a rate of 5000 scfm. The liquid has a SG of 1.15 and a viscosity of 0.8 cP. Determine

(a)  The flow rate of the liquid in gpm that would be 50% of the flow rate at which the column would flood

(b)  The pressure drop through the column in psi

(c)  The horsepower of the blower required to move the gas through the column, if it is 60% efficient

FILTRATION

11.  A fine aqueous suspension containing 1 lb_m of solids per ft³ of suspension is to be filtered in a constant pressure filter. It is desired to filter at an average rate of 100 gpm, and the filter cake must be removed when it gets 2 in. thick. What filter area is required? Data: ΔP = 10 psi, p (wet cake) = 85 lb_m/ft³, K (permeability) = 0.118 darcies, a = 2 × 10⁹ ft⁻¹

12.  An aqueous slurry containing 1.5 lb_m of solid per gallon of liquid is pumped through a filter cloth by a centrifugal pump. If the pump provides a constant pressure drop of 150 psig, how long will it take for the filter cake to build up to a thickness of 2 in.? The density of the filter cake is 30 lb_m/ft³ and its permeability is 0.01 darcies.

13.  A packed bed that consists of the same medium as that in Problem 3 is to be used to filter solids from an aqueous slurry. To determine the filter properties, you test a small section of the bed, which is 6 in. in diameter and 6 in. deep, in the lab. When the slurry is pumped through this test model at a constant flow rate of 30 gpm, the pressure drop across the bed rises to 2 psia after 10 min. How long will it take to filter 100,000 gal of water from the slurry in a full-size bed, which is 10 ft in diameter and 2 ft deep, if the slurry is maintained at a depth of 2 ft over the bed and drains by gravity through the bed?

14.  A slurry containing 1 lb_m of solids per gallon of water is to be filtered in a plate-and-frame filter having a total filtering area of 60 ft². The slurry is fed to the filter by a centrifugal pump, which develops a head of 20 psig. How long would it take to build up a layer of filter cake 4 in. thick on the filter medium? Laboratory data were taken on the slurry using a positive displacement pump operating at 5 gpm and 1 ft² of the filter medium. It was found that the pressure drop increased linearly with time from an initial value of 0.2 psi, reaching a value of 50 psi after 1 min. The density of the dry filter cake was found to be 0.85 g/cm³.

15.  A rotary drum filter that is 6 ft in diameter and 8 ft long is to be used to filter a slurry. The drum rotates at 0.5 rpm, and one-third of the drum’s surface is submerged under the slurry. A vacuum is drawn in the drum so that a constant pressure drop of 10 psi is maintained across the drum and filter cake. You test the slurry in the lab by pumping it at a constant filtrate rate of 20 gpm through 1 ft² of the drum filter screen and find that after 1 min the pressure drop is 8 psi and after 3 min the pressure drop is 12 psi. How long will it take to filter 100,000 gal of the filtrate from the slurry using the rotary drum?

16.  A plate-and-frame filter press contains 16 frames and operates at a constant flow rate of 30 gpm. Each frame has an active filtering area of 4 ft², and it takes 15 min to disassemble, clean, and reassemble the press. The press must be shut down for disassembly when the pressure difference builds up to 10 psi. What is the total net filtration rate in gpm for a slurry having properties determined by the following lab test. A sample of the slurry is pumped at a constant pressure differential of 5 psi through 0.25 ft² of the filter medium. After 3 min, 1 gal of the filtrate had been collected. The resistance of the filter medium may be neglected.

17.  A rotary drum filter is used to filter a slurry. The drum rotates at a rate of 3 min/cycle, and 40% of the drum surface is submerged beneath the slurry. A constant pressure drop at 3 psi is maintained across the filter. If the drum is 5 ft in diameter and 10 ft long, calculate the total net filtration rate in gpm that is possible for a slurry having properties as determined by a lab test, as follows. A sample of the slurry was pumped at a constant flow rate of 1 gpm through 0.25 ft² of the filter medium. After 10 min, the pressure difference across the filter had risen to 2.5 psi. The filter medium resistance may be neglected.

