Phases in Solids.pdf .

Department of Mechanical & Manufacturing Engineering, MIT, Manipal 1 of 3
CHAPTER 03
PHASES IN SOLIDS
MATERIAL SCIENCE AND METALLURGY

In material field, one may come across certain
questions such as:
• What condition is the material in? – worked, heat
treated., as-cast etc.
• What is the micro-constituent present?
• What is the composition and quantity of each
micro-constituent?
• Is micro-constituent has undesired properties?
• What will happen if temperature is increased or
decreased, pressure is changed or chemistry
varied?
• Answers to such questions can be obtained from
“Phase analysis”.

Phases and Gibb’s Phase Rule
Phase is Microscopically homogeneous body of matter.
•Pure metal- Solid, Liquid or gaseous state.
•Pure Iron- Appears in – BCC-FCC-& BCC
Phase is physically and chemically homogeneous
portion of a matter
•Its smallest parts are indistinguishable with each other.
•Ex- solid copper is homogeneous – with all respect.
Phase is a physically distinct, chemically homogeneous
and mechanically separable portion of substance.
•Phase has a unique structure, uniform composition, and
well defined boundaries.
•Phase can be a pure substance or a solution, provided-
structure and composition uniform throughout.

For most pure elements the term phase is the same with state.
For pure elements, a gaseous, liquid, and solid are different phases.
Some metals are allotropic in the solid state and will have different solid
phases.
When the metal undergoes a change in crystal structure, it undergoes
a phase change since each type of crystal structure is physically
distinct.
In the solid state there are three possible phases: (1) pure metal, (2)
intermediate alloy phase or compound, and (3) solid solution.
Solubility - If an alloy is homogeneous (composed of a single phase) in
the solid
state, it can be only a solid solution or a compound. If the alloy is a
mixture, it is then composed of any combination of the phases possible
in the solid state. It may be a mixture of two pure metals, or two solid
solutions or two compounds, or a pure metal and a solid solution, and so on.
The mixture may also vary in degree of fineness.

Gibb’s Phase Rule or Phase Rule
The Gibb’s Phase Rule is given as: P + F = C + n, explains
the phase transformation in material systems.
Used to find the number of phases present in the metallurgical
system. This rule establishes a definite relationship in an alloy system
under the given variables.
Number of phases, P that can co-exist in equilibrium in a chosen system.
The components, C
e.g., Fe-Ni alloy system is a two component (binary) system
Cu-Ni-Zn in German Silver is a three component (tertiary) system
The degrees of freedom, F is the number of independent variables like
temperature, pressure and composition that may be changed without
causing the disappearance of a phase. Temperature and pressure are
external variables where as composition is an internal variable.
The number of external factors, n is simply an integer number which
represents the number of external factors like temperature and pressure.
Generally n = 2.

Application of Gibb’s phase rule for a one-component system:
Temperature
Pressure
O
Solid phase
One-component system (Cu)
A
B
C
Gaseous phase
Liquid phase
p
pI
q
qI
Pp
Pp
I
Tp
Tp
I
O is the triple point where the three phases (solid,
liquid & gas) co-exit.
On line OA, the solid & gaseous phases of Cu
co-exit,
on line OB, the solid & liquid phases of Cu co-exist
and
on line OC, the gaseous & liquid phases of Cu
co-exit.
The term under equilibrium conditions implies
conditions of extremely slow heating and cooling.
In other words, if any change is to occur sufficient
time must be allowed for it to take place.
For one-component system, C = 1, n = 2 (pressure & temperature).
Hence, the Gibb’s phase rule becomes: P + F = C + n, P + F = 1 + 2,
i.e., F = 3 – P - - - - eq. 1.
In one-component system (may be Cu), temperature and pressure are the
variable parameters. i.e., degrees of freedom may be temperature and/or
pressure.

Gibb’s phase rule applied to single-phase region (P = 1):
The temperature and pressure at ‘p’ are ‘Tp
’ & ‘Pp
’ respectively. Now by
changing the temperature and pressure independently to Tp
I
’ & ‘Pp
I
’, the
system is brought to the point ‘pI
’ which falls in the same single phase region
(gaseous region).
F = 2. This can be verified from eq. 1.
F = 3 – P = 3 – 1 = 2, Hence for single phase region, degrees of freedom,
F = 2. i.e., temperature and pressure independently varied without disturbing
the existing phase.
Gibb’s phase rule applied to two-phase region (P = 2):
Consider a point ‘q’ on the line OA (solid-gaseous co-existence) at
temperature Tq
and pressure Pq
. Now, if the temperature is changed to Tq
1
,
the pressure also must be changed exactly to Pq
1
for the two-phase state of
the system to remain unchanged.
F = 1. This can be verified from eq. 1
F = 3 – P = 3 – 2 = 1, Hence for two phase region, degrees of freedom, F
= 1. i.e., temperature and pressure relatively varied with each other without
disturbing the two phase equilibrium (State of the system falls on the line OA).

