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Chapter 11: 상태도 (Phase Diagrams)

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(1)

학습목표

• 2개의 원소가 결합이 되면 평형상태는 어떤 상태일까?

• 특히, 우리가 다음과 같은 요소를 정한다면..

-- 조성 (e.g., wt% Cu - wt% Ni), -- 온도 (T, temperature)

몇 개의 상이 형성될까 ?

각각의 상은 어떤 조성이 될까 ? 각각의 상의 양은 얼마일까?

Chapter 11: 상태도 (Phase Diagrams)

Phase B Phase A

Nickel atom

(2)

Phase Equilibria: Solubility Limit

Q:

20°C 수용액에서 설탕이 녹을 수 있는 용해한도는?

Answer: 65 wt% sugar .

At 20°C, if C < 65 wt% sugar:

syrup

At 20°C, if C > 65 wt% sugar:

syrup + sugar

65

• 용해한도(Solubility Limit):

단일 상에서 다른 원소가 존재할 수 있는 최대 조성

Sugar/Water Phase Diagram

S uga r

T em p er at u re ( ° C)

0 20 40 60 80 100 C = Composition (wt% sugar)

L

(liquid solution i.e., syrup)

Solubility

Limit L

(liquid)

+ S

(solid sugar) 20

4 0 6 0 8 0 10 0

W at er

Adapted from Fig. 11.1, Callister & Rethwisch 9e.

• 용액 (Solution) – 고체, 액체, 가스, 단일상

• 혼합물 (Mixture) – 1개 이상의 상으로 구성

(3)

• 성분 (Components) :

합금에 존재하는 원소 또는 화합물

(e.g., Al and Cu)

• 상 (Phases) :

물리적 및 화학적으로 서로 달리 구성하고 있는 재료의 영역 (e.g.,

α and β).

Aluminum- Copper Alloy

성분과 상 (Components and Phases)

α (darker phase)

β (lighter phase)

Adapted from chapter- opening photograph, Chapter 9, Callister, Materials Science &

Engineering: An

Introduction, 3e.

(4)

70 80 100 60

40 20

0

T em per atur e ( ° C)

C = Composition (wt% sugar)

L

(

liquid solution i.e., syrup)

20 100

40 60 80

0

L

(liquid)

+ S

(solid sugar)

조성과 온도의 영향

• 온도를 변화하면 상의 개수는 : path A to B.

• 조성을 변화하면 상의 개수는 : path B to D.

water- sugar system

Fig. 11.1, Callister &

Rethwisch 9e.

D (100°C,C = 90)

2 phases

B (100°C,C = 70)

1 phase

A (20°C,C = 70)

2 phases

(5)

고용도( Solid Solubility)의 기준

결정구조 전기음성도 r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

• 결정구조, 전기음성도, 원자 반경이 비슷할 경우, 높은 상호 용해도를 갖는다. (W. Hume – Rothery rules)

Simple system (e.g., Ni-Cu solution)

• 니켈과 구리는 모든 요소에서 서로에 대해 완전히 용해된다.

(6)

상태도 (Phase Diagrams)

• T, C, P 의 함수로서 상을 나타냄. .

• 기본으로:

- 2원계 시스템 (binary systems): 2개의 성분으로 구성.

- 독립적 변수: 온도와 조성 (T and C) (P = 1 atm 대기압. )

Phase Diagram

for Cu-Ni system

Fig. 11.3(a), Callister & Rethwisch 9e.

(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

• 2 phases:

L (liquid)

α (FCC solid solution)

• 3 different phase fields:

L

L + α α

wt% Ni

20 40 60 80 100 1000 0

1100 1200 1300 1400 1500 1600

T(°C)

L (liquid)

α

(FCC solid

solution)

(7)

Cu-Ni phase diagram

전율 고용 2원계 상태도

Isomorphous Binary Phase Diagram

• 상태도:

Cu-Ni system.

• System is:

Fig. 11.3(a), Callister & Rethwisch 9e.

(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

-- 2원계 (binary)

i.e., 2성분 :

Cu and Ni.

-- isomorphous

i.e., 다른 원소에

대하여 완전한 용해도를 보여주는 상태; 0 ~ 100 wt% Ni에서 α상이

연장된다.

