Materials Science
TEGS 3591
Equilibrium phase diagrams
Prof. O. T. Johnson
(Department of Mechanical and Metallurgical Engineering)
Outline
• Lesson Objectives
• Definitions and Basic Concepts
• Binary Phase Diagrams
• The Iron-Carbon System
• Summary
2
Lesson Objectives
3
Lesson Objectives
At the end of this lesson, you should be able to do the following:
1. Schematically sketch simple isomorphous and eutectic phase diagrams.
(a) On these diagrams label the various phase regions,
(b) Label liquidus, solidus, and solvus lines.
2. Given a binary phase diagram, the composition of an alloy, its
temperature, and assuming that the alloy is at equilibrium, determine
(a) what phase(s) is (are) present,
(b) the composition(s) of the phase(s), and
(c) the mass fraction(s) of the phase(s).
4
Lesson Objectives
At the end of this lesson, you should be able to do the following:
3. For some given binary phase diagram, do the following:
(a) locate the temperatures and compositions of all eutectic,
eutectoid, peritectic, and congruent phase transformations; and
(b) write reactions for all these transformations for either heating
or cooling.
5
Lesson Objectives
At the end of this lesson, you should be able to do the following:
4. Given the composition of an iron–carbon alloy containing between
0.022 wt% C and 2.14 wt% C, be able to
(a) specify whether the alloy is hypoeutectoid or hypereutectoid,
(b) name the proeutectoid phase,
(c) compute the mass fractions of proeutectoid phase and pearlite, and
(d) make a schematic diagram of the microstructure at a temperature
just below the eutectoid
6
Why study Phase Diagrams?
The understanding of phase diagrams for alloy
systems is extremely important because there is
a strong correlation between microstructure
and mechanical properties, and the
development of microstructure of an alloy is
related to the characteristics of its phase diagram
(i.e. number of phases, their proportions and the
way they are arranged).
7
Why study Phase Diagrams?
In addition, phase diagrams provide
valuable information about melting,
casting, crystallization, and other
phenomena
8
Definitions and Basic
Concepts
9
Equilibrium
• A system is at equilibrium if its free energy is at a
minimum, given a specified combination of
temperature, pressure and composition.
• The (macroscopic) characteristics of the system do
not change with time — the system is stable.
• A change in T, P or C for the system will result in an
increase in the free energy and possible changes to
another state whereby the free energy is lowered.
10
10
Equilibrium Phase Diagrams
• Phase diagrams show how two or more
different elements can co-exist together
over a range of compositions and
temperatures.
• Elements co-exist after they have either:
• Dissolved in each other to form a Solid
Solution, or
• Chemically combined to form a
Compound.
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Alloys
• When a metal coexists with another
element, it forms an Alloy. Two different
elements form a Binary Alloy.
• Usually an Alloy is made up of a
combination of Solid solutions and a
Chemical Compound.
• When two metals combine, they form an
Intermetallic Compound.
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Solid Solutions
Interstitial Solid solutions
Interstitial solute atom
Substitutional Solid solutions
Substitutional solute atom
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Hume-Rothery Rules
Rules for Substitutional Solid Solutions:
(i)
The solute and solvent atoms must have
similar atomic size (within 15%).
The solute and solvent atoms must have
(ii)
similar crystal structure.
(iii) The solute and solvent atoms must have
similar valence.
(iv) The solute and solvent atoms must have
similar electronegativity (ability of an atom to
attract an electron).
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Rules of Solubility
• Solute atoms dissolve into the solvent atom
structure.
• Usually, solubility increases with increasing
temperature.
• Solute atoms dissolve until a Saturated
Solution is formed.
• If solute atoms dissolve beyond saturation
point, Super-saturated Solution is formed.
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Solubility limit
