Diffusion processes
Introduction
Typical scenario of catastrophic pollution
Accidental release of contaminant
Transport of conatminant by underground water, streams, rivers, lake or
ocean currents, atmosphere
Diliution of contaminant by water or air (diffusion)
Mechanism of contaminant transport and dilution crucial for
understanding and control of polution phenomena
Management of environmental pollution
resources
industry
Wastes:
goods
Accidental realease of
contaminants:
Transformation
Accumulation
Dispersion in
environment
Transport
Environmental Quality
Effects on
health
Effects on
climate
Effects on
ecosystem
Possible strategies for pollution control
Control sources of contaminants
Prevention – the most recommended
Waste treatment
Less recommended choice, only if
unavoidable
Dispersion in environment
Impact of waste storage
Accidental release of contaminants
Technical necessity e.g. heat, flue gas emission
Related problems:
Conservation of mass
Point sources/distributed sources
Continuous/accidental release
Basic definitions
Concentration:
Mass M of contaminant in volume V gives concentration
c  lim
M
V
V 0
in general:
Relative concentration:
Averages:
Ensemble average:
Time average:
Space average:
 kg 
 m3 
 

c  c x , t 
(2)
%;  ppm
c
 of solution

c x, t 

1
c  x , t0  
T

1
cV x, t  
V
(1)
t 0 T
 c dt
(3)
(4)
(5)
t0
 c dV
V
(6)
Basic definitions
cf 
Flux average:
Where:
1
1
cU
dA

q dA


Q A
Q A
(7)
Q   U dA - volumeflux oversurface A
A
 kg 
q  2   specificmass flux
m s
Dilution:
1
p
where : p relative volume concentration
S
p
volumeof contaminant
totalvolume
(8)
(9)
Mass conservation equation for contaminant
Conservation of mass of conatminant in volume V with no sources within the
volume:

c dV

t V
 
  q  n dA  0
chanegof mass
cont ainedwit hin
volume
A
mass flux
(10)
Mass conservation equation for contaminant
Due to Gauss-Ostrogradski transformation
 c  
qi  dV  0
 


t

x
i

V 
(11)
Since V is an arbitrary volume:
c 

qi  0
t xi
(12)
Flux of contaminant:
qi  U i c  qDi
advection
(13)
diffusion
Substitution of Eq.(13) into Eq. (12) leads to
c 
U i c     qDi 

t xi
xi
(14)
Mass conservation equation for contaminant
If fluid is at rest:
Ui  0
c

qDi 

t
xi
(15)
U i
0
xi
If fluid is incompressible:
c
c

qDi 
 Ui

t
xi
xi
(16)
If flow is turbulent (Reynolds decomposition):
U i  U i  ui
(17)
C C c
U i c  U i C  ui c
Then mass conservation equation becomes:

C
C

 Ui

qDi  ui c
t
xi
xi
Molecular diffusion

(18)
Turbulent flux – closure problem
Recommended strategies for turbulent transport analysis
Basic mechanisms involved into transport phenomena:
• Advection by flow
• Naturl convection
• Evaporation/condensation
• Entrainment/deposition
• Molecular/turbulent diffusion
• Disperision in shear flows
Separation of dominant mechanisms
Splitting of the area of analysis into subdomians
Recommended strategies for turbulent transport analysis
Experimental versus computer modeling
Type of analysis
accuracy
Time required
Cost
In-situ ivestigations
(full scale)
+
--
--
Experimental (model)
investigations
+-
--
-
Computer modeling
+-
++
+
Dimensional analysis
Order of magnitude
+++
+++
Conclusions:
•Numerical modeling is nowadays the most recommended, however keep in mind
that: garbage
computer
garbage
•Accuracy depends on physical features assumptions but their validity is often
doubtful e.g. diffusivity in ocean varies in the range 10-9 (molecular) to 105 m2/s
(turbulent)
•Dimensional analysis often useful in rough approximations
Molecular diffusion
Molecular diffuison plays minor role in environmental transport processes but is a basis
for understanding other types of diffusion
According to Fourier’s (1822) law of heat transfer
Fick’s law-thermal analogy
T
heat flux   k
x
where : k  thermaldiffusivity coeffient
(19)
Fick’s (1855) analogy for mass flux
qi   D
c
xi
where :
q  mass flux per unit surface and unit t ime
D- diffusion const ant[L2 /T ]
Note on the units:
 m2 
 kg 
 kg 
q  2 ; D  ; c  3 
m s
m 
 s 
(20)
Fick’s law
General form of diffusion equation:
gradient of transported
 

quantity


i.e. it is gradient/flux relationship for mass ( D), momentum( )
or heat (k ) diffusion process
 flux
D - physical constant depending on the properties of fluid and contaminant
Example:
For water:
D~10-9
[m2/s]
Schmidt number Sc 

