Post on 26-Jan-2022
Diffusion and Phase Transformation簡 朝 和
• Course Outline
• 1. Diffusion• (a) Diffusion equation solutions – steady and transient states• (b) Diffusion with moving boundary• (c) Diffusion in heterogeneous systems – controlling kinetics• (d) Atomic theory of diffusion • (e) Diffusion in dilute and concentrated solutions
• 2. Phase Transformation• (a) Nucleation - homogeneous and heterogeneous• (b) Growth with and without composition change
(c) Overall transformation kinetics• (d) Spinodal decomposition•• Textbooks and References• 1. P.G. Shewmon, Diffusion in Solids, 2nd ed., McGraw-Hill• 2. D.R. Poirier and G.H. Geiger, Transport Phenomena in Materials Processing, 2nd ed., TMS publication, 1994.• 3. J. Crank, Mathematics of Diffusion, 2nd ed., Oxford University Press, 1975. • 4. H.S. Carslaw and J.C. Jaeger, Conduction of Heat in Solids, 2nd, ed., Oxford University Press, 1959.• 5. A.K. Jena and M.C. Chaturvedi, Phase Transformations in Materials, Prentice Hall, NJ, 1992.• 6. D.A. Porter, K.E. Easterling and Y. Sherif, Phase Transformations in Metals and Alloys, 3rd ed., CRC • Press, 2009.• 7. J.W. Christian, The Theory of Transformations in Metals and Alloys, Pergamon Press, 1975.• 8 R. W. Balluffi, S. M. Allen and W. C. Carter, Kinetics of Materials, Wiley, 2005.
• Grading• Quizzes (2) 30%• Midterm test 25%• Final test 40% No aid sheet, No open book and No take home• Class attendance –random quizzes 5%
• Thermodynamics:– To study the direction of a reaction, or if a reaction can
take place. (ΔG<0)– To study the equilibrium states in which state variables of
a system do not change with time.
• Kinetics:– To study the rates and paths of a reaction adopted by
the systems approaching equilibrium.– To study the rate-limiting steps of a reaction– To study the controlling factors of the rate-limiting steps
Thermodynamics vs. Kinetics
Input OutputKinetic Processes• Rate-limiting steps• Controlling factors
Thermodynamics vs. Kinetics
Reaction Rate α (Kinetic factor) x (Thermodynamic factor)* Kinetic factor relates to Q (activation energy), while the thermodynamic factor relates to the driving force, ΔG=G2-G1.
* The thermodynamic factor decides the direction of a reaction, while the kinetic factor, the rate of reaction.
G1
G2
Initial Activated FinalState State State
Q
ΔG
Kinetic theory: The reaction rate is proportional to the probability to reach activated state that follows the Arrheniusrate equation, exp(-Q /RT).
* The activation energy (Q) can be obtained from the slope of curve plotted as ln (reaction rate) vs. 1/T
Example: For diamond growth by CVD from reaction of methane and hydrogen
- Q/R
Kinetic factor increased by changing temperature or adding catalysts.
Examples: (1) N2 + 3H2 = 2NH3 using iron as a catalyst(2) 2CO + 2NO = 2CO2 + N2 using Pt and Rh as
catalysts for catalytic converters used in automobile
Activation energy (Q) without catalyst
Activation energy (Q) with catalystReactants
Products
Free
Ene
rgy
Progress of Reaction
ΔG<0
Examples: Thermodynamically favorable but kinetically unfavorable phase changes
(1) Is a diamond forever?
Diamond GraphiteDiamond
Liquid
Vapor
Graphite
0 2000 4000 6000Temperature (K)
1011
109
107Pres
sure
(Pa
)
Diamond
Graphite
Free
Ene
rgy
Progress of Reaction
ΔG=-2.9KJ/mol
Very large Q
(2) Crystallization of glasses
Supercooledliquid
Shrinkage due to freezing
50m
1 h
2 h
8 h
CaO-SrO-BaO-B2O3-SiO2 glass-ceramics annealed at 875oC
Pseudowollastonite(Ca,Ba,Sr)SiO3
Cristobalite SiO2
Cooling Rate
Diffusion driven by decrease in chemical potential
Free energy of diffusion couple = G3Diffusion taking place to homogenize to obtain G4
μ1B>μ2
BB diffusing from (1)→(2)
μ2A>μ1
A A diffusing from (2)→(1)
Down-hill Diffusion
* Down-hill diffusion
G1G2 G3
G4
A 2 1 B
μ2A
μ1A
μ1B
μ2B
2 1
AB
μ1B<μ2
BB diffusing from (2)to (1)
μ2A<μ1
A A diffusing from (1)to (2)
Up-hill Diffusion
* Up-hill Diffusion
A 2 1 B
G3
G4
G1G2
μ2A
μ1A μ2
B
μ1B
2 1A
BFree energy of diffusion couple = G3Diffusion taking place to homogenize to obtain G4
( )
v ( ) ( )
: , :
CDriving force notx x
J C C B F C Bx
B Mobility F Force
Diffusion:Process by which matter is transported through matter as a result of molecular motions
General scheme for transport phenomena
Flux α Driving force α Gradient in potentialMatter J α dC/dx α Concentration potentialHeat q α dT/dx α Temperature potentialElectricity I α dψ/dx α Electrical potential
: thermal c
: electrical conduc
: diff
onductivi
tivity
t
usivit
y
y
( ' )( ' )
( ' )
D
k
J D C Fickq k T Fourier s LI Oh
s La
m s Law
w
aw
time
To increase the rate:
Fick’s First Law
Rate of particles reaching surface (2): number of particles/time =1
(a) Reducing thickness (Δx) (b) Increasing number of particles (ΔC)(c) Increasing area (A)(d) Reducing particle size (D)(e) Increasing temperature (D)
v ( )
v ( )
( D : D i f f u s i v i t y )
n CR a t e At x
n CJ F l u xA t A x
CDx
(a)
(b)
(c)
(d)
(1) (2)
Δx
(D=Doexp(-Q/RT))
Fick’s First Law:Species migrates from a region of high concentration to a region of low concentration；in general the rate of diffusion is proportional to the concentration gradient
CJ Dx
* Flux (J) : Mass/(area‧time), e.g., g/(cm2‧sec)
* Minus (-): Matter moves from high to low concentration.
* Diffusivity (D): Diffusivity related to atomic mobility and crystal structure, e.g., cm2/sec (independent of concentration gradient)
* Concentration gradient ( ): Gradient in ”Mass Potential,”e.g., g‧cm-3/cm
Cx
Magnitude of Diffusivity
10-1Gases
10-4-10-5Liquids
10-12Elemental Semiconductors
10-8Solids (metals near M.P.)
Diffusivity (cm2/sec)Species
Note: Liquids and gases in liquids and gases generally dominatedby convection flow, rather than diffusion.
Diffusion << Convection
Diffusion(Slower)
Convection(Faster)
0( )xCt
( )C C x
Steady State:
Concentration at a given point is invariant with time
i.e.,
J≠J(x,t) when area is fixed
Key: to describe C(x,t) quantitatively
H
Time
Transient State
t = ∞
X=0 X=LCo
ncen
trat
ion
Steady State
Ceq
x x xoo
CSteady State Solution :C( ) = A + B = C -L
Steady State: Constant flux if the area is fixed.
Equilibrium State: No Flux
μ: Chemical Potential
( ) 0 ( )t tC
x x
( ) 0 ( )x xC
t t
RT ln ( ) RT ln( )
ln( ) ln( )( ) RT( ) RT( )t t t
a X
a X
x x x
Steady State:
area is fixed
Under any conditions
when D is constantC Cx x
22( ) 0t
Cx
( ) 0xC
t
( ) 0tJ
x
x
x
J1
No Depletion or Accumulation
C
C
J
t
J1
( ) 0xCt
Fick’s Second LawTransient State: C=C(x,t), or J=J(x,t)
2 2
2 ........2!x x x
J J dxJ J dxx x
From Taylor Series
JXJX+ΔX
X X+ΔX
A (area)accumulation or depletion of concentration exists
Mass Conservation
x x xm CJ A J A A xt t
Flux · Area = Rate of change of concentration · volume(g cm-2sec-1· cm2 = g cm-3sec-1 · cm3)
(A is fixed)
x x+Δx
Jx Jx+Δx
( )
x xJ CJ A J dx A A xx t
J Cdx A A xx t
Fick’s Second Law (cont.)
