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© 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
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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
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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
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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
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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)
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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.
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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
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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
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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( )
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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.
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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
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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
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Fick’s Second Law (Linear Diffusion Equation)
− ∇⋅ −D ∇C( ) = ∂ C∂t
or ∇2 C =1D
∂C∂t
D ∇⋅∇C( ) = ∂C
∂tFick’s 2nd law
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Fick’s Second Law (Cartesian Coordinates)
∇2 ≡
∂2
∂x2 +∂2
∂y2 +∂2
∂z2
D
∂2 C∂x2 +
∂2C∂y2 +
∂2C∂z2
=
∂C∂t
Laplacian
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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
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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
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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)
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Important Diffusion Symmetries
Axisymmetric (radial) Flow
D ∂2 C∂r2
+1r
∂C∂r
= ∂C
∂t
1r
∂∂r
rD ∂C∂r
= ∂C
∂t
(D = constant)
(D = variable)
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Important Diffusion Symmetries
Spherical symmetry
D ∂2 C∂r2
+2r
∂ C∂r
= ∂C
∂ t
1r2
∂∂r
r2D ∂C∂r
= ∂C∂ t
(D = constant)
(D = variable)
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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
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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!
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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
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• 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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
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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
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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
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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
Binary Diffusivity Tensor
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
Binary Diffusivity Tensor
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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
Binary Diffusivity Tensor
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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
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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:
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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!
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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!
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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
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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
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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
Symmetry Operations for Diffusivity Tensors
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
Symmetry Operations for Diffusivity Tensors
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
Symmetry Operations for Diffusivity Tensors
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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
In[48]:= InverseLaplaceTransform @%, s, t D
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
In[7]:= InverseLaplaceTransform @%, s, t D
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
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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.
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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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
In[2]:= InverseLaplaceTransform @%, s, t D
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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
Instantaneous line source in 3-D
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
Instantaneous line source in 2-D
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
Instantaneous line source in 2-D
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
Instantaneous ring source in 2-D
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
Instantaneous ring source in 2-D
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
Instantaneous ring source in 2-D
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
Distance, r, units of 2(Dt)1/2
C(r,t
)/Mri
ng
B
C(r,
t)/M
ring
Distance, r, units of 2(Dt)1/2
(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
Continuous point sources
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
.
Continuous point sources
30 303 3
2(Dt) 1/2=0.1
1 1
1/r1/r C(r)/C
0
.
© meg/aol ‘02
Continuous point sources
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
Time-continuous line source in 2-D
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
Time-continuous line source in 2-D
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
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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
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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 ]
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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.
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Module 7: Diffusion–Reaction
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
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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.
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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
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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.
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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
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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:
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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
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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
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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.
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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
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Reaction-Diffusion Schemes
reaction control
mixed control
diffusion control
C0 C0
C2
Ceq
O l x
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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:
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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
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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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
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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
Fictitious Image Sources
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
Fictitious Image Sources
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
Fictitious Image Sources
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
Application of Fourier Series
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
Application of Fourier Series
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
Application of Fourier Series
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
Application of Fourier Series
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
Application of Fourier Series
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
Application of Fourier Series
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
Application of Fourier Series
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
Dimensionless Distance, (x/l)
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
Dimensionless Distance, (x/l)
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
Codastefano’s Method
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
Codastefano’s Method
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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
Spherical Diffusion Solution
∂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
Spherical Diffusion Solution
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
Quasi-Static Growth and Dissolution
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
Quasi-Static Growth and Dissolution
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
− π erfc 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,
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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
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The Characteristic Equation
∂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
− π erfc Ksphere( )
e−η2
η 2
The concentration derivative with respect to η,
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The Characteristic Equation
dCα
dη
η= Ksphere
= −Cα /β − C∞
e− K sphere2
Ksphere
− π erfc 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
− π erfc Ksphere( )
e− K sphere2
K sphere2 = 0
Diffusion flux at interface Interface “rejection”
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The Characteristic Equation
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
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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
Dimensionless Supersaturation, S
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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
∞
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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
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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
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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
Dimensionless Supersaturation, S
β
β
Λ
Dimensionless Supersaturation, S
Rel
ativ
e B
ound
ary
Laye
r Thi
ckne
ss Λ
/ K
sphe
re
Note how boundary layer decreases with increasing saturation
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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
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Exercise
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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
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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
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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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
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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
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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)]
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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
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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
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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
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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
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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.
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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
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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
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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
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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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
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
Boltzmann’s Transformation
∂∂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
Boltzmann’s Transformation
−
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
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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.
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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
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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 ×
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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
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
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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
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Combining the above equations, and consider independent Markov sequences
Nk
k=0
n
∑ = 2n
One-Dimensional Random Walks
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,
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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.
