introduction to materials science and engineering · 2021. 2. 17. · introduction to materials...

Introduction to Materials Science and Engineering

Eun Soo Park

Office: 33-313 Telephone: 880-7221Email: [email protected] 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.

SummaryContents for previous class

Chapter 9 Phase Diagrams

Metatectic reaction: β ↔ L + α Ex. Co-Os, Co-Re, Co-Ru 3

Syntectic reaction

Liquid1+Liquid2 ↔ α

L1+L2

α

K-Zn, Na-Zn,K-Pb, Pb-U, Ca-Cd

4

5

L+αL+β

α +β

200

C, wt% Sn20 60 80 1000

300

100

L

α βTE

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 μmeutectic micro-constituent

Fig. 11.13, Callister & Rethwisch 9e.

hypereutectic: (illustration only)

β

βββ

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

ααα

hypoeutectic: C0 = 50 wt% Sn

Fig. 11.16, Callister & Rethwisch 9e.

T(°C)

61.9eutectic

eutectic: C0 =61.9wt% Sn

6

• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn• Result: α phase particles and a eutectic microconstituent

Microstructural Developments in Eutectic Systems

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1-Wα) = 0.50

Cα = 18.3 wt% SnCL = 61.9 wt% Sn

SR + S

Wα = = 0.50

• Just above TE :

• Just below TE :Cα = 18.3 wt% SnCβ = 97.8 wt% Sn

SR + S

Wα = = 0.73

Wβ = 0.27Fig. 11.15, Callister & Rethwisch 9e.

Pb-Snsystem

L+β200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L

α βL+α

40

α+ β

TE

L: C0 wt% Sn LαLα

α ironFerrite(BCC)

soft & ductile

γ ironaustenite

(FCC)

Cementite (Fe3C)hard & brittle

A; eutecticB; eutectoid

B

A

c. Fe-C phase diagram

C concentration 0.008w% 2.14w% 6.7w%iron steel cast iron

7

Hypoeutectoid Steel

Proeutectoid

ferrite

Pearlitedark – ferritelight - cementite

8

Hypereutectoid Steel

Proeutectoid cementite

Pearlite

9

10

Alloying with Other Elements

• Teutectoid changes:

Fig. 11.33, Callister & Rethwisch 9e. (From Edgar C. Bain, Functions of the Alloying Elementsin Steel, 1939. Reproduced by permission of ASMInternational, Materials Park, OH.)

T Eut

ecto

id(º

C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

• Ceutectoid changes:

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 elementsC

eute

ctoi

d(w

t% C

)

Ni

Ti

Cr

SiMnWMo

Ternary Eutectic System(with Solid Solubility)

11


12

13


13


14


15

16


16

17


17


18


T= ternary eutectic temp.

L+β+γ

L+α+γL+α+β

19


http://www.youtube.com/watch?v=yzhVomAdetM20

• Isothermal section (TA > T > TB)

THE EUTECTIC EQUILIBRIUM (l = α + β + γ)

(a) TA > T >TB (b) T = e1 (c) e1 > T > e3 (d) T = e3

(g) TA = E(e) T = e2 (f) e2 > T > E (h) E = T 21

Vertical section Location of vertical section


22



23

25



< 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 components6 binary systems4 ternary systemsA quarternary system

ISSUES TO ADDRESS...• Transforming one phase into another takes time.

• How does the rate of transformation depend ontime and T?

• How can we slow down the transformation so thatwe can engineering non-equilibrium structures?

• Are the mechanical properties of non-equilibriumstructures better?

Feγ

(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 0G 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

Backgroundto understand

phase transformation

RepresentativePhase

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 MetalsTm : GL = GS

- Undercooling (supercooling) for nucleation: 250 K ~ 1 K

<Types of nucleation>- Homogeneous nucleation - Heterogeneous nucleation

Solidification: Liquid Solid

31

Driving force for solidification

4.1.1. Homogeneous Nucleation

mTTLG ∆

=∆

L : ΔH = HL – HS

(Latent heat)T = Tm - ΔT

GL = HL – TSL

GS = HS – TSS

ΔG = Δ H -T ΔS ΔG =0= Δ H-TmΔS

ΔS=Δ H/Tm=L/Tm

ΔG =L-T(L/Tm)≈(LΔT)/Tm

V Variation of free energy per unit volume obtained from undercooling (ΔT)

32

4.1.1. Homogeneous Nucleation

LVLS GVVG )(1 += SLSL

LVL

SVS AGVGVG γ++=2

LV

SV GG ,

SLSLS

VL

VS AGGVGGG γ+−−=−=∆ )(12

SLVr rGrG γππ 23 434

+∆−=∆

: 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 size

Why r* is not defined by ∆Gr = 0?

r* dG=0Unstable equilibrium


+∆−=∆

Calculation of critical radius, r*

embryo nucleus

Gibbs-Thompson Equation

V

SL

Gr

∆=∗ γ2


+∆−=∆

22

23

2

3

)(1

316

)(316*

TLT

GG

V

mSL

V

SL

∆

=

∆=∆

πγγπ

mTTLG ∆

=∆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 1rG L Tγ γ

= = ∆ ∆ 22

23

2

3

)(1

316

)(316*

TLT

GG

V

mSL

V

SL

∆

=

∆=∆

πγγπ

36

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 1rG L Tγ γ

= = ∆ ∆ mTTLG ∆

=∆

∆TN is the critical undercooling for homogeneous nucleation.

