chapter 11: 상태도 (phase diagrams) -...
TRANSCRIPT
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Chapter 11 - 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 Copper atom
-
Chapter 11 - 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
Suga
r
Tem
pera
ture
(°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
Wat
er
Adapted from Fig. 11.1, Callister & Rethwisch 9e.
• 용액 (Solution) – 고체, 액체, 가스, 단일상 • 혼합물 (Mixture) – 1개 이상의 상으로 구성
-
Chapter 11 - 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.
-
Chapter 11 - 4
70 80 100 60 40 20 0
Tem
pera
ture
(°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
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Chapter 11 - 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.
-
Chapter 11 - 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 0 1000
1100
1200
1300
1400
1500
1600 T(°C)
L (liquid)
α (FCC solid solution)
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Chapter 11 - 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. wt% Ni 20 40 60 80 100 0
1000
1100
1200
1300
1400
1500
1600 T(°C)
L (liquid)
α (FCC solid solution)
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Chapter 11 -
wt% Ni 20 40 60 80 100 0 1000
1100
1200
1300
1400
1500
1600 T(°C)
L (liquid)
α
(FCC solid solution)
Cu-Ni phase
diagram
8
상태도 (Phase Diagrams): 존재하는 상의 정의
• Rule 1: 온도(T )와 조성 (Co)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 (
1250
ºC,3
5)
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.)
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Chapter 11 - 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 C0, then we can determine: -- the composition of each phase.
• Examples: TA A
35 C0
32 CL
At TA = 1320°C: Only Liquid (L) present CL = C0 ( = 35 wt% Ni)
At TB = 1250°C: Both α and L present CL = C liquidus ( = 32 wt% Ni) Cα = C solidus ( = 43 wt% Ni)
At TD = 1190°C: Only Solid (α) present Cα = C0 ( = 35 wt% Ni)
Consider C0 = 35 wt% Ni
D TD
tie line
4 Cα 3
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.)
B TB
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Chapter 11 - 10
• Rule 3: T 와 C0를 정의하면: -- 각 상의 질량비를 정의할 수 있다. • Examples:
TA에서 : Only Liquid (L) present WL = 1.00, Wα = 0
TD에서 : Only Solid ( α ) present WL = 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
TA A
35 C0
32 CL
B TB
D TD
tie line
4 Cα 3
R S
TB에서 : Both α and L present
= 0.27
WL = S
R + S
Wα = R
R + S
Consider C0 = 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.)
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Chapter 11 - 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) ML Mα
R S
wt% Ni 20
1200
1300
T(°C)
L (liquid)
α (solid)
3 0 4 0 5 0
B T B
tie line
C0 CL Cα
S R
Adapted from Fig. 11.3(b), Callister & Rethwisch 9e.
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Chapter 11 - 12
wt% Ni 20
120 0
130 0
3 0 4 0 5 0 110 0
L (liquid)
α (solid)
T(°C)
A
35 C0
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. C0 = 35 wt% Ni alloy
Ex: Cu-Ni Alloy의 냉각
46 35 43 32
α: 43 wt% Ni L: 32 wt% Ni
B α: 46 wt% Ni L: 35 wt% Ni
C
E L: 24 wt% Ni
α: 36 wt% Ni
24 36 D
α: 35 wt% Ni
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Chapter 11 - 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.
Tens
ile S
treng
th (M
Pa)
Composition, wt% Ni Cu Ni 0 20 40 60 80 100
200
300
400
TS for pure Ni
TS for pure Cu
Elon
gatio
n (%
EL)
Composition, wt% Ni
Cu Ni 0 20 40 60 80 100 20
30
40
50
60
%EL for pure Ni
%EL for pure Cu
Adapted from Fig. 11.5(b), Callister & Rethwisch 9e.
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Chapter 11 - 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
• TE : TE이하에서는 액상X : TE 에서의 조성 • CE
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 0
200
1200 T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2 779°C
cooling
heating
• 공정 반응 (Eutectic reaction) L(CE) α(CαE) + β(CβE)
-
Chapter 11 - 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 C0
11 Cα
99 Cβ
S R
Answer: Cα = 11 wt% Sn Cβ = 99 wt% Sn
W α = Cβ - C0 Cβ - Cα
= 99 - 40 99 - 11 = 59 88
= 0.67
S R+S
=
W β = C0 - Cα Cβ - Cα
= R R+S
= 29 88
= 0.33 = 40 - 11 99 - 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 International, Materials Park, OH.]
