1. 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

2. 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
Pure
Sugar
Temperature (°C) 20 40 60 80
100 Co
=Composition (wt% sugar) L
(liquid solution
i.e., syrup)
Solubility
Limit
(liquid) + S
(solid
sugar) 4 6 8 10
Water
• Solubility Limit:
Max concentration for
which only a single phase solution occurs.
Question: What is the
solubility limit at 20°C? 65
Answer: 65 wt% sugar.
If Co < 65 wt% sugar: syrup
If Co > 65 wt% sugar: syrup + sugar.

3. Components and Phases • Components: • Phases: b (lighter phase) a
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).
Aluminum-
Copper
Alloy b
(lighter
phase) a
(darker
phase)
Adapted from chapter-opening photograph, Chapter 9, Callister 3e.

4. 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)
1 phase
D (100°C,90)
2 phases 70 80
100 60 40 20
Temperature (°C) Co
=Composition (wt% sugar) L (
liquid solution
i.e., syrup)
(liquid) + S
(solid
sugar)
water-
sugar
system
A (20°C,70)
2 phases
Adapted from Fig. 9.1,
Callister 7e.

5. Phase Equilibria Simple solution system (e.g., Ni-Cu solution) Crystal
Structure
electroneg
r (nm) Ni
FCC
1.9
0.1246 Cu
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.

6. 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).
• 2 phases: L
(liquid) a
(FCC solid solution)
• 3 phase fields:
L +
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
(FCC solid
solution) +
liquidus
solidus
• Phase
Diagram
for Cu-Ni
system
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).

7. Phase Diagrams: # and types of phases
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid) a
(FCC solid
solution) L +
liquidus
solidus
Cu-Ni
phase
diagram
• Examples:
A(1100°C, 60):
1 phase: a B
(1250°C,35) B
(1250°C, 35):
2 phases: L + 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).
A(1100°C,60)

8. Phase Diagrams: composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 30 40 50
Cu-Ni system
• Examples: T A C o
= 35 wt% Ni
tie line 35 C o
At T A
= 1320°C:
Only Liquid (L) C L
= C o
( = 35 wt% Ni) B T 32 C L 4 C a 3
At T D
= 1190°C: D T
Only Solid ( a ) C
= C o
( = 35 wt% Ni
At T B
= 1250°C:
Both a
and L
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.) C L
= C
liquidus
( = 32 wt% Ni here) C a
= C
solidus
( = 43 wt% Ni here)

9. 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%).
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5
Cu-Ni system T A 35 C o 32 B D
tie line R S
• Examples: C o
= 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, W
= 100 wt%
At T B :
Both a
and L
= 27 wt% WL = S R + Wa
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.)

10. The Lever Rule
Tie line – connects the phases in equilibrium with each other - essentially an isotherm
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5 B T
tie line C o S R
How much of each phase? Think of it as a lever (teeter-totter) ML M R S
Adapted from Fig. 9.3(b), Callister 7e.

11. Ex: Cooling in a Cu-Ni Binary
• Phase diagram:
Cu-Ni system.
wt% Ni 20
120
130 3 4 5
110
L (liquid) a
(solid) L +
T(°C) A 35 C o
L: 35wt%Ni
Cu-Ni
system
• System is:
--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.
a: 46 wt% Ni
L: 35 wt% Ni B 46 35 C 43 32 a :
43 wt% Ni
L: 32 wt% Ni D 24 36
L: 24 wt% Ni a :
36 wt% Ni E
• Consider
Co = 35 wt%Ni.
Adapted from Fig. 9.4, Callister 7e.

