Material Science I
Phase diagrams II
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Material Science I
Introduction ( Binary Phase diagrams)
• A different kind of phase diagram which is very common is where
the temperature and composition are varied when the pressure is
constant
• We will focus on binary- alloys
• Binary phase diagrams are helpful in predicting phase
transformations and the resulting microstructure
• Therefore, binary phase diagrams are maps representing the
relationships between temperature and compositions as well as
quantities of phases at equilibrium
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Material Science I
Objectives
• Determine for a given binary phase diagram
• Phase(s) present
• Composition of the phase(s)
• Mass fraction of the phase(s)
• Locate the temperature and compositions of all eutectic,
eutectoid, peritectic and congruent phase transforms
• Be able to write reactions for all the transforms for either heating
or cooling
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Material Science I
Binary Phase diagrams (Isomorphous systems)
• The easiest type of binary phase diagrams to understand and
interpret is a copper-nickel system
• The system is said to be isomorphous because they (components)
are completely soluble in liquid and solid forms
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Material Science I
Binary Phase diagrams (Isomorphous systems)
NOTE:
In the phase diagrams, for
metallic
alloys,
solid
solutions are commonly
designated
by
Greek
letters (, , etc)
Consider the copper –
nickel phase diagram given
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Material Science I
Binary Phase diagrams (Isomorphous systems)
In the phase diagrams,
there are three different
phase regions or fields
() an alpha field
() a liquid field
( + ) two-phase field
Each region is defined by
the phase or phases that
exist over the range of
temperatures
and
compositions
The phases are delimited
by phase boundaries
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Material Science I
Binary Phase diagrams (Isomorphous systems)
The liquid is a
homogeneous
solution
composed of nickel and
copper
phase is a substitutional
solid solution with both
copper and nickel having
an FCC crystal structure
At temperatures below
1080℃, copper and nickel
are mutually soluble in
each other in the solid
state at all compositions
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Why?
Material Science I
Interpretation of Phase diagrams
In a binary system with known composition and temperature at
equilibrium, at least three kinds of information are available
Phase(s) present
Composition of the phases present
Percentages or fractions of the phases
We now show how to determine these three pieces of
information in a nickel-copper system
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Material Science I
Interpretation of Phase diagrams
Phases Present
In order to determine the phase(s) present, one needs to know
the temperature and composition at a point on the phase
diagram e.g. 60wt% − 40% at 1100℃
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase compositions
Locate the temperature –composition point with concentration
in terms of concentration of the components
For a single phase, the composition is just the same as the overall
composition of an alloy.
To determine the phase composition for an alloy with
composition and temperature located in a two-phase region is
not easy but can be determined using a tie line
A tie line, can be thought of as a horizontal lines with same
temperature (isotherm)
They are drawn so as to terminate at phase boundary lines on
either side
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase compositions
Using the tie line we can compute the equilibrium concentrations
of the two phase (only) using the following procedure:
Draw a tie line across the two – phase region at the
temperature of the alloy
Note the intersection points on the phase boundaries
Drop the perpendiculars from intersection points to
composition axis where each respective phase is read
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase compositions
Example:
Determine the phase composition for copper-nickel alloy at
35% and 65% alloy at 1250℃
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Material Science I
Interpretation of Phase diagrams
How to determine the Phases compositions
Solution
The problem is to determine the composition in % of both and for both the and phases
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase compositions
Solution
Liquidus boundary is by the composition axis at 31.5% Ni and
68.5% Cu at Solidus tie line intersection , 42.5% − 57.5%
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
It is important to determine the relative amounts as fractions or
percentage of the phases present at equilibrium
In a single phase, only one phase is present therefore the fraction is
1.0 or 100% as a percentage
If the composition and temperature position is located within a two –
phase region, one needs to think
Here again the tie line in conjunction with a procedure called lever
rule must be applied.
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
In the lever rule, the following procedure is followed
i. Draw a tie line across the two-phase region at the temperature
of the alloy
ii. Overall alloy composition is located an the tie line
iii. To compute the fraction of one phase, take the length of the tie
line from the overall alloy composition to the phase boundary
for the other phase and divide by the total tie length.
iv. Use the above steps to determine the fraction of the other
phase
v. Multiply the phase fractions by 100 to get the phase
percentages
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
How do you determine the tie line segment lengths?
i. Either by direct measurement from phase diagram using a
linear scale graduated in millimeters or
ii. By subtracting compositions as taken from the composition axis
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
Consider the diagram below, we want to compute the fraction of
each of the and liquidus phase
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
Locate the overall alloy composition along the tie line and denoted by
and the mass fraction and for respective phases
From the lever rule
=
"
#$"
=
%& '%(
%& '%)
using direct measurement
by subtraction using compositions
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
NOTE:
Composition need to be specified in terms of only one of the
constituents for a binary alloy: so far we have made our
computations in weight percent nickel i.e. = 35%, =
42.5%, = 31.5%
*
− 42.5 − 35
=
=
= 0.68
=
+ + * − 42.5 − 31.5
For the phase
*
, − 35 − 31.5
=
=
= 0.32
=
+ + * − 42.5 − 31.5
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
We can conclude by stating that the lever rule can be employed to
determine the relative amounts or fractions of phases in any twophase region for a binary alloy if temperature and composition are
known and if equilibrium has been established.