( )C J CDt x x x
2
2
C C DDx x x
2
2 ( )C C D CDx x C x
when ( ) 0DD D CC
2
2
C CDt x
Linear partial differential equationSolutions are additiveSolutions require initial and boundary conditions
Fick’s Second Law
C
x
( )D D CSteady State
Transient State:
( ) 0tJ
x
x
x
2
2
C J CDt x x
22( ) 0t
Cx
If J2<J1 Accumulation
t( ) 0x
Ct
J1
Depletion or Accumulation
C
C
J
J2
2
2( ) 0
Cartesian coordinate in one dimension
xC CDt x
Steady State
2
2
Cylindrical coordinate1( ) ( ) 0D C C Cr D
r r r r r r
if C= f(r)
22
2 2
Spherical coordinate2( ) ( ) 0D C C Cr D
r r r r r r
(D is constant)
- Infinite cylinder- Decarburization or Carburizationin single-phase Fe-C alloy
- No phase transformation involved- Steady state reached ( )- Diffusion controlling process
(Equilibrium concentrations reached on both sides)
- To find D(C) in Fe-C alloy
Application of Steady State
Carb
urizat
ion
Gas
Dec
arbu
riza
tion
Gas
( ) 0xCt
2
22
2
4 2
42
2
[ ] ;[ ]
2[ ]
k
c
k
cH
CO C CO
k COa orCO
CH H Ck CHa
P
: ( ) 0xCSteady Statet
2 2
2 2 2
1 1: ( ) 0C C CCylindrical coordinate rr r r r Z
2
2
1 1 ( ) ln( ) 0 + =0 C C Crr r r r
C r A B rr r
Infinite Cylinder (area is not fixed)For steady radial diffusion through the wall of the hollow cylinder, the Laplace equation in cylindrical coordinate is
If the concentration depends only on the radial coordinate
B.C.: C(r1, t)= C1, C(r2, t)= C2
2 2
11 22
ln( )
ln( )
rC C r
rC Cr
A plot of C versus ln(r) should be a straight line if diffusion is the rate-controlling step and D is constant.
r1r2
C1
C2
The quantity of carbon passing through the tube per unit time (j) is constant and independent of r:
: ( ) 0xCSteady Statet
j (mass/time) = 2πrlJr
Area: 2πrl is not constantr: Radiusl: Length of cylinderJr: Local flux
2 ( )2 ln( )
rCJ Dr
C j C Cj rl D rr lD r r
Diffusion of carbon in γ-Fe
Not linearD=D(C)
Experimental data
The above analysis can not be appliedCa
rbon
con
tent
-log(r)
(Flow Rate)
2
2( ) 0xC CDt x
Steady State (diffusion area is not fixed)
1 1 2 2
1 1 2
1 2 2
2
1
2
1 21
2 2
r
r
r r
r
rr lJ r l
m mj jt
Jr J r J
r J J
t
r
r1
r2
Diffusion area of 2πrlis not fixed
Cylinder
2 ln( )j ClD r
For a given experiment, j can be determined by chemical analysis, and then we can determine D from the slope of the plot of C versus ln(r).
Diffu
sivity
of
carb
on
Carbon content
At 1000oC
2 ln( )j ClD r
Concentration profile of carbon in the wall of a hollow steel cylinder under conditions of steady-state diffusion.
Diffusion coefficient of carbon in iron varying with its concentration at 1000℃
( ) (v
ln ln(1 )ln
)
ln
ln( )
ln ln
I
I
I
C B F CBx
a CBRT C Dx x
a CBRT Dx
XBRT BRTX X
x
lnaD =BRTlnC
J = C
∂∴∂
2 2
2 2
1 21 2
1 2
( )
( )
H H
x H H
C CC k P Px
C DkJ D P Px
P1 P2C1
C2
Glass metal foil
2
equilibrium constant
[ ]metal foil
H
Hk
P
Sievert’s law
2
2
1 11
2 22
Hmetal
Hmetal
C H k P
C H k P
P(H2-1) and P(H2-2) are partial pressures of hydrogen on sides 1 and 2, respectively
Steady State Diffusion (area is fixed)
* Diffusion of hydrogen gas through the metal foil is the rate-limiting step.
*Equilibrium concentrations (C1 and C2) on both sides,determined by the local partial pressures.*Steady State achieved.
2 1 [ ]2 metal foilH H
0 0( :equilibrium constant)
x
x
P t
Pk
dm MV dPdt RT dt
DkJ P
dmJ Adt
DkA MV dPPRT dt
dP DkART dtMVP
Consider an iron tank that contains H2 gas at a pressure of Po at a temperature of To. If the pressure outside the tank is zero, after what time the pressure in the tank has decreased to half of its original value?
Rate of loss of mass from tank (Δm/Δt)= mass diffusion flux through walls (JxA)
Assuming H2 is an ideal gas
(M: molecular weight of H2 and V: volume of tank)
(A: Area of the tank)
(PV=nRT=(m/M)RT)
H2 Permeation through Metal Foil
2adsorptionB
PJMk T
α: Sticking coefficientM : Molecular weight
reaction
nJ kPk: Reaction rate constantP: Partial pressure of H2
Adsorption
Diffusion Chemical Reaction
H2
H
H
CJ Dx
C VJR
IxD
1.Adsorption2.Chemical Reaction3.Diffusion4.Chemical Reaction5.Desorption
Process with steps in seriesthe slowest one controls the whole reaction, and the rest reach their equilibrium states.
H2 Permeation through Metal Foil
R1 R2
R3
R5 R4
J
Steady State
(1) Diffusion controlling process
Since ΔC is fixed by chemical reactions (assuming constant D), is inversely proportional to .
CJ Dx
Jx
(2) Reaction controlling process
eqIIC
eqIC
Time
is independent of , and D is inaccessible.J x
SIIC
(reversible)eqIC
0 (irreversible)
eqIIC
eqIC
Time(3) Mixed controlling process
Time
x
(4) Extreme Cases
eqIIC
eqIC
Reaction control
Mixed control
Diffusion control
x
22
2
( ) 0
( )
( )
xJC CDt x x xC CDt x x
C C DDx x C
D is not necessary to be constant.
(a) steady state and D≠D(C)2
20 0 = constantC C CDt x x
. ., 0 De g D and aC
(b) steady state and D=D(C)2
22 ( ) 0C C CD a
t x x
2
2
2
2
2
2
(2) 0 0 (-)
(3) 0 0 ( )
(1) 0 0 ( )
Ca D straight lin
Ca D co
Ca D conc
nvexx
x
av
e
ex
Steady State
J: constant (fixed A)
X
D=Do+ aC
a>0a=0a<0
C
2
2 2
c o n s ta n t
ln
1
1
CJ Dx
J d x D d Ci f D a x
J d x a x d Cd x d Cax J
ax CJ
C Jx a xC J
x a x
(C) Steady State, D=D(x)
2
2
2
2
* 0 0
* 0 0
Ca DxCa D
x
: constant
( ) 0x
JCt
C CJ D Dx x
0(0 ) ( )i
i oJxC C
J x D CJ x D C C
D
(2) D = D(C) = aC
0
i
o
x C
C
CJ a Cx
J d x C d Ca
2 2
01 ( )2 i
Jx C Ca
I.C. C(0, 0) = C0C(x, 0) = Ci
B.C. C(0,t)=C0
Steady state
(1) D ≠ D(C)
C*
2δ
Transient State --Thin-film solution (Infinite Sink)
A quantity of solute, S, is plated as a thin film on one end of a long rod of solute-freematerial, then a similar solute-free rod is welded to the plated end.
t
2
2C CFick's 2nd Law : =D
x
I.C. C(x,0)= 0 |x|>δC(x,0)= C* |x|≤δ
B.C. C(∞,t)=0C(-∞,t)=0
Annealed for time (t) Determine concentration profile of the solute C(x,t).