Gibb’s phase rule applied to three-phase region (P = 3):
3 phase equilibrium is at TO
and its corresponding pressure PO
.
Any change in either temperature or pressure would change
the three-phase state of the system. Hence,
degrees of freedom, F = 0. This can be verified from eq.
F = 3 – P = 3 – 3 = 0
Hence, for three phase existing condition, degrees of freedom, F =
0. When maxm. number of phases coexisting in the
equilibrium system, the degrees of freedom is zero.
In general, if the atmospheric pressure being constant, only
temperature is used as an external variable factor. Therefore, n
=1.
The phase rule expression now takes the form:
Modified Gibb’s phase rule, P + F = C + 1, Pmax= C+1.

Solid solutions (SS)
A homogeneous mixture of atoms of two or more elements in solid state. It is a
single phase system.
Atoms of different elements in it cannot be either mechanically separated or
physically distinguished.
In a solid solution, the metal in the major portion is called the solvent.
The metal in the minor portion is called the solute.
The amount of solute that can be dissolved by any solvent is function of
temperature with pressure as constant - increases with temperature. 3
conditions exists in SS depending on the solubility like, unsaturated, saturated
and super saturated SS (SSS).The supersaturated condition is an unstable
one, and given enough time or a little extra energy, the solution tends to
become stable or saturated by rejecting or precipitating the excess solute.
Types of solid solutions:
Interstitial Solid Solutions
Substitutional Solid Solutions

Interstitial Solid Solution:
These are formed when atoms of small atomic radii fit
into the interstitial spaces of larger solvent atoms.
Atoms of elements such as carbon, nitrogen, boron,
hydrogen, etc. which have radii less than 1 Ao
are likely
to form interstitial solid solutions.
Solvent atoms
Solute atoms
Interstitial solid solution
Favorable conditions.
1. Solvent is metal (W, Fe, Mo, Co)
Solute is non metal (C, N, B)
2. The atomic radii differences is very
high ( > 15%)
3. Solubility of solute in solvent is
very less to get solid solution.

4. Solute occupies at interstitial space in solvent unit
cell.
E.g. In α-Fe, C atoms having smaller size (C atoms
radius being 0.7 Ao
) occupy the interstitial sites of Fe
atoms (Fe atoms radius being 1.4 Ao
). The solubility of
C atoms is very low in α-Fe, as low as 0.025 (wt%) at
724o
C and as low as 0.008 (wt%) at room temp.

Substitutional Solid Solution:
Solute atoms substitute the atoms of the solvent in the crystal
structure of the solvent.
For example, the Au-Ag solid solution, the Ag (silver) atoms
substitute for the Au (gold) atoms in the FCC structure of gold.
Solvent and solute atoms have almost the same size with a slight
difference. This results in only a slight distortion of the solvent
crystal lattice structure.
The crystal structure of the solute and the solvent are the same.
The solubility of the solute in solvent is usually much higher
when compared to that of interstitial solid solution.
There are 2 types of substitutional solid solutions:
• Ordered Substitutional Solid Solutions
• Disordered Substitutional Solid Solution

Subtitutional atom
Parent
atom
Disordered Substitutional Solid Solution Ordered Substitutional Solid Solution
Ordered Substitutional Solid Solution: In this type,
the solute atoms occupy some of solvent atom positions
in a orderly manner (Fixed positions). Generally
obtained on very slow cooling & only particular
composition ranges. Ex.: Super lattices.
Disordered Substitutional Solid Solution: Atoms do not
occupy any fixed positions but are distributed at random in
the lattice structure of the solvent. In this type, the
concentration of solute atoms may vary considerably
throughout the lattice structure.

Applicable to substitutional solid solutions only . The following factors are
considered in the selection of alloy
• Crystal structure factor: For complete solid solubility of two elements, both should
have the same type of crystal structures. For example, in Au – Ag, Cu – Ni, Sb-Bi
both components forming the solid solutions have same crystal structures.
• Relative Size Factor: The atoms of solute and solvent should have approximately
the same atomic size. This factor is satisfied if the difference in the atomic radii of the
two elements is less than 15%. Less than 8% higher solubility, 8 to 15% lesser
solubility and greater than 15% least or no solubility.
0.1%
Ag and Pb-----FCC, rd=20%, Sb and Bi----Rhombohedral, rd=7%
1.5% Complete solubility.
Sb- Al 0.1%, rd=2%
Hume - Rothary Rules
Al crystal structure - FCC

