Diffusion
Chapter 9 (old book)
General process of flow
Heat as the example
Fick’s First law
Use of diffusion Coefficients
personal passive samplers,
SO2 accommodation coef.
Particles
Probability distribution using random movement
Diffusion distances
Fick’s Second Law
Diffusion in a GC column
Diffusion in a sphere
Estimating Diffusion Coefficients
gases
water; liquids
Turbulent Diffusion
Lake system
Atmospheric System
1
Heat
Templow2
Temphigh1
x
The heat that flows thru a slab of material is proportional to the
cross sectional area, A, of the slab and the time, t, for a given
Temp
Heat flow is also  to Temp/x for a given A and time if
Temp/x is small
Δq
ΔT
A
Δt
Δx
if we think about really small thickness of x
dq
dT
  kA
dt
dx
 dq/dt = the rate of heat transfer with respect
to time
 dT/dx = the temperature gradient
 k = thermal conductivity
2
dq
dT
  kA
dt
dx
k has the units of
Kcal / sec
meter 2 oC / meter
material
k
Al
Steel
Pb
air
glass
4.9x10-2
1.1x10-2
8.3x10-3
5.7x10-6
2.0x10-4
Temphigh1
Templow2
L
at steady state for a const. temp gradient across
the rod
Δq
T T
 k 1 2
tA
L
3
Diffusion
1b. cool
1a. hot
2b. low voltage
2a. high voltage
3b. low mass
3a. high mass
4b. low pressure
4a. high hydrostatic pressure
 flux = flow area-1 time-1
 a gradient drives the flow
4
page 184 table 9.1
Page 184 Table 9.1
5
Let us consider a gas diffusing in into a zone where it is constantly
collected or removed
O3
O3 O3
O3
O3
O3
[O3]
x
FluxO   D
3
inlet
x
O3 
 (moles O3 )
=
;
 (time )  area
x
(moles O 3 )
 [O 3 ]
D
( time)area
x
If we measure the # of moles of O3 collected over a period of
time; know the diffusion coef. for O3 in air, [O3] can be
calculated
6
Diffusion and sticking coefficients
the average speed of gas molecules is given by
c
8RT
Mw
The rate of collisions per unit time with a wall of surface A in a
given volume is
rate =
1/4 c x area x concgas
rate 
RT
A  Cgas
2 Mw
rate/A = # molecules time-1 area-1 = flux
if we think about the # of effective collisions, i.e. the ones that
actually stick to the wall, a factor is introduced called
“sticking” factor
surface recombination
accommodation coefficient
removal rate 
RT
A  Cgas  
2 Mw
7
Judeikis et al. were interested in the effectiveness of coal surfaces
in the uptake of SO2 gas.
rate/A = # molecules time-1 area-1 = flux
flux = radial velocity Cgas D C/r
r
SO2
measure SO2



coal soot coating
on surface

-D d [SO2]/d r = rad. vel x [SO2] 


ln {[SO2]/[SO2,o]} = rad. vel x  r /D
ln {[SO2]/SO2,o]} = krate t
8
Removal of SO2 along a tube reactor coated with
fly ash. The accommodation coef. = 4.4x10-4. The
total pressure was 55 torr, with O2 = 6 torr and
SO2 = 9m torr; % RH= 0
9
The Stokes-Einstein Equation (particles)
Let us now think of diffusion in terms of chemical potential
We can think of the driving force of diffusion as the negative
gradient of the chemical potential, i.e.
d/dx = free energy/mol /dx
The frictional force resisting the flow, due to an imbalance in
chemical potential, is the frictional coef. f (force/velocity) on
each molecule x the velocity, v, of the flow,
for a mole this force is f vNo
f v No= -d/dx
Flux has the units of moles, molecules or mass per area per
time
Flux =
moles/(cm2 time)
Conc x velocity = moles/cm3- x cm/time =
moles/(cm2 time)
We can define a diffusion velocity caused by a driving force or
chemical potential, or the concentration gradient such that:
C v = Flux
recall
10
RT
d 
dp
P
substituting for P from PV=nRT, and n/V = C
P= CRT
RT
d 
dC
C
f  Nov = -d/dx
RT
v
dC / dx
f  N0  C
C v = Flux =
RT