Assuming D ≠ D(x) (Constant diffusivity)
' *( , ) ( 2 )C x t dx S C
2
( , ) exp: ( )4
A xC xGeneral So tDtt
lution ' 2
(Note: 2 ): ( , ) exp( ) 42
DtS xC x t
DtDtParticular Solution
'
(0, )2
SC tDt
2
( , ) (0, ) exp( )4xC x t C tDt
2Thus a graph of ln( ( , ) (0, ) ) against x should yield a straight line with a slope of 1 4 .
C x t C tDt
Conservation of mass
S’= g/cm2 (Surface Concentration)C*= g/cm3
Taking the natural logarithm of both sides yields2( , )ln( )
(0, ) 4C x t xC t Dt
(A: constant)
-3 -2 -1 0 1 2 3
X (Distance)
C/S’
2.0
1.5
1.0
0.5
0
Dt=00.025
0.05
0.075
0.25
1.0
3.0
' 2
( , ) exp( )42
S xC x tDtDt
Transient State --Thin-film solution (Infinite Sink)
Time
2( , )ln( ) .(0, )
C x tif vs xC t
is not a linear relation, D is a function of concentration. If it is linear, D is independent of concentration.
' 2
( , ) exp( )42
S xC x tDtDt
0(1) | 0 Impermeable Boundary
' 1(2) (0, ) (0, )2
x
SC t
C
C tDt
x
t
2
(3) 1 24
' (0, )( , ) exp( 1) this plane is determined.2
( 2 ) 0.
xwhen x DtDtS C tC x t
eDtx t this plane x Dt moves away from x
-1/4Dt
x2
ln[C
(x,t)
/C(0
,t)]
Determine D
0| 0 Impermeable BoundaryxCx
Examples:
(1)
Al2O3 coating
(2)
Setter
' 2
( , ) exp( )42
S xC x tDtDt
xConcentration-distance curves for an instantaneous plane source.
' *
'
(1) Constant Area
( , ) ( 2 )
(2) (0, )2
C x t dx S C
SC tDt
C/S’
1.0
0.8
0.6
0.4
0.2
0-5 -4 -3 -2 -1 0 1 2 3 4 5
123
312
(1)
(2)
(3)
Thin-film Solution:
' 2
* ( , ) e x p ( )4
S xC x tD tD t
0| 0 (Impermeable Boundary)xCx
(4) Reflection at a boundary (e.g. metal coated with oxides)
* Reflection and superposition are mathematically sound* Linear sum of the two solutions is itself a solution
2x
Thin Film Solution Gaussian Concentration Distribution' 2
2
exp( ) 24
(
( , ) exp( )42
, ) '
x dx
C x t dx
DtD
S xC x tDtDt
S
t
Superposition and Reflection
x=0∂C | = 0∂x
x x
x x
x
x
2
2
S 'C( , t)
S 'C( , t) =
= exp(- ) < 04Dt2 πD
exp(- ) > 04Dt2 πDt
t
xx x2S'C( , t) = exp(- ) > 0
4DtπDt
Fictitioussource +
ll
0
0.1
0.2
0.3
0.4
0.5
0.6
-4 -3 -2 -1 0 1 2 3 4
xx x2S'C( , t) = exp(- ) > 0
4DtπDt
x x
x x
x
x
2
2
S'C( , t)
S'C( , t) =
= exp(- ) < 04Dt2 πD
exp(- ) > 04Dt2 πDt
t
Superposition and Reflection
xx2S' ( + 1)C( , t) = exp(- )
4DtπDt
xx2S' ( - 1)C( , t) = exp(- )
4DtπDt
0-1 +1
0-1 +1+
0-1 +1
xxx xx2 2( +1)C( , t) exp(- )
4DS'C( , t) = = [ + ( -1)C( , t) exp(- )
4tπ t Dt]
D
Thin Film Solution
CJ=-D x
22
∂C ∂ C=D∂t ∂x
* Please derive (b) and (c)
∂C∂x
Accumulation
Depletion
22
∂ C∂x
C
X=0 X
(a)
(b)
(c)
Thin Film Solution
CJ=-D x
22
∂C ∂ C=D∂t ∂x
∂C∂x
MaximumDepletion
22
∂ C∂x
C
X=0 X
(a)
(b)
(c)
ZeroAccumulation
-J +JJ=0
Maximum Flux
Thin Film Solution
Flux to the left is negativeFlux to the right is positive
2
2
( , ) ( 0 , ) e x p ( )4
2 ( 0 , ) e x p ( )4 4
0 0
0 0 0
0
xC x
C
t C tD t
C x xC tx
Cx Jx
x J
D t D t
x
∂C∂x
C
X=0 X
-J+JJ
+-
-3 -2 -1 0 1 2 3
X (Distance)
C/S’
2.0
1.5
1.0
0.5
Dt=00.025
0.05
0.075
0.25
1.0
3.0
' 2
( , ) exp( )42
S xC x tDtDt
Leak Test
B.C. C(∞,t)≠0B.C. C(-∞,t)≠0
Leak Test (Superposition and Reflection)
Concentration at a given point will be higher than that of “Infinite Case”because of reflection.
C**
C**
C*
C*
+
Leak Test
0
0.2
0.4
0.6
0.8
1
1.2
-6 -4 -2 0 2 4 6-∞ - l 0 l ∞
Sample Dimension
The above analyses are only good for a thin film in the middle of an “Infinite Bar”. If it is not infinite, the diffusion will be reflectedback into the specimen when it reaches the end of the bar, and concentration in that region will be higher than the above solution.Q: How long is long enough to be considered infinite?Leak TestArbitrarily taking 0.1% as a sufficiently insignificant concentration
' 2
0
' 2
3
0( , ) exp( )
4
exp( )( , ) 420.1
2
,2 2
% 1
2
0l
l
x dxLet u duDt Dt
x ulx l uDt
S
S x dxC x
xC x t dx dxDtD
t
t
t dx DtD
'2
3 2'
2
0
exp( )10
exp( )
( )2 1 ( )1 2
lDt
S u du
S u du
lerfclDt erfDt
4.6l Dt
4.6l Dt The bar is considered to be long enoughto use a thin-film solution with 99.9% accuracy.
(Check the Table of Error Function)
Error Function
2 1
0
2
3
20
2
2
2
( ) 1 (0) 0( ) ( )
2 ( 1)Small ( )(2 1) !
exp( ) 1 1Large ( 3) ( ) ( ........)2
2( ) exp( )
21 ( ) ( ) exp( )
( ( )) 2 exp( )
(
n n
n
z
z
erf erferf z erf z
zz erf zn n
zz z erfc zz z
erf z u du
erf z erfc z u du
d erf z zdz
d
22( )) 4 exp( )erf z z z
dz
ComplementaryError Function
Error Function 20
2( ) exp( )z
erf z u du
0.9999 2.8 0.7421 0.80
0.9998 2.6 0.7112 0.75
0.9993 2.4 0.6778 0.70
0.9981 2.2 0.6420 0.65
0.9953 2.0 0.6039 0.60
0.9928 1.9 0.5633 0.55
0.9891 1.8 0.5205 0.50
0.9838 1.7 0.4755 0.45
0.9763 1.6 0.4284 0.40
0.9661 1.5 0.3794 0.35
0.9523 1.4 0.3286 0.30
0.9340 1.3 0.2763 0.25
0.9103 1.2 0.2227 0.20
0.8802 1.1 0.1680 0.15
0.8427 1.0 0.1125 0.10
0.8209 0.95 0.0564 0.05
0.7969 0.90 0.0282 0.025
0.7707 0.85 0 0
erf(z)zerf(z)z
Table 1-1. The Error Function
erf(
z)
z0 1.0 2.0 3.0
1.0
0.5
0
21( ) ( ) exp( ) [1 ( )]x
ierfc x erfc u du x x erf x
x
Superposition(1) No interaction from adjacent slabs(2) Superposition of the distributions
from the individual slabs since the diffusion equation is linear and additive.
