Introduction to Materials Science and Engineering
Eun Soo Park
Office: 33-313
Telephone: 880-7221 Email: espark@snu.ac.kr
Office hours: by appointment
2020 Fall
11. 03. 2020
1
2
• Phase diagrams are useful tools to determine:
-- the number and types of phases present, -- the composition of each phase,
-- and the weight fraction of each phase
given the temperature and composition of the system.
• The microstructure of an alloy depends on
-- its composition, and
-- whether or not cooling rate allows for maintenance of equilibrium.
• In phase diagram, important phase transformations include eutectic, eutectoid, and peritectic.
Summary
Contents for previous class
Chapter 9 Phase Diagrams
Metatectic reaction: β ↔ L + α Ex. Co-Os, Co-Re, Co-Ru
3Syntectic reaction Liquid
1+Liquid
2↔ α
L
1+L
2α
K-Zn, Na-Zn,
K-Pb, Pb-U, Ca-Cd
4
5
L+α
L+β α +β
200
C, wt% Sn
20 60 80 100
0 300
100
L
α β
TE
40
(Pb-Sn System)
Hypoeutectic & Hypereutectic
Fig. 11.7, Callister & Rethwisch 9e.[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B.
Massalski (Editor-in-Chief), 1990.
Reprinted by permission of ASM International, Materials Park, OH.]
160 μm eutectic micro-constituent
Fig. 11.13, Callister &
Rethwisch 9e.
hypereutectic: (illustration only)
β β β
β β
β
Adapted from Fig. 11.16, Callister & Rethwisch 9e.
(Illustration only)
(Figs. 11.13 and 11.16 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, 1985.
Reproduced by permission of ASM International, Materials Park, OH.)
175 μm
α α
α α α
α
hypoeutectic: C0= 50 wt% Sn
Fig. 11.16, Callister &
Rethwisch 9e.
T(°C)
61.9 eutectic
eutectic: C0= 61.9 wt% Sn
6
• For alloys for which 18.3 wt% Sn < C
0< 61.9 wt% Sn
• Result: α phase particles and a eutectic microconstituent
Microstructural Developments in Eutectic Systems
18.3 61.9
S R
97.8 R S
primary α eutectic α
eutectic β
W
L= (1- W α) = 0.50 C
α= 18.3 wt% Sn C
L= 61.9 wt% Sn
S R + S
W
α= = 0.50
• Just above T
E:
• Just below T
E: C
α= 18.3 wt% Sn C
β= 97.8 wt% Sn
S R + S
W
α= = 0.73
W
β= 0.27
Fig. 11.15, Callister &
Rethwisch 9e.
Pb-Sn system
L+ β
200
T(°C)
C, wt% Sn
20 60 80 100
0
300
100
L
α β
L+ α
40
α + β
TE
L: C0 wt% Sn
α
L Lα
α iron Ferrite
(BCC)
soft & ductile γ iron
austenite (FCC)
Cementite (Fe3C) hard & brittle
A; eutectic B; eutectoid
B
A
c. Fe-C phase diagram
C concentration 0.008w% 2.14w% 6.7w%
iron steel cast iron
7
Hypoeutectoid Steel
Proeutect ferriteoid
Pearlite
dark – ferrite light - cementite
8
Hypereutectoid Steel
Proeutectoid cementite Pearlite
9
10
Alloying with Other Elements
• T
eutectoidchanges:
Fig. 11.33, Callister & Rethwisch 9e.
(From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.)
T Eu te c to id (º C)
wt. % of alloying elements
Ti
Ni
Mo Si
W
Cr Mn
• C
eutectoidchanges:
Fig. 11.34,Callister & Rethwisch 9e.
(From Edgar C. Bain, Functions of the Alloying Elements in Steel, 1939. Reproduced by permission of ASM International, Materials Park, OH.)
wt. % of alloying elements C eut ec toi d (w t% C )
Ni
Ti
Cr Si
W Mn Mo
Ternary Eutectic System
(with Solid Solubility)
11
Ternary Eutectic System
(with Solid Solubility)
12
13
Ternary Eutectic System
(with Solid Solubility)
13
Ternary Eutectic System
(with Solid Solubility)
14
Ternary Eutectic System
(with Solid Solubility)
15
16
Ternary Eutectic System
(with Solid Solubility)
16
17
Ternary Eutectic System
(with Solid Solubility)
17
Ternary Eutectic System
(with Solid Solubility)
18
Ternary Eutectic System
(with Solid Solubility)
T= ternary eutectic temp.