1000 0 20 40 60 80 100 wt% Ni

1100 1200 1300 1400 1500 1600

T(°C)

L (liquid)

α

(FCC solid

solution)

(8)

wt% Ni

20 40 60 80 100 1000 0

1100 1200 1300 1400 1500 1600

T(°C)

L (liquid)

α

(FCC solid

solution)

Cu-Ni phase diagram

상태도 (Phase Diagrams) : 존재하는 상의 정의

• Rule 1: 온도(T )와 조성 (C o )를 알면:

-- 상태도에서 어떤 상이 존재할 수 있는지를 정의할 수 있다.

• Examples:

A(1100°C, 60 wt% Ni):

1 phase: α

B (1250°C, 35 wt% Ni):

2 phases: L + α

B ( 1250º C ,35)

A(1100ºC,60)

Fig. 11.3(a), Callister & Rethwisch 9e.

(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

(9)

wt% Ni

20 1200 1300

T(°C)

L (liquid)

α (solid)

30 40 50

Cu-Ni system

Phase Diagrams :

Determination of phase compositions

• Rule 2: If we know T and C 0 , then we can determine:

-- the composition of each phase.

• Examples:

T A A

35 C 0 32 C L

At T

A

= 1320°C:

Only Liquid (L) present C

L = C 0

( = 35 wt% Ni)

At T

B

= 1250°C:

Both

α and L present

C

L

= C liquidus

( = 32 wt% Ni) C α

= C solidus

( = 43 wt% Ni)

At T

D

= 1190°C:

Only Solid (

α) present

C

α = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni

T D D

tie line

4 C 3 α

Fig. 11.3(b), Callister & Rethwisch 9e.

(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

T B B

(10)

• Rule 3: T 와 C 0 를 정의하면:

-- 각 상의 질량비를 정의할 수 있다.

• Examples:

T A 에서

: Only Liquid (L) present

W L = 1.00, W

α

= 0

T D 에서

: Only Solid (

α ) present

W L

= 0, W

α

= 1.00

Phase Diagrams :

Determination of phase weight fractions

wt% Ni

20 1200 1300

T(°C)

L (liquid)

α (solid)

3 0 4 0 5 0

Cu-Ni system

T A A

35 C 0 32 C L T B B

T D D

tie line

4 C 3 α R S

T B 에서

: Both

α and L present

= 0.27

W L = S R + S

W α = R R + S

Consider C 0 = 35 wt% Ni

Fig. 11.3(b), Callister & Rethwisch 9e.

(Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.)

(11)

• 타이 라인 (Tie line) –서로 열적 평형상태의 상을 연결하는 선– 때로는 등온선이라고도 한다.

The Lever Rule

각 상의 비율은 ?

지랫대 법칙을 생각하자.

(teeter-totter)

M L M α

R S

wt% Ni

20 1200 1300

T(°C)

L (liquid)

α (solid)

3 0 4 0 5 0

T B B

tie line

C 0

C L

C α S R

Adapted from Fig. 11.3(b),

Callister & Rethwisch 9e.

(12)

wt% Ni

20 120 0 130 0

3 0 4 0 5 0

110 0

L (liquid)

α (solid) T(

°

C)

A

35 C 0

L: 35 wt%Ni

Cu-Ni system

• Phase diagram:

Cu-Ni system.

Adapted from Fig. 11.4, Callister & Rethwisch 9e.

• A 조성의 합금에 대하여 냉각을 동반 미세 구조 변화를 고려한다.

C 0 = 35 wt% Ni alloy

Ex: Cu-Ni Alloy의 냉각

35 46 32 43

α: 43 wt% Ni L: 32 wt% Ni α: 46 wt% Ni L: 35 wt% Ni B

C

E L: 24 wt% Ni

α: 36 wt% Ni

24 D 36

α: 35 wt% Ni

(13)

기계적 성질 : Cu-Ni System

• 고용 강화(solid solution strengthening)에 따른 효과:

-- Tensile strength (TS) -- Ductility (%EL)

Adapted from Fig. 11.5(a), Callister & Rethwisch 9e.

T ens ile S tr engt h (M P a)

Composition, wt% Ni

Cu Ni

0 20 40 60 80 100 200

300 400

TS for pure Ni

TS for pure Cu

E longat ion (% EL )

Composition, wt% Ni

Cu 0 20 40 60 80 100 Ni 20

30 40 50 60

%EL for pure Ni

%EL for pure Cu

Adapted from Fig. 11.5(b),

Callister & Rethwisch 9e.

(14)

2 components

has a special composition with a min. melting T.

Fig. 11.6, Callister & Rethwisch 9e

[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.].

