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Phase Diagrams II
Cu-Ni
Phase
Diagram
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Binary Eutectic Systems
Examples
Cu Ag
Pb Sn
Important Features
Cu Ag
3 Single phase fields Cu rich solid solution
Ag rich solid solution
Liquid 3 two phase fields
+ liquid
+ liquid
+
Cu Ag
The phase is a solid
solution rich in Cu andhas Ag as the solute
component
The phase is a solid
solution rich in Ag and
has Cu as the solute
component
Solubility in each of
these solid phases is
limited. For example;
For the , at any
temperature below the
BEG line the only a
limited concentration of
Ag will dissolve in Cu.
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Cu Ag
Boundary between the
Liquid region and the
+ L is called the
LIQUIDUS.
The maximum
solubility the phase
corresponds to line
CBA. It increases to amaximum at B and
then decreases to zero
at A, the melting point
of Cu.
The solid solubility limit
between and + is
called the SOLVUS
The boundary between
the and + L is
called the SOLIDUS.
Point at which the
LIQUIDUS lines meet
called the EUTECTIC
POINT.
The Eutectic Reaction
Point E is the Invariant orEutectic Point
Composition, CE, 71.9 wt% Ag
Temperature, TE, 779C
3 phases exist in equilibrium
Cu Ag Phase Diagram
Important Reaction for an
alloy of composition, CE, as
it changes temperature andpasses, TE, is:
)()()( EEE CCCL +
Eutectic Point
Why is it called the eutectic
point?
Simply because eutectic
means EASILY MELTED!
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Pb Sn Binary Eutectic Phase Diagram
Lead Tin is
another common
eutectic system.
It has a eutectic composition
of 61.9 wt% Sn
In this case
represents a Pb
rich solid solution
with Sn as the
solute
represents
a Sn rich
solid solution
with Pb as
the solute.
Notice the phase
diagram for Pb Sn is
slightly different from
Cu Ag
Pb Sn Binary Eutectic Phase Diagram
The eutectic invariant
point is located at:
69.9 wt% Sn
183C
You are all familiar
with SOLDER, it is an
alloy of: 60 wt% Sn
40 wt% Pb
It is completely molten
about 185C.
185 C
This makes it useful as
a low-temperature
solder!
Remember EUTECTIC
means EASILY melted!
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Binary Eutectic Example
Problem Consider an Sn-Pb binary eutectic alloy:
40 wt% Sn-60 wt% Pb
At a temperature of 150C
Determine:
The phases that are present at 150C.
The composition of the phases.
The relative amount of each phase present in terms of:a)mass fraction
b)volume fraction
What phases are present at 150C, andwhat is their composition?
Point B lies at
40 wt% Sn
60 wt% Pb
150C
Because point B lies
in the + region,
both and
phases will exist.
What are the
compositions of the
phases?
C C
2 phases present,
therefore we need
to construct a tie
line across + field
at 150C
Composition of
phase, C = 10 wt%
Sn-90 wt% Pb
Composition of
phase, C = 98 wt%
Sn-2 wt% Pb
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Since the alloy consists of two
phases it is necessary to employ
the LEVER RULE.
34.01098
1040
66.01098
4098
1
1
=
=
=
=
=
=
CC
CCW
CC
CCW
If C1 denotes the overall
compositions, mass fractions
can be calculated by subtracting
compositions, in terms of
weight percent Sn:
C CC1
wt% Sn
Determine the relative amount of each
phase present in terms of mass fraction.
To compute the volume fraction of each phase it is necessary to determine the
density of each phase taking the densities of Pb and Sn to be 11.23 and 7.24 g/cm3.
Determine the relative amount of eachphase present in terms of volume fraction.
Pb
Pb
Sn
Sn CC
)()(
100
+
=
where CSn() and CPb() denote the concentrations in wt% of Sn and Pb in the phase
CSn() = 10 wt%
CPb() = 90 wt%
3/29.7
23.11
2
24.7
98
100cmg=
+
=
3/64.10
23.11
90
24.7
10
100cmg=
+
=
Pb
Pb
Sn
Sn CC
)()(
100
+
=
Similarly for the phase
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Now we can determine the volume fractions, V and V
Determine the relative amount of each
phase present in terms of volume fraction.
WW
W
V
+
= 57.0
29.7
34.0
64.10
66.064.10
66.0
=
+
=V
WW
W
V+
= 43.0
29.7
34.0
64.10
66.0 29.7
34.0
=+
=V
Microstructural Development in Eutectic
Alloys
Depending on composition, several different types
of microstructure are possible for slow cooling of
binary eutectic systems.
Lets consider some of these possibilities for the
Pb Sn system.
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Microstructural
Development During
Cooling, C1 For compositions ranging from pure
Pb to maximum solid solubility of
Pb at room temperature.
Consider alloy of composition C1slowly cooled from liquid region
(350C)
Totally liquid until it passes through
liquidus (330C).
forms in narrow +L region.
Solidification complete at solidus.
Result is a polycrystalline alloy ofuniform C1 composition
a Pb rich alloy with 0 2 wt% Sn
No compositional changes will
occur if further cooled.
Microstructural
Development During
Cooling, C2
For compositions ranging from room
temp. solid solubility to maximum
solid solubility (eutectic temp.)
Alloy of composition C2 slowly
cooled from liquid region (350C)
Changes are similar to previous case
until we reach the SOLVUS. Upon crossing the SOLVUS the
solid solubility is exceeded.
Small particles form and will grow
on cooling because the mass fraction
slightly increases as temp. decreases.
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Microstructural Development During Cooling,
C3 (Eutectic)
For the eutectic composition.
Alloy of composition C3 slowly
cooled from liquid region
(250C)
No changes occur until the
eutectic temperature (183C) is
reached!
Upon crossing the eutectic
isotherm the liquid is
transformed to and phases.
Redistribution of Pb and Sn byatomic diffusion.
Resultant microstructure
consists of alternating layers
(lamellae) of and phases.
Microstructure at the Eutectic
Formation of eutectic structure by
atomic diffusion, directions are
diffusion of Pb and Sn atoms!
Resultant microstructure consists
of alternating layers (lamellae) of
and phases.
86 m
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Solidification and Cooling at Composition C4
For all other compositions
other than the eutectic that
pass the eutectic isotherm.
Alloy of composition C4.
Cooled through the liquidus
into the +L region.
Tie lines can be drawn to find
the composition of the and
L phases just before the
eutectic isotherm (point l).
phase = 18.3 wt% Sn
L phase = 61.9 wt% Sn
Just below the eutectic theliquid with transform to +.
phase will be present in the
eutectic structure and also as
the ' formed during cooling
through +L region.
87.5 m
Solidification and Cooling at Composition, C4
Determine compositions and wt
% of phases at
1.eutectic temperature +1 oC
(184 oC)
2.eutectic temperature -1 oC
(182 oC)
At E+1oC (solid + Liquid)
solid = 18.3 wt% Sn
liquid = 61.9 wt% Sn Find fractions of the eutectic
microconstituents, (W',We=WL)
100' +
=QP
QW
100+
=QP
PWL
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Solidification and Cooling at Composition, C4
At E-1oC (solid + solid)
solid = 18.3 wt% Sn Solid = 97.8 wt% Sn
Find total fractions of the
microconstituents, (WW)
100++
+=
RQP
RQW
100++
=RQP
PW
Phase Diagrams with Intermediate Phases,Cu Zn (Brass)
More complicated
Six solid solutions
2 terminal - and
4 intermediate , ,
and
ordered solid
solution `
Some new invariant
points
Commercial cartridge
brass
70 wt% Cu
30 wt% Zn
single phase
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Other Invariant Points
Eutectoid (Eutectic Like)
note instead of liquid
phase transforming into
two solid phases its a
solid phase!
important in the heat
treatment of steels
+
Peritectic
solid and liquid
transform into one
solid
+L