glicksman slides

© meg/aol ‘02

Module 1: Laws of Diffusion

DIFFUSION IN SOLIDS

Professor Martin Eden Glicksman Professor Afina Lupulescu

Rensselaer Polytechnic Institute Troy, NY, 12180

USA

© meg/aol ‘02

Outline • Preliminary definitions

• Phenomenological laws

• Units

• Fick’s first law

• Vector form of Fick’s first law

• Mass conservation

• Fick’s second law

• Fick’s second law in different coordinate systems

• Diffusion symmetries

• Exercises

© meg/aol ‘02

Definition

Diffusion: a molecular kinetic process that leads to homogenization, or mixing, of the chemical components in a phase.

water

adding dye partial mixing homogenization

time

© meg/aol ‘02

Kinetic Characteristics

D = D0 ⋅e− ∆E / RgT

Diffusion is also a thermally activated process that follows the well-known relationship,

where

D = Diffusion coefficient D0 = Pre-factor ∆E = Activation energy Rg = Universal gas constant T = Temperature

© meg/aol ‘02

Phenomenological Laws

ϕ = −Akµ

d Pd x

Darcy’s Law

Q = −kdTd z

Fourier’s Law

J = −Dd Cd x

Fick’s First Law

Porous flow

Heat flow

Mass flow

© meg/aol ‘02

Fick’s First Law

Jx = − D ∂C∂x

y, z,t

Jy = − D∂C∂y

z ,x ,t

Jz = − D ∂C∂z

x ,y ,t

J x, y,z( ) = Jxex + Jyey + Jzez Mass flux vector

Component form of Fick’s 1st law: (Isotropic system)

Jx Jz Jy

y

z

x

J P (x, y, z)

© meg/aol ‘02

Mass-flux Vector

x

y

z

J

Jxex

Jzez

Jyey

Flux vectors may be visualized at each point in a medium by specifying their directions and magnitudes.

© meg/aol ‘02

1 0.5 0 0.5 1

1

0.5

0

0.5

1

Gradient Field

,M N x

y

negative divergence

zero divergence

positive divergence

Gradient Vector Field

dC/dy

dC/dx

© meg/aol ‘02

Divergence Operators

divF =

∂Fx

∂x

+

∂ Fy

∂y

+∂ Fz

∂z

Cartesian Coordinates:

Cylindrical Coordinates:

Spherical Coordinates:

divF =

1r

∂∂r

r ⋅Fr( )+1r

∂ Fθ

∂θ

+

∂ Fz

∂z

divF = 1

r2∂∂r

r2 ⋅Fr( )+1

rsinθ∂

∂θsinθ ⋅Fθ( )+

1rsin θ

∂ Fφ

∂φ

© meg/aol ‘02

Fick’s First Law

∇C (x, y, z,t) ≡

∂C∂x

y ,z,t

ex +∂C∂y

z ,x ,t

e y +∂C∂z

x ,y ,t

ez

Vector form of Fick’s first law:

J = − D∇C

© meg/aol ‘02

Units

Ji units[ ]=g

s ⋅ cm2

Ji = −D∂C∂x

Ji units?[ ] = Dcm2

s

⋅

∂C∂x

gcm 4

, i = x, y, z( )

© meg/aol ‘02

Mass-flow Control Volume

P

x

y

z

∆y

∆z

∆x

Jy(P-∆y/2) Jy(P+∆y/2)

Control volume indicating mass flows in the y-direction occurring about an arbitrary point P in the Cartesian coordinate system, x,y,z.

© meg/aol ‘02

Mass Conservation

Inflow − Outflow = Accumulation Rate ( A.R.)

Jy P −∆y2

− Jy P + ∆y

2

∆x∆z = A.R.( )y

Jy −∂Jy

∂y

⋅∆y2

− Jy +∂Jy

∂y

⋅∆y2

∆x∆z = A.R.( )y

− ∂Jy

∂y

∆y∆x∆z = A.R.( )y

© meg/aol ‘02

Mass Conservation

Total Inflow − Total Outflow = A.R.( )total

− ∂Jx

∂x+

∂Jy

∂y+

∂Jz

∂z

∆x∆y∆z = ∂C∂t

∆x∆y∆z

−∇ ⋅J = ∂C

∂t

∇⋅J = ex ∂ / ∂x( )+ ey ∂ / ∂y( )+ ez ∂ / ∂z( )[ ]⋅J x, y,z( )

∂C∂t

+ ∇ ⋅J = 0or continuity equation

© meg/aol ‘02

Fick’s Second Law (Linear Diffusion Equation)

− ∇⋅ −D ∇C( ) = ∂ C∂t

or ∇2 C =1D

∂C∂t

D ∇⋅∇C( ) = ∂C

∂tFick’s 2nd law

© meg/aol ‘02

Fick’s Second Law (Cartesian Coordinates)

∇2 ≡

∂2

∂x2 +∂2

∂y2 +∂2

∂z2

D

∂2 C∂x2 +

∂2C∂y2 +

∂2C∂z2

=

∂C∂t

Laplacian

© meg/aol ‘02

Fick’s Second Law (Cylindrical Coordinates)

∇2 ≡ 1r

∂∂r

r ∂∂r

+

∂∂θ

1r

∂∂θ

+

∂∂z

r ∂∂z

Dr

∂∂r

r ∂C∂r

+

∂∂θ

1r

∂C∂θ

+

∂∂z

r ∂C∂z

= ∂C∂ t

Laplacian

© meg/aol ‘02

Fick’s Second Law (Spherical Coordinates)

∇2 ≡ 1r 2

∂∂r

r2 ∂∂r

+

1sinθ

∂∂θ

sin θ∂

∂θ

+

1sin2 θ

∂2

∂2φ

Dr 2

∂∂r

r2 ∂C∂r

+

1sinθ

∂∂θ

sinθ∂C∂θ

+

1sin2 θ

∂2 C∂ 2φ

= ∂C∂t

Laplacian

© meg/aol ‘02

Important Diffusion Symmetries

Linear Flow (1-D flow)

D ∂2 C∂x 2

= ∂ C

∂ t(D = constant)

∂∂x

D ∂C∂x

= ∂C∂ t

(D = variable)

© meg/aol ‘02


Axisymmetric (radial) Flow

D ∂2 C∂r2

+1r

∂C∂r

= ∂C

∂t

1r

∂∂r

rD ∂C∂r

= ∂C

∂t

(D = constant)

(D = variable)

© meg/aol ‘02


Spherical symmetry

D ∂2 C∂r2

+2r

∂ C∂r

= ∂C

∂ t

1r2

∂∂r

r2D ∂C∂r

= ∂C∂ t

(D = constant)

(D = variable)

© meg/aol ‘02

Exercises

1. Given the concentration field C=sin(πx) [g/cm3], plot the gradient field, and then determine the flux field using Fick’s 1st law, assuming that the diffusion coefficient, D=1 [cm2/s]. The use of a mathematical software package or spreadsheet provides an invaluable aid in solving diffusion problems such as suggested by these exercises, and many others introduced in later chapters of this book. In one dimension, the gradient is defined as

∇C ≡

∂C∂x

⋅ex = πcos πx( )⋅ex

© meg/aol ‘02

Exercises The flux field may be found using Fick’s 1st law.

J = −D∇C = −1⋅ πcos πx( )( )⋅ex

1 0.5 0 0.5 1

1

0.5

0

0.5

1

Gradient Field

,M N

y

x

A

1 0.5 0 0.5 1

1

0.5

0

0.5

1

Gradient Field

,M N x

yy

BGradient Plot Flux Plot

Note: These plots differ by a minus sign!

© meg/aol ‘02

1 0.5 0 0.5 1

1

0.5

0

0.5

1

Gradient Field

,M N

Exercises 2. A two–dimensional concentration field is given by C(x,y)=-sin(πx)cos(πy). Calculate the gradient field, and make plots of it and the given concentration field.

∇C ≡

∂C∂x

⋅ex +

∂ C∂y

⋅ey = −π cos πx( )cos πy( )⋅ex − sin πx( )sin πy( )⋅ey( )

Concentration field Gradient field

© meg/aol ‘02

• Diffusion is a molecular scale mixing process, as contrasted with convective, or mechanical mixing. Mixing in solids occurs by diffusive motions of atoms or molecules.

• Fick’s 1st and 2nd laws are based on fundamental physics (mass conservation) and empirical observations (experiments).

• Diffusion involves scalar fields, C(r, t), which have associated vector gradient fields, ∇ C(r, t).

• Fick’s 2nd law is in general non-linear, but is usually approximated as a linear, 2nd-order PDE.

• Special symmetries are often usefully evoke to obtain closed-form, analytical solutions for estimation purposes.

• Modern computational software packages make it easier to visualize the behavior of diffusion fields.

Key Points

© meg/aol ‘02

Module 2: Diffusion In Generalized Media

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline • Diffusivity tensor • Principal directions

– Antisymmetric contribution – Symmetric contribution

• Diffusion in generalized media • Cauchy relations • Influence of imposed symmetry: Neumann’s Principle

– Rotational symmetry operations – Isotropic materials – Cubic crystals – Orthotropic materials – Orthorhombic crystals – Monoclinic crystals – Triclinic crystals

decreasing symmetry

© meg/aol ‘02

Diffusivity Tensor

Ji = − Dij ∂C∂x j

i, j =1,2,3( )

The diffusivity is defined operationally as the ratio of the flux magnitude to the magnitude of the concentration gradient. Equivalently, the diffusivity is the constant of proportionality between flux and gradient.

Tensor form for Fick’s 1st law.

J r( ) = J1e1 + J2e2 + J3e3

The vector flux may be expanded as e3

e2 e1

Triad of unit vectors

© meg/aol ‘02

Diffusivity Tensor

Dij[ ]=

D11 D12 D13

D21 D22 D23

D31 D32 D33

Elements comprising the matrix diffusivity.

J1 = −D11∂C∂x1

− D12∂C∂x2

− D13∂C∂x3

J2 = −D21∂ C∂x1

− D22

∂C∂x2

− D23

∂C∂x3

J3 = −D31∂C∂x1

− D32

∂ C∂x2

− D33

∂C∂x3

Fick’s 1st law in component form Cartesian coordinates

© meg/aol ‘02

Binary Diffusivity Tensor

A = antisymmetric component

S = symmetric component Dij[ ]S≡ 1

2 Dij + Dji( )= Dji[ ]S

Dij[ ]A≡ 1

2 Dij − Dji( )= - Dji[ ]A

Dij[ ]S+ Dij[ ]A

Dij[ ]=

Any square matrix may be decomposed into the sum of a symmetric part and an antisymmetric part.

© meg/aol ‘02


A = antisymmetric matrix:

S = symmetric matrix:

Dij[ ]S=

D1112 D12 + D21( ) 1

2 D13 + D31( )12 D21 + D12( ) D22

12 D23 + D32( )

12 D31 + D13( ) 1

2 D32 + D23( ) D33

Dij[ ]A=

0 12 D12 − D21( ) 1

2 D13 − D31( )12 D21 − D12( ) 0 1

2 D23 − D32( )12 D31 − D13( ) 1

2 D32 − D23( ) 0

Dij[ ]S+ Dij[ ]A

Dij[ ]=

© meg/aol ‘02

S = symmetric matrix:

Dij[ ]S=

D1112 D12 + D21( ) 1

2 D13 + D31( )

12 D21 + D12( ) D22

12 D23 + D32( )

12 D31 + D13( ) 1

2 D32 + D23( ) D33

Dij= Dji


© meg/aol ‘02

A = antisymmetric matrix:

Dij[ ]A=

0 12 D12 − D21( ) 1

2 D13 − D31( )

12 D21 − D12( ) 0 1

2 D23 − D32( )

12 D31 − D13( ) 1

2 D32 − D23( ) 0

Dij= -Dji


© meg/aol ‘02

Diffusion in Generalized Media

∂C∂t

= −∇⋅J

∂C

∂t= ∇⋅ Dij[ ]S ∂ C

∂xj

e j + Dij[ ]A ∂C∂x j

e j

Mass conservation

∂C

∂t= −∇⋅ − Dij[ ]⋅

∂C∂x j

e j

symmetric response antisymmetric response

© meg/aol ‘02

where the gradient operator is expressible as a column matrix

Diffusion in Generalized Media

∂C∂t

= ∇⋅ Dij[ ]S⋅∇C+ ∇⋅ Dij[ ]A

⋅∇C

∇⋅ Dij[ ]S=

∂∂x1

∂∂x2

∂∂x3

⋅ S

∇⋅ Dij[ ]A=

∂∂x1

∂∂x2

∂∂x3

⋅ A

In general:

© meg/aol ‘02

Diffusion in Generalized Media Antisymmetric response

∂ C∂t

A

=∂2 C∂x1

2

0( ) +

∂2 C∂x1∂x2

12 D12 − D21( )[ ]+

∂2 C∂x1∂x3

1

2 D13 − D31( )[ ]

+∂2 C

∂x2∂x1

1

2 D21 − D12( )[ ]+∂2 C∂x2

2

0( )+∂2 C

∂x2∂x3

12 D23 − D32( )[ ]

+∂2 C

∂x3∂x1

1

2 D31 − D13( )[ ]+∂2 C

∂x3∂x2

12 D32 − D23( )[ ]+

∂2 C∂x3

2

0( )

∂2 C∂xi∂x j

=

∂2 C∂x j∂xi

Order of differentiation is inconsequential!

© meg/aol ‘02

Cauchy relations

∂ C∂t

A

= 0

Dij[ ]=

D11 D12 D13

D12 D22 D23

D13 D23 D33

The diffusivity is a symmetric tensor containing at most 6 elements:

So the antisymmetric part contributes nothing!

© meg/aol ‘02

Neumann’s Symmetry Principle

Dij'[ ]=

∂xi'

∂xk

⋅∂x j

'

∂xl

Dkl[ ] Tensor transformation rule (i, j, k, l=1, 2, 3)

Dij'[ ]= αikα jl Dkl[ ]= α[ ] Dkl[ ]α[ ]T

Dij'[ ]= Dij[ ]

Neumann’s principle states that after any symmetry operation on the coordinate system

Direction cosines

transpose

© meg/aol ‘02

3-D Matrix Rotation: Implemented by Mathematica®

• For an arbitrary rotation, θ, in 3-D about axis x1

α Dij

• Other arbitrary rotations about axes x2 and x3 must then be applied.

αT

© meg/aol ‘02

Symmetry Operations for Diffusivity Tensors

Four-fold rotation by π/2 about x1-axis

α[ ] =1 0 00 0 10 −1 0

x

3

x

( a)

2

x

1x1'

x2'

x3'

π/2

x2

x3

x1

( b)

x1'

x2'

x3'

π / 2

© meg/aol ‘02


Two-fold rotation by π about x1-axis

α[ ] =1 0 00 −1 00 0 −1

x

3

x

( a)

2

x

1x1'

x2'

x3'

1

( c)

π

x2

x3

x

x3'

x1'

x2'

π

© meg/aol ‘02


Three-fold rotation by 2π/3 about x1-axis

α[ ] =1 0 00 − 1

23

2

0 − 32 − 1

2

Important in hexagonal and rhombohedral systems.

x

3

x

( a)

2

x

1x1'

x2'

x3'

2π/3

( d)

x3

x2

x1

x3'

x1'

x2'

2π / 3

© meg/aol ‘02


No rotation Orthogonal coordinates

Identity matrix:

α[ ] =1 0 00 1 00 0 1

x

3

x

( a)

2

x

1x1'

x2'

x3'

x

3

x

( a)

2

x

1x1'

x2'

x3'

x1′

x2′

x3′

© meg/aol ‘02

Isotropic Materials

D is a scalar.

—“Isotropy” is the lack of directionality—

Flux vector, J, remains antiparallel to the applied concentration gradient, ∇C, and is invariant with respect to the gradient’s orientation within the material.

Dij[ ]=D 0 00 D 00 0 D

= D .

© meg/aol ‘02

Cubic Crystals

x1< 1 0 0 >

< 0 1 0 >

< 0 0 1 >x3

x2

Typical structure of many engineering materials. Includes FCC and BCC metals and alloys, and many cubic ceramic and mineralogical

systems.

© meg/aol ‘02

Neumann’s principle applied to cubic symmetry

Diffusivity tensor for cubic symmetry, where D11 = D22

Dij'[ ]=

D11 D13 −D12

D31 D33 D32

−D21 −D23 D22

Dij'[ ]=

D11 0 00 D22 00 0 D22

D22 = D33

D12 = D13

D13 = −D21 = 0D32 = −D23 = 0.

Element-by-element comparison shows

π/2

x2

x3

x1

( b)

x1'

x2'

x3'

π / 2

= D (scalar)

© meg/aol ‘02

Diffusivity tensor for orthotropic materials (Tetragonal, Hexagonal, Rhombohedral)

Dij = 0 i ≠ j( )

Dij[ ]=

D11 0 00 D11 00 0 D33

These crystals require two independent diffusivity elements.

© meg/aol ‘02

Diffusivity Tensor: orthorhombic, monoclinic, triclinic crystals

Triclinic Symmetry arguments fail to reduce the number of independent elements in the diffusivity tensor of triclinic crystals. 6 elements are needed to describe diffusion responses in such low symmetry materials.This symmetry, although rare in engineering systems, exists in nature.

Orthorhombic Dij[ ]=

D11 0 00 D22 00 0 D33

Monoclinic Dij[ ]=

D11 0 D13

0 D22 0D13 0 D33

© meg/aol ‘02

Exercise

J1 = −D11∂C∂x1

− D12∂C∂x2

J2 = −D12∂ C∂x1

− D22

∂C∂x2

.

1. The general diffusion response for a two dimensional lattice is

Determine the forms of the diffusivity tensor for the following lattices

(a) Square lattice (b) Rectangular lattice

© meg/aol ‘02

Matrix Transformations in 2-D

α[ ] ≡α11 α12

α21 α22

=cosθ cos 90 − θ( )

cos 90 + θ( ) cosθ

x1

x2

θ

α[ ]T ≡α11 α21

α12 α22

=cosθ cos 90 + θ( )

cos 90 − θ( ) cosθ

α[ ]⋅ Dij ⋅ α[ ]T =

α11 α11D11 + α12D21( )+ α12 α11D12 + α12 D22( ) α21 α11D11 + α12D21( )+ α22 α11D12 + α12D22( )α11 α21D11 + α22D21( )+ α12 α21D12 + α22D22( ) α21 α21D11 + α22D21( )+ α22 α21D12 + α22 D22( )

Matrix transformation rule:

direction cosines

© meg/aol ‘02

Exercise

Dij[ ]=D11 D12

D21 D22

α[ ] =0 1−1 0

Dij'[ ]=

D22 −D21

−D12 D11

x2

x1

(a) In 2-dimensions:

The transformation matrix for an axis rotation of +π/2 is

The diffusivity tensor in the rotated coordinate system

Dij[ ]=D11 00 D11

= D11

Square lattice

© meg/aol ‘02

Exercise

αij[ ]=1 00 −1

Dij[ ]=D11 00 D22

x2

x1

Rectangular lattice Dij[ ]=

D11 D12

D21 D22

(b) In 2-dimensions:

The transformation matrix for a mirror reflection is

The diffusivity tensor in the transformed coordinate system

Dij'[ ]=

D11 −D12

−D21 D22

2 independent elements remain

© meg/aol ‘02

Exercise

α[ ] =cosθ sinθ 0

−sin θ cosθ 00 0 1

2. Use the general transformation properties of the diffusivity tensor and show that in the cases of hexagonal, tetragonal, and rhombohedral crystals the mass flux and diffusivity are independent (orthotropic) of the orientation of the concentration gradient, providing that the gradient lies in the x1–x2 plane. The transformation matrix for an arbitrary rotation, θ, about the x3–axis is given by

© meg/aol ‘02

Exercise

x1

x2

x3

x2'

∇C

θ

x1'

θ

Chemical gradient, ∇C, lying in the x1- x2 plane, applied at angle θ to the x1 axis. The x1

′, x2′ , x3

′ , axes are rotated to make x1

′ parallel to ∇C.

© meg/aol ‘02

Exercise

D11' = α11α11D11 + α12α12 D22 ,

D11' = cos2 θD11 + sin 2 θD22 .

The element D11 in the rotated coordinate system is

Orthotropic materials by definition have D11= D22 , ′ D 11 = cos2 θ + sin 2 θ( )D11

′ D 11 = const.

© meg/aol ‘02

Exercise: Implemented by Mathematica®

3) A 2-D trapezoidal lattice with vectors a and b has the matrix diffusivity, [Dij], given by

Dij =1 0.5

0.5 2.5

×10−9 cm2/s

• Find the flux response, J.

Given the trapezoidal lattice with lattice vectors a and b and a diffusivity tensor,Dij.

881 * 10 ^ - 9 Cm 2 ê S, 0.5 * 10 ^ - 9 Cm 2 ê S <,80.5 * 10 ^ - 9 Cm 2 ê S , 2.5 * 10 ^ - 9 Cm 2 ê S << êê N

: :1.¥ 10- 9 Cm2

S,

5.¥ 10- 1 0Cm2

S>, :

5.¥ 10- 1 0Cm2

S,

2.5¥ 10- 9 Cm2

S>>

MatrixForm @%D

i

k

1 .¥1 0- 9 C m2

S5 .¥1 0- 1 0C m2

S

5 .¥1 0- 1 0C m2

S2 . 5¥1 0- 9 C m2

S

y

© meg/aol ‘02

Exercise: Implemented by Mathematica®

A chemical gradient is applied in the form of the vector

gradC = 882 * 10 ^ 5 Gram ê Cm 4<, 80<<

The flux response is the vector J,

J = Dij . gradC

Dij.gradC

i

k

1. * ^-9 Cm2

S5. * ^-10 Cm2

S5. * ^-10 Cm2

S2.5`* ^-9 Cm2

S

y

. 882 * 10 ^ 5 Gram ê Cm 4<, 80<<

: :0.0002Gram

Cm2 S>, :

0.0001Gram

Cm2 S>>

MatrixForm @%D

i

k

0 . 0 0 0 2G r a mC m2 S

0 . 0 0 0 1G r a mC m2 S

y

© meg/aol ‘02

Exercise The angle of the flux is

j = ArcTan @H1 * 10 ^ - 4L ê H2 * 10 ^ - 4LD

ArcTanB1

2F

ArcTan A12

E êê N

0.463648

0.4636476090008061` * 180 ê Pi

26.5651

26.56505117707799`

26.56505117707799` Degrees

The flux magnitude is

MagJ = $ ik

0.0002` GramCm2 S

y

^ 2 + ik

0.0001` GramCm2 S

y

^ 2

0.000223607$Gram2

Cm4 S2

ScientificForm @%D

2.3607¥ 10- 4

" 2.3607 " ¥ 10 " - 4" Gê HCm 2 * SL* 10

J

∇C ~26.5°

© meg/aol ‘02

Key Points • The diffusion coefficient, D, is in general a tensor quantity

expressible in matrix form, [Dij]. • Physical and mathematical arguments shown that in 3-D, the

diffusion matrix has at most 6 independent elements. In 2-D, at most 4 independent elements occur.

• Neumann’s principle may be applied to reduce the maximum number of diffusivity elements on the basis of crystallographic symmetry operations.

• Many engineering materials fortuitously often exhibit isotropic diffusion behavior.

• Crystal structure and texture have profound influences on the diffusion response of a material.

© meg/aol ‘02

Module 3: Solutions To The Linear Diffusion Equation

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline • Transform methods

• Linear diffusion into semi-infinite medium – Boundary conditions

– Laplace transforms

• Behavior of the concentration field

• Instantaneous planar diffusion source in an infinite medium

• Conservation of mass for a planar source – Error function and its complement

– Estimation of erf(x) and erfc(x)

• Thin-film configuration

© meg/aol ‘02

Transform Methods

where K t, p( ) Kernel of the transform

e− pt Laplace kernel

t Independent variable (time).

L F t( ) ≡ F t( )K t, p( )d t

a

b

∫ = ˜ F p( )

L F t( ) Laplace transform converts an object function, F(t), to its image function, F(p). ~

© meg/aol ‘02

Laplace Transforms: Mathematica© Implementation

In[44]:= Cos @a t D

Out[44]= Cos@a tD

In[45]:= LaplaceTransform @%, t , sD

Out[45]=s

a2 + s2

In[46]:= InverseLaplaceTransform @%, s, t D

Out[46]= Cos@a tD

In[47]:= HCoê sL Exp A-è

s ê D xE

Out[47]=Co „

- $ sD x

s


Out[48]= -

CoD x

i

k

- 1+ ErfB$ x2

D

2 " tF

y

èD3 $ x2

D

In[5]:= Exp @a t D

Out[5]= „ at

In[6]:= LaplaceTransform @%, t , sD

Out[6]=1

- a +s


Out[7]= „ at

Object function

Object function

Image function

Object function

Image function

Object function

© meg/aol ‘02

Linear diffusion into a semi-infinite

medium

time

Initial state

Final state

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

© meg/aol ‘02

Boundary Conditions

1) Initial state: C = 0, for x > 0, t = 0.

2) Left-hand boundary: At x = 0, C0 is maintained for all t > 0.

∂C∂t

= D ∂2 C∂x2

• The diffusion equation is a 2nd-order PDE and requires two boundary or initial conditions to obtain a unique solution.

© meg/aol ‘02

Laplace Transform of the Diffusion Equation

object function C x, t( )

e− ptLaplace kernel

e− pt

0

∞

∫∂2 C∂x2 dt =

∂2

∂x2 Ce− pt

0

∞

∫ dt =d2 ˜ C d x2

˜ C x, p( )≡ C x,t( )e− pt d t0

∞

∫∂2 C∂x 2

Linear Diffusion Equation

˜ C image function

Laplace transform of C(x,t)

˜ C x, p( )

−1D

∂C∂t

= 0

e− pt

0

∞

∫∂2 C∂x2 dt

Laplace transform of the spatial derivative.

© meg/aol ‘02

Laplace Transforms

The Laplace transform of Fick’s second law when C(x,0)=0

d2 ˜ C d x2

−pD

˜ C = 0

−1D

e− pt

0

∞

∫∂ C∂t

dt = −

1D

Ce− pt0

∞− p −Ce− pt

0

∞

∫ dt

= −pD

˜ C

integration by parts

Ce− pt0

∞= 0 − C t =0 = 0 boundary

condition

−1D

e− pt

0

∞

∫∂ C∂t

dt

© meg/aol ‘02

Laplace Transforms

˜ C 0( ) = C0e− pt

0

∞

∫ dt = −C0

pe− pt

0

∞

˜ C 0( ) =C0

p

Transform of the boundary condition:

˜ C = C0

pe

±pD

xGeneral transform solution:

˜ C x( ) =C0

pe

−pD

x

The particular transform solution for the image function arises from the negative root, because the positive root leads to non-physical behavior, . ˜ C (x) → ∞

© meg/aol ‘02

Laplace Transforms: Mathematica© Implementation

HCoê sL Exp A-è

s ê D xE

Co „- $ s

D x

s

InverseLaplaceTransform @%, s, t D

-

CoD x

i

k

- 1+ ErfB$ x2

D

2 " tF

y

èD3 $ x2

D

C x,t( ) = C0 1− erf x2 Dt

© meg/aol ‘02

Laplace Transforms

C x,t( ) = C0 erfc x2 Dt

The concentration field associated with the image field is found by inverting the transform either by formal means, a look-up table, or using a computer-based mathematics package.

erfc z( ) = 1− erf z( ) = 1−2π

e−η 2

dη0

z

∫

The error function, erf (z) , and its complement, erfc (z) , are defined

© meg/aol ‘02

Estimation of the Error Function

erf(z) = − erf(−z) =2π

z −z3

3 ⋅1( )!+

z5

5 ⋅2( )!−

z7

7 ⋅3( )!+ ...

, ( z <1)

• For small arguments:

erf(z) = 1−e−z 2

z π1−

12z2 + ...

, (z → +∞)

• For very large arguments:

© meg/aol ‘02

Estimation of the Error Function

erf(z) =

1.12838z, (0 ≤ z ≤ 0.15)−0.0198+ z 1.2911− 0.4262z( ), (0.15 ≤ z ≤1.5)0.8814 + 0.0584z, (1.5 ≤ z ≤ 2)1, (2 ≤ z)

• Piecewise approximations for restricted ranges of the argument:

erf z( ) =1−1+ 0.278393z + 0.230389z2 +

0.000972z3 + 0.078108z4

−4

+ ε z( ) < 5 ×10−5

• Rational approximation for positive arguments, z > 0:

Useful for spreadsheet calculations.

© meg/aol ‘02

Error Function

-1

-0.5

0

0.5

1

-3 -2 -1 0 1 2 3

Err

or F

unct

ion,

erf

(z)

zz

Erf(z

)

Antisymmetric: erf(z)=-erf(-z)

© meg/aol ‘02

Complementary Error Function

0

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

erfc

(z)

zz

Erfc

(z)

Non-antisymmetric: erfc(z) ≠-erfc(-z)

© meg/aol ‘02

Linear diffusion into a semi-infinite

medium

time

Initial state

Final state

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

t ∞

t =4

t =16

C0

t =1

t =0

x

© meg/aol ‘02

Concentration versus the similarity variable

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3

Rel

ativ

e co

ncen

tratio

n, C

/C0

Space-time similarity variable, x/2(Dt) 1/2

C x,t( ) = C0 erfc x2 Dt

Rel

ativ

e C

onc.

C/C

0

Similarity Variable, x/2(Dt)1/2

© meg/aol ‘02

Concentration field versus distance

λ = 2(Dt)1/2 is the “time tag”

(Note: λ has the units of distance!)

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10Rela

tive

Conc

entra

tion,

C(x

)/C0

Distance, x [ in units of λ=1]

0.20.5

1

23

510

20=λ100

Rel

ativ

e C

onc.

C/C

0

Distance, x [units of λ=1]

© meg/aol ‘02

Diffusion Penetration X*

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10Rela

tive

Conc

entra

tion,

C(x

)/C0

Distance, x [ in units of λ=1]

0.20.5

1

23

510

20=λ100

X* = K t1/2

Rel

ativ

e C

onc.

C(x

)/C0

Rel

ativ

e C

onc.

C/C

0

Distance, x [units of λ=1]

© meg/aol ‘02

Penetration versus square-root of time

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

X*(0.8)X*(0.6)X*(0.4)X*(0.9)

Pen

etra

tion

Dis

tanc

e, X

* (C/C

0)

Time tag, 2(Dt)1/2

Pene

tratio

n D

ista

nce,

X*(

C/C

0)

0.4 0.6

0.8

0.9

© meg/aol ‘02

Instantaneous planar diffusion source in an

infinite medium

These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid.

Initial state

Final state

Tim

e

0 x ∞

M

t = 0

∞-

t →

t =1

t =10

© meg/aol ‘02

0 x ∞

M

t = 0

∞-

t →

t =1

t =10

0 x ∞

M

t = 0

∞-

t →

t =1

t =10

0 x ∞

M

t = 0

∞-

t →

t =1

t =10

0 x ∞

M

t = 0

∞-

t →

t =1

t =10

Instantaneous planar diffusion source in an

infinite medium

These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid. Ti

me

© meg/aol ‘02

Instantaneous planar diffusion source

pD

˜ C x, p( )−d2 ˜ C x, p( )

d x2 =C x, 0( )

D

C x,t( )−∞

∞

∫ dx = M

Application of the Laplace transform to Fick’s second law gives:

The diffusion process is subject to the mass constraint for a unit area:

t = 0, C (x, 0), for all x ≠ 0

Initial condition

C (±∞, t) = 0 Boundary condition

© meg/aol ‘02

Instantaneous planar source

pD

˜ C x, p( )−d2 ˜ C x, p( )

d x2 = 0

˜ C = Ae p D( )x + Be− p D( )x

Reduction of the Laplace transform:

The general solution for which is:

˜ C = Ae p Dx , x < 0, B = 0( )

˜ C = Be − p D( )x, x > 0 , A = 0( )

© meg/aol ‘02

Instantaneous planar source

or e− pt

0

∞

∫ C x,t( )dx dt0

∞

∫ =M20

∞

∫ e− ptdt

or −Bp

De

− pD

x

0

∞

=M2p

and so B =M

2 pD

C x,t( )0

∞

∫ dx =M2

Mass constraint for the field:

L C x,t( )d x

0

∞

∫

= L M

2

Laplace transform the mass constraint:

˜ C x, p( )d x0

∞

∫ =M2p

The integral constraint for the image function is:

© meg/aol ‘02

Instantaneous planar source solution

˜ C x, p( ) =M

2 De

−xD

p

p

Laplace transforms table shows, L-1 e−a p

p

=1π t

e−a24t

The transform solution

where a = x / (D)1/2.

L-1 ˜ C = C x,t( )= M2 D

L-1 p− 12e−a p

Inverting the transform solution

C x,t( )= M

2 πDte

−x2

4 Dt .

Diffusion solution

© meg/aol ‘02

0

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

C(x)

/M [

leng

th]-1

x [distance]

.075

.05

Dt=0

0.025

0.25

13

x [distance]

Normalized plot of the planar source solution

time

© meg/aol ‘02

Conservation of mass for a planar source

ρ x( )−∞

∞

∫ dx = 1C x,t( )M

≡ ρ(x), thus

C x,t( )M

d x−∞

+∞

⌠ ⌡ = 1

1π

e−u2

du−∞

∞

∫ = 1Gauss’s integral

© meg/aol ‘02

Thin-film configuration

Thin-film diffusion configuration is used in many experimental studies for determining tracer diffusion coefficients. It is mathematically similar to the instantaneous planar source solution.

C x,t( )= M

2 πDte

−x2

4 Dt .C x,t( ) =Mthin film

πDte − x 2

4 Dt

where Mthin-film represents the instantaneous thin-film source “strength.”

2Mthin-film=M

© meg/aol ‘02

Procedure for Analysis of Thin-Film Data

C x,t( ) =Mthin film

πDte − x 2

4 DtTake logs of both sides of

ln C x,t( )[ ]= lnMthin film

πDt

−x 2

4Dt

A plot of lnC versus x2 yields a slope=-1/4Dt

© meg/aol ‘02

Thin-Film Experiment

0

1 105

2 105

3 105

4 105

5 105

0 10 20 30 40 50

Counts 100s

Cou

nts

in 1

00 s

Distance, [microns]

Geiger counter data after microtoning 25 slices from the thin-film specimen.

© meg/aol ‘02

Log Concentration versus x2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.5 1 1.5 2 2.5 3 3.5 4

log e R

adio

activ

ty, l

nA*

x2, [cm2 × 25×104]

Slope=-1/4Dt

© meg/aol ‘02

Exercise 1. Show by formal integration of the concentration distribution, C(x,t), given by eq.(3.31), that the initial surface mass, M, redistributed by the diffusive flow is conserved at all times, t>0.

