Chapter 9: Phase Diagrams

Report
Chapter 9: Phase Diagrams
ISSUES TO ADDRESS...
• When we combine two elements...
what is the resulting equilibrium state?
• In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and
-- the temperature (T)
then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
Phase B
Phase A
Nickel atom
Copper atom
Chapter 9 - 1
Phase Equilibria: Solubility Limit
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase
Adapted from Fig. 9.1,
Callister & Rethwisch 8e.
Sugar/Water Phase Diagram
• Solubility Limit:
(liquid)
60
L
40
(liquid solution
i.e., syrup)
20
Answer: 65 wt% sugar.
At 20ºC, if C < 65 wt% sugar: syrup
At 20ºC, if C > 65 wt% sugar:
syrup + sugar
L
0
+
S
(solid
sugar)
20
40
6065 80
100
C = Composition (wt% sugar)
Sugar
solubility limit for sugar in
water at 20ºC?
Solubility
Limit
80
Water
Question: What is the
Temperature (ºC)
Maximum concentration for
which only a single phase
solution exists.
100
Chapter 9 - 2
Components and Phases
• Components:
The elements or compounds which are present in the alloy
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that form (e.g., a and b).
AluminumCopper
Alloy
Adapted from chapteropening photograph,
Chapter 9, Callister,
Materials Science &
Engineering: An
Introduction, 3e.
b (lighter
phase)
a (darker
phase)
Chapter 9 - 3
Effect of Temperature & Composition
• Altering T can change # of phases: path A to B.
• Altering C can change # of phases: path B to D.
B (100ºC,C = 70) D (100ºC,C = 90)
1 phase
watersugar
system
Adapted from Fig. 9.1,
Callister & Rethwisch 8e.
Temperature (ºC)
100
2 phases
L
80
(liquid)
60
L
(liquid solution
40
i.e., syrup)
+
S
(solid
sugar)
A (20ºC,C = 70)
20
2 phases
0
0
20
40
60 70 80
100
C = Composition (wt% sugar)
Chapter 9 - 4
Criteria for Solid Solubility
Simple system (e.g., Ni-Cu solution)
Crystal
Structure
electroneg
r (nm)
Ni
FCC
1.9
0.1246
Cu
FCC
1.8
0.1278
• Both have the same crystal structure (FCC) and have
similar electronegativities and atomic radii (W. Hume –
Rothery rules) suggesting high mutual solubility.
• Ni and Cu are totally soluble in one another for all proportions.
Chapter 9 - 5
Phase Diagrams
• Indicate phases as a function of T, C, and P.
• For this course:
- binary systems: just 2 components.
- independent variables: T and C (P = 1 atm is almost always used).
T(ºC)
Phase
Diagram
for Cu-Ni
system
• 2 phases:
1600
1500
L (liquid)
a (FCC solid solution)
L (liquid)
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
80
• 3 different phase fields:
L
L+a
a
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
100
wt% Ni
Chapter 9 - 6
Isomorphous Binary Phase Diagram
• Phase diagram:
Cu-Ni system.
• System is:
T(ºC)
1600
1500
L (liquid)
-- binary
i.e., 2 components:
Cu and Ni.
-- isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
Cu-Ni
phase
diagram
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
80
100
wt% Ni
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
Chapter 9 - 7
Phase Diagrams:
Determination of phase(s) present
• Rule 1: If we know T and Co, then we know:
-- which phase(s) is (are) present.
A(1100ºC, 60 wt% Ni):
1 phase: a
B(1250ºC, 35 wt% Ni):
2 phases: L + a
1600
L (liquid)
B (1250ºC,35)
• Examples:
T(ºC)
1500
1400
1300
A(1100ºC,60)
1100
1000
a
(FCC solid
solution)
1200
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
Cu-Ni
phase
diagram
0
20
40
60
80
100
wt% Ni
Chapter 9 - 8
Phase Diagrams:
Determination of phase compositions
• Rule 2: If we know T and C0, then we can determine:
-- the composition of each phase.
• Examples:
Consider C0 = 35 wt% Ni
At TA = 1320ºC:
Only Liquid (L) present
CL = C0 ( = 35 wt% Ni)
At TD = 1190ºC:
Only Solid (a) present
Ca = C0 ( = 35 wt% Ni)
At TB = 1250ºC:
Both a and L present
CL = C liquidus ( = 32 wt% Ni)
Ca = C solidus ( = 43 wt% Ni)
Cu-Ni
system
T(ºC)
A
TA
1300
TB
1200
TD
20
tie line
L (liquid)
B
D
3032 35
CL C0
a
(solid)
4043
50
Ca wt% Ni
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).
Chapter 9 - 9
Phase Diagrams:
Determination of phase weight fractions
• Rule 3: If we know T and C0, then can determine:
-- the weight fraction of each phase.
• Examples:
Consider C0 = 35 wt% Ni
At TA : Only Liquid (L) present
WL = 1.00, Wa = 0
At TD : Only Solid ( a) present
WL = 0, Wa = 1.00
At TB : Both a and L present
WL 
Wa 
S
R +S