ν ~ 10-6 [m2/s]
for water
D
Sc  103
k ~ 10-5 [m2/s]
for air
Sc  1
For air:
D ~ ν ~k ~10-5 [m2/s]
A simple model for gradient/flux relationship
Suppose a 1D pipe with concentration gradient in x-direction, two boxes
created in a pipe i.e. left (L) and right (R) containing different numbers of
molecules NL=10 and NR=20
L
R
Δx
Δx
L
Δx
R
ΔNL
Δx
ΔNR
(A)
(B)
Concentration of molecules in the initial (A) and consecutive moment (B) for
1D diffusion process
A simple model for gradient/flux relationship
If each molecule has propability p to pass from one box to the another,
then if (say) p=1/5 after time t+Δt, because
1
1
N R  20  4; N L  10  2
5
5
in a new timestep
N L  10  2  4  12; N R  20  4  2  18
After each time step
N L( n1)  N L( n) ; N R( n1)  N R( n)
We obtain
N 
n1
R
 
 N Ln1  N Rn   N Ln 

And the process continues until the uniform distribution of the molecules is
approached – the concentration difference dimnishes with time and the process
tends toward uniform concentration
Analytical description of gradient/flux relationship
Each molecule has mass m
M L  N L m; M R  N R m
The mass flux during time Δt in positive x-direction is
mass

  m p N L  m pNR  p M L  M R   negativemass flux
 flux 
Concentration in each box:
cL 
ML
;
1 x
cR
MR
1 x
Mass flux per unit area and unit time

x
x 2  c x  2c
 

qx  p cL  cR    p


2
t
t  x 2 x

Analytical description of gradient/flux relationship
Mass transfer should not be dependent on the size of the box
 x 2 
  const  D diffusivity 
lim  p
x 0 

t

t 0 
And hence neglecting higher order terms
qx   D
c
x
We obtained a simple 1D model of Fick’s law
Conclusions:
•Boxes must be large enough, so that we can apply probability analysis
N L ; N R  1
•Boxes must be small enough to apply first order approximation in Taylor series
 2c
c
x 2 
x
x
•Both the above conditions are always satisfied in normal conditions for
molecular diffusion because the mass of a single molecule is small an their
number is very large even in small volumes
Similarity solutions and properties of the 1D diffusion equation
Assume 1D diffusion in fluid at rest:
c
 2c
D 2
t
x
(21)
And uniform concentration along y and z axes
c c

0
y z
Coordinates for 1D diffusion problem
(22)
Similarity solutions and properties of the 1D diffusion equation
Total mass of the contaminant:

M   c dV  A  c dx
V

(23)
That means c will be sought in the form
c
M
1
M 1



A Length A L
(24)
Let us introduce:
L  referencelengthscale
T - referencetimescale
Dimensional analysis yields:
c
c
D 2
T
L
 L2  DT
(25)
Time scale and the length scale in the diffusion process are naturally related
Similarity solutions and properties of the 1D diffusion equation
Example:
Consider a diffusion process after initial injection of the the contaminant.
The time scale is time t elapsed after injection.
After substitution into Eq.(25) the concentration can be evaluated as:
c
M 1

A Dt
(26)
i.e. concentration may be obtained for a similarity solution
c
M 1

f  x, t 
A
Dt
(27)
The only possibility for non-dimensional function f is:




m
 x  

f  f
 
1
2
 Dt    m 2  
s
  s  
(28)
That finally yields:
c
M 1

f  ;  
A
Dt
x
Dt
(29)
Analytical solution of 1D diffusion equation for initial impulse injection of
contaminant
Introducing Eq.(29) into the r.h.s of diffusion equation:
df

f 
;
d
d2 f
f  
d 2
 2c M 1


f 
32
2
x
A Dt 
(30)
and introducing Eq.(29) into the l.h.s. of the diffusion equation
c M  1
1
1  

f
f 

3
t
A  2 D t
Dt 2t 
(31)
Hence the diffusion equation takes the following form

1
 f   f   f 
2
(32)
And finally:
 f   2 f 
(33)
Note that similarity solution allowed to transform the p.d.e into the ordinary
differential equation
Analytical solution of 1D diffusion equation for initial impulse injection of
contaminant
After integration of Eq.(33) we obtain:
 f  2 f   C
(34)
Determination of the constant C
  x    odd
C(x,t)
C(x,t)
(b)
(a)
Initial injection at t=0
At t >0
x
x
Expected evolution of the concentration. Concentration of contaminant at the
initial injection (a) and after time elapsed t (b)
c  even
Analytical solution of 1D diffusion equation for initial impulse injection of
contaminant
c f

f  even 
f   odd    f  odd even  odd
And due to symmetry requirements
C 0
(35)
And finally:
f

