Alternative interpretations of the diffusion process Diffusion flux as

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					BENG 221, Fall 2009
M. Intaglietta
Lecture 13

Alternative interpretations of the diffusion process

Diffusion flux as velocity times concentration

According to the first law of diffusion:

         ∂c
J = −D      and for c = co − sx J = Ds
         ∂x

For the boundary condition c = c0 @ x = 0 and c = c1 @ x = L

     c0 − c1
s=
        L

                         c1
If c0 = 0 then c( x) =      x and the flux is in the negative x-axis.
                         L

Note that J is defined as the rate of passage of N particles per unit time through a unit
area A or:

     1 dN
J=               dN = c dVol = c d ( A × x) = c Adx
     A dt

Therefore for A = 1:

     1 c( x) Adx          dx
J=               = c ( x ) = c ( x )v ( x )
     A     dt             dt

Note that according to this treatment the particle velocity changes, and in fact accelerates
as the diffusion process takes place, since for the linear diffusion case derived:

     c1                             JL 1
J=      xv( x) = constant ∴ v( x) =
     L                              c1 x


The thermodynamic interpretation of diffusion

        Derivation of the mechanics of the diffusion process implicitly introduces the
concept that the particles undergoing the random walk process are subjected to a
“viscous” retarding force. This is a fundamental observation, since it implies the motion
of the moving particles relative to a medium that exerts the viscous drag. In this context


                                                                                             1
thermodynamic analysis defines as “diffusion” the relative motion of particles with
respect to a suspending medium. Absence of relative motion places the phenomena in
the realm of convective processes and the evolution of systems in which solutes and
solvents do not separate, each volume composed of a variety of species moving without
causing differences in species concentration.

        As a consequence of viscosity, relative motion requires the expenditure of energy.
This can occur in several modes. In spontaneously occurring phenomena such as the
diffusion of a species from a high to low concentration along the concentration gradient,
the energy expended is intrinsic to that due to the presence of differences of concentration
in a region.

       Membranes can be used to separate solutes from solutions by the application of a
pressure difference on the solution across the membrane. The ensuing separation
between solute and solvent is a diffusion process, caused by a mechanical effect, the
applied pressure difference.

Application of Gauss’s theorem in deriving the diffusion equation

       The first law of diffusion, also known as Fick’s law of diffusion, identical to
Fourier’s law of conduction is an empirical law:

J = − D∇C                                                                                1

Where the bold character denotes a vector quantity. Conservation of mass states that the
rate of change of mass inside an arbitrary control volume is equal to the amount leaving
this volume, which occurs through its defining surface, therefore:

d
dt ∫
     CdV = − ∫ J ⋅ n dA


Where n is a unit vector perpendicular to the surface A. Using 1 and Gauss’s theorem:

  ∂C
∫ ∂T dV = − ∫ J ⋅ ndA = − ∫ ∇ ⋅ J ⋅ n dV = ∫ ∇ ⋅ D∇CdV
or:

 ⎛ ∂C             ⎞       ⎛ ∂C          ⎞
∫ ⎜ ∂T
  ⎝
         = ∇ ⋅∇DC ⎟dV = ∫ ⎜
                  ⎠       ⎝ ∂T
                               = ∇ 2 DC ⎟ dV
                                        ⎠

Which for constant D reduces to:

 ∂C
    = D∇ 2C
 ∂t


                                                                                             2
  Non-dimensionalization of the diffusion equation

         Given the diffusion equation in one dimension (4) over a one dimensional region
of total length L we introduce a non-dimensional spatial coordinate as the ratio x = x’L
so that the diffusion equation becomes:

∂C      ∂ 2C     D ∂ 2C
   =D           = 2
∂t    ∂ ( Lx′) 2 L ∂x′2

We can also define a non-dimensional time as t = t’t0 where t0 is an arbitrary time scale;
the new equation is:

  ∂C      D ∂ 2C              ∂C     Dt ∂ 2C
         = 2          ∴             = 20
∂ (t0t ′) L ∂x′2             ∂ (t ′) L ∂x′2