• Chemical Affinity Factor (Electro negativity): : The two metals should have very
less chemical affinity. Higher electro negativity of two elements, greater is the
chance of forming a compound rather than a solid solution. Greater the chemical
affinity between the two metals, lesser is the chances of forming a solid solution.
Generally, the farther apart the elements are in the periodic table, the greater is their
chemical affinity. Mg – Pb, Mg – Sn, Cu-Al
The Atomic no. of Ni is 28, Mg is 12, Cu is 29, Al is 13, Pb is 82, Sn is 50, Zn is
30. Ex. Mg3
Zn3
Al2
, Mg2
Pb, Mg2
Sn, CuAl2,
Al3
Ni -Compounds
• Relative Valence Factor: Lower valence is favourable for solid solubility. Among two
metals, which have satisfied all the above rules, the metal with lower valency to
dissolves more amount of higher valence metal and vice-versa.
Cu-Zn system, solubility of Zn (valency 2) in Cu (valency 1) is about 40%, solubility of
Cu in Zn is negligible.
5%
Al-Ni (lower valence) ---system, rd=14%, Al valence 3, Ni valence 2, both has FCC
0.04%

Intermediate Phases: It a compound and is made up of two or
more elements of which at least one of them is a metal. A
compound is a chemical combination of positive and negative
valence elements. These phases are expressed by chemical
formulae like Fe3
C, Mg2
Pb, Mg2
Sn, CuAl2
etc. Molecules, which
are smallest units of a compound. The atoms are held together by
covalent or ionic bond. Most of the intermediate phases are
crystalline in nature, exhibit definite melting points and have the
same cooling curves as pure metals. Intermediate phases are
having definite compositions and possess crystal structure
different from those of solvent metals. Types of intermediate alloy
phases are:
• Intermetallic or Valence Compounds
• Interstitial Compounds
• Electron Compounds

Intermetallic or Valence Compounds: Exclusively metal-metal
systems with narrow range of composition they are called
intermetallic compounds. These are formed between chemically
dissimilar metals and are combined by following the rules of
chemical valence. The property is usually non-metallic and
show poor ductility and poor electrical conductivity and have
complex crystal structure. Examples for intermetallic
compounds: Mg2
Pb, Mg2
Sn, CuAl2
, Mg2
Si, Mg3
Zn3
Al2
Interstitial Compounds: Similar to interstitial solid solutions
except that they have more or less a fixed composition. Fe3
C
contains 6.67% carbon by weight when it solidifies and
maintains same composition at all temp. Example: W2
C, Fe3
C,
Fe4
N . The interstitial compounds are metallic in nature, have
high melting points and are extremely hard.

Electron Compounds: A number of alloy systems formed
between Cu, Au, Ag, Fe, Ni as one group of elements with
Cd, Mg, Sn, Zn, Al as another group of elements show similarity
and three different sets of compounds are formed whose crystal
structure depends on the ratio between total number of free
electrons to total number of atoms (e/a).
If ratio is 3/2 BCC structure, 21/13 – Complex cubic and 7/4
HCP structure.
These are of variable compositions and do not obey the valence
law.
Electron compounds have properties same as those of solid
solutions – wide range of compositions, high ductility and low
hardness.

e/a=3/2 BCC e/a=21/13 complex cubic e/a=7/4 HCP
AgCd Ag5
Cd8
AgCd3
AgZn Au5
Zn8
AuZn3
CuZn (Beta Brass)
Valence of Cu is 1& Zn is 2.
Cu5
Zn8
(Gamma Brass) CuZn3
(Epsilon Brass)
Cu3
Al Cu9
Al4
Cu3
Sn
Cu3
Si Cu31
Sn8
Cu3
Si
EXAMPLES OF ELECTRON COMPOUNDS

Phase Diagrams/Equilibrium Diagrams
Recording data (T, P and COMPOSITION) about phase changes in many
alloying systems.
Phase diagrams are graphical representations of phases present in a material
system at various temperatures, pressures and compositions.
These diagrams are constructed by taking temperature on the ordinate (Y
– axis) and percentage composition on the abscissa (X – axis) EXCEPT
UNARY DIAGRAM.
These diagrams are drawn at equilibrium conditions (ideal case).
The equilibrium condition is a state where there is no change in the
system with passage of time.
Since this is not practically feasible, conditions close to equilibrium can be
attained upon extremely slow cooling or extremely slow heating so that
sufficient time is allowed for the change to take place.

Phase diagrams are constructed to provide the following
information:
1. Different phases existing in the alloy for any combination of
temperature and composition (pressure is assumed constant at
atmospheric value)
2. Temperatures at which different compositions of alloy begin and
end solidification.
3. To study and control processes such as solidification and heat
treatment of metals and alloys.
4. It helps to know the amount of different phases of the system
present at room/any temperature by which approximate
properties of the alloy can be estimated.

Classification of phase diagrams:
A. Based on number of components in the system:
1. Unary phase diagram (one component – pure metal system)
2. Binary phase diagram (two components system)
3. Ternary phase diagram (three components system)
4. Quaternary phase diagram (four components system)
Taking only 45 of the most common metals, any combination
of two gives 990 binary systems. Combinations of three give over
14,190 ternary systems.