43  35
43  32
R
= 0.27
R +S
 0 . 73
Cu-Ni
system
T(ºC)
A
TA
1300
TB
1200
TD
20
tie line
L (liquid)
B
R S
D
3032 35
CL C0
a
(solid)
40 43
50
Ca wt% Ni
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).
Chapter 9 - 10
The Lever Rule
• Tie line – connects the phases in equilibrium with
each other – also sometimes called an isotherm
T(ºC)
What fraction of each phase?
Think of the tie line as a lever
(teeter-totter)
tie line
1300
L (liquid)
B
TB
a
(solid)
1200
S
R
20
30CL
C0 40 Ca
wt% Ni
WL 
ML
ML  Ma
Ma
ML

Adapted from Fig. 9.3(b),
Callister & Rethwisch 8e.
S
R S
R
50

Ca  C0
Ca  CL

S
Ma x S  ML x R
Wa 
R
R S

C0  CL
Ca  CL
Chapter 9 - 11
Ex: Cooling of a Cu-Ni Alloy
• Phase diagram:
Cu-Ni system.
• Consider
microstuctural
changes that
accompany the
cooling of a
T(ºC) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
C0 = 35 wt% Ni alloy
L: 35wt%Ni
Cu-Ni
system
A
35
32
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
30
Adapted from Fig. 9.4,
Callister & Rethwisch 8e.
35
C0
40
50
wt% Ni
Chapter 9 - 12
Cored vs Equilibrium Structures
• Ca changes as we solidify.
• Cu-Ni case: First a to solidify has Ca = 46 wt% Ni.
Last a to solidify has Ca = 35 wt% Ni.
• Slow rate of cooling:
Equilibrium structure
Uniform Ca:
35 wt% Ni
• Fast rate of cooling:
Cored structure
First a to solidify:
46 wt% Ni
Last a to solidify:
< 35 wt% Ni
Chapter 9 - 13
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Ductility (%EL)
400
TS for
pure Ni
300
TS for pure Cu
200
0 20 40
Cu
60 80 100
Ni
Composition, wt% Ni
Adapted from Fig. 9.6(a),
Callister & Rethwisch 8e.
Elongation (%EL)
Tensile Strength (MPa)
-- Tensile strength (TS)
60
%EL for pure Cu
%EL for
pure Ni
50
40
30
20
0 20
Cu
40
60
80 100
Ni
Composition, wt% Ni
Adapted from Fig. 9.6(b),
Callister & Rethwisch 8e.
Chapter 9 - 14
Binary-Eutectic Systems
has a special composition
with a min. melting T.
2 components
Ex.: Cu-Ag system
Cu-Ag
system
T(ºC)
1200
• 3 single phase regions
L (liquid)
1000
(L, a, b)
a L + a 779ºC
• Limited solubility:
TE 800 8.0
a: mostly Cu
b: mostly Ag
600
• TE : No liquid below TE
ab
400
• CE : Composition at
temperature TE
200
• Eutectic reaction
20
a(CaE) + b(CbE)
L(CE)
L ( 71 . 9 wt% Ag)
0
cooling
heating
40
L+b b
71.9 91.2
60 CE 80
100
C, wt% Ag
Adapted from Fig. 9.7,
Callister & Rethwisch 8e.
a ( 8 .0 wt% Ag)  b ( 9 1.2 wt% Ag)
Chapter 9 -
15
EX 1: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine:
-- the phases present
Answer: a + b
-- the phase compositions
Answer: Ca = 11 wt% Sn
Cb = 99 wt% Sn
-- the relative amount
of each phase
Answer:
S
=
W =
R+S
a
99 - 40
=
99 - 11
Wb = R =
R+S
40 - 11
=
99 - 11
T(ºC)
300
200
59
= 0.67
88
C0 - Ca
Cb - Ca
=
=
29
= 0.33
88
L (liquid)
a
L+ a
100
L+b b
183ºC
18.3
150
Cb - C0
Cb - Ca
Pb-Sn
system
61.9
R
97.8
S
a+b
0 11 20
Ca
40
C0
60
80
C, wt% Sn
99100
Cb
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
Chapter 9 - 16
EX 2: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine:
-- the phases present:
Answer: a + L
-- the phase compositions
Answer: Ca = 17 wt% Sn
CL = 46 wt% Sn
-- the relative amount
of each phase
Pb-Sn
system
T(ºC)
300
a
220
200
L (liquid)
L+a
R
L+b b
S
183ºC
Answer:
CL - C0
46 - 40
=
Wa =
CL - Ca
46 - 17
6
=
= 0.21
29
C0 - Ca
23
=
= 0.79
WL =
CL - Ca
29
100
a+b
0
17 20
Ca
40 46 60
C0 CL
80
C, wt% Sn
