Mathematical modeling of particle nucleation and growth in...

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Delft Institute of Applied MathematicsDelft University of Technology

Mathematical modeling of particlenucleation and growth in metallicalloys

Dennis den Ouden

14-12-2009

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Delft Institute of Applied Mathematics

Contents

• Something about metallurgy• A model for nucleation• A model for deformations• Combining the models• Results• Conclusion

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Something about alloys

Binary Ternary Quaternary Complex

Quasi-Binary

Quasi-Binary

Quasi-Binary Quasi-Ternary

Quasi-Ternary

Quasi-Quaternary

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Something about alloys

Binary Ternary Quaternary Complex

Quasi-Binary

Quasi-Binary

Quasi-Binary Quasi-Ternary

Quasi-Ternary

Quasi-Quaternary

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Something about deformations

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Something about deformations

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Models for nucleation

Two models, with small differences• Myhr and Grong (2000)• Robson et al. (2003)

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Models for nucleation

Two models, with small differences• Myhr and Grong (2000)• Robson et al. (2003)

Comparison:• Basic model == Myhr and Grong (2000)• Adapted using Robson et al. (2003)

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Governing DE

Changein time

=Change

by growth+

Productionrate

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Governing DE

Changein time

=Change

by growth+

Productionrate

⇓ ⇓ ⇓

∂N

∂t= −

∂ (Nv)

∂r+ S

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Unknowns

• Growth rate v

• Production term S

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Growth of spherical particles

Directly influence by

• Mean concentration C

• Interface concentrationCi

• Internal concentrationCp

• Diffusion coefficient D

Cp C

Ci

Matrix

Particle

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Growth of spherical particles

v(r, t) =C − Ci

Cp − Ci

D

r Cp C

Ci

Matrix

Particle

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A special particle

A particle with radius r∗ that will neither grow ordissolve:

v(r∗) = 0

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A special particle

A particle with radius r∗ that will neither grow ordissolve:

v(r∗) = 0

Solved for r∗:

r∗ =2γαβVm

RT

(

ln

(

C

Ce

))−1

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A special particle

A particle with radius r∗ that will neither grow ordissolve:

v(r∗) = 0

Solved for r∗:

r∗ =2γαβVm

RT

(

ln

(

C

Ce

))−1

Definition: Critical particle radius

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Production term

• Indicates the number of particles thatnucleate over the whole domain

• Influenced by critical radius r∗

• Influenced by nucleation rate j

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Production term

• Indicates the number of particles thatnucleate over the whole domain

• Influenced by critical radius r∗

• Influenced by nucleation rate j

Kampmann et al. (1987):

S(r, t) =

{

j(t) if r = r∗ + ∆r∗,0 otherwise.

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Nucleation rate

The number of particles that nucleate with radiusr∗ + ∆r∗:

• Influenced by diffusion• Only if some barrier has been overcome

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Nucleation rate

The number of particles that nucleate with radiusr∗ + ∆r∗:

• Influenced by diffusion• Only if some barrier has been overcome

j = j0 exp

(

−∆G∗

het

RT

)

exp

(

−Qd

RT

)

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Nucleation energy barrier

• Chemical composition• Misfit strain energy

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Nucleation energy barrier

• Chemical composition• Misfit strain energy

∆G∗het =

A30

(∆Gv + ∆Gm

s)2

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Model overview

• Governing DE:

∂N

∂t= −

∂ (Nv)

∂r+ S

• Source term:

S(r, t) =

{

j(t) if r = r∗ + ∆r∗,0 otherwise.

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Model overview

• Governing DE:

∂N

∂t= −

∂ (Nv)

∂r+ S

• Growth rate:

v =C − Ci

Cp − Ci

D

r

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Elastic deformations

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Elastic deformations

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Assumptions

Rotation symmetry:• No deformations in tangential direction• No deformation at center axis in radial

direction• All derivatives in tangential direction vanish

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Assumptions

Rotation symmetry:• No deformations in tangential direction• No deformation at center axis in radial

direction• All derivatives in tangential direction vanish

uθ = 0 uη(0, θ, z) = 0∂(.)