C(x,t)d x−∞

∞

∫ =M

2 πDte

−x 2

4 Dt d x−∞

∞

⌠

⌡

The mass conservation integral is given by:

C(x,t)d x−∞

∞

∫ = 2 M2 πDt

e−

x2

4Dt d x0

∞

⌠

⌡

Symmetry of diffusion flow allows:

© meg/aol ‘02

Exercise

Introduce the variable substitution u= x /2 (Dt )1/2 and obtain:

C(x,t)d x−∞

∞

∫ =MπDt

e−u2

2 Dt du0

∞

∫

Simplification yields:

C(x,t)d x−∞

∞

∫ = M 2π

e−u2

du0

∞

∫

The diffusant mass is conserved according to:

C(x,t)d x−∞

∞

∫ = M erf ∞( ) = M

© meg/aol ‘02

Exercise 2a) Two instantaneous planar diffusion sources, each of “strength” M, are symmetrically placed about the origin (x=0) at locations =±1, respectively, and released at time t=0. Using the linearity of the diffusion solution develop an expression for the concentration, C(x,t), developed at any arbitrary point in the material at a fixed time t>0. Plot the concentration field as a function of x for several fixed values of the parameter Dt to expose its temporal behavior. 2b) Find the peak concentration at x=0 and determine the time, t*, at which it develops, if D=10-11 cm2/sec, M=25 µg/cm2, and the sources are both located 1µm to either side of the origin.

2c) Plot the concentration at the plane x=0 against time.

© meg/aol ‘02

Exercise

2a) C x,t( ) =M

2 πDte− x−1( )2 4 Dt + e− x +1( )2 4 Dt( )

0

0.5

1

1.5

2

-4 -3 -2 -1 0 1 2 3 4

C(x

,t)/M

x

0.5

0.05

2

0.025

Dt=0

0.1

0.25

© meg/aol ‘02

Exercise

C 0,t;± ˆ x ( ) =MπDt

e− ˆ x 2 4 Dt

2b) For two sources of strength M the concentration is:

∂ C 0,t;± ˆ x ( )∂t

=MπDt

e− ˆ x 24 Dt ˆ x 2

4Dt 2 −12t

Differentiate with respect to t :

Dt∗ =ˆ x 2

2

The concentration reaches its maximum at t *:

Cmax 0,t∗( )=M

ˆ x eπ / 2≅ 0.484( ) M

ˆ x

The maximum concentration reached at x = 0

© meg/aol ‘02

Exercise

2c)

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 1 2 3 4 5 6 7

C(0

,t) [g

/cm

3 ]

Time [103 sec]

D=10-11 cm/sM=5.10-4 g/cm2

© meg/aol ‘02

Key Points • Solutions to the linear diffusion equations require two initial or boundary

conditions. Examples of problems with constant composition and constant diffusing mass are demonstrated.

• Laplace transform methods were employed to obtain the desired solutions.

• Solutions are in the form of fields, C(r, t). Exposing the behavior of such fields requires careful parametric description and plotting.

• Similarity variables and time tags are used, because they capture special space-time relationships that hold in diffusion.

• Diffusion solutions in infinite, or semi-infinite, domains often contain error and complementary error functions. These functions can be “called” as built-in subroutines in standard math packages, like Maple® or Mathematica® or programmed for use in spreadsheets.

• The theory for the classical “thin-film” method of measuring diffusion coefficients was derived using the concept of an instantaneous planar source in linear flow.

© meg/aol ‘02

Module 4: Diffusion Couples

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Superposition of instantaneous sources

• Diffusion couples

• Diffusion with a fixed boundary concentration

• Slab couples

• Exercises

© meg/aol ‘02

Diffusion couples

C( x,0)= Cl

xd xx 0

dM

∞

Alloy Pure solvent

welded interface

C(x,0)=0

differential source

© meg/aol ‘02

Field and Source Coordinates

ˆ x

Source point

x

Field point

x − ˆ x

A field point denotes the location at which the concentration field C(x, t) is desired.

A source point denotes the location at which a source can release diffusant.

Nevertheless, the concentration, C(x,t), depends on both the field point and the source points.

© meg/aol ‘02

Diffusion couples

C x,t( ) = Cle

−x − ˆ x ( ) 2

4 Dt

4πDtd ˆ x

−∞

0

⌠

⌡

The concentration response of a classical diffusion couple is given by superposition, through the 1-D source integral:

u =x − ˆ x 2 Dt

du =−dˆ x

2 Dt x = const .( )

If the diffusion couple is of “infinite” thickness, use the following substitution:

The total differential of u is:

x, the field point, is considered a constant!

© meg/aol ‘02

Diffusion couples

C x,t( ) = −Cle−u2

πdu

∞

x2 Dt⌠ ⌡ = Cl

e−u2

πx

2 Dt

∞

⌠ ⌡ duSource integral in terms of

new variable, u ˆ x ,t( )

C x,t( ) =Cl

22π

e−u2

du0

∞

∫ −2π

e−u2

du0

x2 Dt

∫

Rearranging the right-hand integral

C x,t( ) =Cl

2erf ∞( ) − erf x

2 Dt

The integrals expressed as error functions

C x,t( ) =Cl

2erfc x

2 Dt

The Grube-Jedele solution

© meg/aol ‘02

Grube-Jedele solution

0.00

0.250

0.500

0.750

1.00

-4 -3 -2 -1 0 1 2 3 4

C/C

0

Distance, x [units of 2(Dt) 1/2]

2(Dt) 1/2 =

5

32

1 0.5

10

0.1

∞

C x,t( ) =Cl

2erfc x

2 Dt

At x=0, Cl/2=0.5

© meg/aol ‘02

Diffusion with a fixed boundary concentration Compare

C x,t( ) =Cl

2erfc x

2 Dt

C x,t( ) = C0 erfc x

2 Dt

The penetration curves of a classical diffusion couple slowly “rotate” around the point (C(0, t )/ Cl)=1/2. The concentration level Cl/2 becomes a fixed point of the diffusion field.

Grube-Jedele Fixed surface concentration

© meg/aol ‘02

Grube-Jedele solution

0.00

0.250

0.500

0.750

1.00

-4 -3 -2 -1 0 1 2 3 4

C/C

0

Distance, x [units of 2(Dt) 1/2]

2(Dt) 1/2 =

5

32

1 0.5

10

0.1

∞

C x,t( ) =Cl

2erfc x

2 Dt

Fixed point

© meg/aol ‘02

Slab couples

C x,t;h( ) =C0

2 πDte

−x − ˆ x ( ) 2

4 Dt d ˆ x −h

+h

∫

C(x,0)=0 C(x,0)= C0 C(x,0)=0h-h

0- +x

distributed sources This is the “formal” integral solution.

© meg/aol ‘02

Slab couples

C x,t( ) =C0

πe−u2

dux−h

2 Dt

x+h2 Dt

∫

Standard form of the integral:

C x,t( ) =C0

22π

e−u2

dux −h

2 Dt

0

∫ +2π

e−u2

du0

x +h2 Dt

∫

Splitting into two integrals:

C(x,t) =C0

2erf x + h

2 Dt

− erf x − h

2 Dt

Expressed as error functions:

C(x / h,t)C0

=12

erfx

h +12 Dt h2

− erf

xh −1

2 Dt h2

Diffusion solution:

© meg/aol ‘02

Exercises

1. Plot the diffusion field, given below, in a composite slab versus normalized distance, x/h, for fixed values of the parameter (Dt )1/2/h. Determine graphically the value of the scaled time tag, (Dt )1/2/h, below which the composition at the mid–plane of this composite slab remains unaltered.

C(x / h,t)C0

=12

erfx

h +12 Dt h2

− erf

xh −1

2 Dt h2

© meg/aol ‘02

0

0.2

0.4

0.6

0.8

1

-4 -2 0 2 4

C(x

/h,t)

/C0

x/h

2

0.5

1

0.25 0.1

(Dt)1/2/h=0

C(x

/h,t)

/C0

For times larger than 0.25 the concentration at x/h=0 will fall.

Concentration versus distance for a composite slab

© meg/aol ‘02

Exercises 2. A thick diffusion couple consists of two end–member alloys containing an initial concentration, Cl on the left side, and Cr on the right side. Develop the general solution, C(x,t) for such a couple. The required diffusion solution is:

C x,t( ) =Cl − Cr

2

erfc x

2 Dt

+ Cr

C x,t( ) =Cl

2 πDte

− x − ˆ x ( )24 Dt d ˆ x

−∞

0

∫ +Cr

2 πDte

− x− ˆ x ( )24 Dt d ˆ x

0

∞

∫

The formal solution may be written as the sum of source integrals:

© meg/aol ‘02

Exercises Using the variable substitution:

C x,t( ) =Cl

22π

e−u2

dux

2 Dt

∞

∫

+

Cr

22π

e−u2

d u−∞

x2 Dt

∫

Writing integrals as sum of error functions:

C x,t( ) =Cl

2erf ∞( )− erf x

2 Dt

+Cr

2− erf −∞( ) + erf x

2 Dt

Splitting integrals in the standard form:

C x,t( ) =Cl

22π

e−u2

dux

2 Dt

0

∫ +2π

e−u2

du0

∞

∫

+Cr

22π

e−u2

d u−∞

0

∫ +2π

e−u2

du0

x2 Dt

∫

© meg/aol ‘02

Exercises

C x,t( ) =Cl

21− erf x

2 Dt

+

Cr

21+ erf x

2 Dt

Considering that erf(∞) = erf(-∞) = 1,

Substituting 1- erf (z ) ≡ erfc (z ),

C x,t( ) =Cl

2erfc x

2 Dt

+

Cr

21− 2 + 2 + erf x

2 Dt

Yields the solution:

C x,t( ) =Cl

2erfc x

2 Dt

+Cr

22 − erfc x

2 Dt

© meg/aol ‘02

Key Points

• Classical diffusion couples are of “semi-infinite” extent. • End-member compositions must remain unaffected by

diffusion. • Instantaneous planar sources may be integrated to find the

solution to the linear diffusion equation (Fick’s 2nd law). • Grube-Jedele solution is expressed with the error function

complement. • Classic diffusion couples are used to determine diffusion

coefficients by analyzing the penetration curve. • Time-tags for slab sources involve a similarity variable

[2(Dt)1/2](1/h) containing two length scales: a geometric scale, h, and a “diffusion” scale, (Dt)1/2.

© meg/aol ‘02

Module 5: Diffusion Point Sources in Higher Dimensions

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Instantaneous point source in three dimensions – Transform solution for a point source

– Mass constraint in spherical symmetry

• Source solutions in 1, 2, and 3 dimensions

• Behavior of the fields upon release of a source – Choice of concentration, time, and length scale

– Behavior of diffusion responses at early times

• Conservation integrals for 1, 2, and 3 dimensions – Mass centroid in higher dimensions

• Exercises

© meg/aol ‘02

Point source in a spherically symmetric space

∂2 C∂r2

+2r

∂C∂r

−

1D

∂ C∂t

= 0

Fick’s second law, spherical symmetry:

r = the spherical polar coordinate

r = three-dimensional position vector t = time

M = mass of diffusant

The diffusant is released at r=0, t=0

r=0^ rM

C(r,t )

x

z

y

© meg/aol ‘02

Transform solution for a point source The Laplace transform operator applied to spherically symmetric linear diffusion equation:

= 0

˜ C = Laplace transform of the resulting field

where

C (r, t ) = object function which is a function of radial position and time

(r, p) = image function that depends on the radial position and the transform parameter , p. ˜ C

e− pt ∂2 C∂r2

d t0

∞

⌠ ⌡ +

2r

e− pt ∂C∂r

dt

0

∞

⌠ ⌡ −

1D

e− pt ∂C∂t

dt

0

∞

⌠ ⌡

= 0−1D

e− pt ∂C∂t

dt

0

∞

⌠ ⌡

d2 ˜ C d r2

+2r

d ˜ C d r

© meg/aol ‘02

Transform solution for a point source

1D

e− pt ∂C∂t

d t

0

∞

⌠ ⌡ =

C r,t( )D

e− pt

0

∞

+pD

Ce− pt d t0

∞

∫

1D

e− pt ∂C∂t

d t

0

∞

⌠ ⌡ =

pD

˜ C

Integration by parts is used to transform the time derivative.

U ≡ ˜ C ⋅ r˜ C is temporarily replaced by a

new function, U,

Laplace-transformed Fick’s second law for spherical symmetry =

pD

˜ C d2 ˜ C d r2

+2r

d ˜ C d r

© meg/aol ‘02

Transform solution for a point source

1r

d2Ud r2

=2r

d ˜ C d r

+

d2 ˜ C d r2

1r

d2Ud r2

=pD

˜ C

Substituting into the Laplace-transform,

d2Ud r2

−pD

U = 0

Replace by U/r, ˜ C

Fick’s law in the variable U

Differentiating U=C r twice with respect to r, then dividing by r yields

~

© meg/aol ‘02

Transform solution for a point source in 3-D

U ≡ ˜ C ⋅ r = AepD r + Be − p

D r

The general solution to this PDE is the sum of growing and decaying exponentials:

˜ C r, p( ) =Br

e − pD r

The solution to the transformed diffusion equation is the image function

(A must be zero)

˜ C r, p( )=Ar

epD r +

Br

e − pD r

The Laplace transformed “concentration” field, C(r, p) is therefore, ~

© meg/aol ‘02

Mass conservation

C r,t( )4π r 2 d r0

∞

∫ = M

The spherically symmetric global mass conservation integral:

C r,t( )4π r 2d r0

∞

∫

0

∞

∫ e− pt d t = M e − pt d t0

∞

∫

The Laplace transform of the mass integral

4π B e − p D rr d r0

∞

∫ =Mp

Substituting for ˜ C r, p( ) =Br

e − p D r

4π ˜ C r, p( ) r 2d r0

∞

∫ =Mp

Integrating both sides:

Laplace transform

© meg/aol ‘02

Mass conservation

B =M

4π p e− pD r r d r

0

∞

∫

e− pDrr d r

0

∞

∫ =Dp

The relation between the mass diffusing, M, and the constant B.

e − pD r r d r

0

∞

∫ = −1p D

re − pD r

0

∞

− e− pD r d r

0

∞

∫

The definite integral is:

B =M

4πDThe constant B is:

© meg/aol ‘02

Mass conservation

˜ C = M4πDr

e − pD r = be − a p

The Laplace transform of the diffusion solution for the case of spherical symmetry is:

C r,t( ) = M e − r 24 Dt

4π Dt( )32

The solution is:

a=r/√D b=M/4πDr

In[1]:= HMê H4 Pi * D* r LL* Exp A-è

s ê D r E

Out[1]=„

- r $ sD M

4 D p r


Out[2]=„ - r2

4Dt M

8 p3ê2 èD3 t3

Mathematica® Implementation

© meg/aol ‘02

Source solution in 1, 2, and 3 dimensions

C rd ,t( )= Mde −

rd2

4 Dt

4π Dt( )d / 2 , d =1,2,3( )

where

rd d - component vector extending from the source point

Md initial source strength

C rd ,t( ) concentration field

t time

Homologous form

© meg/aol ‘02

Conservation integral for 1, 2, and 3 D

Conservation integrals express the global redistribution of diffusant in each spatial setting

C x,t( )d x−∞

∞

∫ = M11-D

2π r2 C r2 ,t( )d r20

∞

∫ = M22-D

4π r32 C r3,t( )d r3

0

∞

∫ = M3.3-D

© meg/aol ‘02

Conservation integrals: d=1,2,3 Probability densities for different spatial settings:

F (x, t) ≡C x, t( )

M 11-D

F(r 2, t) ≡2π r 2C r 2, t( )

M2

2-D

F(r3 , t) ≡4π r 3

2 C r 3, t( )M 3

3-D

© meg/aol ‘02

Moments of a Point Source in 1-D

x F x, t( ) d x−∞

∞

∫ = x e− x 2

4 Dt

4π Dt−∞

∞

⌠

⌡

d x = 0

The first moment of the probability density

x2 F x, t( ) d x−∞

∞

∫ = x 2 e− x 2

4 Dt

4π Dt−∞

∞

⌠

⌡

d x = 2Dt

The 2nd moment of the probability density

© meg/aol ‘02

0

0.05

0.1

0.15

0.2

0.25

0.3

-4 -2 0 2 4

Prob

abilit

y D

ensi

ty, F

(x,t)

Distance, X

Area =1

Probability Density in 1-D

sqrt< x2> sqrt< x2>

<x>

Area=1

Dt=1

Prob

abilit

y D

ensi

ty, F

(x,t)

© meg/aol ‘02

0

0.5

1

1.5

2

2.5

3

-2 -1 0 1 2

d=1

d=2

d=3

C(r d,t)

/Md

rd

λ=0.4 [units]

Diffusion field at early times

C (r

d,t)/M

d

rd

© meg/aol ‘02

Diffusion field when t = 1/4πD

0

0.2

0.4

0.6

0.8

1

-2 -1 0 1 2

C(r d,t)

/Md

rd

λ=(π)-1/2

-0.5642 [units]d=1,2,3C

(rd,t

)/Md

rd

© meg/aol ‘02

Diffusion field when t > 1/(4πD)

0

0.2

0.4

0.6

0.8

1

-2 -1 0 1 2

d=1

d=2

d=3

C(r d,t)

/Md

rd

λ=1 [units]

C (r

d,t)/M

d

rd

© meg/aol ‘02

Diffusion field for long diffusion times

0

0.2

0.4

0.6

0.8

1

-2 -1 0 1 2

d=1

d=2

d=3

C(r d

,t)/M

d

rd

λ=2 [units]

C (r

d,t)/M

d

rd

© meg/aol ‘02

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5

d=1

d=2

d=3

F(r d,t)

abs(x), r2, r

3

Dt=1

Prob

abilit

y D

ensi

ty, F

(rd,t

)

Abs(x), r2, r3

Dt=1

Probability densities in all dimensions

© meg/aol ‘02

Mass centroid in 3 dimensions

P

θ

∆θ

φ ∆ φ

r

y

x

z

(r sin φ ) ∆θ r ∆φ

∆r

r sin φ

P

x

y

z

r

For a spherically symmetric release of diffusant at the origin, the mass centroid remains at (0,0,0).

© meg/aol ‘02

Mass centroid in 3 dimensions

x = r3 F r3,t( )d r3( ) 14π

sin φ cosθ dθ dφθ =0

2π

∫φ=0

π

∫

r=0

∞

∫

x = µ11

4πsinφ cosθ dθ dφ

φ=0

π

∫θ=0

2π

∫

= 0

x component of the centroid is given by the triple integral

The centroid along x–, y–, and z– axes remain zero for all time, as required by spherical symmetry.

First moment

= 0

© meg/aol ‘02

0

0.5

1

1.5

2

0 1 2 3 4 5

F(r 3,t)

radial distance, r3 [same unit as (Dt)1/2]

0.05

Dt=0.25

31Pr

obab

ility

Den

sity

F(r

3,t) Areas under probability curves

equal unity for all times.

Probability density: spreading in 3-D

© meg/aol ‘02

Exercise

1. In two dimensions the probability density surrounding a point source increases linearly with distance near the source location. The probability peaks, and then decays towards zero. Find the location of the peak in the two–dimensional probability density. In two dimensions the probability density is given by,

F(r 2, t) = 2π r 2

C(r2 , t)M2

.

F(r2 , t) =

r2 Dt

e−

r 22

4 Dt.

Substituting for concentration field in 2-d we obtain

© meg/aol ‘02

Exercise

d

d r2

F(r2 ,t)( )=e

−r2

2

4 Dt

2Dt1−

2r22

4Dt

= 0

r2peak = 2 Dt

The peak probability location may be found by differentiating F(r2,t), and setting the derivative to zero.

Solving for the value of the radial distance for the peak probability, we obtain the answer:

© meg/aol ‘02

Key Points • Point source solution in three dimensions provides the “basis”

solution for finding the concentration field for many practical diffusion problems.

• Similar point-source solutions exist in 1-D and 2-D. • Mass conservation integrals define probability density functions

for the diffusion field. • Each spatial setting has an unique response to the release of

solute. • The spreading in space of the probability density distribution has

a characteristic square-root dependence on the diffusion time. • The mass centroid of diffusant released from an instantaneous

point source remains invariant at the source location.

© meg/aol ‘02

Module 6: Generalized Sources

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Generalizing the point source in 3 dimensions – Relationship with Green functions

• Instantaneous line source releasing diffusant in 3 dimensions

• Generalizing the instantaneous point source in two dimensions

• Instantaneous line source in two dimensions

• Instantaneous ring source

• Continuous point source

• Time-continuous line source in two dimensions

• Exercises

© meg/aol ‘02

Point source in 3-D: Green’s functions

C r − ˆ r , t − ′ t ( ) =M ˆ r , ′ t ( )

4π D t − ′ t ( )[ ]3 2 ⋅ e − r− ˆ r 2 4 D t− ′ t ( )

C r,t( ) = M ˆ r , ′ t ( )G r, ˆ r ,t, ′ t ( )d ˆ r d ′ t a

b

∫0

t

∫ + C0

Relationship to Green’s functions:

C x,y, z,t; ˆ x , ˆ y , ˆ z , ′ t ( ) =M ˆ x , ˆ y , ˆ z , ′ t ( )

4π D t − ′ t ( )[ ]3 2 ⋅e−

x − ˆ x ( ) 2 + y − ˆ y ( ) 2+ z− ˆ z ( ) 2

4D t − ′ t ( )

In Cartesian coordinates:

Diffusion solution for a instantaneous point source in 3 dimensions released at the point r, at time t′ ^

© meg/aol ‘02

Instantaneous line source in 3-D

C x,y, z,t( ) =Mline

4π Dt[ ]32

⋅ e−

x( )2 + y( )2 + z− ˆ z ( )2[ ]4 Dt d ˆ z

−∞

∞

⌠ ⌡ Concentration field:

C x,y, t( ) =Mline

4π( )32

e−

x 2 + y2

4 Dt

e−

z− ˆ z ( )2

4 Dt

Dt( )32

d ˆ z

−∞

∞

⌠

⌡

Regrouped:

^

Transforming an instantaneous point source in 3-D to an instantaneous line source on the z-axis, so x=y=0 ^

© meg/aol ‘02


C R,t( ) =Mline e

−R 2

4 Dt

4Dt π32

e −ζ 2

dζ∞

−∞

∫ C R,t( ) = Mline ⋅e

− R 2

4 Dt

4π Dt

C R,t( ) =M 2

4π Dte

−R 2

4 D t

C R,t( ) =Mline e

−R 2

4 Dt

4π Dt[ ]32

e−

z− ˆ z ( ) 2

4 Dt d ˆ z −∞

∞

∫

Introducing the cylindrical coordinate R ≡ (x 2 +y 2)1/2

ζ =z − ˆ z

2 DtVariable substitution dζ =

−dˆ z 2 Dt

and thus

Point source in 2-D an example Hadamard’s method of descent

(√π ) —

© meg/aol ‘02

Instantaneous point source in 2-D

C R − ˆ R ,t − ′ t ( )=

M2

4π D t − ′ t ( )e − R− ˆ R 2 4 D t − ′ t ( )

where R is the field vector

R is the source vector

C x,y, t; ˆ x , ˆ y , ′ t ( )=M2

4π D t − ′ t ( )e − x− ˆ x ( )2 + y − ˆ y ( )2[ ] 4 D t − ′ t ( )

In Cartesian coordinates the equation becomes,

© meg/aol ‘02


C x,t( ) =Mline

4πDte

−x 2

4 Dt e−

y − ˆ y ( ) 2

4 Dt d ˆ y −∞

∞

⌠ ⌡

x

y Setting x=0, -∞< y <+∞, distributes 2-D point sources along the y-axis.

^ ^

Integrating these sources along the y-axis yields the field

© meg/aol ‘02


C x,t( ) =Mline

4π Dte

− x 2

4 Dt

C x,t( ) =Mline

2π Dte

− x2

4 Dt e −ψ 2

dψ−∞

∞

∫yields

e −ψ 2

dψ−∞

∞

∫ = Gauss' s integral = πwhere

Inserting the value for Gauss’s integral, the field reduces to

ψ =y − ˆ y ( )2 Dt

Substituting dˆ y = −2 Dt dψand

© meg/aol ‘02

Instantaneous ring source

All the point sources that continuously comprise a ring give the following response at the origin, r=0.

C 0,0, t( ) =e−a2 4Dt

4πDtmθdθ

0

2π

∫ =Mring

4πDte−a2 4 Dt

Mring = 2πmθwhere

x

y

a ^ ^ x, y

θ

© meg/aol ‘02

Instantaneous ring source

dC 0,0, t( )d t

=

Mring a 2

π 4D( ) 2 t 3 e − a 2 4 Dt −Mring

4π Dt 2 e −a 2 4 Dt

Cmax 0,0,t ∗( )=Mring

e πa 2 ≅ 0.117Mring

a 2

t * =a 2

4D

Differentiating with respect to time,

Cmax at the origin occurs at a time t* when dC/dt = 0. The time derivative vanishes when a 2/ (4Dt *) = 1.

© meg/aol ‘02

Instantaneous ring source in 2-D

C(0

,0,t)

/Mrin

g (c

m-2

)

a=1 t*=1

© meg/aol ‘02


C(r,t) =dm θ

4π Dte − x− ˆ x ( )2 + y− ˆ y ( )2[ ]/ 4 Dt

0

2π

∫

The general response from a two-dimensional ring source at any point x,y is

dmθ =Mring

2πd ˆ θ where

x

y

C r,t( ) =Mring

8π 2Dte − l 2

4 D t d ˆ θ 0

2π

∫Making the substitution gives:

a ^ ^ x, y

x,y r

θ

l = a 2+ r 2− 2ar cos θ

l

© meg/aol ‘02


l = a 2+ r 2− 2ar cos θApply the law of cosines

C r,t( ) =Mring

8π2Dte

−1+r 2 −2r cosˆ θ [ ]

4 Dt d ˆ θ 0

2 π

⌠ ⌡ For a=1

C r,t( ) =Mring

8π2Dte

− 1+r 2( )4 Dt e

−2r cosˆ θ [ ]

4 Dt d ˆ θ 0

2π

⌠ ⌡

The solution for an instantaneous ring source in 2 dimensions

© meg/aol ‘02


Distance, r, units of 2(Dt)1/2 3 2 1 0 1 2 3

0

0.1

0.2

0.3

0.4

0.5

Distance,r

0.412575

.4.83553 10 10

da

bxf1 x

da

bxf2 x

da

bxf3 x

da

bxf4 x

33 r

Distance, r, units of 2(Dt)1/2

C(r,t

)/Mri

ng

A

C(r,

t)/M

ring

(early times)

Distance, r, units of 2(Dt)1/2 3 2 1 0 1 2 3

0

0.05

0.1

0.15

0.2

0.25

Distance,r

0.234199

.1.943539 10 3

da

bxf1 x

da

bxf2 x

da

bxf3 x

da

bxf4 x

33 r


C(r,t

)/Mri

ng

B

C(r,

t)/M

ring


(late times)

© meg/aol ‘02

Continuous point sources

τ ≡r

2 D t − ′ t ( )1 2 dτ =r t − ′ t ( )−3 2

4 Dd ′ t Introducing the variable

τ and its differential

C r,t( ) =Ý M

2π 32 D

e −τ 2

rdτ

r2 Dt

∞

⌠ ⌡

The integral representation of the concentration field from a time–continuous source

C r,t( ) = Ý M e−

r 2

4 D t − ′ t ( )

4π D t − ′ t ( )[ ]32

d ′ t

0

t

⌠

⌡

The concentration field at a distance r from a continuous point source, M operating at the origin, r =0 is

.

d ′ t =4 D t − ′ t ( ) 3

2

rdτsubstitute

© meg/aol ‘02


C r,t( ) =Ý M

4π rD2π

e−τ 2

d τr

2 Dt

0

∫ + e−τ2

dτ0

∞

∫

C r,t( ) =

Ý M 4π rD

erfc r2 Dt

Splitting the previous equation into 2 integrals

2π

e−τ 2

dτ0

z

∫ ≡ erf z( ) 1− erf z( ) ≡ erfc z( )Recalling, and

The field for a time-continuous diffusion source

© meg/aol ‘02

.


30 303 3

2(Dt) 1/2=0.1

1 1

1/r1/r C(r)/C

0

.

© meg/aol ‘02


lim t → ∞ erfc r2 Dt

=1

The quasi-static point source solution, in the limit of long diffusion times

C r,∞( ) =Ý M

4π Dr

The quasi–static form of the continuous point source solution

C r,t( ) =

Ý M 4π rD

erfc r2 Dt

© meg/aol ‘02

Time-continuous line source in 2-D

C x,t( ) =Mline

4πD t − t '( )e

− x2

4D t− t'( )The generalized solution for a time–continuous line source

Consider the line source as a steady succession of infinitesimal releases

C x,t( ) =Ý m line

4πD t − t'( )e

− x 2

4 D t− t'( )dt '

0

t

⌠

⌡

τ = t − t ' dτ = −d t 'Substituting the variable, so

C x,t( ) =Ý m line

4πDτe

−x 2

4 Dτ dτ

t

0

⌠

⌡ yields

© meg/aol ‘02


C x,t( ) =Ý m line

πe

−x 2

4 Dτ

4Dτdτ

t

0

⌠

⌡

=Ý m line x4D

⋅2π

e −u 2

u 2 dux

2 Dt

∞

⌠

⌡

Rewriting in standard form,

C x,t( ) = Ý m lineDtπ

e− x 2

4 Dt −x2

erfc x2 Dt

Considering only the positive values of x,

C x,t( )⋅ DÝ m line

= Dt e− x 2

4 Dt

π−

x2 Dt

erfc x2 Dt

Rearranging with the concentration scaled by the diffusivity and the source strength

© meg/aol ‘02


0

0.5

1

1.5

2

2.5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

(D/m

)C(x

,t)

.

x/2sqrt(Dt)

Dt=400

10

1

0.01

100

© meg/aol ‘02

Exercises

1. Four lines of diffusant forming a “tic–tac–toe” pattern are deposited on a planar substrate. Each line of diffusant initially contains 2×1014[atoms/cm]. The lines are spaced apart a distance w=10µm as sketched in Fig.6-8. The system is quickly heated to the processing temperature where the diffusion coefficient for two–dimensional spreading over the substrate becomes D=3×10-8[cm2/sec]. a) Develop a general solution for the concentration field, C(x,y,t) and plot it for visualization as diffusion progresses. b) Develop explicit expressions for the concentration and vector flux at points O and P. c) Plot the variations with time for the concentration at O and P, and determine the peak concentrations and the times at which they occur.

© meg/aol ‘02

Exercises

M line

y

xO(0,0)

P (w/2,-w/2)

M line

w

C x,t( ) =Mline

4πDte

−x 2

4 Dt

C y,t( ) =Mline

4πDte

−y2

4Dt

a) For y–axis, when x = 0

ˆ y = 0For x–axis, when

© meg/aol ‘02

Exercises

C x,t( ) =Mline

4πDte

− x −w 2( ) 2

4 Dt + e− x +w 2( ) 2

4 Dt

The tic-tac-toe arrangement from the y – axis

C y,t( ) =Mline

4πDte

− y −w 2( ) 2

4 Dt + e− y +w 2( ) 2

4 Dt

The tic-tac-toe arrangement from the x – axis

C x,y, t( ) =Mline

4πDte

− x − w 2( ) 2

4 Dt + e− x + w 2( ) 2

4 Dt + e− y − w 2( ) 2

4 Dt + e− y + w 2( ) 2

4 Dt

The solution for the tic-tac-toe field pattern at any arbitrary location, x,y is

© meg/aol ‘02

0 40 - 40

0

40

-40

t = 4 s

0

40 - 40

0

-40

40

t = 1 s

0 40 -40

0

40

-40

t = 20 s

0 40 -40

0

40

-40

t = 100 s

Exercises

© meg/aol ‘02

Exercises b)

C 0,0, t( ) =2Mline

πDte

− w 2

16 Dt

C w2 , −w

2 ,t( ) =Mline

πDt1 + e

− w 2

4 Dt

J x, y,t( ) = −D ∂ C x,y,t( )∂x

ex +

∂ C x,y,t( )∂y

e y

At location O, the concentration is

At a specific location P, the concentration becomes

Applying Fick’s first law to the field,

© meg/aol ‘02

Exercises

J x, y,t( ) =Mline

4 πDt1t

x − w2( ) e

− x −w2( )2

4 Dt + x + w2( ) e

− x +w2( ) 2

4 Dt

ex

+ y − w2( )e

− y − w2( )2

4 Dt + y + w2( ) e

− y + w2( ) 2

4 Dt

ey

J w2 , −w

2 ,t( ) =Mline

4 πDtwt

e

−w 2

4Dt e x − ey( )

The vector flux field is

At location O, the flux is J x, y,t( ) = 0

At location P, the flux is

© meg/aol ‘02

0

5 1016

1 1017

1.5 1017

2 1017

0 5 10 15 20 25 30 35 40

C(0

,0,t)

[a

tom

s/cm

2 ]

Time [sec]

c)

∂ C 0,0,t( )∂t

=Mline

πDte

−w 2

16 Dt w 2

8Dt 2 −1t

= 0 w 2

8Dt max2 −

1tmax

= 0

The peak concentration occurs at point O when,

where

Exercises

© meg/aol ‘02

Exercises

0

1 1019

2 1019

3 1019

4 1019

5 1019

0 0.1 0.2 0.3 0.4 0.5

C(w

/2,-

w/2

,t)

[ato

ms/

cm2 ]

Time [sec]

C(w

/2,-w

/2,t)

[ato

ms/

cm2 ]

Time [s]

© meg/aol ‘02

Exercises

Solving for tmax ,

tmax =w 2

8D=

10−3cm( ) 2

8 ⋅ 3 ⋅10−8 cm 2 / sec( )= 4.17[sec]

The maximum concentration at point O is found to be:

Cmax 0,0,tmax( )=Mline

π w 2

8

e−

12 =1.94 ⋅1017[atoms / cm2 ]

© meg/aol ‘02

Key Points • Source methods rely on the linearity of the diffusion equation. • Source integrals are closely related to Greens functions

regarding their mathematical properties. • Integral transforms that remove dependence on a spatial variable

is an application of Hadamard’s method of descent. – 3-D point source integrates to a line source in a cylindrical 3-D space. – 3-D line source is mathematically equivalent to a point source in 2-D. – 2-D point source integrates to a line source in 2-D. – 2-D line source is mathematically equivalent to a point source in 1-D – 1-D point source is equivalent to a planar source in 1-D linear flow.

