Chapter 3 phase diagram

PROF. MAYUR S MODI
ASSISTANT PROFESSOR
MECHANICAL ENGINEERING DEPARTMENT
SHREE SWAMI ATMANAND SARASWATI INSTITUTE OF TECHNOLOGY,SURAT
PHASE ANDPHASE AND
PHASE DIAGRAMPHASE DIAGRAM
Material Science and Metallurgy
(2131904)

INTRODUCTION
• One of the important objectives of engineering
metallurgy is to determine the properties of material.
• The properties of material is a function of the
microstructure which is further dependent on the
overall composition and variables such as temp.,
pressure and composition.
• Equilibrium diagram or phase diagram is a graphical
representation of various phases present in the
material system at various temp. and compositions.
01/03/19 PROF.MAYUR S MODI 2

USEFUL TERMINOLOGY

• A pure substance, under equilibrium conditions,
may exist as either of a phase namely vapor, liquid
or solid, depending upon the conditions of
temperature and pressure.
• A phase can be defined as a homogeneous portion
of a system that has uniform physical and chemical
characteristics i.e. it is a physically distinct from
other phases, chemically homogeneous and
mechanically separable portion of a system.
• Other words, a phase is a structurally homogeneous
portion of matter.

It is a region that differs in it’s microstructure and composition
from another region.
For the same composition, different crystal structures represent
different phases.
A solid solution has atoms mixed at atomic level thus it represents
a single phase.
A single-phase system is termed as homogeneous, and systems
composed of two or more phases are termed as mixtures or
heterogeneous. Most of the alloy systems and composites are
heterogeneous

• System
Thermodynamically, a system is an isolated body of
matter. It refers to a specific portion of a object
within specified boundaries subjected to specified
variables.
OR
This refers to any portion of objective space within
specified boundaries subject to specified variables.
An alloy system is a combination of two or more
elements forming alloys which are considered
within a specified range of temp. , pressure and
concentration.

• Phase : It is a physically and chemically
homogeneous composition of a system, separated
from other portions by surface and an interface.
• But each portion have different composition and
properties.
• In an equilibrium diagram, liquid is one phase and
solid solution is another phase.

• Variable :
• A particular phase exists under various conditions of
temp., pressure and concentrations.
• These parameter are known as the variables of the
phase.

• Component – These are the substances, either
chemical elements or chemical compounds whose
presence is necessary and sufficient to make a
system.
• Pure metal is a one-component system whereas an
alloy of two metal is a two-component system.
• H2o 1 component = Ice, Water and Steam( 3
phase)
• Cu-Ni = 2 component

• Alloy :
• It is a mixture of two or more elements having
metallic properties.
• In the mixture, metal is in the large proportion and
the others can be metals or non-metals.
• The element in the largest amount is called as base
metal(Parent metal) or solvent and the other
elements are called as alloying elements or solute.

• Phase equilibrium –
• it refers to the set of conditions where more than
one phase may exist.
• It can be reflected by constancy with time in the
phase characteristics of a system.
• In most metallurgical and materials systems, phase
equilibrium involves just solid phases.

PHASE DIAGRAM OF WATER

(A) SCHEMATIC REPRESENTATION OF THE ONE-COMPONENT PHASE
DIAGRAM FOR H2O. (B) A PROJECTION OF THE PHASE DIAGRAM
INFORMATION AT 1 ATM GENERATES A TEMPERATURE SCALE LABELED
WITH THE FAMILIAR TRANSFORMATION TEMPERATURES FOR H2O
(MELTING AT 0°C AND BOILING AT 100°C).

(A) SCHEMATIC REPRESENTATION OF THE ONE-COMPONENT PHASE
DIAGRAM FOR PURE IRON. (B) A PROJECTION OF THE PHASE DIAGRAM
INFORMATION AT 1 ATM GENERATES A TEMPERATURE SCALE LABELED
WITH IMPORTANT TRANSFORMATION TEMPERATURES FOR IRON. THIS
PROJECTION WILL BECOME ONE END OF IMPORTANT BINARY
DIAGRAMS, SUCH AS THAT SHOWN IN FIGURE

HUME-ROTHERY’S RULES FOR SOLID
SOLUTION
• Solid solution is an alloy of two or more elements
wherein the atomic crystal structure of the alloying
element (solute) is same as that of the base metal
matrix(solvent).
• The solubility limit of the solute in the solvent is
govern by certain factors.
• These governing factors are known as Heme-
Rothery’s rules for solid solubility.

1.Atomic size or Relative Size Factor
2.Chemical-affinity Factor
3.Relative Valence (Valency) Factor
4.Crystal Structure Factor

GIBB’S PHASE RULE
• Dr.Gibbs studied the relationship between the
number of phases and the effect of variables such as
pressure, temperature and composition.
P + F = C + 2
P = Number of phase in system
F = Number of variables that can be changed independently
without affecting the number of phase.= DOF
C = Number of element.
2 = It represents any two variable amongst three

In general, all equilibrium diagrams are studied at
constant pressure, hence rule is modified to.
P + F = C + 1
•The phase rule helps determine maximum number of
phase present in an alloy system under equilibrium
conditions at any point in phase diagram.

COOLING CURVE FOR PURE METAL
• used to determine phase transition temperature
record T of material vs time, as it cools from its molten state through
solidification and finally to RT (at a constant pressure!!!)
Latent heat
B1

GIBBS FREE ENERGY THEORY

NUCLEATION
• Nucleation is the beginning of a phase
transformation.
• It is marked by the appearance in the molten metal
of tiny regions called nuclei of new phase which
grow to solid crystals until the transformation is
complete.
1.Homogeneous or self Nucleation.
2.Heterogeneous Nucleation.

Homogeneous or self Nucleation. Heterogeneous Nucleation.
Interiors of a uniform substance Start at the nucleation sites on the
surface contacting liquid or vapor.
Slower process Faster process
It occurs with much more
difficulties.
Occurs easily.
Nuclei are formed from atoms of
solidifying metals.
At impurity atoms or container
surface acts as a nucleating agent.
Lower temperature. Higher temperature.
It requires supercoiling to form first
nuclei.
It requires little or no supercoiling.

COOLING CURVE FOR PURE METAL
Region AB,
P + F = C + 1
1 + F = 1 + 1
F=1(T without
changing liquid phase)
Region BC,
P + F = C + 1
2 + F = 1 + 1
F=0
(No variable)
Region CD,
P + F = C + 1
1 + F = 1 + 1
F=1(T without
changing a solid
phase)

COOLING CURVE FOR BINARY SOLID
SOLUTION
Region AB,
P + F = C + 1
1 + F = 2 + 1
F=2(T and C without
changing liquid phase)
Region BC,
P + F = C + 1
2 + F = 2 + 1
F=1
(T without changing
liquid-solid phase)
Region CD,
P + F = C + 1
1 + F = 2 + 1
F = 2
(T and C without
changing solid phase)

COOLING CURVE FOR BINARY
EUTECTIC ALLOY

COOLING CURVE FOR OFF-EUTECTIC
BINARY ALLOY

PLOTTING OF EQUILIBRIUM OR
PHASE DIAGRAM
Sample 1 2 3 4 5 6 7 8 9 10 11
% Cu 100 90 80 70 60 50 40 30 20 10 0
% Ni 0 10 20 30 40 50 60 70 80 90 100

LEVER RULE
• It is the method used to find out the exact amount
of a particular phase existing in a binary system for
a given alloy at any temperature under
consideration.
• Let us consider an alloy A and B.
• Z be the composition of alloy under consideration
and T be the temperature at which phase content
is to be found.

• According to lever rule,
• % of liquid =Intercept distance between Z% B alloy
under consideration and the solidus line/Distance
between solidus and liquidus line
So, % of liquid = L(FD) X 100
L(CD)
Similarly,
• % of Solid = Intercept distance between Z% B alloy
under consideration and the solidus line/Distance
between solidus and liquidus line
So, % of liquid = L(CD) X 100
L(CD)

NOW, Amount of solid = L(CF) / L (CD) = L(CF)
Amount of liquid L (FD)/ L (CD) L(FD)
Therefor,
amount of solid x L(FD) = amount of liquid x L(CF)
It means, the line CD acts as a liver
arm and the point F acts as a
fulcrum point as shown in fig,
hence it is called as lever rule OR
lever arm principle.

SR
NO
TYPE OF
EQUILIBRIUM
SYSTEM
SOLUBILITY REACTION EXAMPLE
1 Isomorphous
system
Two metals
completely
soluble in solid
and liquid
state.
Cu-Ni, Au-Ag,
Au-Cu, Au-Ni,
Bi-Sb etc.
2 Eutectic system Two metals
completely
soluble in liquid
and insoluble
in solid state.
L S1 + S2 Pb-As, Bi-Cd,
Au-Si etc.
3 Partial eutectic
system
Two metals
completely
soluble in liquid
and partially
soluble in solid
state.
Ag-Cu, Pb-Sn,
Pb-Sb etc.
Constant temp.

SR
NO
TYPE OF EQUILIBRIUM
SYSTEM
REACTION EXAMPLE
4 Eutectoid
transformation
S1 S2+S3 Fe-C,Cu-Sn
5 Peritectic
transformation
L1+ S1 S2 Fe-C, Cu-Zn
6 Monotectic
transformation
L1 L2 + S1 Cu-Pb, Zn-Pb
7 Peritectoid
transformation
S1 + S2 S3 Ni-Zn, Cu-Sn
8 Layer type system. Two metals
completely
insoluble in liquid
and solid state.
Cu-Mo,Ag-Fe

37
Metals – Phase diagram
The following things are known from
(1) Pure copper melts at 1084 C
(2) Pure nickel melts at 1455 C
(3) The system is a solid solution throughout
(4) Below solidus line – solid
(5) Above liquidus line – liquid
(6) Between, two phases: solid and liquid

Metals – Phase diagram
copper-nickel alloy system
01/03/19 Prof. MAYUR S. MODI

PHASE DIAGRAMS
• Indicate phases as function of T, Compos, and Press.
• 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
• 2 phases:
L (liquid)
α (FCC solid solution)
• 3 phase fields:
L
L + α
α
wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
α
(FCC solid
solution)
L + αliquidus
solidus

wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
α
(FCC solid
solution)
L
+ α
liquidus
solidus
Cu-Ni
phase
diagram
PHASE DIAGRAMS:
# AND TYPES OF PHASES
• Rule 1: If we know T and Co, then we know:
--the number and types of phases present.
• Examples:
A(1100°C, 60):
1 phase: α
B(1250°C, 35):
2 phases: L + α
B(1250°C,35) A(1100°C,60)

wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L +α
liquidus
solidus
30 40 50
L +α
Cu-Ni
system
PHASE DIAGRAMS:
COMPOSITION OF PHASES
--the composition of each phase.
• Examples:
TA
A
35
Co
32
CL
At TA = 1320°C:
Only Liquid (L)
CL = Co ( = 35 wt% Ni)
At TB = 1250°C:
Both α and L
CL = Cliquidus ( = 32 wt% Ni here)
Cα = Csolidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( α)
Cα = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
B
TB
D
TD
tie line
4
Cα
3

--the amount of each phase (given in wt%).
• Examples:
At TA: Only Liquid (L)
WL = 100 wt%, Wα = 0
At TD: Only Solid ( α)
WL = 0, Wα = 100 wt%
Co = 35 wt% Ni
PHASE DIAGRAMS:
WEIGHT FRACTIONS OF PHASES
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L +α
liquidus
solidus
30 40 50
L +α
Cu-Ni
system
TA
A
35
Co
32
CL
B
TB
D
TD
tie line
4
Cα
3
R S
At TB: Both α and L
%73
3243
3543
wt=
−
−
=
= 27 wt%
WL
= S
R +S
Wα
= R
R +S

• Tie line – connects the phases in equilibrium with
each other - essentially an isotherm
THE LEVER RULE
How much of each phase?
Think of it as a lever (teeter-totter)
ML
Mα
R S
RMSM L ⋅=⋅α
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L +α
liquidus
solidus
30 40 50
L +α
B
TB
tie line
CoCL Cα
SR
=Must balance

wt% Ni
20
1200
1300
30 40 50
1100
L (liquid)
α
(solid)
L +
α
L +
α
T(°C)
A
35
Co
L: 35wt%Ni
Cu-Ni
system
• 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 100 wt% Ni.
• Consider
Co = 35 wt%Ni.
EX: COOLING IN A CU-NI BINARY
4635
43
32
α: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
α: 36 wt% Ni
Bα: 46 wt% Ni
L: 35 wt% Ni
C
D
E
24 36

CONSTRUCTION OF EQUILIBRIUM
DIAGRAM

Examples : Pb-As, Bi-Cd, Au-Si

EUTECTIC TRANSFORMATION
1
2

HYPOEUTECTIC TRANSFORMATION
1
2
3

At just below point 3
The already separated pro-eutectic A remains unchanged.
% Amount pro-eutectic A = l(3E)/l(DE) ……….at point 3.
% Amount eutectic = l(D3)/l(DE) ……….at point 3.
At point 4
Total amount of metal A at point 4
=[amount of pro-eutectic A] + [amount of eutectic A]
=[amount of pro-eutectic A] + [ A in the eutectic x amount of eutectic]
=[l(3E)/l(DE)] + [l(EF)/l(DF) x l(D3)/l(DE)]

HYPEREUTECTIC TRANSFORMATION
1
2
3

At just below point 3
The already separated pro-eutectic A remains unchanged.
% Amount pro-eutectic B = l(E3)/l(EF) ……….at point 3.
% Amount eutectic = l(3F)/l(EF) ……….at point 3.
At point 4
Total amount of metal A at point 4
=[amount of pro-eutectic B] + [amount of eutectic B]
=[amount of pro-eutectic B] + [ B in the eutectic x amount of
eutectic]
=[l(E3)/l(EF)] + [l(DE)/l(DF) x l(3F)/l(EF)]

PARTIALY EUTECTIC SYSTEM AT 6% OF CU
Examples : Ag-Cu, Pb-Sn, Sn-Bi, PB-Sb, Cd-Zn, Al-Si

1
2
3 4

At point 5.
% Amount of α = l(5K)/l(HK)
= (98-6/98-2) = 95.8 %....................at room temp.
% Amount of = l(H5)/l(HK)
= (6-2/98-2) = 4.2%...........................at room
temp.

PARTIALY HYPOEUTECTIC SYSTEM AT 20% OF CU

1
2
3 % Amount pro-eutectic = l(3E)/l(DE)
= (28.1-20)/(28.1-8.8)
= 42% ….at point 3.
% Amount eutectic = l(D3)/l(DE)
= (20-8.8)/(28.1-8.8)
= 58 % …...at
point 3.

PARTIALY HYPEREUTECTIC SYSTEM AT
70% OF CU

1
2
3 % Amount pro-eutectic β = l(E3)/l(EF)
= (70—28.1)/(92.1-28.1)
= 65.5% ….at point 3.
% Amount eutectic = l(3F)/l(EF)
= (92.1-70)/(92.1-28.1)
= 34.5 %
…...at point 3.

LEAD-TIN(PB-SN)

: Min. melting TE
2 components
has a ‘special’ composition with a min.
melting Temp.
BINARY-EUTECTIC (POLYMORPHIC)
SYSTEMS
• Eutectic transition
L(CE) α(CαE) + β(CβE)
• 3 single phase regions
(L, α, β)
• Limited solubility:
α: mostly Cu
β: mostly Ag
• TE : No liquid below TE
• CE
composition
Ex.: Cu-Ag system
Cu-Ag
system
L (liquid)
α L + α
L+ββ
α + β
Co , wt% Ag
20 40 60 80 1000
200
1200
T(°C)
400
600
800
1000
CE
TE 8.0 71.991.2
779°C

L+α
L+β
α + β
200
T(°C)
18.3
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α 183°C
61.9 97.8
β
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: Pb-Sn
system
EX: PB-SN EUTECTIC SYSTEM (1)
α + β
--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase (by lever rule):
150
40
C
o
11
Cα
99
Cβ
SR
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
Wα=
Cβ - CO
Cβ - Cα
=
99 - 40
99 - 11
=
59
88
= 67 wt%
S
R+S
=
Wβ =
CO - Cα
Cβ - Cα
=
R
R+S
=
29
88
= 33 wt%=
40 - 11
99 - 11
Adapted from Fig. 9.8,
Callister 7e.

L+β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α β
L+α
183°C
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...
--the phases present: Pb-Sn
system
EX: PB-SN EUTECTIC SYSTEM (2)
α + L
--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase:
Wα =
CL - CO
CL - Cα
=
46 - 40
46 - 17
=
6
29
= 21 wt%
WL =
CO - Cα
CL - Cα
=
23
29
= 79 wt%
40
C
o
46
CL
17
Cα
220 SR
Cα = 17 wt% Sn
CL = 46 wt% Sn

• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of α grains
i.e., only one solid phase.
MICROSTRUCTURES IN EUTECTIC SYSTEMS:
I
0
L+ α
200
T(°C)
Co, wt% Sn
10
2%
20
Co
300
100
L
α
30
α+β
400
(room Temp. solubility limit)
TE
(Pb-Sn
System)
α
L
L: Co wt% Sn
α: Co wt% Sn

• 2 wt% Sn < Co < 18.3 wt% Sn
• Result:
 Initially liquid
 Then liquid + α
 then α alone
 finally two solid phases
 α polycrystal
 fine β-phase inclusions
II
Pb-Sn
system
L + α
200
T(°C)
Co , wt% Sn
10
18.3
200
Co
300
100
L
α
30
α+ β
400
(sol. limit at TE)
TE
2
(sol. limit at Troom)
L
α
L: Co wt% Sn
α
β
α: Co wt% Sn

• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of α and β crystals.
III
160µm
Micrograph of Pb-Sn
eutectic
microstructure
Pb-
Sn
syste
m
L+β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
183°C
40
TE
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn
CE
61.9
L: Co wt% Sn
45.1% α and
54.8% β -- by
Lever Rule

• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: α crystals and a eutectic microstructure
IV
18.3 61.9
SR
97.8
SR
primary α
eutectic α
eutectic β
WL = (1-Wα) = 50 wt%
Cα = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
R + S
Wα= = 50 wt%
• Just above TE :
• Just below TE :
Cα = 18.3 wt% Sn
Cβ = 97.8 wt% Sn
S
R + S
Wα= = 72.6 wt%
Wβ = 27.4 wt%
Pb-Sn
system
L+β200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
40
α+β
TE
L: Co wt% Sn Lα
L
α

L+α
L+β
α+β
200
Co, wt% Sn20 60 80 1000
300
100
L
α βTE
40
(Pb-Sn
System)
HYPOEUTECTIC & HYPEREUTECTIC
COMPOSITIONS
160 µm
eutectic micro-constituent
hypereutectic: (illustration only)
β
β
β
β
β
β
from Metals Handbook, 9th
ed., Vol. 9, Metallography and
Microstructures, American
Society for Metals, Materials
Park, OH, 1985.
175 µm
α
α
α
α
α
α
hypoeutectic: Co = 50 wt% Sn
T(°C)
61.9
eutectic
eutectic: Co =61.9wt% Sn

 Eutectoid: solid phase in equilibrium with two solid phases
S2 S1+S3
γ α + Fe3C (727ºC)
intermetallic compound - cementite
cool
heat
 Peritectic: liquid + solid 1 in equilibrium with a single solid
2 (Fig 9.21)
S1 + L S2
δ + L γ
(1493ºC)
cool
heat
• Eutectic: a liquid in equilibrium with two solids
L α + β
EUTECTOID & PERITECTIC – SOME
DEFINITIONS
cool
heat

PERITECTIC TRANSFORMER
FE-C,PT-AG,CU-ZN,SB-SN
• LIQUID + SOLID 1
SOLID2(1492⁰C)

EUTECTOID REACTION
FE-C,CU-SN,CU-AL,ZN-AL
SOLID1 + SOLID2 SOLID3 (727⁰C)

INTRODUCTION TO IRON
CARBON DIAGRAM

ALLOTROPY OF IRON
In actual practice it is very difficult to trace the cooling of iron from
1600°C to ambient temperature because particular cooling rate is
not known.
Particular curve can be traced from temperature, time and
transformation (TTT) curve.
However allotropic changes observed during cooling of pure iron are
depicted in Fig.

Molten-Fe (Liquid state of iron)
Delta-Fe (Body centered)
austenite
structure

TRANSFORMATION DURING HEATING AND COOLING OF STEEL

Principal phases of steel and their
Characteristics
Phase
Crystal
structure
Characteristics
Ferrite BCC Soft, ductile, magnetic
Austenite FCC
Soft, moderate
strength, non-
magnetic
Cementite
Compound of Iron
& Carbon Fe3C Hard &brittle

1. Austenite
Austenite is a solid solution of free carbon (ferrite) and
iron in gamma iron.
On heating the steel, after upper critical temperature,
the formation of structure completes into austenite
which is ductile and non-magnetic.
It is formed when steel contains carbon up to 1.8% at
1130°C.
On cooling below 723°C, it starts transforming into
pearlite and ferrite.
Structures in Fe-C-diagram

Ferrite
Ferrite contains very little or no carbon in iron.
It is the name given to pure iron crystals which are soft and ductile.
The slow cooling of low carbon steel below the critical temperature
produces ferrite structure.
Ferrite does not harden when cooled rapidly.

Cementite
Cementite is a chemical compound of carbon with iron and is known
as iron carbide (Fe3C).
Cast iron having 6.67% carbon is possessing complete structure of
cementite.
It is extremely hard.
It is magnetic below 200°C.

Pearlite
Pearlite is a eutectoid alloy of ferrite and cementite.
As the carbon content increases beyond 0.2% in the temperature at
which the ferrite is first rejected from austenite drop until, at or
above 0.8% carbon, no free ferrite is rejected from the austenite.
This steel is called eutectoid steel, and it is the pearlite structure in
composition.

SYMBOL TEMP.⁰ C SIGNIFICANCE
A0 (curie
temp.)
210 Above temp. Cementite losses its magnetism
A1 (LCT) 727 Above temp. perlite gets transformed in
austenite
A2 (curie
temp.)
768 Above temp. Ferrite losses its magnetism
A3
(critical
hypo-
eutectoid
steeeel)
727-910 Above temp. Free ferrite gets dissolved to 100
% ferrite.
ACM(critic
al hyper-
eutectoid
steeeel)
727-1147 Above temp. Free cementite gets dissolved to
100 % austenite.
A4 (UCT) 1400-
1492
Above temp. Austenite gets transformed into δ
– ferrite.

Peritectic reaction
0.55 % C
0.55 % C
1400⁰C
1493⁰C
1539⁰C

The Austenite to ferrite / cementite transformation in
relation to Fe-C diagram

Nucleation & growth of pearlite

IRON-CARBON (FE-C) PHASE DIAGRAM
• 2 important
points
-Eutectoid (B):
γ ⇒ α +Fe3C
-Eutectic (A):
L ⇒ γ +Fe3C
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ+Fe3C
α+Fe3C
α+γ
L+Fe3C
δ
(Fe)
Co, wt% C
1148°C
T(°C)
α 727°C = Teutectoid
A
SR
4.30
Result: Pearlite =
alternating layers of
α and Fe3C phases
120 µm
(Adapted from Fig. 9.27, Callister 7e.)
γ γ
γγ
R S
0.76
CeutectoidB
Fe3C (cementite-hard)
α (ferrite-soft)
Max. C solubility in γ iron =
2.11 wt%

HYPOEUTECTOID STEEL
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.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ+Fe3C
α+Fe3C
L+Fe3C
δ
(Fe)
Co, wt% C
1148°C
T(°C)
α
727°C
(Fe-C
System)
C0
0.76
proeutectoid ferritepearlite
100 µm
Hypoeutectoid
steel
RS
α
wα =S/(R+S)
wFe3C =(1-wα)
wpearlite = wγ
pearlite
r s
wα =s/(r+s)
wγ =(1- wα)
γ
γ γ
γ
α
α
α
γγ
γ γ
γ γ
γγ

HYPEREUTECTOID STEEL
Fe3C(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ+Fe3C
α +Fe3C
L+Fe3C
δ
(Fe)
Co, wt%C
1148°C
T(°C)
α
adapted from Binary
Alloy Phase Diagrams,
2nd ed., Vol. 1, T.B.
Massalski (Ed.-in-Chief),
ASM International,
Materials Park, OH,
1990.
(Fe-C
System)
0.76
Co
proeutectoid
Fe3C
60 µmHypereutectoid
steel
pearlit
e
R S
wα =S/(R+S)
wFe3C =(1-wα)
wpearlite = wγ
pearlite
sr
wFe3C =r/(r+s)
wγ =(1-w Fe3C )
Fe3C
γγ
γ γ
γγ
γ γ
γγ
γ γ

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)

SOLUTION:
g3.94
g5.7CFe
g7.5100
022.07.6
022.04.0
100x
CFe
CFe
3
CFe3
3
3
=α
=
=
−
−
=
−
−
=
α+ α
α
x
CC
CCo
b) the amount of carbide
(cementite) in grams that
forms per 100 g of steel
a) composition of Fe3C and ferrite (α)
CO = 0.40 wt% C
Cα = 0.022 wt% C
CFe C = 6.70 wt% C
3
FeC(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
Co, wt% C
1148°C
T(°C)
727°C
CO
R S
CFe
C
3Cα

SOLUTION, CONT:
c) the amount of pearlite and proeutectoid ferrite (α)
note: amount of pearlite = amount of γ just
above TE
Co = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 0.76 wt% C
γ
γ+α
=
Co −Cα
Cγ−Cα
x 100 =51.2 g
pearlite = 51.2 g
proeutectoid α = 48.8 g
FeC(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
Co, wt% C
1148°C
T(°C)
727°C
CO
RS
CγC
α
Looking at the Pearlite:
11.1% Fe3C (.111*51.2 gm = 5.66 gm) & 88.9% α (.889*51.2gm = 45.5 gm)

Chapter 3 phase diagram

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