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
Chapter 9 - 17
100
Microstructural Developments
in Eutectic Systems I
• For alloys for which
C0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of
a phase having
composition C0
T(ºC)
400
L
a
L
300
200
TE
100
Adapted from Fig. 9.11,
Callister & Rethwisch 8e.
L: C0 wt% Sn
a
L+ a
a: C0 wt% Sn
(Pb-Sn
System)
a+ b
0
10
20
30
C0
C , wt% Sn
2
(room T solubility limit)
Chapter 9 - 18
Microstructural Developments
in Eutectic Systems II
• For alloys for which
400
2 wt% Sn < C0 < 18.3 wt% Sn
• Result:
at temperatures in a + b range 300
-- polycrystalline with a grains
and small b-phase particles 200
L: C0 wt% Sn
T(ºC)
L
L+a
a
TE
a: C0 wt% Sn
a
b
100
Adapted from Fig. 9.12,
Callister & Rethwisch 8e.
L
a
a+ b
0
10
20
Pb-Sn
system
30
C0
C, wt%
2
(sol. limit at T room )
18.3
(sol. limit at TE)
Sn
Chapter 9 - 19
Microstructural Developments
in Eutectic Systems III
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of a and b phases.
T(ºC)
L: C0 wt% Sn
300
Pb-Sn
system
a
200
L+ a
L
Lb b
183ºC
TE
100
ab
0
20
18.3
Adapted from Fig. 9.13,
Callister & Rethwisch 8e.
Micrograph of Pb-Sn
eutectic
microstructure
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
160 m
Adapted from Fig. 9.14,
Callister & Rethwisch 8e.
100
97.8
C, wt% Sn
Chapter 9 - 20
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
& Rethwisch 8e.
Chapter 9 - 21
Microstructural Developments
in Eutectic Systems IV
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result: a phase particles and a eutectic microconstituent
L: C0 wt% Sn
L
T(ºC)
300
L
Pb-Sn
system
a
200
a
L+ a
R
TE
L+
S
S
R
a
b b
0
20
18.3
Adapted from Fig. 9.16,
Callister & Rethwisch 8e.
primary a
eutectic a
eutectic b
40
60
61.9
• Just above TE :
Ca = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
= 0.50
=
Wa
R+S
WL = (1- Wa) = 0.50
• Just below TE :
a+b
100
L
80
C, wt% Sn
100
97.8
Ca = 18.3 wt% Sn
Cb = 97.8 wt% Sn
Wa = S = 0.73
R+S
Wb = 0.27
Chapter 9 - 22
Hypoeutectic & Hypereutectic
300
L
T(ºC)
Adapted from Fig. 9.8,
Callister & Rethwisch 8e.
(Fig. 10.8 adapted from
Binary Phase Diagrams,
2nd ed., Vol. 3, T.B.
Massalski (Editor-in-Chief),
ASM International,
Materials Park, OH, 1990.)
a
200
L+ a
a+b
20
40
hypoeutectic: C0 = 50 wt% Sn
a
a
(Pb-Sn
System)
100
0
(Figs. 9.14 and 9.17
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
L+b b
TE
a
60
80
eutectic
61.9
hypereutectic: (illustration only)
b
b
a
Adapted from
Fig. 9.17, Callister &
Rethwisch 8e.
C, wt% Sn
eutectic: C0 = 61.9 wt% Sn
a a
175 m
100
b
b b
b
160 m
eutectic micro-constituent
Adapted from Fig. 9.14,
Callister & Rethwisch 8e.
Adapted from Fig. 9.17,
Callister & Rethwisch 8e.
(Illustration only)
Chapter 9 - 23
Intermetallic Compounds
Adapted from
Fig. 9.20, Callister &
Rethwisch 8e.
Mg2Pb
Note: intermetallic compound exists as a line on the diagram - not an
area - because of stoichiometry (i.e. composition of a compound
Chapter 9 - 24
is a fixed value).
Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L cool a + b
(For Pb-Sn, 183ºC, 61.9 wt% Sn)
heat
• Eutectoid – one solid phase transforms to two other
solid phases
intermetallic compound
- cementite
S2
S1+S3
 cool a + Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
heat
• Peritectic - liquid and one solid phase transform to a
second solid phase
S1 + L
S2
 +L
cool
heat

