Diffusion and Transport 9. Diffusion In isolated room temperature fluids, we observe that concentration and temperature gradients
disappear with time, and the properties of the system become spatially uniform. Diffusion refers
to the transport of mass and energy that leads toward equilibrium, and as we will see it is closely
related to Brownian motion. Thermodynamically, we consider the driving force for diffusion as a
gradient in the free energy or chemical potential of the system.
Continuum Diffusion Let’s discuss the basic concepts governing diffusive transport. These are formalized in two laws
that were written by Adolf Fick (1855). In this case, we will describe the time evolution of
spatially varying concentration distributions as they evolve toward equilibrium.
Fick’s First Law This is the “common sense law” that is in line with everyone’s physical intuition. The flux of
molecules through a surface, J, is proportional to the concentration gradient across that surface.
J D
C
x
(1)
The flux has units of concentration or number density per unit area and time. The proportionality
constant between flux J (mol m-2 s-1) and concentration gradient (mol m-4) is the diffusion
constant D (m2 s-1). The negative sign assures that the flux points in the direction of decreasing
concentration. This relationship follows naturally, when we look
at the two concentration gradients in the figure. Both C and C
have a negative gradient that will lead to a flux in the positive
direction. C will give a bigger flux than C because there is more
probability for flow to right. The gradient disappears and the
concentration distribution becomes constant and time invariant at
equilibrium. Note, in a general sense, C / x can be considered
the leading term in an expansion of C in x.
Fick’s Second Law Fick’s second law is based on the first law, and adds an additional
constraint based on the conservation of mass, resulting in an
expression purely in terms of spatial concentration profiles. Let’s
consider a flow along x for a pipe with cross-sectional area a. In a
time Δt you sweep out a distance Δx, corresponding to a volume V =
a Δx. The concentration in this volume is a ∆C ∆x, where ΔC is the
concentration difference {C(x+Δx) ‒ C(x)}. For every increment in
space there is a corresponding increment in time governed by the
flux, so we can write a ∆C ∆x = ‒{J(x+Δx) ‒ J(x)} a ∆t, or
Andrei Tokmakoff 6/21/2016 C
J
t
x
(2)
This important relationship is known as a continuity expression. Substituting eq. (1) into this
expression leads to
C
2C
D 2
t
x
(3)
This is the diffusion equation in one dimension, and in three dimensions:
C
D2C
t
(4)
If the diffusion constant is a function of space: C DC .1 Equation (4) can be used to solve
diffusive transport problems in a variety of problems, choosing the appropriate coordinate
system and applying the specific boundary conditions for the problem of interest.
As our first example of how concentration distributions evolve diffusively, we consider
the time-dependent concentration profile when the concentration is initially all localized to one
point in space, x = 0. The initial condition is
C ( x, t 0) C0 ( x)
and the solution to eq. (3) is
C ( x, t )
2
C0
e x 4 Dt
4 Dt
(5)
The concentration profile has a Gaussian form which is centered on the origin with the mean
square displacement broadening with time as:
x 2 2 Dt
1.
In three dimensions, Fick’s First Law and the continuity expression are: J(r,t) vC(r,t) DC(r,t) and
dC(r,t) / dt J(r,t) where v is the velocity of the fluid.
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Diffusive transport has no preferred direction. Concentration profiles spread evenly in the
positive and negative direction, and the highest concentration observed will always be at the
origin. We sill shortly relate this concentration profile to a process by which each molecule
executes a random walk.
When we solve for 3D diffusion from a point source:
C ( x, y, z, t 0) C0 ( x) ( y) ( z)
the radial concentration distribution is
C (r , t ) 4 r 2
C0
4 Dt
3/2
e r
2
4 Dt
(6)
and the mean square radial displacement is. r 2 6 Dt . In d dimensions, r 2 2d Dt . For
diffusive behavior we look for rrms ~ t . If rrms ~ t v , and 12 , we refer to the behavior as
anomalous diffusion.
Typical Numbers for Diffusion of a Protein Typical diffusion constants for biological molecules in water are shown in the graph below. For a
typical protein,
in water
in cells
in lipids
r2
1/ 2
D ~ 10–10 m2/s
D ~ 10–12 m2/s
D ~ 10–14 m2/s
1 µm, t ~ 0.4 sec in cells
10 µm, t ~ 40 sec in cells
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Solving the Diffusion Equation Solutions to the diffusion equation, such as eq. (5) and (6), are commonly solved with the use of
Fourier transforms. If we define the transformation from real space to reciprocal space as
C k , t
C ( x) eikx dx
one can express the diffusion equation in 1D as
dC k , t
Dk 2C k , t
dt
[More generally one finds that the Fourier transform of a differential equation in x can be
expressed as in polynomial form: F ( n f / x n ) (ik ) n f (k ) ]. This manipulation converts a partial
differential equation into an ordinary one, which has the straightforward solution
C k , t C k , 0 exp Dk 2t . We do need to express the boundary conditions in reciprocal
space, but then, this solution can be transformed back to obtain the real space solution using
C ( x, t ) (2 ) 1 C k , t e ikx dk .
