Phase Diagrams i i

8/13/2019 Phase Diagrams i i

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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

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