Chapter 9: Phase Diagrams
ISSUES TO ADDRESS...
• When we combine two elements...
what equilibrium state do we get?
• In particular, if we specify...
--a composition (e.g., wt% Cu - wt% Ni), and
--a temperature (T )
then...
How many phases do we get?
What is the composition of each phase?
How much of each phase do we get?
Phase B
Phase A
Nickel atom
Copper atom
Chapter 9 - 1
Phase Equilibria: Solubility Limit
Introduction
– Solutions – solid solutions, single phase
– Mixtures – more than one phase
Adapted from Fig. 9.1,
Callister 7e.
Sucrose/Water Phase Diagram
• Solubility Limit:
Question: What is the
solubility limit at 20°C?
Answer: 65 wt% sugar.
Temperature (°C)
Max concentration for
which only a single phase
solution occurs.
100
Solubility
Limit
80
L
(liquid)
60
L
40
(liquid solution
i.e., syrup)
20
+
S
(solid
sugar)
Pure
Water
Pure
Sugar
If Co < 65 wt% sugar: syrup
0
20
40
6065 80
100
If Co > 65 wt% sugar: syrup + sugar.
Co =Composition (wt% sugar)
Chapter 9 - 2
Components and Phases
• Components:
The elements or compounds which are present in the mixture
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that result (e.g., a and b).
AluminumCopper
Alloy
b (lighter
phase)
a (darker
phase)
Adapted from
chapter-opening
photograph,
Chapter 9,
Callister 3e.
Chapter 9 - 3
Effect of T & Composition (Co)
• Changing T can change # of phases: path A to B.
• Changing Co can change # of phases: path B to D.
B (100°C,70) D (100°C,90)
1 phase
watersugar
system
Adapted from
Fig. 9.1,
Callister 7e.
Temperature (°C)
100
2 phases
L
80
(liquid)
60
L
(liquid solution
40
i.e., syrup)
+
S
(solid
sugar)
A (20°C,70)
20
2 phases
0
0
20
40
60 70 80
100
Co =Composition (wt% sugar)
Chapter 9 - 4
Phase Equilibria
Simple solution 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 miscible in all proportions.
Chapter 9 - 5
Phase Diagrams
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
T(°C)
• 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 phase fields:
L
L+a
a
Adapted from Fig. 9.3(a), Callister 7e.
(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
Phase Diagrams:
# and types of phases
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
A(1100°C, 60):
1 phase: a
B(1250°C, 35):
2 phases: L + a
1600
L (liquid)
B (1250°C,35)
• Examples:
T(°C)
1500
1400
1300
1200
Adapted from Fig. 9.3(a), Callister 7e.
(Fig. 9.3(a) is adapted from Phase
Diagrams of Binary Nickel Alloys, P. Nash
(Ed.), ASM International, Materials Park,
OH, 1991).
1100
1000
Cu-Ni
phase
diagram
a
(FCC solid
solution)
A(1100°C,60)
0
20
40
60
80
100
wt% Ni
Chapter 9 - 7
Phase Diagrams:
composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
T(°C)
Cu-Ni
system
A
TA
Co = 35 wt% Ni
tie line
1300 L (liquid)
At T A = 1320°C:
Only Liquid (L)
B
TB
CL = Co ( = 35 wt% Ni)
a
At T D = 1190°C:
(solid)
1200
D
Only Solid ( a)
TD
Ca = Co ( = 35 wt% Ni)
20
3032 35 4043
50
At T B = 1250°C:
CLCo
Ca wt% Ni
Both a and L
Adapted from Fig. 9.3(b), Callister 7e.
9.3(b) is adapted from Phase Diagrams
CL = C liquidus ( = 32 wt% Ni here) (Fig.
of Binary Nickel Alloys, P. Nash (Ed.), ASM
Ca = C solidus ( = 43 wt% Ni here) International, Materials Park, OH, 1991.)
Chapter 9 - 8
Phase Diagrams:
weight fractions of phases
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Co = 35 wt% Ni
At T A : Only Liquid (L)
W L = 100 wt%, W a = 0
At T D: Only Solid ( a)
W L = 0, Wa = 100 wt%
At T B : Both a and L
WL 
Wa 
S  43  35  73 wt %
R + S 43  32
R
= 27 wt%
R +S
Cu-Ni
system
T(°C)
A
TA
tie line
L (liquid)
1300
B
R S
TB
1200
D
TD
20
3032 35
CLCo
a
(solid)
40 43
50
Ca wt% Ni
Adapted from Fig. 9.3(b), Callister 7e.
(Fig. 9.3(b) is adapted from Phase Diagrams of
Binary Nickel Alloys, P. Nash (Ed.), ASM
International, Materials Park, OH, 1991.)
Chapter 9 - 9
The Lever Rule
• Tie line – connects the phases in equilibrium with
each other - essentially an isotherm
T(°C)
How much of each phase?
Think of it as a lever (teeter-totter)
tie line
1300
L (liquid)
B
TB
a
(solid)
1200
R
20
S
R
30C C
40 C
a
L o
wt% Ni
WL 
Ma
ML
50
S
M a S  M L R
Adapted from Fig. 9.3(b),
Callister 7e.
C  C0
ML
S