• Time continuous sources allow for the steady release of diffusant.

• Time continuous sources may eventually produce bounded steady-state solutions for the concentration field.

© meg/aol ‘02

Module 7: Diffusion–Reaction

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline • Diffusion–Reaction: examples • Linear diffusion–reaction

– One-dimensional diffusion–reaction – Boundary conditions for diffusion–reaction

• Imposing steady–state conditions • General kinetic limits

– Reaction limit – Diffusion limit

• Mixed diffusion–reaction kinetics • Time-dependent reaction–diffusion • Nonlinear diffusion–reaction • Exercises

© meg/aol ‘02

Diffusion–Reaction • Diffusion has been discussed to this point without involving additional chemical or nuclear reaction. Thus Fick’s law was limited to binary interdiffusion medium.

• Diffusional transport may be accompanied by reactions occurring within the bulk material or upon its surfaces.

• The rate–controlling step of the overall diffusion-reaction process can become either diffusion–limited or reaction–limited.

• When neither diffusion nor reaction are the rate–controlling factors, so called “mixed kinetics” obtains.

© meg/aol ‘02

Examples of Diffusion–Reaction • Aspirin dissolution: Aspirin molecules hydrate at the solid-liquid interface and then diffuse away into the solution. The case is treated as heterogeneous hydration followed by diffusion.

• Diffusion–reaction in a solid/liquid diffusion–couple

Solid particles

Liquid

Penetrating Liquid

Interface Thin oxide layer as result of solid–liquid reaction at the interface

© meg/aol ‘02

Diffusion–Reaction in a Porous Catalyst (after Cussler)

Step 5. The product returns to surroundings.

Step 1. The gaseous reagent diffuses from the surroundings onto the catalyst surface.

Step 2. diffuses into the porous structure.

Step 3. reacts at the pore walls.

Step 4. The product, , diffuses out of the pores.

© meg/aol ‘02

One–Dimensional Diffusion–Reaction

d2 Cdx 2 =

1D

∂ C∂t

= 0

At steady-state, Fick’s second law reduces to a second-order ODE:

dCdx

= k1

• The first integration yields:

C x( ) = k1x + k2

• A second integration shows the steady-state concentration field in the linear form

reaction control

mixed control

diffusion control

C0 C0

C2

Ceq

O l x

© meg/aol ‘02

Boundary Conditions for Diffusion-Reaction

k2 = C0

Assuming that at x = 0, C = C0, the second integration constant, k2 may be found

C x( ) = k1x + C0

The diffusion solution now becomes:

k1 =C2 − C0

l

The second boundary condition is: x = l, C(l ) = C2. Thus the first integration constant, k1, becomes:

C x( ) = C0 +C2 − C0

l

x

The diffusion solution is:

© meg/aol ‘02

Imposing Steady-State Conditions

Jss = − D dCd x

The Fick’s first law for steady-state diffusion flux is:

α C2 − Ceq( )=D C0 − C2( )

l

The steady-state relationship between the reaction velocity and diffusivity is:

Jss =D C0 − C2( )

l

Evaluating the gradient by differentiating C (x) shows that the diffusion flux

The reaction flux may be written

Jout = α C2 − Ceq( ) (α ≡ “reaction velocity”)

© meg/aol ‘02

Imposing Steady-State Conditions

C2 = Ceq +C0 − Ceq

1+α lD

After several steps of algebra we obtain:

Jss =α C0 − Ceq( )

1+α lD

The steady-state flux during diffusion–reaction is:

Péclet number

© meg/aol ‘02

General Kinetic Limits

The Péclet number, Pe, may be either defined as:

Pe ≡l

D α

2) the ratio between the physical thickness of the material, l, and the characteristic length, D/α.

physical lengthdiffusion length

Pe ≡αD l

1) the ratio between the reaction “velocity”, α, and the diffusion “speed,” D/l.

reaction velocitydiffusion speed

© meg/aol ‘02

Reaction Limit versus Diffusion Limit

Jss ≅ α C0 − Ceq( )

Jss ≅D C0 − Ceq( )

l

• Diffusion limit: When Pe >>1, the steady–state flux becomes

• Reaction limit: When Pe <<1, the steady–state flux becomes

Note: The steady-state flux is independent of D.

Jss =D C0 − C2( )

lIn this limit, C2→ Ceq, and therefore

Note: The steady-state flux is dependent on D.

© meg/aol ‘02

Mixed Diffusion-Reaction Kinetics

Jss =α C0 − Ceq( )

1+ Pe

• Mixed Kinetics, Pe = 1,

Jss =Dα C0 − Ceq( )

D + α l

• If the concentration gradient and flux are not linearly related to the diffusivity, then

Jss ≅ α C0 − Ceq( )

• If D >> αl, the flux becomes

© meg/aol ‘02

Reaction-Diffusion Schemes

reaction control

mixed control

diffusion control

C0 C0

C2

Ceq

O l x

© meg/aol ‘02

Mixed Diffusion-Reaction Kinetics

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

J ss/ α

(C0-C

eq)

Peclet Number, Pe

J ss /α

(Co -

Ceq

)

Peclet Number, Pe

Flux dependence on Pe

© meg/aol ‘02

Time-Dependent Reaction-Diffusion The average concentration after diffusion-reaction may be found for various shapes:

C slab − C2

C0 − C2

=8

π 21

2n +1( )2 exp − 2n +1( )2π 2

4Dth 2

n=0

∞

∑Slabs:

C sph − C2

C0 − C2

=6

π 21

n 2 exp −n 2π 2 DtR s

2

n=1

∞

∑Spheres:

C cyl − C2

C0 − C2

=1

ξn2 exp −ξn

2 DtR c

2

n=1

∞

∑Solid cylinders:

© meg/aol ‘02

Average concentration ratio, f, versus dimensionless diffusion time

0.0001

0.001

0.01

0.1

1

0 0.25 0.5 0.75 1

Ave

rage

Con

cent

ratio

n (N

orm

aliz

ed)

f=(<

C>-C

2)/(C 0-C

2)

Time Parameter, Dt/(h) 2, or Dt/R2

Slabs

Cylinders

Spheres

Time Parameter, Dt/h2, or Dt/R2

Slabs

Cylinders

Spheres f =

(Cav

- C

2) /

(C0

- C2)

© meg/aol ‘02

Exercise 1. A thick slab of graphite is in contact with a 1mm thick sheet of steel. Carbon steadily diffuses through the steel at 925C. The carbon reaching the free surface reacts with CO2 gas to form CO, which is then rapidly pumped away. Determine the carbon concentration, C2, adjacent to the free surface, and the find the carbon flux in the steel, given that the reaction velocity for C+CO2→2CO is α=3.0×10-6cm/sec. At 925C, the solubility of carbon in the steel in contact with graphite is 1.5wt% and the diffusivity of carbon through steel is D=1.7×10-7cm2/sec. The equilibrium solubility of carbon in steel, Ceq, is 0.1wt% for the CO/CO2 ratio established at the surface of the steel.

© meg/aol ‘02

Exercise

Pe =α lD

= 1.76The Péclet number is

Note: The value of the Péclet number suggests mixed kinetic behavior is expected.

C2 = Ceq +C0 − Ceq

1+ α lD

= 0.1+1.5 − 0.11+1.76

, [wt%]

C2 = 0.61wt%.

The carbon concentration in the steel at the free surface, C2, is

The steady-state flux is Jss = 1.51 × 10-6 [wt% C × cm/s]

Jss = 1.18 × 10-7 [g/ cm2-s]

Divide by the density of steel, ρ=12.8 cm3/100g to obtain the steady-state flux of carbon

© meg/aol ‘02

Exercise

2. Two steel billets—a slab and a solid cylinder—contain 5000ppm residual H2 gas. These billets are vacuum annealed in a furnace at 725C for 24 hours to reduce the gas content. Vacuum annealing is capable of maintaining a surface concentration in the steel of 10ppm H2 at the annealing temperature. Estimate the average residual concentration of H2 in each billet after vacuum annealing, given that the diffusivity of H in steel at 725C is DH=2.25×10- 4 cm2/sec.

© meg/aol ‘02

Exercise

15 cm

2h=1

0 cm

2h=10 cm

15 cm

10 cm

2h = 10 cm

2h =

10

cm

Rectangular and cylindrical slabs of steel

10 cm

© meg/aol ‘02

Exercise Dt

R cyl2 =

2.25 ⋅10−4 × 86400(5) 2 = 0.78• Cylindrical billet (short dimension)

Dth short

2 =2.25 ⋅10−4 × 86400

(5) 2 = 0.78• Slab billet (short dimension)

Dth 2 =

2.25 ⋅10−4 × 86400(7.5) 2 = 0.346• Cylindrical billet (long dimension)

Dth long

2 =2.25 ⋅10−4 × 86400

(7.5)2 = 0.346• Slab billet (long dimension)

© meg/aol ‘02

Shape f long f short

Slab 0.65 0.12

Cylinder 0.65 0.007

0.34

0.094

Exercise

© meg/aol ‘02

Exercise

The average residual gas concentration <C> reported as molecular hydrogen, H2, may be calculated as follows:

C cyl = C0 − C2( ) flong × fshort( )+ C2 ,

C cyl = 5000 −10( ) 0.094 × 0.007( ) +10 = 13 ppm.Solid cylinder:

C slab = C0 − C2( ) flong × fshort × fshort( )+ C2 ,

C slab = 5000 −10( ) 0.34 × 0.12 × 0.12( ) + 10 = 34 ppm.Slab billet:

© meg/aol ‘02

Key Points

• Diffusion-reaction is a commonly encountered situation in many fields of engineering and science.

• Diffusion-reaction is currently a high-level research topic, particularly non-linear diffusion reaction.

• 1-D steady-state diffusion-reaction is a highly simplified, but representative class of problems, indicating parameter ranges for – Diffusion control – Reaction control – Mixed kinetics

• The Péclet number is a dimensionless group that determines the type of diffusion-reaction.

• Time-dependent linear diffusion-reaction occurs in simple degassing problems, where average concentrations are of interest.

© meg/aol ‘02

Module 8: Linear Flow in Finite Systems

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Diffusion in a single–phase composite slab – No-flow boundary conditions – Fictitious image sources

• Fourier’s method • Application of Fourier series • Numerical Approximations

– Error function series approximation – Fourier series approximation – Numerical comparison

• Codastefano’s method • Exercises

© meg/aol ‘02

Diffusion in a single–phase composite slab

0

0 x

h l

C

h

C(x,t)

x

(∂C/∂x)=0

(∂C/∂x)=0C0

l

Bilayer single-phase composite:

h /l =1/3 thickness ratio

C(x,0) = C0 , left slab initial concentration

C(x,0) = 0, right slab initial concentration

No-flow No-flow

© meg/aol ‘02

No–Flow Boundary Condition

∂C x,t( )∂x

0,l

= 0 = −Jbndry

D, t ≥ 0( )

External boundaries in the composite located at x = 0 x = l

Internal boundary in the composite is located at x = h.

To satisfy the external boundary condition Jbndry = 0, Fick’s first law requires a zero gradient at that plane.

© meg/aol ‘02

Fictitious Image

Sources

Sequence of mirror reflections provide no–flow boundary conditions.

x

0-2 l -l 2 l-3 l 3 l

E. Reflection @ x=l

h l-l 3 lx

D. Reflection @ x=0

xh l0

B. Reflection @ x=0

-l

xl0

A. Physical domain

C. Reflection @ x=l

l

0 2 llh

h

h

-h

-h

-h x

0-2 l -l 2 l-3 l 3 l

E. Reflection @ x=l

h l-l 3 lx

D. Reflection @ x=0

xh l0

B. Reflection @ x=0

-l

xl0

A. Physical domain

C. Reflection @ x=l

l

0 2 llh

h

h

-h

-h

-h

x

0-2 l -l 2 l-3 l 3 l

E. Reflection @ x=l

h l-l 3 lx

D. Reflection @ x=0

xh l0

B. Reflection @ x=0

-l

xl0

A. Physical domain

C. Reflection @ x=l

l

0 2 llh

h

h

-h

-h

-h x

0-2 l -l 2 l-3 l 3 l

E. Reflection @ x=l

h l-l 3 lx

D. Reflection @ x=0

xh l0

B. Reflection @ x=0

-l

xl0

A. Physical domain

C. Reflection @ x=l

l

0 2 llh

h

h

-h

-h

-h

© meg/aol ‘02

Fictitious Image Sources

C x,t;h( )=C0

4πDte

−x− ˆ x ( )2

4Dt d ˆ x

−h

h

⌠

⌡

The diffusion solution after a single reflection at x = 0 is formulated by the linear source integral:

The diffusion solution for two reflections becomes the source integral sum:

C x,t;2l ,h( )=C0

4πDte

−x− ˆ x ( )2

4Dt d ˆ x

−h

h

⌠

⌡

+C0

4πDte

−x− ˆ x ( )2

4Dt d ˆ x

2l −h

2l +h

⌠

⌡

© meg/aol ‘02


C x,t;2l,h( ) =C0

4πDte

−x − ˆ x ( ) 2

4 Dt d ˆ x −h

h⌠

⌡ +

C0

4πDte

−x− ˆ x ( ) 2

4 Dt d ˆ x 2l−h

2l+h⌠

⌡

+C0

4πDte

−x − ˆ x ( )2

4 Dt d ˆ x −2l−h

−2l+h⌠

⌡

The diffusion solution after the third reflection step is the sum of three source integrals:

C x,t;l,h( ) =C0

4πDte

−x − ˆ x ( ) 2

4 Dt d ˆ x 2nl −h

2nl +h⌠

⌡

n=−∞,

∞

∑

The diffusion solution for a continuing sequence of steps consists an infinite series of source integrals: I(n, l, h)

© meg/aol ‘02


I(n,l, h) =C0

πe −u 2

dux −2 nl−h( )/ 2 Dt

x −2 nl+h( )/ 2 Dt

∫

Evaluate the definite integrals I(n, l, h) on the right-hand side and sum them as n →∞. The variable substitutions needed for evaluating the integrals I(n, l, h), are:

u =x − ˆ x 2 Dt

du =−d ˆ x 2 Dt

and

yielding

© meg/aol ‘02

Fictitious Image Sources The previous equation may be split into standard integrals as:

I(x,t;n,l,h) =C0

πe −u 2

dux−2nl−h( )/ 2 Dt

0

∫ + e −u 2

du0

x −2 nl+h( )/ 2 Dt

∫

•

I x,t;n,l,h,( ) =C0

2erf x - 2nl + h

2 Dt

− erf x - 2nl - h

2 Dt

and

•

© meg/aol ‘02


C x,t;h,l( ) = C0

2erf x - 2nl + h

2 Dt

− erf x - 2nl - h2 Dt

n=0

±∞

∑

,

(0 ≤ x ≤ l)

C x/ l ,t;h/ l( )C0

=12

erf x/ l - 2n+ h/ l2 Dt / l

− erf x / l - 2n- h/ l

2 Dt / l

n=0

±∞

∑ ,

(0 ≤ x ≤1)

The normalized diffusion solution to the composite slab is:

The diffusion solution for a composite slab is the error function series:

© meg/aol ‘02

Fourier’s Method

C x,t( ) = X x( )⋅T t( )

Solutions to Fick’s second law in finite media may be separated into spatial, X(x ), and time, T(t ), dependent functions:

∂C∂t

= D ∂2 C∂x 2

Fick’s second law for linear flow:

∂C∂t

= X x( )⋅dT t( )

dt

Substituting the product function solution, X(x) • T(t), into Fick’s 2nd law, The left-hand side becomes

D ∂2 C

∂x 2

= D T t()⋅

d2 X x( )dx 2

Similarly the right-hand side becomes

© meg/aol ‘02

Fourier’s Method

d T t( )dt

= −Dλ 2T t( )This part of the equation is a first-order ODE, in t:

d2 X x( )dx 2

= −λ 2X x( )This part of the equation is a second-order ODE, in x:

Combining the two substitutions:

X x( )⋅

dT t()dt

= D T t( )⋅

d2 X x( )dx 2

DT t()[ ]−1⋅

dT t()dt

The variables now may be separated as:

= X x( )[ ]−1

⋅d2 X x( )

dx 2

-λ2 -λ2

© meg/aol ‘02

Fourier’s Method

T t( )T 0( )

= exp −λ 2Dt( )

Solutions for the two ODE are:

Note: This is not a similarity solution,

(temporal function)

X x( )= A sin λx( )+ B cos λx( )and (spatial function)

The assumed product–function solution is:

A T 0( )sin λ x( )+ B T 0( )cos λ x( )[ ] exp −λ 2Dt( ) C x,t( )=

© meg/aol ‘02

Application of Fourier Series

C x,t( )= An sin λnx( )+ Bn cos λn x( )[ ]

n=1

∞

∑ exp −λ n2 Dt( )+ E

The Fourier series solution to the problem of diffusion between two slabs joined at the plane x = h is a sum of periodic eigenfunctions, plus a constant, E. The eigenfunctions used are sine and cosine functions in the spatial variable.

∂C x,t( )∂x

x= 0

= An λn cos 0( )− Bn λ n sin 0( )[ ]n=1

∞

∑ exp −λ n2 Dt( )= 0

Consider the no-flow boundary at x=0. Partially differentiate with respect to x

An

n=1

∞

∑ λn exp −λn2 Dt( )= 0 t > 0( )

The zero-gradient boundary condition at x=0 is satisfied only if:

© meg/aol ‘02


C x,t( ) = Bn cos λnx( )n=1

∞

∑ exp −λn2 Dt( )+ E

The concentration field for this example consists of a sum of cosine eigenfunctions with their individual amplitudes set by coefficients Bn. The diffusion solution takes the form:

∂ C x,t( )∂x x = l

= − Bnλn sin λnl( )n=1

∞

∑ exp −λn2 Dt( )= 0

The second no-flow boundary condition, at x = l, requires that:

λn =nπl

n = 1,2,3,...( )

The corresponding gradients are represented by sine functions, which vanish when:

sin λnl( )= sin nπ( ) = 0, n = 1,2,3,...( )

Substituting for the gradients,

© meg/aol ‘02


C x,t( ) = Bn cos nπxl

n=1

∞

∑ exp −nπl

2

Dt

+ E, 0 ≤ x ≤ l( )

The diffusion solution for a composite is:

C x,t( ) = Bn cos nπxl

n=0

∞

∑ exp −nπl

2

Dt

, 0 ≤ x ≤ l( )

The diffusion solution may be rewritten as the cosine-exponential product-function series:

C x,0( ) = Bn cosnπx

l

n=0

∞

∑ , 0 ≤ x ≤ l( )

Writing as a sum of cosine eigenfunctions with amplitudes Bn:

© meg/aol ‘02


C x,0( ) ≡1l

Bn cos nπxl

d x

n=0

∞

∑0

l⌠ ⌡

The average concentration, , may be defined over a solution interval, 0 ≤ x ≤ l, as:

C x,0( )

C x,0( ) = B0 , 0 ≤ x ≤ l( )

The average value of the concentration over the composite slab is equal to the leading coefficient, B0.

a) C x,0( )= C0, 0≤ x < h( ),The value of the average composition can be found using the initial data for the composite slab:

b) C x,0( )= 0, h < x ≤ l( ).

© meg/aol ‘02


B 0≡1l

C x, 0( )d x0

l

∫ =1l

C00

h

∫ d x +1l

0h

l

∫ ⋅d x

The average composition, B0, may be found from the initial distribution of composition as:

B 0= C0hl

Solving the integrals yields:

cos nπxl

⋅cos pπx

l

dx

0

l

⌠

⌡ =

0 l 2

n ≠ p

n = p

Further: • evaluate the unknown amplitude coefficients, Bn (n = 1, 2, 3,……), • consider the orthogonality of the eigenfunctions, a • b = 0.

© meg/aol ‘02


C x,0( )⋅cos pπxl

d x

0

l⌠ ⌡ = Bn cos nπx

l

⋅cos pπx

l

d xn=1

∞

∑0

l⌠ ⌡

The initial data, , must also satisfy the orthogonality condition, so at t = 0: C x,0( )

cos x,0( )⋅cos pπxl

d x

0

l

∫ = Bnl2

δnp n = 1,2,3( )n=1

∞

∑

Now evaluate the definite integrals,

δnp ≡0 n ≠ p( )1 n = p( )

where is the Kronecker delta, or “pickout function.”

© meg/aol ‘02


Bn =

2l

C0 cos nπxl

d x +

2l

0⋅cos nπxl

dx

h

l

∫0

h

∫

Substituting for the initial concentration, provides expressions to determine all the values of the amplitudes, Bn.

Bn =

2C0

nπsin nπh

l

n = 1,2,3,...( )

Evaluate these integrals and obtain the individual amplitudes:

© meg/aol ‘02


C x,t( ) =C0h

l+

2C0

π1nn=1

∞

∑ sin nπhl

cos nπxl

exp −nπl

2

Dt

The Fourier series solution for the composite slab diffusion problem yields:

C x,t( )C0

=hl

+2π

1nn=1

∞

∑ sin nπ hl

cos nπ x

l

exp −

nπ2

2 2 Dt

l

2

The dimensionless solution is:

© meg/aol ‘02

Error Function Series Approximation

C x / l,t;1/ 3( )C0

≅12

erf x / l +1/32 Dt / l

− erf x / l -1/3

2 Dt / l

+erf x / l + 2 +1/32 Dt / l

− erf x / l + 2 -1/32 Dt / l

+erf x / l - 2 +1/32 Dt / l

− erf x / l - 2 -1/32 Dt / l

A three-term approximation is now evaluated using a polynomial expression to approximate the error function.

© meg/aol ‘02

Error Function Series Approximation (3 terms)

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Composite Slab: h/l=1/3

C/C

0

Dimensionless Distance, (x/l)

0.1 0.25 0.50.025

0.75

∞2(Dt)1/2/l=0

(error function series: 3 terms)

1

© meg/aol ‘02

Fourier Series Approximation (15 terms) C x,t( )

C0

=13

+2π

1nn=1

∞

∑ sin nπ3

cos nπx

l

exp −

nπ2

2 2 Dtl

2

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

C/C

0


2(Dt)1/2/l=0

0.10.025

0.250.5

0.751

∞Composite slab: h/l=1/3

(Fourier series: 15 terms)

Gibbs “ringing”

© meg/aol ‘02

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1

Error function series (n=0, +1,-1)Fourier series (n=0 . . .14)

C/C

0


2(Dt)1/2/l=0.025

Composite Slab: h/l=1/3

Comparison using error function complements and sums of Fourier terms

Negative concentrations!

C/C0>1!

© meg/aol ‘02

Codastefano’s Method Measuring tracer diffusivity with Codastefano’s method requires a solution that is identical to the one-dimensional bilayer slab.

Codastefano’s method requires continuous monitoring of the concentration at two fixed planes along x-axis:

C2 x = 5l / 6,t( ) = C0hl

+2

nπsin nπ h

l

cos n5π

6

e− (n π ) 2 Dt

l 2

n=1

∞

∑

x2 = 5l /6

C1 x = l / 6,t( ) = C0hl

+2

nπsin nπ h

l

cos nπ

6

e−(nπ )2 Dt

l 2

n=1

∞

∑

x1 = l /6,

© meg/aol ‘02

Codastefano’s Method

C1 t( ) − C2 t( )C0

=2π

sin π hl

cos π

6

− cos 5π

6

e

−π 2Dtl 2

ln C1 t( ) − C2 t( )C0

= ln 2 3π

sin π hl

−

π 2Dtl 2

The difference signal may be written as a one-term Fourier approximation:

The key result is obtained by taking the logarithms of both sides and replacing the cosines with their values.

© meg/aol ‘02


1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

0.02

0.04

0.06

0.08

0.10.1

0

C1 t

C2 t

0.40 tDimensionless Time, Dt/l2

Dim

ensio

nless

Con

cent

ratio

n

C2(t)

C1(t)

C odaste fano 's M ethod 5-te rm Fourie r A pp roxim a tion

h/l= 1/30

Dim

ensi

onle

ss C

once

ntra

tion

0

0.02

0.35

0.08

0.06

0.04

0.1 0.40

Dimensionless Time, π2D t / l2

0.05 0.20 0.25 0.30 0.15

C1 (t )

C2 (t )

© meg/aol ‘02

0 0.05 0.1 0.15 0.20

0.05

0.1

0.15

0.20.2

0

S t

0.20 t

Codastefano'sMethod

Log Difference Signal

h/l=1/30

Dimensionless Time, π2Dt/l2

Diff

eren

ce S

igna

l ln

(C1-C

2)

1

1

0

0.05

0.10

0.15

0.20

0.15 0.05 0 0.10 0.20

Diff

eren

ce S

igna

l ln

(C1 - C

2)

Dimensionless Time, π2D t / l 2


Linear response

© meg/aol ‘02

Exercises

1. Sketch the individual forms of the first five Fourier modes (n=0…4) contributing to the finite–slab diffusion solution, eq.(8.47). Then plot the concentration field that results from summing these modes. Set the parameter 2(Dt)1/2/l=0.1, as representative of conditions for a “short” diffusion time.

Cn x,t( ) =1n

sin nπ3

cos nπxl

e−

n2π 2

16 , n =1,2,3,...( )

C x,t( )C0

=13

+2π

1nn=1

∞

∑ sin nπ3

cos nπx

l

exp −

nπ2

2 2 Dtl

2

The finite–slab diffusion solution:

The subsequent terms are:

© meg/aol ‘02

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.2 0.4 0.6 0.8 1

Fou

rier

Mod

e, C

n(x,t)

x/l

n=1

n=2

n=3n=4

2(Dt)1/2/l=0.1

n = 1

n = 3 n = 4

n = 2

Exercises

© meg/aol ‘02

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

C(x

,t)/C

0

x/l

5-term Fourier Approximation

2(Dt)1/2/l=0.1

Sum of 5-term Fourier Approximation

Exercises

© meg/aol ‘02

Exercises

2. Repeat the procedures used in exercise 1, but now set the condition that the parameter 2(Dt )1/2/l = (2)1/2, as representative of conditions for a relatively long diffusion time.

-0.004

-0.002

0

0.002

0.004

0 0.2 0.4 0.6 0.8 1

Fou

rier

Mod

e, C

n(x,t)

x/l

n>1

n=1

2(Dt)1/2/l=(2)1/2

n = 1

n >1

© meg/aol ‘02

Exercises

0.328

0.33

0.332

0.334

0.336

0.338

0 0.2 0.4 0.6 0.8 1

C(x

,t)/C

0

x/l

5-term Fourier Approximation

4(Dt)/l2=2

Sum of 5-term Fourier Approximation

© meg/aol ‘02

Exercises

3. The semiconductor industry uses a two–step diffusion sequence to form active p-n junctions in single crystal silicon wafers by distributing controlled amounts of electronically active diffusants, or “dopants.” The starting wafer initially contains a uniform but miniscule concentration of dopant, C∞. • In the first step, called “pre–deposition diffusion,” a controlled amount of dopant is introduced into the semiconductor near the free surface. • In the second step, called “drive–in diffusion,” the initially pre–deposited source of dopant is diffused much more deeply into the semiconductor.

© meg/aol ‘02

Exercises

Semiconductor Wafer

(1) Pre-deposition diffusion

SemiconductorWafer

(2) Drive-in diffusion

dopant source

sqrt(Dt)1

sqrt(Dt)2

x

0

x

0

Step 1: Predeposition

Step 2: “Drive-in”

Semiconductor Wafer

(1) Pre-deposition diffusion

SemiconductorWafer

(2) Drive-in diffusion

dopant source

sqrt(Dt)1

sqrt(Dt)2

x

0

x

0

© meg/aol ‘02

Exercises

C x,t( ) = C∞ + C0 − C∞( )2π

Dt( )1

Dt( )2

e−

x 2

4 Dt

The final distribution of the dopant with the thin-film solution, for no-flow boundary condition is:

C x,t( )1 − C0

C∞ − C0( )= erf ˆ x

2 Dt( )1

, x ≥ 0( )

The error function distribution is:

© meg/aol ‘02

Exercises

C x,t( )1∗ − C0

C∞ − C0( )= erf ˆ x

2 Dt( )1

, ˆ x ≤ 0( )

The fictitious image source is given by:

The solution is:

C x,t( ) − C0

C∞ − C0( )=

14πDt

erfˆ x

2 Dt1

−∞

0

∫ e−( x− ˆ x ) 2/ 4( Dt )2 d ˆ x

+ erf ˆ x

2 Dt1

0

∞

∫ e −( x− ˆ x ) 2/ 4(Dt )2 d ˆ x

fictitious part

physical source

© meg/aol ‘02

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Distance, [microns]

0.423332

.2.199772 10 14

da

bxf3 x

da

bxf2 x

da

bxf1 x

da

bxf4 x

60 ,,,Q3 Q2 Q1 Q4

Distance, x, [microns]

(Dt )2=1

(Dt )2=5

(Dt )2=0.4

(Dt )2=0.2D

opan

t Con

cent

ratio

n, [C

(x,t)

-C0]/

[C∞-C

0](Dt ) 2 =

0.2

(Dt ) 2 = 0.4

(Dt ) 2 = 1

(Dt ) 2 = 5

Distance, x [microns]

Dop

ant C

once

ntra

tion

Exercises

Dopant distributions for different drive-in annealing times.

© meg/aol ‘02

Exercises

0 1 2 3 4 5 60

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.220.201317

.2.484447 10 5

da

bxf1 x

da

bxf2 x

2π

0.11

12

e

Q2

4 1

2π

0.110

12

e

V2

4 10

60 ,,,Q1 Q2 Q V

Distance, x, [microns]

Dop

ant C

once

ntra

tion,

[C(x

,t)-C

0]/[C

∞-C

0]

Dt=1

Approximate solutionExact source integral

Dt=10

D t = 1

D t = 10

Distance, x [microns]

Dop

ant C

once

ntra

tion

Comparison of exact (integral) and approximate (thin film) solutions.

C x,t( ) = C∞ + C0 − C∞( )2π

Dt( )1

Dt( )2

e−

x 2

4 Dt

© meg/aol ‘02

Key Points • Diffusion in thin materials often involves “no-flow” boundary conditions,

where the diffusant is reflected back into the interior. • Two methods may be used to solve diffusion problems:

– Fictitious sources (method of images) – Fourier’s method

• Fictitious sources satisfy the b.c.’s using error eigenfunctions. • Fourier’s method satisfy the b.c.’s using sine/cosine eigenfunctions. • Error eigenfunctions don’t support the bc’s, whereas Fourier modes do.

– Fourier series converge rapidly for large time tags. – Fictitious sources converge rapidly for small time tags.

• Codestefano’s method allows continuous monitoring of the diffusion signal--a unique experimental characteristic.

• Semiconductor dopant processing can be analyzed exactly with fictitious source integral methods.

© meg/aol ‘02

Module 9: Spherical Bodies

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline • Diffusion around spheres

– Boundary conditions – Spherical diffusion solution

• Behavior of the spherical concentration field • Interfacial fluxes • Quasi-static growth and dissolution • Moving boundaries • Evolution of a spherical particle

– Conditions on the phases – Conditions at the α–β interface – Moving boundary solution

• The characteristic equation • Diffusion boundary layers • Exercise

© meg/aol ‘02

Diffusion around spheres Fick’s second law for the α (matrix) phase in spherically symmetric form

∂Cα

∂t=

1r 2

∂∂r

Dα ⋅ r 2 ∂Cα

∂r

∂Cα

∂t= Dα

∂2 Cα

∂r 2 +2r

∂Cα

∂r

, (r > R0)

If the diffusivity of the α phase is a constant, Dα, then Fick’s second law becomes:

∂Cβ

∂t= Dβ

∂2 Cβ

∂r 2 +2r

∂Cβ

∂r

, (r < R0)

Fick’s second law for the β particle, at constant Dβ is:

© meg/aol ‘02

Boundary conditions

Cβ r( )= Cβ α = const. r ≤ R0( )

Concentration field within the β–phase

∆Ceq ≡ Cα β − Cβ α

The composition “jump” across the α–β phase (phase diagram)

r

C

βα

R0

C∞

Cβ/α

Cα r ,t( )

∆Ceq

Cα/β

Physical situation within and around a β–phase particle embedded in a matrix of α–phase.

Dissolving particle

Note that the matrix contains a lower average concentration of solute, C∞, than the equilibrium value Cαβ.

© meg/aol ‘02

C∞

C∞

α

β

Cα β

Cα β

Cβ α

Cβ α

r0

J(R0)

R0r R0

Spherical β particle growing in a supersaturated α matrix.

Boundary conditions

Note that the matrix contains a higher average concentration of solute, C∞, than the equilibrium value Cαβ.

Growing Particle

© meg/aol ‘02

Boundary conditions

αβ

T

C

∆Ceq

Cα/β Cβ/α

C∞

Equilibrium solubility relationships between α and β.

Tie line defines the solubility “jump” between the phases at equilibrium.

C∞ for alloy is set arbitrarily

© meg/aol ‘02

Spherical Diffusion Solution

2r

∂C∂r

=

2r 2

∂u∂r

−

2ur 3

∂2 C∂r 2 =

1r

∂2u∂r 2

−

2r 2

∂u∂r

+

2ur 3

∂C∂t

=1r

∂u∂t

Substituting for u into the spherically symmetric diffusion equation:

Fick’s second law for spherical symmetry transformed in u

∂u∂t

= Dα∂2u∂r 2

A spherical transform variable u=C • r, is used to solve for the concentration field.

