Particle nucleation and coarsening in aluminum...
Transcript of Particle nucleation and coarsening in aluminum...
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Delft Institute of Applied MathematicsDelft University of Technology
Particle nucleation and coarseningin aluminum alloysA literature study
D. den Ouden
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Contents
This presentation will be about:
• Alloys• Diffusion• Particle nucleation and coarsening• Elastic deformations• Conclusions• Future work
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Alloys
Alloys can be characterized by• the solvent metal, such as
• Aluminum• Lead• Iron
• the number of elements• by the apparent number of elements
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By apparent number
Binary Ternary Quaternary Complex
Quasi-Binary
Quasi-Binary
Quasi-Binary Quasi-Ternary
Quasi-Ternary
Quasi-Quaternary
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Diffusion
• Interstitial
⇒
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Diffusion
• Interstitial• Substitutional
⇒
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Nucleation of particles
Directly influenced by
• Mean concentration C̄
• Equilibrium concentration Ce
• Temperature T
• Activation energy for diffusion Qd
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Nucleation of particles
Directly influenced by
• Mean concentration C̄
• Equilibrium concentration Ce
• Temperature T
• Activation energy for diffusion Qd
Formula:
j(t) = j0 exp
(
−
(
A0
RT
)3(
1
ln(C̄/Ce)
)2)
exp
(
−Qd
RT
)
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Growth of particles
Directly influenced by
• Mean concentration C̄
• Interface concentration Ci
• Internal concentration Cp
• Diffusion coefficient D
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Growth of particles
Directly influenced by
• Mean concentration C̄
• Interface concentration Ci
• Internal concentration Cp
• Diffusion coefficient D
Formula:
v(r, t) =C̄ − Ci
Cp − Ci
D
r
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Critical particles
Particles with no growth:
v(r∗, t) = 0
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Critical particles
Particles with no growth:
v(r∗, t) = 0
or equivalent:
r∗(t) =2σVm
RT
(
ln
(
C̄
ce
))−1
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Number of particles
Directly influenced by• Growth rate v
• Nucleation rate j
• Critical radius r∗
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Number of particles
Directly influenced by• Growth rate v
• Nucleation rate j
• Critical radius r∗
Continuity equation:
∂N
∂t= −
∂Nv
∂r+ S
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Source term
Influenced by• Critical radius r∗
• Nucleation rate j
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Source term
Influenced by• Critical radius r∗
• Nucleation rate j
Definition
S(r, t) =
{
j(t) if r = r∗ + ∆r∗,0 otherwise.
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Relations
Define φ by
φ(r, t) =N(r, t)
∆r
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Relations
Define φ by
φ(r, t) =N(r, t)
∆r
Then the particle volume fraction equals
f(t) =
∫
∞
0
4
3πr3φ(r, t) dr
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Relations
Define φ by
φ(r, t) =N(r, t)
∆r
Then the particle volume fraction equals
f(t) =
∫
∞
0
4
3πr3φ(r, t) dr
and the mean concentration
C̄(t) =C0 − Cpf(t)
1 − f(t)
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Simulation
Material: Aluminum alloy AA 6082• 0.9 wt% Silicon• 0.6 wt% Magnesium
Temperature: 180◦ C
Initial condition: No particles
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Distribution function φ
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Derived quantities
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Derived quantities
100
102
104
106
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Particle volume fraction
part
icle
vol
ume
frac
tion
(%)
time (s)
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Derived quantities
100
102
104
106
0
10
20
30
40
50
60
70
80
90Mean radius vs critical radius
time (s)
radi
us (
Å)
Mean radiusCriticial radius
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Derived quantities
100
102
104
106
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7Mean concentration vs equilibrium concentration