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase Sugar/Water Phase Diagram
• Solubility Limit:
Maximum concentration for
which only a single phase
solution exists.
Question: What is the
solubility limit for sugar in
water at 20°C?
C = Composition (wt% sugar)
Sugar
65
Water
Answer: 65 wt% sugar.
At 20°C, if C < 65 wt% sugar: syrup
At 20°C, if C > 65 wt% sugar: syrup + sugar
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Phases and Grains
• Since an alloy consists of solid solutions and
compounds, these will appear as regions of
different structures under the microscope.
• Regions of different structures are called phases,
and these are made up of Crystals.
• A Phase is a distinct part of a material with the
same physical and chemical properties as well as
the same chemical composition.
• Groups of Crystals with the same orientation are
called Grains.
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Grains of Ferrite (solid solution) and Pearlite
(solution + compound) in a 0.2%C Steel,x500
Grain
Boundaries
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Phase Diagrams
Components and Phases
• Components:
The elements or compounds that are mixed initially (Al and Cu).
• Phases:
A phase is a homogenous, physically distinct and mechanically separable
portion of the material with a given chemical composition and structure
( and ).
AluminumCopper
Alloy
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Components & Phases
21
Types of Phase
Diagrams
22
One Component Phase
Diagram
23
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Outline
• Lesson Objectives
• Definitions and Basic Concepts
• Binary Phase Diagrams
• The Iron-Carbon System
• Summary
2
Binary Phase
Diagrams
3
Binary Phase Diagrams
Binary phase diagrams involving two components
are maps that represent the relationships between
temperature and the compositions and quantities
of phases at equilibrium, which influence the
microstructure of an alloy.
•
•
•
•
Binary Isomorphous System
Binary Eutectic System
Binary Eutectoid System
Binary Peritectic System
4
Binary Isomorphous System
Alloys of Copper and Nickel
• Nickel and Copper satisfy all Hume-Rothery rules. Ni
and Cu have complete solubility in each other (100%).
Ni and Cu show complete miscibility in solid and
liquid states.
• Alloys of copper and nickel are called monels.
Element
Atomic Radius
(nm)
Crystal
Structure
Valence
Copper (Cu)
0.128
fcc
1+, 2+
Nickel (Ni)
0.125
fcc
2+
5
Binary Isomorphous System
Cu-Ni system:
• The liquid L is a homogeneous liquid solution composed of
Cu and Ni.
• The α phase is a substitutional solid solution consisting of
Cu and Ni atoms with an FCC crystal structure.
• At temperatures below 1080 C, Cu and Ni are mutually
soluble in each other in the solid state for all compositions.
• The complete solubility is explained by their FCC structure,
nearly identical atomic radii and electro-negativities, and
similar valences.
• The Cu-Ni system is termed isomorphous because of this
complete liquid and solid solubility of the 2 components.
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6
Cu-Ni Alloys/System
7
Cu-Ni Alloys/System
8
Cu-Ni Alloys/System
9
Cu-Ni Alloys/System
10
Cu-Ni Alloys/System
11
Cooling Curves
Cooling curves of a Pure Metal and a Binary Alloy
(a) Pure Metal
Temp
(b) BinaryAlloy
Temp
Liquid (L)
Liquid state
L+ S
Liquidus
Liquid +
Solid
Tm
Solid (S)
Solidus
Solid state
Time
Time
A Phase Diagram is constructed by plotting Liquidus and
Solidus Temperatures at different Chemical Compositions.
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Cooling Curves
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Construction of phase diagrams
Equilibrium Phase Diagram of a System with Complete Solubility
Temperature C
LIQUID (L)
Liquidus
a
L+ 
b
Tie line
TA
TB
c
L+ 
Solidus
SOLID ()
A
Clf
Cl
Co
Cs
Csi
B
Composition (wt % B)
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Binary Isomorphous System
• Phase diagram:
Cu-Ni system.
• System is:
T(°C)
1600
1500
L (liquid)
-- binary
2 components:
Cu and Ni.
-- isomorphous
i.e., complete
solubility of one
component in
another;  phase
field extends from
0 to 100 wt% Ni.
Cu-Ni
phase
diagram
1400
1300