When the diffusion constant in independent of x and t, the general solution to the
diffusion equation can also be expressed as a Fourier series. If we separate the time and space
variables, so that the form of the solution is C x, t X x T t we find that we can write
1 T 1 2 x
2
2
DT t x x
Where α is a constant. Then T e
form:
2
Dt
and x A cos x B sin x . This leads to the general
C ( x, t ) An cos n t Bn sin n t e n Dt
2
n 0
(7)
Steady‐State Solutions One class of diffusion equation we will encounter often is the steady-state solution, in which the
concentration gradient does not change with time, C / t 0 . For instance, we will see it in the
diffusion-to-encounter problem where we assume a steady-state flux of molecules to an
absorbing sphere. Under those conditions, the diffusion eq. (4) simplifies to 2C 0. Solutions
to this equation with specific boundary conditions are then related to the flex using Ficks first
law, eq. (1).
Examples 1) Diffusion across interface Integrate (x – x0) solution from x0 ∞ (over box). Defining y 2 ( x x0 ) 2 / 4 Dt ,
4
erfc( x) 1 erf ( x)
C ( x, t )
C0
( x x0 )
2
x
2
e y dy
dy e y
2
4 Dt
C0
( x x0 )
erfc
2
4 Dt
2) Diffusion into “hole” A concentration “hole” of width 2a is inserted into a
box of length 2L. Let’s take L = 2a.
Concentration profile solution:
const
space(x) time(t)
L a
n2 Dt
C x, t C0
An cos n x e
L n1
An
2sin n a
n
n
n
L
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3) Fluorescence Recovery after Photobleaching (FRAP) This solution to this problem is used to describe the diffusion of fluorescently labeled
molecules into a photobleached spot. Usually observe the increase of fluorescence with
time from this spot.
a
N FRAP (t ) C ( x, t )dx
a
2a
C0 ( L 1) L An2e n Dt
n 1
L
integrate over spot
Diffusion through a Membrane Steady-state solution to the diffusion equation in one dimension can be used to describe small
molecule (i.e., O2) diffusion through a cell plasma membrane. The membrane resists flow more
than the fluid.
Membrane thickness, h; concentrations
of small molecule in the fluid on left and right
side of membrane, Cl and Cr; The partition
coefficient between membrane and fluid:
K
C membrane
Cfluid
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Applying steady-state approximation for the diffusion equation inside the membrane:
2C / x2 0 . Solutions will take the form C( x) A1 x A2 . Applying boundary conditions for
the concentration of small molecule in the membrane at the two boundaries, we find
A1
K Cr Cl
h
A2 KCl
Then we can write the flux of the small molecule across the membrane as
J Dmol
KD C
C KDmol
C Cr mol
h
h
x
J
KDmol
. Membrane resistance to flow R = 1/P. The
C
h
rate of transport dn/dt = J a, where a is area. We can also treat this as diffusion through a solid
membrane with holes.
The membrane permeability is: P
J D
One pore
n pores
b
C
2r 2
dN
Ja bD C
dt
dN
n aDC
dt
Not proportional to pore area, just pore radius!
Diffusion with Drift If there is an external force acting on a diffusing system (for instance, electrophoresis and
sedimentation), or if the diffusion occurs within a moving fluid, the time-dependent
concentration profiles will be influenced by the local velocity of the fluid. Considering this
diffusion with drift in one-dimension, if the fluid is moving with a velocity vx, there is an
additional flux term in Fick’s first law proportional to vx. vx is often referred to as the drift
velocity. Equation (1) becomes
J D
C
vx C x
Now using the continuity expression, eq. (2), and assuming a constant drift velocity
C
2C
C
D 2 vx
t
x
x
(8)
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This equation is the same as the normal diffusion equation in the inertial frame of reference. If
we shift to a frame moving at vx, we can define the relative displacement
x x – vx t Remember C is a function of x and t, and expressing eq. (8) in terms of x via chain rule,
we find that we can recast it as the simple diffusion equation:
C
2C
D 2 t
x
Then the solution for diffusion from a point source becomes
C x,t
C x, t
2
1
e x 4 Dt 4Dt
2
1
e ( x vx t )
4Dt
4 Dt
So the peak of the distribution moves as x =vxt and the variance in the width grows as x2
2Dt.
1/2
=
Fokker–Planck Equation We will see shortly, that the drift velocity and the external force are related by the friction
coefficient f ext vx , with k BT / D . If the concentration is instead expressed in terms of a
probability density P( x, t ) , eq. (8) becomes
J
2 P f x
P
D 2
x
x
x
J P
With the continuity expression x t , we obtain
P
2P f x
D 2
P
t
x
x
This is known as a Fokker–Planck equation.
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