time (s)
conc
entr
atio
n (w
t%)
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Derived quantities
100
102
104
106
10−120
10−100
10−80
10−60
10−40
10−20
100
1020
Nucleation rate
time (s)
nucl
eatio
n ra
te (
#/m
3 s)
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Time integration
Three methods:• θ-method
(
I − θ∆t
∆rAn
)
~Nn+1 =
(
I + (1 − θ)∆t
∆rAn
)
~Nn + ∆t~Sn
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Time integration
Three methods:• θ-method• First DIRK-method
~Nn+1 = ~Nn +∆t
2An(
~Nn1 + ~Nn2
)
+ ∆t~Sn
~Nn1 = ~Nn + ∆tγ(
An ~Nn1 + ~Sn)
~Nn2 = ~Nn + ∆tAn(
(1 − 2γ) ~Nn1 + γ ~Nn2
)
+ ∆t(1 − γ)~Sn
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Time integration
Three methods:• θ-method• First DIRK-method• Second DIRK-method
~Nn+1 = ~Nn + ∆tAn(
b1~Nn + b2
~Nn2 + γ ~Nn3
)
+ ∆t~Sn
~Nn2 = ~Nn + ∆tγAn(
~Nn + ~Nn2
)
+ 2∆tγ ~Sn
~Nn3 = ~Nn + ∆tAn(
b1~Nn + b2
~Nn2 + γ ~Nn3
)
+ ∆t~Sn
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Time integration
OperationMethod Vector addition Matrix addition Matrix multiplication Matrix inversion
θ-method 1 2 1 1DIRK-1 6 2 2 2DIRK-2 9 2 3 2
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Time integration
Results:• θ-method large differences
Second order for θ = 1/2
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Time integration
Results:• θ-method large differences
Second order for θ = 1/2
• DIRK-methods small differencesSecond order for all γ
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Time integration
Results:• θ-method large differences
Second order for θ = 1/2
• DIRK-methods small differencesSecond order for all γ
• θ = 1/2- and DIRK-methods small differences
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Time integration
Results:• θ-method large differences
Second order for θ = 1/2
• DIRK-methods small differencesSecond order for all γ
• θ = 1/2- and DIRK-methods small differences
Conclusion:θ = 1/2-method favorable
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Recap
Results:• Model correctly predicts nucleation and
coarsening
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Recap
Results:• Model correctly predicts nucleation and
coarsening• Second order accuracy at low costs
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Elastic deformation
Strain in a deformed block:
εij =1
2
(
∂ui
∂xj
+∂uj
∂xi
)
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Elastic deformation
Strain in a deformed block:
εij =1
2
(
∂ui
∂xj
+∂uj
∂xi
)
Stress in a deformed block:
σij = λδij
3∑
k=1
εkk + 2µεij
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Elastic deformation
Force balance:
∇ · σ = −b
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Elastic deformation
Force balance:
∇ · σ = −b
Or equivalent:
λ∂
∂x1
(∇ · u) + µ
(
∇ ·
(
∇u1 +∂u
∂x1
))
= −b1
λ∂
∂x2
(∇ · u) + µ
(
∇ ·
(
∇u2 +∂u
∂x2
))
= −b2
λ∂
∂x3
(∇ · u) + µ
(
∇ ·
(
∇u3 +∂u
∂x3
))
= −b3
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Finite Element Mesh
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Simulation
Before deformation:
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Simulation
After deformation:
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Simulation
Displacements:
−0.5
0
0.5
−0.5
0
0.5−0.5
0
0.5
x1
x2
x 3
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Simulation
Update:
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Conclusions
Conclusions:• Realistic results with both models• High accuracy• Easy to derive data
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Future work
• Coupling of the two models
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Future work
• Coupling of the two models• Non-elastic deformations
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Future work
• Coupling of the two models• Non-elastic deformations• Coupling with elastic deformations
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Future work
• Coupling of the two models• Non-elastic deformations• Coupling with elastic deformations• Coupling of the three models
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Future work
• Coupling of the two models• Non-elastic deformations• Coupling with elastic deformations• Coupling of the three models• Extension to multi-component alloys
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Questions?