L+ β+γ
L+ α+γ L+ α+β
19
Ternary Eutectic System
(with Solid Solubility)
http://www.youtube.com/watch?v=yzhVomAdetM 20
• Isothermal section (T
A> T > T
B)
THE EUTECTIC EQUILIBRIUM (l = α + β + γ)
(a) T
A> T >T
B(b) T = e
1(c) e
1> T > e
3(d) T = e
3(g) T
A= E
(e) T = e
2(f) e
2> T > E (h) E = T
21Vertical section Location of vertical section
THE EUTECTIC EQUILIBRIUM (l = α + β + γ)
22
Vertical section Location of vertical section
THE EUTECTIC EQUILIBRIUM (l = α + β + γ)
23
Vertical section Location of vertical section
THE EUTECTIC EQUILIBRIUM (l = α + β + γ)
25
Vertical section Location of vertical section
THE EUTECTIC EQUILIBRIUM (l = α + β + γ)
< Quaternary phase Diagrams >
Four components: A, B, C, D
A difficulty of four-dimensional geometry
→ further restriction on the system Most common figure:
“ equilateral tetrahedron “ Assuming isobaric conditions, Four variables: XA, XB, XC and T
4 pure components 6 binary systems 4 ternary systems
A quarternary system
ISSUES TO ADDRESS...
• Transforming one phase into another takes time.
• How does the rate of transformation depend on time and T ?
• How can we slow down the transformation so that we can engineering non-equilibrium structures?
• Are the mechanical properties of non-equilibrium structures better?
Fe
γ
(Austenite)
Eutectoid transformation
C FCC
Fe3C
(cementite)
α
(ferrite)
+
(BCC)
Chapter 10. Phase Transformations
: Development of Microstructure and Alteration of Mechanical Properties
Contents for today’s class
27
2 1 0
G G G
∆ = − <
Microstructure-Properties Relationships
Microstructure Properties
Alloy design &
Processing Performance
“ Phase Transformation ”
“Tailor-made Materials Design”
down to atomic scale
28
(Ch1) Thermodynamics and Phase Diagrams (Ch2) Diffusion: Kinetics
(Ch3) Crystal Interface and Microstructure
(Ch4) Solidification: Liquid → Solid
(Ch5) Diffusional Transformations in Solid: Solid → Solid (Ch6) Diffusionless Transformations: Solid → Solid
Background to understand
phase
transformation
Representative Phase
transformation
Contents _Phase transformation course
29
4 Fold Anisotropic Surface Energy/2 Fold Kinetics, Many Seeds
Solidification
: Liquid SolidPhase Transformations
30
- casting & welding - single crystal growth - directional solidification - rapid solidification
4.1. Nucleation in Pure Metals
Tm : GL = GS- Undercooling (supercooling) for nucleation: 250 K ~ 1 K
<Types of nucleation>
- Homogeneous nucleation - Heterogeneous nucleation
Solidification
: Liquid Solid31
Driving force for solidification
4.1.1. Homogeneous Nucleation
T
mT G L ∆
=
∆
L : ΔH = H
L– H
S(Latent heat) T = T
m- ΔT G
L= H
L– TS
LG
S= H
S– TS
SΔG = Δ H -T ΔS ΔG =0= Δ H-T
mΔS ΔS=Δ H/T
m=L/T
mΔG =L-T(L/T
m)≈(LΔT)/T
mV Variation of free energy per unit volume obtained from undercooling (ΔT)
32
33 33
4.1.1. Homogeneous Nucleation
L V L
S
V G
V
G
1= ( + ) G
2= V
SG
VS+ V
LG
VL+ A
SLγ
SLL V S
V G
G ,
SL SL
S V L
V
S
G G A
V G
G
G = − = − − + γ
∆
2 1( )
SL V
r
r G r
G π
34 π
2γ
3
4 ∆ +
−
=
∆
: free energies per unit volume
for spherical nuclei (isotropic) of radius : r
Surface Free Energy- destabilizes the nuclei (it takes energy to make an interface)
Volume (Bulk) Free Energy–
stabilizes the nuclei (releases energy)
r < r* : unstable
(lower free E by reduce size)r > r* : stable
(lower free E by increase size)r*
: critical nucleus sizeWhy r* is not defined by ∆Gr = 0?
