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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
Assist. Prof. Dr. İlkay KALAY
Phase Diagrams
Phase diagrams display the state of a substance
at various pressures and temperatures and the
places where equilibria exist between phases.
Chapter 9 - 2
Assist. Prof. Dr. İlkay KALAY
Phase Diagrams
•  The AB line is the liquid-vapor interface. Eqm btw liquid and gas phases.
•  It starts at the triple point (A), the point at which all three states are in
equilibrium.
•  Each point along this line is the boiling point of the substance at that
pressure.
Chapter 9 - 3
Assist. Prof. Dr. İlkay KALAY
Phase Diagrams
•  The AD line is the interface between liquid and solid.
•  An increase in pressure usually favors the more compact solid
phase, thus, higher temperatures are required to melt the solid at
higher pressures.
•  The melting point at each pressure can be found along this line.
Chapter 9 - 4
Assist. Prof. Dr. İlkay KALAY
Phase Diagrams
•  Below A the substance cannot exist in the liquid state.
•  Along the AC line the solid and gas phases are in
equilibrium; the sublimation point at each pressure is along
this line.
Chapter 9 - 5
Phase Equilibria: Solubility Limit
Introduction
–  Solutions – solid solutions, single phase
–  Mixtures – more than one phase
Adapted from Fig. 9.1,
Callister 7e.
Max concentration for
which only a single phase
solution occurs.
Question: What is the
solubility limit at 20°C?
Answer: 65 wt% sugar.
10
0
Temperature (°C)
• Solubility Limit:
Sucrose/Water Phase Diagram
Solubility Limit
8
0
6
0
L
4
0
(liquid solution i.e., syrup)
20
L (liquid) + S (solid sugar)
Pure Sugar
Pure Water
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 - 6
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., α and β).
AluminumCopper
Alloy
β (lighter
phase)
α (darker
phase)
Adapted from
chapter-opening
photograph,
Chapter 9,
Callister 3e.
Chapter 9 - 7
Nomenclature / Definitions / Basic Concepts
(III)
•  A system is at equilibrium if at constant temperature,
pressure and composition the system is chemically and
structurally stable, not changing with time.
•  Equilibrium is the state that is achieved given sufficient
time. But sometimes it will take too much time to achieve
equilibrium due the kinetics. This is called a meta-stable
state.
• 
In thermodynamics the equilibrium is described as the state of
system that corresponds to the minimum of thermodynamic
function called the free energy.
–  Thermodynamics tells us that
•  under conditions of a constant temperature and pressure and
composition, the direction of any spontaneous change is toward a
lower free energy.
•  the state of stable thermodynamic equilibrium is the one with
minimum free energy.
•  a system at a metastable state is trapped in a local minimum of
free energy that is not the global one.
Chapter 9 -
Free Energy
Unstable
Metastable
Stable
Chapter 9 -
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
L
80
(liquid)
60
L (liquid
solution
40
i.e., syrup)
+
S
(solid
sugar)
A (20°C,70)
20
0
2 phases
2 phases
0
20
40
60 70 80
100
Co =Composition (wt% sugar)
Chapter 9 - 10
Assist. Prof. Dr. İlkay KALAY
One Component (Uniary) Phase Diagram
Phase diagrams display the state of a substance
at various pressures and temperatures and the
places where equilibria exist between phases.
Chapter 9 - 11
Binary Phase Diagram
2 components
Ex.: Cu-Ag system Cu-Ag system
T(°C) 1200
• 3 single phase regions L (liquid)
1000
(L, α, β
) α
L + α
779°C
• Limited solubility: L +β
β
800
TE
α
: mostly Cu 8.0
71.9 91.2
β
:
mostly Ag 600
α+
β
400
200
0
20
40
60
80
100
Co , wt% Ag
Adapted from Fig. 9.7,
Callister 7e.
