2.10 Diffusion in Solids
2.10.1 Steady-State Diffusion in Solids
With a few exceptions, it is possible to describe steady-state transfer of solute through solids by an equation equivalent to Fick’s law. If diffusivity is independent of concentration and there is no bulk flow, the rate of diffusion NA of component A per unit cross section of the solid is given by
NA = -DA (2.74)
where, DA is the diffusivity of A through the solid and (dcA/dz) is the concentration gradient of the solute in the direction of diffusion. If DA is constant, Eq. (2.74) may be integrated for steady-state diffusion through a flat slab of thickness z to give
NA = (2.75)
where, cA1 and cA2 are the concentrations of the solute at the two faces of the slab.
EXAMPLE 2.17 (Steady-state diffusion of a gas through a solid wall): Hydrogen gas at 25°C and 1 atm pressure is diffusing through an unglazed neoprene rubber wall 12.5 mm thick. The solubility of hydrogen in the rubber has been estimated to be 0.053 cm3 (at STP) per cubic centimetre. The diffusivity of hydrogen through the rubber wall is 1.8 # 10-6 cm2/s.
Estimate the rate of diffusion of hydrogen per square metre of the wall.
Solution: Concentration of hydrogen at the inner side of the rubber wall is
cA1 = = 2.37 # 10-3 kmol/m3.
Assuming the resistance to diffusion of hydrogen at the outside surface of the wall is negligible, the concentration of hydrogen at the outer surface of the wall is cA2 = 0.
Given: DA = 1.8 # 10-6 cm2/s = 1.8 # 10-10 m2/s.
The rate of diffusion of hydrogen per square metre = 3.41 # 2 # 10-8
= 6.82 # 10-8 kg/m2$s 2.10.2 Transient or Unsteady-State Diffusion
Unsteady-state operations, in general, are those in which the operating conditions at a particular point change with time. Unsteady-state mass transfer are those where the concentration at a given point varies with time.
Most of the industrial mass transfer operations such as gas-absorption, distillation, extraction, drying, etc. are conducted in the steady-state. However, during start-up and sometimes during shut down all of them undergo unsteady-state operation.
Majority of mass transfer operations involve transfer between two phases one of which is dispersed as thin film or droplets or bubbles in the other phase. On the basis of various theoretical studies, it has been established that these droplets or bubbles assume spherical shape in which mass transfer takes place by transient or unsteady-state molecular diffusion.
Transient diffusion plays very important role in case of mass transfer in solids. In view of the difficulties in transporting solids through continuous equipment, they are frequently handled in batch or semi-batch equipment. Some common examples are leaching or adsorption in packed beds, drying of porous solids, etc. Moreover, even in steady-state mass transfer equipment, each individual piece of solid often experiences unsteady-state diffusion. For example, during drying of a porous solid bar through one large surface only, the movement of moisture within the solid is by unsteady-state diffusion.
From the above discussion, it is evident that knowledge about unsteady-state diffusion is indispensable in the study of mass transfer, particularly in the study of diffusion through solids.
Rate of unsteady-state diffusion
Unsteady-state diffusion in a slab becomes important during drying of some colloids or gel-like materials where it is necessary to know the distribution of moisture in the slab as a function of position and time or to know the relation between the average moisture content of the slab and the drying time.
Unsteady-state diffusion in sphere plays an important role in the study of mass transfer to droplets or bubbles of the dispersed phase which assume spherical shape and through which mass transfer takes
place by unsteady-state diffusion.
The three common geometries involved in industrial operations are plane, spherical and cylindrical surfaces. For each of them, equations for unsteady-state diffusion can be developed by making mass balance over a selected differential volume (Crank 1956). In this section, we shall deal with the working equations for the above three geometries in the order of plane, spherical and cylindrical surfaces.
Unsteady-state diffusion in a slab through both of the large surfaces
The section of a slab in which mass transfer is taking place by unsteady-state diffusion, is shown in Figure 2.10.
Figure 2.10 Unsteady-state molecular diffusion in a slab.
The following assumptions are made to facilitate further calculations:
(i) Uniform concentration of component A, cA0 throughout the slab at i = 0.
(ii) Constant concentration cA* at the two large surfaces of the slab which are both permeable to solute A.
(iii) Diffusion of solute A only in the direction normal to the two large faces of the slab.
(iv) Constant physical properties.
The co-ordinates originate at the mid-plane of the slab which has area A normal to z. Let us consider a control volume defined by the elements of the slab at z, having thickness dz as shown in Figure 2.10.