dC / dx
f  N0
11
RT
Flux  
dC / dx
f  N0
diffusion coef D= RT/(f No)
Stokes (including Cunningham’s slip factor) showed that for
unit spheres and nonturbulent viscous flow that the resisting
force on a particle flowing through a fluid is
f = 6r/Cc
where = is the viscosity of the medium (poise)
air(20oC) = 1.83x10-4 g/(cm sec) r is the
radius of the particles and
Cc = 1+/r(A+ Qe-rb/) where  is the mean free
path of air = 0.067 m
particle
Cc
size (m)
0.01
22.2
0.05
4.97
0.1
2.87
0.25
1.69
0.5
1.33
1
1.16
5
1.03
12
The Randomness of Diffusion
Consider 17 boxes arranged in a row PAGE 185 FIGURE 9.2
13
page 186 figure 9.3
fit to random walk distribution
2
pn (m)   
 n
1/ 2
 m/ 2 2 
exp

n
/
2


Normal Gaussian Distribution
2


x




p( x)   1/ 2  exp 2 
 2  
 2 
1
14
Where we would like to go with this is relate the sigma, ,
which is a basic feature of the normal distribution to the
diffusion coefficient D
  x 2 
 1 
p( x)   1/ 2  exp 2 
 2  
 2 
2
pn (m)   
 n
1/ 2
 m/ 2 2 
exp

n
/
2


22= n/2
 n / 2 
if we multiply this by x, the distance across a box,  can
be related to an actual distance

 We will then calculate the flux across from one box to
another
 The concentration gradient which caused the flux
Substitute into Flux = -D dC/dt
15
BOX #
3
4
x
5
6
40
0
8
60
0
24
0
0
42
0
14
0
28
0
5t
4
6t
7t
7
8t
If we look at 5th box in seventh time step (7t),
12 particles are entering box #5 from the left and
2 from the right
at the eighth (8t) step, 7 particles leave box #5
and go back to box #4
this gives a net flux of 5 particles between the
7th and 8th steps or
Flux = 5/2t
We now define an average change in concentration per length
between adjacent boxes because for every step one box is
emptied and the adjacent one filled
C = N/x
the spatial conc. gradient dC/dx is
dC/dx = C/x = N/x2
16
If we look at time step 6, the gradient dC/dx driving the diffusion
between boxes for time step 7 is
dC/dx = -N/x2 = -(24-4)/x2
recalling that F = -D dC/dx and F= 5/(2t)
 5 /(2t )
( x ) 2
D

2
 20 /(x ) 8t
t= nt and x = n1/2/2 x
x = (2Dt)1/2
for three dimensional movement
s = (4/Dt)1/2
17
Example How long does it take for a gas molecule of
biphenyl and 0.25 m particle to diffuse from the center of a
5 cm sphere to the a walls of the sphere? Assume a
diffusion coef of 0.06 cm2/sec for biphenyl and 1.6x10-6
cm2/sec for the particle.
2.5 cm
for biphenyl
s = (4/Dt)1/2
S2
3.14x 2.5 cm2
t

4D 4x0.06 cm2sec  1
t = 82 seconds
for the aerosol, D= 1.62x10-6 cm2/sec
(d= 0.25 m)
t = 35.5 days
in water biphenyl diffusion is much slower than in air
D10-6
so
t = tens of days
18
Fick’s second law
Fick’s second law attempts to express the change in
concentration with respect to time with the change in Flux
dC/dt = f (flux)
Consider an elemental volume (box) with a flux of material
in and out
FX
FO
x
A mass balance on the elemental volume per unit time (both
in and out)
mass= conc x vol); flux = mass/(area time)
flux x area= mass/time
19
mass= conc x vol); flux = mass/(area time)
mass/time =  (conc x vol)/time
d(mass) d(conc vol)

d(time)
d (time)
= -area
 flux
V dC/dt = -area  flux
V = area  x
division by V
dC/dt = -flux/x;
as x --> zero
1
dflux

flux


lim x
dx
x o
it appears that
dC/dt = - dflux/dx
Fx
C

t x  cons tan t
x t  const
20
Fx
C

t x  cons tan t
x t  const
C

C
 2C
  ( D )  D 2
t
x
x
x
If we think of three dimensions
Fx Fy Fz
C



t
x
y
z
C 
C

C

C

(D x
)  (D y
)  (D z
)
t x
x
y
y  z
z
  2C  2C  2C 
C

 D


t
  x 2 y 2 z 2 
21
1. diffusion in the x direction
A solution in the x direction for a long tube, where diffusion in the
y and z direction is insignificant, is
  x 2 