Chapter 9 - 12
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 - 13
CHAPTER 9: PHASE DIAGRAMS
•  When we combine two elements, create an alloy or a
compound what is the equilibrium structure like ?
•  For a given composition and temperature
•  How many phases do we get ?
•  What is the composition of each phase ?
•  How much of each phase do we get ?
Chapter 9 - 1
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).
• Phase
Diagram
for Cu-Ni
system
T(°C)
1600
1500
• 2 phases:
L (liquid)
1400
1300
1200
α
(FCC solid solution)
1100
1000
0
20
40
60
L (liquid)
α
(FCC
solid solution)
• 3 phase fields: L
L +
α
α
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).
80
100
wt% Ni
Chapter 9 - 15
Interpretation of Phase Diagrams
• 
For a given temperature, T, and composition,
CO, we can use the phase diagram to determine:
1.  The # of phases that are present
2.  Compositions of these phases
3.  The relative fractions of the phases
• 
Finding the composition in a two phase region:
1.  Locate composition and temperature in the diagram
2.  In a two phase region draw the tie line or an isotherm
3.  Note the intersection with phase boundaries. Read
compositions at the intersections.
The liquid and/or solid phases have these compositions.
LETS TRY
Chapter 9 -
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: α
B
(1250°C, 35): 2 phases: L + α
1500
L (liquid)
B
(1250°C,35)
• Examples:
T(°C)
1600
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
0
Cu-Ni
phase
diagram
α
(FCC solid solution)
A(1100°C,60)
20
40
60
80
100
wt% Ni
Chapter 9 - 17
Phase Diagrams:
composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
C
o
= 35 wt% Ni
At T
A
= 1320°C: Cu-Ni
system
T(°C)
T
A
1300
A
L (liquid)
Only Liquid (L) T
B
CL
= Co
(
= 35 wt% Ni)
At T
D
= 1190°C:
1200
α
Only Solid (
) T
D
Cα
= Co
(
= 35 wt% Ni)
20
At T
B
= 1250°C:
Both α
and L C
L
= C
liquidus
(
= 32 wt% Ni here) C
α
= C
solidus
(
= 43 wt% Ni here)
tie line
B
D
α
(solid)
40
4
3
50
C
LC
o
C
α
wt% Ni
3032 35
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 - 18
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:
C
o
= 35 wt% Ni
At T
A
: Only Liquid (L) W
L
= 100 wt%, W
α
= 0
At T
D
:
Only Solid (
α
) W
L
= 0, W
α
= 100 wt%
At T
B
:
Both α
and L
S = 43 ! 35 = 73 wt %
=
WL
R + S 43 ! 32
R
= 27 wt%
Wα =
R +S
Cu-Ni
system
T(°C)
A
T
A
1300
tie line
L (liquid)
T
B
B
R
S
T
D
D
1200
20
α
(solid)
4
0
4
3
5
0
C
LC
o
C
α
wt% Ni
3 032 35
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 - 19
Interpretation of Phase Diagrams :
Lever Rule
• 
• 
Finding the amounts of phases in a two phase
region:
1.  Locate composition and temperature in
diagram
2.  In two phase region draw the tie line or
isotherm
3.  Fraction of a phase is determined by taking
the length of the tie line to the phase
boundary for the other phase, and dividing by
the total length of tie line
The lever rule is a mechanical analogy to the
mass balance calculation. The tie line in the twophase region is analogous to a lever balanced on
a fulcrum.
Chapter 9 -
THE LEVER RULE: A PROOF
How much of each phase?
Think of it as a lever (teeter-totter)
• Sum of weight fractions: WL + W! = 1
• Conservation of mass (Ni): C o = WL CL + W! C !
• Combine above equations:
C! " Co = S
Co " CL = R
=
WL
W! =
+
C! " CL R + S
C! " CL R S
• A geometric interpretation:
Co
CL
C!
