NH Phase Diagrams.ppt

Phase Diagrams
• What is a phase?
• What is the equilibrium state when different elements are
mixed?
• What phase diagrams tell us.
• How phases evolve with temperature and composition
(microstructures).

1. Different physical states:
vapor, liquid, solid
Phases
Phase: A homogeneous portion of a system that has uniform physical
and chemical characteristics.
e.g.
2. Different chemical composition
e.g.
• Solubility Limit:
Max concentration for
which only a solution
occurs.
Question: What is the solubility limit at 20C?
Answer: 65wt% sugar.
If Co < 65wt% sugar: syrup
If Co > 65wt% sugar: syrup + sugar.
• Solubility limit increases with T:
e.g., if T = 100C, solubility limit = 80wt% sugar.

Aluminum-
Copper
Alloy
• Components:
elements or compounds which are mixed initially
(e.g., Al and Cu) i.e., chemically recognizable species
• Phases:
A portion of a system that has uniform physical and
chemical characteristics. Usually denoted in Greek letters
e.g., a and b. A phase may contain one or more components.
Plain carbon steel

Temperature and composition
A
B C
A: 2 phases present (liquid solution and solid sugar)
B: Single phase
C: back to 2 phases but at different composition
Note: this is an equilibrium phase diagram
(What does it mean to have phase equilibrium?)

Phase Equilibrium
Equilibrium: minimum energy state for a given T, P, and composition
(i.e. equilibrium state will persist indefinitely for a fixed T, P and
composition).
Phase Equilibrium: If there is more than 1 phase present, phase
characteristics will stay constant over time.
Phase diagrams: A graphical representation of the combination of
temperature, pressure and composition of equilibrium phases.
•A phase diagram show that what phases exist at equilibrium and
what phase transformations we can expect when we change one of
the parameters of the system.
•(here, we’ll always keep P constant for simplicity).
Phase equilibrium is the study of the equilibrium which exists
between or within different states of matter namely solid, liquid and
gas. Equilibrium is defined as a stage when chemical potential of any
component present in the system stays steady with time.

Phase equilibrium is the study of the equilibrium which exists
between or within different states of matter namely solid, liquid and
gas. Equilibrium is defined as a stage when chemical potential of any
component present in the system stays steady with time.

Unary Systems
Single component system
Consider 2 metals:
Cu has melting T = 1085oC
Ni has melting T = 1453oC (at standard P = 1 atm)
1085oC
Cu
solid
liquid
T
1453oC
Ni
solid
liquid
T
What happens when Cu and Ni are mixed?

Binary Isomorphous Systems
2 components
Complete liquid and solid solubility because both
Cu and Ni have the same crystal structure, FCC,
Similar radii, electro negativity and valence
1085oC
Cu
solid
liquid
T
1453oC
Ni
solid
liquid
T
wt% Ni
0 100
Expect Tm of solution to lie in between Tm of two pure components
For a pure
component,
complete melting
occurs before T
increases (sharp
phase transition).
But for
multicomponent
systems, there is
usually a
coexistence of L
and S.
L
S

Solid Solutions
What is a solid solution?
When foreign atoms are incorporated into a crystal structure, whether in
substitutional or interstitial sites, the resulting phase is a solid solution of the
matrix material (solvent) and the foreign atoms (solute)
• Substitutional Solid Solution: Foreign (solute) atoms occupy “normal” lattice
sites occupied by matrix (solvent) atoms, e.g. Cu-Ni;Ge-Si
• Interstitial Solid Solutions: Foreign (solute) atoms occupy interstitial sites,
e.g., Fe-C

Solid solutions
• A solid solution is a single phase which exists over a range of chemical
compositions. Some minerals are able to tolerate a wide and varied chemistry,
whereas others permit only limited chemical deviation from their ideal chemical
formulae. In many cases, the extent of solid solution is a strong function of
temperature, with solid solution being favoured at high temperatures and
unmixing and/or ordering favoured at low temperatures.
• Types of solid solution:
• Substitutional solid solution: chemical variation is achieved simply by substituting
one type of atom in the structure by another.
• Coupled substitution: this is similar to the substitutional solid solution, but in a
compound cations of different valence are interchanged. To maintain charge
balance, two coupled cation substitutions must take place.
• Omission solid solution: chemical variation is achieved by omitting cations from
cation sites that are normally occupied.
• Interstitial solid solution: chemical variation is achieved by adding atoms or ions
to sites in the structure that are not normally occupied.

