mse 09 phase diagrams(39) 091116

8/10/2019 MSE 09 Phase Diagrams(39) 091116

1/39

Introduction to

.

1http://bp.snu.ac.kr


2/39

Contents

1 IntroductionIntroduction

Solubility LimitSolubility Limit2

Phase DiagramsPhase Diagrams3

Microstructural Evolution durin CoolinMicrostructural Evolution durin Coolin4

2

--

http://bp.snu.ac.kr


3/39

Issues to address

en we com ne two e ements... What equilibrium state do we get?

In particu ar, i we speci y...

Composition (e.g., atomic % Ge atomic % Sn), andTemperature (T )

Then...

How many phases do we get? What is the composition of each phase?

How much of each phase do we get?

ase

3Nickel atomCopper atom

http://bp.snu.ac.kr


4/39

Phase: chemically and structurally homogeneous region of material.

Components: chemically distinct and essentially indivisible

substance.

Solubility limit - maximum concentration of solute that may

dissolve in a solvent at a given tem erature to form a solid solution.

Precipitate - a solid phase that forms from the original matrix

hase when the solubilit limit is exceeded.

Phase diagram - graphical representation of the phases present

and the ran es in com osition tem erature and ressure over

which the phases are stable.

: = +

C : # components

4

F : degree of freedom (temperature, pressure, composition.)http://bp.snu.ac.kr


5/39

Binary phase diagram - A phase diagram for a system with two

componen s. Ternary phase diagram - A phase diagram for a system with three

.

Isomorphous phase diagram - A phase diagram in which components

display unlimited solid solubility.

Liquidus temperature - The temperature at which the first solidbegins to form during solidification.

Solidus temperature - The temperature below which all liquid hascompletely solidified.

-metals that has its own unique composition, structure, and

properties. Eutectic - A three-phase invariant reaction in which one liquid

phase solidifies to produce two solid phases.

5

Peritectic - A three-phase reaction in which a solid and a liquidcombine to produce a second solid on cooling.http://bp.snu.ac.kr


6/39

Contents





6

--

http://bp.snu.ac.kr


7/39

Phases & Solubility

(a) The three forms of water

gas, iqui so i are

each a phase.

(b) Water and alcohol have

unlimited solubility.

(c) Salt and water have limited

solubility.

(d) Oil and water have virtually

7 no solubility.http://bp.snu.ac.kr


8/39

Solubility Limit

-Liquid copper-nickel are

completely soluble.

- Solid copper-nickel are

com letel soluble, with

copper and nickel atoms

occupying random lattice

sites.

8

- In copper-zinc alloys containing more than 30 at. % Zn,

a second phase forms because of the limited solubility of zinc in copper.http://bp.snu.ac.kr


9/39

Solubility Limit

Solubility limit - maximum concentration of solute that may dissolve in asolvent at a given temperature to form a solid solution.

Precipitate - a new solid phase that forms when the solubility limit is

exceeded.

(Fig. 9-1)


C12H22O11 H2O


10/39

Solubility Limit

Phase Diagram of) Solubility

100

Water - Sugar System

ture(

L

Limit (liquid)

+60

80

emper

(liquid solutioni.e., syrup)

S(solidsu ar20

40

solubility limit at 20C?

ar

0 20 40 60 80 100

65 r

.

If Co < 65 wt. % sugar: syrupPur

Sug

Pur

Wat

o > wt. sugar: syrup + sugar

Solubilit limit increases with T :

10

e.g., if T= 100C, solubility limit = 80 wt. % sugarhttp://bp.snu.ac.kr


11/39

Components and Phases

om onenThe elements or compounds which are mixed initially

(e.g., Al and Cu).

Phases:The physically and chemically distinct material regions

that result.