2
(36)
f
That yields the solution:
 2 
M
f  C1 exp    cx, t   C1
A
 4
 x2 
1

exp 
Dt
 4 Dt 
Note that C1 the only unknown quantity
(37)
Analytical solution of 1D diffusion equation for initial impulse injection of
contaminant
From the contaminant mass conservation equation Eq. (23)

 x 2  dx

C1  exp  
1
4
Dt
Dt



(38)
From tables of definite integrals we find



2 2
exp

a
x dx 



a
(39)
Substitution into eqaution (38) leads to:
C1 
1
4
(40)
And finally we obtain the equation for concentration evolution due to molecular
diffusion
M
cx, t  
A
 x2 
1

exp 
4
Dt
4Dt


(41)
Statistical measures of concentration distribution establidhed due
to diffusion processs
First order moment – expected value:

[E] 
 x cx dx

x  odd, c  even,
x  c   odd,
i.e. :

E    x cx dx  0

Second order moment – variance:
A
 
M
2


 x2 
x cxdx  4 Dt exp  4Dt dx
1
2
Having known that the following definite integral is:



2
2 2
 x exp  a x dx 


2a
; and a 
3
1
2 Dt
Variance can be expressed as:
 2  2Dt
Variance is a measure of the contaminant displacement from the
initial positiion i.e. is a measure of teh size of the cloud
Diffusion and random walk model
Definition of radnom walk process:
Step ξ of the length λ randomly to the right or to the left (drunkard’s walk
   t   
(42)
Probablity density function (pdf) of the random walk is:
1
p        
2
∞
(43)
∞
p(ξ)
-λ
Δ – Dirac delta
function
+λ
ξ
Diffusion and random walk model
Consider n successive and independent (i.e. no memory of previous step) ξi steps;
i=1…n and the random walk variable X(t):
X t   1  2   n
(44)
Where: X(t) is the particle position after n steps
Problem: what is the pdf of the process X(t)?
Central Limit Theorem (CLT):
The probability density function of the sum of n independent random
variables tends towards the normal distribution with variance equal to the
variance of the sum whatever the individual distributions of these variables
provided n tends towards infinity
Pdf of normal (Gaussian) distribution process is:
 x2 
1
px  
exp  2 
 2
 2 
(45)
Diffusion and random walk model
Variance of the radnom walk process:
 2  1   2  i     n 2  12   22   i2    n2  2i j
(46)
Since the steps are independent:
(47)
 i j  0
Then for random walk process:
 2  n2
(48)
If τ is teh time required for every step and V is velocity such that:
  V 
(49)
Then the time t to make n steps is:
t  n 
(50)
Adn the variance of random walk process:
 2  n 2  nV  
2
2
t2
 t
 n V   V   V  t
n
 n
(51)
Diffusion and random walk model
 2   V  t
(52)
Close analogy of random walk process to molecular diffusion with λ analogous to mean
free path and V analogous to molecular agitation (≈ temperature), which finally leads to
the expression for diffusivity:
 V   D
Example: N molecules of mass m injected into infinitely thin layer
(53)
Diffusion and random walk model
Total mass of injected molecules:
N  m  M0
(54)
If the molecules move according to random walk due to collisions then
according to Central Limit Theorem probability p(x)dx that one given molecule
is in the slice (x,x+dx) is the Gaussian pdf and the number of molecules in
slice (x,x+dx) yields
N  px dx
(55)
Mass of the contaminant in volume Adx:
N m pxdx
(56)
Concentration of the contaminant:
cx, t  
m ass Nm px dx M 0


p x 
volum e
Adx
A
(57)
Since the pdf is Gaussian then the concentration reads:
 x2 
M0 1
cx, t  
exp  2 , where  2   V t
A  2
 2 
(58)
Diffusion and random walk model
If diffusion process is described by Fick’s law:
c
 2c
D 2
t
x
the solution for initial mass M0 injected at x=0:
 x2 
M0 1
c  x, t  
exp  2 , where  2  2 Dt
A  2
 2 
From comparison of Fick’s law and random walk:
D
1
V
2
Conclusion:
Any random walk process produces the diffusion with
diffusivity
1
V
2
Diffusion and random walk model
Example: Diffusion 1024 particles injected at x=0 and t=0 and then
dispersed due to random walk
Illustration of particles diffusion by random walk process
Diffusion and random walk model
In order to smooth the distribution let us calculate the average distribution for the
intermediate step n=6.5 and compare it with the Gaussian distribution, i.e.:
 x2 
1
N  N0
exp  2 
 2
 2 
for   1,   1   2  n
n  6.5; N 0
(59)
 x2 
N  X , n  6.5  160.2 exp  
 13 
(60)
x
-7
-6
-5
-4
-3
-2
-1
0
1
2
Random
walk
4
8
28
48
84
128
140
160
140
128
Gaussian
process
3.7
23.4 46.8 80.1
117.8
148.3 160.2 148.3
117.8
10.1
Conclusion: after only 6.5 steps the distribution obtained from random
walk is already close to the Gaussian process
Brownian motion
R. Brown (1826) – observation of motion of small particles
A. Einstein (1905) – rigorous explanation
practical application: motion of aerosol particles
Starting point for analysis: macroscopic particle suspended in fluid,
radius a ≈ 1 μm, particle subjected to molecular agitation i.e. collisions with molecules of
fluid (much smaller than suspended particle)
Numbers of collisions do not balance
Finite probability to have more
impulses on one side
Particle will move
Creation of impulse in brownian
motion of macroscopic particles
Brownian motion
Suppose that the particle is spherical and obeys Stokes law:

drag force  6   r U
where : r  radius of part icle
  viscosit y
(59)

U  part iclevelocit y
Equation of motion for particle:

 
dU
m
 6   rU  F
dt
Inertia force
(60
Random impulse (+ or -)
Form fluid molecules
Due to random impulse the motion of particle as in random walk
Brownian motion
Problem: what is the diffusivity of this process?
D
1
 V
2
To evaluate the diffusivity of random walk we need two scales
  linear scale
V  velocityscale


 or 


  timescale
V  velocityscale
From dimensional analysis the time scale can be estimated:
accelerati on - a 

F 6   rU

m
m
 
U
m U
m


a 6   rU 6   r
U
a
(61)
Brownian motion
Between the impulses the motion equation is homogeneous:
dU
dU
6  r
dt
 6  rU 

dt  
dt
U
m

t
 t
ln U    ln U 0  U t   U 0 exp  

 
m
(62)
Velocity scale U0 is determined by the equipartition of energy
1
m U 02  k T
2
where : k  1.3810 23 J / K
(63)
(Boltzmanconstant)
Hence the velocity scale is:
V 2  U 02 
2kT
m
(64)
And diffiusivity due to brownian motion can be evaluated as:
Dbm  V 2  
2kT
m
1 kT


m 6  r 3   r
(65)
Brownian motion
Example: diffisuivity for spherical particles in air
R [μm]
0.1
Dbm [m2/s]
1.3x10-10
1
1.3x10-11
10
1.3x10-12
Note: compare these values with kinematic viscosity for air
equal ν= 1.5x10-5 [m2/s]
i.e. diffusiivity of brownian motion is small compared with
diffusivity of air
Dispersion by turbulent motion
Growth rate of the contaminant cloud due to moelcular diffusion:
 2  2 Dt 
d D

dt 
Conclusion:
the larger the cloud
The smaller growth rate:
(66)

d

dt
Extemely large time scales needed to achieve efficinet mixung by moldeular
diffusion, for example in water:


D

1m
 10 9 s  10 4 days
2
m
10 9
s
Note: time constant for reaction of human nerves –diffusion of ions through the
cellular membrane of the thickness h=10-6 m, τ=1ms
Dispersion by turbulent motion
How to intensify diffusion ?
A possible way to speed up diffusion
Break up the cloud into smaller fragments
If the size of the cloud is σ2, then the diffusion time scale is: 2
 
 
n
t0 
, if t hecloud is split int o n smalleerpart st hen t het imescale is t n   
D
D
2
t 1
and t hereduct ionof t hediffuison t imescale is : n   
t0  n 
2
Dispersion by turbulent motion
What can be the role of turbulence in mixing enhancement ?
Turbulence -> superposition of eddies
Case 1: the size of the turbulent eddy L is much larger than the size of the cloud σ
Cloud transported by turbulent eddy without any change of shape
No speed-up of the molecular diffusion
Dispersion by turbulent motion
Case 2:
L 
Cloud of contaminant distorted by turbulent eddy (random)
Speed-up of the molecular diffusion by break-up of the mixing volume
Dispersion by turbulent motion
Case 3:
L  
Small turbulent eddies distributed within the cloud of contaminant
Molecular mixing augmented by turbulent eddies
Dispersion by turbulent motion
Possible mechanisms of turbulence interaction with molecular diffusion
Vortex stretching
Full range of eddies (from largest to smallest) appears in turbulent flow
Even if initially only largest eddies exist (case 1) after a while eddies of sizes corresponding
to cases 2 and 3 will be developed
Vortex stretching
Vortex stretching
Steeper contration gradients inside the vortex
Since {flux} ~ {concentration gradients}
Enhanced diffusion in radial direction
Summary of turbulent diffusion
•Turbulence cannot directly enhance mixing and homogenisation at
molecular level because fo disparity of scales
smallest scales

molecular

  

of turbulenteddies
scales

•Turbulence can make molecular mixing more efficient by
Reducing the local size of the contaminant volume cases 2+3
Making local concentration gradients steeper thus enhamcing
the diffusive mass flux