(For Fe-C, 1493ºC, 0.16 wt% C)
Chapter 9 - 25
Eutectoid & Peritectic
Peritectic transformation  + L
Cu-Zn Phase diagram
Eutectoid transformation 
+
Adapted from Fig. 9.21,
Callister & Rethwisch 8e.
Chapter 9 - 26

Iron-Carbon (Fe-C) Phase Diagram
T(ºC)
1600

L   + Fe3C
- Eutectoid (B):
  a + Fe3C
L
1400
1200
 +L

(austenite)
 
 
1000
a
800
A
1148ºC
 +Fe3C
727ºC = T eutectoid
B
a+Fe3C
600
120 m
Result: Pearlite =
alternating layers of
a and Fe3C phases
(Adapted from Fig. 9.27,
Callister & Rethwisch 8e.)
400
0
(Fe)
L+Fe3C
1
0.76
2
3
4
4.30
5
6
Fe3C (cementite)
• 2 important
points
- Eutectic (A):
6.7
C, wt% C
Fe3C (cementite-hard)
a (ferrite-soft)
Adapted from Fig. 9.24,
Callister & Rethwisch 8e.
Chapter 9 - 27
Hypoeutectoid Steel
T(ºC)
1600

L
 
 
 
 
a


a
 a
 +L

1200
(austenite)
1000
 + Fe3C
800
727ºC
a
a + Fe3C
600
pearlite
1
0.76
a
400
0
(Fe)C0
L+Fe3C
1148ºC
2
3
4
5
6
6.7
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
C, wt% C
100 m
pearlite
(Fe-C
System)
Fe3C (cementite)
1400
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30, Callister & Rethwisch 8e.
Chapter 9 - 28
Hypoeutectoid Steel
T(ºC)
1600

L
a
 +L

1200
(austenite)
a

 a
1000
 + Fe3C
Wa = s/(r + s)
800 r s
W =(1 - Wa)
727ºC
aRS
a + Fe3C
600
pearlite
Wpearlite = W
400
0
(Fe)C0
1
0.76
a
L+Fe3C
1148ºC
Wa’ = S/(R + S)
WFe3C =(1 – Wa’)
pearlite
2
3
4
5
6
(Fe-C
System)
Fe3C (cementite)
1400
6.7
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
C, wt% C
100 m
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30, Callister & Rethwisch 8e.
Chapter 9 - 29
Hypereutectoid Steel
T(ºC)
1600