Aluminum-Copper Alloy

1 component (H2O)

phase)

ar e

phase)



12/39

Contents





12

--

http://bp.snu.ac.kr


13/39

Fundamental Concepts

Phase diagram: graphical representation of the phases

resent and the ran es in com osition tem erature

and pressure over which the phases are stable. Gi s p ase ru e: F= + 2 P

C: # components, P: # phases in equilibrium

(Eq. 9-16)

F: degree of freedom (temperature, pressure, composition.)

ex 2 , = , = + = -

1 phase F= 2

2 phase F= 1

3 phase F= 0 (invariant)

13 * pressure constant F= C+ 1 Phttp://bp.snu.ac.kr


14/39


15/39

Isomorphous Phase Diagram

Complete liquid and solid solutions

Constant ressure:(Fig. 9-3)

15 F= C+ 1 P, C= 2, F= 3 - Phttp://bp.snu.ac.kr

- 2009-10-28


16/39

Phase Diagrams

Tell us about the phases as a function of T, Co, and P For this course:

- Binary systems: just 2 components

- Inde endent variables: Tand Co (at P= 1 atm)

1600

1400

1500T(C)

L (liquid)

Cu-Ni

(FCC solid solution)1300

1100

1200

(FCC solid

16 20 40 60 80 10001000

so u on

wt% Ni http://bp.snu.ac.kr


17/39

Phase Diagrams: Number and Types of Phases

Rule 1: If we know Tand Co, then we know:- the number and t es of hases resent.

Examples:

1600

L li uid

Cu-Ni

,

1 phase:

1400

50,35)

,2 phases: L + 1300

(FCC solidB(1

1100 A(1100,60)

17 wt% Ni20 40 60 80 10001000http://bp.snu.ac.kr


18/39

Phase Diagrams: Composition of Phases

Rule 2: If we know Tand Co, then we know:- the composition of each phase.

Examples: Cu-Ni systemT(C) (Fig. 9-3)

At TA:

Only Liquid (L)

o

130 0 L (liquid)

TA tie line

CL= Co( = 35wt% Ni)At TD:TB

B

At TB:

C= Co( = 35wt% Ni)

120 0

(solid)D Both and L

CL= Cliquidus( = 32wt% Ni here)

TD

18

= solidus = w ere

wt% NiC oC L Chttp://bp.snu.ac.kr


19/39

Phase Diagrams: Fractions of Phases

Rule 3: If we know Tand Co, then we know:- the amount of each phase (given in wt. or at. %).

Cu-Ni system Examples:

Co= 35wt%Ni

T(C)

A At TA: Only Liquid (L)

WL= 100wt%, W= 0

1300 L (liquid)B

e ne

At TB: Both and L

WL= 0, W= 100wt%

1200

DB R S

20 30 40 50443532CC C

What would be WL and W?

44 35S wt% Ni= =44 32

wL R + S

19

= =

25 %

44 32

wtW =R + S

http://bp.snu.ac.kr


20/39

The Lever Rule (Proof)

Sum of weight fractions:

WL + W = 1

=

(Example 9-1)

Combine above equations:

= R

R +SW =

Co CLC C

= S

R+ SWL

= CCo

C C

: CoCL C

W

L

R = W

S

WWL 1 W

20 solving gives Lever Rulehttp://bp.snu.ac.kr


21/39

Contents





21

--

http://bp.snu.ac.kr


22/39

Cooling in a Cu-Ni Binary (equilibrium)

Phase diagram: Cu-Ni system. System is:

(Fig. 9-4 incorrect)

- binary

2 components: Cu and Ni

- isomorphous

com lete solubilit

.

A phase field extendsFrom 0 to 100 wt. % Ni

.

ConsiderCo = 35 wt. % Ni

What would be the

22 microstructures?http://bp.snu.ac.kr


23/39

Coolin in Cu-Ni(nonequilibrium) .

(skip)

F g. -5

ConsiderCo = 35 wt. % Ni. X



24/39

Contents





24

--

http://bp.snu.ac.kr


25/39

Binary Eutectic Systems

A special compositionwith an eas meltin T

Greek - Easily melting (Fig. 9-7)

2 components

utectic reactionL

____



26/39

Binary Eutectic Systems

2 components as a spec a compos onwith a minimum melting T.