Therefore setting t0 = L2/D the diffusion equation becomes:

 ∂C ∂ 2C
     =
 ∂t ′ ∂x′2

        This result indicates that all diffusion problems are the same. This requires
scaling the geometry so that the basic dimension ranges from zero to one. The
combination of the size and the diffusivity yield the appropriate time unit. On the scaled
domain and in the proper time units, problems of different size and diffusion constants
will have the same solution. Thus solving the diffusion equation for one set of boundary
conditions solves it for all cases. As an example the time that it takes for diffusion to
change concentration by a given amount is directly proportional to the size of its principal
dimension. Thus doubling its size quadruples the time.


The diffusion coefficient

        The rms distance achieved by a particle subjected to Brownian motion along the x-axis
which from (14) is given by:

          2kT
x rms =       t
           f

Since concentration in a semi infinite system that is changing composition by diffusion
changes so that the distance at which c = co/2 increases in proportion to:




                                                                                                3
x1 = 0.96 Dt                                                                                       27
     co
 2


It appears that diffusion and Brownian motion are different aspects of the same phenomenon,
and that the diffusion coefficient, which is an experimentally found quantity is related to:

           kT
D=B                                                                                                28
            f

where B is numerical factor related to the fact that diffusion deals with the flux of a very large
number of particles, while Brownian motion refers to only one particle. This factor is shown to
be equal to one, and therefore using (1) the diffusion coefficient for a spherical particle in a liquid
is:

           kT         RT 1
D=                =                                                                                 29
          6πµ R       N 6πµ R

R is the gas constant and N is Avogadro’s number.

indicating that the diffusion coefficient is the ratio of kinetic vs. viscous effects.
Furthermore, for a particle the rms displacement from its origin is given by (27), namely:

xrms = 2 Dt                                                                                          30

Note that:

    2
   xrms           1 xrms       1
D=                       xrms = cλ
    2t            2 t          2

Where c is the average velocity and λ is the mean free path.

Diffusion and chemical reactions

        The presence of either a source or sink of material go(x,t) in the region described by
equation (20) is accounted for by adding this term, namely:

∂c    ∂ 2c
   = D 2 + g o ( x, t )                                                                           31
∂t    ∂x

         Equation (31) is useful to analyze the diffusion of oxygen in tissues where metabolism
causes the consumption of the material at a rate -go. A more accurate representation of the events
taking place in tissue is to make the consumption a function of the concentration of the diffusing
species, which may be expressed as -go(x,c).

Diffusion and variable diffusion constant



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The equation:

∂c         ∂ 2c
   = D (t ) 2
∂t         ∂x

I solved in general by making the change of variable:

dT = D ( t ) dt

Which reduces the one dimensional diffusion equation to:

∂c ∂T ∂c              ∂ 2c               ∂c ∂ 2 c
     =   D(t ) = D(t ) 2           or      =
∂T ∂t ∂T              ∂x                 ∂T ∂x 2

Where:

      t
T = ∫ D(t ')dt '
      0


For D a function of c(x), and therefore a function of x set:

   1 −1
η = xt 2
   2

Then the differential equation:

∂c ∂      ∂c
  = D( x)
∂t ∂x     ∂x

Has the differentials:

  ∂c    1 − 2 ∂c               ∂c 1 − 1 ∂c
                   3

     = − xt              and     = t 2
  ∂t    4     ∂η               ∂x 2     ∂η

Therefore:

∂        ∂c ∂ ⎛ D − 1 ∂c ⎞ 1 ∂ ⎛ ∂c ⎞
   D ( x) = ⎜ t 2        ⎟=      ⎜D   ⎟
∂x       ∂x ∂x ⎝ 2    ∂η ⎠ 4t ∂η ⎝ ∂η ⎠

Which reduces the original equation to the ordinary differential equation:

      dc   d ⎛       dc ⎞
−2η      =   ⎜ D(η )    ⎟
      dη dη ⎝        dη ⎠




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