1200
(FCC solid
1100
1000
solution)
0
20
40
60
80
100 wt% Ni
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15
Binary Isomorphous System
• Indicate phases as a function of Temp, Comp and Pressure.
• Focus on:
- binary systems: 2 components.
- independent variables: T and C (P = 1 atm is almost always used).
T(°C)
1600
• 2 phases:
L (liquid)
1500
Cu-Ni
system
 (FCC solid solution)
L (liquid)
• 3 different phase fields:
L
L+ 
1400
1300

1200
(FCC solid
1100
1000

solution)
0
20
40
60
80
100 wt% Ni
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16
Effect of Temperature &
Composition (Co)
• Changing T can change # of phases: path A to B.
• Changing Co can change # of phases: path B to D.
T(°C)
1600
1500
L (liquid)
1400
Cu-Ni
system
D
B
1300

1200
A
1100
1000
Cu 0
20
40
(FCC solid solution)
60
80
100
wt% Ni
39
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Binary Isomorphous System
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Determination of phase(s) present
• Rule 1: If we know T and Co, then we know:
--how many phases and which phases are present.
• Examples:
Cu-Ni
phase
diagram
Melting points: Cu =
1085°C, Ni = 1453 °C
Solidus - Temperature where alloy is completely solid. Above this line, liquefaction begins.
Liquidus - Temperature where alloy is completely liquid. Below this line, solidification begins.
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Determination of composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
Cu-Ni
system
At TA = 1320°C:
Only Liquid (L) present
CL = C0 ( = 35 wt% Ni)
At TD = 1190°C:
Only Solid () present
C = C0 ( = 35 wt% Ni)
At TB = 1250°C:
Both  and L present
CL = C liquidus ( = 32 wt% Ni)
C = C solidus ( = 43 wt% Ni)
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Lever Rule
• Relative amounts of phases present in an alloy at a
given temperature may be calculated using the lever
rule, which states:
fs =
Co - Cl
Cs - Cl
fl
=
Cs - Co
Cs - Cl
• Where fs is the proportion of solid phase and fl is the
proportion of the liquid phase
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Illustration of the Lever Rule
Illustration of the Lever Rule
Cs - Cl
Cl
Cs
Co - Cl
Cs - Co
Co
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Determination of weight fractions of
phases
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Cu-Ni system
WL
W
=
43 − 35
= 73 wt %
43 − 32
= 27wt %
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Development of Microstructure in
Isomorphous Systems
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Limited Solubility with  AND 
• The binary alloy formed from limited solid
solubility of say element A in element B
has two different solid phases designated
as  and  phases.
• The  phase is the solid solutions of B
atoms in element A.
• The  phase is the solid solution of A
atoms in element B.
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Invariant Reactions
• Invariant points in a binary alloy system define points
where three different phases coexist in equilibrium.
• When temperature is reduced slightly below the
invariant point, the high temperature phase will fully
transform to the low temperature phases.
• Transformations that occur at the invariant points are
isothermal and reversible. They are called invariant
reactions.
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6
Typical Invariant Reactions
(a)
EutecticReaction
Liquid
(b)
solid + solid 
[  +  ]
PeritecticReaction
Liquid + solid 
(d)
[ L +  ]
EutectoidReaction
Solid 
(c)
solid + solid 
solid 
[ L +  ]
solid 
[  +  ]
PeritectoidReaction
Solid  + solid 
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Eutectic
• A eutectic or eutectic mixture is a mixture of two or more
phases at a composition that has the lowest melting point.
• It is where the phases simultaneously crystallize from molten
solution.
• The proper ratios of phases to obtain a eutectic is identified by
the eutectic point on a binary phase diagram.
• The term comes from the Greek 'eutektos', meaning 'easily
melted.‘
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Binary Eutectic Phase Diagrams
• In a binary eutectic phase diagram, there
exists an invariant point called the
eutectic point at which a liquid of
eutectic composition CE solidifies
isothermally to produce two distinct solid
phases  and .
• The temperature at which the eutectic
transformation takes place is called the
eutectic temperature T .
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E
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Hypothetical Binary Eutectic
Phase Diagram
TB
LIQUID (L)
Liquidus
Temperature
C
TA
Liquidus
Solidus
L+

E
TE
Solvus
L +
Solidus

TE
E is the
Eutectic Point
and CE
the Eutectic
Composition
Solvus
 +
RT
0
CE
(100% A)
100%B
(0% A)
Composition (wt % B)
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Binary Eutectic System
has a special composition
with a min. melting T.
2 components
Cu-Ag system
• 3 single phase regions
(L, , )
• Limited solubility:
: mostly Cu
: mostly Ag
• TE : No liquid below TE
• CE : Composition at
temperature TE
T(°C)
1200
L (liquid)
1000