M
C( x,t)  
exp
1/ 2 
2
2
(

Dt
)
4
Dt




where = (2Dt)1/2
Figure 9.5 Page 193
22
What kind of diffusion can we expect for a compound
traveling down a 30 m fused silica column (0.25mm id);
assume a flow of 1 cm3/min?
id vol of the col = (0.025cm/2)2x 30m x100cm/m;
vol =1.47cm3 ; Flow = ??
The flow time = 1.47cm3/ 1cm3 per min = 1.47 min
 A typical diffusion coef= 0.07 cm2/sec,
 (2Dt)1/2= 
 so our peak would broaden by
 4 x = 4(2x0.07x1.47*60)1/2 = 14.06 cm; why 4
 the carrier travels 30 m or 30x100 cm in 1.47 min
 this equals 3,000 cm/1.47 min = 34 cm/sec
our peak would broaden in this time
 14.06 cm /34cm/sec = ~0.4 sec
23
2. Diffusion in the Radial Direction
PAH
 2C 2C 2C 
C

 D

)
2
2
2
t
y
 x
z 
converting to polar (radial) coordinates
x= r sincos, y = r sinsin  , z = r cos 
  2C
C
C
 2C
1
C 1  2C 

 D 
2

 2 cot 
 2
2
2
2
t
rr r sin  r
 r  
 r
if diffusion is only in the radial direction
  2C
C
C
 D 
 2r
t
r
 r 2



  2 (C  r ) 
(C  r )
 D

2
t
 r

24
PAH
dr
These types of systems can be solved with numerical techniques
to calculate the C at successive depths of dr into the particle
over time
U= Cr
U(p) t  t  U(p) t  D 
t
 U(p 1)2U(p)U(p 1)
2 
r
Figure 3. Modeling the uptake of deuterated pyrene by
diesel exhaust aerosol at 298K.
25
Diffusion between two parallel plates
Cout
W
L
Cin
H
Let’s say that we wanted to strip a gas and not particles
Solutions to the partial differential equations take on the form
 
C out 
  An exp n
C in n 1
C/Co= 1 - 1.52652/3 +1.5  +0.0342 4/3
where = 8 x D x L x W/(H x flow)
26
Large Stripper System
Transmission Properties
100
% Transmission
80
particles D=.00005
60
40
organic gas D=.05
20
0
0
10
20
30
40
Flow Rate (cfm)
Organic Gases (D=0.05cm2/s)
50
60
Particles (D=0.0005cm2/s)
27
Estimating Diffusion Coefficients
Factors that influence diffusion


average distance traveled between collisions,
i.e. mean free path, 

more collisions for a given distance translates
into a lower mean free path
organic
air
 org,air 
where
1
(1 z)1/ 2 Nair  2org,air
N= # air molecules/vol
= collision diameter of air and organic
z= molecular wt ratio of organic/air
 We would predict that diffusion coefficients would decrease
with increasing (molecular weight)1/2 and effective collision
diameters squared.
Diameter2  radius2  cross sect. area of molecule
V  r3or r V1/3; so r2 or area  V2/3
If we assume the volume of a molecule is  molar vol V
28
and molar volume =
molecular weight
in cm/mol
liquiddensity
Figure 9.6 top page 195
See Figure 9.6 page 195
29
Molecular weight vs Diffusion Coefficients
Page 195 Fig 9.6 bottom
30
Diffusivities in water page 196 Figure 9.7
31
Estimating Gas Phase Diffusion Coefficients (page 197)
equation on page 197 with definitions
32
Estimating molar volumes
1. molar volume, V =


molecular weight
in cm3 /mol
liquiddensity
for benzene
Mw= 78, density =0.88gcm-3=89 cm3/mole
2. sum of atom size---> diffusion
for benzene
V= 6(C) +6(H) +ring
V= 6x16.5+6x2.0-20.2 = 90.8 cm3mol-1
33
35
Calculating liquid Diffusion Coefficients
Dw
13.26x10 5
2
1
 1.14
(
cm
sec
)
0.589
 xV
V = molar volume
 = solution viscosity in centipoise (10-2g cm1
sec-1) at the temperature of interest. The units
of poise refer to the property of fluids that
requires a shearing force of one dyne (g
cm/sec2) of two parallel layers of one cm2 at
velocity of 1 cm/sec over a gradient of one cm
log  = A / T+B
For other liquids Wilke-Chang (1955) give
Dliq
7.4x10 8  (   Mw )1/ 2 T
2
1

(
cm
sec
)
0.6
 xV
where  is the solvent association term, Mw is
the molecular weight of the solvent,  is the
solvent viscosity, T is the temp.

water
CH3-OH
ethanol
Heptane
Benzene
2.6
1.9
1.5
1
1
36
page 409 In Barrow Physical Chemistry 1962
Table 12.5
37
Reynolds Numbers
A body moving through a fluid creates
1. Inertial forces, i.e. forces due to the
acceleration or deceleration of small fluid
masses near the body
2. Viscous frictional forces due to the
viscosity of the medium
Inertial forces/viscous forces = Re#
Re# = m x vel x d/
/m = = (kinomatic for air) =
Re# = vel x d/