Classification of Binary phase diagram
B) Based on solubility of solute
1. Isomorphous system
2. Eutectic System ( Type-1 Eutectic, Type-2 Eutectic)
3. Pertectic System
4. Monotectic System
5. Eutectoid transformation
6. Peritectoid transformation

Construction of a Phase Diagram:
TA
TB
L1
L2
L3
L4
S1
S2
S3
S4
1 2 4
3 5 6
Time
Composition by weight, %
A B
20%B 40%B 60%B 80%B
Temperature
L
α
L + α
Alloy 1 = 100% A & 0 % B (Pure
Metal A)
Alloy 2 = 80 % A & 20 % B
Alloy 3 = 60 %A & 40 % B
Alloy 4 = 40 % A & 60 % B
Alloy 5 = 20 % A & 40 % B
Alloy 6 = 0 % A & 100 % B (Pure
Metal B)
Liquidu
s
Solidu
s
Phase diagrams for a binary alloy
system can be obtained from the
cooling curves of the
system at various compositions of the
two components.

•Consider two pure metals A & B in the liquid states.
•Let them be mixed in different weight proportions, in the liquid
state, to obtain alloys 1 to 6 as shown.
•Each of these alloys is taken separately in their liquid state and
allowed to slow cool (Equilibrium condition).
•For each alloy, the temperatures at start of solidification and end
of solidification are noted and the cooling curves are plotted.
•Cooling curve is a plot of temperature of the alloy versus time.
Pure metals solidify isothermally
•Alloys 2 to 5, solidification of takes place over a range of
temperatures, i.e., temperature at start of solidification is greater
than the temperature at end of solidification.
Construction procedure for simple binary phase
diagram-Isomorphous system

• These temperature points are re-plotted (projected from the cooling curves) on
a temperature versus composition diagram by taking temperature on the Y-axis
and composition of the alloys on X-axis.
•L1
, L2
, L3
& L4
are the temperature of start of solidification and S1
, S2
, S3
& S4
are the temperatures at end of solidification of the alloys 2, 3, 4 & 5
respectively.
•1 & 6 are pure metals with melting points temperatures TA
& TB
respectively.
•A smooth curve passing through TA
-L1
-L2
-L3
-L4
-TB
represents the temperature
at start of solidification of any alloy of A & B. It is called the “Liquidus”.
•A smooth curve passing through TA
-S1
-S2
-S3
-S4
-TB
represents the temperature
at end of solidification of any alloy of A & B. It is called the “Solidus”.
•The diagram so obtained is called the “Phase diagram for a binary solid solution
system” or “Phase diagram for an Isomorphous system”.

•There are minimum of 3 regions. The uppermost region above the
liquidus is the single phase liquid region. The region between the
liquidus and the solidus is the two phase region and known as the
liquid + α region (Mushy region).
•The lowermost region below the solidus is the single phase solid
region - the α region.
•Minimum of two lines known as liquidus & solidus
•This is a reversible process and can be represented by the
reaction:
L α
where, α is a solid solution of metals A and B.

Simple binary Phase Diagram-
Isomorphous system

For a given alloy, at a given temperature T in 2 phase
region in equilibrium condition, It is possible to
determine:
1. The number and types phases that are present
2. The chemical composition of each phase
3. The amount of each phase
Lever rule- : It states that the relative amounts of phases
are directly proportional to the opposite lever arms.
Relative amounts means the weight percentage or
amount of solid that has already been solidified and the
weight percentage of liquid that is still remaining.
Interpretation of Phase Diagram

Lever Rule
22 55
40
Two phases present, Type: Liquid and solid alloy phase.
Composition of 2 phases existing at T:
Liquid: Alloy containing 40% A (by wt.)
Solid: Alloy containing 22% B (by wt.)

Applying Lever Rule to the tie line CD, with pivot at Q
Relative amount of solid phase α =
(QD is the lever arm touching liquidus)
Relative amount of liquid phase = (CQ is
the lever arm touching solidus)
Lever Rule

A solid solution is a combination of two metals
which are completely soluble in the liquid state as well
as in the solid state.
The figure depicts a typical phase diagram for an
isomorphous system made of two metallic elements A
and B. As cited earlier, any phase diagram can be
considered as a map.
Solid solution (Isomorphous System) phase
diagram

Isomorphous System phase diagram
In binary system, two components are completely soluble in liquid and solid
states
For any composition of alloy, the system gives solid solution type of alloy,
mutually soluble with each other.

wt% Ni
2
0
1200
1300
30 40 50
1100
L
(liquid)
α
(solid
)
L +
α
L +
α
T(°
C)
A
35
C0
L:
35wt%Ni Cu-Ni
system
• Part of phase diagram:
Cu-Ni system.
4
6
3
5 40
29
α: 40 wt% Ni
L: 29 wt% Ni
B
α: 46 wt% Ni
At T1, L: 35 wt% Ni
C
E
L: 24 wt% Ni
α: 35 wt% Ni
2
4
35
D
Equilibrium Cooling of a Cu-Ni alloy-isomorphous system
At T3, solidification is
complete to obtain
equilibrium phase α
solid solution with
composition 35 wt% Ni
throughout.
α: 35 wt% Ni
Equilibrium
phase
T1
T2
T3
T4
At T1, the
precipitated solid α
has composition 46
wt% Ni
At T2, the precipitated solid α has
composition 40 wt% Ni
Example 1