TE 800
• Eutectic reaction
L(CE)
L(71.9 wt% Ag)
(CE) + (CE)
cooling
heating
L + 
779°C
8.0
L + 
71.9 91.2
600
 + 
400
200
0
20
40
100
60 CE 80
C , wt% Ag
(8.0 wt% Ag) + (91.2 wt% Ag)
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Copper-Silver Phase Diagram
12
Copper-Silver Phase Diagram
•
•
•
•
•
Solvus – (solid solubility line) BC, GH
Solidus – AB, FG, BEG (eutectic isotherm)
Liquidus – AEF
Maximum solubility: α = 8.0 wt% Ag, β = 8.8 wt %Cu
Invariant point (where 3 phases are in equilibrium) is at E; CE =
71.9 wt% Ag, TE = 779C (1434F).
• An isothermal, reversible reaction between two (or more) solid
phases during the heating of a system where a single liquid
phase is produced.
Eutectic reaction
L(CE)
L(71.9 wt% Ag)
(CE) + (CE)
cooling
heating
(8.0 wt% Ag) + (91.2 wt% Ag)
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Lead-Tin Phase Diagram
Temperature C
350
327
300
LIQUID (L)
L+
250

200
232C
183C
61.9%
(Eutectic)
19.2%
150
100
L +
95%

 +
50
0
0
Pb
20
40
60
80
100%
Sn
(0% Pb)
(100% Pb)
Composition (wt% Sn)
(Adopted from Metals Handbook, volume 8: Metallography, Structures and Phase
Diagrams, 8th edition, ASM International, 1986)
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Lead-Tin Eutectic Alloy
• Eutectic alloys have many engineering uses since
these alloys have melting points much lower than
the melting points of the pure elements.
• The lead-tin eutectic system produces the material
for joining metals called solder.
• The eutectic alloy has a composition of 61.9% Sn
(38.1%Pb) and a melting point of 183C.
• The maximum solubility of tin in lead is 19.2%.
This occurs at the eutectic temperature of 183C.
• The maximum solubility of lead in tin is about 5%.
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Exercise 1
(1)Referring to the given Pb-Sn binary equilibrium
phase diagram, calculate the proportion of α-phase and
β-phase in a lead-tin alloy of eutectic composition just
below the eutectic temperature.
(2)Referring to the given Pb-Sn binary equilibrium
phase diagram, calculate the proportion of α and β
phases in an alloy of lead and tin that contains 40% Sn
at a temperature of 182C.
Use the Lever rule.
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Lead-Tin Equilibrium Phase Diagram: Tie Line (red) and
Composition lines (blue) for the given Exercise
Temperature
C
350
327
300
LIQUID (L)
L+
250

200
232 C
183C
19.2%
150
100
61.9%
(Eutectic)
L +
95%

 +
50
0
0
Pb
20
40
60
(100% Pb)
80
100%
Sn
(0% Pb)
Composition (wt% Sn)
(Adopted from Metals Handbook, volume 8: Metallography, Structures and Phase
Diagrams, 8th edition, ASM International, 1986)
FPL Kavishe
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Lead-Tin Phase Diagram
Liquidus
Solidus
Solidus
Solidus
Solvus
Solvus
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Solidification of Eutectic Mixtures
• Mixtures of some metals, such as copper & nickel, are completely soluble in
both liquid and solid states for all concentrations of both metals. Copper &
nickel have the same crystal structure (FCC) and have nearly the same
atomic radii. The solid formed by cooling can have any proportion of copper
& nickel. Such completely miscible mixtures of metals are called
isomorphous.
• By contrast, a mixture of lead & tin that is eutectic is only partially soluble
when in the solid state. Lead & tin have different crystal structures (FCC
versus BCT) and lead atoms are much larger. No more than 18.3 weight %
solid tin can dissolve in solid lead and no more than 2.2% of solid lead can
dissolve in solid tin (according to previous phase diagram).
• The solid lead-tin alloy consists of a mixture of two solid phases, one
consisting of a maximum of 18.3 wt% tin (the alpha phase) and one
consisting of a maximum of 2.2 wt% lead (the beta phase).
19
Pb-Sn Phase Diagram
Ex. 1
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine:
-- the phases present
Answer:  + 
-- the phase compositions
Answer: C = 11 wt% Sn
C = 99 wt% Sn
-- the relative amount
of each phase
T(°C)
300
200
L (liquid)

L+ 

=
C - C0
C - C
= 99 - 40 = 59 = 0.67
88
99 - 11
W  = C0 - C
C - C
29 = 0.33
= 40 - 11 =
99 - 11
88
L + 
183°C
18.3
61.9
150
Answer:
W
Pb-Sn
system
100
97.8
 +
0 11 20
C
40
C0
60
80
C, wt% Sn
99100
C
20
Pb-Sn Phase Diagram
Ex. 2
For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:
-- the phases present: T(°C)
Answer:  + L
•
-- the phase compositions
Answer: C = 17 wt% Sn
CL = 46 wt% Sn
-- the relative amount
of each phase
300
L+ 
L (liquid)