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One-Dimensional Random Walks
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
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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
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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
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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
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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
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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!
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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
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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
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1-D Random Walk Java Applet
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2-D Random Walk Java Applet
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Random Walks in Higher Dimensions
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
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Random Walks in Higher Dimensions
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
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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
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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
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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
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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
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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
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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
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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.
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Module 13: Structure and Diffusion
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
Constraints Imposed by Crystal Structure
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
Constraints Imposed by Crystal Structure
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
Constraints Imposed by Crystal Structure
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
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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
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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
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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:
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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
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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:
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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( )
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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:
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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:
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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
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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
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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
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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)
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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:
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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
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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.
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Module 14: Correlation Effects in Diffusion
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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Outline
• Markov sequences and correlated walks
– Jumping probability estimate
• Estimates of the correlation factor: pure materials
• Computation of correlation factors in pure crystals
• Exercises
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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
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Jumping Probability Estimate
Correlated exchanges between “tagged” tracer atom and a lattice vacancy on an adjacent crystal plane
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Jumping Probability Estimate
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
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Jumping Probability Estimate
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
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Jumping Probability Estimate
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
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Jumping Probability Estimate
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( )
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Jumping Probability Estimate
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
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Jumping Probability Estimate
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
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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
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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
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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.
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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.
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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.
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Module 15: Vacancy–Assisted Diffusion
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
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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
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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
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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*
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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
Distribution Function for E & p
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
Distribution Function for E & p
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
Distribution Function for E & p
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
Distribution Function for E & p
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
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Empirical Methods in Diffusion
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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:
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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
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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
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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
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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
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LeClaire’s Correlation
Concentration dependencies of D for isomorphous systems (Birchenall)
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LeClaire’s Correlation
Correlation of D and Q with composition in γ–Fe-C and Fe–Mn alloys (Birchenall)
© meg/aol ‘02
LeClaire’s Correlation
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.)
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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.
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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
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Module 16: Diffusion in Dilute Alloys
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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Outline
• Five–frequency exchange model
• Isotope effect
• Diffusion of H+ and D+ with trapping
• Saturable diffusion traps
• Irreversible traps
• Exercises
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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
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Five-Frequency Exchange Model
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
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Five-Frequency Exchange Model
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
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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
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Five-frequency Exchange Model
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,
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Five-frequency Exchange Model
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
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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
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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
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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.
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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)
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Diffusion of H+ and D+ with Trapping
H+
H2H2
H+
Lattice saturated with H+
H2 trapped in a 3-D void (unsaturable trap)
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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
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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
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Fick’s Law with Trapping
∂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
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Fick’s Law with Trapping
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
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Saturable Diffusion Traps
Schematic of H+ Trapped
α β
At an internal interface
H+
At the core of an edge dislocation
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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
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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.
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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
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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
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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.
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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
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• 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
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Module 17: Kirkendall Effect
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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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
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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
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Kirkendall’s Diffusion Couple
70-30 Brass
2.5mm Mo wires0.13mm diameter
Cu
h(t)
Schematic of the Smigelskas–Kirkendall diffusion couple
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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
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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
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Kirkendall Experiment Kirkendall–Smigelskas experiment showing inert Mo
wire–marker positions versus time
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Significance of Kirkendall Marker Motion
x marker frame
Um
marker
observer X-scale
crystal lattice
Marker motions observed during Kirkendall Diffusion
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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.
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Control volume for Kirkendall mass advection Um
Jnet Ω∆t
unit area
marker plane
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Nonreciprocal Atomic Fluxes
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,
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Nonreciprocal Atomic Fluxes
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
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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
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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
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-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
Atomic fluxes during uncompensated diffusion DA = 1.33 × DB
© 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
Atomic fluxes during uncompensated diffusion DA = 0.67 × DB
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Intrinsic Diffusivities:Vacancy Wind
(dCV/dt)<0
(a)
(dCV/dt)>0
(b)
Vacancies and lattice planes are created
Vacancies and lattice planes are annihilated
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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
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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.
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Diffusion–Advection
Z
X
Y
z
x
y
P
u t→
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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
Mass Conservation with Diffusion–Advection
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
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Mass Conservation with Diffusion–Advection
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
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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
Darken’s Analysis of the Kirkendall Effect
∂∂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′
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Darken’s Analysis of the Kirkendall Effect
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
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Darken’s Analysis of the Kirkendall Effect
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
Darken’s Analysis of the Kirkendall Effect
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
Darken’s Analysis of the Kirkendall Effect
∂∂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
Darken’s Analysis of the Kirkendall Effect
∂∂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
Darken’s Analysis of the Kirkendall Effect
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
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Strain Effects
Kirkendall porosity
Cu6Sn5
Cu3Sn
Sn–Ag Solder
Cu
© 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.