Dendritics segregation
(Coring)-
Inhomogeneous
chemical composition
from one zone to other.
Nonuniform properties
to the microscopic level
Development of microstructure during the non-equilibrium
solidification of a 35 wt% Ni-65 wt% Cu alloy outcome
46 S47
L29Ni
S47Ni
S42Ni
S
S
L24Ni
S47Ni
S42Ni
S38Ni
S47Ni
S42Ni
S38Ni
S35Ni
S47Ni
S47Ni
S42Ni
S38Ni
S35Ni
T1
T2
T3
T4
In Nonequilibrium cooling, at
T4, solidification is complete
to obtain Non equilibrium
phase S solid solution with
composition with wide
composition variation S47Ni,
S42Ni, S38Ni to S35Ni
T5
At T1
At T2
At T3
At T4
At T5
Above T1
Non-equilibrium
solidus
Equilibrium solidus
Cu
Example 1

Fast rate of cooling
Cored structure
Non-equilibrium phases
Slow rate of cooling
Equilibrium structure
equilibrium phases
Non-equilibrium phases & equilibrium cooling
S47Ni
S38Ni
S42Ni
S35Ni
S44Ni
L α46 α 40 α 35
In equilibrium cooling, at T3,
solidification is complete to obtain
Equilibrium phase α solid solution
with composition 35 wt% Ni
In Non-equilibrium cooling, at T4,
solidification is complete to obtain Non
equilibrium phase S solid solution with
wide composition variation like, S47Ni,
S42Ni, S38Ni to S35Ni.
In equilibrium cooling, diffusion rate of
atoms is in pace with crystal growth
(cooling rate)
Cooling rate (Crystal growth rate)
is faster than diffusion rate of
atoms (cooling rate).
Weaker in Higher
MP metal atoms
Stronger in Higher
MP metal atoms

L30 wt%B, SS5 wt%B
L30 wt%B Alloy
L45 wt%B, SS10 wt%B
L58 wt%B, SS20 wt%B
L65 wt%B, SS30 wt%B
Homogeneous SS30 wt%B
FIG.:
EQUILIBRIUM COOLING OF AN ALLOY
For 30 wt% B alloy, at
temp. T3
,
a3
( in wt.%) = (T3
L3
/ a3
L3
)X100
=(58-30)/(58-20)X100 =
74
L4
(in wt.%) = (a4
T4
/ a4
L4
)X100
=(30-20)/(58-20)X100 =
Example 2

NON-EQUILIBRIUM COOLING OF AN ALLOY
Wt% of B in α2
1
>α2
and so on
68
65
Composition of the last liquid to solidify;
Under eqbm. Cooling: L65 wt%B, solidfcn. temp. range is less (R)
Under non-eqbm. Cooling: L68wt%B, solidfcn. temp. range is less(R1
)
c
c
R1 R
16 58
20
Example 2

At T4
Temperature,
In non-equilibrium cooling.
a4
1
(in wt.%) = (T4
L4
/ a4
1
L4
)X100 =75
L4
(in wt.%) = (a4
1
T4
/ a4
1
L4
)X100 =25.
In Equilibrium cooling.
a4 =
a ( in wt.%) = (T4
L4
/ a4
L4
)X100 =100
L4
(in wt.%) = (a4
T4
/ a4
L4
)X100 =00.
Similarly at T3
Temperature,
In non-equilibrium cooling.
a3
1
( in wt.%) = (T3
L3
/ a3
1
L3
)X100 =(58-30)/(58-16)X100 = 66.7
L4
(in wt.%) = (a4
1
T4
/ a4
1
L4
)X100 =(30-16)/(58-16)X100 = 33.3.
In equilibrium cooling.
a3
( in wt.%) = (T3
L3
/ a3
L3
)X100 =(58-30)/(58-20)X100 = 74
L4
(in wt.%) = (a4
T4
/ a4
L4
)X100 =(30-20)/(58-20)X100 = 26.

Problem: Two pure metals A & B with melting points 900
and 400o
C respectively are completely soluble in their molten
state. Upon solidification the binary system gives rise to a single
homogeneous solid.
Alloy of
composition
Temperature (o
C)
at start of
solidification
at end of
solidification
90%A-10%B 890 790
80%A-20%B 870 700
70%A-30%B 840 630
60%A-40%B 810 570
50%A-50%B 770 525
40%A-60%B 715 485
30%A-70%B 650 450
20%A-80%B 580 425
10%A-90%B 500 405
Details of start and end of
solidification of various alloys in the
series are as follows:
1. Assuming that there are no
solid-state reactions taking place,
draw the phase diagram of the series
and label all the regions.
2. Find out the number, type,
composition and relative amounts of
the phases present in an alloy of
60%A-40%B at 700 o
C.
3. Compositions of 1st
solid
precipitates & last liquid to be
precipitated.