220
200
L + 
183°C
Answer:
CL - C0
46 - 40
=
W =
CL - C 
46 - 17
6
=
= 0.21
29
C0 - C
WL =
CL - C 
23
=
= 0.79
29
100
 +
0
17 20
C
40 46 60
80
C0 CL C, wt% Sn
100
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Pb-Sn Phase Diagram
• For lead & tin the eutectic composition is 61.9 wt%
tin and the eutectic temperature is 183ºC -- which
makes this mixture useful as solder.
• At 183ºC, compositions of greater than 61.9 wt% tin
result in precipitation of a tin-rich solid in the liquid
mixture, whereas compositions of less than 61.9 wt%
tin result in precipitation of lead-rich solid.
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Microstructural Developments in
Eutectic System - 1
T(°C)
L: C0 wt% Sn
400
•For alloys where
C0 < 2 wt% Sn
•Result at room temperature is a
polycrystalline with grains of 
phase having composition C0
L

L
300
200

TE
: C0 wt% Sn
+ 
100
Pb-Sn
system
L+
0
C0
10
20
30
C , wt% Sn
2
(room T solubility limit)
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Microstructural Developments in
Eutectic System - 2
Pb-Sn
system
L: C0 wt% Sn
T(°C)
400
L
L
2 wt% Sn < C0 < 18.3 wt% Sn
•Results in polycrystalline
microstructure with  grains and
small -phase particles at lower
temperatures.
300

L +
: C0 wt% Sn

200
TE


100
 +
0
10
2
(sol. limit at T room )
20
C0
30
C , wt% Sn
18.3
(sol. limit at TE)
24
Microstructural Developments in
Eutectic System - 3
•Co = CE
•Results in a
eutectic
microstructure
with alternating
layers of  and
 crystals.
Pb-Sn
system
L(61.9 wt%Sn)
cooling
heating
 (18.3 wt% Sn) +  (97.8 wt% Sn)
25
Lamellar Eutectic Structure
❑A 2-phase microstructure
resulting from the
solidification of a liquid
having the eutectic
composition where the
phases exist as a lamellae
that alternate with one
another.
❑Formation of eutectic layered
microstructure in the Pb-Sn
system during solidification at
the eutectic composition.
Compositions of α and β
phases are very different.
Solidification involves
redistribution of Pb and Sn
atoms by atomic diffusion.
Pb-rich
Sn-rich
26
Pb-Sn Microstructures
26
Ni-Al
Pb-Sn
Copper phosphorus eutectic
20mol% CeO2-80mol% CoO.
Ir-Si
28
Microstructural Developments in
Eutectic System - 4
• For alloys with18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result:  phase particles and a eutectic microconstituent
T(°C)
L: C0 wt% Sn
L
300
L

200
 L

L +
L+  
TE
Primary α
eutectic 
eutectic 
0
20
18.3
• Just above TE :
C = 18.3 wt% Sn
CL = 61.9 wt% Sn
C -C
W  = L 0 = 0.50
CL - C 
WL = (1- W  ) =0.50
• Just below TE :
+ 
100
Pb-Sn
system
40
60
61.9
80
C, wt% Sn
100
97.8
C  = 18.3 wt% Sn
C  = 97.8 wt% Sn
Cβ - C0
W =
Cβ - C  = 0.727
W  = 0.273 wt% Sn
29
Binary Peritectic Phase Diagrams
• In the binary peritectic diagram, an alloy of
composition Co undergoes a peritectic reaction in
such a way that a mixture of liquid of composition ClP
and solid  of composition CP transforms at the
peritectic temperature TP to form solid  of
composition CP.
•
The invariant reaction is:
Liquid + solid 
solid 
FPL Kavishe
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Binary Peritectic Phase Diagrams
LIQUID (L)
Temperature C
TA
a
Solidus
TP
+L
Liquidus
b

TP
+L
TB
+
Solidus

Solvus
Solvus
A
CP
Ci
Co
CP
Composition (wt% B)
FPL Kavishe
ClP
B
31
Cu-Zn (Brass) Phase Diagrams
Cartridge brass:
70 wt% Cu
32
Section of Cu-Zn Phase Diagrams
1200
LIQUID (L)
1084
Temperature
C
1000
+ L
+