-
Chapter 11 - 16
Answer: Cα = 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 = CL - C0 CL - Cα
= 46 - 40 46 - 17
= 6 29
= 0.21
WL = C0 - Cα CL - Cα
= 23 29
= 0.79
40 C0
46 CL
17 Cα
220 S R
Answer: α + L
CL = 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.]
-
Chapter 11 - 17
• 예를 들어, C0 < 2 wt% Sn • 결과적으로: 실온에서는 -- 조성 C0를 갖는 α상이 다결정으로 존재한다.
공정 상태도를 이용한 미세 구조 개발 I
0
L + α 200
T(°C)
C, wt% Sn 10
2
20 C0
300
100
L
α
30
α + β
400
(room T solubility limit)
TE (Pb-Sn System)
α L
L: C0 wt% Sn
α: C0 wt% Sn
Fig. 11.10, Callister & Rethwisch 9e.
-
Chapter 11 - 18
• 예를 들어, 2 wt% Sn < C0 < 18.3 wt% Sn • Result: α + β 의 온도에서는 -- α 결정립의 다결정이 생성되고 작은 β-상의 입자들이 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 C0
300
100
L
α
30
α + β
400
(sol. limit at TE)
TE
2 (sol. limit at T room )
L α
L: C0 wt% Sn
α β
α: C0 wt% Sn
-
Chapter 11 - 19
• 합금 조성이 공정 조성일 때 (C0 = CE) • 결과: 공정 미세구조를 형성 (lamellar structure) -- α 와 β 상이 판상 구조로 교대로 반복됨.
Fig. 11.12, Callister & Rethwisch 9e.
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)
C, wt% Sn 20 60 80 100 0
300
100
L
α β L + α
183°C
40
TE
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn
CE 61.9
L: C0 wt% Sn
-
Chapter 11 - 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.)
-
Chapter 11 - 21
• 예를 들어, 18.3 wt% Sn < C0 < 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
S R
primary α eutectic α
eutectic β
WL = (1- W α ) = 0.50
Cα = 18.3 wt% Sn CL = 61.9 wt% Sn
S R + S
Wα = = 0.50
• TE 의 바로 위:
• TE 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 & 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 L α L α
-
Chapter 11 - 22
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)
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 & 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
-
Chapter 11 - 23
중간 화합물 (Intermetallic Compounds)
Mg2Pb
Note: 금속 간 화합물은 상태도에 라인으로 존재한다 – 화학양론적조성 이므로 면이 아님 - (i.e. 화합물의 조성은 고정 값이다)
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.]
-
Chapter 11 - 24
• 공석 (Eutectoid) – 하나의 고상은 두 개의 다른 고상으로 변환 S2 S1+S3 γ α + Fe3C (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 α + β (For Pb-Sn, 183°C, 61.9 wt% Sn)
cool heat
cool heat
• 포정 (Peritectic) - 액체 및 하나 고체상이 제2의 고상으로 변환 S1 + L S2 δ + L γ (For Fe-C, 1493°C, 0.16 wt% C)
-
Chapter 11 - 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
-
Chapter 11 - 26
Iron-Carbon (Fe-C) 상태도 • 2 important points
- Eutectoid (B): γ ⇒ α + Fe3C
- Eutectic (A): L ⇒ γ + Fe3C
Fig. 11.23, 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.]
Fe3C
(cem
entit
e)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ +Fe3C
α +Fe3C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C = T eutectoid
4.30 Result: Pearlite = alternating layers of α and Fe3C phases
120 μm
Fig. 11.26, Callister & Rethwisch 9e. (From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, 1985. Reproduced by permission of ASM International, Materials Park, OH.)
0.76
B γ γ γ γ
A L+Fe3C
Fe3C (cementite-hard) α (ferrite-soft)
-
Chapter 11 - 27 Fe
3C (c
emen
tite)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0.76
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.]
Adapted from Fig. 11.29, Callister & Rethwisch 9e. (Photomicrograph courtesy of Republic Steel Corporation.)
초석 페라이트 pearlite
100 μm Hypoeutectoid steel
α
pearlite
γ γ γ
γ α α α
γ γ γ γ
γ γ γ γ
-
Chapter 11 - 28 Fe
3C (c
emen
tite)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
0.76
아공석 강 (Hypoeutectoid Steel)
γ γ γ
γ α α α
s r Wα = s/(r + s) Wγ =(1 - Wα)
R S α
pearlite
Wpearlite = Wγ Wα’ = S/(R + S) W =(1 – Wα’) Fe3C
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.]