The concentration gradient at (z + dz) at any given instant is
\ (2.76)
and the flow rate of solute into the control volume is given by
(NAz)z+dz$A = D$A (2.77)
The flow rate of solute out of the control volume at the same instant across the face at z is
(NAz)z$A = D$A (2.78)
The net flow rate into the control volume, as obtained by subtracting Eq. (2.78) from Eq. (2.77), is given by
DA (2.79)
The rate of accumulation of solute in the control volume is given by
Adz (2.80)
Equations (2.79) and (2.80) may therefore be equated, and solved for dcA/di to obtain Fick’s second law of molecular diffusion, namely
(2.81)
For general three-dimensional diffusion with constant D the expression corresponding to Eq. (2.81) may be obtained in a similar manner by considering the rates at which solute enters, leaves and accumulates in a cubic control volume of differential dimensions (dxdydz). The resulting expression becomes,
=
= D 2cA 2.82)
where, the Laplace operator.
For diffusion through two opposite faces of a slab, (d2cA/dx2) and (d2cA/dy2) are zero and from the initial assumptions the boundary conditions become
cA(a, i) = c* A cA(z, 0) = cA0
(0, i) = 0 Let y′ = (cA - cA*), y′ = y′ (z, i). Then
(2.83) and the boundary condition becomes
y′(a, i ) = 0, y′(z, 0) = (cA0 - cA*) = y′0 and (0, i) = 0
Equation (2.83) may be solved both by the method of separation of variables and by using Laplace transformation. The first method is suitable for large diffusion times because the series rapidly converges under such conditions. The second method yields results suitable for small diffusion times.
Solving Eq. (2.83) by the method of separation of variables, the following expression is obtained for the local concentration:
(2.84) and for average concentration:
(2.85) Diffusion through a single large surface of a slab
In some cases, for instance during the tray drying of some colloidal or gel-like substances, diffusion may take place only through one large surface of the slab, the other being impermeable to transfer.
The concentration gradient (dcA/dz) is zero at the impermeable surface which therefore coincides with the mid-plane of the slab in which diffusion takes place symmetrically through the two large surfaces of the slab having identical concentration, cA*.
The solution for symmetrical diffusion is therefore applicable here with the permeable surface at z = a, and the impermeable surface at z = 0.
Unsteady-state diffusion in a sphere
Let us consider a sphere of radius rs through which unsteady-state diffusion takes place.
The following assumptions are made to facilitate calculations:
(i) The concentration of solute is uniform at cA0 throughout the sphere at the start of diffusion (i = 0).
(ii) The resistance to transfer in the medium around the sphere is negligible, so that surface concentration of the sphere is constant at cA*.
(iii) The diffusion is radial, there being no variation in concentration with angular position.
(iv) The physical properties of the material for sphere are constant.
Making a shell balance within a control volume bounded by the surfaces at r and (r + dr) as shown in Figure 2.11.
Figure 2.11 Unsteady-state molecular diffusion in a sphere.
The rate of flow of solute into the control volume = -D(4rr2) (2.86) and the rate of flow out of the control volume is
(2.87)
The net flow rate of solute into the control volume as obtained by subtracting Eq. (2.87) from Eq.
(2.86), and neglecting second and higher order differentials, is
(2.88)
The rate of accumulation of solute in the control volume is given by (2.89)
Equating Eqs. (2.88) and (2.89) and solving for dcA/di, we get
(2.90)
From the initial assumptions, the boundary conditions are:
cA(r, 0)= cA0 cA(rs, i) = cA* cA (r, i) = bounded
Solving Eq. (2.90) by the method of separation of variables, the following expression may be obtained for the local concentration
(2.91) and for the average concentration:
(2.92) Unsteady-state diffusion in a cylinder
A section of a cylinder is shown in Figure 2.12.
Figure 2.12 Unsteady-state molecular diffusion in a cylinder.
The following assumptions are made to facilitate mathematical analysis:
(i) The plane ends of the cylinder are sealed so that diffusion takes place only in the radial direction.
(ii) The physical properties of the material for cylinder are constant.
By making a mass balance for component A entering, leaving and accumulating in an annular cylindrical volume element, we get
(2.93) The boundary conditions are:
cA(r, 0)= cA0
cA(a, i)= cA*
cA(r, i)= bounded.
By adapting the solution of the differential equation for diffusion into a fluid in plug flow through a cylindrical tube having the walls at constant solute concentration, Eq. (2.93) may be solved for local concentration as
(2.94)
where, bn’s are roots of J0(bna) = 0, and J0(br) is the Bessel function of the first kind of order zero.
The average concentration is given by
(2.95)
Solutions to Eqs. (2.85), (2.92) and (2.95) for slab, sphere and cylinder respectively are quite complex and time consuming. However, Newman (1931) had worked out graphical solutions to these
equations by plotting = which represents solute unremoved as ordinate against Fourier Number, Fo as the abscissa as shown in Figure 2.13.
The Fourier number, the dimensionless number, is expressed as and for slab, sphere and cylinder respectively.
Figure 2.13 Plot of F vs Fo for unsteady diffusion.
Newman’s work has much simplified the procedure for unsteady-state diffusion through some specific solids.