R
S
WL
W!
moment equilibrium:
WLR = W"S
1! W"
Chapter 9 - 9
solving gives Lever Rule
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)
TB
1200
20
R
Mα
ML
B
S
α
(solid)
R
3 0C C
40 C
α
L o
wt% Ni
50
Adapted from Fig. 9.3(b),
Callister 7e.
Cα − C0
ML
S
WL =
=
=
ML + Mα R + S Cα − CL
S
M α ⋅S = M L ⋅R
C0 − CL
R
Wα =
=
R + S Cα − CL
Chapter 9 -
Consider the sugar––water phase diagram.
(a)  How much sugar will dissolve in 1500 g water at 90°C (363 K)?
(b)  If the saturated liquid solution in part (a) is cooled to 20°C (293
K), some of the sugar will precipitate out as a solid. What will
be the composition of the saturated liquid solution (in wt%
sugar) at 20°C?
(c) How much of the solid sugar will come out of solution upon
cooling to 20°C?
Temperature (°C)
100
L
80
(liquid)
60
L (liquid
solution
40
i.e., syrup)
+
S
(solid
sugar)
20
0
0
20
40
60 70 80
100
23
Chapter 9 Co =Composition (wt% sugar)
Consider the sugar––water phase diagram.
(a)  How much sugar will dissolve in 1500 g water at 90°C (363 K)?
D (90°C)
Temperature (°C)
100
L
80
(liquid)
60
L (liquid
solution
40
i.e., syrup)
+
S
(solid
sugar)
20
0
0
20
40
60
80
100
Co =Composition (wt% sugar)
Chapter 9 -
Consider the sugar––water phase diagram.
(b) If the saturated liquid solution in part (a) is cooled to 20°C (293 K),
some of the sugar will precipitate out as a solid. What will be the
composition of the saturated liquid solution (in wt% sugar) at 20°C?
Temperature (°C)
100
L
80
(liquid)
60
L (liquid
solution
40
i.e., syrup)
+
S
(solid
sugar)
20
0
0
20
40
60
80
100
Co =Composition (wt% sugar)
(c) How much of the solid sugar will come out of solution upon
Chapter 9 cooling to 20°C?
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
α: 46 wt% Ni
i.e., complete
solubility of one
component in
another; α phase
field extends from
0 to 100 wt% Ni.
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
32
--isomorphous
L: 35wt%Ni
35
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
α:
43 wt% Ni
E
L: 24 wt% Ni
α:
36 wt% Ni
α
(solid)
110 0
20
30
Adapted from Fig. 9.4,
Callister 7e.
35
Co
40
50
wt% Ni
Chapter 9 - 26
SUMMARY
• 
Development of microstructure in isomorphous alloys during
equilibrium (very slow) cooling
1.  Solidification in the solid + liquid phase occurs
gradually upon cooling from the liquidus line.
2.  The composition of the solid and the liquid change
gradually during cooling (as can be determined by the
tie-line method.) => DIFFUSION !
3.  Nuclei of the solid phase form and they grow to
consume all the liquid at the solidus line.
Chapter 9 -
Development of microstructure in
isomorphous alloys
Non-equilibrium (fast) cooling
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
--isomorphous
i.e., complete
solubility of one
component in
another; α phase
field extends from
0 to 100wt% Ni.
• Consider
Co = 35wt%Ni & fast
cooling
Adapted from Fig. 9.3,
Callister 6e.
Chapter 9 -
• 
• 
• 
• 
SUMMARY
Development of microstructure in isomorphous alloys during non-equilibrium (fast) cooling
Compositional changes require diffusion in solid and liquid phases
–  Diffusion in the solid state is very slow.
–  The new layers that solidify on top of the existing grains have the equilibrium
composition at that temperature but once they are solid their composition does
not change.
–  Formation of layered (cored) grains and the invalidity of the tie-line method to
determine the composition of the solid phase.
The tie-line method still works for the liquid phase, where diffusion is fast.
Average Ni content of solid grains is higher.