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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
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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
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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.
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Module 18: Influence of Solution Ideality
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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Outline
• Generalized forces
• Atom mobilities
• Chemically inhomogeneous systems
• Limiting cases
• Tracer diffusion
• Exercise
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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
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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
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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.
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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
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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
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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.
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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,
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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
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Chemically Inhomogeneous Systems
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
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Chemically Inhomogeneous Systems
∇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,
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Ji = −Di∇Ci = −RgT 1+∂ lnγ i
∂ ln Ni
Bi Di
⋅∇CiThe intrinsic diffusivity appears in flux equation
Chemically Inhomogeneous Systems
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
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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
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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
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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
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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
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Interdiffusion Coefficient
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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
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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
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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
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-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
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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
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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.
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Module 19: Diffusional Anelasticity
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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Outline
• Background
• Standard linear solid
• Internal friction
• Interstitial sub–populations in BCC
• Reorientation of atom pairs in FCC materials
• Internal friction methods
• Exercises
• Summary
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Standard Linear Solid
Spring and dashpot model of a three–element standard linear solid
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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
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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
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Cyclic Stress and Strain
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
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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.
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Cyclic Stress and Strain
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
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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
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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
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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
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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!
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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)
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Geometry of Interstitial Sites in BCC
X
Z
Y
X-site
Y-site
Z-site
Assume that a uniaxial stress σzz is applied.
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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( )
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Interstitial Subpopulations in BCC
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
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Interstitial Subpopulations in BCC
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
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Interstitial Subpopulations in BCC
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
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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
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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.
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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.
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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
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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.
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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
Module 20: Field–Assisted Diffusion
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
© 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
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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
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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
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Guard–ring circuit for ionic conductivity measurements
Recor der
Amp.
Speci men+
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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
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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
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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
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Silicate Network with Mobile Cations
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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
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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
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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
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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
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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
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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
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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( )
Vacancies in Thermal Equilibrium
or
Simplifying
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Di = kBTLii
Ci
DA = kBTLAA
CA
,
DB = kBTLBB
CB
,
DA = DB.
Vacancies in Thermal Equilibrium
The intrinsic diffusivities are
The intrinsic diffusivities can be expressed as
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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.
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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?
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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
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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
Module 21: Multiparticle Diffusion: Capillary Effects
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
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Outline
• Length scales in microstructures
• Kelvin’s equation
• Dimensionless diffusion potential
• The mean–field kinetic equation
• Experimental observations
• Exercise
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Length Scale in Microstructure
Early Microstructure Later coarsened stage
Note: Microstructural phase coarsening is a process generically termed Ostwald ripening.
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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
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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).
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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
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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
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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
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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
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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
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.
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
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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
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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
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The Mean–Field Kinetic Equation
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
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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
The Mean–Field Kinetic Equation
© meg/aol ‘02
Experimental Observations (After Meloro)
Sn-Bi Alloy
t = 0 min
t = 15 min
t = 2 h
t = 240 h
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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
Experimental Observations (After Mani)
2 h 13 h
31 h 120 h
Sn–Pb Alloys
© 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?
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Exercise
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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
Module 22: Population Dynamics
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
© meg/aol ‘02
Outline
• Continuity equation
• Scale–factor population dynamics
• Particle distribution function
• Volume fraction constraint
• Coarsening kinetics
• Exercise
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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
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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
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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
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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”
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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
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Population Dynamics Three behaviors of the normalized kinetic equation
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
Particle Distribution Function
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
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Particle Distribution Function
′ 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
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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
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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
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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
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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
Lifshitz–Slyozov Theory
Lifshitz–Slyozov Theory
99% ρ<1.33
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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
Experimental Observations
© 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
Module 23: Multicomponent Diffusion
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
© 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
Multicomponent Fick’s Law in 1–D
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
Multicomponent Fick’s Law in 1–D
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
Darken’s multicomponent diffusion experiment
© 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
Multicomponent Intrinsic Diffusivities and Interdiffusion Coefficients
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
Multicomponent Intrinsic Diffusivities and Interdiffusion Coefficients
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
Fick’s Second Law for Multicomponent Diffusion
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
Fick’s Second Law for Multicomponent Diffusion
−∂∂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
Module 24: Multicomponent Diffusion: Profiler Program
DIFFUSION IN SOLIDS
Professor Martin Eden Glicksman Professor Afina Lupulescu
Rensselaer Polytechnic Institute Troy, NY, 12180
USA
© 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
Application to Ternary Alloy
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
Application to Ternary Alloy
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
Recommended