Some hints on use of the curve (Figure 2.13) depicting vs. Fourier number, Fo have been presented below:
(i) Unsteady-state diffusion from two large faces of a solid slab of thickness ‘2a’
= f(Fo)a = Fa
where, Fa is the ordinate of curve-I corresponding to (Fo)a expressed as (Di/a2)
(ii) Unsteady-state diffusion through one large face of a slab with the other large face made impermeable to the solute.
(iii) Unsteady-state diffusion in a rectangular bar of thickness 2a1 and width 2a2 with both ends sealed
F = f(Fo)a1 f′(Fo)a2 = Fa1$Fa2
Fa1 and Fa2 being ordinates of the curve-I corresponding to (Fo)a1 and (Fo)a1 being expressed as and respectively.
(iv) Unsteady-state diffusion in a brick shaped solid of thickness 2a1, width 2a2 and length 2a3 F = f(Fo)a1f′(Fo)a2 f′′(Fo)a3 = Fa1$Fa2$Fa3
Fa1, Fa2 and Fa3 being ordinates of the curve-I corresponding to (Fo)a1, (Fo)a2 and (Fo)a3 being expressed as , and respectively.
(v) Unsteady-state diffusion through a sphere of radius r F = f(Fo)r = Fr
where, Fr is the ordinate of curve II corresponding to (Fo)r expressed as (Di/r2)
(vi) Unsteady-state diffusion through a cylinder of radius a and length 2b with one end sealed F = f(Fo)cyl$f′(Fo)2b = Fcyl$F2b
where, Fcyl is the ordinate of curve-III corresponding to (Fo)cyl expressed as (Di/a2) and F2b is the ordinate of curve-I corresponding to (Fo)2b expressed as (Di/(2b)2
(vii) Unsteady-state diffusion through a cylinder of radius a and length 2b with both ends exposed F = f(Fo)cyl f′(Fo)b = FcylFb
where, Fcyl is the ordinate of curve-III corresponding to (Fo)cyl expressed as (Di/a2) and Fb is the ordinate of curve-I corresponding to (Fo)b expressed as (Di/b2).
EXAMPLE 2.18 (Diffusion from both the large faces of a solid slab): A slab of clay 40 mm thick with the four thin edges sealed, is being dried from the two flat faces by exposure to dry air. The initial moisture content is 18%. Drying takes place by internal diffusion of liquid water followed by evaporation at the surface. The surface moisture content is 2.5%. The average moisture content has fallen to 9.75% in 6 hr.
Assuming the diffusivity to be independent of moisture content and uniform in all directions, calculate (a) the diffusivity in m2/s (Assume Fo = 0.47 F)
(b) how much time will be required to reduce the average moisture content to 6% under the same drying conditions. (Assume Fo = 2.3 F)
Solution:
(a) cA0 = 0.18, = 0.0975, cA* = 0.025, a = 0.02 m, i = 6 hr = 21600 s
Figure 2.14 Schematic representation of Poiseuille flow and Knudsen flow.
Hence,
F = = 0.468
Fo = = 0.47F = 0.47 # 0.468 = 0.22
D = = 4.07 # 10-9 m2/s
(b) In this case,
F = = 0.226
Hence, Fo = = 2.3F = 2.3 # 0.226 = 0.52
or, i = = 51106 s = 14.2 hr
Time required = 14.2 hr.
EXAMPLE 2.19 (Diffusion from a porous solid sphere): Spheres of porous clay 12 mm diameter were thoroughly impregnated with an aqueous solution of sodium chloride, concentration being 0.15 g/cm3. When exposed to a running supply of fresh water at 18°C, the spheres lost 60% of their salt content in 2.25 hr.
At 18°C, the average diffusivity of NaCl in water is 1.36 # 10-9 m2/s Assume Fo = 0.1175F.
Estimate the time for removal of 80% of the dissolved solute if the spheres were impregnated with an aqueous solution of ethanol, concentration being 0.17 g/cm3 when exposed to a running supply of water containing 0.01 g/cm3 of ethanol at 18°C.
The average diffusivity of ethanol in water at 18°C is 0.53 # 10-9 m2/s.
Assume Fo = 0.933F.
Solution: For NaCl solution
cA0 = 0.15, = 0.15 # (1 - 0.6) = 0.06, cA* = 0 a = 6 mm = 0.006 m, i = 2.25 # 3600 s = 8100 s
Hence, F = = 0.40,
or, Fo = = 0.1175F = 0.1175 # 0.40 = 0.047
Def = = 2.09 # 10-10 m2/s
= = 6.507
For ethanol solution:
Deff = = 8.15 # 10-11 m2/s
cA0 = 0.17, = 0.17 # (1 - 0.8) = 0.034, cA* = 0.01, a = 6 mm = 0.006 m
whence, F = = 0.15
Fo = Deff = 0.933F = (0.933 # 0.15) = 0.14
whence, i = = 61840 s = 17.18 hr
Time required for removal of 80% solute = 17.18 hr