>Upon cooling from liquid state, the temperature of the pure metai (A or B) drops continuously till melting point at which solidification starts. Solidification happens at a constant temperature (line PQ) as F =0 (F = 1 - 2 +1 = 0). The temperature drops again on completion of solidification.>For any alloy (1, 2, 3 etc.) temp, drops till the liquidus (L15 L2, L3). However, in this case, solidification proceeds over a range of temperature as F = 1 ( 2 - 2 + 1 = 1). Once solidification completes at the solidus (S15 S2, S3) the temp, drops again.

Phase Equil ibria

S im p le so lu tion sys tem (e.g., N i-Cu so lu tion)

1 Crysta lS tructu re

T " "

Ni FCC 1.9 0 .1246

Cu FCC 1.8 0 .1278

Both have the sa m e crystal s truc ture (F-CC) and have s im ila r e lec t ronega t iv i t ies and a tom ic radii (W. H um e R o th e iy rules) sugges t ing high m utua l solubil ity.

Ni ana Cu a io to ta lly m 'sc ib ie in all p iopo r i ions .

AV.'

Phase Diagrams• ind ica te phases as funct ion of T: C 0, and P.• For th is course:

-b inary systems: just 2 components .- independen t variables: T and Co (P= 1 aim is almost always used;.

7 (C )

for Cu-N i sys tem

Vf/i

Phase Diagrams: # and types of phases

R u le 1: If we know 7 and C 0. then we know: - th e # and types of phases present

Exam ples : ^ ^

■ .i-i: z J frc; : " r'r.i:?.:'

' i \ .C-■ ; / ■ '.C /s3 V .!’i:v m y. o i ! W ; j t o ■' 3■

: \ y

Phase Diagrams: composition of phases

R u le 2: If we know T and C0l then we know: - th e composition of each phase.

E xam ples :

At Ta = 1320 C:

Only Liquid (/_) C[ = C 0

At 7 [) ~ 11 SO C:

Only Solid { u)

Cu - Cq

At T q = 1250 C:

Both u and L C/_ = C l iq u e sCu = C so |idus = 4 j wl% Ni here

Cu-N i

3 .J :

wt'Vo Nl' ' X I r.?,

iT k ; 9 2 ; : ; s a cc ;p :cd r - ^ r r ilA c 1' /V '/u y i. r ; \ : i s r ( i l

M a 'v ' cjis Par*. C " '.

Phase Diagrams:w e igh t frac t ions o f phases

Rule 3: If we know T and C0 then we knov\ --the amount of each phase ig.ven m vvtvey

E xam ples : r , C)

At TA- Only Liquid (L)UV/l = 100 wt%. Wu = 0

At Tq \ Only Solid ( u)

\Ni - 0. W(, = 100 wi% At Tg- Both u and L

S 43-35+ S 43-32

... = 27 wt% + 6

73 lv/%

Cu-Nisys tem

wt% Ni

The Lever Rule

Tie line - connec ts the phases in equ il ib r ium with each o the r - essentia l ly an iso therm

How m uch of each p h a s e 9T h ink o f it as a lever ( teeter-to t ter)

M, M

M ■ S M R

wt% Ni

M, S C -C-Vv ■---------- - - - - -------- -

M. • M R - S C - Cw - A . , c ,--C:

R - S C - C

Ex: Cooling in a Cu-I\!i BinaryP hase d iagram : T{ C)Cu-Ni system.

System is:

i.o . 2 c o m p o n e n ts Cu anu Ni.

r e . com p le te so lub i l i ty o f one c o m p o n e n t in ano ther ; u. phase fie ld ex te nd s from 0 to 100 w t% Ni.

C ons ide r

Cu-Nisys tem

w t% NiA V '

Cored vs Equil ibrium Phases• Ca c h an ge s as w e solidify.■ Cu-N i case: First u. to solidify has Cu - 46 wt% Ni.

Last to solidify has Ca = 35 wt% Ni.• Fast rate of cooling: • S low rate o f cooling

Cored structure Equil ibr ium structure

Mechanical Properties: Cu-Ni System• Effect of solid so lu t ion s treng then ing on:

- T e n s i le strength ITS) -D u c t i l i ty (% E L % A R )

3CG -

60 i cr pi:rc vjLiI

o \

40 -

30 -

20Cr ' " 20- 1 ■ 1 ■J 10 l-J Aj o 'j c j 1 -J ^ 3 20 40 6C 60 *00Cu Ni Cu Ni

C om pos it ion , w t% Ni C om pos it ion , w t% Ni

-P eak as a function of C0 -M in . as a function of Co

Binary-Eutectic Systems

Ex.: C u-Ag sys tem• 3 s ingle phase regions

(L. cUS)• Limited solubility:

(/: m o s t l y C u

M: m o s t l y A g

• Tc: . N o l i qu i d b e l o w

• Cr- Min. melting 're­composit ion

• Eutectic transitionL (C e) ' - ‘ w (C ;,■-) + |5(C;;e)

W

EX: Pb-Sri Eutectic System (1)For a 40 w t% S n-60 w t% Pb alloy at 150 C, find, - th e phases present: +

- th e relative am ourt o f each phase:

Pb-S i

w t% Sn

.....

EX: Pb-Sn Eutectic System (2)• For a 40 w t% Sn-60 w t% Pb alloy at 200 C. find...