r* dG=0
Unstable equilibrium
SL V
r
r G r
G π
34 π
2γ
3
4 ∆ +
−
=
∆
Calculation of critical radius, r*
embryo nucleus
Gibbs-Thompson Equation
V SL
r G
= ∆
∗
2 γ
SL V
r r G r
G π 3 4π 2 γ
3
4 ∆ +
−
=
∆
2 2
2 3 2
3
) (
1 3
16 )
( 3
* 16
T L
T G G
V m SL V
SL
∆
=
= ∆
∆ π γ πγ
Tm
T G = L∆
∆ V
Critical ΔG of nucleation at r*
nuclei < r* shrink; nuclei>r* grow (to reduce energy)
35
* SL SL m
V V
2 2 T 1
r G L T
γ γ
= ∆ = ∆
2 22 3 2
3
) (
1 3
16 )
( 3
* 16
T L
T G G
V m SL V
SL
∆
=
= ∆
∆ π γ πγ
36
Fig. 4.5 The variation of r* and rmax with undercooling ∆T
∆ T↑ → r* ↓
→ r
max↑
The creation of a critical nucleus ~ thermally activated process
of atomic cluster
* SL SL m
V V
2 2 T 1
r G L T
γ γ
= ∆ = ∆
Tm
T G = L∆
∆
∆TN is the critical undercoolingfor homogeneous nucleation.
The number of clusters with r
*at ∆T < ∆T
Nis negligible.
370 500 1000 1500 2000 2500 3000 3500 4000 0
200 400 600 800 1000
Os
W Re
Ta Mo
Nb Ir Ru Hf Rh Zr Pt Fe
Pd Ti Co Ni
Cu Mn Au Ag Ge
Al Sb Te Pb
Cd Bi
Sn
Se In Ga
Hg
Melting temperature > 3000K : Ta, Re, Os, W measured by ESL
Slope : 0.18
JAXA(2007)
JAXA(2005) JAXA(2003)
JAXA(2003)
Δ = 55.9 + 1.83 x 10-1T Measuring high temperature properties!
Maximum undercooling vs. Melting temperature
38
39
* Copper Homogeneous nucleation
ΔT = 230 K → r* ~ 10-7 cm < 4 *
(Diameter of Cu atom)
If nucleus is spherical shape,
V = 4.2 * 10-21 cm3 ~ 360 atoms (∵one Cu atom 1.16 *10-23 cm3)
“Typically in case of metal” ΔT * ~ 0.2 TE / σSL ~ 0.4 L r* (critical nucleus for homogeneous nucleation) of metal ~ 200 atoms
But,
if cluster radius~ (only
4 * atom diameter),
“no spherical shape”
(large deviation from spherical shape) →
→ Possible structure for the critical nucleus of Cu : bounded only by {111} and {100} plane
- σSL may very with the crystallographic nature of the surface.
- The faces of this crystal are close to their critical size for 2D nucleation at the critical temp for the nucleus as a whole.
4.1.2. The homogeneous nucleation rate - kinetics
How fast solid nuclei will appear in the liquid at a given undercooling?
C
0: atoms/unit volume
C
*: # of clusters with size of C* ( critical size )
) exp(
* hom
0
kT
C G
C
∗= − ∆
clusters / m3
The addition of one more atom to each of these clusters will convert them into stable nuclei.
) exp(
* hom 0
hom
kT
C G f
N
o∆
−
=
nuclei / m3∙s fo ~ 1011 s-1: frequency ∝ vibration frequency energy
of diffusion in liquid surface area (const.)
Co ~ typically 1029 atoms/m3
2 2
2 3
) (
1 3
* 16
T L
G T
V m SL
∆
=
∆ πγ
- hom
* 3 1
1 cm s when G ~ 78 k
N ≈
−∆ T
Homogeneous Nucleation rate
임계핵 크기의 cluster 수
한 개 원자 추가로 확산시 핵생성
Reasonable nucleation rate 40
kT L A T
V
m SL 2
2 3
3 16 πγ
} =
) exp{ (
20
hom
T
C A f
N
o− ∆
≈
A = relatively insensitive to Temp.
Fig. 4.6 The homogeneous nucleation rate as a function of undercooling ∆T. ∆TN is the critical undercooling for homogeneous nucleation.