2원계 공정시스템

(Binary-Eutectic Systems)

• 3 개의 단일상 지역 존재 (L, α, β)

• 제한된 용해 한도:

α: mostly Cu β: mostly Ag

• T E : T E 이하에서는 액상X : T E 에서의 조성

• C E

Ex.: Cu-Ag system

Cu-Ag system

L (liquid)

α L + α L + β

β

α + β

C, wt% Ag

20 40 60 80 100

200 0 1200

T(°C)

400 600 800 1000

C E

T E 8.0 71.9 91.2

779°C

cooling

• 공정 반응 ( Eutectic reaction)

L(C E ) α(C αE ) + β(C βE )

(15)

L + α

L + β

α + β

200

T(°C)

18.3

C, wt% Sn

20 60 80 100

0 300

100

L (liquid)

α 183°C

61.9 97.8 β

• 150°C에서 40 wt% Sn-60 wt% Pb alloy: 상을 정의하시오.

Pb-Sn system

EX 1: Pb-Sn 공정 상태도

Answer:

α

+ β -- 각 상의 조성은?

-- 각상의 상대적으로 존재하는 양은?

150

40 C 0 11

C α

99 C β

R S

Answer: C α = 11 wt% Sn C β = 99 wt% Sn

W α = C β - C 0 C β - C α

= 99 - 40

99 - 11 = 59

88 = 0.67 S

R+S =

W β = C 0 - C α C β - C α R =

R+S

= 29

= 0.33

= 40 - 11 Answer:

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

(16)

Answer: C α = 17 wt% Sn -- 각 상의 조성은?

L + β

α + β

200

T(°C)

C, wt% Sn

20 60 80 100

0 300

100

L (liquid)

α β

L + α

183°C

• 220°C에서 40 wt% Sn-60 wt% Pb alloy: 상을 정의하시오.

Pb-Sn system

EX 2: Pb-Sn 공정 상태도

-- 각상의 상대적으로 존재하는 양은?

W α = C L - C 0

C L - C α = 46 - 40 46 - 17

= 6

29 = 0.21

W L = C 0 - C α

C L - C α = 23

29 = 0.79

40 C 0

46 C L 17

C α

220 R S

Answer: α + L

C L = 46 wt% Sn

Answer:

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.]

(17)

• 예를 들어,

C 0 < 2 wt% Sn

• 결과적으로: 실온에서는 -- 조성 C 0 를 갖는 α상이 다결정으로 존재한다.

공정 상태도를 이용한 미세 구조 개발 I

0

L + α

200

T( ° C)

C, wt% Sn

10 2

20 C

0

300

100

L α

30

α + β

400

(room T solubility limit)

T

E

(Pb-Sn

System)

α L L: C

0

wt% Sn

α : C

0

wt% Sn

Fig. 11.10, Callister &

Rethwisch 9e.

(18)

• 예를 들어,

2 wt% Sn < C 0 < 18.3 wt% Sn

• Result:

α + β 의 온도에서는 -- α 결정립 의 다결정이 생성되고

작은 β-상의 입자 들이 석출된다.

Fig. 11.11, Callister &

Rethwisch 9e.

공정 상태도를 이용한 미세 구조 개발 II

Pb-Sn system

L + α

200

T(°C)

C , wt% Sn

10

18.3 20 0

C

0

300

100

L

α

30

α + β

400

(sol. limit at T

E

) T

E

2

(sol. limit at T room )

L α L: C

0

wt% Sn

α β

α : C

0

wt% Sn

(19)

• 합금 조성이 공정 조성일 때 (C 0 = C E )

• 결과: 공정 미세구조를 형성 (lamellar structure) -- α 와 β 상이 판상 구조로 교대로 반복됨.

공정 상태도를 이용한 미세 구조 개발 III

Fig. 11.13, Callister & Rethwisch 9e.

(From Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985.

Reproduced by permission of ASM International, Materials Park, OH.)

160 μm

Micrograph of Pb-Sn eutectic

microstructure

Pb-Sn system

L + β

α + β

200

T(°C)

20 60 80 100

0

300

100

L

α β

L + α

183°C

40 T

E

18.3

α : 18.3 wt%Sn

97.8 β: 97.8 wt% Sn

C

E

L: C

0

wt% Sn

(20)

Lamellar Eutectic Structure

Figs. 11.13 & 11.14, Callister & Rethwisch 9e.

(Fig. 11.13 from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.)

(21)

• 예를 들어, 18.3 wt% Sn < C 0 < 61.9 wt% Sn

• 결과: α 상의 초정상와 공정 미세구조가 형성 된다.