= 0.00121 g/cm-1sec-1 /1.82x10-4 g/cm3= 0.151cm2sec-1
for flow in a pipe
laminar region
intermediate region
turbulent flow
1-2000
2100 - 4000
>4000
A 1 m aerosol is flowing in a 16” duct with a velocity of 3500
ft/min. What is the Re# in the duct?
Re# = duct diameter x rel. velocity of air to duct/v
Re# = vel x d/ = 3500x12x2.54/60 x 15x2.54/.151= 479000
38
Turbulent Diffusion
for molecular diffusion
x = (2Dt)1/2
L2
td 
2D
If we look a molecular diffusion times
(page 201 Table 9.5)
39
For transport by advection
Flux = C v
Fad = Cv (m L-2 t-1)
C
C 
 C
 C


divF


v

v
)

v
 
 x

adv
y
z
 t  adv
 x
y
z 
(M L-3 T-1)
For advection the time scale is simply
L2
td 
(diffusion)
2D
L
t a  (advection)
v
L L2

;
v 2D
L crit
2D

v
For distances larger than L advection is more important than diffusion.
In your homework, you will calculate the critical distances for typical
air and water advection and then explain what it means.
An expression for Turbulent Flux
It is possible to develop a turbulent diffusion coef. analogous to a
molecular diffusion coef.
Let’s assume that turbulent fluctuations cause a change in
concentration along with an associated flow Qex of volume between C1
and C2 over some distance Lx.
40
Lx
C2
C1
x
If Lx is small, the concentration difference C1 -C2 can be described by
the product of Lx and the slope of the curve at the point it intercepts
the plane between C1 and C2:
C1-C2 = - Lx C/x
Since this moves across some area, a, at a flow of Qex the velocity vx
= Qex/ a
Since Flux =  C  vx= Qex/ a ( - Lx C/x)
Fturb,x 
L xQ x  C
a  x
Fturb,x   Ex
C
 C vx
x
and
 Ex 
Cvx
 C/  x
This is property of the fluid motion and not the substance
described by C
41
Effect of eddies on dispersion
Consider a patch of ink on a water surface that is turbulent. It is
characterized by the size of its concentration variance 2 about its
center of mass. The patch after time t will grow and be displaced
 = 2Et
and
2/t = 2E
Figure 9.9 page 206
42
The growth of 2 becomes faster with increasing 2.
Figure 9.10 page 207
43
Table 9.6 page 206
44
How do we determine turbulent diffusion
coefs in vertical mixing?
1) from the change in heat content 
Tz 
dz ; from Fickian flux
t
z
Cp = heat capacity
  c p  Az Ez ( T /  z )
zB
  c p   Az 
substituting and solving for Ez
Tz 
dz
 Az 
t

z
Ez  

c p  Az ( T /  z )
Az ( T /  z )
zB
45
2) Radioactive tracers
Radon-222 gas is released from
Radium-226 in sediments
the radon diffuses into the water, where
it is transported upward by turbulence
Broecker sets up a Fick’s second law flux expression for
excess radon, i.e. radon beyond the “conc.” of that which
should be in equilibrium with existing Ra (hence your book
uses the word activity)
 Rn/t  Ez
 2Rn
h
2
  Rn
if there is little change in Rn/t after some time
 Rn/t  Ez
 2Rn
h
2
  Rn = zero
a solution to this at Rn/t = zero
Rn (h) = Rn(h=0) exp[-(/Ez )1/2 h]
so if we plot
ln Rn(h) vs h we get a slope of ........
46
Figure 9.13 page 212
47
Parameters used to characterize vertical mixing in Lakes
In a lake water is usually stratified with the most
dense water at the bottom.
A stability time N, is introduced which describes how
how long it will take to “restore” a water parcel which is
moved vertically to a different density regime.
 g d 
N 

  dz 
1/ 2
g= gravity
acceleration
 = density
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
the coef. of thermal expansion is related to the
change in density with temperature by
d/dT = -  
This gives



N2 = -g dT/dz
Finally
N2 is related to the vertical eddy diffusion coef.
Ez = a (N2)-q
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Turbulent Diffusion in the
Atmosphere
50
Fisk’s 2nd law
Solutions in 1,2,3 dimensions
If we just think in two dimensions; ie there is no
diffusion in the x direction compared to y and z, and
substitute an emissions strength for X (emissions in
g/time normalized fro velocity)
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Historically
Ky and Kz have been related to a Gausian  such that
substituting
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