–  Application of the lever rule gives us a greater proportion of liquid phase as
compared to the one for equilibrium cooling at the same T.
–  Solidus line is shifted to the right (higher Ni contents), solidification is
complete at lower T, the outer part of the grains are richer in the lowmelting component (Cu).
Upon heating grain boundaries will melt first. This can lead to premature
mechanical failure at high temperatures.
Chapter 9 -
Cored vs Equilibrium Phases
• Cα changes as we solidify.
• Cu-Ni case: First α to solidify has Cα = 46 wt% Ni.
Last α to solidify has Cα = 35 wt% Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First α
to solidify:
46 wt% Ni
Last α
to solidify:
< 35 wt% Ni
Uniform C α
:
35 wt% Ni
Chapter 9 - 30
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 60 80 100
Cu
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 - 31
Binary-Eutectic Systems
has a special composition
with a min. melting T.
2 components
Ex.: Cu-Ag system Cu-Ag
system
T(°C) 1200
• 3 single phase regions L (liquid)
1000
(L, α, β
) α
L + α
779°C
• Limited solubility: L +β
β
800
TE
α
: mostly Cu 8.0
71.9 91.2
β
:
mostly Ag 600
• TE : No liquid below TE α+
β
400
• CE : Min. melting TE
composition 200
• Eutectic transition
L(CE)
0
α(CαE) + β(CβE)
20
40
60 CE 80
100
Co , wt% Ag
Adapted from Fig. 9.7,
Callister 7e.
Chapter 9 - 32
More Information on Eutectic Systems
(I)
3 phases are in eqm
Eutectic isotherm
Chapter 9 -
More Information on Eutectic Systems
(II)
•  Eutectic reaction – transition between liquid
and mixture of two solid phases, α + β, at
eutectic concentration CE. –  The melting point of the eutectic alloy is lower than
that of the components (eutectic = easy to melt in
Greek).
•  At most two phases can be in equilibrium
within a phase field. Three phases (L, α, β) may
be in equilibrium only at a few points along the
eutectic isotherm.
•  Single phase regions are separated by 2-phase
regions.
Chapter 9 -
EX: Pb-Sn Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: α + β
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
--the relative amount
of each phase:
Cβ - CO
S
=
Wα
=
R+S
Cβ - Cα
300
L (liquid)
L+
α
α
200 100
40 - 11
29
=
= 33 wt%
99 - 11
88
183°C
18.3
150
99 - 40
59
=
=
= 67 wt%
99 - 11
88
C - Cα
Wβ
= R = O
Cβ - C α
R+S
=
Pb-Sn
system
61.9
R
0 11 20
Cα
L +β
β
97.8
S
α
+
β
40
Co
Adapted from Fig. 9.8,
Callister 7e.
60
80
C, wt% Sn
99100
Cβ
Chapter 9 - 35
EX: Pb-Sn Eutectic System (2)
• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...
--the phases present: α + L
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Cα = 17 wt% Sn
CL = 46 wt% Sn
--the relative amount
of each phase:
CL - C O
46 - 40
=
Wα
=
CL - C α
46 - 17
6
=
= 21 wt%
29
300
220
200
Pb-Sn
system
L (liquid)
L+ α
α
R
S
L +β
β
183°C
100
CO - Cα
23
=
WL =
= 79 wt%
CL - C α
29
0
α
+
β
17 20
Cα
40 46 60
Co CL
Adapted from Fig. 9.8,
Callister 7e.
80
C, wt% Sn
Chapter 9 - 36
100
Microstructures
in Eutectic Systems: I
• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of α grains
i.e., only one solid phase.
T(°C)
L: Co wt% Sn
400
L
α
L
300
α
200
α: Co wt% Sn
TE
100
Adapted from Fig. 9.11,
Callister 7e.