Solution for a pair of Semi-infinite Solids
00
x
CC’
i
. . ( ,0) 0 0 ( ,0) ' 0 . . ( , ) '
( , ) 0
I C C x xC x C x
B C C t CC t
2'
1
2'
0
( , ) exp( )42
0
( , ) exp( )42
ni
i
i
xCC x tDtDt
n
xCC x t dDtDt
C(x,t)
0α2
X
C
Δα
0
* 22( ) /4
2x Dtc e
Dt
2'
'
( , ) exp( )42
:
ii
xCC x tDtDt
Note S C
Sum the solution of all thin slabs
2'
1
2'
0( , ) exp(
( , ) exp )
)
0
4
(
2
42
n
i
i
i
xCC x tDt
xCC x t dDtDt
Dtn
Substituting , 22xu d Dtdu
Dt
C(x,t)
0α2
X
C
Δα
0
* 22( ) /4
2x D tc e
D t
'2
2
02
( , ) exp( )xDt
xuDt
uCC x t u du
reverse limits of integration and split integral
'0 220
( , ) ( ) exp( )xDt CC x t u du
By definition of error function2
0
2( ) exp( )
( ) 1 (0) 0( ) ( )
zerf z u du
erf erferf z erf z
' 0 2 220
'
'
( , ) [ exp( ) exp( ) ]
[ ( )]2 2 2
[1 ( )]2 2
xDtCC x t u du u du
C xerfDt
C xerfDt
1
0 0
0.50
CC
0.5
X
0.50.5 ( )2
xerfDt
0.5 ( )2
xerfDt
(2)
0
0
0
0
0
0
1 [1 ( )]2 2
1(1) 0 0.5 ( )2 21 0 0.5 ( )2 2
(2) 1 0.921 2
which varies with time as 2
(all compositions except 0.5)
(4)
.
1(3) 0 (implicit B.C.)2
x
o
C xerfC Dt
C xx erfC Dt
xCx erfC Dt
x CCDt
x DtC
C
J D
CxC
0
00
0
(5) Total mass crosses the plane a
| (
:to the lef
t 0
(A:a
t)
re
2
a)
x
t
C
M Dt
C D
C
x
tA
tx
Jd
-
Note: If the concentration is fixed (C=C*), the term of is then also fixed. This means that the penetration distanceis a function of the square root of
the diffusion time. For example, if a diffusion penetration of 0.1mm develops in one hour, it will take 4 hours to develop a penetration of 0.2 mm.
/ 2x Dt
Error Function Short-time Solution
C/C o
1.0
0.5
0
x/l
Time
The above analyses are only good for an “Infinite Slab”. If it is not infinite, the diffusion will be reflectedback into the specimen when it reaches the end of the bar, and concentration in that region will be higher than the above solution.
Q: How long is long enough to be considered infinite?Leak TestArbitrarily taking 0.1% as a sufficiently insignificant concentration
0
3
00
01 [1 ( )
1( , ) [1 ( )]2
](
2
, ) 2 20.1% 10l
l
xC erf dxC x tx
dx DtC x t dx C erf dx
Dt
Co
0 l
time
time
0 X0
1
0.5CCo
t∞
'( , )'( , ) 0
2'( , ) [1 ( )]
2 2
( , ) ( )2
C t C A BCC t A B A B
x
C xC x t erf
C x t A
D
e
t
B rfDt
Using B.C. to solve the problem
Examples
(1) if C(0, t) = 0 '( , ) ( )2
xC x t C erfDt
(2) if C(0, t) = C”=A, C(∞,t)=C’"( , ) [1 ( )], 0
2xC x t C erf xDt
" ( ), 0' " 2
C C xerf xC C Dt
(next page)
C''
0 X>0X<00
C' "
'
(0, )( , )( , ) 0
C t CC t CC t
B.C.
"
"
"
( , ) ( )2
(0, )( ,
( , ) [1 (
0
)]2
)
0
xC x t C er
xC x t A BerfDt
C t C AC t A B C
fDt
x
"
' '
" ' "
' "
( , ) ( ) ( )2
(0, )( , )
0C t C AC t C A B
xC x t C C C erfDt
C B C C
x
'"
' ''
'
'
2
0, ( , ) ( ) ( )2
0, ( , ) (1 ( )
2 2
(1 ( )2 2
2 2
Cif C
C C xx C x t C
C xx C x t erfDt
erfDt
C xerfDt
C'/2
0 X>0X<00
C'
Error Function & its Derivatives
Note: (2) is the thin film solution
xx
x
xx
xx
≡
≡3
3
2
2d (er
(1) erf(
f( ))(3) Accumulation
d(erf( ))(2) (-)Fl
d (erf(
u
))(4)
xd
d
)
d
(1)
(1)(2)
(3)
(4)
Semi-infinite Solution
2
( , ) ( )2
e x p ( )4
xC x t A B e r fD t
C B xx D tD t
0 0 J < 0CBx
∂C∂x
C
X=0 X
+ +- -J Flux to the left
is negative
∂C∂x
C
X=0 X
+ +- -
JFlux to the right is positive
n-type p-type
Donors: Sb, As, P Acceptors: B, Al, Ga
Doping of Semiconductors
Diffusion from a Limited Source (thin film)
00
20
( , ) constant
( , ) exp( )4
(0, ) constant
C x t dx S
S xC x tDtDt
C t
:Gaussian function
Example: p-n junctionC’= Background Concentration
'
2' 0
( )
junction distance
exp( )4
j
j
C C p n or n p
x
xSCDtDt
t1 t2 t3 t4
X
Co
C'
t1<t2<t3<t4
x-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 0.5 1 1.5 2 2.5 3
C’
x
Time
xj
p-type SiP
Diffusion from a Constant Source
t1 t2 t3 t4
Co
C’
x
t1< t2< t3< t4
00
0
0
( , ) ( )2
(0,
( )
)
, 2
xC x t C e
DtM A C x t dx AC
C
cD
t
rft
C
: the amount of dopant entering the base
xj
p-type Sip-type SiP
P
Example: p-n junctionC’= Background Concentration
'
'0
( )
: junction distancee ( ); 2 j
j
C C p n or n p
xx
C C rfcDt
'
' ' ' 2
0 1
'
' ' 2'
. . ( ,0) ( ') . . ( , ) 0, (- , ) 0
( ) ( )( , ) lim exp( )42
( ')0
( ) ( )( , ) exp( )42
( )(
: initial distribution of concentration
x n
b
I C C x f x a x bB C C t C t
f x x x xC x tDtDt
f xwhen x
f x x xC x t dxDtDt
' ' 2
'
' ' 2'
'
' 2'
( ) ( )exp( ))42
( ) ( )exp( )42
( )
( )( , ) exp( )42
a b
a
b
a
i
b ia
f x x x dxDtDt
f x x x dxDtDt
when f x C
C x xC x t dxDtDt
Example
x
C/Co
Time
a b
1.0
0.5
0
x’
f(x’)
''
' '
22
2
22
2
22
,2 2
( , ) exp( )( 2 )2
exp( )
[ ( ) ( )]2 2 2
[ ( ) ( )]2 2 2
x biDt
x aDt
x bi Dt
x aDt
i
i
x xLet u dx DtduDtx a x bx a u x b u
Dt DtCC x t u Dt du
Dt
C u du
C x b x aerf erfDt Dt
C x a x berf erfDt Dt
-( )-( ) -
- -[ ( ) - ( )]2 2
2 22 2
2
i
i
i
C a berf erfD
C bC ae err fD t
f
D
t
t t
Dx x
x x
. . ( , ) ( , ) 0
B CC t C t
0 a b
Ci
Ci
0 a b
-( , ) - [ - ]( )2
)2
-(2
oo
CC ae berfrfD
tD
Ct t
xx x
0 a b x
C0
0 a b x
C0 Time
1 -( , ) [ ( )]2 2oC bC t erf
Dtxx
1 -( , ) [ ( )]2 2oC aC t erf
Dtxx
+
ll
- -( , ) - [ ( ) - ( )]2 2 2o
oC a bC t C erf erf
Dt Dtx xx
0 a b x
C0
0 a b x
C0 Time
Time
C0-
B.C. C(-a,t)=C(a,t)=Co
-a 0 a
-(( ) ( ) ( )2
[ (,2
) )]oa aererf fca b C t C
Dc
tDtxx x
( , ) ( )2o
xC t C er fcD t
ax
-( , ) ( )2oaC t C erfc
Dtxx
-a 0 a
-a 0 a+
(a)
(b)
Semi-infinite
'
' ' 2 ' ' 20 ' '
0
' '
( ,0) ( )
( ) ( ) ( ) ( )( , ) exp( ) exp( )4 42 2
(0)
)( (
0)
,C x f x
f x x x f x x xC x t dx dxDt DtDt Dt
f x f x odd functi nC t
o
(2)
'
' ' 2 ' ' 20 ' '
0
' '
( , 0) ( )
( ) ( ) ( ) ( )( , ) exp( ) exp( )4 42 2
( ) (
, )
)
( 0
0
C x f x
f x x x f x x xC x t dx dxDt DtDt Dt
f x f x even functi
C tx
on
(1)f(x')
f(-x')
x = ∞x = -∞
f(x')f(-x')
∞-∞ 0
)
-[ ( ) - ( )]2 2 2
- - -(
( , )
-[2 ( ) - ( ) - ( )]
- [ ( ) - ( )]
2
2
2 2
2 2
2
o
o
o
C x x aerf e
C x t
C x x a x aerf e
rfD
rf erfDt D
C x x aerf er
t Dt
D
fDt
t
Dt
t
Equivalent I.C.
