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