 a
ML  Ma R  S Ca  CL
Wa 
C  CL
R
 0
R  S Ca  CL
Chapter 9 - 10
Ex: Cooling in a Cu-Ni Binary
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(°C) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
35
32
--isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
L: 35wt%Ni
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 7e.
35
Co
40
50
wt% Ni
Chapter 9 - 11
Cored vs Equilibrium Phases
• 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.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First a to solidify:
46 wt% Ni
Last a to solidify:
< 35 wt% Ni
Uniform C a:
35 wt% Ni
Chapter 9 - 12
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
--Ductility (%EL,%AR)
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 7e.
--Peak as a function of Co
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 7e.
--Min. as a function of Co
Chapter 9 - 13
Binary-Eutectic Systems
has a special composition
with a min. melting T.
2 components
Cu-Ag
system
T(°C)
Ex.: Cu-Ag system
1200
• 3 single phase regions
L (liquid)
1000
(L, a, b)
a L + a 779°C
• Limited solubility:
L+b b
800
T
a: mostly Cu
8.0
71.9 91.2
E
b: mostly Ag
600
• TE : No liquid below TE
ab
400
• CE : Min. melting TE
composition
200
• Eutectic transition
L(CE)
0
a(CaE) + b(CbE)
20
40
60 CE 80
100
Co , wt% Ag
Adapted from Fig. 9.7,
Callister 7e.
Chapter 9 - 14
EX: Pb-Sn Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: a + b
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Ca = 11 wt% Sn
Cb = 99 wt% Sn
--the relative amount
of each phase:
Wa =
C - CO
S
= b
R+S
Cb - Ca
Pb-Sn
system
300
200
L (liquid)
a
L+ a
18.3
150
100
99 - 40
59
=
= 67 wt%
99 - 11
88
C - Ca
Wb = R = O
Cb - Ca
R+S
L+b b
183°C
61.9
R
97.8
S
a+b
=
=
40 - 11
29
=
= 33 wt%
99 - 11
88
0 11 20
Ca
40
Co
60
80
C, wt% Sn
99100
Cb
Adapted from Fig. 9.8,
Callister 7e.
Chapter 9 - 15
EX: Pb-Sn Eutectic System (2)
• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...
--the phases present: a + L
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Ca = 17 wt% Sn
CL = 46 wt% Sn
--the relative amount
of each phase:
CL - CO
46 - 40
=
Wa =
CL - Ca
46 - 17
6
=
= 21 wt%
29
Pb-Sn
system
300
a
220
200
L (liquid)
L+a
R
L+b b
S
183°C
100
CO - Ca
23
=
WL =
= 79 wt%
CL - Ca
29
a+b
0
17 20
Ca
40 46 60
Co CL
80
C, wt% Sn
Adapted from Fig. 9.8,
Callister 7e.
Chapter 9 - 16
100
Microstructures
in Eutectic Systems: I
• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of a grains
i.e., only one solid phase.
T(°C)
L: Co wt% Sn
400
L
a
L
300
a
200
(Pb-Sn
System)
a: Co wt% Sn
TE
a+ b
100