The number of clusters with r* at ∆T < ∆TN is negligible. 37

0 500 1000 1500 2000 2500 3000 3500 4000

0

200

400

600

800

1000

Os

W

Re

TaMo

NbIr

Ru

Hf

RhZr

Pt

Fe

TiPd

Co

Ni

MnCu

AuAg

Ge

Al

SbTe

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 for2D 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?

C0 : atoms/unit volumeC* : # of clusters with size of C* ( critical size )

)exp(*hom

0 kTGCC ∆

−=∗

clusters / m3

The addition of one more atom to each of these clusters will convert them into stable nuclei.

)exp(*hom

0hom kTGCfN o

∆−=

nuclei / m3∙s

fo ~ 1011 s-1: frequency ∝ vibration frequency energy of diffusion in liquid surface area (const.)

Co ~ typically 1029 atoms/m322

23

)(1

316*

TLTG

V

mSL

∆

=∆

πγ

-hom

*3 11 cm when G ~ 78s kN T− ∆≈

Homogeneous Nucleation rate

임계핵 크기의 cluster 수

한 개 원자 추가로 확산시 핵생성

Reasonable nucleation rate40

kTLTA

V

mSL2

23

316πγ

=})(

exp{ 20hom TACfN 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


2hom1~T

N∆

→ 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

41

0* exp( * / )C C G kT= −∆

Homogeneous Nucleation Rate

hom 0*exp expmG GN C

kT kTω ∆ ∆ = − −

Concentration of Critical Size Nuclei per unit volume

C0 : number of atoms per unit volume in the parent phase

hom *N f C= ( )kTGexpf m∆−= ωvibration frequency, area of critical nucleusω ∝

:mG 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


kT kTω ∆ ∆ = − −

= Qd


43

Under suitable conditions, liquid nickel can be undercooled (or supercooled) to 250 K below Tm (1453oC) 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

0hom kTGCfN o

∆−=*

( ) ( )

3 3 2SL SL m

2 2 2V V

16 16 T 1G3 G 3 L T

πγ πγ ∆ = = ∆ ∆

Why this happens? What is the underlying physics?

Real behavior of nucleation: metal ΔTbulk < ΔTsmall drop

Ex) liquid

containeror

liquid

Solid thin film (such as oxide)

44

Introduction to Materials Science and Engineering

Eun Soo Park

Office: 33-313 Telephone: 880-7221Email: [email protected] hours: by appointment

2020 Fall

11. 05. 2020

1

2


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 components6 binary systems4 ternary systemsA 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

Backgroundto understand

phase transformation

RepresentativePhase

transformation

Contents_Phase transformation course

2 1 0G G G∆ = − <A B

Time

Chapter 10. Phase Transformations: Development of Microstructure and Alteration of Mechanical Properties

Liquid Undercooled Liquid Solid

<Thermodynamic>


• Interfacial energy ΔTN

Tm

Melting and Crystallization are Thermodynamic Transitions

6

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 1rG L Tγ γ

= = ∆ ∆ mTTLG ∆

=∆

∆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

TaMo

NbIr

Ru

Hf

RhZr

Pt

Fe

TiPd

Co

Ni

MnCu

AuAg

Ge

Al

SbTe

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 for2D nucleation at the critical temp for the nucleus as a whole.

kTLTA

V

mSL2

23

316πγ

=})(

exp{ 20hom TACfN 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


2hom1~T

N∆

→ 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


kT kTω ∆ ∆ = − −

= Qd


12

Under suitable conditions, liquid nickel can be undercooled (or supercooled) to 250 K below Tm (1453oC) 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

0hom kTGCfN o

∆−=*

( ) ( )

3 3 2SL SL m

2 2 2V V

16 16 T 1G3 G 3 L T

πγ πγ ∆ = = ∆ ∆

Why this happens? What is the underlying physics?

Real behavior of nucleation: metal ΔTbulk < ΔTsmall drop

Ex) liquid

containeror

liquid

Solid thin film (such as oxide)

13

Nucleation becomes easy if ↓ by forming nucleus from mould wall.SLγ

From

Fig. 4.7 Heterogeneous nucleation of spherical cap on a flat mould wall.

4.1.3. Heterogeneous nucleation

22

23

)(1

316*

TLTG

V

mSL

∆

=∆

πγ

SMSLML γθγγ += cos

SLSMML γγγθ /)(cos −=

het S v SL SL SM SM SM MLG V G A A Aγ γ γ∆ = − ∆ + + −

3 24 4 ( )3het V SLG r G r Sπ π γ θ ∆ = − ∆ +

where 2( ) (2 cos )(1 cos ) / 4S θ θ θ= + −

In 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)

* *hom( )hetG S Gθ∆ = ∆ )(

316*2* 2

3

θπγγ SG

GandG

rV

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)

* *hom( )hetG S Gθ∆ = ∆

Fig. 4.8 The excess free energy ofsolid clusters for homogeneous andheterogeneous nucleation. Note r* isindependent of the nucleation site.