A B
α
L + α
L
P R
Q Liquidus
Solidus
Plot the phase diagram, Draw
the tie line at 700o
C, Draw the
vertical line at 40 wt % of B.
Composition of liquid
phase at 700o
C is
38%A-62%B.
5. Composition of solid
phase at 700o
C is
80%A-20%B.
Compositions of : 1st
solid
precipitates: 90%A-10%B.
Last liquid to be precipitated
20%A-80%B.
Relative amount of solid phase α = (QR is the lever arm touching
liquidus)
Relative amount of liquid phase
=
(PQ is the lever arm touching
solidus)
-------------------
--------------------------
80(Comp.of last drop)
10(Comp. Nucleus)
20 62
Number of phases: 2, type: Liquid and solid
solution

A B
α
L + α
L
Liquidus
Solidus
Composition of liquid phase
at 500o
C is 10%A-90%B.
Composition of solid phase
at 500o
C is
45%A-55%B.
Relative amount of solid phase, Wt.%α = no/mn X 100 =29% (Apprx.)
Relative amount of liquid phase, Wt.% L= mo/mn X 100 =71% (Apprx.)
-------------------
--------------------------
m
55 90
O
n
Type of phase: Liquid (L) and solid solution (α). Number: 2 phases
Wt.%α Wt.%L
Problem 2: In the same
problem, find out the
number, type, composition
and relative amounts of the
phases present in an alloy
of 80%B at 500o
C.
Alloy of 80%B

• Eutectic mixture is a mixture of two or more
phases at a composition that has the lowest
melting point.
• Phases simultaneously crystallize from molten
solution.
• The proper ratios of phases to obtain a eutectic
is identified by the eutectic point.
Eutectic

The Eutectic Phase Diagram
A Eutectic System is one where two components
which are completely soluble in the liquid state are
either:
(i) Completely insoluble in the solid state or ⇒ Type 1
Eutectic
(ii) Partially soluble in the solid state. ⇒ Type 2
Eutectic

These can be expressed as reactions which are
reversible in nature:
where, A & B are pure metals
α is a solid solution rich in ‘A’
β is a solid solution rich in ‘B’
The Eutectic Phase Diagram

• For alloys where
C0
< 2 wt% Sn
• Result at room
temperature is a
polycrystalline with grains of
α phase having composition
C0 0
L+ α
200
T(°C)
C , wt% Sn
10
2
20
C0
300
100
L
α
30
α + β
400
(room T solubility limit)
TE
α
L
L: C0
wt% Sn
α: C0
wt% Sn
Pb-Sn
system
Microstructural Developments
in Eutectic Systems
18.3

2 wt% Sn < C0
< 18.3
wt% Sn
• Results in
polycrystalline
microstructure with α
grains and small β-phase
particles at lower
temperatures.
L + α
20
0
T(°C)
C
, wt%
Sn
1
0
18.3
2
0
0
C0
30
0
10
0
L
α
3
0
α + β
40
0
(sol. limit at TE
)
TE
2
(sol. limit at
T
roo
m
)
L
α
L: C0
wt% Sn
α
β
α: C0
wt% Sn
Microstructural Developments
in Eutectic Systems - II
18.3

Pb-Sn Phase Diagram
Liquid
us
Solid
us
Solid
us
Solid
us
Solvu
s
Solvu
s

The Peritectic Reaction and Phase Diagram

Eutectoid Reaction
It is the reversible isothermal reaction in which a
single solid phase converts into two solids up on
cooling. It can be expressed as
(Eutectoid mixture)

Melting temperatures of pure metals ‘A’ & ‘B’ are 1000o
C and 800o
C respectively. The metals
‘A’ and ‘B’ are mutually soluble in the liquid state and completely insoluble in the solid state. A
liquid phase alloy containing 40% A completely transforms into a mixture of two metals at
600o
C. Assuming the curves to be linear, draw phase diagram to scale and label the regions.
For 40% B alloy determine the following:
• Composition of the nucleus.
• Composition of the last drop of the liquid precipitating
• Weight percentage of eutectic formed at 400o
C.
• Weight ratio of the solid phases in the eutectic mixture
• Temperature at which there is 80 weight % of liquid phase present.
• Temperature at which there is 80 weight % of solid phase present.
• Composition of the liquid phase undergoing eutectic reaction
Numerical on Type 1- Eutectic system