© meg/aol ‘02


∂u∂t

= Dα∂2u∂r 2

u = Cα β ⋅ R0, (r = R0, t ≥ 0)

Boundary condition for u at α–β interface

u = C∞ ⋅ r, (t = 0, r > R0 )

Boundary condition on u for α–phase composition

˜ u = R0 Cα β − C∞( ) e−r −R0

Dα

⋅ p

p

+C∞r

p

The Laplace transform solution is:

˜ u ≡ ue− pt dt

0

∞

∫where

PDE for u in t and r

© meg/aol ‘02


Cα r ,t( )= C∞ + Cα β − C∞( ) R0

r

erfc r − R0

2 Dαt

, (r ≥ R0)

The spherical diffusion solution by inverting the transform is:

C ≡

Cα r,t( ) − C∞

Cα β − C∞

CThe dimensionless concentration, , can be defined as:

C r,t( ) =

R 0

rerfc

r R 0( )−14Dαt R 0

2

, r R0 ≥ 1( )

and therefore

© meg/aol ‘02

Behavior of the spherical concentration field

∂ Cα

∂r

= − Cα β − C∞( ) R0

r2 erfcr − R0

2 Dαt

+

R0

r πDαte

− r − R0( )2

4 Dαt

The concentration gradient in the α matrix surrounding the spherical β particle is:

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5r/R

0

102.5

0.50.250.1

(Dαt/R2)1/2=∞

10

[C(r

,t)-C

∞]/(

Cα

/β-C

∞)

(Dα t/R2)1/2 =∞

C(r,

t)-C

∞/(C

αβ-

C∞)

© meg/aol ‘02

Interfacial Fluxes

J r ,t( )= −Dα

∂Cα

∂r

, r ≥ R0( )

The flux flowing in the α–phase at any radial distance, r, is

J r ,t( )=Dα R0

rCα β − C∞( ) 1

rerfc r − R0

2 Dαt

+

1πDαt

e− r − R0( )2

4Dα t

The radial flux is found by substituting for the gradient function

J R0,t( )= Dα Cα β − C∞( ) 1

R0

+1

πDαt

The flux at the α–β interface is:

© meg/aol ‘02

Interfacial Fluxes

J R0,t( )= Dα Cα β − C∞( ) 1

R0

+1

πDαt

At short times,

1R0

<<1

πDαt

For long times,

1R0

>>1

πDαt

J t( ) ≅Dα

πtCα β − C∞( )

J R0( )≅

Dα Cα β − C∞( )R0

short

long

© meg/aol ‘02

Quasi-Static Growth and Dissolution

dVdt

= 4πR 2 d Rdt

The rate of the volume change

The quasi-static approximation Ý R t( ) =Dα

R t( )Cα β − C∞

Cα β − Cβ α

dimensionless saturation, S

Cα β − C∞

Cα β − Cβ α

4πR2 ∆Ceq( ) dR

dt

− 4πR 2J R( )= 0Stefan’s interface condition

RdRR 0( )

R t( )

∫ = Dα

Cα β − C∞

Cα β − Cβ α

0

t⌠

⌡ d ′ t Direct integration of the ODE

Rate of mass change Rate of inflow

(Kinematic law)

© meg/aol ‘02


R t( ) ≅ R02 + 2Dα S ⋅ tThe quasi–static evolution equation

τ ≡R 0

2

2Dα

Non-dimensionalized by the characteristic time, τ.

R t( )R0

= 1+ S⋅ t / τ( )Non-dimensional quasi–static evolution equation for a spherical particle:

Ý R ≡ d R

dt=

SDα

R 02+ 2SDαt

The quasi–static interfacial speed:

© meg/aol ‘02

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10

R(t

)/R 0

Dimensionless time, (t/τ)

S=0.50.35

0.2

S=-0.5 -0.3 -0.2

0.1

-0.05

-0.1-0.14

0

τ=(R0)2/2D

α

τ = (R0) 2 / 2Dα


Undersaturated S<0

Supersaturated S >0

© meg/aol ‘02

Evolution of a Spherical Particle

Condition on the phases: The β-phase is assumed to be uniform with a composition fixed at the equilibrium interfacial value Cβ/α.

Cβ(r,t) = Cβ / α = constant, for all r < R(t)

∆Ceq

d Rdt

+ Dα

∂Cα r ,t( )∂r

= 0

Condition on the moving interface: Stefan’s interface mass balance applies at every point on the α–β interface.

Rate of mass increase Rate of diffusion

© meg/aol ‘02

Moving Boundaries

Cα r − R0,t( )= Cα η( )The quasi-static diffusion solution is

η ≡

r − R0

4Dαt

The argument, η, denotes the similarity variable introduced by the Laplace transform solution for spherical diffusion,

Replacing each term in Fick’s second law, using η and chain differentiation yields

∂Cα r ,t( )∂t

=∂η∂t

dCα

dη

=

−η2t

dCα

dη

• first time derivative:

∂Cα r ,t( )∂r

=∂η∂r

dCα

dη

=

14Dαt

dCα

dη

• first spatial derivative:

© meg/aol ‘02

Moving Boundaries

∂2 Cα r ,t( )∂r 2 =

∂∂η

∂η∂r

⋅

∂Cα

∂r

=

14Dαt

d2Cα

dη2

• the second spatial derivative:

−η2Dαt

dCα

dη

=

14Dαt

d2Cα

dη2

+

2r − R0

14Dαt

dCα

dη

Substitution yields Fick’s law in the similarity variable, η.

−η

dCα

dη

=

12

d2Cα

dη2

+

1η

dCα

dη

The second order ODE with non-constant coefficients is

d2Cα

dη2 = −2 η +1η

dCα

dη

or

© meg/aol ‘02

Moving Boundaries

w ≡

dCα

dηIntroducing the new variable, w,

d2Cα

dη2 = −2 η +1η

dCα

dη,

dwdη

= −2 η +1η

w

Transforming the second-order ODE into a first-order ODE in w,

dww

⌠ ⌡ = −2 η +

1η

dη

⌠

⌡ + const

The variables in the first-order ODE may be separated and integrated.

ln w= −2lnη − η2 + ln aSolving for w, yields the solution

w =

aη2 e−η2

or

© meg/aol ‘02

Moving Boundaries

dCα

dη=

aη2 e−η2

Combining w ≡

dCα

dηand

w =

aη2 e−η2

yields

dCα

Cα η( )

C∞

∫ = a e−η2

η2η

∞

∫ dη

C∞ − Cα η( )= a⋅ I η( )The solution is where

I η( )=e−z2

z2 dzη

∞

⌠

⌡

The result may be separated and integrated

© meg/aol ‘02

Moving Boundaries

udv∫ = uv− vdu∫ u = e−z 2

dv =

d zz2

The functional I(η) may be evaluated using integration by parts

where

I η( )= −e− z 2

zη

∞

− −1z

−e−z 2( )2zd z

η

∞

⌠

⌡

Inserting the choices for u and dv

I η( )=

e−η 2

η− 2 e−z 2

d zη

∞

∫or

© meg/aol ‘02

Moving Boundaries

I η( ) =e− η 2

η− 2 e−z 2

d zη

∞

∫ + e−z 2

d z0

η

∫ − e−z 2

d z0

η

∫

Adding and subtracting the integral 2 e− z 2

0

η

∫ gives

I η( )=

e−η2

η− 2 e− z 2

dz0

∞

∫ + 2 e−z 2

d z0

η

∫

The integral may be rearranged in standard form,

I η( )=

e−η 2

η− πerf ∞( )+ π erf η( )

or written as the error functions: 1

© meg/aol ‘02

Moving Boundaries

Cα η( )= C∞ − a e− η2

η− πerfc η( )

or

Ksphere ≡

R t( )− R0

4DαtThe position of the interface is tracked by the interface characteristic, Ksphere:

C∞ − Cα η( )= a e− η2

η− π + π erf η( )

The solution around a growing or dissolving spherical particle:

© meg/aol ‘02

Moving Boundaries

Cα Ksphere( )= Cα / β = C∞ − ae− KS( )2

Ksphere

− π erfc Ksphere( )

a = −Cα /β − C∞

e− Ksphere( )2

Ksphere

− πerfc Ksphere( )

Inserting the equilibrium boundary condition that C(Ksphere,t) = C α/β,

Solving for a

© meg/aol ‘02

Moving Boundaries

Cα η( ) = C∞ +Cα / β − C∞

e− Ksphere( )2

Ksphere


e−η2

η− π erfc η( )

Inserting a yields

Cα r ,t( )= C∞ − C∞ − Cα / β( )

e−

r − R0( )2

4Dαt

r − R0

4Dαt

− π erfc r − R0

4Dαt

e−

R t( )− R0( )2

4Dα t

R t()− R0

4Dαt

− π erfcR t()− R0

4Dαt

Reintroducing the definition of η,

© meg/aol ‘02

Moving Boundaries

C r,t( ) =

e−

r− R0( )2

4 Dαt

r − R0

4Dαt

− π erfcr − R0

4Dαt

e−

R t( )−R0( )2

4Dα t

R t( ) − R0

4Dαt

− π erfc R t( ) − R0

4Dαt

In terms of dimensionless concentration, the diffusion solution is

© meg/aol ‘02

The Characteristic Equation

The characteristic Ksphere ≡

R t( )− R0

4Dαtis a key diffusion parameter.

d Rdt

= Ksphere

Dα

t

The speed of the interface, dR/dt, according to the definition of Ksphere is

± Ksphere

Dα

t

+ Dα

∂Cα r ,t( )∂r

R

= 0

Substituting into Stefan’s local mass balance condition, gives

Ksphere ∆Ceq( ) ±

Dα

t

+ Dα

∂Cα r ,t( )∂η

∂η∂r

R

= 0

Considering only the positive root for a supersaturated matrix

© meg/aol ‘02


∂η∂r

=

12 Dαt

The partial differentiation gives

2Ksphere∆Ceq +

∂Cα

∂η

η=ηα / β

= 0Substituting back into Stefan’s local mass balance

∂C∂η

= −

Cα / β − C∞

e− Ksphere( )2

Ksphere


e−η2

η 2

The concentration derivative with respect to η,

© meg/aol ‘02


dCα

dη

η= Ksphere

= −Cα /β − C∞

e− K sphere2

Ksphere


e− K sphere2

K sphere2

The interfacial gradient can be determined by setting η = Ksphere

Substituting yields

2Ksphere∆Ceq −Cα /β − C∞

e− K sphere2

Ksphere


e− K sphere2

K sphere2 = 0

Diffusion flux at interface Interface “rejection”

© meg/aol ‘02


2 Ksphere( )2

1− π KsphereeKsphere( )2

erfc Ksphere( )

=

Cα /β − C∞

∆Ceq

= S

After a few steps of algebra and rearranging:

Ý R = d R

dt= Ksphere Dα( )⋅t

−12 =

κt

The interfacial speed is Characteristic equation.

Kinetic equation

© meg/aol ‘02

Moving Boundary Solution

Comparison of moving boundary solution with quasi-static result

0.0001

0.001

0.01

0.1

1

10

100

0.0001 0.001 0.01 0.1 1

2(K

Sphe

re)2

Dimensionless Supersaturation, S

Quasi-staticapproximation

Moving Boundary SolutionMoving Boundary Solution

Quasi-static approximation

Non-linear behavior 2(

Ksp

here

)2


© meg/aol ‘02

Moving Boundary Solution

10

100

1000

0.7 0.8 0.9 1

2(K

Sphe

re)2

Dimensionless Supersaturation, SDimensionless Supersaturation, S

2(K

sphe

re)2

Behavior at large supersaturations

∞

© meg/aol ‘02

Diffusion Boundary Layers

Λ BL ⋅

∂C∂η

η= Ksphere

≡ C∞ − Cα / βThe diffusion boundary layer, , may be defined as Λ BL

Λ BL =

K sphere2

e− K sphere2

e−K sphere2

Ksphere

− πerfc Ksphere( )

Substituting for the interfacial gradient

Λ BL = Ksphere1− π Kspheree

K sphere2

erfc Ksphere( )( )=S

2Ksphere

In the view of the characteristic equation, the relation becomes

ΛBL

Ksphere

=S

2K sphere2

Comparing the boundary layer with the interface characteristic

© meg/aol ‘02

A Spherical Diffusion Boundary Layer

Cα/ β

C∞

α

Λ

Cβ/ α

com

posit

ion

r

Λ

β

boundary layer

com

posi

tion

boundary layer

© meg/aol ‘02

Boundary Layer Thickness

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Rel

ativ

e Bo

unda

ry L

ayer

Thi

ckne

ss, Λ

/Ksp

here


β

β

Λ


Rel

ativ

e B

ound

ary

Laye

r Thi

ckne

ss Λ

/ K

sphe

re

Note how boundary layer decreases with increasing saturation

© meg/aol ‘02

Exercise

1. An Al–10 at.% Mg forging alloy contains widely–spaced, uniform, spherical particles of the brittle intermetallic, β–Al3Mg2, the mean diameter of which is 15 micrometers. A heat treatment is desired for homogenization of the alloy to achieve a better forging response. If the alloy is annealed at 450C, a) Determine the minimum time, tmin, it would take for the particles of β–Al3Mg2 to totally dissolve into the Al-Mg 19 at.% matrix, given that the diffusion coefficient, D, at 450C for Mg mixing with Al in this alloy is 4.5×10-10cm2/sec. b) Compare the estimates for tmin using each of the two following theoretical approaches: a) quasi–static linear diffusion theory b) moving boundary theory

© meg/aol ‘02

Exercise

© meg/aol ‘02

Exercise

S ≡Cα / β − C∞

Cα / β − Cβ / α

= 19 −10

19 − 38.5,

S = - 0.461.

a) The (under)saturation, S , in the matrix under a specified temperature and concentration is

R(t) = R 02+ 2Dα S ⋅ t

The quasi-static linear diffusion theory shows the following radius–time relationship

tdis = − R 02

2DαS= -

7.5 × 10-4 cm( )2

2 × 4.5 ⋅10-10 cm2/s × (-0.461),

tdis = 1356 sec, or 22.6 min .

The dissolution time tdis at R(t) = 0 is

© meg/aol ‘02

Exercise

Ksphere =R(t) − Ro

4Dαt

b) Applying moving boundary theory, the characteristic equation is

tdis =R o

2

(Ksphere ) 2 4Dα

=7.5 ×10−4cm( )2

(−1)2 × 4 × 4.5 ⋅10−10 cm2/s,

tdis = 312.5 sec, or 5.2 min.

The dissolution time tdis at R(t ) = 0, for moving boundary solution

© meg/aol ‘02

Key Points • Spherical diffusion in isotropic materials involves only one

spatial variable and time. • Diffusion in a matrix surrounding a second-phase particle

requires interface equilibrium information (constitutive data) extracted from the phase diagram.

• Kinematic considerations can be added to approximate quasi-static growth of a spherical particle.

• Quasi-static diffusion allows a convenient description of precipitate growth and dissolution, limited to small saturation.

• The moving boundary solution involves non-linear behavior caused by the changing size of the field domain, Cα(r,t).

• The concept of diffusion boundary layers is introduced, as an approximate method to estimate the interaction distances around a particle undergoing growth or dissolution.

© meg/aol ‘02

Module 10: Steady-State Diffusion

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Permeation in linear flow – Molecular dissociation at surfaces

– Steady-state permeation

– Permeability units

• Steady flow with variable D – Boundary Conditions

• Steady-state diffusion in cylinders: constant D

• Steady-state diffusion in spheres

• Exercise

© meg/aol ‘02

Permeation in Linear Flow The reaction for the dissociation of hydrogen on a solid surface is

12

H2 (gas) ⇔ H(solid soln.) Keq =

H[ ]H2[ ]

12

=S

PH2

12

with the equilibrium constant

Fick’s first law relates the flux and the gradient as

Jss = −D ∂C

∂x

= −D C2 − C1

l

Jss = −DKeq

P2 − P1( )l

= −D S1− S 2

l

The flux is proportional to

the solubility gradient

Π ≡ DKeqgas permeability

C2 at x = l , for the steady-state

C1 at x = 0

© meg/aol ‘02

Permeation in Linear Flow

Gaseous diffusant Metal Π0

[cm3(STP)/sec-cm-atm1/2]QΠ

[cal/mole]H2 Ni 1.2×10-3 13,850H2 Cu 1.9×10-4 17,400H2 α-Fe 2.9×10-3 8,400H2 Al 3.7×10-1 30,800N2 α-Fe 4.5×10-3 23,800

Permeability data, gases in metals [ Π = Π0 exp (-QΠ/kBT)]

© meg/aol ‘02

Steady Flow with Variable D

∂C∂t

= 0 =∂∂x

D ∂C∂x

The general form of Fick’s second law for steady flow is

D ⋅

dCdx

= A const.( )Within the steady-state diffusion field

Jss = −D ∂C

∂x

= −AConsequently, Fick’s first law reduces to

© meg/aol ‘02

Steady Flow with Variable D

D C( ) = D 0( ) 1+ f C( )[ ]

The diffusivity is considered a function of the local concentration.

D 0( )1+ f C( )[ ]⋅ dC

d x

= A

Inserting the expression for the variable diffusivity yields

D 0( ) 1+ f C( )[ ]∫ dC = Ax + B

Separating the variables

© meg/aol ‘02

Steady Flow with Variable D: Boundary Conditions

C = 0, x = l( ) C = C0, (x = 0)Boundary conditions for a steady-state

diffusion through a slab of thickness, l .

f C( )= K C

C0

Select a form for the concentration dependence of the diffusivity.

D 0( ) 1+KCC0

C0

C

⌠

⌡ dC = A d x

0

x

∫ , 0≤ x ≤ l( )Substituting for f(C ),

D 0( ) C − C0 +

K2C0

C2 − C02( )

= Axwhich when evaluated yields

© meg/aol ‘02

Steady Flow with Variable D: Boundary Conditions

D 0( ) −C0 +

K2C0

−C02( )

= AlInserting the boundary condition

C = 0 at x = l ,

A =

− D 0( )l

C0 +KC0

2

=

− D 0( )C0

l1+

K2

Solving the equation for

constant A.

Jss = − A=

D 0( )C0

l1+

K2

Inserting the value of A into Fick’s

first law

C − C0 +

K2

C2 − C02( )

C 0

=−xl

1+K2

C0

The steady-state concentration field is

© meg/aol ‘02

Steady Flow with Variable D: Boundary Conditions CC0

−1+K2

CC0

2

−1

− 1+K2

=xl

In dimensionless variables the concentration equation becomes:

C x l( )C0

= −2K

± 2 1K 2 +1+

2K

−2K

+1

xl

Inverting via the quadratic formula

where K is the sensitivity factor.

C x / l( )C0

= 1−xl

For K = 0, the solution becomes

© meg/aol ‘02

Linear flow with a variable diffusivity, D(C).

Steady-State Concentration Field

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Conc

entra

tion,

C/C

0

Distance, x/l

K=0

-1-0.8

-0.5

0.51.5

3 7.5100

D(C)=D(0)[1+K(C/C0)]

Distance, x / l

D (C) = D (0) [ 1+K (C/C0)]

Con

cent

ratio

n, C

/ C

0

Note that solutions do not exist for K<-1

Solution for D=constant.

© meg/aol ‘02

Steady-State Gradient Field

Linear flow with a variable diffusivity, D(C )

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0 0.2 0.4 0.6 0.8 1

Gra

dien

t, d(

C/C 0)/d

(x/l)

Distance, x/l

-0.5-0.8

-1

0.5

1.5-0.5

3

31.5

K=0

100

0.5

100

-1

D=D(0)(1+K(C/C0))D (C) = D (0) [ 1+K (C/C0)]

Distance, x / l

Gra

dien

t, d

(C/C

0) /

d (x

/l)

Note that solutions do not exist for K<-1

© meg/aol ‘02

Steady-State Diffusion in Cylinders: Constant D

D d

drr dC

d r

= 0

Fick’s law reduces to an ODE (Laplace’s Equation) in a single spatial variable, r .

C r( )= A + B ln r

The steady-state solution takes the form

C = Cb; r = b

C = Ca; r = aThe boundary conditions are

© meg/aol ‘02

Steady-State Diffusion in Cylinders: Constant D

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

C(r

)/Ca

r/a

b=2 3 4 5 6 7 8 9 10

C r( )=Ca ln b

r( )+ Cb ln ra( )

ln ba( )

Steady–state diffusion solution for cylinders, a =1.

Thick shells

Thin shells

© meg/aol ‘02

Steady-State Diffusion in Spheres

baO

R

D dd R

R 2 dCd R

= 0

C = Cb; R = b

Fick’s second law in spherical coordinates

C R( )=

AR

+ B

The Laplace equation in spherical coordinates

C = Ca; R = aThe boundary conditions over a pair of 2 radial positions are

© meg/aol ‘02

Steady-State Diffusion in Spheres

C R( ) =aCa b − R( )+ bCb R − a( )

R b − a( )The steady-state diffusion solution for spheres, a =1

Thin shells

Thick shells

© meg/aol ‘02

Exercise 1. Two nearby edge dislocations of opposite sign initially occupy a common slip plane (y=0). At the surface of the dislocation core the vacancy concentration is close to its equilibrium value, Ceq=0.5, but far from the dislocation cores the surrounding crystal has a substantial supersaturation of lattice vacancies, Cv=2Ceq=1. The dislocations absorb the excess vacancies at their cores and, consequently, climb up, or down, as indicated in Fig.10-7. Assume that the dislocation climb is diffusion limited, and ignore any influence of stress on diffusion. Sketch the steady–state vacancy saturation fields: a) in the initial state, when the dislocations cores occupy a common slip plane at locations (-1, 0) and (1, 0); b) when their cores have climbed to the positions (-1, -1) and (1, 1); and c) when the dislocations become more widely separated, with their cores at the positions, (-1, -4) and (1, 4).

© meg/aol ‘02

Exercise

Pair of edge dislocations undergoing climb by absorbing vacancies from the surrounding crystal.

up-climb

down-climb

y

x

Cv=1

Ceq=0.5

ra=0.5

rb=10

2

© meg/aol ‘02

Exercise

C(x,y) =0.25

ln 10( )ln 100

x −1( )2+ y − ˆ y ( )2[ ]

12

+0.5

ln 10( )ln x −1( )2+ y − ˆ y ( )2[ ]

12

+0.25

ln 10( )ln 100

x +1( )2+ y − ˆ y ( )2[ ]

12

+0.5

ln 10( )ln x +1( )2+ y + ˆ y ( )2[ ]

12

.

Vacancy concentration field

© meg/aol ‘02

Exercise Dislocation cores at (-1, 0) and (1, 0)

0 2 -2 -4 4 -4

0

2

-2

4

Dislocation cores at (-1, - 4) and (1, 4)

0

0

2

2

-2

-2 -4

-4

4

4

Field visualizations

0

0 2

2

-2

-2

-4

4

-4

Dislocation cores at (-1, -1) and (1, 1)

4

© meg/aol ‘02

Key Points • Some diffusion problems involve the passage of gas molecules

through solids. It is more convenient to formulate diffusion in terms of the gas solubility and the permeability, Π=DKeq, where Keq is the Sieverts law constant. Watch out for the units!

• Steady-state solutions with variable diffusivity were investigated. Such problems are non-linear, and may not always have physical solutions.

• Steady-state solutions were developed for fixed concentration boundary conditions: – Cylinders – Slabs – Spheres

• Application to dislocation climb was discussed.

© meg/aol ‘02

Module 11: Inverse Methods

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Time-dependent diffusion with variable diffusivity

• Boltzmann’s transformation

• Matano’s geometry – Matano’s interface

• Application of the Boltzmann–Matano Method

• Application of the method of Sauer, Freise, and den Broeder

• Exercise

© meg/aol ‘02

Boltzmann’s Transformation

∂C∂t

=∂∂x

D C( )∂C∂x

Fick’s second law with variable diffusivity D (C ) shows:

∂C∂x

=∂C∂ξ

∂ξ

∂x

=

12 t

∂C∂ξ

Introducing Boltzmann’s similarity variable, ξ ≡

x − XM

2 t, we obtain the first spatial

derivative.

∂C∂t

=∂C∂ξ

∂ξ∂t

= −x − XM

4t32

∂C∂ξ

The corresponding time derivative is

∂C∂t

= −ξ2t

∂C∂ξ

or equivalently,

© meg/aol ‘02


∂∂x

D C( ) ∂C∂x

=

∂∂ξ

∂ξ∂x

D C( )2 t

∂C∂ξ

Writing in terms of ξ and using the chain rule, we obtain

∂ξ∂x

=

12 t

Simplifying the equation by substituting for

∂∂x

D C( ) ∂C∂x

=

14t

⋅∂∂ξ

D C( ) ∂C∂ξ

The simplified result is

© meg/aol ‘02


−

ξ2t

dCdξ

=

14t

⋅d

dξD C( ) dC

dξ

The Fick’s second law, in its nonlinear form, may be reduced to a nonlinear ODE

−2ξ

dCdξ

=

ddξ

D C( ) dCdξ

or

© meg/aol ‘02

Matano’s Geometry

C = CR , (x > 0, t = 0)

The Boltzmann–transformed Fick’s second law, and the experimental configuration for finding D(C) is called the Boltzmann–Matano method.

−2 ξ dCCR

′ C

∫ = d D C( )⋅dCdξ

CR

′ C

⌠

⌡

The first integral with the established boundary conditions is

C = CL , (x < 0, t = 0)The initial conditions are

−2 ξ dCCR

′ C

∫ = D C( )⋅dCdξ

CR

′ C Carrying out the integration, we obtain

© meg/aol ‘02

Matano’s Geometry

The solute distribution provides a gradient, dC/dξ, that vanishes at C= CR .

−2 ξdC

CR

′ C

∫ = D ′ C ( ) dCdξ

C= ′ C

The boundary condition simplifies as

D ′ C ( )= −2 dξ

dC

′ C

⋅ ξdCCR

′ C

∫Solving gives the result

D ′ C ( )= −

12t

dxdC

′ C

⋅ x − XM( )dCCR

′ C

∫The Boltzmann–Matano solution is

© meg/aol ‘02

Matano’s Interface

Boltzmann–Matano geometry for a diffusion couple

x

"Matano Interface"

time=const.

C

CR

CL

′x

A1A2

A3

A4

A5

C'(x)

C=0xM

C = 0

“Matano Interface” at XM

time = constant

© meg/aol ‘02

Matano’s Interface The “Matano Interface” is determined through a mass–conservation condition

= C x( )− CR[ ]d x

x0

∞

∫gain

CL − C x( )[ ]d x

−∞

x0

∫loss

CL − C x( )[ ]x

−∞

XM− xdC

CL

CM

∫ = C x( )− CR[ ]xXM

∞− xdC

CM

CR

∫

When integrated by parts

CL − CM( )XM − xdC

CL

CM

∫ + CM − CR( )XM − xdCCM

CR

∫ = 0

Evaluating the integrals by applying Matano boundary conditions

© meg/aol ‘02

Matano’s Interface

CL − CR( )XM − xdC

CL

CM

∫ − xdCCM

CR

∫ = 0

Simplifying

xdC

CR

CM

∫ + xdCCM

CL

∫ = 0

When X M = 0, the equation becomes

xdC

CR

CL

∫ = ξdCCR

CL

∫ = 0

The diffusion time is fixed, so implicitly

© meg/aol ‘02

Application of the Boltzmann–Matano Method

Determine the D(C) function for two pairs of alloys by applying the Boltzmann–Matano method.

C(x,t ) = 0.75− 0.25erfc x

4Dt

• Apply Grube–Jedele solution to one pair of alloys, (A1 –B), exhibiting constant diffusivity

C(x, 0.5) = 0.25 tanh(x) + 2[ ]

• The second alloy pair, (A2–B), yields the solution diffusion in the form

© meg/aol ‘02

Concentration versus Distance for Boltzmann–Matano Analysis

0.2

0.3

0.4

0.5

0.6

0.7

0.8

-3 -2 -1 0 1 2 3

C=0.25[tanh(x/l)+2]C=0.75-0.25erfc(x/l)

Com

posi

tion,

C(x

/λ)

Distance, x/λ

CL=0.25

CR=0.75CR = 0.75

CL = 0.25

Distance, x/λ

Com

posi

tion,

C (x

/λ)

C = 0.25 [ tanh( x/λ )+2 ]

C = 0.75 - 0.25 erfc( x/λ )

© meg/aol ‘02

Numerical Results for Boltzmann–Matano Analysis

0

0.2

0.4

0.6

0.8

1

1.2

0.3 0.4 0.5 0.6 0.7

Diff

usiv

ity, D

(C)(

2t/λ

2 )

Composition, C

C=0.75-0.25erfc[x/(4Dt)1/2]

C=0.25[tanh(x/λ)+2]

Boltzmann-Matano

numerical solutions

C = 0.25 [ tanh ( x/λ) +2 ]

C = 0.75 - 0.25 erfc[ x /(4Dt)1/2 )]

Composition, C

Diff

usiv

ity, D

(C) (

2t /

λ2 )

Boltzmann–Matano

numerical solutions

© meg/aol ‘02

Application of Sauer, Freise, and den Broeder Method

Sauer, Freise, and den Broeder showed that Boltzmann–Matano analysis could be modified by mathematical procedures.

Ψ ≡

′ C − CR

CL − CR

In finding the Matano plane a concentration ratio, Ψ , may be defined as

A 2 = Ψ A 2+ A3 + A 4( )The definition of Ψ permits the relation

A1 ≡ ΨA 1+ 1− Ψ( )A1The left- and right-hand sides of the identity, can be added to yield the area relationship

A 1+ A 2 = 1− Ψ( )A 1+ Ψ A 1+ A 2+ A 3 + A 4( )Adding A1 and A2

© meg/aol ‘02

Sauer, Freise, den Broeder

x

"Matano Interface"

time=const.

C

CR

CL

′x

A1A2

A3

A4

A5

C'(x)

C=0xM

“Matano Interface” at XM

time = constant

′ C − CRΨ ≡ CL − CR

© meg/aol ‘02

Application of Sauer, Freise, and den Broeder Method

A 1+ A 2+ A3 = A5

The definition of the Matano interface is consistent now with the geometric statement

A 1+ A 2 = 1− Ψ( )A 1+ Ψ A 5+ A 4( )The relationship between the areas becomes

D ′ C ( )=1

2t dCdx

′ x

1− Ψ( ) ′ C − CR( )dx + Ψ CL − ′ C ( )dx−∞

′ x

∫′ x

∞

∫

The Sauer–Freise–den Broeder inverse solution is the integrodifferential expression

D x( )= cosh2 x( )⋅ ln 2cosh x( )( )− 2x sinh 2x( )The Sauer–Freise–den Broeder inverse solution in the form of a hyperbolic tangent

© meg/aol ‘02

Comparison of analytical results using den Broeder’s solution with numerical results using Boltzmann–Matano: • constant diffusivity, • variable D(C)

0

0.5

1

1.5

2

0.3 0.4 0.5 0.6 0.7

D(C) analytical

D[C=0.25(tanh(x/l)+2)]

D[C=0.75-0.25erfc(x/l)]

Diff

usiv

ity, D

(C)

Composition, C

D C = 0.25 [ tanh( x/λ )+2 ]

D[C = 0.75 - 0.25 erfc( x/λ )]

D (C) analytical

Composition, C

Diff

usiv

ity, D

(C)

Variable D(C)

Constant D

© meg/aol ‘02

Comparison of the analytical form of D (C) with the numerical results obtained using: × Boltzmann–Matano • Sauer–Freise–den Broeder

0.0

0.50

1.0

1.5

2.0

0.2 0.3 0.4 0.5 0.6 0.7 0.8

D(C)-den BroederD(C)-Boltzmann-MatanoD(C)-Analytical

D(C

)

ConcentrationComposition, C

Diff

usiv

ity, D

(C)

D (C) – den Broeder

D (C) – Boltzmann–Matano

D (C) – Analytical

×

© meg/aol ‘02

Exercise

1. Interdiffusion experiments carried out in Ta-W alloys by Romig and Cieslak involved the use of pure Ta and pure W as end-members. The diffusion couple was annealed at 2100C for 16 hours. At the end of the anneal the diffusion couple was removed from the furnace and quenched. The couple was sectioned for analyzing the concentration. Electron microprobe analysis yielded the concentration profile shown next. The Boltzmann–Matano and the Sauer, Freise, den Broeder solutions were applied to these concentration data, and diffusivity as a function of concentration was calculated. The diffusivity profile is plotted as a function of composition. The concentrations are in the couple range from pure Ta to pure W. Careful inspection of the results reiterates the earlier observation that the calculated diffusivities using either inverse method provide accurate results so long as the concentrations are not too far from the average composition of the two end-member alloys.

© meg/aol ‘02

Experimental Concentration Profile for Ta in a Ta–W Diffusion couple

0.00

20.0

40.0

60.0

80.0

100

0 20 40 60 80 100 120

Tant

alum

Con

cent

ratio

n [a

t.%]

Distance [µm]

Ta-W alloys

Distance , [µm]

Tant

alum

Con

cent

ratio

n, [a

t %]

Ta–W alloys

© meg/aol ‘02

Numerical Calculations Applied to the Experimental Profile

2.0 10-11

4.0 10-11

6.0 10-11

8.0 10-11

1.0 10-10

0 20 40 60 80 100

D(C)-Boltzmann-MatanoD(C)-SFdB

Diff

usiv

ity [c

m2 /s

]

Tantalum Concentration (at.%)Tantalum Concentration, [at%]

Diff

usiv

ity, [

cm2 /s

]

D (C) – Boltzmann–Matano

D (C) – Sauer–Freise–den Broeder ×

© meg/aol ‘02

Key Points • Most real alloy systems have D varying with concentration,

whereas linear diffusion theory requires that D=constant. • If D is only a function of concentration—not distance—, then the

Boltzmann transformation may be used. • This transformation introduces a similarity variable, ξ=x-XM/2t1/2

that depends on a new coordinate called the Matano interface, XM. • The Matano interface, is the balance point, found using the

Matano geometry that specifies compatible boundary conditions. • The Boltzmann-Matano equation extracts D(C) from C(x) data. • An alternative procedure is the Sauer, Freise, den Broeder

method. • The two methods yield results of comparable accuracy.

© meg/aol ‘02

Module12: Random Walks and Diffusion

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Background

• Diffusive motions

• Random walks

• One-dimensional random walks

• Combinatoric formulation

• Random walks in higher dimensions

• Diffusion viewed as a stochastic process

• Diffusivity: a microscopic transport property

• Exercises

© meg/aol ‘02

One-Dimensional Random Walks

-1 +1

+1-1 -1

Start x=0

+1+1-1-3

000 000 +4-4 -2 -2 -2 +2 -2 +2+2+2

+200-2

+3

A simple coin toss can be used as the random number generator • “heads” correspond to steps of +1 • “tails” correspond to steps of -1

© meg/aol ‘02

Combining the above equations, and consider independent Markov sequences

Nk

k=0

n

∑ = 2n


Pn,k ≡

Nk

2n

The probability, Pn,k, of a displacement, |kλ|, is defined as the ratio of the degeneracy, N k, to the total number of step sequences, 2n.