∂θ= 0

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Strain and deformation

Chau and Wei (2000):

εηη =∂uη

∂ηεθθ =

η

εzz =∂uz

∂zεηθ = 0

εηz =1

2

(

∂uη

∂z+

∂uz

∂η

)

εθz = 0

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Stress and strain

Hook’s Law:

σαβ = δαβλ (εηη + εθθ + εzz) + 2µεαβ

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Stress and strain

Hook’s Law:

σαβ = δαβ λ (εηη + εθθ + εzz) + 2 µ εαβ

Stiffness matrix Shear modulus

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Force balance

Jaeger et al. (2007):

∂σηη

∂η+

∂σηz

∂z+

σηη − σθθ

η+ bη = 0

∂σηz

∂η+

∂σzz

∂z+

σηz

η+ bz = 0

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Boundary conditions

• Symmetry condition:

uη(0, θ, z) = 0

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Boundary conditions

• Symmetry condition:

uη(0, θ, z) = 0

• Fixed boundaries:

uα(η, θ, z) = 0

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Boundary conditions

• Symmetry condition:

uη(0, θ, z) = 0

• Fixed boundaries:

uα(η, θ, z) = 0

• Moving boundaries:(

σ(η, θ, z))

α· n = fα(η, θ, z)

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Coupling the models

Remember the nucleation energy barrier:

∆G∗het =

A30

(

∆Gv + ∆Gms

)2

Misfit strain energy

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Coupling the models

Remember the nucleation energy barrier:

∆G∗het =

A30

(

∆Gv + ∆Gms

)2

Misfit strain energy

Question:

Is there something like elastic strain energy?

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Coupling the models (2)

Answer:YES!!!

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Coupling the models (2)

Answer:YES!!!

Solution:

∆Gels =

1

2σ : ε

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Coupling the models (2)

Answer:YES!!!

Solution:

∆Gels =

1

2σ : ε

and:

∆G∗het =

A30

(

∆Gv + ∆Gms +∆Gel

s

)2

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Coupling the models (3)

Question:Is there also a reverse coupling?

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Coupling the models (3)

Question:Is there also a reverse coupling?

Answer:YES !!!

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Coupling the models (3)

Question:Is there also a reverse coupling?

Solution by Pal (2005):

µ = µm +

(

15(1 − νm)(µp − µm)

2µp(4 − 5νm) + µm(7 − 5νm)

)

µm f

E = Em + (10β1(1 + νm) + β2(1 − 2νm)) Em f

λ = µE − 2µ

3µ − E

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Recap

Two models:• Nucleation model• Elastic model

Two couplings:• From elastic to nucleation• From nucleation to elastic

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Numerical methods

Nucleation model:• Upwind scheme

• IMEX-θ method with θ = 1

2

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Numerical methods

Nucleation model:• Upwind scheme

• IMEX-θ method with θ = 1

2

(

I −1

2

∆t

∆rAn

)

~Nn+1 =

(

I +1

2

∆t

∆rAn

)

~Nn + ∆t~Sn

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Numerical methods (2)

Elastic model:• Finite Element Method• Linear elements• Use of rotation symmetry

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Numerical methods (2)

0 0.5 1 1.5 2 2.5 3

x 10−3

0

0.005

0.01

0.015

0.02

0.025

0.03z

(m)

η (m)

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Numerical methods (2)

Equation:[

Sηη Sηz

Szη Szz

][

uz

]

=

[

qz

]

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Numerical methods

Algorithm:

1. Set all constants;

2. Set all initial values;

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Numerical methods

Algorithm:

1. Set all constants;

2. Set all initial values;

3. For each time step:(a) Calculate elastic parameters;(b) Build matrices for elastic deformation;(c) Calculate elastic deformations;(d) Calculate elastic strain energy;

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Numerical methods

Algorithm:. . .

3. For each time step:. . .

(e) For each point:i. Calculate nucleation parameters;ii. Calculate matrices for nucleation;iii. Calculate nucleation.

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Simulation

Material:• Aluminum alloy AA 6082• Mg2Si particles

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Simulation

Material:• Aluminum alloy AA 6082• Mg2Si particles

Shape:• Cylindrical• Height 30 millimeter• Radius 3 millimeter

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Simulation (2)

Time• Total of 3000 seconds• Time step of 0.5 seconds

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Simulation (2)

Time• Total of 3000 seconds• Time step of 0.5 seconds

Test:• Tensile test• Bottom axial and radial fixed• Top radial fixed

• Axial force at top of 6 million N/m2

• Sides free

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Typical deformations: Axial

0 0.5 1 1.5 2 2.5 3

x 10−3

0

0.005

0.01

0.015

0.02

0.025

0.03

0

1

2

3

4

5

6

7

8x 10

−6Deformation uz (m)z

(m)