C(x,0)=Co 0≤x≤aC(x,0)=-Co –a≤x≤0C(x,0)=0,x>a, x<-a
I.C.C(x,0)=Co 0≤x≤aC(x,0)=0 x>a, x<0
FictitiousSource
B.C. C(0,t)=0
y
-y
≣
0 a
Co
a-a
Co
-Co
Example
' ' 20 '
' ' 2'
0
' '
. .( ,0) 0
( ,0) 0 0
. .(0, ) 0
( , )
( ) ( )( , ) exp( )42
( ) ( )exp( )42
( ) , ' 0; ( ) ' 0
. .
( ,0) 0
( ,0) 0
( , ) (2
i
i
i i
i
i
i
I CC x C x
C x x
B CC t
C t C
f x x xC x t dxDtDt
f x x x dxDtDt
f x C x f x C x
Equivalent I C
C x C x
C x C x
xC x t C erf
( 0)) xDt
0 x
Ci
Fictitious Source
Ci
-Ci
Example
' ' 2'
' 22 '
2
. .( ,0)
( ,0) 0 ,
. .( , ) 0 ,
( ) ( )( , ) exp( )42( )exp( )
42(2 )[ ( )
2 2
. .( ,0) 2 2 ,
i
nl h inl h
i
i
I CC x C h x h
C x x h x h
B CC x t x l x l
x
f x x xC x t dxDtDt
C x x dxDtDt
C x nl herfDt
Equivalent I CC x C nl h x nl h n
(2 )( )]2
x nl herfDt
0 h-h l-l
Ci
0 h-h l-l 2l-h 2l 2l+h 3l
-l -h h l
-l -h h l
Example . .( , ) 0 ,
B CC x t x l x l
x
-l -h h l
-l -h h l
-l -h h l
-l -h h l
. .( ,0) 2 2 ,i
Equivalent I CC x C nl h x nl h n
( ,0)(0, )
( ,0)
(0, ) 0
( , ) ( )2
( ) ( )2
( )2
i
s
s
i s i
i
s i s
s
si
C x CC t CDefine C C
x C C
t
xx t erfDt
xC C C C erfDt
C C xerfC C Dt
Example
0 x
Cs
Ci
Time
Example: CarburizationA piece of AISI 1020 steel is heated to 1255K and subjected to acarburizing atmosphere such that
is in equilibrium with 1 wt%C in solution at the surface2 ( )2 (1)sCO CO C
Cs is fixed by eq. (1) =1 wt%CC2=0.2 wt%C
7 2 1
2
2 10 1255
( )2
c
s
s
D x cm s at K
C C xerfcC C Dt
When t = 3600 sC=0.35 wt%C at x = 5x10-2 cm
1.0
0.2
TimeC
(wt%)
Fe-C Phase Diagram
Series SolutionsSmall system + long timeReal solution to all systems without assumptions of “Infinite System”
Assuming the solution can be represented by
2
2
'
2
2
( , ) ( ) ( )
assum ing ( )
( ) ( ) "
dT D Td
C x t x T tC CD D D Ct x
x T t Dt x
T
Separation of Variables
x/h
C/Co
Time
-5 -4 -3 -2 -1 0 1 2 3 4 5
1.0
0.5
0
' "
'
"
"
:
:
( , )
TDT
Divide both sides by C x tdT
DTdtT T
TDT
functio
function onl
n o
y o
nly
f di
of
st
time
ance
20
2
ex
1
p( )
dT DT dtT T Dt
where T0 is a constant, -λ2 is chosen because one deals with the systemin which any inhomogeneities disappear as time passes, i.e., T approaches zero as time increases.
Since they vary independently, both sides must be equal to a constant, designated as -λ2 where λis a real number
22 0d
dx
x
The equation in is
the solution to this equation is of the form'( ) sin( ) cos( )x A x B x
where A’ and B’ are functions of λ
2 '0
2
( , ) ( ) ( )exp( )( sin cos )
( sin cos ) exp( )
C x t x T tT Dt A x B x
A x B x Dt
But if this solution holds for any real value ofλ, then a sum of solutions with different values ofλ is also a solution. Thus in its most general formthe product solution will be infinite series of the form
2
1( , ) [( sin cos )exp( )]n n n n n
nC x t A x B x Dt
2
1
0
. . ( ,0) 0
(0, ) 0 0 ( , ) 0 the argument of sin is equal to zero
where is a positive integer
( , ) ( sin )(
. . ( , )
exp( ))
( ,0
0
)
0 o
n
n
n
n n nn
B C C x tI C C x C x h
C t Bx and
C h t xn nh
C x t A x
x
Dt
C C
h
x
1
1
sin (0 )
sin( ) ?
n nn
n nn
A x x h
nA x Ah
00 01
Multiplying both sides by s
in( ) and integrate over the rang
sin
e of 0 to determine
( ) sin
( ) s
(
in )h h
n
n
n
p
p x n x p xC dx A dxh h
x x x h Ah
h
0
01
1
sin ( ) sin( )
s
2
in ( ) sin( ) 0h
nn
h
n nn
n x p x
n x p x hn
n p A dx
p A d Ah h
h h
x
Example :”Diffusion out of a Slab”
2
1( , ) [( sin cos ) exp( )]n n n n n
nC x t A x B x Dt
Time
Co
0 h
00 0
000
0 0
0
0
0
0
2 22 sin( ) ( ) cos( ) cos( )
2 20 [cos( ) cos( )] [1 cos( )]
: 04 4: 2 1
4 1( ,
0,1,2......(2 1)
)(2 1)
hh
nh
n
n
j
C Cn x h n x n xA C dxh h h n h n hC Cn n h n
n h h nn even A
C Cn odd j A jn j
The solution Cx tij
s C
2(2 1) (2 1)sin( ) exp( ( ) )j x j Dth h
Note: Each successive term is smaller than the preceding one, and the percentage decreases between terms and increases exponentially with time. Thus after a shorttime has elapsed, the infinite series can be satisfactorily represented by only a fewterms. To determine the error, we compare the ratio of the maximum values of thefirst and second terms (R) 2
2
2
2
83exp( )
100 4.75
(4.75)
R whe
DtR
n h Dt
htD
h
The error in using the first term to represent C(x,t) is less than 1% at all points.
D
t
is not linear
1o
CnC
C
Degassing of MetalsIt is difficult to measure the concentration of gas at various depths, and what is
experimentally determined is the quantity of gas which has been given off or the quantity remaining in the metal. Therefore, the average concentration ( ) is used.