→
critical value for detectable nucleation- critical supersaturation ratio - critical driving force
- critical supercooling
How do we define ∆T
N?
where
4.1.2. The homogeneous nucleation rate - kinetics
hom 2
~ 1 N T
∆
→ for most metals, ΔT
N~0.2 T
m(i.e. ~200K)
Changes by orders of magnitude from essentially zero to very high values over a very narrow
temperature range
41
*
0exp( * / ) C = C −∆ G kT
Homogeneous Nucleation Rate
hom 0
exp G
mexp G *
N C
kT kT
ω ∆ ∆
= − −
Concentration of Critical Size Nuclei per unit volume
C0 : number of atoms
per unit volume in the parent phase
hom
*
N = f C f = ω exp ( − ∆ G
mkT )
vibration frequency, area of critical nucleus ω ∝
m :
G activation energy for atomic migration
∆
Homogeneous Nucleation in Solids
If each nucleus can be made supercritical at a rate of f per second,
: f depends on how frequently a critical nucleus can receive an atom from the α matrix.
: This eq. is basically same with eq (4.12) except considering temp. dependence of f.
42
hom 0
exp G
mexp G *
N C
kT kT
ω ∆ ∆
= − −
= Q
d4.1.2. The homogeneous nucleation rate - kinetics
43
Under suitable conditions, liquid nickel can be undercooled (or
supercooled) to 250 K below T
m(1453
oC) and held there indefinitely without any transformation occurring.
Normally undercooling as large as 250 K are not observed.
The nucleation of solid at undercooling of only ~ 1 K is common.
The formation of a nucleus of critical size can be catalyzed by a suitable surface in contact with the liquid. → “Heterogeneous Nucleation”
Which equation should we examine?
) exp(
* hom 0
hom
kT
C G f
N =
o− ∆
* ( ) ( )
3 3 2
SL SL m
2 2 2
V V
16 16 T 1
G 3 G 3 L T
πγ πγ
∆ = ∆ = ∆
Why this happens? What is the underlying physics?
Real behavior of nucleation: metal ΔT
bulk< ΔT
small dropEx) liquid container
or
liquid
Solid thin film (such as oxide)
44
Introduction to Materials Science and Engineering
Eun Soo Park
Office: 33-313
Telephone: 880-7221 Email: espark@snu.ac.kr
Office hours: by appointment
2020 Fall
11. 05. 2020
1
2
Ternary Eutectic System
(with Solid Solubility)
2
< Quaternary phase Diagrams >
Four components: A, B, C, D
A difficulty of four-dimensional geometry
→ further restriction on the system Most common figure:
“ equilateral tetrahedron “ Assuming isobaric conditions, Four variables: XA, XB, XC and T
4 pure components 6 binary systems 4 ternary systems
A quarternary system
Microstructure-Properties Relationships
Microstructure Properties
Alloy design &
Processing Performance
“ Phase Transformation ”
“Tailor-made Materials Design”
down to atomic scale
4
Contents for previous class
5
(Ch1) Thermodynamics and Phase Diagrams (Ch2) Diffusion: Kinetics
(Ch3) Crystal Interface and Microstructure
(Ch4) Solidification: Liquid → Solid
(Ch5) Diffusional Transformations in Solid: Solid → Solid (Ch6) Diffusionless Transformations: Solid → Solid
Background to understand
phase
transformation
Representative Phase
transformation
Contents _Phase transformation course
2 1 0
G G G
∆ = − <
A B
Time
Chapter 10. Phase Transformations
: Development of Microstructure and Alteration of Mechanical Properties
Liquid Undercooled Liquid Solid
<Thermodynamic>
Solidification:
Liquid Solid• Interfacial energy ΔTN Tm
Melting and Crystallization are Thermodynamic Transitions
6
7 7
Fig. 4.5 The variation of r* and rmax with undercooling ∆T
∆T↑ → r* ↓
→ rmax↑
The creation of a critical nucleus ~ thermally activated process
of atomic cluster
* SL SL m
V V
2 2 T 1
r G L T
γ γ
= ∆ = ∆
Tm
T G = L∆
∆
∆TN is the critical undercooling for homogeneous nucleation.