18.  You must filter 1000 lb_m/min of an aqueous slurry containing 40% solids by weight using a rotary drum filter with a diameter of 4 m and a length of 4 m, which operates at a vacuum of 25 in. Hg with 30% of its surface submerged beneath the slurry. A lab test is run on a sample of the slurry using 200 cm² of the same filter medium, under a vacuum of 25 in. Hg. During the first minute of operation, 300 cm³ of the filtrate is collected, and during the second minute, an additional 140 cm³ is collected.

(a)  How fast should the drum be rotated?

(b)  If the drum is rotated at 2 rpm, what would the filter capacity be in lb_m of the slurry filtered per minute?

19.  A rotary drum filter is to be used to filter a lime slurry. The drum rotates at a rate of 0.2 rpm, and 30% of the drum surface is submerged beneath the slurry. The filter operates at a constant ΔP of 10 psi. The slurry properties were determined from a lab test at a constant flow rate of 0.5 gpm using 1/2 ft² of the filter medium. The test results indicated that the pressure drop rose to 2 psi after 10 s and to 10 psi after 60 s. Calculate the net filtration rate per unit area of the drum under these conditions, in gpm/ft².

20.  A plate-and-frame filter press operating at a constant ΔP of 150 psi is to be used to filter a sludge containing 2 lb_m of solids per ft³ of water. The filter must be disassembled and cleaned when the cake thickness builds up to 1 in. The frames have a projected area of 4 ft², and the downtime for cleaning is 10 min/frame. The properties of the sludge and cake were determined in a lab test operating at a constant flow rate of 0.2 gpm of the filtrate and a filter area of 1/4 ft². The test results show that the pressure drop rises to 3 psi after 20 s and to 8 psi after 60 s. Calculate the overall net filtration rate per frame in the filter in gpm of the filtrate, accounting for the downtime. The density of the cake was found to be 150 lb_m/ft³.

21.  A packed bed composed of crushed rock having a density of 175 lb_m/ft³ is to be used as a filter. The size and shape of the rock particles is such that the average surface area to volume ratio is 50 in.²/in.³ and the bed porosity is 0.3. A lab test using the slurry to be filtered is run on a small bed of the same particles, which is 6 in. deep and 6 in. in diameter. The slurry is pumped through this bed at a constant filtrate rate of 10 gpm, and it is found that after 5 min the pressure drop is 5 psi, while after 10 min it is 8 psi. Calculate how long it would take to filter 100,000 gal of filtrate from the slurry in a full-scale bed that is 10 ft in diameter and 2 ft deep, if the slurry is maintained at a depth of 2 ft above the bed and drains through it by gravity. Assume the slurry densities to be the same as water.

22.  A rotary drum filter has a diameter of 6 ft and a length of 8 ft and rotates at a rate of 30 s/cycle. The filter operates at a vacuum of 500 mm Hg, with 30% of its surface submerged. The slurry to be filtered is tested in the lab using 0.5 ft² of the drum filter medium in a filter funnel operating at 600 mm Hg vacuum. After 5 min of operation, 250 cm³ of the filtrate was collected through the funnel, and after 10 min, a total of 400 cm³ is collected. What would be the net (average) filtration rate of this slurry in the rotary drum filter, in gpm?

23.  A rotary drum filter, 10 ft in diameter and 8 ft long, is to be used to filter a slurry of incompressible solids. The drum rotates at 1.2 rpm, and 40% of its surface is submerged beneath the slurry at all times. A vacuum in the drum maintains a constant pressure drop of 10 psi across the drum and filter cake. The slurry is tested in the lab by pumping it at a constant rate of 5 gpm through 0.5 ft² of the drum filter screen. After 1 min, the pressure drop is 9 psi, and after 3 min it has risen to 15 psi. How long will it take to filter 1 million gal of the filtrate from the slurry using the rotary drum? How long would it take if the drum rotated at 3 rpm?

24.  A slurry is being filtered at a net rate of 10,000 gal/day by a plate-and-frame filter with 15 frames with an active filtering area of 1.5 ft² per frame, fed by a positive displacement pump. The pressure drop varies from 2 psi at start-up to 25 psi after 10 min, at which time it is shut down for cleanup. It takes 10 min to disassemble, clean out, and reassemble the filter. Your boss decides that it would be more economical to replace this filter by a rotary drum filter, using the same filter medium. The rotary filter operates at a vacuum of 200 mm Hg with 30% of its surface submerged and rotates at a rate of 5 min/rev. If the drum length is equal to its diameter, how big should it be?