L + α
(Pb-Sn
System)
α
+ β
0
Co
10
20
30
Co, wt% Sn
2
(room T solubility limit)
Chapter 9 - 37
Microstructures
in Eutectic Systems: II
L: Co wt% Sn
T(°C)
• 2 wt% Sn < Co < 18.3 wt% Sn 400
• Result:
§  Initially liquid + α
§  then α alone
§  finally two phases
Ø  α polycrystal
Ø  fine β-phase inclusions
L
300
L+α
α
200
α: Co wt% Sn
TE
α
β
100
α
+
β
0
Adapted from Fig. 9.12,
Callister 7e.
(sol.
L
α
10
20
Pb-Sn
system
30
Co
Co , wt%
2
limit at T room )
18.3
(sol. limit at TE)
Sn
Chapter 9 - 38
Microstructures
in Eutectic Systems: III
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of α and β crystals.
T(°C)
L: Co wt% Sn
300
Pb-Sn
system
L+ α
200
α
100
TE
0
L
183°C
α + β
20
18.3
Adapted from Fig. 9.13,
Callister 7e.
40
Micrograph of Pb-Sn
eutectic
microstructure
L + β β
β: 97.8 wt% Sn
α: 18.3 wt%Sn
60
CE
61.9
80
160 µm
Adapted from Fig. 9.14, Callister 7e.
100
97.8
C, wt% Sn
Chapter 9 - 39
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
7e.
Chapter 9 - 40
Microstructures
in Eutectic Systems: IV
• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: α crystals and a eutectic microstructure
T(°C)
300
Pb-Sn
system
200
TE
100
L: Co wt% Sn
L
α
L
• Just above TE :
α
L
Cα
= 18.3 wt% Sn
CL = 61.9 wt% Sn
Wα
= S = 50 wt%
R + S
WL = (1- Wα
) = 50 wt%
L +
α
α
R
β
S
R
L+β
S
• Just below TE :
α
+
β
primary α
eutectic α
eutectic β
0
20
18.3
Adapted from Fig. 9.16,
Callister 7e.
40
60
61.9
80
Co, wt% Sn
100
97.8
Cα
= 18.3 wt% Sn
C
β
= 97.8 wt% Sn
Wα
= S = 73 wt%
R + S
Wβ
= 27 wt%
Chapter 9 - 41
Hypoeutectic & Hypereutectic
300
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.)
200
TE
100
α
L
L+
α
20
40
hypoeutectic: Co = 50 wt% Sn
(Figs. 9.14 and 9.17
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
α
α
L+β
β
α
60
80
Co, wt% Sn
hypereutectic: (illustration only)
eutectic: Co = 61.9 wt% Sn
β
β
α
Adapted from
Fig. 9.17, Callister 7e.
100
eutectic
61.9
α
α
175 µm
(Pb-Sn
System)
α
+
β
0
β
β
β
β
160 µm
eutectic micro-constituent
Adapted from Fig. 9.14,
Callister 7e.
Adapted from Fig. 9.17,
Callister 7e. (Illustration
only)
Chapter 9 - 42
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 - 43
Eutectoid & Peritectic
•  Eutectic - liquid in equilibrium with two solids
L cool α + β
heat
•  Eutectoid - solid phase in equation with two solid
phases
intermetallic compound
S2
S1+S3
- cementite
γ cool α + Fe3C
(727ºC)
heat
•  Peritectic - liquid + solid 1 à solid 2 (Fig 9.21)
S1 + L
S2
cool
δ + L heat γ
(1493ºC)
Chapter 9 - 44
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition δ
γ+ε
Peritectic transition γ + L
Adapted from
Fig. 9.21, Callister 7e.
Chapter 9 - 45
δ
Iron-Carbon (Fe-C) Phase Diagram
L ⇒ γ + Fe3C
-Eutectoid (B):
T(°C)
1600
δ
L
1400
1200
γ ⇒ α + Fe3C
γ +L
γ
(austenite)
1000
600
120 µm
Result: Pearlite =
alternating layers of
α and Fe3C phases
(Adapted from Fig. 9.27, Callister 7e.)