Ex.: Cu-Ag system 1200

T(C) (Fig. 9-7)

(L,, )

Limited solubility: L + L+

1000

: mostly Cu: mostly Ag

600E

8.0 71.9 91.2

CE: Min. melting T +

400

20 40 60 80 1000 CE

Cu Composition (wt. %) Ag


- 2009-11-02


27/39

Example: Pb-Sn Eutectic System

For a 40 wt. % Sn 60 wt. % Pb alloy at 150C, find...

- e a e re en

- the compositions of the phases:

(Fig. 9-8)

(Fig. 9-9) = wt. n

C = 99 wt. % Sn

T C

300

- the relative amountsof each phase: L + L+200

qu

=59

=

18.3 61.9 97.8150

R S 88

=29

=

+

27

88

http://bp.snu.ac.krPb Composition (wt. %) Sn


28/39

Microstructures in Eutectic Systems - I

Co < 2 wt. % Sn

T(C) L: Co wt. % Sn4 00

(Fig. 9-11)

esu

polycrystal of grains 3 00 L

L +

2 0 0 : C o wt%SnT E

- n

System)

1 00 +

1 0 2 00 3 0

28

C o , wt% Sn2

C o

(room T solubility limit) http://bp.snu.ac.kr


29/39

Microstructures in Eutectic Systems - II

2 wt. % Sn < Co < 18.3 wt. % Sn

T C L: C o wt.% Sn(Fig. 9-12)

with fine crystals. L400

L

: C owt%SnL +

300

200

TE

1 00 + XX

Pb-Snsystem

C wt. % n1 0 200

C30

29 (solubility limit at TE)

(solubility limit at Troom

18.3.

)

http://bp.snu.ac.kr


30/39

Microstructures in Eutectic Systems - III

o = E Result Eutectic microstructure --- alternating.

-(Fig. 9-13) (Fig. 9-14)T(C)

300L: Cowt%Sn

L + 200 L +183C

100 +160m

: 18.3wt%Sn

0

: 97.8wt%Sn

30 Co, wt% SnC

E18.3 97.8

61.9http://bp.snu.ac.kr


31/39

Formation of Eutectic Lamellar Structure

(Fig. 9-15)


E


32/39

Microstructures in Eutectic Systems

18.3 wt. % Sn < Co < 61.9 wt. % Sn Result: crystals and an eutectic microstructure

Just above TE : L(Fig. 9-16)

C = 18.3 wt. % Sn

C L = 61.9 wt. % Sn300 L

L

WL = (1-Wa) =50 wt. %R + S

W

==50 wt. %L +

200 TEL + SR

-

system

Just below TE :C = 18.3 wt. % Sn

1 00

+ C = 97.8 wt. SnS

R + SW = =73 wt. %

20 400 600

80 100Co18.3 61.9

r mary

97.8

eutectic eutectic

32

W = 27 wt. %o, w . n

http://bp.snu.ac.kr

(Fig. 9-17)

I i R i


33/39

Invariant Reactions

___

____


O h E l


34/39

Other Examples

Fi . 9-20


O h E l


35/39

Other Examples

. -


Other Examples


36/39

Other Examples

(Fig. 9-22)


C t t


37/39

Contents




Microstructural Evolution duringMicrostructural Evolution during4

oo ngoo ng

37

--

http://bp.snu.ac.kr

F C h s di m


38/39

Fe-C phase diagram

iron (Fig. 9-24)

(FCC)

A

iron Cementite Fe CB

err te(BCC)

soft & ductile

hard & brittle

A; eutectic

38

B; eutectoidconcen ra on . w . w . w

iron steel cast ironhttp://bp.snu.ac.kr

S


39/39

Summary

Phase diagrams are useful tools to determine: The number and types of phases.

The at. % or wt. % of each phase.

.

For the given Tand composition of the system.

Binary alloys allow various ranges of microstructures.

Problems from Chap. 9 http://bp.snu.ac.kr

. - . - . - . -

Prob. 9-9 Prob. 9-11 Prob. 9-12 Prob. 9-17

39

ro . - ro . - ro . - ro . -

http://bp.snu.ac.kr

mse 09 phase diagrams(39) 091116

Documents