공정 상태도를 이용한 미세 구조 개발 IV

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

• T E 의 바로 위:

• T E 의 바로 아래 : 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 &

Pb-Sn system

L + β

200

T(°C)

20 60 80 100

0

300

100

L

α β

L + α

40

α

+

β

T

E

L: C 0 wt% Sn α L

L α

(22)

L + α

L + β α + β

200

C, wt% Sn

20 60 80 100

0 300

100

L

α β

T

E

40

(Pb-Sn System)

아공정(Hypoeutectic) & 과공정(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 &

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: C

0

= 50 wt% Sn

Fig. 11.16, Callister &

Rethwisch 9e.

T(°C)

61.9 eutectic

eutectic: C

0

= 61.9 wt% Sn

(23)

중간 화합물

(Intermetallic Compounds)

Mg 2 Pb

Note: 금속 간 화합물은 상태도에 라인으로 존재한다 – 화학양론적조성 이므로 면이 아님

Fig. 11.19, Callister &

Rethwisch 9e.

[Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A.

Nayeb-Hashemi and J. B.

Clark (Editors), 1988.

Reprinted by permission of ASM International, Materials Park, OH.]

(24)

• 공석 (Eutectoid) – 하나의 고상은 두 개의 다른 고상으로 변환

S 2 S 1 +S 3

γ α + Fe 3 C (For Fe-C, 727°C, 0.76 wt% C)

intermetallic compound - cementite

cool heat

공정, 공석& 포정

Eutectic, Eutectoid, & Peritectic

• 공정 (Eutectic): 액체는 두 개의 고체 상으로 변환

L cool α + β (For Pb-Sn, 183°C, 61.9 wt% Sn)

heat

cool heat

• 포정 (Peritectic) - 액체 및 하나 고체상이 제2의 고상으로 변환

S 1 + L S 2

δ + L γ (For Fe-C, 1493°C, 0.16 wt% C)

(25)

Eutectoid & Peritectic

Cu-Zn Phase diagram

Fig. 11.20, Callister & Rethwisch 9e.

[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in- Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

Eutectoid transformation δ γ + ε

Peritectic transformation γ + L δ

(26)

Iron-Carbon (Fe-C) 상태도

• 2 important points

- Eutectoid (B):

γ ⇒ α + Fe 3

C

- Eutectic (A):

L ⇒ γ + Fe 3

C

Fig. 11.23, Callister

& Rethwisch 9e.

[Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted

Fe 3 C ( c em ent it e)

1600 1400 1200

1000 800

600

400 0 1 2 3 4 5 6 6.7

L γ

(austenite)

γ +L

γ +Fe 3 C

α +Fe 3

C

δ

(Fe) C, wt% C

1148°C

T(°C)

α 727°C = T eutectoid

4.30

Result: Pearlite = alternating layers of α and Fe

3

C phases

120 μm

Fig. 11.26, Callister & Rethwisch 9e.

(From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985.

0.76

B

γ γ γ γ

A L+Fe

3 C

Fe 3 C (cementite-hard)

α (ferrite-soft)

(27)

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800

600

400 0 1 2 3 4 5 6 6.7

L γ

(austenite)

γ +L

γ + Fe 3 C

α + Fe 3 C

L+Fe 3 C

δ

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

C0

0 .7 6

아공석 강 (Hypoeutectoid Steel)

Adapted from Figs. 11.23 and 11.28, Callister &

Rethwisch 9e.

[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.

Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

초석 페라이트 pearlite

100 μm Hypoeutectoid steel α

pearlite

γ γ γ α γ

α α

γ γ γ γ

γ γ γ γ

(28)

Chapter 11 - 28

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800

600

400 0 1 2 3 4 5 6 6.7

L γ

(austenite)

γ +L

γ + Fe 3 C

α + Fe 3 C

L+Fe 3 C

δ

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

C0

0 .7 6

아공석 강 (Hypoeutectoid Steel)

γ γ γ α γ

α α

s r

W α = s/(r + s) W γ =(1 - W α

)

R S

α

pearlite

W pearlite = W γ W α

= S/(R +

S) W Fe =(1 – W α’

)

3 C

Adapted from Fig. 11.29, Callister & Rethwisch 9e.

(Photomicrograph courtesy of Republic Steel Corporation.)

초석 페라이트 pearlite

100 μm Hypoeutectoid steel

Adapted from Figs. 11.23 and 11.28, Callister &

Rethwisch 9e.