η (m)

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Typical deformations: Radial

0 0.5 1 1.5 2 2.5 3

x 10−3

0

0.005

0.01

0.015

0.02

0.025

0.03

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3x 10

−7

z(m

)

η (m)

Deformation uη (m)

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Typical deformations: Energy

0 0.5 1 1.5 2 2.5 3

x 10−3

0

0.005

0.01

0.015

0.02

0.025

0.03

2

4

6

8

10

12

14x 10

4Elastic strain energy (N/m2)z

(m)

η (m)

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Nucleation results: Nucleation rate

0

1

2

3

x 10−3

0

0.01

0.02

0.0310

0

101

102

103

0.5

1

1.5

2

2.5

3

x 10−7Particle nucleation rate

time

(s)

Per

cent

age

(%)

z (m) η (m)

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Nucleation results: Number density

0

1

2

3

x 10−3

0

0.01

0.02

0.0310

0

101

102

103

0.5

1

1.5

2

2.5

3

x 10−7Particle number density

time

(s)

Per

cent

age

(%)

z (m) η (m)

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Nucleation results: Concentration

0

1

2

3

x 10−3

0

0.01

0.02

0.0310

0

101

102

103

−2

−1.8

−1.6

−1.4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0x 10

−7Mean concentration

time

(s)

Per

cent

age

(%)

z (m) η (m)

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Reflection

Are the results anomalies during simulation?

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Reflection

Are the results anomalies during simulation?

Increase force to test for similar behavior.

F = 6 × 109N

m2

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Reflection

Are the results anomalies during simulation?

Increase force to test for similar behavior.

F = 6 × 109N

m2

Physically no longer elasticity

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New elastic stain energy

0 0.5 1 1.5 2 2.5 3

x 10−3

0

0.005

0.01

0.015

0.02

0.025

0.03

1

2

3

4

5

6

7

8

9

10

11

12x 10

10Elastic strain energy (N/m2)z

(m)

η (m)

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Nucleation results: Number density

0

1

2

3

x 10−3

0

0.01

0.02

0.0310

0

101

102

103

0.005

0.01

0.015

0.02

0.025

0.03

0.035

Particle number density

time

(s)

Per

cent

age

(%)

z (m) η (m)

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Conclusions

• Two separate nucleation models combined

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Conclusions

• Two separate nucleation models combined• Formulated model for elastic deformations

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Conclusions

• Two separate nucleation models combined• Formulated model for elastic deformations• Coupling between nucleation and

deformations

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Conclusions

• Two separate nucleation models combined• Formulated model for elastic deformations• Coupling between nucleation and

deformations• Simulations show influence of deformations

on nucleation

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Recommended future work

• Extension to multiple particle configurations

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys• Improving numerical techniques

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys• Improving numerical techniques• Comparison with experimental data

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys• Improving numerical techniques• Comparison with experimental data• Including plastic deformations

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys• Improving numerical techniques• Comparison with experimental data• Including plastic deformations• Including homogeneous nucleation

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Delft Institute of Applied Mathematics

Recommended future work

• Extension to multiple particle configurations• Adaption to other alloys• Improving numerical techniques• Comparison with experimental data• Including plastic deformations• Including homogeneous nucleation• Including grain prediction models

References

Chau, K.T. and Wei, X.X. . Finite solid circular cylinders sub-

jected to arbitrary surface load. Part I – analytic solution.

International Journal of Solids and Structures, 37(40):5707–

5732, 2000.

Jaeger, J.C. , Cook, N.G.W. , and Zimmerman, R.W. .

Fundamentals of rock mechanics. Wiley, Blackwell, 2007.

Kampmann, R. , Eckerlebe, H. , and Wagner, R. . In:

Materials Research Society Symposium Proceedings, vol-

ume 57, page 525. MRS, 1987.

Myhr, O. R. and Grong, Ø. . Modelling of non-isothermal trans-

formations in alloys containing a particle distribution. Acta

Materialia, 48(7):1605–1615, 2000.

Pal, R. . New models for effective Young’s modulus of partic-

ulate composites. Composites Part B: Engineering, 36(6-7):

513–523, 2005.

Robson, J.D. , Jones, M.J. , and Prangnell, P.B. . Extension of

the N-model to predict competing homogeneous and hetero-

geneous precipitation in Al-Sc alloys. Acta Materialia, 51(5):

1453–1468, 2003.

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