 +L
Y

800
600
X



+

400
’
200
 +
 +L

(Adopted from Metals
Handbook, volume 8:
Metallography, Structures and
Phase Diagrams, 8 t h edition,
A S M International, 1986)
0
Cu
20
40
60
80% Zn
( 100% Cu)
Composition (wt% Zn)
X: Alloy of composition 36.8% Zn at a temperature of 903  C.
Liquid (37.5%Zn) + Solid  (32.5%Zn)  Solid  (36.8%Zn).
FPL Kavishe
33
Eutectoid & Peritectic
Peritectic transformation  + L 
Cu-Zn Phase diagram
Eutectoid transformation

+
34
Intermetallic Compounds
19 wt% Mg-81 wt% Pb
Mg2Pb
Note: intermetallic compounds exist as a line on the diagram - not a phase
region. The composition of a compound has a distinct chemical formula.
35
Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L cool  + 
(For Pb-Sn, 183C, 61.9 wt% Sn)
heat
• Eutectoid – one solid phase transforms to two other solid phases
Solid1 ↔ Solid2 + Solid3

•
cool 
heat
+ Fe3C
(For Fe-C, 727C, 0.76 wt% C)
Peritectic - liquid and one solid phase transform to a 2nd solid
phase Solid1 + Liquid ↔ Solid2
 +L
cool
heat
ε
(For Cu-Zn, 598°C, 78.6 wt% Zn)
36
37
Outline
• Lesson Objectives
• Definitions and Basic Concepts
• Binary Phase Diagrams
• The Iron-Carbon System
• Summary
2
The Iron-Carbon
System
3
Iron - Carbon System
• Pure iron when heated experiences 2 changes
in crystal structure before it melts.
• At room temperature the stable form, ferrite
( iron) has a BCC crystal structure.
• Ferrite experiences a polymorphic
transformation to FCC austenite ( iron) at 912
˚C (1674 ˚F).
• At 1394˚C (2541˚F) austenite reverts back to
BCC phase  ferrite and melts at 1538 ˚C
(2800 ˚F).
• Iron carbide (cementite or Fe3C) an
intermediate compound is formed at 6.7 wt%
C.
• Typically, all steels and cast irons have carbon
contents less than 6.7 wt% C.
• Carbon is an interstitial impurity in iron and
forms a solid solution with the    phases
4
Iron - Carbon System
5
Though carbon is present in relatively low concentrations, it
significantly influences the mechanical properties of ferrite: (a) α
ferrite, (b) austenite.
6
Four Solid Phases
7
Iron Carbide (Cementite / Fe3C)
• Forms when the solubility limit of carbon in  ferrite is
exceeded at temperatures below 727 ˚C.
• Mechanically, cementite is very hard and brittle.
• For ferrous alloys there are 3 basic types, based on
carbon content:
❑ Iron (ferrite phase): <0.008 wt% C room temp
❑ Steel ( + Fe3C phase): 0.008 to 2.14 wt% C
❑ Cast iron: 2.14 to 6.70 wt% C
8
9
Iron-Iron Carbide Phase Diagrams
1600
1537
1495
 + L
L I Q U I D (L)
(P e ri te c tic)
1394C
 -ir o n
 + L
1300
Austenite ()
1200
1133
2% C
4.3%C
(E u te c tic )
1000
Austenite ()
910
800
 +
(Fe3C)
723C
723
0.8%C
(Eutectoid)
600
+ Cementite
-iron
 + Fe3C
C e m e n t i te
400
STEELS
CAST IRONS
RT
0
1
2
3
4
5
6
6.67
Composition (wt% C)
100% Fe
100% Fe3C
FPL Kavishe
10
Steel Portion of Fe-C System
Temperature C
1133
Acm
Austenite ()
910
Austenite +
Cementite
( + Fe3C)
A3
+
A1
723
Ec

Ferrite () + Cementite (Fe3C)
500
+P
RT
0
.2
.4
P
.6
Pearlite (P) + Fe3C
.8
1.0
1.2
1.4
1.6
1.8
2.0
(100% Fe)
Weight % Carbon
FPL Kavishe
11
Eutectoid Reaction in Steels
• The eutectoid reaction takes place at a
temperature of 723C (A1 temperature) for
steels that contain 0.8%C by weight.
• Reaction involves isothermal decomposition of
the high temperature solid phase called
austenite () to produce two solid phases called
ferrite () and cementite (Fe3C).
• For all temperatures below the eutectoid
temperature, the microstructure consists of a
eutectoid mixture of ferrite and cementite,
which is called pearlite.
FPL Kavishe
12
Eutectoid Reaction in Steels

Austenite

 + Fe3C
[at 723C]
Pearlite
• In the eutectoid reaction, the austenite
contains 0.8%C in solution, the ferrite
formed contains 0.025%C in solution and
the cementite contains 6.67% combined
carbon.
• A steel consisting of 100% pearlite (a fully
pearlitic steel) is called a eutectoid steel.
FPL Kavishe
13
Iron - Carbon System
T(°C)
• 2 important points
1600
- Eutectic (A):
1400
↔
L
+ Fe3C
- Eutectoid (B):
 ↔  + Fe3C
1200
 +L

A
 
 
1000
800
 +Fe3C
B
727°C = T eutectoid
+Fe3C
600
400
120 m
Result: Pearlite =
alternating layers of
 and Fe3C phases,
not a separate phase.
0
(Fe)
L+Fe3C
1148°C
(austenite)
Fe3C (cementite)
L

1
0.76
2
3
4
4.30
5
6
6.7
C, wt% C
Fe3C (cementite-hard)
 (ferrite-soft)
14
Eutectoid reaction:
↔
 + Fe3C
Pearlite
Redistribution of carbon by diffusion
Austenite – 0.76 wt% C
Ferrite - 0.022 wt% C
Cementite - 6.70 wt% C
15
Hypoeutectoid Steel
T(°C)
1600

L
1400

 
W =
1200
C - C 0

W =(1 - W)
 + Fe3C
800
727°C

 + Fe3C
600

pearlite 400
W’ =
CFe3C - C0
CFe3C - C
W pearlite = (1 – W’)
1
C0
0.76
 + Fe3C
0
(Fe)
L+Fe3C
1148°C
(austenite)
1000
C - C 
pearlite =
 +L
Fe3C (cementite)


2
3
4
5
6
C, wt% C
6.7
Microstructures for iron-iron carbide alloys that are below
the eutectoid with compositions between 0.022 and 0.76
wt% Carbon are hypoeutectoid.
16
Hypoeutectoid Steel
T(°C)
1600

L
1400
 
 

 

1200

 +L
1000
 + Fe3C
800
727°C

 + Fe3C
600
pearlite
1
0.76

400
0
(Fe) C0
L+Fe3C
1148°C
(austenite)
Fe3C (cementite)
 
 
2
3
4
5
6
6.7
C, wt% C
17
Proeutectoid
• Formed before the eutectoid
• Ferrite that is present in the pearlite is called eutectoid ferrite.
• The ferrite that is formed above the Teutectoid (727°C) is proeutectoid.
18
Hypereutectoid Steel
T(°C)
1600

L
1400
1200
 

 +L

1000
 +Fe3C
W =x/(v + x)

600
pearlite
400
0
(Fe)
Wpearlite = W
W = X/(V + X)
WFe3C’=(1 - W)
v x
800
V
X
0.76
WFe3C =(1-W)
L+Fe3C
1148°C
(austenite)
Fe3C (cementite)
Fe3C
1 C0
 +Fe3C
2
3
4
5
6
6.7
C, wt%C
Microstructures for iron-iron carbide alloys that have
compositions between 0.76 and 2.14 wt% carbon are
hypereutectoid (more than eutectoid).
19
Hypereutectoid Steel
1600

L
1400
 

1200

 +L
L+Fe3C
1148°C
(austenite)
1000
Fe3C (cementite)
Fe3C




 +Fe3C
800

 +Fe3C
600
400
0
(Fe)
pearlite
0.76




1 C0
2
3
4
5
6
6.7
C, wt%C
20
Hypereutectoid Steel (1.2 wt%C)
pearlite
Proeutectoid: formed above the Teutectoid (727°C)
21
Hypoeutectic & Hypereutectic
300
L
T(°C)

200
L +
L + 
TE
(Pb-Sn
System)
 +
100
0
20



60
80
eutectic
61.9
hypoeutectic: C0 = 50 wt% Sn

40
175 m
hypereutectic: (illustration only)
eutectic: C0 = 61.9 wt% Sn



C, wt% Sn
100

160 m




eutectic micro-constituent
22
Microstructure of Annealed
Plain Carbon Steels
• Steels with 0.8%C (Eutectoid steels)
Microstructure is 100% pearlite.
• Steels with less than 0.8%C (Hypoeutectoid steels)
Microstructure is a mixture of primary ferrite and
pearlite.
• Steels with more than 0.8%C, but up to 2%C
(Hypereutectoid steels)
Microstructure is a mixture of primary cementite and
pearlite
FPL Kavishe
23
Exercise
Calculate the proportions of ferrite and pearlite in a
0.5%C steel just below 723C.
Solution:
Note: Just below 723oC, ferrite contains 0.025%C and
pearlite contains 0.8%C.
% Ferrite = 0.8 – 0.5 x 100% = 38.7%
0.8 – 0.025
% Pearlite = 0.5 – 0.025 x 100% = 61.3%
0.8 – 0.025
FPL Kavishe
24
Exercise
Calculate the proportions of ferrite and cementite in a
0.5%C steel just below 723oC.
Solution:
The ferrite present is the primary ferrite plus the ferrite
contained in pearlite, i.e. the eutectoid ferrite. It contains
0.025%C in solution. The cementite is contained in the
pearlite, i.e. it is eutectoid cementite. Cementite contains
6.67%C.
% Ferrite
% Cementite
x 100% = 92.85%
= 6.67 – 0.5
6.67 – 0.025
= 0.5 – 0.025_ x 100% = 7.15%
6.67 – 0.025
FPL Kavishe
25
26
Exercise
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 ().
b) The amount of
cementite (in grams)
that forms in 100 g of
steel.
27
Exercise
C = 0.022 wt% C
CFe3C = 6.70 wt% C
Using the lever rule with
the tie line shown
WFe 3C =
=
R
R +S
=
C0 − C
CFe 3C − C
0.40 − 0.022
= 0.057
6.70 − 0.022
1600

L
1400
T(°C)
1200

 +L
1000
 + Fe3C
800
727°C
R
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
L+Fe3C
1148°C
(austenite)
S
 +Fe3C
600
4000
C C0
1
2
3
C, wt% C
4
5
6
Fe3C (cementite)
b)
6.7
CFe C
3
28
Gibbs Phase Rule
• Phase diagrams and phase equilibria are subject to the laws of thermodynamics.
• Gibbs phase rule is a criterion that determines how many phases can coexist within a
system at equilibrium.
P + F = C +N
P: # of phases present
F: degrees of freedom (temperature, pressure, composition)
C: components or compounds
N: noncompositional variables
For the Cu-Ag system @ 1 atm for a single phase P:
N=1 (temperature), C = 2 (Cu-Ag), P= 1 ( , L)
F = 2 + 1 – 1= 2
This means that to characterize the alloy within a single phase
field, 2 parameters must be given: temperature and composition.
If 2 phases coexist, for example, +L  +L, + then according to GPR, we have 1
degree of freedom: F = 2 + 1 – 2= 1. So, if we have Temp or composition, then we can
completely define the system.
If 3 phases exist (for a binary system), there are 0 degrees of freedom.
29
Gibbs Phase Rule
30
Gibbs Phase Rule
31
Gibbs Phase Rule
32
Gibbs Phase Rule
33
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
For a given temperature and composition of
the system.
• The microstructure of an alloy depends on
-- its composition, and
-- rate of cooling equilibrium
34
Classwork
1.
A 0.8% C eutectoid plain-carbon steel is slowly cooled from 750oC to a temperature
just slightly below 723oC. Assuming that the austenite is completely transformed to α ferrite
and cementite:
(a) Calculate the weight percent eutectoid ferrite formed.
(b) Calculate the weight percent eutectoid ferrite formed
2.
A 0.4% C hypoeutectoid plain-carbon steel is slowly cooled from 940oC to a
temperature just slightly above 723oC:
(a) Calculate the weight percent austenite present in the steel.
(b) Calculate the weight percent proeutectoid ferrite present in the steel.
35
Classwork
3.
A 0.4% C hypoeutectoid plain-carbon steel is slowly cooled from 940oC to a
temperature just slightly below 723oC:
(a)Calculate the weight percent austenite proeutectoid ferrite present in the steel.
(b)Calculate the weight percent eutectoid ferrite and eutectoid cementite present in the
steel.
4.
A hypoeutectoid plain-carbon steel that was is slowly cooled from the austenitic
region to room temperature contains 9.1 wt% eutectoid ferrite. Assuming no change in
structure on cooling from just below the eutectoid temperature to room temperature, what is
the carbon content of the steel?
36