Diffusion – And Its Role In
Material Property Control
R. Lindeke, Ph. D.
Engr 2110
DIFFUSION
IN LIQUIDS
IN GASES
Carburization
IN SOLIDS
Surface coating
Diffusion
• Interdiffusion: In an alloy, atoms tend to migrate
from regions of high conc. to regions of low conc.
Initially
After some time
Adapted from
Figs. 5.1 and
5.2, Callister
7e.
This is called a DIFFUSION COUPLE a
sketch of Cu – Ni here
Diffusion Mechanisms
Vacancy Diffusion:
• atoms exchange with vacancies
• applies to the atoms of substitutional impurities
• rate depends on:
-- number of vacancies
-- activation energy to exchange – a function of temp. and
size effects.
increasing elapsed time
Diffusion Mechanisms
• Interstitial diffusion – smaller
atoms can diffuse between atoms.
More rapid than vacancy diffusion
Adapted from Fig. 5.3 (b), Callister 7e.
Processing Using Diffusion
• Case Hardening:
• Diffuse carbon atoms
into the host iron atoms
at the surface.
• Example of interstitial
diffusion to produce a
surface (case) hardened
gear.
Adapted from
chapter-opening
photograph,
Chapter 5,
Callister 7e.
(Courtesy of
Surface Division,
Midland-Ross.)
The carbon atoms (interstitially) diffuse from a carbon
rich atmosphere into the steel thru the surface.
Result: The presence of C atoms makes the iron
(steel) surface harder.
Diffusion
• How do we quantify the amount or rate of
diffusion? – we define a “mass flux” value
moles (or mass) diffusing
mol
kg
J  Flux 

or
2
surface area time 
cm s m2s
• Flux is Commonly Measured empirically
– Make thin film (membrane) of known surface area
– Impose concentration gradient (high conc. On 1 side low
on the other)
– Measure how fast atoms or molecules diffuse through
the membrane
M=
J  slope
mass
M
l dM
diffused
J

At A dt
time
Simplest Case: Steady-State Diffusion
The Rate of diffusion is independent of time
dC
 Flux is proportional to concentration gradient =
dx
This model is captured as
Fick’s first law of diffusion
C1 C1
C2
x1
x
C2
dC
J  D
dx
x2
dC C C2  C1
if linear


dx
x
x2  x1
D  diffusion coefficient
which is a function of diffusing
species and temperature
For steady state diffusion concentration gradient = dC/dx is linear
F.F.L. Example: Chemical Protective
Clothing (CPC)
• Methylene chloride is a common ingredient in paint removers.
Besides being an irritant, it also may be absorbed through skin.
When using this paint remover, protective gloves should be
worn.
• If butyl rubber gloves (0.04 cm thick) are used, what is the
diffusive flux of methylene chloride through the glove?
• Data:
– diffusion coefficient in butyl rubber:
D = 110 x10-8 cm2/s
– surface concentrations:
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
Example (cont).
• Solution – assuming linear conc. gradient
glove
C1
J -D
paint
remover
tb 
2
6D
skin
Data:
D = 110 x 10-8 cm2/s
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
x2 – x1 = 0.04 cm
C2
x1 x2
J   (110 x 10
-8
dC
C  C1
 D 2
dx
x2  x1
(0.02 g/cm 3  0.44 g/cm 3 )
g
cm /s)
 1.16 x 10-5
(0.04 cm)
cm2s
2
What happens to a Worker?
• If a person is in contact with the irritant and more than about 0.5 gm of the
irritant is deposited on their skin they need to take a wash break
• If 25 cm2 of glove is in the paint thinner can, How Long will it take before they
must take a wash break?
Flux  1.16 105 g
2
cm -s
if the exposed area of the gloves are 25 cm
2
how long will it take to get 0.5g of M-C onto the hands?
Flux  25cm  1.16  10
2
Exposure time= 0.5g
5
g
2
cm -s
 2.9 104 g
2.9 104 g
s
 1724s  0.48 hr
s
Another Example: Chemical Protective
Clothing (CPC)
• If butyl rubber gloves (0.04 cm thick) are used, what
is the breakthrough time (tb), i.e., how long could the
gloves be used before methylene chloride reaches
the hand?
• Data (from Table 22.5)
– diffusion coefficient in butyl rubber:
D = 110 x10-8 cm2/s
Example (cont).
• Solution – assuming linear conc. gradient
glove
C1
paint
remover
2
tb 
6D
skin
C2
  x2  x1  0.04 cm
x1 x2
tb 
Equation 22.24
D = 110 x 10-8 cm2/s
(0.04 cm) 2
( 6)(110 x 10
-8
2
 240 s  4 min
cm /s)
Time required for breakthrough ca. 4 min
Diffusion and Temperature
• Diffusion coefficient increases with increasing T
 Qd 

D  Do exp
RT


D = diffusion coefficient [m2/s]
Do = pre-exponential [m2/s]
Qd = activation energy [J/mol or eV/atom]
R = gas constant [8.314 J/mol-K]
T = absolute temperature [K]
Diffusion and Temperature
300
600
1000
10-8
1500
D has exponential dependence on T
T(C)
So Note:
D (m2/s)
Dinterstitial >> Dsubstitutional
C in a-Fe
C in g-Fe
10-14
10-20
0.5
1.0
1.5
1000 K/T
Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A.
Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th
ed., Butterworth-Heinemann, Oxford, 1992.)
Al in Al
Fe in a-Fe
Fe in g-Fe
Example: At 300ºC the diffusion coefficient and
activation energy for Cu in Si are
D(300ºC) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient at 350ºC?
transform
data
D
Temp = T
1
  and
 T2 
Q
D
 lnD2  lnD1  ln 2   d
D1
R
Qd
lnD2  lnD0 
R
ln D
1/T
Qd
lnD1  lnD0 
R
 1 1
  