The number of clusters with r* at ∆T < ∆TN is negligible. 8
0 500 1000 1500 2000 2500 3000 3500 4000 0
200 400 600 800 1000
Os
W Re
Ta Mo
Nb Ir Ru Hf Rh Zr Pt Fe
Pd Ti Co Ni
Cu Mn Au Ag Ge
Al Sb Te Pb
Cd Bi
Sn
Se In Ga
Hg
Melting temperature > 3000K : Ta, Re, Os, W measured by ESL
Slope : 0.18
JAXA(2007)
JAXA(2005) JAXA(2003)
JAXA(2003)
Δ = 55.9 + 1.83 x 10-1T Measuring high temperature properties!
Maximum undercooling vs. Melting temperature
9
10
* Copper Homogeneous nucleation
ΔT = 230 K → r* ~ 10-7 cm < 4 *
(Diameter of Cu atom)
If nucleus is spherical shape,V = 4.2 * 10-21 cm3 ~ 360 atoms (∵one Cu atom 1.16 *10-23 cm3)
“Typically in case of metal” ΔT * ~ 0.2 TE / σSL ~ 0.4 L r* (critical nucleus for homogeneous nucleation) of metal ~ 200 atoms
But,
if cluster radius~ (only
4 * atom diameter),
“no spherical shape”
(large deviation from spherical shape) →
→ Possible structure for the critical nucleus of Cu : bounded only by {111} and {100} plane
- σSL may very with the crystallographic nature of the surface.
- The faces of this crystal are close to their critical size for 2D nucleation at the critical temp for the nucleus as a whole.
kT L A T
V
m SL 2
2 3
3 16 πγ
} =
) exp{ (
20
hom
T
C A f
N
o− ∆
≈
A = relatively insensitive to Temp.
Fig. 4.6 The homogeneous nucleation rate as a function of undercooling ∆T. ∆TN is the critical undercooling for homogeneous nucleation.
→
critical value for detectable nucleation- critical supersaturation ratio - critical driving force
- critical supercooling
How do we define ∆TN?
where
4.1.2. The homogeneous nucleation rate - kinetics
hom 2
~ 1 N T
∆
→ for most metals, ΔTN~0.2 Tm (i.e. ~200K)
Changes by orders of magnitude from essentially zero to very high values over a very narrow
temperature range
11
hom 0
exp G
mexp G *
N C
kT kT
ω ∆ ∆
= − −
= Qd
4.1.2. The homogeneous nucleation rate - kinetics
12
Under suitable conditions, liquid nickel can be undercooled (or
supercooled) to 250 K below T
m(1453
oC) and held there indefinitely without any transformation occurring.
Normally undercooling as large as 250 K are not observed.
The nucleation of solid at undercooling of only ~ 1 K is common.
The formation of a nucleus of critical size can be catalyzed by a suitable surface in contact with the liquid. → “Heterogeneous Nucleation”
Which equation should we examine?
) exp(
* hom 0
hom
kT
C G f
N
o∆
−
=
* ( ) ( )
3 3 2
SL SL m
2 2 2
V V
16 16 T 1
G 3 G 3 L T
πγ πγ
∆ = ∆ = ∆
Why this happens? What is the underlying physics?
Real behavior of nucleation: metal ΔT
bulk< ΔT
small dropEx) liquid container
or
liquid
Solid thin film (such as oxide)
13
Nucleation becomes easy if ↓ by forming nucleus from mould wall
γ
SL. From
Fig. 4.7 Heterogeneous nucleation of spherical cap on a flat mould wall.
4.1.3. Heterogeneous nucleation
2 2
2 3
) (
1 3
* 16
T L
G T
V m SL
∆
=
∆
πγ
SM SL
ML
γ θ γ
γ = cos +
SL SM
ML
γ γ
γ
θ ( ) /
cos = −
het S v SL SL SM SM SM ML
G V G A γ A γ A γ
∆ = − ∆ + + −
3 2
4 4 ( )
het
3
V SLG π r G π γ r S θ
∆ = − ∆ +
where S ( ) θ = (2 cos )(1 cos ) / 4 + θ − θ
2In terms of the wetting angle (θ ) and the cap radius (r) (Exercies 4.6)
14
S(θ) has a numerical value ≤ 1 dependent only on θ (the shape of the nucleus)
* *
( )
homG
hetS θ G
∆ = ∆
r * 2 γG and G* 163 πγG32 S(θ)V SL V
SL ⋅
= ∆
∆ ∆
=
( ) S θ
θ = 10 → S(θ) ~ 10-4 θ = 30 → S(θ) ~ 0.02 θ = 90 → S(θ) ~ 0.5
15
S(θ) has a numerical value ≤ 1 dependent only on θ (the shape of the nucleus)
* *
( )
homG
hetS θ G
∆ = ∆
Fig. 4.8 The excess free energy of solid clusters for homogeneous and heterogeneous nucleation. Note r* is independent of the nucleation site.