Adapted from Fig. 9.11,
Callister 7e.
L+ a
0
Co
10
20
30
Co, wt% Sn
2
(room T solubility limit)
Chapter 9 - 17
Microstructures
in Eutectic Systems: II
L: Co wt% Sn
• 2 wt% Sn < Co < 18.3 wt% Sn 400T(°C)
• Result:
 Initially liquid + a
 then a alone
 finally two phases
 a polycrystal
 fine b-phase inclusions
L
300
L+a
a
200
TE
a: Co wt% Sn
a
b
100
a+ b
0
Adapted from Fig. 9.12,
Callister 7e.
(sol.
L
a
10
20
Pb-Sn
system
30
Co
Co , wt%
2
limit at T room )
18.3
(sol. limit at TE)
Sn
Chapter 9 - 18
Microstructures
in Eutectic Systems: III
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
T(°C)
L: Co wt% Sn
300
Pb-Sn
system
a
200
L+ a
L
Lb b
183°C
TE
100
ab
0
20
18.3
Adapted from Fig. 9.13,
Callister 7e.
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 7e.
100
97.8
C, wt% Sn
Chapter 9 - 19
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
7e.
Chapter 9 - 20
Microstructures
in Eutectic Systems: IV
• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
L: Co wt% Sn
T(°C)
300
L
a
L
Pb-Sn
system
a
200
a L
L+a
R
TE
L+b b
S
S
R
primary a
eutectic a
eutectic b
0
20
18.3
Adapted from Fig. 9.16,
Callister 7e.
40
60
61.9
Ca = 18.3 wt% Sn
CL = 61.9 wt% Sn
Wa = S = 50 wt%
R+S
WL = (1- Wa) = 50 wt%
• Just below TE :
a+b
100
• Just above TE :
80
Co, wt% Sn
100
97.8
Ca = 18.3 wt% Sn
Cb = 97.8 wt% Sn
Wa = S = 73 wt%
R+S
Wb = 27 wt%
Chapter 9 - 21
Hypoeutectic & Hypereutectic
300
L
T(°C)
Adapted from Fig. 9.8,
Callister 7e. (Fig. 9.8
adapted from Binary Phase
Diagrams, 2nd ed., Vol. 3,
T.B. Massalski (Editor-inChief), ASM International,
Materials Park, OH, 1990.)
a
200
L+ a
a+b
20
40
hypoeutectic: Co = 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 7e.
Co, wt% Sn
eutectic: Co = 61.9 wt% Sn
a a
175 m
100
b
b b
b
160 m
eutectic micro-constituent
Adapted from Fig. 9.14,
Callister 7e.
Adapted from Fig. 9.17,
Callister 7e. (Illustration
only)
Chapter 9 - 22
Intermetallic Compounds
Adapted from
Fig. 9.20, Callister 7e.
Mg2Pb
Note: intermetallic compound forms a line - not an area because stoichiometry (i.e. composition) is exact. Chapter 9 - 23
Eutectoid & Peritectic
• Eutectic - liquid in equilibrium with two solids
L cool a + b
heat
• Eutectoid - solid phase in equation with two solid
phases
intermetallic compound
S2
S1+S3
- cementite
 cool a + Fe3C
(727ºC)
heat
• Peritectic - liquid + solid 1  solid 2 (Fig 9.21)
S1 + L
S2
cool
 + L heat 
(1493ºC)
Chapter 9 - 24
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition 
+
Peritectic transition  + L
Adapted from
Fig. 9.21, Callister 7e.
Chapter 9 - 25