Factors Affecting the Extent of Solid Solution:
• Atomic/ionic size: If the atoms or ions in a solid solution have similar ionic radii, then the
solid solution is often very extensive or complete. Generally, if the size difference .is less
than about 15%, then extensive solid solution is possible. For example, Mg2+ and Fe2+
have a size mismatch of only about 7%, and complete solid solution between these two
elements is observed in a wide range of minerals. However, there is a 32% size
difference between Ca2+ and Mg2+, and we expect very little substitution of Mg for Ca to
occur in minerals.
• Temperature: High temperatures favour the formation of solid solutions, so that
endmembers which are immiscible at low temperature may form complete or more
extensive solid solutions with each other at high temperature. High temperatures promote
greater atomic vibration and open structures, which are easier to distort locally to
accommodate differently-sized cations. Most importantly, solid solutions have a higher
entropy than the endmembers, due to the increased disorder associated with the
randomly distributed cations, and at high temperatures, the -TS term in the Gibb's free
energy stabilises the solid solution.
• Structural flexibility: Although cation size is a useful indicator of the extent of solid solution
between two endmembers, much depends on the ability of the rest of the structure to
bend bonds (rather than stretch or compress them) to accommodate local strains.
• Cation charge: Heterovalent substitutions (i.e. those involving cations with different
charges) rarely lead to complete solid solutions at low temperatures, since they undergo
complex cation ordering phase transitions and/or phase separation at intermediate
compositions. These processes are driven by the need to maintain local charge balance
in the solid solution as well as to accommodate local strain.

Types of Solid Solubility
•Unlimited Solid Solubility: Solute and solvent are mutually
soluble at all concentrations, e.g., Cu-Ni system
– Meets the requirements of the Hume-Rothery Rules
– Result is a “single phase alloy”
•Limited or Partial Solid Solubility: There is a limit to how
much of the solute can dissolve in the solvent before
“saturation” is reached, e.g., Pb-Sn and most other systems
– Does not meet the requirements of the Hume-Rothery Rules
– Results in a “multi-phase alloy”

Hume RotheryRules
1. Relative Size Ratio ±15%
2.Crystal Structure-must be the same
3.Electronegativity Difference –within ±0.4 e.u.
4. Valence must be the same

Binary Isomorphous Systems
What can we learn from
this phase diagram?
1. Phase(s) present.
A: solid (a) only
B: solid and liquid
2. Composition of
those phases
A: 60 wt% Ni
B: 35 wt% Ni overall (how about
in L and S separately?)
3. Amount of the
phases.
A: 100% a phase
B: % solid and % liquid?
Solid-liquid
coexistence
region
Three phase region can be identified on the phase diagram:
Liquid(L), Solid+Liquid (α + L) and Solid(α)

Determining phase composition
in 2-phase region:
1. Draw the tie line.
2. Note where the tie line intersects
the liquidus and solidus lines (i.e.
where the tie line crosses the
phase boundaries).
3. Read off the composition at the
boundaries:
Liquid is composed of CL amount
of Ni (31.5 wt% Ni).
Solid is composed of Ca amount of
Ni (42.5 wt% Ni).

Interpretation of Phase Diagrams
For a given temperature and composition we can use phase
diagram to determine:
1) The phases that are present
2) Compositions of the phases
3) The relative fractions of the phases
Finding the composition in a two phase region:
1. Locate composition and temperature in diagram
2. In two phase region draw the tie line or isotherm
3. Note intersection with phase boundaries. Read
compositions at the intersections. The liquid and solid
phases have these compositions.

The Lever Rule
Finding the amounts of phases in a two phase region:
1. Locate composition and temperature in diagram
2. In two phase region draw the tie line or isotherm
3. Fraction of a phase is determined by taking the length of the
tie line to the phase boundary for the other phase, and dividing
by the total length of tie line
The lever rule is a mechanical analogy
to the mass balance calculation. The tie
line in the two-phase region is analogous
to a lever balanced on a fulcrum

Determining phase amount in the
2-phase region:
1. Draw the tie line.
2. Determine the “distance from the
point of interest (B) to each of
the phase boundaries.
R = Co – CL
S = Ca - Co
3. Mass fractions (wt%) of each
phase:
Lever Rule
L
o
L
C
C
C
C
S
R
S
W





a
a
L
L
o
C
C
C
C
S
R
R
W





a
a
68
.
0
5
.
31
5
.
42
35
5
.
42




32
.
0
5
.
31
5
.
42
5
.
31
35




Liquid:
Solid:
i.e. 68% of the mass is liquid and 32% of the mass is solid.

Lever Rule: Derivation
Since we have only 2 phases:
1

 a
W
WL
Conservation of mass requires that:
Amount of Ni in a-phase + amount of Ni in liquid phase = total amount of Ni
or
o
L
L C
C
W
C
W 

a
a
(1)
(2)
From 1st condition, we have: L
W
W 
1
a
Sub-in to (2): o
L
L
L C
C
W
C
W 

 a
)
1
(
Solving for WL and Wa gives :
L
o
L
C
C
C
C
W



a
a
L
L
o
C
C
C
C
W



a
a

• A geometric interpretation:
moment equilibrium:
1 Wa
solving gives Lever Rule
WLR  WaS
Lever Rule: Derivation

We can also express in terms of volume fractions:
L
v
v
v
vol
total
of
vol
V



a
a
a
a
_
_
_
Since
j
j
j
v
mass
total
W
j
of
vol
j
component
of
mass _
_
_
_
_
_ 



L
L
W
W
W
V



a
a
a
a
a
/
/
/


L
L
L
L
L
W
W
W
V



a
a /
/
/


j
total
j
j
M
W

  substitute this relation (for each component) into
volume fraction definition – Mtotal cancels out.