[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.

Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

(29)

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800

600

400 0 1 2 3 4 5 6 6.7

L γ

(austenite)

γ +L

γ + Fe 3 C

α + Fe 3 C

L+Fe 3 C

δ

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

C0

과공석 강 (Hypereutectoid Steel)

0. 76

C

0

Fe

3

C

γ γ γ γ γ γ γ γ γ γ γ γ

proeutectoid Fe 3 C

60 μm Hypereutectoid steel

pearlite

pearlite

Adapted from Figs. 11.23 and 11.31, Callister &

Rethwisch 9e.

[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.

Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

(30)

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800

600

400 0 1 2 3 4 5 6 6.7

L γ

(austenite)

γ +L

γ + Fe 3 C

α + Fe 3 C

L+Fe 3 C

δ

(Fe) C, wt% C

1148°C

T(°C)

α 727°C

(Fe-C System)

C0

Adapted from Fig. 11.32, Callister & Rethwisch 9e.

proeutectoid Fe 3 C

60 μm Hypereutectoid steel

pearlite

Adapted from Figs. 11.23 and 11.31, Callister &

Rethwisch 9e.

[Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B.

Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

과공석 강 (Hypereutectoid Steel)

0. 76 C

0

pearlite Fe

3

C

γ γ γ γ

v x V X

W pearlite = W γ W α = X/(V + X)

W Fe =(1 - W α )

3

C’

W =(1-W γ )

W γ =x/(v + x)

Fe

3

C

(31)

Example 11.4

99.6 wt% Fe-0.40 wt% C 강은 공석반응선 바로 밑에 존재한다. 다음을 정의하시오:

a) Fe 3 C 및 ferrite ( α)의 조성은?.

b) 100g의 강에서 cementite (Fe 3 C) 는 몇 gram?

c) 100g의 강에서 pearlite 및 공석 페라이트 ( α)

는 몇 gram?

(32)

Solution to Example Problem

b) 지랫대 법칙 사용

a) Using the RS tie line just below the eutectoid

C α = 0.022 wt% C C Fe

3 C = 6.70 wt% C

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800 600

400 0 1 2 3 4 5 6 6.7

L

γ

(austenite) γ +L

γ + Fe

3

C α + Fe

3

C

L+Fe

3

C

δ

C , wt% C 1148°C

T(°C)

727°C

C 0

R S

C Fe C C α 3

100 g 에서 Fe 3 C의 양

= (100 g)W Fe

3 C

= (100 g)(0.057) = 5.7 g

Fig. 11.23, Callister & Rethwisch 9e.

[From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T.

B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

(33)

Fe 3 C ( c em ent it e)

1600 1400 1200 1000 800 600

400 0 1 2 3 4 5 6 6.7

L

γ

(austenite) γ +L

γ + Fe

3

C α + Fe

3

C

L+Fe

3

C

δ

C , wt% C 1148°C

T(°C)

727°°C

Solution to Example Problem (cont.)

c) Using the VX tie line just above the eutectoid and realizing that

C 0 = 0.40 wt% C C α = 0.022 wt% C

C pearlite = C γ = 0.76 wt% C

C 0 V X

C γ C α

100 g에서의 펄라이트양

= (100 g)W pearlite

= (100 g)(0.512) = 51.2 g

Fig. 11.23, Callister & Rethwisch 9e.

[From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T.

B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]

공석 페라이트= 펄라이트 양 – Fe3C 의 양 = 51.2-5.7 = 45.6g

(34)

Alloying with Other Elements

• T eutectoid changes:

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 eutectoid 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 elements

C eut ec toi d (w t% C )

Ni

Ti

Cr Si

W Mn

Mo

(35)

• 상태도(Phase diagrams) 다음을 결정하는데 유용한 데이터:

-- 존재하는 상의 종류 및 숫자, -- 각 상의 조성( composition),

-- 각 상의 존재 질량비 (weight fraction)

(어떤 시스템에서 온도 및 조성 주어진 상태에서)

• 합금의 미세 구조에 대한 영향

-- 합금의 조성

-- 열적 평형 상태를 유지하는 냉각 속도의 여부

• 상태도에서 중요한 상 변태(phase transformation)는

공정(eutectic), 공석(eutectoid) 그리고 포정 (peritectic) 이다 .

Summary

참조

관련 문서

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Phase Equilibria and Cage Occupancy of Binary Hydrates : Its Application to Gas Hydrate Refrigeration