Iron-Carbon (Fe-C) Phase Diagram
T(°C)
1600

1200
  a + Fe3C
 +L

(austenite)
 
 
1000
a
800
600
120 m
Result: Pearlite =
alternating layers of
a and Fe3C phases
(Adapted from Fig. 9.27, Callister 7e.)
S
 +Fe3C
727°C = Teutectoid
R
S
1
0.76
L+Fe3C
R
B
400
0
(Fe)
A
1148°C
2
3
a+Fe3C
4
5
6
Fe3C (cementite)
L   + Fe3C
-Eutectoid (B):
L
1400
C eutectoid
• 2 important
points
-Eutectic (A):
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
a (ferrite-soft)
Adapted from Fig. 9.24,Callister 7e.
Chapter 9 - 26
Hypoeutectoid Steel
T(°C)
1600

L
 
 
 
 
a


a
 a
 +L

1200
(austenite)
1000
800
 + Fe3C
r s
727°C
aRS
w a =s/(r +s) 600
w  =(1- wa )
400
0
(Fe)
pearlite
a + Fe3C
1
C0
w pearlite = w 
2
3
4
5
6.7
100 m
w a =S/(R+S)
w Fe3 =(1-w a )
C
6
Adapted from Figs. 9.24
and 9.29,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
Co , wt% C
0.76
a
L+Fe3C
1148°C
(Fe-C
System)
Fe3C (cementite)
1400
pearlite
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30,Callister 7e.
Chapter 9 - 27
Hypereutectoid Steel
T(°C)
1600

L
Fe3C


 +L

1200
(austenite)


1000
 
 
r
800
w Fe3C =r/(r +s)
w  =(1-w Fe3C )
a R
600
400
0
(Fe)
pearlite
L+Fe3C
1148°C
 +Fe3C
0.76




(Fe-C
System)
s
S
1 Co
w pearlite = w 
w a =S/(R+S)
w Fe3C =(1-w a )
a +Fe3C
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
6.7
Co , wt%C
60 mHypereutectoid
steel
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33,Callister 7e.
Chapter 9 - 28
Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature
just below the eutectoid, determine the
following
a) composition of Fe3C and ferrite (a)
b) the amount of carbide (cementite) in grams
that forms per 100 g of steel
c) the amount of pearlite and proeutectoid
ferrite (a)
Chapter 9 - 29
Chapter 9 – Phase Equilibria
Solution: a) composition of Fe3C and ferrite (a)
b) the amount of carbide
(cementite) in grams that
forms per 100 g of steel
CO = 0.40 wt% C
Ca = 0.022 wt% C
CFe C = 6.70 wt% C
3
1600

1200

0.4  0.022
x 100  5.7g
6.7  0.022
L

a  94.3 g
L+Fe3C
1148°C
(austenite)
1000
 + Fe3C
800
Fe 3C  5.7 g
 +L
Fe C (cementite)
Fe 3C
Co  Ca
1400

x100 T(°C)
Fe 3C  a CFe3C  Ca
727°C
R
S
a + Fe3C
600
400
0
Ca CO
1
2
3
4
5
6
Co , wt% C
6.7
CFe
3C
Chapter 9 - 30
Chapter 9 – Phase Equilibria
c. the amount of pearlite and proeutectoid ferrite (a)
note: amount of pearlite = amount of  just above TE
1600

L
1400
T(°C)
 +L

Co Ca


x 100  51.2 g 1200
(austenite)
  a C Ca
L+Fe3C
1148°C
1000
 + Fe3C
800
727°C
RS
pearlite = 51.2 g
proeutectoid a = 48.8 g
a + Fe3C
600
400
0
1
Ca CO C
2
3
4
5
6
Co , wt% C
Chapter 9 - 31
Fe C (cementite)
Co = 0.40 wt% C
Ca = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
6.7
Alloying Steel with More Elements
Ti
Mo
Si
W
Cr
Mn
Ni
wt. % of alloying elements
Adapted from Fig. 9.34,Callister 7e. (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 (°C)
• Teutectoid changes:
Ni
Cr
Si
Ti Mo
W
Mn
wt. % of alloying elements
Adapted from Fig. 9.35,Callister 7e. (Fig. 9.35
from Edgar C. Bain, Functions of the Alloying
Elements in Steel, American Society for Metals,
1939, p. 127.)
Chapter 9 - 32
Summary
• Phase diagrams are useful tools to determine:
--the number and types of phases,
--the wt% of each phase,
--and the composition of each phase
for a given T and composition of the system.
• Alloying to produce a solid solution usually
--increases the tensile strength (TS)
--decreases the ductility.
• Binary eutectics and binary eutectoids allow for
a range of microstructures.
Chapter 9 - 33
ANNOUNCEMENTS
Reading:
Core Problems:
Self-help Problems:
Chapter 9 - 34