 T2 T1 
 1
 
 T1 
Example (cont.)
 Qd
D2  D1 exp 
 R
 1 1 
  
 T2 T1 
T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K
D2  (7.8 x 10
11
  41,500 J/mol  1
1 
m /s) exp 



 8.314 J/mol - K  623 K 573 K 
2
D2 = 15.7 x 10-11 m2/s
Non-steady State Diffusion
• If the concentration of diffusing species is a
function of both time and position that is
C = C(x,t)
• In this case Fick’s Second Law is used
C
 2C
Fick’s Second Law
D 2
t
x
Concentration (C) in terms of time and position can be
obtained by solving above equation with knowledge of
boundary conditions
 The solution depends on the specific case we are
treating
One practically important solution is for a semi-infinite solid in which the
surface concentration is held constant. Frequently source of the diffusing
species is a gas phase, which is maintained at a constant pressure value.
A bar of length l is considered to be semi-infinite when
l  10 Dt
The following assumptions are implied for a good solution:
1.
Before diffusion, any of the diffusing solute atoms in the solid are
uniformly distributed with concentration of C0.
2.
The value of x (position in the solid) at the surface is zero and
increases with distance into the solid.
3.
The time is taken to be zero the instant before the diffusion
process begins.
Non-steady State Diffusion
• Copper diffuses into a bar of aluminum.
Surface conc.,
Cs of Cu atoms
bar
pre-existing conc., Co of copper atoms
Adapted from
Fig. 5.5,
Callister 7e.
Cs
Notice: the concentration
decreases at increasing
x (from surface) while it
increases at a given x as
time increases!
Boundary Conditions:
at t = 0, C = Co for 0  x  
at t > 0, C = CS for x = 0 (const. surf. conc.)
C = Co for x = 
Solution:
C x , t   Co
 1  erf
Cs  Co
C(x,t) = Conc. at point
x at time t
erf (z) = error function
2 z y 2

e dy

0

erf(z) values are given
in Table 5.1
 x 


 2 Dt 
CS
C(x,t)
Co
Non-steady State Diffusion
• Sample Problem: An FCC iron-carbon alloy initially containing
0.20 wt% C is carburized at an elevated temperature and in an
atmosphere that gives a surface carbon concentration (C0 )
constant at 1.0 wt%. If after 49.5 h the concentration of carbon
is 0.35 wt% at a position 4.0 mm below the surface, determine
the temperature at which the treatment was carried out.
• Solution: use Eqn. 5.5
C( x, t )  Co
 x 
 1  erf 

Cs  Co
 2 Dt 
Notice that the solution requires the use of the erf function which was
developed to model conduction along a semi-infinite rod as we saw earlier
erf  z  
2
 
z
0
e
 y2
dy
here, consider that z replaces:
x
2 Dt
Solution (cont.):
C( x , t )  Co
 x 
 1  erf 

Cs  Co
 2 Dt 
– t = 49.5 h
– Cx = 0.35 wt%
– Co = 0.20 wt%
x = 4 x 10-3 m
Cs = 1.0 wt%
C( x, t )  Co 0.35  0.20
 x 

 1  erf 
  1  erf ( z )
Cs  Co
1.0  0.20
 2 Dt 
 erf(z) = 0.8125
Solution (cont.):
We must now determine from Table 5.1 the value of z for which the
error function is 0.8125. An interpolation is necessary as follows
z
erf(z)
0.90
z
0.95
0.7970
0.8125
0.8209
Now By LINEAR Interpolation:
z  0.90
0.8125  0.7970

0.95  0.90 0.8209  0.7970
z  0.93
Now solve for D
x
z
2 Dt
D
x2
4 z 2t
3
2
 x2 
(
4
x
10
m)
1h

D  
 2.6 x 10 11 m2 /s
 4z 2t  ( 4)(0.93)2 ( 49.5 h) 3600 s


Solution (cont.):
• To solve for the temperature at
which D has above value, we use a
rearranged form of Equation (5.9a):
Qd
T
R(lnDo  lnD)
from Table 5.2, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

T
148,000 J/mol
(8.314 J/mol - K)(ln 2.3x10 5 m2 /s  ln 2.6 x10 11 m2 /s)
T = 1300 K = 1027°C
Following Up:
• In industry one may wish to speed up this
process
– This can be accomplished by increasing
• Temperature of the process
• Surface concentration of the diffusing species
• If we choose to increase the temperature,
determine how long it will take to reach the
same concentration at the same depth as
in the previous study?
Diffusion time calculation:
• X and concentration are equal therefore:
• D*t = constant for non-steady state diffusion!
• D1300 = 2.6x10-11m2/s (1027C)
D1250C  D0 e
 QD
RT
5
5
 2.3 10 e
148000
8.311523
6
D1250C  2.3 10  8.344 10  1.92 10
 D1250C t1250C  D1027 C t1027 C  t1250C 
t1250C
2.6 1011  49.5hr

 6.71 hr
10
1.92 10
10
m2
s
D1027 C t1027 C
D1250C
Summary
Diffusion FASTER for...
Diffusion SLOWER for...
• open crystal structures
• close-packed structures
• materials w/secondary
bonding
• materials w/covalent
bonding
• smaller diffusing atoms
• larger diffusing atoms
• lower density materials
• higher density materials