L
Fe3C


 +L

1200
(austenite)


1000
 
 
L+Fe3C
1148ºC
 +Fe3C
800
a
a +Fe3C
600
400
0
(Fe)
pearlite
0.76




(Fe-C
System)
1 C0
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)
6.7
C, wt%C
60 mHypereutectoid
steel
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33, Callister & Rethwisch 8e.
Chapter 9 - 30
Hypereutectoid Steel
T(ºC)
1600

L
 +L

1200
(austenite)
 
 
W =x/(v + x)
WFe3C =(1-W)
 +Fe3C
v x
800
a V
600
pearlite
400
0
(Fe)
Wpearlite = W
X
1 C0
Wa = X/(V + X)
WFe
3C’
L+Fe3C
1148ºC
1000
0.76
Fe3C
(Fe-C
System)
a +Fe3C
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister &
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)
6.7
C, wt%C
60 mHypereutectoid
steel
=(1 - Wa)
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33, Callister & Rethwisch 8e.
Chapter 9 - 31
Example Problem
For a 99.6 wt% Fe-0.40 wt% C steel at a
temperature just below the eutectoid,
determine the following:
a) The compositions of Fe3C and ferrite (a).
b) The amount of cementite (in grams) that
forms in 100 g of steel.
c) The amounts of pearlite and proeutectoid
ferrite (a) in the 100 g.
Chapter 9 - 32
Solution to Example Problem
a) Using the RS tie line just below the eutectoid
Ca = 0.022 wt% C
CFe3C = 6.70 wt% C
W Fe 3 C 

R
RS

C0  Ca
C Fe 3 C  C a
0 .40  0 .022
6 .70  0 .022
 0 .057
1600

L
1400
T(ºC)
1200

 +L
1000
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
L+Fe3C
1148ºC
(austenite)
Fe C (cementite)
b) Using the lever rule with
the tie line shown
 + Fe3C
800
727ºC
R
S
a + Fe3C
600
400
0
Ca C0
1
2
3
4
5
6
C, wt% C
6.7
CFe
3C
Chapter 9 - 33
Solution to Example Problem (cont.)
c) Using the VX tie line just above the eutectoid and
realizing that
C0 = 0.40 wt% C
Ca = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
1600


V
V X

C0  Ca
C  Ca
0 .40  0 .022
0 .76  0 .022
 0 .512
Amount of pearlite in 100 g
= (100 g)Wpearlite
T(ºC)
1200

 +L
L+Fe3C
1148ºC
(austenite)
1000
 + Fe3C
800
727ºC
VX
a + Fe3C
600
400
0
1
Ca C0 C
= (100 g)(0.512) = 51.2 g
2
3
4
5
6
C, wt% C
Chapter 9 - 34
Fe C (cementite)
W pearlite 
L
1400
6.7
VMSE: Interactive Phase Diagrams
Microstructure, phase compositions, and phase fractions respond interactively
Change alloy composition
Chapter 9 - 35
Alloying with Other Elements
Ti
Mo
Si
W
Cr
Mn
Ni
wt. % of alloying elements
Adapted from Fig. 9.34,Callister & Rethwisch 8e.
(Fig. 9.34 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
• Ceutectoid changes:
Ceutectoid (wt% C)
T Eutectoid (ºC)
• Teutectoid changes:
Ni
Cr
Si
Ti Mo
W
Mn
wt. % of alloying elements
Adapted from Fig. 9.35,Callister & Rethwisch 8e.
(Fig. 9.35 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
Chapter 9 - 36
Summary
• 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.
• Important phase diagram phase transformations include
eutectic, eutectoid, and peritectic.
Chapter 9 - 37
ANNOUNCEMENTS
Reading:
Core Problems:
Self-help Problems:
Chapter 9 - 38

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