Development of Microstructure in
Isomorphous Alloys Equilibrium
(Very Slow) Cooling
1. Solidification in the solid + liquid phase
occurs gradually upon cooling from the liquidus
line
2. The composition of the solid and the liquid
change gradually during cooling (as can be
determined by the tie-line method.)
3. Nuclei of the solid phase form and they grow
to consume all the liquid at the solidus line

Microstructures in Isomorphous Alloys
Microstructures will vary on the cooling rate (i.e. processing conditions)
1. Equilibrium Cooling: Very slow cooling to allow phase equilibrium to
be maintained during the cooling process.
a (T>1260oC): start as
homogeneous liquid solution.
c (T= 1250oC): calculate composition
and mass fraction of each phase.
d (T~ 1220oC): solidus line reached.
Nearly complete solidification.
Ca = 35 wt% Ni; CL = 24 wt% Ni
e (T<1220oC): homogeneous solid
solution with 35 wt% Ni.
b (T ~ 1260oC): liquidus line
reached. a phase begins to
nucleate. Ca = 46 wt% Ni; CL = 35 wt% Ni

Example problem
• 65 wt% Ni – 35 wt% Cu alloy is heated to
T within the a+L region. If a-phase
contains 70 wt% Ni, determine:
a. Temperature of the alloy.
b. Composition of the liquid phase.
c. Mass fraction of both phases.

Temperature=1345 oC
Given:
C0= 65 wt% Ni
Ca=70 wt% Ni
CL Ca
C0
Temperature=1345 oC
CL= 59 wt% Ni
WL= 0.45
Wα= 0.55

4.3 Example problem
mass
by
amount
phase
Liquid %
33
.
33
%
100
18
72
18
36





mass
by
amount
phase
Solid %
67
.
66
%
100
18
72
36
72





.
%
%
18
%
72
%
36
%
36
:
2
2
mass
by
amounts
phase
solid
and
phase
liquid
Compute
B
of
S
B
of
L
B
of
a
B
of
X
data
Given





Non-equilibrium cooling
Fast cooling, but how fast?
Fast w.r.t. diffusion
Since diffusion rate is especially low in solids, consider case where:
Cooling rate >> diffusion rate in solid
Cooling rate << diffusion rate in liquid
(equilibrium maintained in liquids phase)

Non-equilibrium cooling
a’ (T>1260oC): start as
homogeneous liquid solution.
c’ (T= 1250oC): solids that formed at pt b’
remain with same composition (46wt%) and
new solids with 42 wt% Ni form around the
existing solids (Why around them?).
d’ (T~ 1220oC): solidus line reached. Nearly
complete solidification.
•Previously solidified regions maintain original
composition and further solidification occurs at 35
wt% Ni.
e (T<1220oC): Non-equilibrium solidification
complete (with phase segregation).
b’ (T ~ 1260oC): liquidus line
reached. a phase begins to nucleate.
Ca = 46 wt% Ni; CL = 35 wt% Ni

Cored vs. Equilibrium phases
• Ca changes as we solidify.
• Cu-Ni case:
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First a to solidify has Ca = 46wt%Ni.
Last a to solidify has Ca = 35wt%Ni.

Binary Eutectic Systems
“easily melted”
Cu-Ag phase diagram
Single phase regions:
a-phase (solid solution rich in
Cu).
b-phase (solid solution rich in
Ag).
L-phase (liquid solution).
2-Phase coexistence regions:
a+b phase (limited solubility of
Ag in Cu and vice versa lead
to 2 different solid solution
phases).
a+L and b+L phase regions.
Tie lines and Lever Rule can be
applied in the 2-phase
regions.
A
B
Calculate composition and mass
fractions of each phases at pts A and B.
Ca = 3wt% Ag Cb = 97wt% Ag
Tie line
Tie line
Ca =7wt% Ag CL = 50wt% Ag
Alloy with Limited Solubility

Eutectic Point
Eutectic point: Where 2 liquidus lines
meet (pt. E).
Sometimes also referred to as invariant
point.
Eutectic Reaction:
cool
L(CE) a(CaE) + b(CbE)
similar to one component (pure) system
except 2 solid phases.
Eutectic Isotherm: Horizontal solidus
line at T = TE.
heat
Cu-Ag phase diagram

Eutectic System: Example
Pb-Sn phase diagram
At 150oC for a 40wt% Sn/ 60
wt% Pb alloy:
1. What phases are present?
2. What are the compositions of
the phases present?
3. What are the mass fractions of
the phases?
4. What are the volume
fractions?
For a 10wt% Sn/ 90wt% Pb alloy:
1. At what T, can a state with 50%
liquid be achieved?
2. At 250oC, how much Sn must
be added to achieve the same
state (50% liquid)?