• Plot the phase diagram by taking temperature on Y axis and composition (%
weight) of A or B on X axis.
• Locate the points TA
= 1000o
C on the ordinate through ‘A’ and TB
= 800o
C on the
ordinate through ‘B’. Draw a horizontal line (isotherm) FG at 600o
C.
• Mark the Eutectic point ‘E’ on the FG at 60%B. Draw the Liquidus as a straight lines
connecting TA
-E-TB
.
• Draw solidus as straight lines connecting TA
-F-E-G-TB
. Mark the two-phase regions
‘L + A’ and ‘L + B’ between the liquidus and solidus. Also denote the A + (A + B), B
+ (A + B) and (A + B) regions as shown in the fig.
Solution

(2) The alloy of 60%A-40%B under consideration is represented by drawing an ordinate
at 40%B.
(a) Composition of the nucleus is nothing but the composition of the first particle of solid
formed. Here, the first solid formed is pro-eutectic phase ‘A’. To determine the
composition of the first nucleus we draw an isotherm at the point of intersection of
the alloy with the liquidus. The intersection of this isotherm with the solidus will give
us the composition of the solid phase. From fig. 4.12 it is 100%A.
(b) Composition of the last drop of the liquid precipitating is obtained where the alloy
intersects the solidus, i.e., at ‘H’. At H we draw an isotherm to intersect the liquidus
at E which gives us the required composition of the last drop of liquid precipitating,
i.e., 60%B. It may be noted that this is the Eutectic temperature.

(c) Weight percentage of eutectic formed at 400o
C is obtained by applying lever rule to
the tie line JK through M.
(d) Weight ratio of the solid phases in the eutectic mixture
We need to find the ratio A/B or B/A in the Eutectic mixture.
wt. % of A in (A+B) = Total wt. % of A after Eutectic reaction – wt. % of Pro-eutectic A
Note: The weight ratio of the two phases in the Eutectic mixture is found to be constant for any alloy undergoing Eutectic reaction.

(e) Temperature at which there is 80 weight % of liquid phase present. This can be done
by a trial and error procedure. A tie line is drawn in the L + A region in such a way that
the lever arm touching the solidus is 80% of the total tie-line length. At approximately
670o
C when such a tie line (xz) is drawn (refer fig. 4.12), the total tie line length is xz =
50%B, the lever arm length touching solidus is xy = 40%B and 40/50 = 80%.
∴Temperature where there is 80 weight % of liquid phase present is 670o
C
(f) Temperature at which there is 80 weight % of solid phase present.
This means that we need to determine the temperature at which 80% of the solidification
is complete. From the Fig., up till H pro-Eutectic phase A is formed whose amount is:
i.e., 33.33% of solidification is complete and remaining 66.66% of solidification will take
place during Eutectic reaction. From this we understand that when 80% of solid is
formed (80% solidification is completed), Eutectic reaction will still be in progress, i.e.,
the corresponding temperature is Eutectic temperature = 600o
C.

(g) Composition of the liquid phase undergoing eutectic reaction.
During slow cooling, the liquid of the alloy under consideration, will reach Eutectic
Composition and Eutectic Temperature at H. So, the tie line FE at H will be useful to find
the liquid composition. The tie line FE intersects the liquidus at E. An ordinate through E
dropped on the composition axis will give the composition of the liquid phase undergoing
Eutectic reaction.
∴ Composition of ‘L’ to undergo eutectic reaction is 40%A+60%B

TA
: M. P. of ‘A’ Eutectic Point: E
TB
: M. P. of ‘B’ Eutectic Composition: 40%A-60%B
Liquidus: TA
-E-TB
Eutectic Temperature: 600o
C
Solidus: TA
-F-E-G-TB
Eutectic Mixture: (A + B)
Numerical on Type 1 Eutectic system
Composition by weight of B, %
A B
20%B 40%B 60%B 80%B
L
L + A
E
Tem
pera
ture,
o
C
L + B
(A
+
B)
A + (A + B) B + (A + B)
F
Hypo Eutectic Alloys Hyper Eutectic Alloys
G
100
200
300
400
500
600
700
800
900
TA
= 1000
100
200
300
400
500
600
700
TB
= 800
900
1000
H
J
K
M
z
y
x

Melting temperatures of pure metals ‘A’ & ‘B’ are 1000o
C and 800o
C respectively.
The metals ‘A’ and ‘B’ are mutually soluble in the liquid state and partly soluble in the
solid state. A liquid phase alloy containing 40% A completely transforms into a mixture of
two solid solutions at 600o
C. Maximum solubility of ‘A’ in ‘B’ and ‘B’ in ‘A’ are 10% and
20% respectively at 600o
C, 5% and 10% respectively at 0o
C. Assuming the curves to be
linear, draw phase diagram to scale and label the regions. For 40% B alloy determine the
following:
Composition of the nucleus.
Composition of the last drop of the liquid precipitating
Weight percentage of eutectic formed at 400o
C.
Weight ratio of the solid phases in the eutectic mixture
Temperature where there is 80 weight % of liquid phase present.
Composition of the liquid phase undergoing eutectic reaction.
Numerical on Type II Eutectic system