IMPORTANT
Compositions of phases are expressed in terms of % of
components i.e. % , %
Fractional phase amounts (e.g. mass fraction of or liquid phase) of
a two –phase alloy
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
Home work
Derive the lever rule for determining mass fraction of a two-phase
region
(see example 9.1 ) William D Callister Jr.
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
For multiphase alloys, it is often more convenient to specify relative
phase amounts in terms of volume fraction rather than mass fraction.
Phase volume fractions cab be examined from examination of the
microstructure
Similarly, properties of multiphase alloy may be estimated on the
basis of volume fractions
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Material Science I
Interpretation of Phase diagrams
How to determine the Phase amounts
For an alloy consisting of phase - is defined as
.
- =
./ + .
Where
. and ./ denote the volumes of the respective phases in
the alloy
We also know that for an alloy with two phases
- + -/ = 1
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Material Science I
Interpretation of Phase diagrams
How to convert from mass fraction to volume fraction and vice versa
Converting from mass fraction to volume fraction
0
- =
0
+
/
0/
/
0/
-/ =
0
+
/
0/
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Material Science I
Interpretation of Phase diagrams
How to convert from mass fraction to volume fraction and vice versa
Converting from volume fraction to mass fraction
- 0
=
- 0 + -/ 0/
/
-/ 0/
=
- 0 + -/ 0/
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Material Science I
Interpretation of Phase diagrams
How to convert from mass fraction to volume fraction and vice versa
Note :
0 and 0/ are the densities of respective phases and their
appropriate values can be found by using the equation:
0123
0123
100
=
4 5
+
04 05
46 74 + 56 75
= 6
4 74 56 75
+
04
05
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Material Science I
Development of microstructure in Isomorphus alloys
Consider the figure below
We are interested in determining
(uniform equilibrium cooling)
the
development
of
microstructure that occurs for
isomorphous
alloys
during
solidification
Solidification occurs very slowly
and continuously maintained
The overall composition remains
unchanged during cooling even
though there is a redistribution
of copper and nickel between
phases and is along the dashed
solid line
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Material Science I
Development of microstructure in Isomorphus alloys
Non - equilibrium cooling (self study)
William D callister Jr (2009), Material Science and Engineering, an
Introduction, 7th Edition, chapter 9
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
For the copper-silver system shown in the phase diagram called
eutectic phase diagram, there are a number of features which are
important and worth noting
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
The three single phase regions are found i.e. , and liquid
phase is a solid solution rich in copper and has silver as the
solute component (an FCC crystal structure)
phase is a solid solution rich in silver with copper as the solute (
FCC crystal structure)
and are pure copper and pure silver respectively
Solubility of these solid phases is limited since at any temperature
below line BEG only a limited concentration of silver will dissolve
in copper (for the phase)
Similarly, for copper in silver ( for phase)
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
Solubility limit for the phase corresponds to the boundary line
CBA between the ⁄( + ) and ⁄( + ) phase regions
The solubility increases with temperature to a maximum (
8.0%7DE779℃) at point B and decreases back to zero at
the melting temperature of pure copper at point A (1085℃)
The maximum solubility of copper in the phase point G
(8.8%) also occurs at 779℃
Line BEG represents the lowest temperature at which a liquid
phase may exist for any copper-silver alloy that is at equilibrium
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
Note:
There are three two-phase regions for the copper-silver system,
i.e. + , + and α + compositions and relative amounts for the phases may be
determined using tie lines and the lever rule
Point E is called the invariant point designated by H (71.6%)
and temperature IH (779℃)
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
At the invariant point H (71.6%), as it changes temperature
in passing through IH , the reaction is expressed as
This equation is called eutectic reaction in which H and IH
represent the eutectic composition and temperatures
respectively
H and /H are the respective compositions of the and phases at IH
For this reaction
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
The horizontal solidus line at IH is called eutectic isotherm
Note:
The eutectic reaction upon cooling is similar to solidification and
is always two-phases , where as for pure component only a
single phase forms
Phase diagrams similar to the one discussed here are termed
eutectic phase diagrams and components which exhibit this
behaviour comprise a eutectic system
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Material Science I
BINARY EUTECTIC (EASILY MELTED) SYSTEMS
When constructing a binary phase diagram, it is important to
understand that one or at most two phases may be in equilibrium
within a phase field
For a eutectic system, three phases (, and ) may be in
equilibrium but only at a point along the eutectic isotherm
The other general rule is that a single-phase regions are always
separated from each other by a two phase region that consists of
two single phases that is supports
JKLMKNO(PQELRSM9.2ETU9.3)
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