Eutectic System: Example
Pb-Sn phase diagram
At 150oC for a 40wt% Sn/ 60 wt%
Pb alloy:
1. What phases are present?
2. What are the compositions of
the phases present?
3. What are the mass fractions of
the phases?
4. What are the volume fractions?
For a 10wt% Sn/ 90wt% Pb alloy:
1. At what T, can a state with 50%
liquid be achieved?
2. At 250oC, how much Sn must
be added to achieve the same
state (50% liquid)?
L
L
W
W
W
V



a
a
a
a
a
/
/
/


L
L
L
L
L
W
W
W
V



a
a /
/
/


Where ρα and ρL are densities of the respective phases

Microstructures in Eutectic Alloys
1. One component rich composition.
a: start with homogeneous liquid.
b: a-phase solids with liquid.
Compositions and mass fractions can be
found via tie lines and lever rule.
c: a-phase solid solution only.
Net result: polycrystalline a solid.
Cooling at this composition is similar to
binary isomorphous systems.
Partial Pb-Sn phase diagram

2. One-component rich but cooling to a + b
coexistence.
d: homogeneous liquid.
e: a + L phase (same as previous but at
different compositions and mass fractions).
f: all a-phase solid solution.
g: a + b phase (passing through solvus line
leads to exceeding solubility limit and b phase
precipitates out).
Net result: polycrystalline a-solid with fine b
crystals.

3. Cooling through eutectic point.
h: homogeneous liquid.
i: solidification via eutectic reaction
cool
L(CE) a(CaE) + b(CbE)
note: a and b phases have very
different compositions than the
original composition of the liquid (e.g.
Ca = 18.3 wt% Sn; Cb = 97.8 wt% Sn
whereas CL = 61.9 wt% Sn).
Eutectic Structure: layered (lamellar)
structure.
Why does this structure form?
heat

Eutectic Structure
Pb-Sn Eutectic microstructure
L
Sn
Pb
b
a Pb rich
Sn rich
In order to achieve large
homogeneous regions, long
diffusion lengths are required.
Lamellar structure forms because
relatively short diffusion lengths
are required.

Redistribution of
alloying elements
during lamellar
growth of a lead-tin
eutectic. Tin atoms
from the liquid
preferentially
diffuse to the b
plates, and Lead
atoms diffuse to
the a plates.

4. Cooling through eutectic
isotherm.
j: homogeneous liquid.
k: a + L phases: use tie lines
and lever rule.
l: just above eutectic isotherm
compositions given but what
about mass fraction?
m: remaining liquid transforms
to eutectic structure upon
crossing eutectic isotherm.
Microconstituent: an element
of a microstructure with
identifiable and characteristic
structure (at pt. m there are 2
microconstituents: primary a
and eutectic structures)
Hypoeutectic

Just above the eutectic isotherm
at C4’:
Q
P
P
We

 All of this
amount will
turn into
eutectic
structure (We).
Q
P
Q
W


'
a
Primary a structure: Total a (primary + eutectic):
R
Q
P
R
Q
W




a
Total b (all in eutectic):
R
Q
P
P
W



b
Note: this is for equilibrium cooling. Non-equilibrium cooling will lead to:
Cored primary phases
Increased fraction of eutectic microconstituent
Calculation of Amounts of Microconstituents

Adapted from Fig. 9.7,
Callister 6e. (Fig. 9.7
adapted from Binary Phase
Diagrams, 2nd ed., Vol. 3,
T.B. Massalski (Editor-in-
Chief), ASM International,
Materials Park, OH, 1990.)
(Figs. 9.12 and 9.15
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
Adapted from
Fig. 9.15, Callister 6e. Adapted from Fig. 9.12,
Callister 6e.
Adapted from Fig. 9.15,
Callister 6e. (Illustration
only)

Intermediate phases
Intermediate solid solutions (intermediate phases): Solid solutions that do
not extend to pure components in the phase diagram.
Cu-Zn
Terminal solutions: a
and h.
Intermediate solutions:
b, g, d and e.
Tie lines and lever rule can
be used to determine
compositions and wt% of
phases.
e.g. at 800oC with 70 wt%
Zn
CL = 78 wt% Zn
Cg = 67 wt% Zn

Eutectoid and peritectic reactions
Eutectoid reaction: one solid phase turning into two other solid phases upon cooling
e.g. d g + e
heat
cool
Peritectic reaction: one solid phase transforms into liquid and a different solid
phases upon heating
e.g. d + L e
heat
cool

Congruent phase transformation
Congruent transformation: no change in composition upon phase
transformation.
Incongruent transformation: phase transformation where at least one of the
phases go through composition change.
At what overall
composition does Mg2Pb
melt congruently?

Intermetallic compounds
Intermediate (intermetallic) compounds: discrete metal compounds rather
than solutions (i.e. distinct chemical formula with stoichiometric ratio or range).
AxBy: in solution x and y can vary
in compounds x and y are fixed (always fixed composition of A and B).
Mg-Pb phase diagram
Can be considered as two
phase diagrams put
together back to back.
Mg-Mg2Pb phase diagram Mg2Pb-Pb phase diagram

Intermetallic compounds: Example
A 50wt% Pb/50wt% Mg
alloy is heated to
300oC. The mass
fraction of each phase
is 0.5.
1. What are the phases
present?
2. What are the
compositions of the
phases?