S
γ +Fe3C
B
727°C = Teutectoid
R
400
0
(Fe)
S
1
0.76
C eutectoid
α
L+Fe3C
R
γ γ
γ γ
800
A
1148°C
2
3
α+Fe3C
4
5
6
Fe3C (cementite)
• 2 important
points
-Eutectic (A):
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
α (ferrite-soft)
Adapted from Fig. 9.24,Callister 7e.
Chapter 9 - 46
Hypoeutectoid Steel
T(°C)
1600
δ
L
γ γ
γ γ
γ +L
γ
1200
(austenite)
γ γ
γ γ
1000
800
γ + Fe3C
r s
727°C
αRS
w α =s /(r +s) 600
w γ = (1- wα )
400
0
α
(Fe)
pearlite
α + Fe3C
1
C0
2
3
4
5
0.76
α
γ
γ
α
γ αγ
L+Fe3C
1148°C
w pearlite = w γ
6
(Fe-C
System)
Fe3C (cementite)
1400
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.)
6.7
Co , wt% C
100 µm Hypoeutectoid
steel
w α = S/(R + S)
w Fe3C =(1- w α )
pearlite
proeutectoid ferrite
Adapted from Fig. 9.30,Callister 7e.
Chapter 9 - 47
Hypereutectoid Steel
T(°C)
1600
δ
L
γ +L
γ
1200
(austenite)
γ
γ
γ
γ
γ +Fe3C
r
800
w Fe3C =r /( r +s)
w γ
=(1- w Fe3C )
α R
600
400
0
(Fe)
pearlite
L+Fe3C
1148°C
1000
0.76
Fe3C
γ
γ
γ
γ
γ
γ
γ
γ
(Fe-C
System)
s
S
1 Co
w pearlite = w γ
w α
=S/( R +S)
w Fe3C =(1- w α
)
α +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 µmHypereutectoid
steel
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33,Callister 7e.
Chapter 9 - 48
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 (α)
b)  the amount of carbide (cementite) in grams
that forms per 100 g of steel
c)  the amount of pearlite and proeutectoid
ferrite (α)
Chapter 9 - 49
Chapter 9 – Phase Equilibria
Solution: a) composition of Fe3C and ferrite (α)
b)  the amount of carbide
(cementite) in grams that
forms per 100 g of steel
CO = 0.40 wt% C
Cα = 0.022 wt% C
CFe C = 6.70 wt% C
3
1600
δ
1200
0.4 " 0.022
=
x 100 = 5.7g
6.7 " 0.022
L
γ
(austenite)
! = 94.3 g
γ + Fe3C
727°C
R
S
α + Fe3C
600
400
0
L+Fe3C
1148°C
1000
800
Fe3C = 5.7 g
γ +L
Fe C (cementite)
Fe3C
Co " C!
1400
=
x100 T(°C)
Fe3C + ! CFe3C " C!
Cα CO
1
2
3
4
5
6
Co , wt% C
6.7
CFe
3C
Chapter 9 - 50
Chapter 9 – Phase Equilibria
c.  the amount of pearlite and proeutectoid ferrite (α)
note: amount of pearlite = amount of γ just above TE
1600
δ
L
1400
T(°C)
γ +L
"
Co $ C#
γ
=
x 100 = 51.2 g 1200
(austenite)
" + # C" $ C#
1000
γ + Fe3C
800
pearlite = 51.2 g
proeutectoid α = 48.8 g
727°C
RS
600
400
0
L+Fe3C
1148°C
α + Fe3C
1
Cα CO Cγ
2
3
4
5
6
Co , wt% C
Chapter 9 - 51
Fe C (cementite)
Co = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 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 - 52
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 - 53
ANNOUNCEMENTS
Reading:
Core Problems:
Self-help Problems:
Chapter 9 - 54