)(3

16*2* 2

3

θπγγ SG

GandG

rV

SL

V

SL ⋅∆

=∆∆

=

16

θ

AV

BV

Barrier of Heterogeneous Nucleation

4)coscos32(

316)(

316*

3

2

3

2

3 θθπγθπγ +−⋅

∆=⋅

∆=∆

V

SL

V

SL

GS

GG

θ θ − +∆ = ∆

3* 2 3cos cos

4*

sub homoG G

32 3cos cos ( )4

A

A B

V SV V

θ θ θ− += =

+

How about the nucleation at the crevice or at the edge?

* *hom( )hetG S Gθ∆ = ∆

17

*homo

34∆G *

homo12∆G *

homo14∆G

Nucleation Barrier at the crevice

contact angle = 90groove angle = 60

What would be the shape of nucleus and the nucleation barrier for the following conditions?

*homo

16∆G

18

How do we treat the non-spherical shape?

∆ = ∆ +

* * Asub homo

A B

VG GV V

Substrate

Good Wetting

AV

BV Substrate

Bad Wetting

AV

BV

Effect 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?

SVLVSL γγγ <+

In general, wetting angle = 0 No superheating required!

3.7 The Nucleation of Melting

(commonly)In the case of gold,

SLγLVγSVγ

20

Liquid Undercooled Liquid Solid

<Thermodynamic>


• Interfacial energy ΔTN

Melting: Liquid Solid

• Interfacial energy

SVLVSL γγγ <+

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 1hom 1 cmN s−≈

Plot ∆G*het & ∆G* hom vs ΔTand N vs ΔT.

*( ) ( )

3 3 2SL SL m

2 2 2V V

16 16 T 1G3 G 3 L T

πγ πγ ∆ = = ∆ ∆

)(3

16* 2

3

θπγ SG

GandV

SL ⋅∆

=∆

n1 atoms in contactwith the mold wall

)exp(*

1*

kT

Gnn het∆

−=

)exp(*

11 kT

GCfN het

het

∆−=

22

입자 성장은 장거리 원자 확산에 의함

성장속도 G의 온도의존성은 확산계수 경우와 동일.

23

5.4 Overall Transformation Kinetics – TTT DiagramThe 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 timefor different transformation temperatures.

If isothermal transformation,

24

Transformation Kinetics

Avrami proposed that for a three-dimensional nucleation and growth process kinetic law

( )nktf −−= exp1

specimen of Volumephase 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

tτ

δτ

t-τ

f(t)

specimen of Volume during formed

nuclei ofnumber tat time measured during

nucleated particle one of Vol.

×

=ττ dd

dfe

( )[ ] ( )

0

03

34

V

dIVtvdfe

ττπ ×−=

( )[ ]∫ −⋅=t 3

34)(

oe dtvItf ττπ

( ) 43

0

43

31

41

34 tIvtvI

t

πτπ =

−−⋅=

- 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

( ) edffdf −= 1f

dfdfe −=

1( )ffe −−= 1ln

( )

−−=−−= 43

3exp1)(exp1)( tIvtftf e

π

Johnson-Mehl-Avrami Equationf

t

1

4t∝

beforeimpingement

1-exp(z)~Z (z ≪1 )

( )nktf −−= exp1k: 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 7.05.0 =nkt nk

t /15.07.0

= 4/34/15.09.0vN

t =

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 (-ktn)

– k & n fit for specific sample

All out of material - done

log tFrac

tion

trans

form

ed, y

Fixed T

fraction transformed

time

0.5

By convention rate = 1 / t0.5

Adapted 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/t0.5 = 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)3t4/3where f is the fractional volume of crystals formed, typically taken to be 10-6, a barely observable crystal volume.

∆−

∆

∆=

RTE

TT

RTH

nI Dmcryst exp81

16exp

2πν

∆

∆−

=

m

m

TT

RTH

TaNfRTT exp1

)(3)( 2ηπ

µ

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)

Nu

clea

tion

Rat

e (s

ec-1

)

0

5E-26

1E-25

1.5E-25

2E-25

2.5E-25

Gro

wth

Rat

e (m

-sec

-1)

Nucleation rate (sec-1) Growth rate (m/sec)

31

Time Transformation Curves for Silica

T-T-T Curve for Silica

200400600800

100012001400160018002000

1 1E+13 1E+26 1E+39 1E+52 1E+65 1E+78

Time (sec)

Tem

per

atu

re (K

)

32

33

* Time-Temperature-Transformation diagrams

TTT diagrams

33

34

* Continuous Cooling Transformation diagrams

CCT diagrams

34

introduction to materials science and engineering · 2021. 2. 17. · introduction to materials...

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