Ceramic phase diagrams
Al2O3-Cr2O3 MgO-Al2O3

Iron-Carbon System
Iron
Typical metal (e.g. Cu)
Solid
Liquid
Tm
T
T(oC)
912
1394
1538
Ferrite (BCC)
Austenite (FCC)
Ferrite (BCC)
Liquid

Iron-Carbon System
FCC g-phase has highest C concentration (2.14 wt% C) whereas BCC a-phase has low
solubility (0.022 wt% C). Recall FCC is close packed (i.e. larger APF). Why is C more soluble
in FCC?
Note: only goes out to 6.7
wt% C (100 wt% Fe3C
intermediate compound)
Eutectic point
Eutectoid

Eutectoid Cooling
g (0.76 wt% C)
cool
heat
a (0.022 wt% C)
+ Fe3C (6.7 wt% C)
Layered structure forms due to the same reason as eutectic structure formation.
Pearlite structure

Hypoeutectoid Alloys
Cooling below eutectoid
composition.
c: homogeneous g solid.
d: a  g coexistence. a-phase
nucleate at the grain boundaries
(Why?).
e -> f:
- crossing eutectoid isotherm will
cause all remaining g-phase into
eutectoid structure.
- a-phase that formed prior to
eutectoid isotherm are called
proeutectoid ferrite.
Fraction of pearlite =
022
.
0
76
.
0
022
.
0


 o
p
C
W
022
.
0
76
.
0
76
.
0
'


 o
C
Wa
Fraction of proeutectoid a =

Hypereutectoid Alloys
Cooling above the
eutectoid composition
Compositions and wt% can be
found similarly as hypoeutectoid
cooling.
Instead of proeutectoid a,
proeutectoid cementite appears.

Bainite
coarse fine
Austenite
Martensite
Moderate cooling (AS)
Isothermal treatment (PCS)
Tempered
Martensite
Pearlite
AS: Alloy Steel
PCS: Plain-carbon Steel
Slow
Cooling
Rapid
Quench
Spheroidite
Re-heat
Re-heat

Example problem
For 0.35 wt% C, at T just below
eutectoid isotherm,
determine:
a) Fractions of total ferrite and
cementite phases.
b) Fractions of proeutectoid
ferrite and pearlite.
c) Fraction of eutectoid ferrite.

Influence of other alloying elements
Changes eutectoid T
Changes eutectoid composition
Useful processing info to control microstructure.

Determination of Phase
Diagrams
• Cooling Curves
• Differential Scanning Calorimetry, ΔH
• Thermomechanical Analysis,ΔV
• Differential Thermal Analysis, ΔT
• Dilatometry, ΔL
• Metallography/Petrography
• Energy Dispersive X-ray Spectroscopy, SEM
• Electron Microprobe Analyzer, EMPA
• X-ray Diffraction
• Transmission Electron Microscopy

Experimental measurement of
∆Hm from DSC

Gibbs Phase Rule
P + F = C + N
Number of
phases present
Degree of freedom (externally controllable
parameters: i.e. T, P, and C)
Number of
components
Number of non-
compositional variables
(Temperature & Pressure)
e.g. Cu-Ag phase diagram
Cu and Ag are the only components
-> C = 2
Temperature is the only non-compositional variable
here (i.e. fixed pressure).
-> N = 1 (but in general N = 2)
When 2 phases are present
-> P = 2 which leads to F = C+N-P = 2+1-2 = 1
When only 1 phase is present.
-> P = 1 which leads to F = 2
What does this mean? Why should you care?

Gibbs Phase Rule
In the previous example of Cu-Ag phase diagram, when F = 1, only one
parameter (T or C) needs to be specified to completely define the system.
e.g. (for a+L region)
If T is specified to be 1000oC,
compositions are already determined
(Ca and CL).
Or
If composition of the a phase is
specified to be Ca then both T and CL
are already determined.
Ca CL

Gibbs Phase Rule
When F = 2, both T and C have to be specified to completely define the
state of the system.
e.g.(for a region)
If T is specified to be 800oC, Ca can be
any where between 0 to ~8 wt% Ag)
Or
If composition of the a phase is
specified to be Ca = 3 wt%, then T and
can be any where between ~600 to
1100oC.
Ca

Gibbs Phase Rule
Where in the Cu—Ag diagram, is there a 0 degree of freedom?
(i.e. T, P, and C are all fixed)

TERNARY PHASE DIAGRAMS
Course on Phase Diagrams, 2007 NED University Karachi

What are Ternary Phase
Diagrams?
• Diagrams that represent the equilibrium between the
various phases that are formed between three
components, as a function of temperature.
• Normally, pressure is not a viable variable in ternary
phase diagram construction, and is therefore held
constant at 1 atm.

Ternary Phase Diagrams
• Utility of Ternary Phase Diagrams
• Glass compositions
• Refractories
• Aluminum alloys
• Stainless steels
• Solder metallurgy
• Several other applications

Ternary Phase Diagrams
Three component systems A, B and C
– Requires that we know the three binary systems for
the 3 components
– – AB, BC, CA
– Ternary diagrams present a map of the Liquidus
surface which is contoured with respect to
Temperature.
– • Fields indicated on the ternary diagram represent
the primary phase fields present on the Liquidus
surface.