0
202 2
0
1 ( , )
8 1 (2 1)exp( ( ) )(2 1)
h
j
C C x t dxhC j Dt
j h
0( ) 0.8C t CWhen the first term is a good approximation to the solution or when t issufficiently large
2
2
20
: relaxation time
8 exp( )
Large slow process
hD
C tC
(1st term is not a good approximation)
Fourier SeriesExample: Initial concentration is a sum of sine series
o n nn=1
nn=1
C = A sin (λ )
nπ= A sin( )
x
xh
∞
∞
0 h x
y
y
w h e n
xYh
Yx h
n=1
n=2
n=3
1sin( )n n
nA x
Co
n=3
Fourier Series
0
1
2
1 9
8( , 0 ) 2 , ( , 0 ) 2 s in ( )
8 8( , 0 ) 2 s in ( ) s in (3 )3
8 8 8( , 0 ) 2 s in ( ) s in (3 ) . . . . . . s i
8
n (3 9
s in ( 2 1)( , 0
)3 3 9
) 22 1
oC x C x x
C x x x
C x
C x
x
j
x
j x
x
. .( , ) 2
0, , 2 ........
B CC x tx
-π 0 π 2π 3π x
C
C-- j=1-- j=2-- j=3-- j=11
Example:
0
0
20 2 2
0
0
200
0
. .( ,0) 0 0
( ,0) 0,
. .( , ) 0,
8 1 (2 1)[1 exp( ( ) )](2 1)
1 ( , )
4 1 (2 1) (2 1)sin( )exp( ( ) )(2 1)
h
j
j
C C x t
I CC x x h
C x C x x h
B CC x t C x h
jC
dx
t
h
C Dj h
C j x jC C Dtj h h
Time Co
0 h
Example:
0
2
2 2
2 2 20
2 2
20
. .( ,0)
( ,0) 0 ,
. .
(0, ) 0
( , )
8 1 (2 1)exp[ ](
4 ( 1) (2 1)exp( )(2 1)
(2 1)cos( )2
1 ( , )
8 exp( )
4
2 1) 4
i
ns
s
s
ni
L
ni s
s
n x
I CC x C L x L
C x x L x L
BCC tx
C L t C
C C n DtC C n
L
C
L
C x
C C n DtC C
t dx
L
L
t
n
2
2
(only the first term is taken)
4LwhereD
Time
Ci
-L 0 L
Cs
Phase Transformation in Pb-Sn Alloy System
Solidification of an off-eutectic Alloy -- Coring
Transverse section
L0 x
last to solidify
first to solidify
time
0 L
Co
As-cast
completely homogenized
short time
(Ref: Crank’s Book p.63)
Example : Homogenization of AlloysCoring: Diffusion of most alloying element in the solid state is too slow to maintain a
solid if uniform concentration is in equilibrium with the liquid during solidification.
. . : ( ,0) ( )
. . : (0, ) 0
( , ) 0
I C C x f xCB C txC L tx
2 20 2
0
( , ) cos exp( )
2 ( )cos
n
L
n
n x DtC x t C A nL L
n xA f x dxL L
Dendrite
0 0 .C at x and x lx
Numerical ApproximationsError Function Series
Sequence of mirror reflection to provide no-flow boundary conditions
0 l
0 h l
Co
∵∂C/∂x|x=0 =0
∂C/∂x =0
0 h l
Co
-l -h
2( ')( , ) exp( ) '42
h oh
C x xC x t dxDtDt
∂C/∂x =0
2l 3l
Co
l
∂C/∂x|x=l =0
0 h -l -h
∵∂C/∂x|x=l =0
22
2
2( ')( , ) exp( ) ' ( '42
)exp( ) '42
h o l h
l hhoCC x x dx
Dx xC x t dx
Dt DDt tt
( ,0) 0( ,0) 0
0 0,
oC x C x hC x h x l
C x x lx
I.C.
B.C.
22
20
0
0 ........
( ')( , ) exp( ) '42
'2
2 2( , ) ( [ ( ) ( )])2 2 2
0
| 0 | 0.,
nl h onl h
o
x lx
C x xC x t dxDtDt
x xuDt
C x nl h x nl hC x t erf erfDt Dt
x l
Cx
Cx
0
1 13 3[ ( ) ( )]
2 2
1 1 1( , ; ) 2 213 3 3[ ( ) ( )]2 2 2
1 12 23 3[ ( ) ( )]
2 2
x xl lerf erf
Dt Dtl l
x xxC tl l lerf erfC Dt Dt
l lx xl lerf erf
Dt Dtl l
Error Function (3 terms)
Note: When increases, more error function terms would be needed because error functions spread indefinitely and therefore do not provide support of the no-flow boundary conditions.
Time
C/C o
1.0
0.5
0
x/l
2 Dtl
13
hl
2 2
10
( , ) 1 2 1 2sin( )cos( )exp[ ( ) ( ) ]3 3 2n
C x t n n x n DtC n l l
Note: At short times (small ), more Fourier terms would be needed to
eliminate spurious ripples and negative concentrations in the solution.
Fourier Series (15 terms)
2 Dtl
13
hl
Time
0
1
C/C 0
2(Dt)1/2/l=0
0.2
0.4
-0.2
0.6
0.8
0.8
1
1
1.2
0 0.2 0.4 0.6
0.750.50.250.1
∞
Dimensionless Distance (x/l)
0.0250
Short Time Solution2 0.025Dt
l
Fourier Series(15 term)2(Dt)1/2/l=0.025
Error Function
13
hl
C/C 0
0.8 10.2 0.4 0.6
Dimensionless Distance (x/l)
0
0.2
0.4
-0.2
0.6
0.8
1
1.2
0
Method of Laplace TransformLaplace transform to the diffusion equation is to remove the time variable leaving an ordinary differential equation, the solution of which yields the transform of the concentration as a function of the space variables (x,y,z). (Solve p.d.e. by making it an o.d.e.) Definition
0( ) ( )ptF p e f t dt
F(p) is called the Laplace transform of the original function f(t), and denoted by L(f(t))
0( ) ( ) ( )ptL f F p e f t dt
This operation is called Laplace Transform.
1( ) ( )f t L Fis called inverse transform or inverse of F(p)
22 2
23 2 2
1 2 2
2 2
( ) ( ) ( ) ( )111
1 c o s ( )
2 ! s i n ( )
! c o s h ( )
s i n h ( )
a t
nn
f t L f f t L f
ep p apt p w t
p wwt w tp p wpnt a tp p a
aa tp a
'
2
( ) ( ) (0)( ) ( ) (0) (0)
L f pL f fL f p L f pf f
( ,0) 0 0C x x
Example: Semi-infinite SystemI.C.
B.C. 0(0, )C t C
(Note: diffusion in a semi-infinite medium, x>0 when the boundary is kept at a fixed concentration. e.g. doping problem)
2
2C CDt x
2
20 0
1 0pt ptC Ce dt e dtD tx
multiplying both sides by e-pt and integrating with respect to time from 0 to ∞, one obtains
Assuming the orders of differentiation and integration can be interchanged
0
2 2 2
2 2 20pt ptC Ce dt Ce dt
x x x
0
: ( ) ( ) ( , ) ptNote C L C F p C x t e dt
CoTime
x0
0
'
2
2
00
2
0
1 2
0
01
( ) (0 ) 0
N o te : ( ) ( ) (0 )
0
. . : (0 , ) ( )
( , ) 0 0
p t
pD
p pD Dx x
Ce d t p L C C p Ct
L f p L f f
C p Cx D
B CCC
C A e
t C L C Cp
C t A
A
CC A
e
e
0
0
1
0
1
F ro m T a b le (
(
)
)2
pD x
CAp p
CC ep
xC C er fcD t
L C C
I.C.