-
Chapter 11 - Fe
3C (c
emen
tite)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
29
과공석 강 (Hypereutectoid Steel)
0.76
C0
Fe3C
γ γ γ γ
γ γ γ γ γ γ
γ γ
Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.)
proeutectoid Fe3C
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.]
-
Chapter 11 - Fe
3C (c
emen
tite)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
(Fe) C, wt% C
1148°C
T(°C)
α 727°C
(Fe-C System)
C0
Adapted from Fig. 11.32, Callister & Rethwisch 9e. (Copyright 1971 by United States Steel Corporation.)
proeutectoid Fe3C
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.]
30
과공석 강 (Hypereutectoid Steel)
0.76
C0
pearlite
Fe3C
γ γ γ γ
x v
V X
Wpearlite = Wγ Wα = X/(V + X)
W =(1 - Wα) Fe3C’
W =(1-Wγ) Wγ =x/(v + x)
Fe3C
-
Chapter 11 - 31
Example 11.4 99.6 wt% Fe-0.40 wt% C 강은 공석반응선 바로 밑에 존재한다. 다음을 정의하시오:
a) Fe3C 및 ferrite (α)의 조성은?. b) 100g의 강에서 cementite (Fe3C) 는 몇 gram? c) 100g의 강에서 pearlite 및 공석 페라이트 (α) 는 몇 gram?
-
Chapter 11 - 32
Solution to Example Problem
b) 지랫대 법칙 사용
a) Using the RS tie line just below the eutectoid Cα = 0.022 wt% C CFe3C = 6.70 wt% C
Fe3C
(cem
entit
e)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
C , wt% C
1148°C
T(°C)
727°C
C0
R S
CFe C 3 Cα
100 g 에서 Fe3C의 양
= (100 g)WFe3C = (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.]
-
Chapter 11 -
Fe3C
(cem
entit
e)
1600
1400
1200
1000
800
600
400 0 1 2 3 4 5 6 6.7
L
γ (austenite)
γ +L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
C , wt% C
1148°C
T(°C)
727°°C
33
Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and
realizing that C0 = 0.40 wt% C Cα = 0.022 wt% C Cpearlite = Cγ = 0.76 wt% C
C0
V X
Cγ Cα
100 g에서의 펄라이트양 = (100 g)Wpearlite
= (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 – Fe3C의 양 = 51.2-5.7 = 45.6g 초석 페라이트= 전체 페라이트 - 공석 페라이트= 94.3- 45.6= 48.8
-
Chapter 11 - 34
Alloying with Other Elements
• Teutectoid 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 Eut
ecto
id (º
C)
wt. % of alloying elements
Ti
Ni
Mo Si W
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 elements C
eute
ctoi
d (w
t% C
)
Ni
Ti
Cr
Si Mn W Mo
-
Chapter 11 - 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
Chapter 11: 상태도 (Phase Diagrams)Phase Equilibria: Solubility Limit성분과 상 (Components and Phases)조성과 온도의 영향 고용도( Solid Solubility)의 기준 상태도 (Phase Diagrams)전율 고용 2원계 상태도 �Isomorphous Binary Phase Diagram상태도 (Phase Diagrams):�존재하는 상의 정의Phase Diagrams:�Determination of phase compositionsPhase Diagrams:�Determination of phase weight fractionsThe Lever RuleEx: Cu-Ni Alloy의 냉각기계적 성질 : Cu-Ni System2원계 공정시스템 �(Binary-Eutectic Systems)EX 1: Pb-Sn 공정 상태도EX 2: Pb-Sn 공정 상태도공정 상태도를 이용한 �미세 구조 개발 I공정 상태도를 이용한 �미세 구조 개발 II공정 상태도를 이용한 �미세 구조 개발 IIILamellar Eutectic Structure공정 상태도를 이용한 �미세 구조 개발 IV아공정(Hypoeutectic) & 과공정(Hypereutectic)중간 화합물 �(Intermetallic Compounds)공정, 공석& 포정 �Eutectic, Eutectoid, & PeritecticEutectoid & PeritecticIron-Carbon (Fe-C) 상태도아공석 강 (Hypoeutectoid Steel)아공석 강 (Hypoeutectoid Steel)과공석 강 (Hypereutectoid Steel)과공석 강 (Hypereutectoid Steel)Example 11.4Solution to Example ProblemSolution to Example Problem (cont.)Alloying with Other ElementsSummary