25.  You want to select a rotary drum filter to filter a coal slurry at a rate of 100,000 gal of the filtrate per day. The filter operates at a differential pressure of 12 psi, and 30% of the surface is submerged in the slurry at all times. A sample of the slurry is filtered in the lab through a 6 in. diameter sample of the filter medium at a constant rate of 1 gpm. After 1 min, the pressure drop across this filter is 3 psi, and after 5 min it is 10 psi. If the drum rotates at a rate of 3 rpm, what total filter area is required?

26.  A slurry containing 40% solids by volume is delivered to a rotary drum filter, which is 4 ft in diameter and 6 ft long and operates at a vacuum of 25 in. Hg. A lab test is run with a 50 cm² sample of the filter medium and the slurry, at a constant flow rate of 200 cm³/min. After 1 min, the pressure across the lab filter is 6 psi and after 3 min it is 16 psi. If 40% of the rotary drum is submerged under the slurry, how fast should it be rotated (rpm) in order to filter the slurry at an average rate of 250 gpm?

27.  A slurry is to be filtered with a rotary drum filter that is 5 ft in diameter, 8 ft long, rotates once every 10 s and has 20% of its surface immersed in the slurry. The drum operates with a vacuum of 20 in. Hg. A lab test was run on a sample of the slurry using 1/4 ft² of the filter medium at a constant flow rate of 40 cm³/s. After 20 s, the pressure drop was 30 psi across the lab filter, and after 40 s, it was 35 psi. How many gallons of the filtrate can be filtered per day in the rotary drum?

28.  A rotary drum filter is to be installed in your plant. You run a test in the lab on the slurry to be filtered using a 0.1 ft² sample of the filter medium at a constant pressure drop of 10 psi. After 1 min, you find that 500 cc of the filtrate has passed through the filter, and after 2 min, the filtrate volume is 715 cc. If the rotary drum filter operates under a vacuum of 25 in. of Hg with 25% of its surface submerged, determine the following:

(a)  The capacity of the rotary drum filter, in gallons of filtrate per square foot of surface area, if it operates at (1) 2 rpm and (2) 5 rpm.

(b)  If the drum has a diameter of 4 ft and a length of 6 ft, what is the total filter capacity in gallons/day for each of the operating speeds of 2 and 5 rpm?

29.  A slurry of CaCO₃ in water at 25°C containing 20% solids by weight is to be filtered in a plate-and-frame filter. The slurry and filter medium are tested in a constant pressure lab filter, having an area of 0.0439 m², at a pressure drop of 338 kPa. It is found that 10^-3 m³ of the filtrate is collected after 9.5 s and 5 × 10^-3 m³ is collected after 107.3 s. The plate-and-frame filter has 20 frames, with 0.873 m² of the filter medium per frame, and operates at a constant flow rate of 0.00462 m³ of slurry per second. The filter is operated until the pressure drop reaches 500 kPa, at which time it is shut down for cleaning. The downtime is 15 min per cycle. Determine how much filtrate passes through the filter in each 24 h period of operation (SG of CaCO₃ is 1.6).

30.  An algal sludge is to be clarified by filtering. A lab test is run on the sludge using an area A of the filter medium. At a constant pressure drop of 40 kN/m², a plot of the time required to collect a volume $\tilde{V}$ of the filtrate times $\Delta P/\left(\tilde{V}/A\right)$ versus $\tilde{V}/A$ gives a straight line with a slope of 1.2 × 10⁶ kN s/m⁴ and an intercept of 6.0 × 10⁴ kN s/m³. A repeat of the data at a pressure drop of 200 kN/m² also gave a straight line on the same type of plot, with the same intercept but with a slope of 2.1 × 10⁶ kN s/m⁴. When a filter aid was added to the sludge in an amount equal to 20% of the algae by weight, the lab test gave a straight line with the same intercept but with a slope of 1.4 × 10⁶ kN s/m⁴.

(a)  What does this tell you about the sludge?

(b)  The sludge is to be filtered using a rotary drum filter, with a diameter of 4 ft and a length of 6 ft, operating at a vacuum of 700 mm Hg with 35% of the drum submerged. If the drum is rotated at a rate of 2 rpm, how many gal of the filtrate will be collected in a day, with and without the filter aid?