Ternary Isomorphous System-1
Three Component system Completely soluble, Only one solid phase.
• All components are completely soluble.
• The ternary system is therefore made up of three binaries that exhibit total solid solubility
• The Liquidus Surface: A plot of the temperatures above which a homogeneous liquid forms
for any given overall composition
• The Solidus Surface: A plot of the temperatures below which a (homogeneous) solid phase
forms for any given overall composition
Each Ternary diagram is constructed using the three binary diagrams for the
three components AB, BC and CA
Fig. 1: Binary phase diagrams.

Liquidus Surface
Solidus Surface
a b c
d e f
Fig. 2: Construction of Ternary Diagram from binary phase diagrams.

Isothermal Section: A “horizontal” section of a ternary phase diagram
obtained by cutting through the space diagram at a specified TemperatureT1
Draw the horizontal, intersecting
the liquidus and solidus surfaces
at points 1, 2, 3 & 4
•Connect points 1 & 2 with curvature
reflecting the liquidus surface
•Connect points 3 & 4 with curvature
reflecting the solidus surface

• The line connecting points 1 & 2
represents the intersection of the
isotherm with the liquidus surface
• The line connecting points 3 & 4
represents the intersection of the
isotherm with the solidus surface
• Area A-B-1-2: homogeneous liquid
phase
• Area C-3-4: homogeneous solid
phase
• Area 1-2-3-4: two phase region -
liquid + solid

Isothermal Section – continued…
At Temperature = T2, below melting points of A & B, but above
melting point of C
•Area A-1-2: homogeneous liquid phase
•Area B-C-4-3: homogeneous solid phase
•Area 1-2-3-4: two phase region - liquid + solid
1

How to Read a Ternary Phase Diagrams?

• Overall Composition -1
• The concentration of each of the three
components Can be expressed as either
“wt. %” or “molar %”
• The pure components are represented
by each corner
• Sum of the concentration of the three
components must add up to 100%
• The Gibbs Triangle is always used to
determine the overall composition
How to Read a Ternary Phase Diagram?-1
Determination of:
Chemical composition of phases present
when the overall composition is in a two phase region

Let the overall composition be
represented by the point X
–Draw lines passing through
X, and parallel to each of the
sides.
–Where the line A’C’ intersects
the side AB tells us the
concentration of component B
in X.
Now wt% of
B = AA’
C = BB”
A = CC”
A’
B’
C’
A”
C”
B”
Determining Chemical Composition Method 1:

Draw lines through X,
parallel to the sides of the
Gibbs Triangle
•A’C’ intersects AB at A’
•B’C” intersects AB at B’
Now wt% of
B = AA’
C = A’B’
A = B’B
Determining Chemical Composition Method 2:
B’
B” C’
C”
A’
A”
100WT%C

Application of the Lever
Rule
– Draw straight lines from
each corner, through X
–%A = MX/MA
–%B = NX/ NB
–%C = LX/LC
Important Note:
Always determine the concentration of the components independently, then check
by adding them up to obtain 100%
Determining Mass Fraction of Phases

How to Read a Ternary Phase Diagram-
Example
Figure 2: Stainless steel phase diagram
at 900 oC (ASM 1-27)
The Ni-Cr-Fe System includes
Stainless Steel and Inconel
(Nickel based Superalloys)
Inconel:
•Very good corrosion resistance
•Used in enviroenment where SS Fails
Exmples of Inconel Uses:
Piping on Nuclear Plant
Steam Generators and
other Power plant applications
Ni-Cr-Fe System

How to Read a Ternary Phase Diagram
Example
Reading the compositions of iron, chromium and
nickel at any point on the stainless steel ternary
phase diagram in Fig. 2 is simple.
– Instead of drawing one tie-line, as in a binary phase
diagram , three lines are drawn, each parallel to a side of
the triangle and going through the point in question.
– Extend the lines so they pass through an axes.
– To find the iron composition, the line drawn parallel to the
axis opposite the Fe vertex is the one needed. The percent
iron is then read off the axis.

Example
Compositions of the 18-8 Stainless Steel (circle point near
the lower left corner of Fig. 2.)
Draw these lines:
– Draw the first line to be parallel with the axis opposite the Fe
vertex, we find that the composition of iron is 74%
– Next draw a line parallel with axis opposite the Ni vertex and
read the composition of nickel to be 8%
– And finally draw a line parallel to the axis opposite the Cr vertex
and we see that there is 18% of chromium at that point.
The point described is then called 18-8 stainless steel,
naming only the percentages of the chromium and the
nickel; the iron content being dependent on the other two.
This 'recipe' for stainless steel is the most common.