Pn,k

k=0

n

∑ = 1

The sum of the individual step probabilities is unity,

© meg/aol ‘02

n Sequence Displacementλk

SquareDisplacement,

λ k2

Root–Mean–Square Displacement

λk2

k∑

1 +1-1

+1-1

11

[(1/2)(2)(1)]1/2=1

2 +1,+1+1,-1-1,+1-1,-1

+200-2

4004

[(1/4)(2(4)+2(0))]1/2=(2)1/2

3 +1,+1,+1+1,+1,-1+1,-1,+1-1,+1,+1-1,-1,+1-1,+1,-1+1,-1,-1-1,-1,-1

+3+1+1+1-1-1-1-3

91111119

[(1/8)(2(9)+6(1))]1/2=(3)1/2

4 +1,+1,+1,+1+1,+1,+1,-1+1,+1,-1,-1+1,-1,-1,-1-1,-1,-1,-1-1,-1,-1,+1-1,-1,+1,+1-1,+1,+1,+1-1,+1,-1,+1-1,+1,+1,-1+1,-1,-1,+1-1,+1,-1,-1+1,-1,+1,+1+1,-1,+1,-1-1,-1+1,-1

+1,+1,-1,+1

+4+20-2-4-20

+2000-220-2+2

164041640400044044

[(1/16)(2(16)+8(4)+6(0))]1/2=(4)1/2

n Sequence Displacement λk

Sq. Displacement λ2

k RMS Displacement

( Σkλ2k )1/2

Markov Sequences: Each step sequence, k, and its associated displacement ± kλ occurs with equal probability.

© meg/aol ‘02


limn→∞

λkk=0

n

∑ = 0

The net displacement for a large number of random walks of equal step length is zero

12n Nkλk

2

k=0

n

∑ ≡ λk2 = nλ2

The mean–square–displacement, λ k2 , is defined as

λ k2 1

2= nλ

Taking the square root of both sides gives RMS displacement for an n–step, one dimensional symmetric random walk

© meg/aol ‘02

Combinatoric Formulation Consider a symmetric one-dimensional random walk comprised of n steps of equal size λ.

λn = p− n− p( )[ ]⋅λ = 2p − n( )⋅λThe net displacement achieved is

a + b( )n

=n !

p ! n− p( )!p=0

n

∑ apbn− pThe binomial theorem states that

2n =

n!p! n− p( )!p=0

n

∑If we choose a = b = 1, then

λn

2 =12n

n! 2p − n( )2

p! n− p( )! ⋅p=0

n

∑ λ2The mean–square–displacement may be written as

© meg/aol ‘02

Combinatoric Formulation

λ n

2 =1

2 nn!

p! n− p( )!p=0

n

∑ 4p 2− 4pn+ n 2( )⋅λ2The equation may be partially expanded

λ n

2 =λ2

2n −4 n!p! n− p( )! p

p=0

n

∑ n− p( )+ n 2 n!p! n− p( )!p=0

n

∑

and regrouped

This relationship simplifies because the first summation vanishes when p = 0, n, and the second summation is just 2n.

λn

2 =λ2

2n −4 n!p! n− p( )! p

p=1

n−1

∑ n− p( )+ n 22n

thus

© meg/aol ‘02

Combinatoric Formulation

λ n

2 =λ 2

2n −4n n−1( ) n− 2( )!p −1( )! n− p −1( )!p=1

n−1

∑ + 2nn2

Regrouping the factorials yields

λn

2 =λ 2

2 n −4n n−1( ) n − 2( )!p −1( )! n− 2( )− p −1( )[ ]!p−1= 0

n−2

∑ + 2nn2

The summation indices may be shifted

λ n2 = −n n−1( )+ n 2( )⋅λ2Substituting yields

λ n2 = nλ2The mean–square displacement is

λ n2 1/ 2

= nλand the RMS becomes

2n-2

© meg/aol ‘02

2-D Lattice Random Walk

-20

-15

-10

-5

0

5

10

15

-10 -5 0 5 10 15 20 25

Dis

tanc

e, y

Distance, x

START

FINISH

Each step has four possibilities: • North • South • East • West

© meg/aol ‘02

Random Walks in Higher Dimensions

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

Dist

ance

, yn

Distance, xn

n=1

2

3

n = 1

Distance, xn

Dis

tanc

e, y

n

Short sequence of a two-dimensional random walk

This random walk has 360 degrees of freedom per step!

© meg/aol ‘02

227-step random walk in two dimensions

-4

-2

0

2

4

6

8

10

12

-10 -8 -6 -4 -2 0 2 4 6

Dist

ance

, yn

Distance, xn

n=227

n=0

λ227

Distance, xn

Dis

tanc

e, y

n

n = 0

n = 227

Net Displacement=8.2 RMS=15.06

© meg/aol ‘02

Multiple Random Walks

-32

-16

0

16

32

-40 -20 0 20 40 60

y n

xn

(<R2>)1/2=31.62

n=1000

xn

y n

n = 1000

(<R2>)1/2 = 31.62

4 × 1000-steps two dimensional random walks with independent “seeds”

RMS=31.62

© meg/aol ‘02

1-D Random Walk Java Applet

© meg/aol ‘02

2-D Random Walk Java Applet

© meg/aol ‘02


The total displacement of the random walker’s net jump displacement is a scalar product of the form:

Ln

2= λ i

2 cosθi , ii =1

n

∑ + 2 λ i2 cosθ i , i+ j

j =1

n − i

∑i =1

n −1

∑

or

Ln2

=l 1 +

l 2 +

l 3 + ... +

l n( )⋅

l 1 +

l 2 +

l 3 + ...+

l n( )

Substituting yields the square displacement

Ln2 ≡ Ln ⋅Ln

Ln2

Ln =l 1 +

l 2 +

l 3 + ...+

l n

The total displacement vector for an n-jump sequence is the vector sum of the individual jumping vectors,

l i

© meg/aol ‘02


Ln

2= nλ2 1+ 2 n − i

n

cosθ

ii =1

n−1

∑

For i fixed, summing over the j index yields

Ln2

= nλ2The mean–square–displacement is

Ln

2= nλ2 1+

2n

cosθi , i + jj =1

n−i

∑i =1

n−1

∑

Ln

2= nλ2 + 2 λi

2 cosθi , i + jj =1

n− i

∑i =1

n−1

∑The sums of scalar products simplify to

or

Ln2

= nλThe correspondent RMS becomes

© meg/aol ‘02

Diffusion as a Stochastic Process

f r ,t( )≡ 4πr 2 e−

r 2

4Dt

4πDt( )32dr

0

r3

⌠

⌡

The cumulative distribution surrounding an instantaneous point source in 3-D

f (r3,t) = erf r3

2 Dt

−

2π

r3

2 Dt

e

−r3

2

4Dt

When evaluated gives the result

r

dr

Spherical diffusion cloud spreading from a point source

© meg/aol ‘02

Diffusion as a Stochastic Process

F r ,t( )≡

∂ f r( )∂r

=r 2

2 π Dt( )3 2 exp −r 2

4Dt

The radial derivative of f (r,t) is defined as F(r,t) , which is the radial probability density

F r ,t( )

0

∞

∫ dr = f (∞,t)= 1

where F(r,t) has units of reciprocal length and exhibits the integral property or normalization

© meg/aol ‘02

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

F(r,t)f(r,t)

Prob

abili

ty d

ensit

y fu

nctio

n, F

(r3,t)

Cum

ulat

ive

prob

abili

ty, f

(r3,t)

Distance from source, r3

Dt=1Dt = 1

F(r, t)

f (r, t)

Distance from source, r3

Pro

babi

lity

dens

ity fu

nctio

n, F

(r3,

t) C

umul

ativ

e pr

obab

ility

, f (

r 3, t

)

cumulative

probability density

Comparison of the probability density function with the cumulative probability in 3-D

© meg/aol ‘02

Diffusivity: As a Microscopic Transport Property

R2 ≡ r 2 F r ,t( )d r0

∞

∫ =1

2 π Dt( )3 2 r 4 exp −r 2

4Dt

dr

0

∞

⌠

⌡

The mean–square radial displacement, <R2>, is defined as the second moment of the radial probability density F(r,t)

R2 =

16Dtπ

z4 exp −z2( )d z0

∞

∫

z2 ≡

r 2

4Dt 2zdz =

r dr2Dt

Substituting for and for

© meg/aol ‘02

Diffusivity: A Microscopic Transport Property

exp −az2( )z2n d z

0

∞

∫ =1⋅3⋅5⋅...⋅ 2n −1( )

2n+1anπa

a > 0( )

Considering a =1, and n = 2, yields the following definite integral

z4 exp −z2( )dz

0

∞

∫ =3 π

8

This definite integral is evaluated as

R2 1

2= 6Dt

R2 = 6Dt

For the diffusion cloud in 3–D

© meg/aol ‘02

Einstein’s Formula

D =

16

nt

λ2

or 6Dt = nλ2

Einstein’s famous result is

D =

16

Γ3 λ2D expressed on the basis of purely microscopic quantities

Diffusivity ranges in different states of matter

Range: D [ cm2/s] Phase/State

10-6 10-25 10-4 10-7

1 10-2

Solids Liquids Gases

© meg/aol ‘02

Exercises 1. The diffusivity of bismuth atoms in lead at 600K is 1×10-9 cm2/sec. The radius of a lead atom, rPb, is 0.175 nm. Calculate the total distance traveled by the Bi atoms after one hour, and compare it to the expected RMS displacement for the same period.

a0 = 2 2 rPb = 0.495nm

The lattice parameter a0 of the FCC Pb host crystal is:

λ =

22

ao = 0.350nm

The jump distance λ of a substitutional Bi atom in FCC Pb is given as:

© meg/aol ‘02

Exercises

Γ =

6Dλ2 =

6× 1⋅10−9cm2/s( )(0.350nm)2

1014nm2

cm2 = 4.9⋅106s−1

Applying the Einstein’s relationship we obtain the jumping frequency

Rtot = nλ = Γ ⋅t( )λ = 4.9⋅106s−1 × 3600s( )× 0.35nm= 6.174m

The total distance Rtot traveled by Bi atoms is equivalent to the length of a directed walk

R212 = 6Dt = nλ,

R212 = 6×1⋅10−9cm 2/ s× 3600s = 4.65⋅10−3cm = 46.5µm

The expected RMS displacement of the Bi atoms can be found

© meg/aol ‘02

Exercises 2. Derive and graph the cumulative probability, f (r2,t), of finding diffusant at a distance r2 at various times released from an instantaneous point source located in two dimensions.

F r2 ,t( )=

2πre−

r 2

4Dt

4πDt

If the point has unit strength, M2 =1 , and releases diffusant at t =0, from the origin r =0, then the probability density function is

f r2 ,t( )=re

−r 2

4Dt

2Dtd r

0

r2

⌠

⌡

The cumulative probability is defined as the integral of the equation

© meg/aol ‘02

Exercises

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

Cum

ulat

ive

prob

abilit

y, f(

r 2,t)

Distance from point source, r2

Dt=0.1 Dt=0.5Dt=1

Dt=4

Dt=2

Dt=8

Dt=20

Distance from point source, r2

Cum

ulat

ive

prob

abili

ty, f

(r2,

t)

Dt = 0.1 Dt = 0.5

Dt = 1

Dt = 2

Dt = 4

Dt = 20

Dt = 8 f r2 ,t( )= 2 ze−z2

dz0

r2

2 Dt

∫

Using the variable substitution, z = r/2 (Dt)1/2,

f r2 ,t( )= 1 − e−

r22

4Dt

Evaluating the integral yields the desired result

© meg/aol ‘02

Exercises

3. Find the equivalent Einstein expression for the diffusivity in 2–D as was derived for 3–D.

R2

2 = r22F r2 ,t( )d r2

0

∞

∫

The mean–square displacement of a random walker in two dimensions could be found by calculating the second moment of the particle density

R22 = r2

2 r2 e−

r22

4Dt

2Dt

dr2

0

∞

⌠

⌡

The second moment of the distribution F(r2, t) may be expressed as

© meg/aol ‘02

Exercises

R2

2 = 8Dt ζ3e −ζ2

dζ0

∞

∫

The variable substitution ζ = (r2)/2(Dt)1/2 and dζ = dr2/2(Dt)1/2 allows the integral to be written as

The definite integral may be evaluated using integration by parts, as follows

ζ2e−ζ2

ζdζ0

∞

∫ = −12

ζ2e−ζ2

0∞ − (2ζe−ζ2

)dζ0

∞

∫

, ζ3e−ζ2

dζ0

∞

∫ =12

D =

14

nt

λ2 =14

Γ2 λ2

Substituting yields

Dd =

12d

Γd λ2or

© meg/aol ‘02

Key Points • Diffusive motion in gases, liquids, and solids occur by a series of

discrete steps, separated, respectively, by elastic collisions, localized vibrations, and short shuffles.

• Such motions may be treated by elementary random walk theory. • Continuum diffusion theory (Fick’s laws) may be connected to

random walk theory leading to Einstein’s microscopic description of the diffusion coefficient.

• Einstein showed that the diffusion coefficient is proportional to the frequency of the step motion and the square of the step length.

• Curiously, diffusion, which is an irreversible macroscopic process, actually is comprised of reversible microscopic steps.

• Irreversibility devolves from the large number of steps involved in macroscopic diffusion and the improbability of reversing these motions.

© meg/aol ‘02

Module 13: Structure and Diffusion

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline • Random walks in crystals

– Primitive (simple) cubic – Body–centered cubic

• Constraints imposed by the structure • Interpreting diffusivities in cubic crystals • Diffusion mechanisms

– Interstitial diffusion – Ring diffusion – Vacancy–assisted diffusion – Interstitialcy diffusion

• Diffusion in FCC crystals • Lattice vacancies • Divacancies • Exercise

© meg/aol ‘02

Random Walks: Primitive (Simple) Cubic Crystals

100[ ], 010[ ], 001[ ],

1 00[ ], 0 1 0[ ], 001 [ ]The individual crystallographic directions along which a diffuser can jump in SC are:

Z = 6 In primitive cubic crystals (SC) there exists one lattice site per unit cell surrounded by 6 neighbors:

Γtot = Γi

i =1

z

∑The total jumping frequency Γtot is related to the specific jumping frequencies for a given structure:

Three of the SC jump vectors, of the type a0u<100> are shown:

© meg/aol ‘02

Random Walk: Body-Centered Cubic Crystals

Z = 8 In body–centered cubic crystals (BCC) there exists one lattice site per unit cell surrounded by 8 nearest neighbors:

111[ ], 1 11[ ], 11 1[ ], 111 [ ],

1 1 1[ ], 11 1 [ ], 1 11 [ ], 1 1 1 [ ]The individual crystallographic directions along which a diffuser can jump in BCC are:

3a0 2( )u 111

Six of the BCC jump vectors, of the type are shown:

© meg/aol ‘02

Random Walks: Face-Centered Cubic Crystals

Z = 12 In face–centered cubic crystals (FCC) there exists one lattice site per unit cell surrounded by 12 nearest neighbors:

The individual crystallographic directions along which a diffuser can jump in FCC are:

110[ ], 101[ ], 011[ ], 1 10[ ],

11 0[ ], 01 1[ ], 01 1 [ ], 1 01[ ],

101 [ ], 1 0 1 [ ], 0 1 1 [ ], 1 1 0[ ]

© meg/aol ‘02

Random Walks: Face-Centered Cubic Crystals

Six of the FCC jump vectors, of the type are shown below: a0 2( )u 110

© meg/aol ‘02

Constraints Imposed by Crystal Structure

ni = Γi ⋅tThe average number of jumps, <ni > over a time interval, t, is:

The total displacement achieved, L, over a time interval, t , is:

L t( )= ni

λ i

i=1

Z

∑

The expected (mean) displacement of such of diffuser after a time, t, is found as :

L t() = ni ⋅

λ i

i =1

Z

∑

The mean displacement averaged over all the diffusers is:

L t() = t Γi ⋅λ i

i =1

Z

∑

© meg/aol ‘02


L t( ) = t ⋅

Γtot

Zλ i

i=1

Z

∑ = 0For the cubic system all the jumping frequencies and jump vectors are equivalent. Consequently, the equation simplifies to:

Ln

2 =λ i

i=1

n

∑ ⋅λ i + 2

λ i

j≠ i=1

n−i

∑ ⋅λ i+ j

i=1

n−1

∑The square displacement may be written as:

Ln2

2

λ i

j≠ i=1

n−i

∑ ⋅λ i+ j

i=1

n−1

∑ = 2λ2 cosθi,i+ jj≠ i=1

n−i

∑i=1

n−1

∑The sum of parallel and non-parallel scalar products may be written in terms of the sum of the cosines:

Ln

2 =λ i ⋅

λ i

i=1

n

∑i

+ 2λ i ⋅

λ j

j≠ i=1

n−i

∑i=1

n−1

∑An expression for the mean–square displacement may be developed:

© meg/aol ‘02


nk λk2

k =1

Z

∑ = nλ2

For the case of cubic crystals, Ln

2 = nk λk2

k =1

Z

∑ + 2 λ i λ j cosΘi ,i+ jj ≠i=1

n− i

∑i=1

n−1

∑

Substituting for the definition of the jumping frequency, we obtain:

Ln2 = nλ2 = Z Γi λ2t

Combining the previous equations:

2 λiλ j cosθi ,i+ jj≠ i=1

n−i

∑i=1

n−1

∑ = 0

Averaging over many independent random walks,

© meg/aol ‘02


If the jump vectors and coordination numbers for each cubic crystal structure are inserted back, then we obtain:

L

2= 6Γi a0

2( )tPrimitive cubic:

L

2= 8Γi

34

a02

tBody-centered cubic:

L

2= 12Γi

12

a02

tFace-centered cubic:

L

2= 6 Γi a0

2 t

In terms of specific jumping frequency, Γi, one formula can represent the mean–square displacement for the cubic structures given above:

© meg/aol ‘02

Interpreting Diffusivities in Cubic Crystals

The diffusivity in cubic structures may also be expressed in terms of the total jumping frequency, Γtot,

D =16

Γtot λ2

D = Γi a02Applying Einstein’s famous result, = 6Dt, yields

L

2

Computer simulation of ten “atoms” executing 100 Monte Carlo steps in SC, BCC, and FCC lattices.

0

2

4

6

8

10

12

0 2 4 6 8 10

FCCSCBCC

1/ 2nn1/2

RM

S di

spla

cem

ent

n1/2

RM

S di

spla

cem

ent

FCC

BCC

SC

© meg/aol ‘02

Interpreting Diffusivities in Cubic Crystals

11, 520 “atoms” executing 100 Monte Carlo steps in SC, BCC, and FCC lattices:

0

2

4

6

8

10

12

0 2 4 6 8 10 12

FCCSCBCC

RM

S di

spla

cem

ent

n1/2n 1/2

RM

S di

spla

cem

ent FCC

SC

BCC

© meg/aol ‘02

Diffusion Mechanisms

• Interstitial diffusion

• Ring diffusion

• Vacancy-assisted diffusion

• Interstitialcy diffusion

© meg/aol ‘02

Interstitial Diffusion: Schematic of Atom Movements

saddle-point planesaddle-point plane

saddle-point planesaddle-point plane

diha0a0 h

diffusive jump

© meg/aol ‘02

Diffusion Mechanisms

ε =

di − ha0

The in–plane local strain, ε, may be approximated as:

h = 1−

32

a0 ≅ 0.134a0

The unstrained interatomic spacing along <100> directions in BCC crystals may be found using the standard formula:

p = e−∆E ∗

kB T

The Boltzmann probability to acquire an energy, ∆E* >> kBT, is:

© meg/aol ‘02

Comparison of calculated saddle–point strain energies with experimentally determined interstitial diffusion activation energies for some BCC materials

Solvent Impurity Strain (ε)

Saddle-point Energy,

∆E* (kcal/mol)

Activation Energy

∆Eexp (kcal/mol)α-iron H

C

0.077

0.406

1.3

25

3.0

20.1tantalum C

N

0.338

0.298

42

35

38.5

33vanadium N

O

0.334

0.302

22

19

34

29niobium N

O

0.298

0.267

20

17

34.2

26.9

Solvent Impurity Strain, ε Saddle-point energy Activation Energy

α–iron

Tantalum

Vanadium

Niobium

© meg/aol ‘02

Ring Diffusion

A: direct exchange

B: cyclic exchange

AB

© meg/aol ‘02

Vacancy–Assisted Diffusion

W = 3−1( )Da = 0.73Da

The FCC lattice geometry requires

a0

δ=a0[110 ]

Da

1

2

3

45

Vacant

δ

(a)

Da

a0

Vacant

δ = a0 [110]

a0

3 4

1 2

Wa0

plan view

© meg/aol ‘02

Interstitialcy Diffusion

W

Dilated configuration of a diffusion jump “window” on the 110 plane in FCC

Interstitialcy diffusion showing the movement of a crowdion effect

crowdioncrowdion

© meg/aol ‘02

Diffusion in FCC Crystals

D =

16

Γtot λ2Random walk theory shows that in 3-D:

λ proj =

a0

2For FCC, the specific jump vector λi = (a0 /2) u<110>, so its projection is:

pv = ZNv = 12Nv

If the thermal vacancy is Nv , the probability of having a vacant nearest-neighbor site adjacent to any lattice atom in FCC is:

© meg/aol ‘02

Diffusion in FCC Crystals

DFCC =

16

Nv Γv λ2

Substituting for total jumping frequency

DFCC = (a02 / 24) Nv Γv

The expression for the diffusivity, DFCC , for vacancy-assisted diffusion in FCC crystals is

Γtot = pv

Γv

12

The total jumping frequency of the atom in FCC crystals is

© meg/aol ‘02

Lattice Vacancies

G = H0 − T S0 + nv ∆ H vf − nv T ∆S v

f − kB T ln W

The Gibbs free energy of a crystal containing monovacancies is:

W=

Nl !nv ! Nl − nv( )!

Combinatoric theory shows that the number of distinguishable ways of distributing nv vacancies on Nl lattice sites is:

ln z! ≅ z ln z− z

Stirling’s approximation for large arguments of the factorial is:

© meg/aol ‘02

Lattice Vacancies

ln W≅ Nl ln Nl − Nl − Nl − nv( ) ln Nl − nv( )The previous equation may be approximated as:

Cv

eq ≡nv

eq

Nl

The equilibrium concentration of lattice vacancies forming at a particular temperature is defined as:

Differentiating with respect to nv, gives after a few steps of algebra:

∂G∂nv

Nl

= ∆Hvf − T∆Sv

f − kBT ×

× −1− lnnv +Nl

Nl − nv( )−nv

Nl − nv( )+ ln Nl − nv( )

© meg/aol ‘02

Lattice Vacancies

∂G∂nv

Nl

eq

= 0 = ∆ Hvf − T ∆ Sv

f + kB T ln nv

Nl − nv

The minimum Gibbs function occurs when:

ln nv

Nl − nv

≅ ln nv

Nl

The last term may be approximated as:

ln nv

Nl

= −∆ Hv

f − T ∆Svf( )

kB T

Substituting gives:

© meg/aol ‘02

Lattice Vacancies

n veq

Nl

≡ Cveq = exp −

∆ Hvf − T ∆Sv

f( )kBT

Solving for the equilibrium lattice monovacancy concentration yields the result:

Cveq= e

− ∆ Gvf

kBT = e∆ Sv

f

kB e− ∆ Hv

f

kB T

or

Cv

eq T( )= Kv exp −∆Hv

f

kBT

The fractional concentration of monovacancies, at any temperature, may be estimated as:

© meg/aol ‘02

Lattice Vacancies

FCC ∆Hf (eV) BCC ∆Hf (eV) Alkali Halide ∆Hf (eV)Al 0.68 Mo 3.0 NaCl 2.02Ag 1.12 Nb 2.65 LiF 2.68Au 0.89 Ta 2.8 LiCl 2.12Cu 1.29 V 2.1 LiBr 1.80Fe 1.4 W 4.0 LiI 1.34Ni 1.78 Fe 1.6 KCl 2.3Pt 1.32 AgCl 1.4*Pd 1.85 AgBr 1.1*Pb 0.57 *Frenkel

Vacancy Formation Enthalpies in Solids

© meg/aol ‘02

Divacancies

p2 = 12⋅ CveqThe probability, p2 , for an FCC material, that one of the nearest

neighbors to this vacant site is a second vacancy

nv 2 = nv ⋅ p2 = nv (12⋅ Cveq)

The total number of divacancies, nv2 , present in the crystal

Cv 2

eq =12

⋅12 nv

Nl

2

= 6 Cveq T( )[ ]2

Dividing both sides by Nl and multiplying by 1/2

Cv 2eq T( )= 6e

− 2∆ Gvf

kB TThe temperature dependence of noninteracting divacancies

© meg/aol ‘02

Divacancies

v + v ⇔ 2v( )One must consider the additional mass-action equilibrium

kBT lnCv 2

eq

Cveq( )2

= −∆Gint

Suggesting

Cv 2eq = 6⋅ Cv

eq( )2e

−∆GintkBT

Applying the mass-action law to the

divacancy-pair orientation in FCC

Cv 2eq T( )= 6⋅ e

−2∆Gv

f +∆Gint( )kBT

The divacancy concentration arising from interacting lattice monovacancies

© meg/aol ‘02

Divacancies

D = D01e−

Q1

RT+ D02 e−

Q 2RT

The diffusivity may be expressed as the sum of independently activated processes.

10 -12

10 -11

10 -10

10 -9

10 -8

10 -7

0.0025 0.003 0.0035 0.004 0.0045 0.005

Sodium (self-diffusion)J. Mundy, Phys. Rev. B3, p. 2431, 1971

D (c

m2 /s

ec)

1/T (K-1)1/T (K-1)

D (c

m2 /s

ec)

© meg/aol ‘02

Exercise

1. a) Find the mean–square displacement in a diamond cubic (DC) crystal in terms of the lattice parameter, a0, and the specific jumping frequency, Γi. b) Show that the diffusivity expressed through the total jumping frequency, Γtot, agrees with eq.(13.19).

Ln2

= Z Γi λ2t = 4⋅Γi ⋅3

16a0

2

⋅ t

Ln2

=34

Γia02t

a. Si, and Ge (diamond cubic crystals) have Z = 4. The mean–square displacement is found:

© meg/aol ‘02

Exercise

L

2= 6Dt =

34

Γi a02t

Using Einstein’ s relationship

D =

18

Γia02

Solving for the diffusivity of atoms in a DC crystal

D =

18

⋅Γtot

4⋅

163

λ2

=16

Γtotλ2

b. When expressed in terms of Γtot= Z Γi the diffusivity is

© meg/aol ‘02

Key Points • The internal (crystal) structure of a material determines the

distances and directions that diffusing atoms jump. The specific jumping frequency, Γi, uniquely determines the dynamics.

• The Einstein equation holds: D=(1/6)Γtotλ2, where Γtot=Z Γi . • Diffusion in solids occurs by several well-known mechanisms:

– Interstitial diffusion – Vacancy-assisted diffusion – Ring mechanism – Interstitialcy (crowdion) mechanism

• Monovacancies and divacancies form equilibrium populations. Quasi-chemical theory and statistical mechanics can be used to treat these populations.

• Diffusivities depend on the Arrhenius behavior of point defects, leading to exponential increases with temperature.

© meg/aol ‘02

Module 14: Correlation Effects in Diffusion

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Markov sequences and correlated walks

– Jumping probability estimate

• Estimates of the correlation factor: pure materials

• Computation of correlation factors in pure crystals

• Exercises

© meg/aol ‘02

Markov Sequences and Correlated Walks

L2corr

= f ⋅nλ2

Reduction in mean-square displacement for non-Markov random walks may be introduced through an exchange efficiency or correlation factor.

f ≡Dcorr

*

Drand* = lim

n→∞

L2corr

nλ2

For self–diffusion in a pure crystal

Ln2

= 1+2n

cosΘ i ,i + jj =1

n−1

∑i =1

n− i

∑

⋅

f

nλ2

The square displacement of an n-step random walk can be described as

f = limn→∞

1+2n

cosΘi ,i+ jj=1

n−1

∑i=1

n−i

∑

Substituting and comparing the result

© meg/aol ‘02

Jumping Probability Estimate

Correlated exchanges between “tagged” tracer atom and a lattice vacancy on an adjacent crystal plane

© meg/aol ‘02


vR = CVΓ PRjPL

j−1

j =1

∞

∑net motion

−

PRPL( )j

j=1

∞

∑no net motion

vR = CVΓ ⋅ PR − PRPL + PR PLPR − PR2PL

2 + ...( )

The effective frequency, νR, of a tracer atom jumping to the right is

The alternating series of conditional exchange probabilities can be gathered as two summations

© meg/aol ‘02


vR = CVΓ 1−PRPL( )j

j=1

∞

∑

PRjPL

j−1

j =1

∞

∑

PRjPL

j −1

j =1

∞

∑ = CVΓG ⋅ f

The correlation factor, f, may be reintroduced

f ∝ 1−PR PL( )j

j =1

∞

∑

PRjPL

j −1

j =1

∞

∑

PRj PL

j −1

j=1

∞

∑

where

Note: G has been termed by Manning as the vacancy-wind factor

© meg/aol ‘02


vR = CVΓPR PL( )j

PLj =1

∞

∑ − PR PL( )j

j=1

∞

∑

The two series can be combined

vR = CVΓ1PL

−1

PRPL( )j

j=1

∞

∑so

PRPL( )j

j =1

∞

∑ =1

1− PR PL

−1The infinite series may be summed

vR = CVΓ1PL

−1

11− PRPL

−1

Combining yields

© meg/aol ‘02


PR =Γ

Γ + Γ'

The probability, PR, can be expressed

vR = CVΓ1PL

−1

11− PRPL

−1

1− PR

1− PR

Exchange probability calculation, multiply, and divide the right-hand side by 1- PR

vR = CVΓ1− PL( ) 1− PR( )

1− PRPL

PR PL

PL 1− PR( )

so

and hence vR = CVΓPR

1− PR( )⋅

1− PL( )1− PR( )1− PRPL( )

© meg/aol ‘02


PR

1− PR( )=

ΓΓ′

In view of the probability factor

vR = CVΓΓ′

Γ

1− PL( ) 1− PR( )1− PR PL( )

Upon substitution

vR = CVGΓ1− PL( )1− PR( )

1− PRPL( )

The expression becomes

© meg/aol ‘02


PL = PR ≅1Z

Introducing the coordination number of the crystal lattice, the jump probabilities are

vR = CVGΓ1− 1

Z( )2

1− 1Z( )2

The equation could be written in terms of Z

vR ≅ CVGΓ 1−2Z

+ ...

For any 3–D lattice, 1/Z << 1, so retaining only the first–order term

f ≅ 1−2Z

Comparison shows that f may be approximated

vR ≅ CVGΓ ⋅ fthus

© meg/aol ‘02

Correlation for Self-diffusion in Pure Crystals

Diffusion mechanism Lattice or structure Correlation factor, f

Vacancy assisted square 2-d 0.46699" hexagonal 2-d 0.56006" diamond cubic 0.5 (exact)" simple cubic 0.65311" BCC 0.72722" FCC 0.78146" HCP (a-axis) 0.78146" HCP (c-axis) 0.78121

Interstitialcy BCC 0.8 (exact)Colinear jumps NaCl 0.66666

Noncolinear , cosθ=±1/3 NaCl 0.96

© meg/aol ‘02

Estimates of f (Compaan &Haven)

Lattice Z 1–2/Z f FCC

BCC SC

DC

12

8 6

4

0.833

0.750

0.667

0.500

0.781

0.727

0.653

0.500

© meg/aol ‘02

Exercises 1. Explain why interstitial diffusion in crystals is usually not a correlated process.

• When interstitial diffusion involves the motion of a dilute population of small atoms, each interstitial atom has a uniform, high probability of being surrounded by vacant interstitial sites. Every interstitial jump is independent of the atom’s prior jumps, because almost all the surrounding interstitial sites are vacant. The random walk would be Markovian. • In a few systems, concentrated interstitial solid solutions do form. Were this to occur, the average interstitial atom experiences a reduced number of available, adjacent interstitial sites. The probability of such an interstitial exchanging sites depends in detail on the arrangement of its available neighboring sites. Its jump probability, moreover, would be reduced from the random value, and depends on the jumping history of the adjacent interstitial atoms.

© meg/aol ‘02

Exercises 2. Tracer diffusivity methods are used to measure self–diffusion coefficients in pure materials. Why does a tracer diffusivity experiment not give the true value of the self–diffusion coefficient in a pure material? Tracer diffusion involves three participating elements: (tracer atoms, vacancies, and host

atoms). The tracer motions, being correlated, disperse these atoms less efficiently than a random walk. Self–diffusion, per se, involves just lattice vacancies exchanging with host atoms. Only two participating elements comprise this process. The diffusion mixing of host atoms in a crystal containing lattice vacancies is equivalent to the mixing of the vacancies themselves. No meaningful distinction can be drawn between the motion of the vacancies and the counter motion of the host atoms with which the vacancies exchange. Self–diffusion is not experimentally detectable, unless a tracer is used to provide a

measurable entity that approximates the exchange between host atoms and lattice vacancies. The tracer diffusivity is an approximation of the true self–diffusion coefficient. Self–diffusion is an uncorrelated process, whereas tracer diffusion is not. As explained, these diffusivities differ slightly from correlation effects, and also from the atomic mass difference and related exchange frequency differences that the isotope atoms have compared to the host atoms. The latter effect, known as the isotope effect, will be discussed in detail later.

© meg/aol ‘02

Key Points

• Random walk theory is based on Markov processes: independent steps that lack “memory.”

• Memory effects reduce the efficiency of random walks. Such effects are termed correlation.

• Generally, three diffusing “elements” are required for correlation effects to be significant.

• The correlation factor represents the efficiency of non-Markovian random walks associated with diffusion. For defect-assisted self-diffusion, the correlation factor depends on the crystal structure.

• Interstitial diffusion in dilute systems is not correlated. • Correlation factors for self-diffusion in pure materials are well

known, and typically may be approximated as f=1-2/Z.