• Plot the phase diagram by taking temperature on Y axis and composition (% weight)
of A or B on X axis. Locate the points TA
= 1000o
C on the ordinate through ‘A’ and TB
= 800o
C on the ordinate through ‘B’.
• Draw a horizontal line (isotherm) at 600o
C. On this isotherm mark ‘F’ to represent
maximum solubility of ‘B’ in ‘A’. Mark ‘G’ to represent the maximum solubility of ‘A’ in
‘B’. Similarly, mark ‘M’ & ‘N’ on the 0o
C line to represent the minimum solubilities of
‘B’ in ‘A’ and ‘B’ in ‘A’ respectively. Mark the Eutectic point ‘E’ on the 600o
C line.
• Draw the Liquidus as a straight lines connecting TA
-E-TB
. Draw solidus as straight
lines connecting TA
-F-E-G-TB
. Draw the solvus lines as straight lines connecting F to
M and G to N. Mark the terminal solid solution regions of ‘α’ and ‘β’ to the left of F
and to the right of G respectively.
• Mark the two-phase regions ‘L + α’ and ‘L + β’ between the liquidus and solidus. Also
denote the α + (α + β), β + (α + β) and (α + β) regions as shown in the fig.

• The alloy of 60%A-40%B under consideration is represented by drawing an ordinate
at 40%B.
• Composition of the nucleus is nothing but the composition of the first particle of solid
formed. Here, the first solid formed is pro-eutectic phase α. To determine the
composition of the first nucleus we draw an isotherm at the point of intersection of
the alloy with the liquidus. The intersection of this isotherm with the solidus will give
us the composition of the solid phase. From fig. 4.18 it is approximately 13%B.
• Composition of the last drop of the liquid precipitating is obtained where the alloy
intersects the solidus, i.e., at ‘H’. At ‘H’ we draw an isotherm to intersect the liquidus
at ‘E’ which gives us the required composition of the last drop of liquid precipitating,
i.e., 60%B.

• Weight percentage of eutectic formed at 400o
C is obtained by applying lever rule to
the tie line PQ through R.
• Weight ratio of the solid phases in the eutectic mixture. We need to find the ratio α/ β
or β/α in the Eutectic mixture. wt. % of α in (α + β) = Total wt. % of α after Eutectic
reaction – wt. % of Pro-eutectic α

Note: The weight ratio of the two phases in the Eutectic mixture is found to be constant
for any alloy undergoing Eutectic reaction.
• Temperature where there is 80 weight % of liquid phase present. This can be done by
a trial and error procedure. A tie line is drawn in the L + α region in such a way that the
lever arm touching the solidus is 80% of the total tie-line length. At approximately
690o
C when such a tie line (xz) is drawn (refer fig. 4.18), the total tie line length is
xz = 31%, the lever arm length touching solidus is xy = 25% and 25/31 = 80.65%
∴ Temperature where there is 80 weight % of liquid phase present is 690o
C
• Composition of the liquid phase undergoing eutectic reaction. During slow cooling, the
liquid of the alloy under consideration, will reach Eutectic Composition and Eutectic
Temperature at H. So, the tie line FE at H will be useful to find the liquid composition.
The tie line FE intersects the liquidus at E. An ordinate through E dropped on the
composition axis will give the composition of the liquid phase undergoing Eutectic
reaction.
∴ Composition of ‘L’ to undergo eutectic reaction is 40%A+60%B

TA
: M. P. of ‘A’ Eutectic Point: E
TB
: M. P. of ‘B’ Eutectic Composition: 40%A-60%B
Liquidus: TA
-E-TB
Eutectic Temperature: 600o
C
Solidus: TA
-F-E-G-TB
Eutectic Mixture: (α + β)
Solvus: FM & GN Terminal Solid Solutions: α & β
α: ‘A’ rich solid solution β: ‘B’ rich solid solution
Fig. Numerical on Type 2 Eutectic system
Composition by weight of B, %
A B
20%B 40%B 60%B 80%B
L
L + α
E
Tem
pera
ture,
o
C
L + β
(α +
β)
α + (α + β) β + (α + β)
F
Hypo Eutectic Alloys Hyper Eutectic Alloys
G
100
N
α
β
200
300
400
500
600
700
800
900
TA
= 1000
100
200
300
400
500
600
700
TB
= 800
900
1000
M
H
Q R
P
z
y
x

5. What is co-ordination number? Write the co-ordination
number for BCC, FCC and HCP structures. Find the
packing factor of HCP unit cell.
4. What is a miller index? Explain the procedural steps to find
the miller indices of atomic planes. Show the following:
[123], (201), [310], (212).
3. What are the various types of Crystal Imperfections?
Explain with neat sketches, edge & screw dislocations with
the Burger’s vector marked.
2. With the help of equilibrium cooling curve and relevant
sketches, explain different stages of solidification in pure
metals.
1. With examples, explain the salient features of solid
solutions.
ASSIGNMENT 1

Phases in Solids.pdf .

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