(c)  What would the answer to (b) be if the drum speed was 4 rpm?

31.  A slurry containing 0.2 kg of solids per kg water is filtered through a rotary drum filter, operating at a pressure difference of 65 kN/m². The drum is 0.6 m in diameter, and 0.6 m long, rotates once every 350 s, and has 20% of its surface submerged below the slurry.

(a)  If the overall average filtrate flow rate is 0.125 kg/s, the cake is incompressible with a porosity of 50%, and the solids SG = 3.0, determine the maximum thickness of the cake on the drum (you may neglect the filter medium resistance).

(b)  The filter breaks down, and you want to replace it with a plate-and-frame filter having the same overall capacity, which operates at a pressure difference of 275 kN/m². The frames are 10 cm thick, and the maximum cake thickness at which the filter will still operate properly is 4 cm. It will take 100 s to disassemble the filter, 100 s to clean it out, and 100 s to reassemble it. If the frames are 0.3 m square, how many frames should the filter contain?

32.  You want to filter an aqueous slurry using a rotary drum filter, at a total rate (of filtrate) of 10,000 gal/day. The drum rotates at a rate of 0.2 rpm, with 25% of the drum surface submerged in the slurry, at a vacuum of 10 psi. The properties of the slurry are determined from a lab test using a Buchner funnel under a vacuum of 500 mm Hg, using a 100 cm² sample of the filter medium and the slurry, which resulted in the data given in the following table. Determine the total filter area of the rotary drum required for this job.

Lab data:

Time (s)	Volume of Filtrate (cc)
50	10
100	18
200	31
400	51

33.  You want to use a plate-and-frame filter to filter an aqueous slurry at a rate of 1.8 m³/8 h day. The filter frames are square, with a length on each side of 0.45 m. The “downtime” for the filter press is 300 s plus an additional 100 s per frame for cleaning. The filter operates with a positive displacement pump, and the maximum operation pressure differential for the filter is 45 psi, which is reached after 200 s of operation.

(a)  How many frames must be used in this filter to achieve the required capacity?

(b)  At what flow rate (in gpm) should the pump be operated?

The following lab data were taken with the slurry at a constant ΔP of 10 psi and a 0.05 m² sample of the filter medium:

After 300 s, the total volume of the filtrate was 400 cc.

After 900 s, the total volume of the filtrate was 800 cc.

34.  An aqueous slurry is filtered in a plate-and-frame filter, which operates at a constant ΔP of 100 psi. The filter consists of 20 frames, each of which have a projected area per side of 900 cm². A total filtrate volume of 0.7 m³ is passed through the filter during a filtration time of 1200 s, and the downtime for the filter is 900 s. The resistance of the filter medium is negligible relative to that of the cake. You want to replace the plate-and-frame filter with a rotary drum filter with the same overall average capacity, using the same filter medium. The drum is 2.2 m in diameter and 1.5 m long and operates at 5 psi vacuum with 25% of the drum surface submerged in the slurry. At what speed (in rpm) should the drum be operated?

35.  You must transport a sludge product from an open storage tank to a separations unit at 1 atm, through a 4 in. sch 40 steel pipeline that is 2000 ft long, at a rate of 250 gpm. The sludge is 30% solids by weight in water and has a viscosity of 50 cP with Newtonian properties. The solid particles in the sludge have a density of 3.5 g/cc. The pipeline contains 4 gate valves and 6 elbows.

(a)  Determine the pump head (in feet) required to do this job. You can select any pump with the characteristics given in Appendix H, and you must find the combination of motor speed, motor horsepower, and impeller diameter that should be used.

(b)  You want to install a long radius venturi meter in the line to monitor the flow rate, and you want the maximum pressure drop to be measured to be equal to or less than 40 in. of water. What should the diameter of the venturi throat be?

(c)  At the separations unit, the sludge is fed to a settling tank. The solids settle in the tank, and the water overflows the top. What should the diameter of the tank be if it is desired to limit the size of the particles in the overflow to 100 μm or less?

(d)  If the sludge is fed to a centrifuge instead of the settling tank, at what speed (rpm) should the centrifuge operate to achieve the same separation as the settling tank, if the centrifuge dimensions are L = D = 1 ft, R₁ = R₂/2 = 0.25 ft.

