Engineering 45
Solid State
Diffusion-2
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-45: Materials of Engineering
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Learning Goals - Diffusion
 How Diffusion Proceeds
 How Diffusion Can be Used in Material
Processing
 How to Predict The Rate Of Diffusion
Be Predicted For Some Simple Cases
• Fick’s first and SECOND Laws
 How Diffusion Depends On Structure
And Temperature
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
 Fick’s 1st Law
dC
J D
dx
• Where
– J  Flux in kg/m2•s
or at/m2•s
– dC/dx = Concentration
GRADIENT in units of
kg/m4 or at/m4
– D  Proportionality
Constant (Diffusion
Coefficient) in m2/s
Engineering-45: Materials of Engineering
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Concen., C
Recall Fick’s FIRST Law.
Cu flux Ni flux
x
C
Position, x
 In the SteadyState
Case J = const
• So dC/dx = const
– For all x & t
 Thus for ANY two
points j & k
dC C Ck  C j


dx x
xk  x j
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
NONSteady-State Diffusion
 In The Steady Case
 In The NONSteady,
or Transient, Case
the Physical
Conditions Require
J , C, dC dx  f x, t 
Engineering-45: Materials of Engineering
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Increasing Time
0.8
Vapor Saturation Fraction, v
J  f t 
 C  f t 
Liquid Source Vapor Bubble Saturation Transient (Calculated)
1.0
n

0.6

v
r , t   1  2   1 
n1
0.4
sin n  r ro 
n  r r0
en
2
2
2
 D v t / ro
Pv(r,t) (t=0.01)
Pv(r,t) (t=0.04 s)
Pv(r,t) (t=0.10 s)
Pv(r,t) (t=0.15 s)
0.2
Pv(r,t) (t=0.2 s)
Pv(r,t) (t=0.3 s)
• Bubble Diameter = 4 mm
• Dv = 5 mm2/s
Pv(r,t) (t=0.35 s)
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
Radial Position Inside Bubble, r (mm)
1.6
• In The Above
Concen-vs-Position
Plot Note how, at x 
1.5 mm, Both C and
dC/dx CHANGE
with Time
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
1.8
2.0
file = BubPv(t)2.xls
NONSteady State Diffusion Math
x
 Consider the
Situation at Right
 Box Dimensions
• Width = x
• Height = 1 m
• Depth = 1 m
– Into the slide
 Box Volume,
V = x•1•1 = x
 Now if x is small
Engineering-45: Materials of Engineering
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J(left)
J(right)
Concentration,
C, in the Box
• Can Approximate
C(x) as
Cavg 
Cleft  Cright
2
 C x 
 The Amount of Matl
in the box, M
kg 3
M  C V  3  m
m
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
NONSteady State Diffusion
cont
dx
 or M  C 11 x  Cx
 Material ENTERING
the Box in time t
M in  J left  Area  t
 J left 11  J left t
J(left)
J(right)
Concentration
C, in the Box
 For NONsteady
Conditions
 Material LEAVING the
J left  J right
Box in time t
 So Matl
M out  J right  Area  t
ACCUMULATES
in the Box
 J right 11  J rightt
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
NONSteady State Diffusion
cont.2
x
 So the NET Matl
Accumulation
M  J left  J right t
J(left)
J(right)
Concentration
C, in the Box
 Adding (or
 With V = 1•1•x
Subtracting) Matl
11 x  C  J left  J right t or
From the Box
CHANGES C(x)


C 
C
C
 
x
  D
   D
M M
Cnew  Cold  C 

t 
x x 
x x  x  
V
V
M
 Partials Req’d as
 C 
 M  V  C
C = C(x,t)
V
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
NONSteady State Diffusion
cont.3
x
 In Summary for
CONSTANT D
 C
C
C
x
 D

t
 x x  x x
J(left)
J(right)


x
Concentration
C, in the Box
 Now, And this is CRITICAL, by TAYLOR’S
SERIES
C / x
C xxx  C xx 
 x
x
x
 so
  C x x
C
C

x
 D 
x 

t
x
x
x

Engineering-45: Materials of Engineering
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 C

 x
x



x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
NONSteady State Diffusion
cont.4
x
 After Canceling
C x x
C
D
t
x
C
D
x 2
J(left)
2
x
J(right)
Concentration
C, in the Box
x
 Now for very short t
C C
 2C
 2C
lim