Isomorphous System
Determination of:
Amount of each phase present
when the overall composition is in a two phase region
1. Locate overall composition using the Gibbs
triangle.
2. Draw tie-line passing through X, to intersect
the phase boundaries at Y and Z.
3. The chemical composition of the liquid
phase is given by the location of the point Y
within the Gibbs Triangle
4. The chemical composition of the solid
phase is given by the location of the point Z
within the Gibbs Triangle.
Amount of each phase present is
determined by using the Inverse
Lever Rule
5. Fraction of solid = YX/YZ 6. Fraction of liquid = ZX/YZ

Ternary Eutectic System
(No Solid Solubility)

(No SolidSolubility)
The Ternary Eutectic Reaction: L = α + β + γ
A liquid phase solidifies into three separate solid phases
Phase Regions: Homogeneous liquid phase (L), Liquid + one solid phase,
Liquid + two solid phases, Three solid phases
Ternary diagram is constructed using the three binary diagrams for the
three components AB, BC and CA, as shown in Fig. 1a-c.
L L
L
L+α
L+α
L+β
α+β β+γ
γ+α
L+γ
L+β
L+γ
Fig. 1: Binary Phase diagram.

d
Simulation of Liquidus Surface
b
L
L
L
P
Ternary Eutectic
a
L
L
L
α+β
Three Binary on
Gibb’s Triangle
c
Liquidus and Solidus Surfaces
Fig. 2: Construction of Ternary Diagram from binary phase diagrams.

Phase regions:
•Homogeneous liquid phase
above the liquidus surface
(AEFJNOP).
• Below the liquidus surface,
represent the existence of two or
more phases.
•Three solid phases below the
Solidus surface
(BCDGHPJKLMO)
S-1
S-2
S-3
S-T
L
Ternary Solidus= DPK

• The liquidus surface “dips down”
somewhere in the middle, to the
ternary eutectic point (P).
• The ternary eutectic point would be
at a temperature lower than all three
binary eutectic temperatures. E
J
O
P
Fig. 1: A projection of the liquidus surface
onto a plane, with indications of isotherms
and phase regions
Isotherm Sections
Liquidus Projection

Example 1: Isothermal Section at
T1, below melting point of A, but
above melting points of B and C
•We have two regions: a region of
“liquid” and a region of “liquid + A”
•The boundary between these two
regions is a line, the curvature of
which is in accordance with the
liquidus surface

Example-2: Isothermal Section at
T2, below the melting points of A
and C, but above the melting point
of B, and above the eutectic
temperature of the system A-C
•This isothermal section has three
regions - L, L+A, and L + C
•The boundary between L and L+A is
determined by where the isothermal
plane cuts the liquidus surface

Example 3: Isothermal section
at temperature T3, below the
eutectic temperature of the
system A-C, but still above the
melting point of B
•The isothermal section now has
four regions: L, L+A, L + C, L + A +
C

Example-4: Isothermal section
at temperature T4, below the
melting points of A, B and C,
and below the eutectic
temperature of the A-C system,
but above the A-B and B-C
eutectic temperatures
•The isothermal section now has
five regions: L, L+A, L+B, L+C,
L+A+C
•Point 2 has moved further into the
Gibbs
Triangle, towards the ternary
eutectic point

Solidification Sequence
in

Solidification Sequence
Given an overall composition, determine the
sequence of solidification, assuming
equilibrium conditions.
• Let the overall composition be given by the
point X
• Imagine a line, orthogonal to the plane of
the liquidus projection, passing through X
• Let this line intersect the liquidus surface at
a temperature T1
Now
• For all temperatures T > T1, there is one
homogeneous liquid phase and
Solidification can only begins when T = T1
The first solid to appear is: A?
d
Simulation of Liquidus Surface
P
E
J
O
O
P
J
E
Solidification Sequence-1

Solidification Sequence…..
• When T < T1, then precipitation of A
occurs as the temperature drops and
the composition of the liquid phase
“travels” along the line XY, on the
liquidus surface, towards Y.
• Let the temperature at Y be T2
• At temperatures of T2 < T < T1, there
are two phases in equilibrium - A
and L
Detremination of Amount of Phases
• The amount of each phase present at
point Z and temperature T = T’,
From Lever Rule
Fraction of L = AX/AZ
Fraction of A = XZ/AZ
• Chemical composition of the liquid
phase at point Z?

Solidification Sequence…
•At point Y, where T = T2, the second solid
phase, B, begins to precipitate
Over the temperature range of T2 > T > TE:
– The solid phases A and B exist in
equilibrium with L
– Both solid phases, A and B,
coprecipitate as the temperature is
lowered
Analysis when T = T”, i.e., at point M
Composition of L is given by the
composition of M within the Gibbs Triangle
•How do we determine the amounts of
A, B and L?
•Let temperature T” correspond to the
point M

Solidification Sequence…
Let temperature T” correspond to the
point M
Now
•Construct the triangle A-B-M
This triangle is a ternary system in
which the overall composition X can be
represented in terms of the three
constituents.
The amount of each phase,
by Lever Rule:
– Fraction of A = PX/PA
– Fraction of B = QX/QB
– Fraction of L = RX/RM
Analysis at Point ‘M’