0
2
2
0
0
0
0
Note: cosh(
. .( , 0) 0. .( , ) ,
(0, ) 0
0 0
0 at 0
at ,
)2
=
4 ( 1)
cosh( )
cosh(
(2 1)
)
x x
n
n
I CC x l x lB CC x t C x l x l
C tx
C p C x lx DC xx
CC x l x l
e
p
expC xDCpp D
CC Cn
l
2 2
20
(2 1) (2 1)exp( ) cos( )4 2
n Dt n xll
Note: please check Crank’s Book pages 22-24 to have the solution
Example
Time Co
-L 0 L
0
0
2
2
1 2 3
0 1
2 3
2
20
2
. . : ( , 0) 0. . : ( , )
( , ) 0
( , )
(0, ) (0, )( ) ( )
(0, )
( ,
(0,
)
)(0, )
p px xD D
p xD
p xD
I C C x C xB C C t C
C CDt x
C k e k e kC t C k
C x p k e k
C t C p AL D D L Ax x p
C p Ax Dp
C x p p k
C tx
ex D
J t A D
CC p CD Dx
2(0, )C p p Akx D Dp
Example: Constant Flux at the surface
Co
x0
C
2 312 2
AkD p
To get k3, go back to partial differential equations
2
20
2
02 2 3
0 03 3
031
2 2
121 4
12
32
( , ) ( , )
[ ]
( , ) (*)
(Ref:Crank Book No.9)
( ) 2 ( )2
multiplied
p px xD D
p xD
xqxDt
p xqx D
CC x p p C x px D D
Cp pk e k e kD D D
C Cp k kD D p
CAC x p epD p
pq D
e Dt xL e xerfcpq Dt
e D epq
p
1 1 1
by
[ .(*)] (Crank No.9) ( )o
AD
CAL eq L LD p
2 3 ( , )p xDC x p k e k
2
2
142
0
142
0
12
0
Check ( ,0)
(0, )
( , ) 22( ) ( ) ( ) ( ( ))4 2 2
(0, ) (
( , ) 2( ) (
0, )
)2
2(0, ) ( )
xD
xDt
t
C x C
C tD Ax
C x t A Dt x x xe erf
A Dt xC x t e x
c x erfcx D Dt xDt Dt
C t A C tA
erfc CD Dt
Dx D x
A DtC t CD
Solutions for Variable D2
2( )C C D C CD Dt x x x x x
Dx
makes this equation inhomogeneous, especially when D=D(C) or D(T) or D(t) or D(x).
The key in solving the above p.d.e. is to simplify the equation with x and t to x or t function.
Boltzman-Matano Analysis (D=D(C))
32
32
12
1
1 ( )2
1 (
(2
)
)
xt
C C x Ct t tC C Cx x t
x C D C
C
x ttCD
Ct
D
and
Therefore
0
00
CxdC
'
' '
' '
''
'
0 0
00
0 0 00
1
( )2
1 |2
| | |2 ( )
C C C
C
C C
C
C
x C dC dCdC D D D tx d
CdC d D
dC
xt d
xdC D
t
tdx
* x=0 is not determined yetIf D≠D(C), C=Co/2 which determines x=0 for an infinite system. However, if D=D(C) the above condition is no longer valid, the x=0 must be determined by
which expresses the equality of the two shaded areas.
Example: Infinite System I.C. C(x,0)=Co x<0η= -∞C(x,0)=0 x>0η= ∞
Original interface
Co
X=0
Co
Equal area
Matanointerface
Time
X=0
Co
Matanointerface
C’
Tangent=(dC/dx)C’
Error-Function SolutionC/
C o1.0
0.5
0
x/l
Gain
Loss
0
0
00
CxdC
Constant Diffusivity
Error-Function SolutionC/
C o1.0
0.5
0
x/l
Gain
Loss
0
0
00
CxdC
Constant Diffusivity
D increases with decreasing concentration
For an infinite system
000 0 | 0C
odC dCwhen C or C Cdx dx
Therefore0
0
0
0
00
0
0
1 |2
| 0
0
C C
C
C
dCxdC Dtdx
dCdx
xdC
which is an additional boundary condition and determines the location of Matano interface.
'
'
0
0
'
0
1( ) ( )2
0C
C
C
dxD C xdCt dC
xdC
x=0 plane (Matano interface) determined by
X=0
Co
Matanointerface
C’
Tangent=(dC/dx)C’
0
''
00
1: |2
C CdCNote xdC Dtdx
:
:
:
S lo p e
D i f fu s iv i t y
A r e a
xx
x x *
1
2
* *
C * *
C * *
C
C
C *
C
*
C
C
>
<
=
∂ C |∂
D
d C
∂ C |∂
D
d C
'
''
0
1( ) ( )2
C
CdxD C xdC
t dC
C*
D2D1
C1
C2 Original interface
t =0
C1
C2
Matano interfacet =t
x=0
Areas equal
C1
C2
0
Slope=dC/dx at C*
C**
1 32
x xC'
C' 0
-1 dD(C'
> D
) = ( ) d
(C'D ( ) >
C2t dC
C ) D' (C')
123C’
t=0
Matano Interface
'
0
'
0
[ ( ) 0[ ( )]
1( ') ( ) ( )2
Integrated by parts ( )
. . : ( , ) ; ( ,
]
[ ( ) 0] |
[ ( )] |
( ) ( 0)
) 0
M
MM
M
LM
M
M
x
x
L
CxL C
C
L
M
C
C M
x C
M
L
M MM C
x xt
dxD C x x dCt dC
udv uv vd
C x dx
C x x xdC
C x
C C x dx
C C x x xdC
C C x x
u
B C C
dC
t C C t
0
0
0
0
( 0) 0
( 0) ( ) 0
L
M M
L
L M
M
M
C
L M C C
C C
L M C
C
C x xdC xdC
C x xdC x
xdC
dC
Loss
Gain
CM
C’(x)
xMx0
CL
x’
x’: Original interfacexM: Matano interface
If the coordinate of the Matano interface, xM, is selected as the origin of the x-axis, i.e., xM=0, then
0
0
0
0
0
L
M
L L
MC C
C
C CxdC dC
xdxdC C
Although the location of xM=0 is not known a priori, it may be found from concentration-distance data by balancing the area denoted Loss against the area of Gain.
Loss
Gain
CM
C’(x)
xMx0
CL
x’
Hardening of Steels
THROUGH HARDENING In order to harden steel, the iron mix must contain a certain amount of carbon. Carbon dissolves in molten iron. In through hardening steel, there is a high level of carbon added to theiron mix. When the component is heat treated, it becomes hard all the way through from the surface to the core, hence the term “through hardened”. Through-hardened steel components are relatively brittle and can fracture under impact or shock load.
The Moving Boundary Problem*Diffusion controlling process along with reaction at phase boundary
?dSdt Kinetic Issue
General Aspects
2
2
2
2
* *
.
* C
C
*
* Diffusion controlling process
:partition
and : equ
ratio between
ilibrium concentrations in phases I and II
phaC C
II
CS Dt x
CS Dt x
k
C C
k
x
x
Ⅰ ⅠⅠ
Ⅱ ⅡⅡ
Ⅰ
* *Ⅰ Ⅱ
* *
sesC* ( ) ( ) ( )x S x S
C dSD D C Cx x dt Ⅰ Ⅱ
Ⅰ Ⅱ Ⅱ Ⅰ
xCs
S
II I
C0
CII*
CI*
To have net diffusion flux between and , the thickness of + has to vanish.
Decarburization
orC C
C Ca a
==
Concentration
Tempe
ratu
reFr
ee E
nerg
yCo
ncen
trat
ion
Micro
stru
ctur
e
α
γα+γ
CαCγ
α γ
αγ
α+γ
CαCγ
α γ
α+γ
x
Cα
Cγ
S
α γ
T1
Carburization: -Fe -FeDecarburization: -Fe -Fe
Decarburizationforming phase at the interface
Carburizationforming phase at the interface
Decarburization
Carburization
T
C (wt%)
γα+γ
Cα* Cγ*
T* γ+Fe3C
910°C
723°C
0.0250.83
Ci1Ci
2
α
xCs
S
α γ
Ci1
Cα*
Cγ*
T=T*
x
Cs
αγ
Ci2
Cα*
Cγ*
T=T*
S
C Cα γμ =μ at x=S
where α and γ coexist
Sx
μc, ac
A: area for diffusion: constant
x x
Mass Conservation
x x
II I
II III I II =S
* *II
I
I
=S
J - J∂C ∂CJ -
( )A dt = (C - C )A dS
J -D ( ) - (-D )∂ ∂
=
S
dS
0
III
CO
C*I
C*II
CS
Fe-C Phase Diagram
CarburizationCs
C*
C*
Co
S0 x
T=T*
γ α γ α
known parameters:
?dSdt
* * C CS oC , C , C , C , D and D
2
22
2
4 2
42
2
[ ] ;[ ]
2[ ]
k
c
k
cH
CO C CO
k COa orCO
CH H Ck CHa
P
Example: Carburization
*Rate of advance of boundary controlled by diffusion of carbon in Fe. Therefore
C C
( C C )
x S k
if at x S reaction controlling process
* *
* *
*
*
C C at x S
C C at x S
* Semi-infinite solid( ) C C
CD D C D D
0V mass flux requiring no density correction
*Chemical activity of carbon at surface can be set up by
*
*
*
Fick’s 2nd law
*
2
2
2
2
0
*
0
*
*
*
*
0
.