D
D 2
2
t 0 t
t
x
x  x
 Finally Fick’s SECOND LAW for Constant
Diffusion Coefficient Conditions
C
C
D 2
t
x
2
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Comments of Fick’s 2nd Law
x
 The Formal
Statement
C
C
D 2
t
x
2
 This Assumes That
D is Constant, i.e.;
D  DCx, t 
 In many Cases
Changes in C also
Change D
Engineering-45: Materials of Engineering
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J(left)
J(right)
Concentration,
C, in the Box
 This Leads to the
GENERAL, and much
more Complicated,
Version of the 2nd Law
C   C 
 D

t x  x 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Example – NonSS Diffusion
 Example: Cu Diffusing into a Long Al Bar
Surface conc.,
Csof Cu atoms
bar
pre-existing conc., Co of copper atoms
 The Copper Concentration vs x & t
 The General Soln is
Gauss’s Error
C(x,t)
Function, “erf”
Cs
Co to
t1
C  x , t   Co
 x 
 1  erf 

C S  Co
 2 Dt 
t3
t2
position, x
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Comments on the erf
 Gauss's Defining
Eqn
erf z  
2
z
e


0
 y2
dy
 z is just a NUMBER
• Thus the erf is a
(hard to evaluate)
DEFINITE Integral
 Treat the erf as any
other special Fcn
Engineering-45: Materials of Engineering
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 Some Special Fcns
with Which you are
Familiar: sin, cos, ln,
tanh
• These Fcns used to
be listed in printed
Tables, but are now
built into Calculators
and MATLAB
 See Text Tab 5.1 for
Table of erf(z)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Comments on the erf cont.
 1-erf(z) appears So
Often in Physics
That it is Given its
Own Name, The
COMPLEMENTARY
Error Function:
C  x , t   Co
 x 
 erfc

C S  Co
 4 Dt 
 Notice the Denom in
this Eqn
4Dt  Ld
 This Qty has SI
2 z  y2
erfcz   1 
e dy Units of meters, and
 0
is called the
“Diffusion Length”
 Recall The erfc
• The Natural Scaling
Diffusion solution
Factor in the efrc

Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
SemiInfite Slab Diffusion vs Time
1.0
ErfC t1; Ld = 0.2236
0.9
ErfC t2; Ld = 0.3873
0.8
ErfC t3; Ld = 0.6708
Relative Concentation
TIME
ErfC t4; Ld = 1.1619
0.7
ErfC t5; Ld = 2.0125
0.6
0.5
PARAMETERS
• D = 0.05 (arb Units)
0.4
0.3
0.2
0.1
0.0
0.0
0.2
0.4
0.6
file = erfc_0401.xls
Engineering-45: Materials of Engineering
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0.8
1.0
1.2
1.4
1.6
x (arbitrary units)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
1.8
2.0
Example  D = f(T)
 Given Cu Diffusing
into an Al Bar
 At given point in the
bar, x0, The Copper
Concentration
reaches the Desired
value after 10hrs at
600 °C
 Get a New Firing
Furnace that is Only
rated to 1000 °F =
538 °C
• To Be Safe, Set the
New Fnce to 500 °C
 Need to Find the
NEW Processing
• The Processing
TIME for 500 °C to
Recipe
yield the desired
C x0   10hrs @ 600C C(x0)
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Example  D = f(T) cont
 Recall the erf
Diffusion Eqn



x0
x0


1  erf
 1  erf 
 4 D 10hrs 
 4D  t
600
500



C  x , t   Co
 x 
 1  erf 

C S  Co
 4 Dt 
 Since CS and Co
have NOT changed,
Need
C x0 ,10hrs 600C  Co C x0 , t 500C  Co

CS  Co
CS  Co
 Since by the erf
Engineering-45: Materials of Engineering
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 For this Eqn to be
True, need Equal
Denoms in the erf
D600 10hrs  D500  t
or
D600 10hrs
t
D500
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt




Example  D = f(T) cont.2
 Now Need to
Find D(T)
 As With Xtal PtDefects, D Follows
an Arrhenius Rln
 Qd 
D  D0 exp  

 RT 
• Where
– D0  Temperature
INdependent
Exponential
PreFactor in m2/s
Engineering-45: Materials of Engineering
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– Qd  Arrhenius
Activation Energy in
J/mol or eV/at
– R  Gas Constant =
8.31 J/mol-K =
8.62x10-5 eV/at-K
– T  Temperature in K
 Find D0 and Qd from
Tab 5.2 in Text
• For Cu in Al
– D0 = 6.5x10-5 m2/s
– Qd = 136 kJ/mol
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Example  D = f(T) cont.3
 Thus D(T) for
Cu in Al
 Thus for the new
500 °C Recipe
 In this Case
46.92 10hrs
t
 113.0hrs
4.152

136000 J / mol 

D  6.5 10 5 exp  
 8.31J / mol  K   T 
• D600 = 4.692x10-13 m2/s
• D500 = 4.152x10-14 m2/s
 Now Recall the
Problem Solution
D600 10hrs
t
D500
Engineering-45: Materials of Engineering
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 This is 10x
LONGER than
Before; Should have
bought a 600C fnce
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Find D Arrhenius Parameters
 Recall The D(T) Rln
y  b  mx
 Qd 
D  D0 exp  

 Applied to the D(T)
RT


Relation
 Take the Natural Log
of this Eqn
y  ln D
Qd
ln D  ln D0 
R
1
 
T 
 This takes the form
of the slope-intercept
Line Eqn:
Engineering-45: Materials of Engineering
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m  -Qd R
x 1 T
b  ln D0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Find D(T) Parameters cont
 And, Since TWO
Points Define a Line
If We Know D(T1)
and D(T2) We can
calc
• D0
• Qd
 Quick Example
• D(T) For Cu in Au at
Upper Right
Engineering-45: Materials of Engineering
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y
x
 Slope, m = y/x
• x = (1.1-0.8)x1000/K
• = 0.0003 K-1
• y = ln(3.55x10-16) −
ln(4x10-13) = −7.023
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Find D(T) Parameters cont.2
 By The Linear Form
m   Qd R or
Qd   R  m   Ry x 
Qd  8.31   7.023 0.0003
Qd  194500 J/mol
 Now, the intercept, b
y  mx  b or
b  y1mx1
 Pick (D,1/T) pt as
• (4x10-13,0.8)
Engineering-45: Materials of Engineering
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 in the (x,y) format
• x1 = 0.0008
• y1 = ln(4x10-13)
= −28.55
 So b
 7.023
b   28.55 
0.0008
0.0003
b  9.819
 Finally D0
D0  eb  e 9.819
D0  5.44 105 m 2 s
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Diff vs. Structure & Properties
 Faster Diffusion for
 Slower Diffusion for
• Open crystal
structures
• Lower melting Temp
materials
• Materials with
secondary bonding
• Smaller diffusing
atoms
• Cations
• Lower density
materials
• Close-packed
structures
• Higher melting Temp
materials
• Materials with
covalent bonding
• Larger diffusing
atoms
• Anions
• Higher density
materials
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
Diffusion Summarized
 Phenomenon: Mass Transport In Solids
 Mechanisms
• Vacancy InterChange by KickOut
• Interstitial “squeezing”
 Governing Equations
• Fick's First Law
• Fick's Second Law
 Diffusion coefficient, D
• Affect of Temperature
• Qd & D0
– How to Determine them from D(T) Data
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt
WhiteBoard Work
 Problem 5.28
• Ni Transient Diffusion into Cu
Prob 5.28: Ni Diffusion into Cu
5%
500hrs @ 1000 °
500hrs @ T2
CNi (Wt% Ni in Cu)
4%
Same Concentration
3%
2%
1%
0%
0.0
0.5
file = erfc_0401.xls
Engineering-45:
Materials of Engineering
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1.0
1.5
2.0
x into Cu (mm)
2.5
3.0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-07_Diffusion_Fick-2.ppt