© meg/aol ‘02

Module 15: Vacancy–Assisted Diffusion

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Dynamical Theory: Flynn’s model • Quasi–Chemical Theory • Empirical Relationships • Correlation with diffuser size • Arrhenius Correlation • Dependence of D on composition • LeClaire’s Correlation • Empirical Observations • Exercises

© meg/aol ‘02

Flynn’s model

Γν =

35

12

νDebyee−

cij Ω ∆2

kBT

Flynn’s showed that a Debye model of lattice vibrations may be expressed

cij =

15µ λ + 2µ( )2 2λ + 7µ( )

For an isotropic FCC crystal this elastic constant may be approximated

1c

≅2

153

c11

+2

c11 − c12

+1

c44

For real FCC crystals–elastically anisotropic–we use single–crystal elastic coefficients, cij

© meg/aol ‘02

Flynn’s model

0 Sq< 1 1 0 >

E(x )

x

( ¦ 2 ) a0 / 2

Lattice potential and displacements according to Flynn’s model

∆=q/S

λ

© meg/aol ‘02

Flynn’s model

FCC metal Emig (calc) eV Emig (exp) eV

Cu 0.84 0.80 - 1.1Ag 0.83 0.83 - 0.88Au 0.82 0.68 - 0.87Ni 1.42 1.25 - 1.50Al 0.83 0.40 - 0.65Pb 0.48 0.56

Flynn’s vacancy migration theory for FCC metals

© meg/aol ‘02

Quasi-Chemical Theory The reaction path as a complex jumps from one equilbrium site to another equilbrium site

X0’s represents a pair of adjacent equilibrium sites X* represents the saddle–plane

X *

∆G vm

∆(width)

X * X 0 X 0 X0 X0 X*

© meg/aol ‘02

E-p Phase–Space for Quasi–Chemical Theory

X0 X0X*

p+dp

p

p

Position

Mom

entu

mSaddle PlaneEq. Site Eq. Site

X*+d X=X*+( p/m )d t

© meg/aol ‘02

Distribution Function for E & p

E X0 − X, p( )= const×e− Etot

kBT = ζe−

12

mv 2+κE2

X0 − X( )2

kBT

The distribution of (positional) potential energy and (momentum) kinetic energy is

E X0 − X, p( )d X d p

− X∗

X ∗

∫−∞

∞

∫ = 2 X0 − X ∗( )N∗

Integrating Boltzmann distribution

ζe

−

12

mv 2+ κE2

X0− X( )2

kBT d X d p− X ∗

+ X ∗

∫−∞

∞

∫ = 2 X0 − X ∗( )N ∗

Substituting gives

distribution function in a 2–D phase-space

© meg/aol ‘02


12

mυ2 =p2

2mThe kinetic energy may be expressed in terms of momentum, p

ζ e

−κE X0− X( )2

2kBT d X e− p2

2mkBT d p−∞

∞

∫− X∗

X ∗

∫ = 2 X∗ − X0( )N∗Substituting back yields

ζ 2mπkBT

2πkBTκ E

= 2 X∗ − X0( )N∗Evaluating each integral

ζ =

1π

κ E

mX∗ − X0( )N∗

kBTSimplifying

© meg/aol ‘02

E X0 − X, p( )=

1π

κ E

mX∗ − X0( )N∗

kBTe

−

p2

2m+

κ E2

X0 − X( )2

kBT


The distribution function

d2 N∗ p( )=

1π

κ E

mX ∗ − X0( )N∗

kBTe

−

p2

2m+

κE2

X0− X∗( )2

kBT d X ⋅d pThe number of activated complexes

d2 N∗ p( )=

1π

κ E

mX ∗ − X0( )N∗

kBTe

−

p2

2m+

κE2

X0− X∗( )2

kBT vdt ⋅d p

Substituting for dX =υd t

© meg/aol ‘02


d2 N∗ p( )=

1π

κ E

mX ∗ − X0( )N∗

kBTe

−

p2

2m+

κE2

X0− X∗( )2

kBT pm

d p⋅dt

Introducing υ = p/m, yields

d d N∗ p( )( )dt

=1π

κ E

mX∗ − X0( )N∗

kBTe

−

p2

2m+

κ E2

X0− X ∗( )2

kBT pm

d p

The rate of passage through the saddle–point position is

d N∗ vdt

=1π

κ E

mX∗ − X0( )N∗

mkBTe

−κE X0− X∗( )2

2kBT e− p2

2mkBT pd p0

∞

⌠ ⌡

The rate of passage at X = X *

= mkBT

© meg/aol ‘02

d N∗ vdt

=1π

κ E

mX∗ − X0( )N∗e

−κE X0− X∗( )2

2kBT


Γv = νE e−

Emig

kBTThe rate of reaction for the activated complexes is

d N∗ v

dt= 2νE X ∗ − X0( )N∗e

−κ E X0 − X ∗( )2

2kBT

Replacing for Einstein oscillatory frequency

υE =

12

κE

m

d f ∗

dt= νE e

−Emig

kBTDividing both sides by 2(X *–X0)N*

Inserting for mkBT yields

© meg/aol ‘02

Self–Diffusion in Crystals

Cveq = e

− ∆Gvf

kBT = e∆Sv

f

kB ⋅e−

∆Hvf

kTThe fractional concentration of lattice vacancies at thermal equilibrium is

DFCC = a0

2νE exp −∆Emig

kBT

⋅exp ∆Sv

f

kB

⋅exp −

∆Hvf

kBT

Combining the expression for self-diffusion with the vacancy exchange frequency

DFCC = a0

2νE ⋅exp∆Sv

f + ∆Smig

kB

⋅exp −

∆Hvf + ∆Hmig

kBT

Expressed as enthalpy and entropy terms

∆Smig ≅ kB ln Πi νi

o

Πi νi*

The entropy change for atom-vacancy migration

© meg/aol ‘02

Empirical Methods in Diffusion

© meg/aol ‘02

Activation Energy for Self-Diffusion

∆E* = H * − ∆HsThe activation energy for self-diffusion, ∆E*

∆E* ≈ β ⋅∆HsAn empirical rule developed to estimate, ∆E*

∆E* = 32⋅Tm [cal/mole]The rule states that

∆Sf =

∆H f

Tm

≅ 2.3 [cal/mole⋅ K]An alternative energy scale

∆E* = η⋅∆H f [cal/moleSubstituting in Bugakov–van Limpt rule

∆E* ≈18⋅ RgTm [cal/mole]Relating ∆E* with Tm

D = e−17(Tm /T )Flynn’s law of corresponding states:

© meg/aol ‘02

Activation Energy for Self-Diffusion

∆Hs

H*

∆Ε

φx 1 x 2 x 3 x 4 x 5

a0

Periodic lattice potential versus distance in a crystal

© meg/aol ‘02

Corresponding States

D = D0 e−Q

RgTThe Arrhenius correlation

Diffuser Radius (Å) BCC(800C)

FCC(1100C)H 0.46 2.7×10

-41.9×10

-4

C 0.77 1.7×10-6

6.7×10-7

N 0.71 7.3×10-7

3.8×10-7

B 0.97 –– 6.1×10-7

Influence of diffuser atom size on diffusion in Fe

Diffuser Radius (Å) BCC(800C)

FCC(1100C)O 0.60 –– 1×10

-9

Fe* 1.28 BCC 3×10-12

9×10-12

Co 1.26 FCC 1.5×10-12

2.5×10-12

Ni 1.25 FCC –– 8×10-12

Mn 1.18 α-cubic –– 2×10-11

Mo - 7×10-12

4×10-12

Substitutional (vacancy/lattice) diffusion, D Correlation with diffuser size

© meg/aol ‘02

Arrhenius Correlation

Element Tm

[K]T-range

[K]D0

[cm2/sec]Q

[kcal/mol]Q/Tm

[cal/mol-K]Al 933 573/923 2.25 34.5 37.0Cu 1357 971/1334 0.78 50.4 37.1Ag 1235 913/1228 0.67 45.2 36.6Au 1337 977/1321 0.09 41.7 31.2Pb 600 473/596 0.99 25.6 42.7Co 1768 1393/1643 0.23 64.0 36.2Ni 1726 1143/1677 1.27 67.2 38.9Fe 1811 1443/1634 0.49 67.9 37.5Pt 2045 1598/1873 0.33 68.1 33.2Ta 3250 2150/3200 0.12 98.7 30.4

Correlation among D0, Q, Tm, for self-diffusion

© meg/aol ‘02

Bugakov–van Limpt correlations

35 Tm

(Kcal/mol)

X

∆Eexp

0 20 40 60 80 100 120 1400

20

40

60

80

100

120

140

15.2 ∆Hf

y=15

.2∆

Hf

y=3

5Tm

© meg/aol ‘02

LeClaire’s Correlation

Concentration dependencies of D for isomorphous systems (Birchenall)

© meg/aol ‘02


Correlation of D and Q with composition in γ–Fe-C and Fe–Mn alloys (Birchenall)

© meg/aol ‘02


Correlation of D and Q with composition in Fe–Ni and Cu–Zn alloys (Birchenall)

© meg/aol ‘02

Exercises

15.1 Plot the activation energies for the pure metals versus their melting points, as listed by the Arrhenius correlations shown in Table 15.3. Show the correspondence of a linear regression of these data to the larger survey, eq.(15.16), by Bokshtein and Zhukhovitskii.

The plot shows Q versus Tm for each of the pure elements listed in Table 15.3. The dashed line shown represents the linear regression of all the data. The Bokshtein and Zhukhovitskii diffusion data survey is indicated on this plot by the gray triangle with the slope 18×Rg, where Rg=1.987 cal/mol-K. is the universal gas constant. As indicated in this graph, the Bokshtein and Zhukhovitskii data correlates well with this linear regression. Note, however, that several of the elements (Au, Ni and Pt) have activation energies for self–diffusion that are well off the average correlation. Nonetheless, this exercise suggests that empirical correlations based on the criterion of lattice stability are usually helpful in estimating diffusivities, especially if the activation energy is not known.

© meg/aol ‘02

Exercise

20

30

40

50

60

70

500 1000 1500 2000 2500

Act

ivat

ion

Ene

rgy,

Q [

kcal

/mol

]

Tm

[K]

Pb

Ag

Au

Cu

NiCo

Fe Pt

18Rg

Linear regression

Bokshtein & Zhukhovitskii

Al

Activation energy for self-diffusion versus melting point for various metals (Bokshtein et al.)

© meg/aol ‘02

Exercises 15.2 Demonstrate that a comparison of Flynn’s law of corresponding states, eq.(15.17), with the standard Arrhenius correlation, eq.(15.18), leads to a relationship for the activation energy, Q, that is consistent with the Bugakov–van Limpt rule. Taking logarithms of eqs.(15.31) and (15.32) and equating the right–hand sides yields

−17 Tm

T

= ln D0 −Q

RgT

Solve eq.(15.33) for Q provides the empirical relationship that

Q = RgTm 17+ ln D0( )

Arrhenius Flynn

© meg/aol ‘02

Exercises

Choosing units for Q of [cal/mol-K], so that Rg=1.987 [cal/mol-K], yields conditions on the Arrhenius parameters:

Q ≅ 34Tm,

D0 ≈1.

The first relation is in excellent agreement with the Bugakov–van Limpt rule. This result shows that the law of corresponding states is consistent with the survey data for self–diffusion. The second condition, that the frequency factor, D0, for self–diffusion at elevated temperatures is of the order of unity, also is in agreement with the data for metals for which the average D0≈0.7.

© meg/aol ‘02

Key Points

• Vacancy-assisted diffusion is modeled by estimating the exchange frequency between host atoms and monovacancies.

• Three theoretical approaches are used to estimate exchange rates: – Dynamical theories: e.g., Flynn’s model for FCC materials – Quasi-chemical theory: Statistical mechanics, Monte Carlo simulations – Molecular dynamics: Standard MD, hyperdynamics, parallel replicas

• In addition, several empirical estimates prove useful: – Lattice stability – Arrhenius correlation – LeClaire’s correlation for alloys – Empirical data correlations

© meg/aol ‘02

Module 16: Diffusion in Dilute Alloys

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Five–frequency exchange model

• Isotope effect

• Diffusion of H+ and D+ with trapping

• Saturable diffusion traps

• Irreversible traps

• Exercises

© meg/aol ‘02

Five-Frequency Exchange Model

Γ0

Γ1

Γ1

Γ2

Γ3

Γ3

Γ3

Γ4

Solute–vacancy exchange model for diffusion in dilute FCC alloys

Γ1 is the rate of non–dissociating solvent atom–vacancy

Γ3 is the rate of dissociating solvent atom–vacancy exchanges

Γ0 is the rate of normal solvent atom–vacancy

Γ4 is the rate of associating solvent atom–vacancy exchanges

Γ2 is the rate of solute atom–vacancy exchange

© meg/aol ‘02


DS = a02 f Γ2 pv

The diffusion coefficient for the motion of vacancy-assisted solute atoms may be expressed by

pv = Cv exp −∆Ev

*

kBT

The probability pv of finding a vacant lattice site adjacent to a solute atom is

where Cv is the fractional concentration of monovacancies in thermal equilibrium with the lattice.

p2 =

Γ2

4Γ1 + Γ2 + 7Γ3

The model predicts for a dilute binary FCC crystals

Binding energy Monovacancy concentration

© meg/aol ‘02


f ≅

2Γ1 + 7Γ3

2Γ1 + 2Γ2 + 7Γ3

The correlation factor may be approximated from Lidiard’s model

Note: The exchange frequency, Γ0, is not included

Γ3 << Γ1,

Γ3 << Γ2.

If strong binding exists between the solute atom and the vacancy, the following approximations hold

f ≅

Γ2

Γ1

+1

−1

The correlation factor for diffusion in an FCC lattice is found to be

© meg/aol ‘02

Five-frequency Exchange Model

DS = a02 Γ1

Γ1

Γ2

+1

pv ≈ a02Γ1 pV

If the solute atom has a favorable valency,

DS ≅ a0

2 Γ1

Γ1 + Γ2

Γ2 pvAn estimate for diffusivity becomes

DS =

a02

2Γ1 + 7Γ3( )2Γ1 + 2Γ2 + 7Γ3( )Γ2 pV

If the dissociating jumps are permitted

DS =Γ1 +

72

Γ3

Γ1

Γ2

+1+72

Γ3

Γ2

a02pV

or

© meg/aol ‘02


Γ2 >> Γ1,

Γ2 >> Γ3.

If the solute atom–vacancy exchange rate is higher

DS ≅ a0

2 Γ1 +72

Γ3

pVSubstituting for these inequalities

DS ≅ a02Γ2 pVIf we assume that the atom-vacancy exchange

frequency is small

Γ1 ≈ Γ2 ≈ Γ3If the solute atom is an isoelectronic “impurity”

DS ≅

911

a02 Γ pVThen,

© meg/aol ‘02


pV ≅ CvThe probability of a lattice vacancy exchanging with a solute atom

DS ≅

911

a02ΓCV = 0.818DselfSubstituting for the probability approximation

D* ≅ 0.781a0

2ΓCVThe tracer diffusivity is given by

© meg/aol ‘02

Isotope Effect

fα =

2Γ1 + 7Γ3

2Γ1 + 2Γα + 7Γ3

The correlation factor according to five-frequency exchange model for Ag crystals

fβ =

2Γ1 + 7Γ3

2Γ1 + 2Γβ + 7Γ3

and

fα 2Γ1 + 2Γα + 7Γ3( )= fβ 2Γ1 + 2Γβ + 7Γ3( )Cross-multiplying and setting their left–hand sides equal gives

fα

fβ

=Γβ + Γ1 +

72

Γ3

Γα + Γ1 +72

Γ3

or

© meg/aol ‘02

Isotope Effect

Γ1 +

72

Γ3 =fα Γα

1− fα

A few steps of algebra show that

fβ = fα

Γα +fα Γα

1− fα

Γβ +fα Γα

1− fα

Solving for fβ

fα Γα +

fα Γα

1− fα

= fβ Γβ +

fα Γα

1− fα

Recombining

fβ =

Γβ

Γα

1− fα( )fα

+1

−1

or

© meg/aol ‘02

Isotope Effect

Dα = a0

2 fαΓα exp ∆S*

kB

exp −∆H *

kBT

The diffusivities of two different solute atoms moving within an FCC host lattice

Dβ = a0

2 fβΓβ exp ∆S*

kB

exp −∆H *

kBT

and

Dα

Dβ

=fα Γα

fβ Γβ

Dividing and canceling the common exponential terms

Γα

Γβ

=mβ

mα

Harmonic oscillatory theory implies that

where mα and mβ are the isotopic nuclear masses.

© meg/aol ‘02

Isotope Effect

Dα

Dβ

=fα

fβ

mβ

mαHence the isotopic diffusivity is given by

fβ =

1− fα

fα

mα

mβ

+1

−1

The isotopic correlation coefficients can be related

fα

fβ

=mα

mβ

+ fα 1−mα

mβ

Rearranging as

Dα

Dβ

= 1+ fα

mβ

mα

−1

yields the isotope effect (Schöen)

© meg/aol ‘02

Diffusion of H+ and D+ with Trapping

H+

H2H2

H+

Lattice saturated with H+

H2 trapped in a 3-D void (unsaturable trap)

© meg/aol ‘02

Diffusion of H+ and D+ with Trapping

The dissociation reaction of H2 is H2 (trapped) ⇔ 2H+ (lattice) + 2e-

Hydrogen and deuterium behave as perfect interstitial diffusers for the unsaturable traps.

K T( )=

H +[ ]2H2[ ]

The law of mass–action states that

K(T) =

CH +( )2

CH2

The ratio of the dissolved lattice hydrogen to molecular hydrogen is

© meg/aol ‘02

Fick’s Law with Trapping

∂Ctot

∂t= DH + ∇2CH +

In a material undergoing diffusion of hydrogen, Fick’s second law is

Note: Total hydrogen (Ctot) = trapped H2 + untrapped H+

Mobile hydrogen = H+

∂∂t

CH+ + CH2( )= DH + ∇2CH +Including each form of hydrogen

∂CH +

∂t

+

∂CH2

∂CH+

∂CH +

∂t

= DH+ ∇2CH +

Carrying out the differentiation and using the chain rule

© meg/aol ‘02


∂CH +

∂t

=

DH +

1+∂CH2

∂CH +

∇2CH +Solving for the time rate of change

∂CH +

∂t

= Deff∇

2CH +Fick’s second law in terms of effective diffusion

Deff ≡DH +

1+∂CH2

∂CH +

The effective diffusion coefficient

∂CH2

∂CH +

=

2CH +

K T( )Differentiation of the trapped hydrogen concentration with respect to the lattice concentration

© meg/aol ‘02


Deff ≡DH +

1+2CH +

K T( )Relating the effective diffusivity to K(T)

Deff =DH +

1+ 2CH2

CH +

Replacing K(T)

Deff =

DH +

2CH +

CH2

When CH2 /C H+ >>1

© meg/aol ‘02

Saturable Diffusion Traps

Schematic of H+ Trapped

α β

At an internal interface

H+

At the core of an edge dislocation

© meg/aol ‘02

Irreversible Traps

Tmax ≈

∆H H2

kB

Irreversible trapping of H+ at dangling bonds within a random network structure (amorphous silicon)

where

∆HH2 is the binding energy

Tmax is the temperature

kB is the Boltzmann constant

© meg/aol ‘02

Exercises 1. Estimate the ratios of the solute diffusion coefficients for a dilute FCC alloy using Lidiard’s 5–frequency model for the case of tight–binding of the vacancy with a solute that has isoelectronic interactions. Apply each of the following additional assumptions: a) All vacancy jumps are non–dissociating, and the vacancy exchanges with solute or solvent atoms occur with equal probability—i.e., with equal specific jumping frequencies. b) Dissociating vacancy exchanges occur with negligible frequency, but the specific jumping frequency between the vacancy and solute, Γ2, is four times larger than the specific jumping frequency, Γ1, for non–dissociating vacancy–solvent exchanges.

© meg/aol ‘02

Exercises

Γ1=Γ2=Γi Γ1/Γ2=0.25

Γ0

Γ1

Γ1

Γ2

Γ3

Γ3

Γ3

Γ4

Γ1 is the rate of non–dissociating solvent atom–vacancy

Γ3 is the rate of dissociating solvent atom–vacancy exchanges

Γ0 is the rate of normal solvent atom–vacancy

Γ4 is the rate of associating solvent atom–vacancy exchanges

Γ2 is the rate of solute atom–vacancy exchange

© meg/aol ‘02

Exercises

f =

14

+1

−1

=45

a) The correlation factor is

Ds

Diso

=f

fiso

=4 5( )9 11( )= 0.98The ratio of solute diffusivities

f =

Γ2

4Γ1

+1

−1

=4Γ1

4Γ1

+1

−1

= 0.5b) For Γ1/Γ2=0.25, the correlation factor is

Ds

Diso

=f

fiso

=1/2( )9 11( )= 0.61

The ratio of the solute diffusivities becomes

© meg/aol ‘02

Exercises

2. Determine the time needed to vacuum degas a cylindrical steel billet at 725C containing trapped hydrogen to the same residual levels as was determined in Exercise 7.7.2. The trapped hydrogen concentration, 2500 ppm, is equal to that of the initial dissolved lattice concentration, for total of 5000 ppm . The lattice diffusivity of H atoms in steel at 725C is DH+=2.25×10-4 cm2/sec.

© meg/aol ‘02

Exercises

Deff =DH +

1+ 2CH2

CH +

=2.25×10−4

1+ 2 25002500

cm2

s

,

Deff = 7.5×10−5 cm2

s

.

The effective diffusivity must be found

DtR2 =

7.5×10−5( )⋅ t(5)2 = 0.78The annealing time is

t = 2.60×105 s= 3daysor

© meg/aol ‘02

• Diffusion in dilute alloys may be treated with Lidiard’s 5 frequency model.

• Several cases are treated, depending on the relative exchange frequency of vacancies with solute and solvent atoms.

• Diffusivity depends on the slowest exchange frequency. • Schöen’s “isotope effect” is explained on the basis of the

diffusing atom mass. • Diffusion of hydrogen is important in many applications

involving corrosion, electroplating, etc. • Hydrogen may become trapped in voids, at interfaces, and at

defects like dislocation cores. Fick’s law must be modified: – Saturable traps – Unsaturable traps – Irreversible traps

Key Points

© meg/aol ‘02

Module 17: Kirkendall Effect

DIFFUSION IN SOLIDS



USA

© meg/aol ‘02

Outline

• Nonreciprocal Diffusion • Kirkendall Effect

– Kirkendall’s Diffusion Couple – Kirkendall Experiment

• Significance of Kirkendall Marker Motion • Nonreciprocal Atomic Fluxes • Intrinsic Diffusivities: Vacancy Wind • Diffusion–Advection

– Mass Conservation with Diffusion–Advection – Steady–State Diffusion–Advection

• Darken’s analysis of the Kirkendall Effect • Strain Effects • Exercise

© meg/aol ‘02

Nonreciprocal Diffusion

Par t ic les

st eel

Scale

Fe

Scale

(a) (b) (c)

O2

scal e

Pfeil’s (1929) observations of the diffusion-controlled overgrowth of an oxide scale

Par t ic les

st eel

Scale

Fe

Scale

(a) (b) (c)

O2

scal e

Par t ic les

st eel

Scale

Fe

Scale

(a) (b) (c)

O2

scal e

© meg/aol ‘02

Kirkendall’s Diffusion Couple

70-30 Brass

2.5mm Mo wires0.13mm diameter

Cu

h(t)

Schematic of the Smigelskas–Kirkendall diffusion couple

© meg/aol ‘02

Kirkendall Experiment

h t()− h 0( )= −Kt12

It was demonstrated that the distance between the inert markers h behaves according to the following kinetic expression

where

h(t) is the marker separation at the annealing time, t

h(0) is the initial separation of the markers

K is the rate constant in µm/(day)1/2

© meg/aol ‘02

Kirkendall Experiment

56 days

0 -0.10

8

16

24

32

Distance from interface [cm]

% Zn

-0.18

wt%

Zn

C0

6 days

wt%

Zn

Zn penetration curves observed for various annealing times Kirkendall–Smigelskas experiment

© meg/aol ‘02

Kirkendall Experiment Kirkendall–Smigelskas experiment showing inert Mo

wire–marker positions versus time

© meg/aol ‘02

Significance of Kirkendall Marker Motion

x marker frame

Um

marker

observer X-scale

crystal lattice

Marker motions observed during Kirkendall Diffusion

© meg/aol ‘02

Nonreciprocal Atomic Fluxes

JB = −DB∂CB

∂x (t = const.)

JA = −DA∂CA

∂x (t = const.)

The nonreciprocal atomic fluxes of species A and B written in terms of the intrinsic diffusion coefficients Di (i =A, B), using Fick’s first law:

Note: The net atom flux in the Kirkendall couple is no longer zero, because of nonreciprocal diffusion implied by the fact that DA≠DB.

© meg/aol ‘02

Control volume for Kirkendall mass advection Um

Jnet Ω∆t

unit area

marker plane

© meg/aol ‘02


uM = −Jnet Ω mol

cm2 ⋅s×

cm3

mol=

cms

The marker speed may be related to the net atomic flux as

Jnet = JA + JB = −DA

∂CA

∂x− DB

∂CB

∂x

The net atomic flux is the algebraic sum of the intrinsic Fickian fluxes specified

CA + CB ≡ Ω−1If the two components have equal partial molar densities then

uM = DA

∂CA

∂x+ DB

∂CB

∂x

ΩSubstituting yields the relation

ΩCA + ΩCB =1or,

© meg/aol ‘02


Ω

∂CA

∂x

t

= −Ω∂CB

∂x

t

Differentiating provides

ΩCA = NA

ΩCB = NB

Defining NA and NB,

∂NA

∂x= −

∂NB

∂xSubstituting gives the relationship

uM = DA − DB( )∂NA

∂xCombining we obtain

uM = DB − DA( )∂NB

∂xor

Implies that gradients are equal and opposite

© meg/aol ‘02

Intrinsic Diffusivities: Vacancy Wind

JV + JA + JB = 0The conservation of lattice sites requires that

JV = − JA + JB( )or

JV = DA

∂CA

∂x+ DB

∂CB

∂xSubstituting for JA and JB

JV = DB − DA( )∂CB

∂xThe concentration gradients are equal and opposite

© meg/aol ‘02

Atomic fluxes during compensated diffusion DA = DB

-40

-20

0

20

40

-4 -2 0 2 4

DA/D

B=1

Flux

(JA, J

B, JV),

dC V

/dt

Distance, x

JB

JA

JV=0 (dC

V/dt)=0

Vacancies at equilibrium

© meg/aol ‘02

-40

-30

-20

-10

0

10

20

30

40

-4 -2 0 2 4

DA/D

B=1.1

Flux

(JA, J

B, JV),

dC V

/dt

Distance, x

JA

JB

JV

(dCV/dt)

Vacancy creation Vacancy annihilation

Atomic fluxes during uncompensated diffusion DA = 1.1 × DB

© meg/aol ‘02

-40

-20

0

20

40

-4 -2 0 2 4

DA/D

B=4/3

Flux

(JA, J

B, JV),

dC V

/dt

Distance, x

JA

JB

JV

(dCV/dt)

Vacancy annihilation Vacancy creation


© meg/aol ‘02

-40

-20

0

20

40

60

-4 -2 0 2 4

DA/D

B=2/3

Flux

(JA, J

B, JV),

dC V

/dt

Distance, x

JB

JV

JA

(dCV/dt)

Vacancy creationVacancy annihilation


© meg/aol ‘02

Intrinsic Diffusivities:Vacancy Wind

(dCV/dt)<0

(a)

(dCV/dt)>0

(b)

Vacancies and lattice planes are created

Vacancies and lattice planes are annihilated

© meg/aol ‘02

Vacancy Concentration in Kirkendall Diffusion

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-3 -2 -1 0 1 2 3

Vaca

ncy

Satu

ratio

n, (

C v-Ceq

L )/Ceq

L

Distance, x

DA/DB=4/3

DA/DB=2/3

DA/DB=0.9

DA/DB=0.5

DA/DB=2

DA/DB=1.1

Annihilation DA/DB>1

Annihilation DA/DB<1

Creation DA/DB>1

Creation DA/DB<1

© meg/aol ‘02

Diffusion-Advection

• During uncompensated (nonreciprocal) diffusion, the lattice planes flow, because of the supersaturation or undersaturation of vacancies..

• Motion of the lattice planes constitutes “solid-state advection.”

• The standard form of Fick’s law holds in a stationary system.

• Fick’s law must be modified to account for lattice flow.

© meg/aol ‘02

Mass Conservation with Diffusion–Advection

dZ

dYdX

UY C[P+dY/2]UY C[P-dY/2]

JY[P+dY/2]JY[P-dY/2]

(X,Y)

P(X,Y,Z)

X

Y

Z

© meg/aol ‘02


JY −

∂JY

∂YdY2

+ uY C X,Y,Z( )+∂C∂Y

dY2

d X dZ

The inflow of B through the shaded faces of the control volume is

JY +

∂JY

∂YdY2

+ uY C X,Y,Z( )−∂C∂Y

dY2

d X dZ

The correspondent outflow of B is

Φ y = −

∂JY

∂Y+ uY

∂C∂Y

dYd X dZ

The component flow, , into the control volume is Φ y

© meg/aol ‘02


Φi

X ,Y,Z∑ = −∇⋅J + u⋅∇C( )dYd X dZThe sum of all the component flows

∂C∂t

= −∇⋅J + u ⋅∇CThe mass conservation rate becomes

J = −D∇CFick’s first law is given by

∂C∂t

= ∇⋅ D∇C+ u⋅∇CCombining the conservation equation with Fick’s first law

∂C∂t

= ∇⋅ D∇CIf u→0,

diffusive advective

© meg/aol ‘02

Steady–State Diffusion–Advection

u⋅∇C = 0

The orthogonality of the velocity and the concentration gradient vectors implies

∇2 C+

uD

⋅∇C = 0

The steady–state diffusion–advection equation

u = const.If an interface moves steadily,

∂C∂t

=∂∂x

D ∂C∂x

+ u∂C∂x

The general PDE for diffusion–advection reduces in 1-D to

∂C∂t

= D ∂∂x

∂C∂x

+uD

C

If D and u are constant

© meg/aol ‘02

Darken’s Analysis of the Kirkendall Effect

XLaboratory frame

Original interfaceUM=0 UM=0

xUM

Marker motion in a Kirkendall diffusion couple needed for Darken’s analysis

CA + CB = Ω−1 = const.Fixed ends!

© meg/aol ‘02


∂∂t

CA+ CB( )= 0For all compositions found throughout the couple

∂∂x

DA∂CA

∂x+ DB

∂CB

∂x− uM CA+ CB( )

=

∂∂t

CA+ CB( )Expanding

∂∂x

DA∂CA

∂x+ DB

∂CB

∂x− uMΩ−1

= 0The right–hand side is zero, so

d DA∂CA

∂x+ DB

∂CB

∂x− uMΩ−1

−∞

x'⌠

⌡ = const.

Integrating from the left end of the couple, x = -∞, to an arbitrary plane, x′

© meg/aol ‘02


DA

∂CA

∂x −∞

x'

+ DB∂CB

∂x

−∞

x '

−uM

Ω −∞

x'

= const.

Integration gives

uM = Ω DA

∂CA

∂x+ DB

∂CB

∂x

Expressing the concentrations as molar fractions

uM =0, ∂CA/∂x =0, ∂CB/∂x =0.

The boundary conditions imposed near both ends

© meg/aol ‘02


uM = DA

∂N A

∂x+ DB

∂N B

∂xIntegration gives Darken’s first equation

uM = DA − DB( )∂N A

∂x,

uM = DB − DA( )∂N B

∂x.

In terms of either A or B gradients

X = x+ uMtInterrelating through the lattice motion, uM, and applying the Galileen transformation

© meg/aol ‘02


dX =

∂X∂x

t

dx +∂X∂t

x

dt

A Galileen transformation implies that X (x, t ;uM), so

∂X∂x

t

=1

∂X∂t

x

= uMThe partial derivatives are and

∂N A

∂x

t

=∂N A

∂X

t

∂N B

∂x

t

=∂N B

∂X

t

The concentration gradients are unaltered by a Galileen transformation

© meg/aol ‘02


∂∂x

DA∂CA

∂x− uMCA

=∂CA

∂tInserting Darken’s first equation we obtain

∂∂x

DA∂CA

∂x− DA − DB( )CA

∂ N A

∂x

=

∂CA

∂tand

∂∂x

DA∂N A

∂x− DA N A

∂N A

∂x+ DB N A

∂N A

∂x

=

∂N A

∂tMultiplying both sides and simplifying

∂∂x

DA∂N A

∂x− DA 1− N B( )∂N A

∂x+ DB N A

∂N A

∂x

=

∂N A

∂tor

© meg/aol ‘02


∂∂x

˜ D ∂CA

∂x

=∂CA

∂tThe identical form of Fick’s second law becomes

D ≡ NBDA + NADBThe identical form of Fick’s second law becomes Darken’s second equation

DA = 0.8+ 0.2NB2 ,

DB = 2+ 0.5NB.

Example: intrinsic diffusivities are defined as linear and quadratic functions of the mole fraction of B.

uM = DA − DB( )∂N A

∂x,

˜ D = NBDA + NADB.

Darken’s two equations are capable of analyzing the Kirkendall effect:

© meg/aol ‘02


0

0.5

1

1.5

2

2.5

0 0.2 0.4 0.6 0.8 1

Diff

usiv

ity [a

rbitr

ary

units

]

Mole fraction, NB

DA*=1

DB*=

DA

DB

~D

Intri

nsic

Diff

usiv

ity,

DA, D

B

Interdiffusion coefficient

Tracer diffusivity of A

Tracer diffusivity of B

© meg/aol ‘02

Exercise 1. An interdiffusion experiment is performed at 1054C for 300 hours with pure copper and pure nickel as end members of a diffusion couple. The concentration profile is measured as shown in Fig.17.14. Inert markers placed within the couple move 200µm during the period of the diffusion anneal. Determine the intrinsic diffusivities, DCu and DNi at 1054C for a concentration near the Matano interface, x=0.

© meg/aol ‘02

Exercise

0

20

40

60

80

100

0 0.05 0.1 0.15 0.2

Conc

entra

tion,

CC

u [at.%

]

Distance, x [cm]

slope, (dNCu

/dx),at C

Cu=71 at.%

© meg/aol ‘02

Exercise

X2 t()

t= k (const.

To determine the interdiffusion coefficient perform a Boltzmann–Matano analysis.

uM ≡

∂X t()∂t

=k

2X t()

The corresponding marker speed is

uM =

X t( )2t

Insert the proportionality constant

© meg/aol ‘02

Exercise

uM =X t( )2t

= 200[µm]2 × (300) × (3600) s[ ]

= 9.3×10−5 [µm/s].

Calculate uM

uM = 9.3×10−9 [cm/s] = DNi − DCu( )⋅12.69 [mol.frac./cm]

˜ D = 3.16×10−10[cm2/s] = 0.71⋅ DNi + 0.29DCu.

Insert the marker speed uM , D, and the local concentration gradient, dNCu /dx

© meg/aol ‘02

Key Points

• The diffusivity from the Boltzmann-Matano analysis represents the effective mixing of A and B atoms. It is called the interdiffusion coefficient, or the chemical diffusivity.

• The motions of individual atoms are controlled by intrinsic diffusivities, because, in general, atoms move through their surroundings at unique speeds. This is nonreciprocal (uncompensated) diffusion, occurring whenever DA≠ DB.

• Kirkendall showed that nonreciprocal diffusion occurs in Cu-Zn couples, and measured the associated lattice marker movements.

• Darken analyzed the Kirkendall experiments. • Manning suggested that a “vacancy wind” is needed to explain

lattice plane advection. • Strains often develop during Kirkendall diffusion.