12. 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
Uniform C :
35 wt% Ni
Last
< 35 wt% Ni

13. Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
--Tensile strength (TS)
--Ductility (%EL,%AR)
Tensile Strength (MPa)
Composition, wt% Ni Cu Ni 20 40 60 80
100
200
300
400
TS for
pure Ni
TS for pure Cu
Elongation (%EL)
Composition, wt% Ni Cu Ni 20 40 60 80
100 30 50
%EL
for
pure Ni
for pure Cu
Adapted from Fig. 9.6(a), Callister 7e.
Adapted from Fig. 9.6(b), Callister 7e.
--Peak as a function of Co
--Min. as a function of Co

14. 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)
(L,
a, b )
1000
• Limited solubility: a L + a L + b
800
779°C b a TE
: mostly Cu
8.0
71.9
91.2 b
: mostly Ag
600
• TE
: No liquid below TE a + b
• CE
: Min. melting TE
400
composition
200 20 40 60 CE 80
100
• Eutectic transition
L(CE) (CE) + (CE) Co ,
wt% Ag
Adapted from Fig. 9.7,
Callister 7e.

15. EX: Pb-Sn Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present:
a + b
Pb-Sn
system
--compositions of phases: L + a b
200
T(°C)
18.3
C, wt% Sn 20 60 80
100
300
L (liquid)
183°C
61.9
97.8
CO = 40 wt% Sn
Ca = 11 wt% Sn
Cb = 99 wt% Sn
--the relative amount of each phase: W a =
C - CO
C - C 59 88
= 67 wt% S
R+S
150 R 11 C 40 Co S 99 C W  =
CO - C
C - C R
R+S 29 88
= 33 wt%
Adapted from Fig. 9.8,
Callister 7e.

16. EX: Pb-Sn Eutectic System (2)
• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...
--the phases present:
a + L
Pb-Sn
system
--compositions of phases: L + b a
200
T(°C)
C, wt% Sn 20 60 80
100
300
L (liquid)
183°C
CO = 40 wt% Sn
Ca = 17 wt% Sn
CL = 46 wt% Sn
--the relative amount of each phase:
220 17 C R 40 Co S 46 CL W a =
CL - CO
CL - C 6 29
= 21 wt% W L =
CO - C
CL - C 23 29
= 79 wt%
Adapted from Fig. 9.8,
Callister 7e.

17. Microstructures in Eutectic Systems: I
• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of a grains
i.e., only one solid phase. L + a
200
T(°C) Co ,
wt% Sn 10 2 20
300
100 30 b
400
(room T solubility limit) TE
(Pb-Sn
System)
L: Co wt% Sn a L
a: Co wt% Sn
Adapted from Fig. 9.11, Callister 7e.

18. Microstructures in Eutectic Systems: II
L: Co wt% Sn
• 2 wt% Sn < Co < 18.3 wt% Sn
• Result:
Initially liquid + 
then  alone
finally two phases
a polycrystal
fine -phase inclusions
Pb-Sn
system L + a
200
T(°C) Co ,
wt% Sn 10
18.3 20
300
100 30 b
400
(sol. limit at TE) TE 2
(sol. limit at T
room ) L a
a: Co wt% Sn a b
Adapted from Fig. 9.12, Callister 7e.

19. Microstructures in Eutectic Systems: III
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
Adapted from Fig. 9.14, Callister 7e.
160 m
Micrograph of Pb-Sn
eutectic
microstructure
Pb-Sn
system
L
a
200
T(°C)
C, wt% Sn 20 60 80
100
300 L a b +
183°C 40 TE
L: Co wt% Sn CE
61.9
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
Adapted from Fig. 9.13, Callister 7e.

20. Lamellar Eutectic Structure
Adapted from Figs & 9.15, Callister 7e.

21. Microstructures in Eutectic Systems: IV
• wt% Sn < Co < 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure WL
= (1- W a )
= 50 wt% C
= 18.3 wt% Sn CL
= 61.9 wt% Sn S R + =
• Just above TE :
Adapted from Fig. 9.16, Callister 7e.
Pb-Sn
system L + b
200
T(°C)
Co, wt% Sn 20 60 80
100
300 a 40 TE
L: Co
wt% Sn L a L a
18.3
61.9 S R
97.8 S R
primary a
eutectic b
• Just below TE : C a
= 18.3 wt% Sn b
= 97.8 wt% Sn S R + W =
= 73 wt%
= 27 wt%

22. 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-in-Chief), ASM International, Materials Park, OH, 1990.) L + a a b
200 L + b
(Pb-Sn TE
System) a + b
100
(Figs and 9.17 from Metals Handbook, 9th ed.,
Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
175 mm a
hypoeutectic: Co = 50 wt% Sn
Adapted from
Fig. 9.17, Callister 7e.
hypereutectic: (illustration only) b
Adapted from Fig. 9.17, Callister 7e. (Illustration only)
Co, wt% Sn 20 40 60 80
100
eutectic
61.9
eutectic: Co = 61.9 wt% Sn
160 mm
eutectic micro-constituent
Adapted from Fig. 9.14, Callister 7e.