) 3 (
* 16
* 2 2
3 θ
πγ
γ S
G G G and
r
V SL V
SL ⋅
= ∆
∆ ∆
=
16
θ
V
AV
BBarrier of Heterogeneous Nucleation
4
) cos
cos 3
2 ( 3
) 16 3 (
* 16
3 2
3 2
3
θ πγ θ θ
πγ − +
∆ ⋅
=
∆ ⋅
=
∆
V SL V
SL
S G G G
θ θ
− +
∆ = ∆
3
*
2 3 cos cos
4
*
sub homo
G G
2 3 cos cos
34 ( )
A
A B
V S
V V
θ θ θ
− +
= =
+
How about the nucleation at the crevice or at the edge?
* *
( )
homG
hetS θ G
∆ = ∆
17
* homo
3
4 ∆G
homo*1
2 ∆G 1
homo*4 ∆G
Nucleation Barrier at the crevice
contact angle = 90 groove angle = 60 What would be the shape of nucleus and the nucleation barrier for the following conditions?
* homo
1
6 ∆G
18
How do we treat the non-spherical shape?
∆ = ∆ +
* * A
sub homo
A B
G G V
V V Substrate
Good Wetting
V
AV
BSubstrate
Bad Wetting
V
AV
BEffect of good and bad wetting on substrate
19
Although nucleation during solidification usually requires some undercooling, melting invariably occurs at the equilibrium melting temperature even at relatively high rates of heating.
Because, melting can apparently, start at crystal surfaces without appreciable superheating.
Why?
SV LV
SL
γ γ
γ + <
In general, wetting angle = 0 No superheating required!
3.7 The Nucleation of Melting
(commonly)
In the case of gold,
γ
SLγ
LVγ
SV20
Liquid Undercooled Liquid Solid
<Thermodynamic>
Solidification:
Liquid Solid• Interfacial energy ΔTN
Melting:
Liquid Solid• Interfacial energy
SV LV
SL
γ γ
γ + <
No superheating required!
No ΔTN Tm
vapor
Melting and Crystallization are Thermodynamic Transitions
21
The Effect of ΔT on ∆G*
het& ∆G*
hom?
Fig. 4.9 (a) Variation of ∆G* with undercooling (∆T) for homogeneous and heterogeneous nucleation.
(b) The corresponding nucleation rates assuming the same critical value of ∆G*
-3 1 hom
1 cm
N ≈ s
−Plot ∆G*het & ∆G* hom vs ΔT and N vs ΔT.
* ( ) ( )
3 3 2
SL SL m
2 2 2
V V
16 16 T 1
G 3 G 3 L T
πγ πγ
∆ = ∆ = ∆
) 3 (
* 16 2
3 θ
πγ S
G G and
V SL ⋅
= ∆
∆
n1 atoms in contact with the mold wall
) exp(
* 1
*
kT n G
n = − ∆
het) exp(
* 1
1 kT
C G f
Nhet = − ∆ het
22
입자 성장은 장거리 원자 확산에 의함
성장속도 G의 온도의존성은 확산계수 경우와 동일
.
23
5.4 Overall Transformation Kinetics – TTT Diagram
The fraction of Transformation as a function of Time and Temperature
Plot the fraction of
transformation (1%, 99%) in T-log t coordinate.
→ f (t,T)
Plot f vs log t.
- isothermal transformation
- f ~ volume fraction of β at any time; 0~1
Fig. 5.23 The percentage transformation versus time for different transformation temperatures.
If isothermal transformation,
24
Transformation Kinetics
Avrami proposed that for a three-dimensional nucleation and growth process kinetic law
( ) ktn
f = 1 − exp −
specimen of
Volume
phase new
of Volume
f : volume fraction transformed
=
Assumption :
√ reaction produces by nucleation and growth
√ nucleation occurs randomly throughout specimen
√ reaction product grows rapidly until impingement
Johnson-Mehl-Avrami equation
25
Constant Nucleation Rate Conditions
Nucleation rate ( I ) is constant.