( ) (
.
)
( ) (
: ( ,0)
. . :
0
( , )
)
( : )
C
C
s
C Cx s x
C CD x St x
C CD x S
t xI C C x C
B C
C C at x S
C C at x
C C at x S
C C at x
C s t C kC k Partitio
in phase
At x S
J J dt C C
in pha
dSC CD Dx x
se
n Ratio
*
* *
*
: ( ) (
)
)
(s
Note J J dt A C C dS A
dSC Cdt
ds
JSolute receiving
Solute entering
J
Cs
C*
C*
Co
S0 x
*
0
*
0
*
*
In phase
( ), 2
In phase
' ' ( ), ; 0 '2
( , ) ( )2
( , ) ' ( )2
At boundary
a
(
' )2
C
sC
C
s C
s C
xC A B erfc x C C AD t
xC A B erf x S C C x C C AD t
xC x t C B erfcD txC x t C B erfD t
x S C C
SC C B erfD
C
t
*
therefore is a constant too (con
nd are constants ( diffusion c
stant)2 2
2 where i
ontrolling
s a constan
process)
( , ) ' ( ) ' ( )2
t
S
s s
C C
C
C
C
SC S t C C B erf C B erf
S SD D
D
t
t
t
S D t
Cs
C*
C*
Co
S0 x
*
1 *2
2
2
0
0 0
' 2( ) exp( ) | (1)4
Similarly
Replace
2
2( ) exp( ) | (242
2( , ) (
( , ) ( )
) ( )2 2
CC
x s x sCC
CC
x s x sC
C
C
C C
C
C
DD
C S t C B e
C D B xDx D tD
D tSC S t C C B erfc C B erfcD t D t
rfc C
t
C D B xDx D tD t
2 2
* *
* *
2
(Note: it is not a c
.(1) (2) ( )( )
'
onst
)
ant)
C
C
S D
SEq C Ct
B e BeC C
t
DdSdt t
2 2
*
* ** * 0
1 12
1 *20
2
solve by trial and err
then we get S
or
( ) (
2 an
'
)
(
(
)
d
)s
s
CC
C C C CC Ce erf e e
C B erf C
C Berf
rfc
DdSdt t
c
D t
C
Note: the only unknown parameter in the above equation is β* *
0( , , , , )C CsC C C C D and D are known parameters
Use them to estimate B’ and B, and then to solve β
2
( )e erfc
β
1
0
Example: Decarburization0.4% Carbon Alloy SteelT=800oCCs=0.01% (equilibrium by CO and CO2)t = 30 min S-Fe?
Answer: Co=0.4%, Cs=0.01%C*=0.24%, C*=0.02% (Obtained from Fe-C Phase Diagram)
2 2
**0
8 6 2
1/2
1
*
2
*
2 2
/
1 1
3x10 ; 2x10 cm /sec
66.6
(0.01 0.02) (0.24 0.4)(0.02 0.24)( ) ( )
0.01 (0.24 0.4)0.
0.1
( )(
44 (
) ( )
22( ) ( )
C C
C
C
s
D D
DD
f f
C CC CC Ce erf e
by trial
erf
and error see
c
fF
f
2 0.0 7
)
1 3CS D t c
i
m
gure
Fe-C Phase Diagram
*
*0
*2
2* 2
0
*2
02 *
( , )
( ) ( )
( )
exp( )
elimina 1 ( ) exp ) (t ( )e s
II II
IIII x s II
s II
II
II
II
C C er
C S t CC dSD C Cx dt
C C B erf
BC C
B fC C
S
CI*
Cs
Co
723<T<910℃
CarburizationS
CI*
Cs
Co723<T<910℃
De-carburization
S
Fe3c
CI*
Cs
Co
T<723℃
De-carburizationS
+Fe3C
CI*
Cs
CoT>723℃
De-carburization
Formation of a single-phase layer from an initial two-phase mixtureIn the two-phase region, the average composition, Co, is assumed to be uniform, which requires, in effect, that the grain size is small and that second-phase dispersion is uniform.
3α+Fe C
*
*0
*2
2* 2
0
*2
02 *
( , )
( ) ( )
( )
exp( )
elimina 1 ( ) exp ) (t ( )e s
II II
IIII x s II
s II
II
II
II
C C er
C S t CC dSD C Cx dt
C C B erf
BC C
B fC C
S
CI*
Cs
Co
723<T<910℃
CarburizationS
CI*
Cs
Co723<T<910℃
De-carburization
S
Fe3c
CI*
Cs
Co
T<723℃
De-carburizationS
+Fe3C
CI*
Cs
CoT>723℃
De-carburization
Formation of a single-phase layer from an initial two-phase mixtureIn the two-phase region, the average composition, Co, is assumed to be uniform, which requires, in effect, that the grain size is small and that second-phase dispersion is uniform.
3α+Fe C
Time
723<T<910oC
Carburization Process
CS
CII*
γ α+γ
S
Co
* *2 2
* *1 1
* *1 2
2 1
1 2and are activity coefficient is a partition coefficient
a k C
a k C
C kCk k kk kk
*1 1
1 1
*2 2
2 2
C CJ Dx
C CJ Dx
Steady state without Interfacial resistance
No interfacial resistance*1*2
( )C k constC
* * 1 1 2 21 2
2 1
( )D C D CC C k kD kD
Note: *
1 2 1 1*
1 2 2 2
D D C C
D D C C
Steady State J1 = J2
(immobile boundary)
Δx Δx
C1
C2
I IIC2*
C1*
a1
I II
a2
a*
Concentration profile Chem. activity profile
a*
I II
CI*
CII*
DI>>DII
Diffusion Control inPhase II
I II
CII*CI
CI*
Mixed diffusion control
I II
CI CII*CI
CI*
Reaction @ interface of Phase I & diffusion control in Phase II
I II
CI*
CII*
DI<<DII
Diffusion Control inPhase I
1
2
0
*
*
. . : ( ,0) 0( ,0) 0 0
. . : (0, ) 0
(0, ) 0
I C C x C xC x x
B C C t C x
C t C x
Infinite composite system without Interfacial Resistance
*2*
1
1 21 2
1 2
0
0
C k at xC
C CD D at xx x
10 2 21 1
2 2 1 1
1
1 1 11
2 2 22
02 1
2 2 2
1
{1 ( ) ( )}21 ( )
( )
( )
02
( ) 0
21 ( )
2
xC A
C D xC k erfD D D
B erf x
tkD
xk
D tx
CC erf
C A B er
cD D tkD
f xD t
No accumulation at the interface
1
2
* 01
2 2
1
* 01
2 2
1
(0, )1 ( )
(0, )1 ( )
CC t DkD
kCC t DkD
x=0
0C2
C1
C0
*1C
*2C
(both of them are fixed)
(k=1, D2/D1=1C1*=C2*=Co/2)
(immobile boundary)
Chemicalactivity
11 1 2
1 21 2 1 2
( ) 0
( )
CD h C C xxC CD D h C Cx x
00
0.5Co
C0
C2C1t
t
Composite system with Interfacial Resistance
Contact Resistance at x=0
The concentrations on either side of the interface are no longer constant, but each approaches the equilibrium value of 1/2C0 relatively slowly. (if D1=D2)
1 20 2 21 1 1 1 1 11
2 2 1 1 1
1
202
11 21
1 2
12 22
2 1
2 2 2 2 212 2 2 2
1
{1 ( ) { ( ) exp( ) [ ]}}2 21 ( )
{ ( ) exp( ) ( )}2 21 ( )
{1 ( ) }
{1 ( ) }
C D x xC erf h x h D t erfc h D tD D D t D tD
x xCC erfc h x h D t erfc h D tD D t D t
DhhD D
DhhD D
D
h: Chemical reaction rate constant at interface
(if D2=D1, C1(0,∞)=C2(0,∞)=Co/2)
(immobile boundary)Chemical activity