23. 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.

24. Eutectoid & Peritectic
Eutectic - liquid in equilibrium with two solids
L  + 
cool
heat
intermetallic compound - cementite
cool
heat
Eutectoid - solid phase in equation with two solid phases
S S1+S3
  + Fe3C (727ºC)
cool
heat
Peritectic - liquid + solid 1  solid 2 (Fig 9.21)
S1 + L S2
 + L  (1493ºC)

25. Eutectoid & Peritectic
Peritectic transition  + L 
Cu-Zn Phase diagram
Eutectoid transition   + 
Adapted from
Fig. 9.21, Callister 7e.

26. Iron-Carbon (Fe-C) Phase Diagram
• 2 important
Adapted from Fig. 9.24,Callister 7e.
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a +
L+Fe3C d
(Fe)
Co, wt% C
1148°C
T(°C)
727°C = T
eutectoid
points
-Eutectic (A): L Þ g +
Fe3C A S R
4.30
-Eutectoid (B): g Þ a +
Fe3C g
Result: Pearlite =
alternating layers of a
and Fe3C phases
120 mm
(Adapted from Fig. 9.27, Callister 7e.)
0.76 C
eutectoid B R S
Fe3C (cementite-hard) a
(ferrite-soft)

27. Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite) a
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
(Fe)
Co , wt% C
1148°C
T(°C)
727°C
(Fe-C
System) C0
0.76 g g
Adapted from Figs and 9.29,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) r s w a = /( + ) g
(1-
Adapted from Fig. 9.30,Callister 7e.
proeutectoid ferrite
pearlite
100 mm
Hypoeutectoid
steel R S a w = /( + )
Fe3C
(1- g

28. Hypereutectoid Steel Fe3C (cementite) L g (austenite) +L a d T(°C) g g
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a
L+Fe3C d
(Fe)
Co , wt%C
1148°C
T(°C)
(Fe-C
System) g g
Adapted from Figs and 9.32,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) s r w
Fe3C = /( + ) g
=(1-
Adapted from Fig. 9.33,Callister 7e.
proeutectoid Fe3C
60 mm
Hypereutectoid
steel
pearlite R S w a = /( + )
Fe3C
(1- g Co
0.76

29. Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following
composition of Fe3C and ferrite ()
the amount of carbide (cementite) in grams that forms per 100 g of steel
the amount of pearlite and proeutectoid ferrite ()

30. Chapter 9 – Phase Equilibria
Solution:
a) composition of Fe3C and ferrite ()
the amount of carbide (cementite) in grams that forms per 100 g of steel
CO = 0.40 wt% C Ca = wt% C CFe C = 6.70 wt% C 3
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
Co , wt% C
1148°C
T(°C)
727°C CO R S
CFe C 3 C

31. Chapter 9 – Phase Equilibria
the amount of pearlite and proeutectoid ferrite ()
note: amount of pearlite = amount of g just above TE
Co = 0.40 wt% C Ca = wt% C Cpearlite = C = 0.76 wt% C
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
Co , wt% C
1148°C
T(°C)
727°C CO R S C C
pearlite = 51.2 g proeutectoid  = 48.8 g

32. Alloying Steel with More Elements
• Teutectoid changes:
• Ceutectoid changes:
Adapted from Fig. 9.34,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
Adapted from Fig. 9.35,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T
Eutectoid
(°C)
wt. % of alloying elements Ti Ni Mo Si W Cr Mn C
eutectoid
(wt%C)

33. 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.

34. ANNOUNCEMENTS
Reading:
Core Problems:
Self-help Problems: