Chapter 3 phase diagram

solidification of metal,allotropy of metal,iron caebon equilibrium diagram, phase transfer,metallurgy,material science

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

2. 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.
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3. USEFUL TERMINOLOGY
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4. •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.
01/03/19 PROF.MAYUR S MODI 4

5. 01/03/19 PROF.MAYUR S MODI 5
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

6. •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.
01/03/19 PROF.MAYUR S MODI 6

7. •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.
01/03/19 PROF.MAYUR S MODI 7

8. •Variable :
•A particular phase exists under various conditions of
temp., pressure and concentrations.
•These parameter are known as the variables of the
phase.
01/03/19 PROF.MAYUR S MODI 8

9. •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
01/03/19 PROF.MAYUR S MODI 9

10. •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.
01/03/19 PROF.MAYUR S MODI 10

11. •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.
01/03/19 PROF.MAYUR S MODI 11

12. 01/03/19 PROF.MAYUR S MODI 12

13. PHASE DIAGRAM OF WATER
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14. (A) SCHEMATIC REPRESENTATION OF THE ONE-COMPONENT PHASE
DIAGRAM FOR H
2
O. (B) A PROJECTION OF THE PHASE DIAGRAM
INFORMATION AT 1 ATM GENERATES A TEMPERATURE SCALE LABELED
WITH THE FAMILIAR TRANSFORMATION TEMPERATURES FOR H
2
O
(MELTING AT 0°C AND BOILING AT 100°C).

15. (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

16. 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.
01/03/19 PROF.MAYUR S MODI 16

17. 01/03/19 PROF.MAYUR S MODI 17
1.Atomic size or Relative Size Factor
2.Chemical-affinity Factor
3.Relative Valence (Valency) Factor
4.Crystal Structure Factor

18. 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
01/03/19 PROF.MAYUR S MODI 18

19. 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.
01/03/19 PROF.MAYUR S MODI 19

20. 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!!!)
01/03/19 PROF.MAYUR S MODI 20
Latent heat
B1

21. GIBBS FREE ENERGY THEORY
01/03/19 PROF.MAYUR S MODI 21

22. 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.
01/03/19 PROF.MAYUR S MODI 22

23. 01/03/19 PROF.MAYUR S MODI 23
Homogeneous or self Nucleation. Heterogeneous Nucleation.
Interiors of a uniform substanceStart 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.

24. COOLING CURVE FOR PURE METAL
01/03/19 PROF.MAYUR S MODI 24
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)

25. COOLING CURVE FOR BINARY SOLID
SOLUTION
01/03/19 PROF.MAYUR S MODI 25
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)

26. COOLING CURVE FOR BINARY
EUTECTIC ALLOY
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27. COOLING CURVE FOR OFF-EUTECTIC
BINARY ALLOY
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28. PLOTTING OF EQUILIBRIUM OR
PHASE DIAGRAM
Sample1 2 3 4 5 6 7 8 9 10 11
% Cu1009080 70 60 50 40 30 20 10 0
% Ni0 1020 30 40 50 60 70 80 90 100
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29. 01/03/19 PROF.MAYUR S MODI 29

30. 01/03/19 PROF.MAYUR S MODI 30

31. 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.
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32. 01/03/19 PROF.MAYUR S MODI 32

33. •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)
01/03/19 PROF.MAYUR S MODI 33

34. 01/03/19 PROF.MAYUR S MODI 34
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.

35. 01/03/19 PROF.MAYUR S MODI 35
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 + S2Pb-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.

36. 01/03/19 PROF.MAYUR S MODI 36
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. 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

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

39. 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)
a (FCC solid solution)
• 3 phase fields:
L
L +a
a
wt% Ni204060801000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
a
(FCC solid
solution)
L + a
liquidus
solidus
01/03/19 Prof. MAYUR S. MODI

40. wt% Ni204060801000
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
a
(FCC solid
solution)
L + a
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: a
B(1250°C, 35):
2 phases: L + a
B

(
1
2
5
0
°
C
,
3
5
)
A(1100°C,60)
01/03/19 Prof. MAYUR S. MODI

41. wt% Ni
20
1200
1300
T(°C)
L (liquid)
a
(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni
system
PHASE DIAGRAMS :
COMPOSITION OF PHASES
• Rule 2: If we know T and Co, then we know:
--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 a and L
CL = C
liquidus ( = 32 wt% Ni here)
Ca = C
solidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid (a)
Ca = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
B
TB
D
TD
tie line
4
Ca
3
01/03/19 Prof. MAYUR S. MODI

42. • Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
At TA: Only Liquid (L)
W
L = 100 wt%, Wa = 0
At TD: Only Solid (a)
W
L = 0, Wa = 100 wt%
Co = 35 wt% Ni
PHASE DIAGRAMS :
WEIGHT FRACTIONS OF PHASES
wt% Ni
20
1200
1300
T(°C)
L (liquid)
a
(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni
system
TA
A
35
Co
32
CL
B
TB
D
TD
tie line
4
Ca
3
RS
At TB: Both a and L
% 73
3243
3543
wt=
-
-
=
= 27 wt%
W
L
=
S
R+S
W
a
=
R
R+S
01/03/19 Prof. MAYUR S. MODI

43. •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)
M
L
M
a
R S
RMSM
L×=×
a
wt% Ni
20
1200
1300
T(°C)
L (liquid)
a
(solid)L + a
liquidus
solidus
30 40 50
L + a
B
TB
tie line
C
o
C
L
C
a
SR
=Must balance
01/03/19 Prof. MAYUR S. MODI

44. wt% Ni
20
1200
1300
30 40 50
1100
L (liquid)
a
(solid)
L + a
L + a
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; a phase
field extends from
0 to 100 wt% Ni.
• Consider
Co = 35 wt%Ni.
EX: COOLING IN A CU-NI BINARY
46
35
43
32
a: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
a: 36 wt% Ni
B
a: 46 wt% Ni
L: 35 wt% Ni
C
D
E
24 36
01/03/19 Prof. MAYUR S. MODI

45. CONSTRUCTION OF EQUILIBRIUM
DIAGRAM
01/03/19 PROF.MAYUR S MODI 45

46. 01/03/19 PROF.MAYUR S MODI 46

47. 01/03/19 PROF.MAYUR S MODI 47
Examples : Pb-As, Bi-Cd, Au-Si

48. EUTECTIC TRANSFORMATION
01/03/19 PROF.MAYUR S MODI 48
1
2

49. HYPOEUTECTIC TRANSFORMATION
01/03/19 PROF.MAYUR S MODI 49
1
2
3

50. 01/03/19 PROF.MAYUR S MODI 50
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)]

51. HYPEREUTECTIC TRANSFORMATION
01/03/19 PROF.MAYUR S MODI 51
1
2
3

52. 01/03/19 PROF.MAYUR S MODI 52
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)]

53. 01/03/19 PROF.MAYUR S MODI 53

54. PARTIALY EUTECTIC SYSTEM AT 6% OF CU
01/03/19 PROF.MAYUR S MODI 54
Examples : Ag-Cu, Pb-Sn, Sn-Bi, PB-Sb, Cd-Zn, Al-Si

55. 01/03/19 PROF.MAYUR S MODI 55
1
2
3 4

56. 01/03/19 PROF.MAYUR S MODI 56
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.

57. PARTIALY HYPOEUTECTIC SYSTEM AT 20% OF CU
01/03/19 PROF.MAYUR S MODI 57

58. 01/03/19 PROF.MAYUR S MODI 58
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.

59. PARTIALY HYPEREUTECTIC SYSTEM AT
70% OF CU
01/03/19 PROF.MAYUR S MODI 59

60. 01/03/19 PROF.MAYUR S MODI 60
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.

61. LEAD-TIN(PB-SN)
01/03/19 PROF.MAYUR S MODI 61

62. : Min. melting T
E

2 components
has a ‘special’ composition with a min.
melting Temp.
BINARY-EUTECTIC (POLYMORPHIC)
SYSTEMS
• Eutectic transition
L(C
E
) a(C
aE
) + b(C
bE
)
• 3 single phase regions
(L, a, b)
• Limited solubility:
a: mostly Cu
b: mostly Ag
• T
E
: No liquid below T
E
• C
E
composition
Ex.: Cu-Ag system
Cu-Ag
system




L (liquid)
aL + a
L+b
b
a + b
C
o
, wt% Ag
204060801000
200
1200
T(°C)
400
600
800
1000
C
E
T
E 8.0 71.991.2
779°C

63. L+a
L+b
a+b
200
T(°C)
18.3
C, wt% Sn
20 60801000
300
100
L (liquid)
a
183°C

61.997.8
b
• 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)
a + b
--compositions of phases:
C
O
= 40 wt% Sn
--the relative amount
of each phase (by lever rule):
150
40
C
o
11
C
a
99
C
b
SR
C
a
= 11 wt% Sn
C
b
= 99 wt% Sn
W
a=
C
b
- C
O
C
b
- C
a
=
99 - 40
99 - 11
=
59
88
= 67 wt%
S
R+S
=
W
b=
C
O
- C
a
C
b
- C
a
=
R
R+S
=
29
88
= 33 wt%=
40 - 11
99 - 11
Adapted from Fig. 9.8,
Callister 7e.

64. L+b
a+b
200
T(°C)
C, wt% Sn
20 60801000
300
100
L (liquid)
a

b
L+ a
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)
a + L
--compositions of phases:
C
O
= 40 wt% Sn
--the relative amount
of each phase:
W
a=
C
L
- C
O
C
L
- C
a
=
46 - 40
46 - 17
=
6
29
= 21 wt%
W
L=
C
O
- C
a
C
L - C
a
=
23
29
= 79 wt%
40
C
o
46
C
L
17
C
a
220
SR
C
a
= 17 wt% Sn
C
L
= 46 wt% Sn

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

66. • 2 wt% Sn < Co < 18.3 wt% Sn
• Result:
 Initially liquid
 Then liquid + a
 then a alone
 finally two solid phases
 a polycrystal
 fine b-phase inclusions
MICROSTRUCTURES IN EUTECTIC SYSTEMS:
II
Pb-Sn
system
L + a
200
T(°C)
C
o
, wt% Sn
10
18.3
200
C
o
300
100
L
a
30
a+b
400
(sol. limit at T
E)
T
E
2
(sol. limit at Troom)
L
a
L: C
o wt% Sn
a
b
a: C
o
wt% Sn

67. • Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
MICROSTRUCTURES IN EUTECTIC SYSTEMS:
III
160 mm
Micrograph of Pb-Sn
eutectic
microstructure
Pb-
Sn
syste
m
L + b
a + b
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
a

b
L+ a
183°C
40
T
E
18.3
a: 18.3 wt%Sn
97.8
b: 97.8 wt% Sn
C
E
61.9
L: C
o
wt% Sn
45.1% a and
54.8% b -- by
Lever Rule

68. LAMELLAR EUTECTIC STRUCTURE

69. • 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
MICROSTRUCTURES IN EUTECTIC SYSTEMS:
IV
18.3 61.9
SR
97.8
SR
primary a
eutectic a
eutectic b
W
L
= (1-Wa) = 50 wt%
Ca = 18.3 wt% Sn
C
L
= 61.9 wt% Sn
S
R + S
Wa = = 50 wt%
• Just above T
E
:
• Just below T
E
:
Ca = 18.3 wt% Sn
Cb = 97.8 wt% Sn
S
R + S
Wa = = 72.6 wt%
Wb = 27.4 wt%
Pb-Sn
system
L+b
200
T(°C)
C
o
, wt% Sn
20 60 80 1000
300
100
L
a
b
L+a
40
a+b
T
E
L: C
o
wt% Sn L
a
L
a

70. L+a
L+b
a+b
200
C
o
, wt% Sn20 60 801000
300
100
L
a b
T
E
40
(Pb-Sn
System)
HYPOEUTECTIC & HYPEREUTECTIC
COMPOSITIONS
160 mm
eutectic micro-constituent
hypereutectic: (illustration only)
b
b
b
b
b
b
from Metals Handbook, 9th
ed., Vol. 9, Metallography and
Microstructures, American
Society for Metals, Materials
Park, OH, 1985.
175 mm
a
a
a
a
a
a
hypoeutectic: C
o
= 50 wt% Sn
T(°C)
61.9
eutectic
eutectic: C
o
= 61.9 wt% Sn

71. Eutectoid: solid phase in equilibrium with two solid phases
S
2
S
1
+S
3
g a + Fe
3
C (727ºC)
intermetallic compound - cementite
cool
heat
Peritectic: liquid + solid 1 in equilibrium with a single solid
2 (Fig 9.21)
S
1
+ L S
2
d + L g
(1493ºC)
cool
heat
•Eutectic: a liquid in equilibrium with two solids
L a + b
EUTECTOID & PERITECTIC – SOME
DEFINITIONS
cool
heat

72. PERITECTIC TRANSFORMER
FE-C,PT-AG,CU-ZN,SB-SN
•LIQUID + SOLID 1
01/03/19 PROF.MAYUR S MODI 72
SOLID2(1492⁰C)

73. EUTECTOID REACTION
FE-C,CU-SN,CU-AL,ZN-AL
01/03/19 PROF.MAYUR S MODI 73
SOLID1 + SOLID2 SOLID3 (727⁰C)

74. 01/03/19 PROF.MAYUR S MODI 74

75. INTRODUCTION TO IRON
CARBON DIAGRAM
01/03/19 PROF.MAYUR S MODI 75

76. 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.

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

78. TRANSFORMATION DURING HEATING AND COOLING OF STEEL

79. 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 Fe
3
C
Hard &brittle

80. 01/03/19 PROF.MAYUR S MODI 80

81. 01/03/19 PROF.MAYUR S MODI 81

82. 01/03/19 PROF.MAYUR S MODI 82

83. 01/03/19 PROF.MAYUR S MODI 83

84. 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

85. 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.

86. Cementite
Cementite is a chemical compound of carbon with iron and is known
as iron carbide (Fe
3
C).
Cast iron having 6.67% carbon is possessing complete structure of
cementite.
It is extremely hard.
It is magnetic below 200°C.

87. 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.

88. 01/03/19 PROF.MAYUR S MODI 88
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)
727-910Above temp. Free ferrite gets dissolved to 100
% ferrite.
ACM(critic
al hyper-
eutectoid
steeeel)
727-1147Above temp. Free cementite gets dissolved to
100 % austenite.
A4 (UCT)1400-
1492
Above temp. Austenite gets transformed into δ
– ferrite.

89. 01/03/19 PROF.MAYUR S MODI 89
Peritectic reaction
0.55 % C
0.55 % C
1400⁰C
1493⁰C
1539⁰C

90. 01/03/19 PROF.MAYUR S MODI 90

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

92. Nucleation & growth of pearlite

93. IRON-CARBON (FE-C) PHASE DIAGRAM
• 2 important
points
-Eutectoid (B):

gÞa+Fe
3
C
-Eutectic (A):

LÞg+Fe
3
C
F
e
3
C

(
c
e
m
e
n
t
it
e
)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 66.7
L
g
(austenite)
g+L
g+Fe
3C
a+Fe
3
C
a
+
g
L+Fe
3
C
d
(Fe)
C
o
, wt% C
1148°C
T(°C)
a
727°C = Teutectoid
A
SR
4.30
Result: Pearlite =
alternating layers of
a and Fe
3
C phases
120 mm
(Adapted from Fig. 9.27, Callister 7e.)
g
g
gg
R S
0.76
C
e
u
t
e
c
t
o
id
B
Fe
3
C (cementite-hard)
a(ferrite-soft)
Max. C solubility in g iron =
2.11 wt%

94. 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.)
F
e
3
C

(
c
e
m
e
n
t
it
e
)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 66.7
L
g
(austenite)
g+L
g + Fe
3
C
a
+ Fe
3
C
L+Fe
3
C
d
(Fe)
C
o
, wt% C
1148°C
T(°C)
a
727°C
(Fe-C
System)
C
0
0
.
7
6
proeutectoid ferritepearlite
100 mm
Hypoeutectoid
steel
RS
a
wa =S/(R+S)
w
Fe
3C =(1-wa)
w
pearlite = wg
pearlite
rs
wa=s/(r+s)
wg=(1-wa)
g
gg
g
a
a
a
gg
gg
gg
gg

95. HYPEREUTECTOID STEEL
F
e
3
C

(
c
e
m
e
n
t
it
e
)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 66.7
L
g
(austenite)
g+L
g +Fe
3
C
a +Fe
3
C
L+Fe
3
C
d
(Fe)
C
o
, wt%C
1148°C
T(°C)
a
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
.
7
6C
o
proeutectoid
Fe
3
C
60 mm
Hypereutectoid
steel
pearlit
e
R S
wa =S/(R+S)
w
Fe
3
C =(1-wa)
w
pearlite= wg
pearlite
sr
w
Fe
3
C =r/(r+s)
wg =(1-w
Fe
3
C)
Fe
3
C
gg
g
g
gg
g
g
gg
g
g

96. 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 Fe
3
C 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)

97. SOLUTION:
g 3.94
g 5.7 CFe
g7.5100
022.07.6
022.04.0

100x
CFe
CFe
3
CFe3
3
3
=a
=
=
-
-
=
-
-
=
a+
a
a
x
CC
CC
o
b)the amount of carbide
(cementite) in grams that
forms per 100 g of steel
a) composition of Fe
3
C and ferrite (a)
C
O
= 0.40 wt% C
C
a
= 0.022 wt% C
C
Fe C
= 6.70 wt% C
3
F
e
3
C

(
c
e
m
e
n
t
it
e
)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 66.7
L
g
(austenite)
g+L
g + Fe
3
C
a
+ Fe
3
C
L+Fe
3
C
d
C
o
, wt% C
1148°C
T(°C)
727°C
C
O
R S
C
Fe
C
3C
a

98. SOLUTION, CONT:
c)the amount of pearlite and proeutectoid ferrite (a)
note: amount of pearlite = amount of g just
above T
E
C
o
= 0.40 wt% C
C
a
= 0.022 wt% C
C
pearlite
= C
g
= 0.76 wt% C

 
g
g+a
=
C
o
-C
a
C
g
-C
a
x 100=51.2 g
pearlite = 51.2 g
proeutectoid a = 48.8 g
F
e
3
C

(
c
e
m
e
n
t
it
e
)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 66.7
L
g
(austenite)
g+L
g + Fe
3
C
a
+ Fe
3
C
L+Fe
3
C
d
C
o
, wt% C
1148°C
T(°C)
727°C
C
O
RS
C
gC
a
Looking at the Pearlite:
11.1% Fe
3
C (.111*51.2 gm = 5.66 gm) & 88.9% a (.889*51.2gm = 45.5 gm)
\ total a = 45.5 + 48.8 = 94.3 gm

99. 01/03/19 PROF.MAYUR S MODI 99

100. 01/03/19 PROF.MAYUR S MODI 100