Growth rate (v) is constant.
No compositional change
f
τ t δτ
t-τ
f(t)
specimen of
Volume
during formed
nuclei of
number t
at time measured
during
nucleated particle
one of
Vol.
×
= d τ d τ
df
e( )
[ ] ( )
0
0 3
3 4
V
d IV t
v df
eτ τ
π − ×
=
( )
[ ]
∫ ⋅ −
=
t 33 ) 4
(
oe
t I v t d
f π τ τ
( )
3 40 3 4
3 1 4
1 3
4 v t Iv t
I
t
π τ
π =
− −
⋅
=
- do not consider impingement& repeated nucleation - only true for f ≪ 1
As time passes the β cells will eventually impinge on one another and the rate of transformation will decrease again.
Constant Nucleation Rate Conditions
consider impingement + repeated nucleation effects
( f ) df
edf = 1 −
f df
edf
= −
( f ) 1 f
e= − ln 1 −
( )
−
−
=
−
−
=
3 4exp 3 1
) ( exp
1 )
( t f t Iv t
f
eπ
Johnson-Mehl-Avrami Equation
f
t 1
t
4∝
before
impingement
1-exp(z)~Z (z ≪1 )
( ) kt
nf = 1 − exp −
k: T sensitive f(I, v)
n: 1 ~ 4 (depend on nucleation mechanism) Growth controlled. Nucleation-controlled.
i.e. 50% transform
Exp (-0.7) = 0.5
kt
0n.5= 0 . 7
nt
0.5= k 0
1.
/7
0.5 10
/4. 9
3/4v t = N
Rapid transformations are associated with (large values of k), or (rapid nucleation and growth rates)
* Short time:
t→∞, f →1
* Long time:
If no change of nucleation mechanism during phase transformation, n is not related to T.
Rate of Phase Transformation
Avrami rate equation → y = 1- exp (-kt
n) – k & n fit for specific sample
All out of material - done
log t
Fraction transformed, y
Fixed T
fraction
transformed
time 0.5
By convention rate = 1 / t
0.5Adapted from Fig. 10.10, Callister 7e.
maximum rate reached – now amount unconverted decreases so rate slows t0.5
rate increases as surface area increases
& nuclei grow
28
Rate of Phase Transformations
• In general, rate increases as T ↑
r = 1/t 0.5 = A e -Q/RT
– R = gas constant – T = temperature (K)
– A = preexponential factor – Q = activation energy
Arrhenius expression
• r often small: equilibrium not possible!
Adapted from Fig.
10.11, Callister 7e.
(Fig. 10.11 adapted from B.F. Decker and D. Harker,
"Recrystallization in Rolled Copper", Trans AIME, 188, 1950, p.
888.)
135°C 119°C 113°C 102°C 88°C 43°C
1 10 102 104
29
Time–Temperature–Transformation Curves (TTT)
How much time does it take at any one temperature for a given fraction of the liquid to transform
(nucleate and grow) into a crystal?
f(t,T) ~πI(T)µ(T)
3t
4/3
where f is the fractional volume of crystals formed, typically taken to be 10
-6, a barely observable crystal volume.
∆−
∆
∆
= RT
E T
T RT
n H
I cryst m exp D
81 exp 16
π 2
ν
∆
− ∆
=
m m
T T RT
H T
a N
T fRT 1 exp
) ( ) 3
( 2
η µ π
Nucleation rates Growth rates
30
Nucleation and Growth Rates
Nulceation and Growth for Silica
0.E+00 2.E+07 4.E+07 6.E+07 8.E+07 1.E+08 1.E+08 1.E+08
1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100
Temperature (K)
Nucleation Rate (sec-1 )
0 5E-26 1E-25 1.5E-25 2E-25 2.5E-25
Growth Rate (m-sec-1 )
Nucleation rate (sec-1) Growth rate (m/sec)
31
Time Transformation Curves for Silica
T-T-T Curve for Silica
200 400 600 800 1000 1200 1400 1600 1800 2000
1 1E+13 1E+26 1E+39 1E+52 1E+65 1E+78
Time (sec)
T em p er at u r e ( K )
32
33
* Time-Temperature-Transformation diagrams
TTT diagrams
33
34
* Continuous Cooling Transformation diagrams
CCT diagrams
34