- th e phases presen;: u + Pb-Sn-com posit ions of pnases: T( C) system

- th e relative amount of each phase:

C - C A \ '}I - ' / r : T-

0 i “ 2 C 6C 3CC, w t% Sn

Iviicrostruciu ros in Eutectic Systems: I

C 0 < 2 w t% Sn Result:

--at ex trem e ends -p o lyc rys ta l ot u grains

i.e.. only one solid phase(Pb-SnSystem)

\ f -

itfscrostruc tu res in Euiectic Systems: II

2 w t% Sn < C0 < 18.3 w t% Sn JX C) Result:

• Initially liquid + ■

■ then i ; alone

* finally two phases

^ u polycrystal 200^ f in e [ ' - -phase in c lu s io n s

P b-Snsys tem

2 0 30C w t% Sn

Microstructuresin Eutectic Systems: II!

’ Cq = C t• R esu lt : Eutectic m cros truc tu re ( lamellar structure)

-a l te rn a t in g layers ( lameilae) of u. and |i crysta ls

T( C)

Microstructures in Eutectic Systems: IV

18 .3 w t% Sn < C 0 < 6 1 9 w t% SnR esu lt : u. crystals arm a eutect ic rr.;crostructure

• Just above T-,7 ( C )

.. C’, = 18.3 w t% Sn *« % - 01.9 w t% Sn

- ~~ = 50 VVtJ/y R + ;■

= (1- W a) = 50 w t%

• Just below Te .Ci/ = 18.3 w t% Sn

■..Y, = 97 8 w t% Sn

L' I ;■ — ' — / 3 VV t \

i'V = 27 Wt%C .. vvt% Sn

Hypoeutectic & Hypereutectic

r( C)

(Pb-S nSys tem )

C.,. wt% Sri

intermstaii ic Compounds

Note, in term eta ll ic co m p ou nd fo rm s a line - not an area - because s to ich iom e try (i.e. com pos it ion ) is exact. . 'Vjv

Eutectoid & Peritectic

- l iquid in equ il ib r ium with two so lids

L :■??.I a + f lrest

- solid phase in equa t ion w ith two solidphases

^ ^ _ . n t e m i u t a i l i c c o m p o u n d<“>2 " ' " ' o - + 0 3 - c e m e n t i t c

V u + r e C (727°C)Meat

- liquid + solid 1 -> solid 2 (Fig 9.21)

S. L < - S

6 + L S 7 (1493°C)

Eutectoid & Peritectic

Cu-Zn Phase diagram /

Iron-Carbon (Fe-C) Phase Diagram2 im por tan t Ti c )

points

+ Fe„C

y —r> u + F e3C

- A :. ; ; '■'■V' • ■’ • . . s .V

eutectoid Steeln o

Hypereutectoid Steeln o

Example: Phase Equil ibria

For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following

a) composition of Fe3C and ferrite (a)b) the amount of carbide (cementite) in grams

that forms per 100 g of steelc) the amount of pearlite and proeutectoid

ferrite (</)

Chapter 9 - Phase hquil ibriaSolution: a) com pos it ion o f F e3C and ferr ite (a)b) the a m o un t of carb ide

(cem entite ) in g ram s that fo rm s per 100 g of steel

1

Fe3C Fe~C ■

CM - 0.022 S .7 - -0.022

x 100 ■ 5.7g

Fe3C = 5.7 g

u - 94.3 g

Chapter 9 - Phase Equilibriac. the am o un t o f pearl i te and proeu tec to id ferr ite (u)

note: am oun t o f pearl i te = am oun t o f y just above 7’

■ ^pearlite - 51.2 g : ,, , re,c !proeutecto id u ~ 48.8 g .. J

C C . v V * c.......

Alloying Steel with More Elements

* Teutectc id changes: * Ce^tectoic i changes.

,! Me n.6

1000- jii

10.4 j

800 / ■■

.. M 'l 0.21 ' .. ... ■ •■f-i r/in 1. Me

vvt % o f a l loy ing e lem en ts

"o 4 o o i o ■ ? m vvt. % of a l loy ing e lem en ts

10

20 40 60 80 100Wt.% Zn

Cu-Zn phase diagram, a and rt are terminal phases and p, v, 5 and are intermediate phases.

Phase diagrams with compounds^Sometimes a crystalline compound called intermetallic compound may form between two metals.>Such compounds generally have a distinct chemical formula or stoichiometry./-Example - Mg2Pb in the Mg-Pb system (appear as a vertical line at 81% Pb ), Mg2Ni, Mg2Si, Fe3C.

Mg - Pb phase diagram

700

r'ji’ if)

LL - Iv'g - Pb

0 0 0u - L P : L

*

400 aL

300 --r

/ T

200 --/ * f /crPb

, /P

1 U’ >

.... 1 ________1 . _ . I. _

/ t / ' o P d

Peritectic Phase diagram>L + a —>> p. An alloy cooling slowly through the peritectic point, P, the a phase will crystallize first just below the liquidus line. At the peritectic temperature, Tp all of the liquid and a will convert to [ I>Any composition left of P will generate excess a and similarly compositions right of P will give rise to an excess of liquid.>Peritectic systems - Pt - Ag, Ni - Re, Fe - Ge, Sn-Sb (babbit).