© meg/aol ‘02

Outline

• Generalized forces

• Atom mobilities

• Chemically inhomogeneous systems

• Limiting cases

• Tracer diffusion

• Exercise

© meg/aol ‘02

Thermodynamics of Irreversible Processes

A phenomenological description of diffusion transport processes is provided by the thermodynamics of irreversible processes.

The thermodynamics of irreversible processes is applied to explain influence of transport and interaction of several flows

Gas transport through a membrane (T, P gradients) Thermal diffusion (T, C gradients) Electron transport (electric potential, C gradients)

Note: Isothermal (and/or isobaric) diffusion pertains to this class of cross effects.

If diffusion proceeds by vacancy mechanism will always involve at least three types of currents

The atoms of the solvent, A

The atoms of dissolved substance, B

Vacancies

© meg/aol ‘02

Thermodynamics of Irreversible Processes

In using the principles of the thermodynamics of irreversible processes it is necessary to introduce a new driving force.

What is it?

This is chemical potential gradient in place of the usual concentration gradient.

Internal or generalized forces (inhomogeneous chemical fields) act similarly to the way that external fields (gravitational, mechanical, electrical, magnetic) act.

Introducing the concept of

© meg/aol ‘02

Analogy of electrical resistance

J = −Dd Cd x

Fick’s First Law Mass flow

Ohm Law Current Flow

I =

VR

Note: Fick’s first law and Ohm’s law are two phenomenological laws. One describes atom movements during diffusion, the other describes current flow.

© meg/aol ‘02

Onsanger Theory: External Forces

One of Onsager’s assumptions for an isothermal process is X = −∇µ i

where X = is the thermodynamic force

µ= is the chemical potential

The work done during the passage of i component from one point to another equals the loss of free energy

Ai = −∆µ i

or Fi ∆x = −∆µ i

or Fi = −∇µ i for ∆x = 0

© meg/aol ‘02

External Forces

If some external force Fi acts on the diffusing atom, then the atom will come under a directional movement with a mean velocity

< vi >= Ui ⋅Fi

where Ui is the mobility expressed as velocity under the action of a unit force.

U i ≡ Bi ⋅Fi

Bi (i = A,B)The molar mobility during binary diffusion is defined as

© meg/aol ‘02

Atom Mobilities: Constrained Equilibrium

FB ≡ −mBg

R. Feynman describes the influence of gravity, g, acting on a homogeneous dilute solution of B atoms dissolved among lighter A atoms

FdiffB − mBg = 0,

FdiffA − mAg = 0.

At equilibrium, in the presence of gravity, the net force acting on the atoms vanishes

where diffusion forces, Fdiff, are example of generalized forces.

© meg/aol ‘02

Atom Mobilities:Diffusion Forces

Fidiff = −∇µ i r( )The corresponding diffusion force can be described as

the spatial gradient of the chemical potential

Ji = Ci ui

A diffusive flux, Ji , of species i arises and could be expressed in terms of its Fickian drift velocity, ui

Ji = Ci Bi ⋅Fidiff

In terms of the molar mobility and the unbalanced diffusion force, Ji becomes

Ji = −Ci Bi ⋅∇µ i r( )

Replacing the diffusion force by the chemical potential gradient,

© meg/aol ‘02

Chemically Inhomogeneous Systems

µ i = µ0 + RgT ln ai

The definition of the chemical potential for an isothermal homogeneous binary alloy in equilibrium

The spatial gradient of the chemical potential may be written

∇µ i r( )= RgT ∇ ln ai r( )[ ] (T = const.),

∇µ i r( )=

RgTai r( )∇ai r( )

or

© meg/aol ‘02


lnai = lnγ i + ln Ni

Taking logarithms

ai ≡ γ i Ni

Relating the thermodynamic activity of a homogeneous alloy to its concentration

∇µ i = RgT ∇ lnγ i + ln Ni( )

Substituting

∇µ i = RgT ∇ lnγ i + ∇ ln Ni( )

Carrying out the gradient operation for the right-hand side with T held constant

© meg/aol ‘02


∇µ i =

RgTNi

Ni∇ lnγ i

∇Ni

+1

∇Ni

Rearranging

∇µ i =

RgTNi

∇ lnγ i

∇ ln Ni

+1

∇Ni

Rearranging in terms of the logarithmic derivative

Ji = −Ci Bi ⋅

RgTNi

∇ lnγ i

∇ ln Ni

+1

∇Ni

Substituting into the flux equation,

© meg/aol ‘02

Ji = −Di∇Ci = −RgT 1+∂ lnγ i

∂ ln Ni

Bi Di

⋅∇CiThe intrinsic diffusivity appears in flux equation


Ni ≡ CiΩ =

Ci

CA + CB

i = A,B( )The mole fraction is defined as

Ji = −RgT Bi ⋅∇Ci 1+

∂ lnγ i

∂ ln Ni

Substituting into flux equation

Bi =

Di

RgT1+

∂ lnγ i

∂ ln Ni

−1

, i = A,B( )The atom mobility could be expressed as

thermodynamic factor

© meg/aol ‘02

Limiting Cases

γ i → 1 as Ni → 1Raoult’s Law for ideal solutions

γ i → kH as Ni → 0Henry’s Law for dilute solutions

Di = RgT Bi The chemical term vanishes in these limits

∂ lnγ i

∂ ln Ni

T ,P

= 0The activity coefficient becomes constant

Di = RgT 1+∂ lnγ i

∂ ln Ni

thermodyamic factor, ϕ

Bi For concentrated solutions

True for dilute solutions

© meg/aol ‘02

1

non–ideal

Raoult's Law

Henry's Law

T=Const.

Ni

ai

0

1

Activity versus mole fraction

Raoult’s Law

Henry’s Law

© meg/aol ‘02

Tracer Diffusion

C* x,t( )=

Mthin film*

πD∗texp −x2

4D∗t

The thin film solution

lnC∗ = ln

Mthin film*

πD* t

−

x2

4D*t

Taking logs of both sides

m= −

1D*

The slope, m, becomes

D* = RgT Bi

* The tracer diffusion is

Bi∗ = Bi Where for isotopes similar to the solute

© meg/aol ‘02

Tracer Diffusion

Di = RgT 1+

∂ lnγ i

∂ ln Ni

Bi Remember that

Di

D∗ = 1+∂ lnγ i

∂ ln Ni

=

ϕRgT

Substituting

˜ D = NADB

* + NBDA*( )⋅ ϕ C,T( )

RgTThe interdiffusion coefficient

Darken–Hartley–Crank equation

© meg/aol ‘02

Exercise

1. Calculate the interdiffusion coefficient, at 700C, as a function of mole fraction of component A, NA, for an A–B alloy which has the following tracer diffusivities: DA

*= 2.5x10-12 cm2/sec, and DB*=5x10-11 cm2/sec.

Na aa Na aa

0.020000 0.11867 0.52000 0.797110.040000 0.22086 0.54000 0.799410.060000 0.30875 0.56000 0.801810.080000 0.38423 0.60000 0.807200.10000 0.44895 0.62000 0.810330.12000 0.50433 0.64000 0.813830.14000 0.55162 0.66000 0.817750.16000 0.59191 0.68000 0.822170.20000 0.65516 0.70000 0.827110.22000 0.67967 0.72000 0.832640.24000 0.70030 0.74000 0.838810.26000 0.71761 0.76000 0.845650.28000 0.73209 0.78000 0.853230.30000 0.74415 0.80000 0.861580.32000 0.75416 0.82000 0.870770.34000 0.76246 0.84000 0.880830.36000 0.76932 0.86000 0.891830.38000 0.77499 0.88000 0.903810.40000 0.77970 0.90000 0.916840.42000 0.78363 0.92000 0.930980.44000 0.78697 0.94000 0.946290.46000 0.78986 0.96000 0.962850.48000 0.79243 0.98000 0.980730.50000 0.79481 0.98000 0.98073

© meg/aol ‘02

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Mole fraction, NA

Act

ivity

, aA

700 C

Activity of A versus mole fraction NA

© meg/aol ‘02

0

1

2

3

4

5

6

7

0 0.2 0.4 0.6 0.8 1

Activ

ity C

oeffi

cien

t, γ A

Mole Fraction, NA

Activity coefficient γA versus mole fraction NA

© meg/aol ‘02

-1

-0.8

-0.6

-0.4

-0.2

0

0 0.2 0.4 0.6 0.8 1

700C

dln(

γΑ

)/dl

n(N

A)

Mole fraction, NA

Logarithmic derivative of the activity coefficient γA versus mole fraction NA

© meg/aol ‘02

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

700C

Ther

mod

ynam

ic fa

ctor

, ϕ(N

A,T

)/RgT

Mole fraction, NA

Thermodynamic factor ϕ(NA) versus mole fraction NA

© meg/aol ‘02

10 -12

10 -11

10 -10

0 0.2 0.4 0.6 0.8 1

700C

Mole fraction, NA

DA*

DB*

Inte

rdiff

usio

n C

oeffi

cien

t, D

[cm

2 /s]

~

Interdiffusion Coefficient

© meg/aol ‘02

Key Points

• The concept of generalized forces is introduced: these forces arise from microscopic interactions within a material. Resistivity is used as a common example of such a generalized force.

• Atomic mobilities may be defined to relate drift speed with the generalized diffusion “force.”

• Chemically inhomogeneous systems (systems with gradients) develop diffusion forces not present in homogeneous systems.

• These forces are related to the thermodynamic properties of the solution. Specifically, how activity changes with concentration.

• Limiting cases are 1) ideal (solution) alloys, 2) dilute alloys, 3) very concentrated alloys. These alloys behave according to the rules developed early in the semester.

• Tracer diffusion is shown to be related to the simple case of a nearly ideal solution.

© meg/aol ‘02

Outline

• Background

• Standard linear solid

• Internal friction

• Interstitial sub–populations in BCC

• Reorientation of atom pairs in FCC materials

• Internal friction methods

• Exercises

• Summary

© meg/aol ‘02

t

t

σ

εelasti c

εelasti cεsat .

εan.

εan.

ε

σzz

σ + τε Ý σ = M R ε + τσ Ý ε ( )

The applied stress–time function

Schematic showing anelastic straining and the mechanical aftereffect.

where

is the stress relaxation time at constant strain

τε

τε

τσis the strain retardation time at constant stress

is the relaxed elastic modulus

M R =σε

© meg/aol ‘02

Cyclic Stress and Strain

σ ωt( ) = σ0 sin ωt( )

Ý σ ωt( ) = σ0ω cos ωt( )

σ0 sin ωt( )+ τεσ0ω cos ωt( ) = M R ε + τσ Ý ε ( )

ε ωt( )= Asin ωt( )+ Bcos ωt( )

Ý ε ωt( ) =ω A cos ωt( )− Bsin ωt( )[ ]

The applied stress–time function is of the form

Its time derivative is

The linear relationship between the five primary mechanical variables is

The more general anelastic time–dependent strain response is the linear strain–time function

The associated strain rate is

© meg/aol ‘02


σ 0

M R

sin ωt( )+σ 0τεω

M R

cos ωt( ) = A − Bτσω( )sin ωt( )+ B + Aτ σω( )cos ωt( )

The response of the standard linear solid to the applied sinusoidal stress–time function is

Matching the coefficients of the sine and cosine terms yields two expressions

A − Bτ σω =σ 0

M RIn–phase

B + Aτσω =σ 0τεω

M ROut–of–phase

© meg/aol ‘02

Normalized stress and strain versus phase angle, ø

-1.5

-1

-0.5

0

0.5

1

1.5

0 0.8 1.6 2.4 3.2 4 4.8 5.6 6.4

stressstrain

Nor

mal

ized

Stre

ss a

nd S

train

, σ

/ σ0, ε

/ε0

angle (ω t) [radians]

ω t*

φ

The value of φ is selected arbitrarily at 0.5.

© meg/aol ‘02


φ ≡ωt * −π2

Ý ε = 0 = ω A cos ωt *( )− Bsin ωt*( )[ ]

AB

= tan ωt *( )

tan Φ( )= tan ωt* −π2

= −cot ωt *( )= −

BA

The phase angle is defined as

φ

Setting the left–hand side equal to zero

The strain amplitude ratio A/B is

The tangent of the phase angle named the loss tangent is

© meg/aol ‘02

Internal Friction

Q−1 ≡Edissip

tot

Estoredmax =

σ ωt( )Ý ε ωt( )out-of -phased ωt( )0

2π

∫

σ ωt( )Ý ε ωt( )in-phased ωt( )0

π2

∫

Q−1 =−B sin2 ωt( )d ωt( )

0

2π

∫

A sin ωt( )cos ωt( )d ωt( )0

π2

∫

Q−1 = −π BA

Q−1 = π tan Φ( )

The internal friction is given by

Substituting

The internal friction Q-1 is

The phase angle relationship is

© meg/aol ‘02

Internal Friction

1−BA

τσω =σ 0

A M R

BA

+ τσω =σ 0τ εωA M R

1A

=M R

σ 0τεωτσω +

BA

1−BA

τσω =1

τεωτσω +

BA

−BA

=ω τ σ − τε( )1+τ ετσω2

−BA

≡ tan Φ( ) =ω τ σ − τε( )1+ω2τσ τ ε

Dividing by A and

A few steps of algebra will lead to

Inserting back

Solving for the strain amplitude ratio

Inserting definition of loss tangent

© meg/aol ‘02

Internal Friction

τ ≡ τετσ

M ≡ M RMU

M =τσ

τε

M R =τε

τσ

MU

tan φ( )=MU − M R

M

ωτ1+ ωτ( )2

The single relaxation time is

Combining the relaxed and unrelaxed moduli

The mean modulus is

A key expression for the frequency–dependent loss tangent

relaxation strength frequency dependence

© meg/aol ‘02

Internal Friction Q-1 and Frequency–Dependent Modulus versus Dimensionless Time

0.01 0.1 1 10 100

Internal frictionModulus

ωτ

Inte

rnal

Fric

tion,

Q-1

=πta

n(Φ

)

Q-1

MR

MU

Freq

uenc

y -d

epen

dent

mod

ulus

, Mω

© meg/aol ‘02

Internal Friction

tan φ( )max =MU − M R

2M The maximum loss tangent is given by

The maximum internal friction occurs when

Q−1 = πMU − M R( )

M ωτ

1+ ωτ( )2

ωτ = 1

The final result of analyzing internal friction and anelastic relaxation in the standard linear solid:

This is the key result needed to analyze internal friction!

© meg/aol ‘02

Occurrence of the Maximum Internal Friction at ωτ=1

0

0.5

1

1.5

2

0 1 2 3 4 5 6

Q-1

angle, ωτ , [radians]

tan(Φ)max=1/2

Inte

rnal

fri

ctio

n, Q

-1=π

tan(

Φ)

© meg/aol ‘02

Geometry of Interstitial Sites in BCC

X

Z

Y

X-site

Y-site

Z-site

Assume that a uniaxial stress σzz is applied.

© meg/aol ‘02

Interstitial Subpopulations in BCC

dnX

dt

= −Γint nX +

Γint

2nY + nZ( )

n0 = nX + nY + nZ

The detailed balance equation in terms of the interstitial jumping frequency

Under isotropic conditions

and

nY + nZ = n0 − nX

Substituting gives

dnX

dt

= −Γint nX +Γint

2n0 − nX( )

© meg/aol ‘02


dnX

dt

= −

32

Γint nX −n0

3

dnX

nX − n0 3nx ( t<0, σ =σ zz )

nx ( t , σ =0)⌠ ⌡ = −

3Γint

2d ′ t

0

t

∫

32

Γint ≡ τ−1

or

The ODE predicting the interstitial population dynamics for a step change in the steady-state stress from σzz→0 can be integrated as

The anelastic relaxation theory coefficient is defined as

The solution is

nX t( )σ =0 =n0

3+ nX 0,σ zz( )−

n0

3

exp −

tτ

© meg/aol ‘02


D =16

Γintλ2

D =1

24Γinta0

2

τ ≡32

Γint

−1

D =a0

2

36τ

ω*τ =2ω*

3Γint= 1

In 3-D the Snoek diffusion

Inserting the interstitial jump distance

The relaxation time implied is

Substituting we obtain

This relation provides the condition for Snoek diffusion at a specified T

© meg/aol ‘02


P ≡12

3nP

n0

−1

P =e

∆UkBT −1

e∆UkBT + 2

D =ω∗ a0

2

36This yields the interstitial diffusivity as

The imbalance of interstitial sites caused by uniaxial strain relaxation, using the microscopic parameter, P, becomes

P can be expressed as

© meg/aol ‘02

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Ord

er p

aram

eter

, P

Orientation Energy, - U/kBT

P=(-1+3np/n)/2

P=0, randomP=1, oriented

Order Parameter, P, versus Orientation Interaction Energy

© meg/aol ‘02

Pair Reordering in

FCC Materials

0

200

400

600

800

1000

1200

1400

1 1.2 1.4 1.6 1.8 2

alpha-brass, 620 Hz

Inte

rnal

Fric

tion,

Q-1 x

105

1000/T

727 227283352441560T, [deg.C]

B =ττ0

e− ∆HRgT

Q−1 = πω ∆ M B+ B−1[ ]−1

• The relaxation times varies with T

• The internal friction may be defined

© meg/aol ‘02

Internal Friction Methods

0

0.5

1

1.5

2

2.5

3

0.5 1 1.5 2 2.5

Inte

rnal

fric

tion,

Q-1

[arb

itrar

y sc

ale]

Reciprocal Temperature, 1/T, [K -1x1000]

ω=const.

1/T *

Spectrum of internal friction

Note: Multiple peaks are difficult to analyze.

© meg/aol ‘02

Diffusional Anelastic Effects • The temperature dependence of a single anelastic relaxation

time, τ, yields the activation enthalpy of the responsible diffusion process.

• As the concentration of interstitials increases, the position of the internal friction peak remains steady, but its amplitude, Q-1, increases.

• Internal friction methods require only microscopic (i.e. atomic) motions of the interstitials.

• Consequently, the technique developed applies for small diffusivities, long relaxation times, low temperature.

• Modern mechanical damping methods detect diffusivities well below 10-22 cm2/s.

© meg/aol ‘02

Exercise 1. The Snoek effect occurs by short–range ordering of interstitial atoms subject to long range strains. Explain why this anelastic diffusion phenomenon occurs commonly in BCC materials, such as iron, niobium, and tantalum, but is not observed in FCC materials, such as aluminum, copper and gold.

1 & 2 octahedralinterstitial sites

tetrahedralinterstitial site

FCC

1

2

© meg/aol ‘02

Solution Consider the relationship of interstitial sites in a BCC crystal for which there is an applied uniaxial tensile stress, σzz, along the z–axis. Nearest neighbors lattice atoms parallel to the z–axis are axially strained, and pulled slightly further away from Z–type interstitials, giving them more space. By contrast, X– and Y–type interstitials have their nearest–neighbor lattice atoms drawn closer, because of the Poisson contraction occurring in the x– and y–directions, which are orthogonal to the assumed applied stress. The resulting imbalance in the local interstitial volumes at the three types of interstitial site orientations leads to the population bias and anelastic Snoek effect. Now, in the parallel case of a uniaxially strained FCC crystal, the volume of the “cage” formed by neighboring host lattice atoms surrounding the octahedral interstitial atoms of type–1, remains unchanged from that for the octahedral interstitial atoms of type–2. A diffusive exchange onto a neighboring site consequently lacks a site preference. The interstitial population therefore remains unbiased by the uniaxial stress, so diffusional anelasticity of the Snoek-effect type will not develop in FCC crystals. The same conclusion holds for the tetrahedral sites in FCC materials.

© meg/aol ‘02

Exercise

2. Snoek internal friction peaks are observed for nitrogen in niobium based on measurements with a 5–second period torsion pendulum operated at 527 K, and on similar measurements with a 60–second pendulum at 493 K. a) Determine the interstitial jumping frequencies of nitrogen atoms in Nb at 527 K and 493 K. b) Calculate the diffusion coefficients for the interstitial motion of nitrogen through Nb at each temperature. c) Derive the values of D0 and ∆H for this binary alloy system, as needed to express the diffusion coefficient in the Arrhenius form: D = D0 exp(-∆H/RgT). a) The average jumping frequencies for the nitrogen atoms, ΓN, may be found from the straining frequencies yielding peak internal friction. The following formula proves useful

τ =2

3ΓN

© meg/aol ‘02

Solution

DN =124

ΓN a02

3a0 = 4RNb = 4 (0.143 ×10−9 m) 100cm1m

,

a0 = 3.3×10−8 cm.

a) The input data are: τ = 5sec at 527K and τ = 60sec at 493K, gives the nitrogen jumping frequency as ΓN = 4.4×10-2sec-1 at T = 527K, and ΓN = 1.1×10-2sec-1 at T = 493K

b) The diffusivity in niobium for nitrogen atoms can be calculated using the formula

Niobium is a BCC metal and has R = 0.143 nm:

Substituting the lattice parameter and the jumping frequency for nitrogen into the diffusivity formula yields the values DN = 2×10-18cm2/sec at T = 527K, and DN = 5×10-19cm2/sec at T = 493K.

© meg/aol ‘02

Solution c) The Arrhenius correlation for these diffusivities may be found as follows:

DN527 K

DN493K =

D0e−

∆HRg⋅527

D0e−

∆HRg⋅493

=e

−∆H [Joule/mole]

(8.31)⋅527

e−

∆H [Joule/mole](8.31)⋅493

Inserting the diffusivities found above into the Arrhenius correlation gives the activation energy as ∆H = 88 KJ/mole. Selecting a diffusivity value at either temperature, and now knowing the activation enthalpy, ∆H, allows the calculation of the frequency factor, D0. This calculation shows that D0 = 1.07×10-9cm2/sec. Choosing the other diffusivity allows a cross–check on the value of D0. Thus, the Arrhenius correlation is

DN T( ) = 1.07 ×10−9e−

88[KJ/mole]RgT

© meg/aol ‘02

Key Points • The “standard linear (visco-elastic) solid” was treated under cyclic

stressing. Internal friction could be measured as the ratio of the dissipated elastic energy to the maximum stored energy, designated as Q-1.

• Interstitial site population balances were derived, assuming a single relaxation time for interstitial diffusion, τ.

• Maximum internal friction occurs when ω τ=1. • Snoek effect in BCC materials was discussion as a method of

measuring very small D values at reduced temperatures. • Solute pair reordering under cyclic stress in Cu-Zn alloys, as

measured by Zener, was discussed. • Internal frictions has many origins in materials, so one must be

judicious in selecting the cause of a given internal friction peak. • Internal friction, in the form of ultrasonics, provides a useful

method of measuring very small diffusivities, because motions of only few atomic spacings are needed to produce the energy absorption.

© meg/aol ‘02

Outline • Background

• Diffusion currents in homogeneous ionic solids

• Measurement of ionic conductivity in solids

• Defects in ionic solids

• Experiments in ionic conductors

• Irreversible thermodynamics and diffusion

• Isothermal binary diffusion

• Vacancies in thermal equilibrium

• Net vacancy flux

• Exercise

© meg/aol ‘02

Background

J i = Ci B i ⋅Fi

J i = −Ci Bi ⋅∇µ i

µ i = µ0 + RgT ln ai , (0 < ai < 1)

J i = −RgT Bi ⋅∇Ci

The flux of a mobile component Ji related to the molar diffusion mobility tensor Bi to a generalized force Fi

The diffusion flux arising from the gradient of chemical potential,

The chemical potential of the ith component of a solution defined relative to unit activity

Substituting

© meg/aol ‘02

Background

J i = −RgT Bi ⋅ ∇Ci +Ciqi∇φ

RgT

Di = RgT Bi

J i = −Di∇Ci − DiCi qi∇φ

RgT

An additional flux term appears that arises from the electrostatic force acting on the ions

For an ideal or highly dilute solutions

Substituting yields a form of Nernst’s equation:

© meg/aol ‘02

Diffusion currents in homogeneous ionic solids

Ji = J iqi = −Di

Ci qi

2

RgT

⋅∇φ

I ext = Di

Ciqi

2

RgT

⋅

AL

V

I ext = Gi V

The ion current flowing through a unit area of a homogeneous specimen is

The electronic current flowing in the detector circuit is given by

Ohm’s law states that

© meg/aol ‘02

Diffusion currents in homogeneous ionic solids

Gi =Ci Di

RgTq

i

2 AL

Gi =Ci Di

RgTZF( )2 A

L

G ≡ σAL

σ =Ci Di

RgT⋅ ZF( )2

General expression for the electrical conductance in a homogeneous conductor

In terms of Faraday’s constant we may rewrite

Introducing the electrical conductivity, σ

Substituting yields a basic formula for the ionic conductivity

© meg/aol ‘02

Measurements of ionic conductivity in solids

I ext = Ji Λe + Λa + Λc( )

Λ ii

∑ = 1

The mobile carriers in most ceramic materials

These transport numbers are defined so

Defects in ionic solids

The most common ion–defect pairs encountered are

Schottky pairs va + vc

Frenkel pairs vc + ic

© meg/aol ‘02

Equilibria Among Charged Point Defects

Nva⋅ Nvc

= exp−∆GSchottky

kBT

The concentration of Schottky pair defects

The concentration of Frenkel pair defects

Nic ⋅ Nvc= exp

−∆GFrenkel

kBT

Nvc= Nva

+ NicOverall

Dic* = Dα Niα

Niα ≡Ciα

Cα

σCi Dα

=Zα F( )2

RgT

The tracer diffusivity of such an interstitial defect is

where

The general conductivity equation

© meg/aol ‘02

Equilibria Among Charged Point Defects

σCiα

Cα( )Dα

=Cα Zα F( )2

RgT

σD

i c

* =Cα Zα F( )2

RgT

Dic

* = σ ⋅RgT

Cα Zα F( )2

Multiplying both sides by the concentrations of the mobile ions

Comparing

The interstitial tracer diffusivity

Considering that

Dva

* = fα Dα Nvα

The expression becomes

σD

vα

* =Cα ZvF( )2

fα RgT

© meg/aol ‘02

Ionic Conductivity and Diffusivity in Soda–lime Glass (J. Kelly’s experiments)

T (C) D [cm2/sec] σ [ohm-cm]-1 f exp

200 4.2x10-13 5.05x10-8 0.23250 4.0x10-12 3.25x10-7 0.29300 2.6x10-11 1.51x10-6 0.38

© meg/aol ‘02

Irreversible Thermodynamics and Diffusion

J1 = L11X1 + L12 X2 + L13 X3 + ... L1nXn,

J2 = L21X1 + L22 X2 + L23 X3 + ... L2n Xn,

J3 = L31 X1 + L32 X2 + L33 X3 + ... L3nXn,

Jn = Ln1 X1 + Ln2 X2 + Ln3 X3 + ... LnnXn.

1) Each transport flux depends linearly on all generalized thermodynamic forces

2) The Onsager matrix of kinetic coefficients [Lik] is comprised of diagonal terms [Lii]

Onsager’s reciprocity theorem Lik = Lki

© meg/aol ‘02

Irreversible Thermodynamics and Diffusion

TdStot

dT

= J i ⋅Fii

∑

3) Each of the thermodynamic forces dissipates free energy and produces entropy

Flux, Ji Heat, q Diffusion, Ji Electron flow, iForce, Xi −T ∇

1T

−T ∇

µi

T

E=-∇φ

List of the generalized forces associated with q, Ji, i

© meg/aol ‘02

Isothermal Binary Diffusion

J1 = −L11∇µ1 − L12∇µ2 − L1v∇µv ,

J2 = −L21∇µ1 − L22∇µ2 − L2 v∇µv ,

Jv = −Lv1∇µ1 − Lv2∇µ2 − Lvv∇µv .

J1 + J2 + Jv = 0

L11 + L21 + Lv1( )∇µ1 = 0,

L12 + L22 + Lv2( )∇µ2 = 0,

L1v + L2 v + Lvv( )∇µv = 0.

For the case of vacancy–assisted diffusion

The fluxes

Each term must vanish column by column

© meg/aol ‘02

Isothermal Binary Diffusion

−Lv1 = L11 + L21,

−Lv2 = L12 + L22 ,

−Lvv = L1v + L2v .

J1 = −L11∇ µ1 − µv( )− L12∇ µ2 − µ v( ),

J2 = −L12∇ µ1 − µv( )− L22∇ µ2 − µ v( ).

Jv = −L1v∇ µ1 − µv( )− L2v∇ µ2 − µ v( )

The kinetic coefficient for the vacancy flux is dependent on those for the component atoms

Combining with Onsager’s reciprocity relationship

The vacancy flux can be written

© meg/aol ‘02

Vacancies in Thermal Equilibrium

JA = −LAA∇ µ A( )− LAB∇ µB( ),

JB = −LAB∇ µA( )− LBB∇ µB( ).

µA = µ0A + kBT lnCA,

µB = µ0B + kBT lnCB,

µv = 0.

JA = −LAAkBT ∇ lnCA( )− LABkBT ∇ lnCB( ),

JB = −LABkBT ∇ lnCA( )− LBBkBT ∇ lnCB( ).

The chemical potential for a binary alloy is

If in the binary alloy the lattice vacancies form an ideal solution

Writing in terms of the component concentrations

© meg/aol ‘02

The equations have a similar form with Fick’s first law

JA = −LAA

CA

kBT ⋅∇CA −

LAB

CB

kBT ⋅∇CB,

JB = −LAB

CA

kBT ⋅∇CA −LBB

CB

kBT ⋅∇CB.

JA = −LAA

CA

kBT ⋅∇CA,

JB = −LBB

CB

kBT ⋅∇CB.

Ji = −Di ⋅∇Ci , i = A,B( )


or

Simplifying

© meg/aol ‘02

Di = kBTLii

Ci

DA = kBTLAA

CA

,

DB = kBTLBB

CB

,

DA = DB.


The intrinsic diffusivities are

The intrinsic diffusivities can be expressed as

© meg/aol ‘02

Net Vacancy Flux

The Gibbs–Duhem relationship provides that

CA∇µA = −CB∇µB

It then follows that

For non–reciprocal binary diffusion, Onsager’s equations show that

DA = kBTLAA

CA

−LAB

CB

,

DB = kBTLBB

CB

−LAB

CA

,

DA ≠ DB.

JA = −LAA

CA

−LAB

CB

kBT

⋅∇CA,

JB = −LBB

CB

−LAB

CA

kBT

⋅∇CB.

© meg/aol ‘02

Exercise

1. When a DC voltage is applied to a bar of γ–Fe containing a homogeneous interstitial solution of carbon, an electric current flows and carbon atoms migrate and collect near the cathode. This “unmixing” process is called electro–diffusion, and is analogous to the ionic diffusion in a homogeneous ceramic solids described in §20.2. a) Develop the linear phenomenological equations for electro–diffusion, assuming that the iron atoms, at the temperature of this experiment, are immobile compared to the highly mobile carbon interstitials, and that the current measured is comprised virtually entirely of electrons. b) Find the ratio of the electron flux to the carbon flux in a homogeneous Fe–C specimen. c) If the applied voltage is zero, show that a charge still flows when the carbon atoms diffuse under their own gradient of chemical potential. What is the nature of this charge?

© meg/aol ‘02

Solution

Je− = −Le−e−

∂φ∂x

− Le−C

∂µC

∂x

,

JC = −LCe−

∂φ∂x

− LCC

∂µC

∂x

.

Je−

JC

∇C=0

=L

e−e−

LCe−

Je

JC

∇φ=0

=LeC

LCC

A) In analogy to the ionic diffusion case, the flux of each mobile quantity depends on all the thermodynamic forces—in this example the Coulomb force and the generalized chemical force. For electrons (e-) and carbon (C) one can write individual flux equations

B) The ratio of the electron flux to the carbon flux when the chemical potential gradient vanishes (a homogeneous alloy) is given by the ratio of the expressions shown in eq.(20.47) and setting (∂µ/∂x)=0.

C) The ratio of the electron flux to the carbon flux when the voltage is absent is given by the ratio of the expressions shown in eq.(20.47), and setting (∂φ/∂x)=0; hence

© meg/aol ‘02

Key Points • In ionic conductors, anions and cations diffusion on their separate lattices. • To preserve electrical neutrality, special defects, in addition to lattice

vacancies are needed: – Frenkel pairs (vc+ ic) – Shottkey pairs (vc + va)

• Atom fluxes are generally composed of a diffusive term and a drift term, caused by an external field.

• In ionic conductors, the flux can be driven by an applied potential, and the net charge flux senses as an external current.

• Tracer studies in soda-lime glasses discussed. • Onsager’s irreversible thermodynamic formalism applied to diffusion in

solids. Intrinsic diffusivities for reciprocal (no net vacancies) and non-reciprocal binary diffusion (vacancy wind) may be expressed in terms of the Onsager kinetic coefficients.

• Electro-diffusion (akin to electrolysis) can be formulated similarly, to show that the flow of electrons moves atoms, and the flow of atoms moves electrons.

© meg/aol ‘02

Outline

• Length scales in microstructures

• Kelvin’s equation

• Dimensionless diffusion potential

• The mean–field kinetic equation

• Experimental observations

• Exercise

© meg/aol ‘02

Length Scale in Microstructure

Early Microstructure Later coarsened stage

Note: Microstructural phase coarsening is a process generically termed Ostwald ripening.

© meg/aol ‘02

Kelvin’s Equation

bulk phase reservoir

P(R), T

droplet reserv oir

1

µ0

µR

P0, T2

3

vaporvapor

∆µtot =µR - µ0

Thermodynamic path for deriving Kelvin’s equation

© meg/aol ‘02

Kelvin’s Equation Consider the free energy changes for the following three–step isothermal path:

∆µ1=0

A) A mole of substance is isothermally evaporated from the bulk phase

P0 is the equilibrium vapor pressure

T is the temperature µ0 is the chemical potential at equilibrium

µ1 is the chemical potential for reversible evaporation

∆µ2 = Vvapor d P

P0

P R( )

∫ , (T const.

B) The mole of vapor from the left reservoir is isothermally compressed to a slightly higher pressure, P(R).

