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Phase Diagrams I

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Phase Diagrams

Need to study phase diagrams to Understand alloying Explain heat treatments and microstructure Understand solidification

Casting and welding Definition of a Phase

A homogeneous portion of a system that has uniform physicaland chemical characteristics

Examples are solid, liquid, gas For a solid, allotropes: eg FCC, BCC, HCP are considered

different phases

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Sugar/Water Phase Diagram

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Binary Isomorphous Systems Binary

Two elements

Isomorphous is defined as The liquid and solid region each

contain only one phase Cu-Ni is a good example

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Cu-Ni System

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homogeneousliquid solution

Substitutionalsolid solutionconsisting ofboth Cu andNi atoms,FCC structureCu, Ni havenearlyidenticalatomic radii,FCC structure,etc.,..

Liquid phase ispresent at alltemperaturesandcompositionsabove this line

Solid phase only existsbelow this line

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Use of Phase Diagrams Three kinds of information available

Determination of phases Go to the temperature and composition of interest and read off phase:

single phase vs two phase regions

Composition of these phases Percentages or fractions of these phases

DEMO

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Determination of Phase Composition

Single Phase Region Composition =

overall composition

Two Phase Region Draw a horizontal

line at temperature ofinterest tie line

Intersection of the tieline and phase

boundaries givescomposition

Overall composition = C 0

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Determination of Phase Amounts

Overall composition = C 0

Lever Rule In two phase region Draw tie line Mark intersection with

phase boundaries Mark the overall

composition Use Lever Rule:

100100

=

+=

L

o

L C C

C C

S R

S W

100100

=

+=

L

Lo

C C C C

S R R

W

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Determination of Phase Amounts(1250 oC)

Overall composition = C 0

100100

=

+=

L

o L C C

C C S R

S W

100100

=

+=

L

Lo

C C C C

S R R

W

C Ni

C NiC Ni L

0 35%

425%315%

=

=

=

;

. ;.

W

mass fraction

l =

=

=

42 5 3542 5 315

0 68

68%

.. .

.

W

mass fraction

=

=

=

35 31542 5 315

0 32

32%

.. .

.

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Derivation of Lever Rule

The sum of the mass fractions ofthe phases must equal 1

The mass of one of the components(Cu or Ni in this case) that is

present in both phases must beequal to the mass of thatcomponent in the total alloyW W L + = 1

W C W C C L L + = 0

Simultaneous solution of these two equations leads to the leverrule expression for this particular case

100

=

L

o L C C

C C W

100

=

L

Lo

C C C C

W

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EquilibriumCooling

Very slow cooling Continuous readjustment

of compositions in both phases

Compositions from tie-line

Amount of solid andliquid at given time can

be calculated from leverrule

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Non-EquilibriumCooling

Complete Mixing inLiquid

No compositionalchanges in solid

Composition is non-uniform in solid Coring segregated

Common in castings Non-optimal