Phase Analysis at a given temperature
in Two Phase Region
For all overall compositions that fall
within the region marked L, the chemical
composition of the liquid phase is the
same as the overall composition.
If the overall composition falls within a
two phase region, e.g., L + C, then:
1. Locate the position of the overall
composition, X, within the Gibbs
Triangle
2. Draw tie lines, in this case connecting
point C to line 2-3, and passing through
X, and intersecting line 2-3 at Y
3. Use the Lever Rule to determine the
amounts of L and C
– Fraction of L = CX/CY
– Fraction of C = YX/CY
Phase Analysis at Temperature T4
Example-4: Isothermal
section at temperature
T4, below the melting
points of A, B and C,
and below the eutectic
temperature of the A-C
system, but above the A-
B and B-C eutectic
temperatures

Phase analysis at a given temperature in
Three Phase Region
The chemical composition of C in this case
is 100% C
The chemical composition of L is given by
determining the composition the point
Y represents within the Gibbs Triangle
If the overall composition falls within a
three phase region, e.g., L + A + C
1. Locate the position of the overall
composition, Y, within the Gibbs
Triangle
2. Construct the following straight
lines: A-Y-Q, M-Y-R and C-Y-P
3. Use the Lever Rule to determine the
amounts of L, A and C
– Fraction of A = QY/QA
– Fraction of C = PY/PC
– Fraction of L = YR/MR
Phase Analysis at Temperature T4

(with Solid Solubility)
After Selvaduray, J. et. al.,

Let us consider three binary systems with solid solubility:
α
β γ
L
α+β β+γ γ+α
γ
β
L L
α

Construction of Ternary Phase Diagram form Three Binary
Systems with Solid Solubility

Construction of Ternary Phase Diagram form Three Binary
Systems with Solid Solubility
Main outline of Ternary Phase Diagram with
Ternary Eutectic (Te)
All Liduidus Surfaces
[Red(α+L), Purple(β+L), Green(γ+L)]
Computer Simulation of Liquidus

Computer Simulation of Ternary Phase Diagram
Ternary Eutectic System (with Solid Solubility)
After Selvaduray, J. et. al.,

Isothermal Section at Temperature T1<TB, But>TA, TC

Isothermal Section at Temperature T2<TA, But T2<TB, TC

Ternary Eutectic System (with Solid Solubility)
Isothermal Section at Temperature
T3<TA, TB, TC, But T3>TE1, TE2 , TE3

Dr. Ashraf Ali Course of Phase
Diagrams, 2007 NED University
Isothermal Section at Temperature T4<TTE2, and T4<TTE1, TE2

CONCLUSIONS
Phase Diagrams
Unary,
Binary
and Ternary

Concepts to remember
• Phases, physical states, chemical composition, phase equilibrium.
• Phase diagrams tell us about:
– Number and types of phases present.
– Composition of each phase.
– Mass fraction (wt%) of each phase.
• Binary isomorphous systems.
• Intermediate phases and compounds.
• Binary eutectic systems.
• Microstructure evolution in cooling (equilibrium vs. non-equilibrium).
• Eutectic, eutectoid, and peritectic reactions.
• Hyper- and hypoeutectic & hyper- and hypoeutectoid alloys
• Fe-C phase diagram.
• Phase diagrams help us to determine the equilibrium
microstructures which in turn determines the properties of
materials!

Figure 11-4 The five most important three-phase
reactions in binary phase diagrams.

Summary
Make sure you understand language and concepts:
• Austenite
• Cementite
• Component
• Congruent transformation
• Equilibrium
• Eutectic phase
• Eutectic reaction
• Eutectic structure
• Eutectoid reaction
• Ferrite
• Hypereutectoid alloy
• Hypoeutectoid alloy
• Intermediate solid solution
• Intermetallic compound
• Invariant point
• Isomorphous
• Lever rule
• Microconstituent
• Pearlite
• Peritectic reaction
• Phase
• Phase diagram
• Phase equilibrium
• Primary phase
• Proeutectoid cementite
• Proeutectoid ferrite
• Solidus line
• Solubility limit
• Solvus line
• System
• Terminal solid solution
• Tie line
• Liquidus line
• Metastable

1. Porter, D.A. and.Easterling, K.E., “Phase Transformations in Metals
and Alloys”, Van Nostrand Reinhold, 2000 or latest edition..
2. Reed-Hill, R. E. and Abbaschian, R., “Physical Metallurgy
Principles”, PWS, 2000 or latest edition.
3. Smallman R.E “Modern Physical Metallurgy”, 4th ed., Butterworths,
2005.
4. Shewmon, “Diffusion in Solids”, Mineral, Metals & Mat. Soc., Latest
edition.
5. Shewmon, “Phase Transformations”, Springer, Latest edition.
6. Honeycombe, R.W.K., Bhadeshia, H.K.D.H., “Steels,
Microstructures and Properties”, Edward Arnold, 2005.
7. Chadwick, G. A., “Metallography of Phase Transformations”,
Butterworths, Latest edition.
8. Wilson, E.A., “Worked Examples and Thermodynamics of Phase
Transformations”, Woodhead, 2004.
9. ASM Metals Handbook, vol.3
10. Various Web sites at Internet
Phase Diagrams: Suggested Readings

Thanks!
Any questions?
You can find me at
• @hamzaahmed
• hamzaahmed0696@mail.me

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