© meg/aol ‘02

Kelvin’s Equation C) The slightly compressed vapor is slowly condensed onto the droplets at their equilibrium pressure

∆µ3=0 No change in the free energy

∆µ1 + ∆µ2 + ∆µ3 = 0+ VvapordP

P0

P R( )

∫ + 0Free energy per mole of pure substance is the sum over the steps of the reversible path

PVvapor = RgTTreat the mole of vapor as an ideal gas

µR − µ0 = RgT

dPPP0

P R( )

∫ = RgT lnP R( )

P0

Substituting

© meg/aol ‘02

Kelvin’s Equation

P R( )P0

= eµ R−µ0

RgT

Rearranging

Pint =

P0 (in the bulk),

Pdrop (in the droplet

At constant temperature

µR − µ0 = VdropdP

P0

Pdrop

∫ ≅ Vdrop Pdrop − P0( )Chemical potential could be approximated as

© meg/aol ‘02

Kelvin’s Equation

P(R)

γ

γ

Pdrop

R

β α

α/β is the interface between the two phases

γ is the interfacial force between the phases

Pdrop is the internal pressure

P(R) is the external pressure

πR2Pdrop − 2πRγ − πR2 P R( )= 0The forces acting on and within a droplet

© meg/aol ‘02

Kelvin’s Equation

Pdrop − P R( )=

2γR

Laplace–Young equation states that

µR − µ0 = Vdrop

2γR

+ P R( )− P0

Substituting

P R( )− P0 <<

2γR

Approximating that

H ≡

κ1 + κ2

2=

1R

(spheresThe mean curvature is defined as

ai H( )a0

≡Pi R( )

P0

= e2Vdropγ

RgT⋅H

Kelvin’s equation becomes

P R( )P0

= eγVdrop

RgT⋅2R

then

© meg/aol ‘02

0

1

2

3

4

5

-2 -1 0 1 2

Activ

ity, a

(H)

Normalized Mean Curvature, H/(RgT/2V

drop)

Concaveinterfaces

Convexinterfaces

Linear approximation

Range of curvatures inmost microstructures

Kelvin’s Equation

© meg/aol ‘02

Influence of Curvature on Solubility

µB H( )= µB0 + RgT lnaB H( )The chemical potential of the solute B

becomes

aB H( )aB

0 ≅C H( )Cplanar

For small changes in concentration the ratios of the activities are

CB H( )= CBe

γ ΩRgT

H

≅ CB 1+γ ΩRgT

H

The solubility of B atoms saturating the matrix α

CB H( )− CB

CB

=2Ω γRgT

⋅ HThe increase in solubility

© meg/aol ‘02

.

α

β

solubilit y increase

0 1XβXα

µBo

µAo

µBH

HµA

G i b b s e n e r g y p e r m o l e

XB

2γΩ/ R

XB

Solubi lityincrease

X B

Gib

bs e

nerg

y pe

r mol

e

Influence of Curvature

© meg/aol ‘02

Dimensionless Diffusion Potential

ϑ R( )≡

CB H( )− CB

CB

The capillary diffusion potential is

∇2ϑ r( )= 0The quasi–static diffusion potential

ϑ(R) =2Ω γRgT

⋅

1R

=lcR

,

ϑ( ˜ R ) = 1˜ R .

A local boundary value for the dimensionless concentration field

© meg/aol ‘02

The Mean–Field Kinetic Equation

ϑ=1/R

ϑ∞ =1/ <R>

α-phase

ϑ

r

β-phase R

Diffusion potential surrounding an isolated spherical particle

© meg/aol ‘02


ϑ r( )R=

1R

,

ϑ r( )lim r →∞

=1R

.

The boundary conditions

ϑ r( )=

1r

1−RR

+

1R

The solution to the quasi–static diffusion equation

dϑd r

r = R

=1R2

RR

−1

The rate of growth or shrinkage of the

particle

© meg/aol ‘02

0

1

2

3

4

5

-4 -2 0 2 4

Diff

usio

n Po

tent

ial, ϑ

(r, R

/<R

>)

R=0.25

0.4

R=<R>=1

0.6

r /<R>

R=2

d Rdτ

=dϑdr

r = R

=1R2

RR

−1


© meg/aol ‘02

Experimental Observations

3.8

4

4.2

4.4

4.6

4.8

5

5.2

2.5 3 3.5 4 4.5 5 5.5 6

Inte

rfac

ial a

rea,

(SV),

log-

scal

e [m

-1]

log time [sec]

Sn-Bi alloys, T=143C(after Meloro)

1

3

© meg/aol ‘02

Exercise

Consider an Al–15wt.% Mg alloy in which Al3Mg2 particles are dispersed in the matrix (Al solid solution). Two nearby–particles have diameters of 10µm and 1µm, respectively. These particles are separated by a center–to–center distance of 8µm. The molar volume of β–phase is 10 cm3/mole, and the interfacial energy of the α–β phase is 450mJ/m2. The alloy is being annealed at 350C.

Determine the concentration of Mg atoms in the α–phase just adjacent to the α/β interface for the two particles, considering the effect of curvature on solubility. If these particles are separated by 100nm, what is the concentration gradient between them? Based on the answer, qualitatively describe the evolution of the particles if they were to be annealed?

© meg/aol ‘02

Exercise

CMg H = CMg 1+

γΩRgT

⋅

1R

Thomson–Freundlich equation

CMg H = 10 1+0.45 J

m2 ×10⋅10−6 m3

mol

8.314 Jmole⋅ K

× 623K

⋅ 15⋅10−6m

,

CMg H = 10.0034 wt.%.

For a spherical particle with a radius of 5 µm

© meg/aol ‘02

Exercise

∇C =10.0034−10.034

0− 2.5= 0.0122 [wt.%/µm],

∇C = 122 [wt.%/cm].

The maximum concentration gradient between these particles

CMg H = 10 1+2× 0.45 J

m2 ×10⋅10−6 m3

mol

8.314 Jmole⋅ K

× 623K

⋅ 10.5⋅10−6m

,

CMg H = 10.034 wt.%.

For a spherical particle with a radius of 0.5 µm

© meg/aol ‘02

Key Points • Ostwald Ripening is the process that leads to the growth of large

particles at the expense of the smaller particles. • The phase separation resulting from curvature–induced solubility

changes and the competitive multiparticle diffusion in solids and liquids is termed microstructural phase coarsening.

• Phase coarsening in solids requires diffusional transport to grow large phase domains and dissolve smaller ones.

• A new length scale called–capillary length–is identified and introduced into spherical diffusion solution. It allows the equilibrium solubility at an interface to be related to its curvature through the Thomson–Freundlich effect.

• Kelvin’s equation is the basic relationship between curvature and thermodynamic activity. It is derived using a reversible thermodynamic cycle.

© meg/aol ‘02

Outline

• Continuity equation

• Scale–factor population dynamics

• Particle distribution function

• Volume fraction constraint

• Coarsening kinetics

• Exercise

© meg/aol ‘02

Continuity Equation

∂F R,t( )∂t

+∂

∂RF R,t( )⋅

∂R∂t

= 0

The continuity equation relates the change of a population’s distribution function F(R,t) as particle grow or shrink

F R,t( )dR= NV

0

∞

∫

The integral of the distribution function, F(R,t), as particle grow or shrink

Time rate of increase of the PSD

Net accumulation due to shrinking and growing particles

© meg/aol ‘02

Continuity Equation

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3

Dist

ribut

ion

Func

tion,

F(R

,t)

Normalized Radius, R/<R>

d(R/<R>)

ŽF(R,t)/Žt

Shrinkingparticles

Growingparticles

© meg/aol ‘02

Population Dynamics

ρ ≡

RRThe normalized radius, ρ, is defined as

∂∂t

= ∂ρ∂t

R

∂∂ρ

+ ∂τ∂t

R

∂∂τ

,

∂∂R

= ∂ρ∂R

t

∂∂ρ

.

The partial derivatives with respect to the normalized variables are

∂ρ∂τ

R

∂ f∂ρ

+∂ f∂τ

+

∂ρ∂R

t

∂∂ρ

f ρ,τ( )⋅∂R∂t

= 0

Substituting

© meg/aol ‘02

Population Dynamics

∂ρ∂R

t

= 1R

,

∂ρ∂τ

R

= − R

R2

d Rdτ

.

Normalizing with respect to the growing radius of the average particle, <R>:

Substituting

Ý ρ ≡dρdτ

=1R

d Rdτ

−R

R2

d Rdτ

,

Ý R ≡ d Rdτ

,

© meg/aol ‘02

Population Dynamics

Ý ρ

∂ f∂ρ

+∂ f∂τ

+f ρ,τ( )

R∂ Ý R ∂ρ

= 0The continuity equation

d Rdτ

=1

R2

ρ −1ρ2

.Writing the kinetic equation in terms of

the normalized particle radius, ρ

Ý ρ

∂ f∂ρ

+∂ f∂τ

+f ρ,τ( )

R3

2 ρ −1( )ρ3 −

1ρ2

= 0

Inserting the kinetic equation in the continuity equation

“master equation”

© meg/aol ‘02

Population Dynamics

Ý ρ =

1

R3

1− ρ( )ρ2 −

ρR

d Rdτ

Solving for the time-derivative of ρ,

Ý ρ =

1

R3

1− ρ( )ρ2 − ρ R

2 d Rdτ

or

Ý ρ =

1

R3

1− ρ( )ρ2 −

ρ3

d Rdτ

3

Thus

© meg/aol ‘02

Population Dynamics

1

R3

1− ρρ2 −

ρ3

d R3

dτ

∂ f∂ρ

+∂ f∂τ

+f ρ,τ( )

R3

2 ρ −1( )ρ3 −

1ρ2

= 0

Substituting yields the following population equation

d R

3

dτ>

49

All the particles have negative growth rates

d R

3

dτ<

49

Some particles grow, some shrink, but a monodispersion arises!

d R

3

dτ=

49

All particles shrink relative to the maximum sized particle, ρ=3/2

A

B

C

© meg/aol ‘02

Particle Distribution Function

f ρ,τ( )= g ρ( )⋅ h τ( )Lifshitz and Slyozov suggested that the distribution function could be expressed in product–function form

∂ f∂ρ

τ

= ′ g ρ( )⋅h τ( ),

∂ f∂τ

ρ

= g ρ( )⋅ ′ h τ( ),This implies

1

R3

1− ρρ2 −

ρ3

d R3

dτ

′ g ρ( )⋅h τ( )( )+ g ρ( )⋅ ′ h τ( )( )+g ρ( )⋅ h τ( )

R3

2 ρ −1( )ρ3 −

1ρ2

= 0

Inserting g(ρ), h(τ), and their derivatives into the affine continuity equation

© meg/aol ‘02


1− ρρ2 −

ρ3

d R3

dτ

′ g ρ( )g ρ( )

+

2 ρ −1( )ρ3 −

1ρ2

= − R

3 ′ h τ( )h τ( )

Dividing through by g(ρ) and h(τ) yields the following ODE

1− ρρ2 −

ρ3

d R3

dτ

′ g ρ( )g ρ( )

+

2 ρ −1( )ρ3 −

1ρ2 = λ,

− R3 ′ h τ( )

h τ( )

= λ.

Separating into two ODEs:

ODE for g(ρ)

ODE for h(τ)

The only acceptable value for the volumetric growth rate is 4/9

© meg/aol ‘02


′ g ρ( )g ρ( ) =

dln g ρ( )[ ]dρ

=

−6+ 3ρ + 3λρ3

−3ρ 1− ρ( )−4ρ4

9

For the volumetric growth rate 4/9, the following ODE for g(ρ) is obtained

© meg/aol ‘02

Volume Fraction Constant

4πR3

3F R,t( )d R

0

∞

∫ = VVAccording to mass conservation

R4

ρ3 g ρ( )h τ( )dρ0

ρmax

∫ = 3VV

4π,

R4h τ( )=

3VV

4π1

ρ3 g ρ( )dρ0

ρmax

∫

.

Expressing as the product of the two functions, g(ρ) and h(τ)

The time–dependent function takes the specific form h τ( )=

const.

R4

© meg/aol ‘02

Volume Fraction Constant

′ h τ( )=

−4

R5

d Rdτ

Differentiating with respect to τ

λ =

43

d R3

dτ

=1627

Given that d<R>/dτ=4/9, the separation constant, λ, can be found

g ρ( )= exp−6+ 3ρ +

169

ρ3

−3ρ 1− ρ( )−4ρ4

9

dρ

0

ρ

⌠

⌡

+ ln(A)

Inserting this result

eigenvalue

© meg/aol ‘02

Volume Fraction Constraint

g ρ( )=

49

ρ2 33+ ρ

73 3

3− 2ρ

113e

−2ρ3−2ρ

, 0 ≤ ρ <

32

,

0, ρ >32

.

Evaluating the integral analytically yields the affine particle size distribution, g(ρ), (PSD) function

Maximum particle size

© meg/aol ‘02

LSW Particle Size Distribution

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5

g(ρ)

ρ

Normalized particle distribution function derived by Lifshitz and Slyozov

© meg/aol ‘02

Coarsening Kinetics

R τ( ) 3

− R 0( ) 3=

49

τThe kinetic equation predicts the growth of the average particle, <R>, with time

R τ( )3

=49

τAt long diffusion times,

lc ≡

2γ αβΩmol

RgTReinserting the capillary length, lc ,

tdiff ≡

lc2

DC∞ΩmolIntroducing the diffusion time, tdiff,

© meg/aol ‘02

Coarsening Kinetics

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

PSD

, F(R

,t) [p

artic

les/

cm4]

Particle Radius, R [cmX10 -4]

τ=1

τ=2

τ=8

Particle size distribution versus time (early times)

© meg/aol ‘02

Coarsening Kinetics

0

0.05

0.1

0.15

0.2

0 1 2 3 4 5

PSD

, F(R

,t) [p

artic

les/

cm4]

Particle Radius, R [cmX10 -4]

τ=1

τ=8

τ=27

τ=64

Areas of PSD’s are proportional to the number of particles per unit volume.

Average particle radius grows as the cube-root of time.

Particle size distribution versus time

© meg/aol ‘02

Time(min) Average size (mm) 5

75

1020

33.2

70.7

143.7

Exercise 1. Use the microstructures to determine the dimensional kinetic rate constant relating the average volume of the Sn–rich particles to time, assuming that cube–root of time diffusional kinetics remains valid.

Data are as follows

4π3

R3(t) − R3(0)( )= K tand

If the volume scales linearly with time V t()− V(0) = K t

© meg/aol ‘02

Exercise

4π3

35.36( )3− 16.6( )3( )= K 70× 60( ),

K = 39.52 µm( )3

sec.

At annealing time of 75 min

4π3

71.88( )3− 16.6( )3( )= K 1020× 60( ),

K = 27.10 µm( )3

sec.

At annealing time of 1020 min

© meg/aol ‘02

Key Points • The multiparticle diffusion equation can be applied to the

“continuity equation in size space” to derive population dynamics under competitive diffusion conditions.

• The method of Lifshitz, Slyozov, and Wagner (LSW) is used to derive a self-similar (affine) particle size distribution (PSD).

• Microstructures coarsening under affine conditions appear unchanged to within a scale-factor that grows with time.

• Diffusion-limited coarsening leads to fewer particles per unit volume, having the same volume fraction.

• Microstructural evolution under diffusion-limited conditions leads to the growth of the average particle size as the cube-root of time.

• Life-time predictions can be based on diffusion-limited phase coarsening theory.

© meg/aol ‘02

Outline

• Fick’s first law for multicomponent diffusion

• Fick’s second law for multicomponent diffusion

• Multicomponent diffusion couples

• Square–root diffusivity method

• Approximation using effective binary diffusion coefficients

• Exercise

© meg/aol ‘02

Fick’s First Law for Multicomponent Diffusion

Ji = −

DiCi

kBT∇µ i r ,N1,N2,N3,,Nn( )

The flux of ith component in a multicomponent system may be expressed through the phenomenological flux equation

µ i = µ0 + kB lnai r,N1,N2 ,N3,...Nn( )

The chemical potential of ith component in a multicomponent alloy system is given in general by

∇µ i =kBTai

∂ai

∂xk

k=1

d

∑ ek ,

∇µ i =kBTai

∂ai

∂N j

j =1

n

∑k=1

d

∑∂N j

∂xk

ek .

Applying the gradient operator in any dimensions (d=1, 2, or 3) shows that

© meg/aol ‘02

Fick’s First Law for Multicomponent Diffusion

∇µ i =

Ω kBTai

∂ai

∂N j

j =1

n

∑ ⋅∇N j

Ω

After multiplying and dividing by the molar volume Ω

Ji = −

Di Ni

ai

∂ai

∂N j

j =1

n

∑ ⋅∇CjSubstituting gives Fick’s first law for an n-component alloy

Ji = − Dij[ ]⋅∇CjEquation may be written more compactly in a matrix form

Dij ≡

Di Ni

ai

∂ai

∂N j

With the individual matrix

elements, Dij , defined as

© meg/aol ‘02

Multicomponent Fick’s Law in 1–D

∇Ci =

∇C1

∇C2

∇C3

∇Cn

In 1–D the intrinsic diffusion fluxes, Ji , and the component gradients ∇Cj become column matrices

Ji =

J1

J2

J3

Jn

© meg/aol ‘02


Dij[ ]=

D11 D12 D13 D1n

D21 D22

D31 D33

Dn1 Dnn

These matrices represent vectors in n–dimensional composition space, so the intrinsic diffusivity becomes [n×n] matrix

Dij[ ]=D11 D12 D13

D21 D22 D23

D31 D32 D33

For a ternary alloy n=3, and the intrinsic diffusivity matrix is

© meg/aol ‘02

Multicomponent Fick’s Law in 1–D The diffusivity matrix could be reduced to its independent components (n-1).

∂Nn

∂x= −

∂N1

∂x−

∂N2

∂x

∂Nn−1

∂xThe following differential form applies

∂Cn

∂x= −

∂C1

∂x−

∂C2

∂x

∂Cn−1

∂x

Dividing by the molar volume, yields the corresponding constraint on the concentration gradients

N1 + N2 + N3 +Nn = 1

Recognizing that the sum of the mole fractions is unity,

© meg/aol ‘02


J1 = −D11

∂C1

∂x

− D12∂C2

∂x

− D1n −

∂C1

∂x−

∂C2

∂x

∂Cn−1

∂x

Substituting into Fick’s first law we may obtain the intrinsic diffusion flux for every component

J1 = − D11 − D1n( ) ∂C1

∂x

− D12 − D1n( ) ∂C2

∂x

− D1,n−1 − D1n( ) ∂Cn−1

∂x

Collecting terms gives

© meg/aol ‘02

Multicomponent Intrinsic Diffusivities and Interdiffusion Coefficients

The coefficients multiplying the n-1 independent gradients are intrinsic diffusivities.

Interdiffusion coefficients obtained by inverse methods involve

Intrinsic flux Advective Flux

Fickian Flux Kirkendall drift Flux

© meg/aol ‘02


J 1 = J1 + C1 ⋅u

The interdiffusion flux, Ji

˜ J 1 = − ˜ D 11

∂C1

∂x

− ˜ D 12

∂C2

∂x

− ˜ D 1,n−1

∂Cn−1

∂x

For component 1, the flux can be expressed purely as a sum of Fickian components

˜ J 1 = − D11 − D1n( ) ∂C1

∂x

− D12 − D1n( ) ∂C2

∂x

− D1,n−1 − D1n( ) ∂Cn−1

∂x

intrinsic flux

+ C1 ⋅u1

advective flux

The interdiffusion flux, Ji is comprised of two terms

© meg/aol ‘02


Dij[ ]=22 7.67.8 12.6

⋅10−11 cm2/sec[ ],

wherei = Al , (5≤ CAl ≤ 9.5at.%),

j = Cr, (8≤ CCr ≤19at.%).

Thompson et al. found that in Ni–Cr–Al system the ternary diffusivity could be expressed as the interdiffusion coefficient of the matrix

Ji = − Dij[ ]∂

∂xCj( )Fick’s first law when rewritten

J1 = −D11∂C1

∂x

− D12

∂C2

∂x

,

J2 = −D21∂C1

∂x

− D22

∂C2

∂x

.

The expanded form for a ternary alloy is

© meg/aol ‘02

Fick’s Second Law for Multicomponent Diffusion

2dx

x-dx

C1(x)

C2(x)

J2(x+dx)

J1(x+dx)

J2(x-dx)

J1(x-dx)

x+dx

J1 x − dx( )− J1 x + dx( )=

∂C1

∂t

⋅2dx ⋅1,Mass balance for component 1

© meg/aol ‘02


J1 x( )−

∂J1

∂x

d x− J1 x( )+

∂J1

∂x

dx

=∂C1

∂t

2dx

Expanded as a first order Taylor series

Collecting terms gives −2 ∂J1

∂x

dx =∂C1

∂t

2dx,

−

∂J1

∂x

=∂C1

∂t

or

−∇ ⋅Ji =

∂Ci

∂tIn 3–D mass conservation becomes

© meg/aol ‘02


−∂∂x

−D11∂C1

∂x

− D12

∂C2

∂x

=∂C1

∂t,

− ∂∂x

−D21∂C1

∂x

− D22

∂C2

∂x

= ∂C2

∂t.

Fick’s second law for ternary diffusion

−D11∂2C1

∂x2

− D12

∂2C2

∂x2

=

∂C1

∂t,

−D21∂2C1

∂x2

− D22

∂2C2

∂x2

=

∂C2

∂t.

If Dij is a constant

∂2

∂x2 Dij[ ]⋅Cj( )=∂∂t

Ci( ), (i = 1,2n−1).Multicomponent form for Fick’s second law

© meg/aol ‘02

Multicomponent Diffusion Couples

C(x,t) =

C0

2erfc ξ( )

Fick’s second law for a diffusion couple consisting of two binary alloys

ξ ≡

x4Dt

Where the similarity variable as previously defined is

ξ i ≡

x4Eit

, (i = n −1)Multicomponent similarity variables are of a similare form

© meg/aol ‘02

Square–root Diffusivity Method

Dij[ ]= r[ ]⋅ r[ ]The square–root diffusivity matrix [r] is related to the diffusivity Dij

D11 D12

D21 D22

=r11 r12

r21 r22

⋅r11 r12

r21 r22

For the case of ternary alloy, expands to the form

Dij[ ]

12 = αij[ ] Eij

12 α ij[ ]−1

= r[ ]

The square–root diffusivity matrix [r] is obtained from the matrix operation

© meg/aol ‘02

Approximation Using Binary Diffusion Coefficients

Ji = −DEBDC

i ∂Ci

∂x

(i =1, 2,n)

The multicomponent diffusion is sometimes described as the effective binary diffusion coefficient (EBDC), usually written as DEBDC.

© meg/aol ‘02

Exercises

1. Plot the locations of the following Ni–Cr–Al alloys in the Al–Cr composition space: 1) Ni–8at.%Cr–5at.%Al; 2) Ni–15at.%Cr–9at.%Al; 3) Ni–8at.%Cr–10at.%Al; 4) Ni–17at.%Cr–5at.%Al. Assuming that alloy (1) is the left–hand member of a ternary diffusion couple, indicate the composition vectors (magnitude and direction) for alloys (2), (3), and (4), when they become the right–hand members of the couple.

© meg/aol ‘02

Exercises

0

2

4

6

8

10

12

0 5 10 15 20

Con

cent

ratio

n of

Al,

[at.%

]

Concentration of Cr , [at.%]

(Alloy 1)

(Alloy 2)

(Alloy 3)

(Alloy 4)29.75o

Composition space plot of the four Ni–Cr–Al ternary alloys

© meg/aol ‘02

Exercises

2. A diffusion couple consisting of alloys (1) and (2) is annealed at 1100C for 100 hours. The resulting penetration curves for the distributions of Cr and Al, in the form of typical error function complements. Plot the diffusion path for this pair of ternary alloys.

4

6

8

10

12

14

16

-0.04 -0.02 0 0.02 0.04

Conc. CrConc. Al

Con

cent

ratio

n, [a

t.%]

Distance, [cm]

Concentration versus distance for multicomponent interdiffusion system

© meg/aol ‘02

Exercises

0

5

10

15

20

0 2 4 6 8 10

Conc

entra

tion

Cr, [

at.%

]

Concentration Al [at.%]

Alloy (1)

Alloy (2)

Diffusion path for interdiffusion in a ternary alloy

© meg/aol ‘02

Exercises 3. A diffusion couple consisting of alloys (1) and (4) is annealed at 1100C for 100 hours. The resulting penetration curves for the distributions of Cr and Al are shown below. Plot the diffusion path for this pair of ternary alloys.

-0.04 -0.02 0 0.02 0.044

6

8

10

12

14

16

18

Distance [cm]

Conc

entra

tion,

[at.%

]

Al

Cr

Cr

Al

Concentration versus distance

© meg/aol ‘02

Exercises

0

5

10

15

20

0 2 4 6 8 10

Concentration Al [at.%]

Conc

entra

tion

Cr, [

at.%

]

Alloy (1)

Alloy (4)

S–shaped diffusion path for interdiffusion in a ternary alloy

© meg/aol ‘02

Key Points

• Multicomponent diffusion is normally the “rule” in commercial alloys, which seldom behave as binary alloys.

• The chemical potential gradients depend on all the components, so the diffusivity adopts a tensorial form, expressed as a matrix.

• A type of “coupling” occurs among the fluxes and gradients of each of the components.

• Darken’s diffusion experiment on Fe-C-Si alloys is the most famous example showing “up-hill” diffusion in a ternary alloy.

• The linear solutions to Fick’s laws in one dimension are similar to binary diffusion, except they involve sums of error functions.

• The square-root diffusivity method of Thompson and Morral is adopted to find solutions to multicomponent diffusion problems.

© meg/aol ‘02

Outline

• Background

• Profiler software

• Application to a ternary alloy

• Composition directions, eigenvalues, eigenvectors

• Multicomponent diffusion paths

• Multicomponent concentration profiles

• Exercise

© meg/aol ‘02

Composition Space

C2

[at.%

]

C1 [at.%]

•

•

CL1 CR1

CR2

CL2

∆C0

CR

CL∆C1

∆C2

ψ

Initial composition difference vector ∆C0 between a pair of ternary alloys

© meg/aol ‘02

Composition Space C

2 [a

t.%]

C1 [at.%]

•

•

CL1 CR1

CR2

CL2

∆C0

CR

CL∆C1

∆C2

ψ

m= tanΨ =

∆C2

∆C1

C2 = mC1 + B

B = CL2− tanΨ( )CL1

C2 = tanΨ( )C1 + CL2− tanΨ( )CL1

This equation relates the compositions of alloys along the direction of the composition vector

© meg/aol ‘02

PROFILER

The program Profiler evaluates the square–root diffusivity matrix from input data and calculates numerical solutions to the linear multicomponent diffusion equation.

∆Ci = ∆C0 f i x,t,Ψk( )The program Profiler yields a difference solution to Fick’s second law in the form

∆Ci = CRi− Ci x,t( )The data tabulated and displayed by Profiler are

in the form

f i x,t,Ψk( )= Aij

j =1

n−1

∑ erfc x2ej t

Specifically, this function is the sum of error function complements

© meg/aol ‘02

PROFILER

Diag r[ ]= αij[ ] r[ ]αij−1[ ]

Aij = α ij

−1 α jk cosΨkk=1

n−1

∑The amplitudes of the basis (complementary error) functions Aij are found by the matrix operation

Diag Dij[ ]= αij[ ] Dij[ ]α ij−1[ ]

The [αij] matrix and its inverse will diagonalize the diffusivity matrix and the square–root diffusivity through the standard matrix operation

and

© meg/aol ‘02

Application to Ternary Alloy

C x,t( )= CRi+ ∆C0 f i x,t,Ψ( )

A diffusion field reduces to its simplest multicomponent form in the case of ternary alloys

f i x,t,Ψ( )= Ai1erfc x

2e1 t

+ Ai2 erfc x

2e2 t

where

Note: The coefficients e1 and e2 are the major and minor eigenvalues of the square–root coefficient matrix

© meg/aol ‘02


A11 =r22 − r11 − r11 − r22( )2

+ 4r12r21

4 r11 − r22( )2+ 4r12r21

cosΨ −

r12sinΨ

2 r11 − r22( )2+ 4r12r21

A12 =r11 − r22 − r11 − r22( )2

+ 4r12r21

4 r11 − r22( )2+ 4r12r21

cosΨ +

r12 sinΨ

2 r11 − r22( )2+ 4r12r21

The four amplitude coefficients Aij are given by the following formulas derived using linear algebra

© meg/aol ‘02


A22 =r22 − r11 − r11 − r22( )2

+ 4r12r21

4 r11 − r22( )2+ 4r12r21

sinΨ +

r21cosΨ

2 r11 − r22( )2+ 4r12r21

A21 =r11 − r22 − r11 − r22( )2

+ 4r12r21

4 r11 − r22( )2+ 4r12r21

sinΨ −

r21cosΨ

2 r11 − r22( )2+ 4r12r21

© meg/aol ‘02

Extrema in the Composition Profile

The turning points in the profiles occur at locations that can be determined by setting the derivative with respect to x, equal to 0.

xi

∗( )2=

4t e1e2( )2

e22 − e1

2 ln − Ai1 e2

Ai2 e1

, 0 <

− Ai1e2

Ai2 e1

< 1

The values for x* for component i where the gradient vanishes and extrema occur may be found

© meg/aol ‘02

0

30

60

90

120

210

240

270

300

330

Max FluxΨ=17.33 deg

Zero FluxΨ=-72.67 deg

Max Grad.Ψ=-24.42 deg

Zero Grad.Ψ=65.6 deg

MajoreigenvalueΨ=21.31 deg

MinoreigenvalueΨ=-54.27 deg

Orientations in composition space

Ni–Al–Cr

In multicomponent alloy diffusion a zero gradient does not correspond to a zero flux plane!

© meg/aol ‘02

Composition Directions, Eigenvalues, Eigenvectors

E1 =12

D11 + D22 + D∆( ),

E2 =12

D11 + D22 − D∆( ),

D∆ ≡ D11 − D22( )2+ 4D12D21.

The eigenvalues may be calculated from the matrix elements

e1 = E1,

e2 = E2.

The eigenvalues, ei, of the r-matrix may be found by taking the square-root of the eigenvalues of the D-matrix.

tanΨmajoreigenvector

=e1 − r11

r12

,

tanΨminoreigenvector

= e2 − r11

r12

.

The orientations of the major and minor eigenvectors in the C1–C2 composition space

© meg/aol ‘02

Properties of the [r] Matrix

tanΨmaxflux (1) =r12

r11

,

tanΨzeroflux (1) =−r11

r12

,

tanΨzerograd(1) = r22

r12

,

tanΨmaxgrad(1) =−r12

r22

.

Relationships for special bearing angles in the 1–2 component composition space of a 1–2–3 ternary alloy.

© meg/aol ‘02

Composition Space: Zero Gradient

0

0.05

0.1

0.15

0.2

0 0.05 0.1 0.15 0.2

Con

cent

ratio

n Al

[At.%

]

Concentration Cr [At.%]

Alloy 1

Alloy 2

Zero Grad Cr : tanψ=r22

/r12

78 deg

© meg/aol ‘02

Penetration Curves

Zero Gradient

8

10

12

14

16

18

20

-0.04 -0.02 0 0.02 0.04

Zero Grad: (10Cr-10Al-80Ni)/(12Cr-19Al-69Ni)

Cr At.%

Al At.%

Con

cent

ratio

n, C

i, [At

.%]

Distance, x, [cm]

t=100 hr.

Gradient of Cr vanishes

© meg/aol ‘02

Fluxes: Zero Gradient

-7

-6

-5

-4

-3

-2

-1

0

1

-0.04 -0.02 0 0.02 0.04

Zero Grad Cr-Al-Ni

Flux(Cr)

Flux(Al)

Flux

, Ji, [

At.%

-cm

/s] x

108

Distance, x, [cm]

t=100 hr.

© meg/aol ‘02

ZERO–FLUX PLANES (ZFP) • A zero–flux plane (ZFP) is the location in the diffusion zone where the flux of a component vanishes.

• The direction of the component’s flux vector reverses on either side of a ZFP.

Flux, Ji

0

ZFP

• ZFPs slow the achievement of thermodynamic equilibrium.

Flux, Ji

0

ZFP

© meg/aol ‘02

Properties of the [r] Matrix

Dij[ ]=

4.0 2.01.1 2.0

×10−11[cm2/sec]

The diffusivity tensor for Ni–8 at. %, Al–10 at.%, Bal. Cr is:

Where 1=CAl 2=CCr

r[ ]=

6.16 1.921.06 4.24

×10−6 [cm/ sec]

The corresponding square–root diffusivity for this ternary system is

© meg/aol ‘02

Multicomponent Diffusion Paths

0

20

40

60

80

100

0 20 40 60 80 100

Ni [

at.%

]

Cu [at. %]

CL

A

α4B

CD

E

F

G

H

I

J

Locus of alloy composition

ψ

© meg/aol ‘02

0

20

40

60

80

100

0 20 40 60 80 100

Ni [

at. %

]

Cu [at.%]

A

α4BCD

E

F

G

H

I

J

minor eigenvector

majoreigenvector

zero gradient Diffusion Paths

ZERO GRADIENT

MAJOR EIGENVECTOR

MINOR EIGENVECTOR

Paths consist of all compositions encountered along the diffusion couple.

MAX GRADIENT

© meg/aol ‘02

Concentration Profiles

25

30

35

40

45

50

55

15

20

25

30

35

40

45

50

-0.02 -0.015 -0.01 -0.005 0 0.005 0.01 0.015 0.02

Cu ProfilerCu D&K

Ni ProfilerNi D&K

Cu [a

t. %

] Ni [at.%

]

Distance [cm]

Cu–Ni–Zn Alloys

© meg/aol ‘02

Exercise

A diffusion couple consisting of alloys Ni–8at.%Cr–5at.%Al and Ni–17at.%Cr–5at.%Al is annealed at 1100C for various time periods (from t=1min. to 20,000hrs). Plot representative Al penetration profiles for this pair of ternary alloys, and determine the rate constant for the spreading of aluminum atoms.

© meg/aol ‘02

Exercise

4.4

4.6

4.8

5

5.2

5.4

5.6

-0.2 -0.1 0 0.1 0.2

Con

cent

ratio

n, A

l [at

.%]

Distance [cm]

2x104 h.

104 h.

104 h.

2x104 h.

2x103 h.

2x103 h.

102 h.

102 h.

Penetration curves for Al in a Ni–Cr-Al ternary

© meg/aol ‘02

Exercise

0.0001

0.001

0.01

0.1

0.01 0.1 1 10 100 1000 104 105

Loca

tion

of e

xtre

mum

in A

l pro

file,

x extr [c

m]

Annealing time, t [h]

xextr

=0.001* t0.5

Positions from Matano interface as a function of annealing time

© meg/aol ‘02

Key Points • Profiler® is a public domain program that evaluates the [r] matrix

from data, and computes the multicomponent diffusion field for each of the independent components.

• The program computes the numerical difference solution by determining the eigenvalues of the [r] matrix, and the amplitudes of the n-1 independent error function complements.

• The Euler angles specifying the “directions” in composition space between the left- and right-hand end members determine the qualitative behavior of the solution.

• These include: – Max gradient – Max flux – Zero gradient – Zero flux – Diffusion path

glicksman slides

Documents