(e)  Suppose the sludge is fed to a rotary drum filter instead, which removes all of the solids from the stream. The drum operates at a vacuum of 6 psi, has dimensions L = D = 4 ft, and operates with 30% of the surface submerged. A lab test is performed on the sludge using 1 ft² of the same filter medium as on the drum, operating at a vacuum of 500 mm Hg. In this test, it is found that 8 gal passes the filter after 2 min, and a total of 20 gal passes through after 10 minutes. What speed (rpm) should the rotary drum filter be operated?

36.  Consider a dilute aqueous slurry containing solid particles with diameters from 0.1 to 1000 μm and a density of 2.7 g/cc, flowing at a rate of 500 gpm.

(a)  If the stream is fed to a settling tank in which all particles with a diameter greater than 100 μm are to be removed, what should the tank diameter be?

(b)  The overflow from the settling tank contains almost all of the water plus the fines not removed in the tank. This stream is fed to a centrifuge, which has a diameter of 20 in., a length of 18 in., and an overflow dam that is 6 in. from the centerline. What speed in rpm should the centrifuge rotate in order to separate all particles with a diameter of 1 μm and larger?

(c)  If the centrifuge rotates at 2500 rpm, what size particles will be removed?

(d)  Instead of the tank and centrifuge, the slurry is fed to a rotary drum filter, which has a diameter of 5 ft and a length of 10 ft. The drum operates under a vacuum of 10 in. Hg, with 35% of its surface submerged in the slurry. A lab test is run on the slurry at a constant flow rate of 100 cc/min, using 50 cm² of the filter medium. In the test filter, the pressure drop reached 10 mm Hg after 1 min and 80 mm Hg after 10 min. How fast should the drum rotate (in rpm) to handle the slurry stream?

NOTATION

A	Area, [L2]
a	Filter medium resistance, [1/L]
as	Particle surface area/volume, [1/L]
C1	Filter parameter = (μW)/(2K), [M/(L3 t)]
C2	Filter parameter = μa, [M/(L2 t)]
D	Diameter, [L]
d	Particle diameter, [L]
ds	Diameter of equal volume sphere [L]
Dh	Hydraulic diameter, [L]
ef	Energy dissipated per unit mass of fluid, [F L/M = L2/t2]
F	Correction factor, Fig. 14.5 [—]
f	Friction factor [—]
fpm	Porous media friction factor, Equation 14.11, [—]
G	Gas mass flux, [M/(L2 s)]
K	Permeability, [L2]
L	Length, [L]
Msolids	Mass of solids [M]
N	Rotation rate, rpm, [1/t]
n	Number of frames, [—]
NRePM	Porous media Reynolds number, Equation 14.13, [—]
P	Pressure, [F/L2 = M/(L t2)]
Q	Volumetric flow rate, [L3/t]
s	Compressibility parameter, Equation, 14.44, [—]
t	Time, [t]
V	Velocity, [L/t]
$\tilde{V}$	Volume of filtrate, [L3]
W	Slurry/cake solids loading parameter $=\left(M_{solids}/\tilde{V}\right)/{\rho }_s\left(1-\text{ε}\right)$, [—]
Wp	Wetted perimeter, [L]
x, y, z	Coordinate directions, [L]

GREEK

Δ()	()2 – ()2
ε	Porosity or void fraction, [—]
Φ	Potential = P + ρgz
μ	Viscosity, [M/(L t)]
ρ	Density, [M/L3]
ψ	Sphericity factor, [—]

SUBSCRIPTS

1,2,3	Reference points
f	Filter frame side
G	Gas
i	Interstitial
L	Liquid
s	Superficial or solids

REFERENCES

Branan, C.R., Rules of Thumb for Chemical Engineers, Gulf Publishing Co., Houston, TX, 1994.

Cheremisinoff, N.P., Liquid Filtration, 2nd edn., Butterworth-Heinemann, Oxford, U.K., 1998.

Coulson, J.M., J.F. Richardson, J.R. Blackhurst, and J.H. Harker, Chemical Engineering, Vol. 2, 5th edn., Butterworth-Heinemann, Oxford, U.K., 2002.

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