bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt1

Bruce Mayer, PE Engineering-45: Materials of Engineering

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 45

PhasePhaseDiagrams Diagrams

(2)(2)



Learning Goals – Phase DiagramsLearning Goals – Phase Diagrams

When Two Elements are Combined, Determine the Resulting MicroStructural Equilibrium State

For Example• Specify

– a composition (e.g., wt%Cu - wt%Ni), and

– a temperature (T)

• Determine Structure



Learning Goals.2 – Phase Dia.Learning Goals.2 – Phase Dia.

• Cont: Determine Structure– HOW MANY phases Result

– The COMPOSITION of each phase

– Relative QUANTITY of each phase

Nickel atom Copper atom

Phase A Phase B



Binary Eutectic SystemsBinary Eutectic Systems Binary → Two

Components • 3 Phases (A, B, Liq)

Eutectic → Easily Melted

Eutectic Point• Composition & Temp

at Which Pure-Liquid and Pure-Solid CoExisit– The Low-Melt Temp

Eutectic Point



Cu-Ag Binary Eutectic SysCu-Ag Binary Eutectic Sys 3 Phases: , , L LIMITED Solubility

• → Mostly Cu– 8.0 wt% Ag

• → Mostly Ag– 8.8 wt% Cu

TE → NO Liquid Below 779 °C

CE → Min. Melting-Temp Composition

• 71.9% Ag

L (liquid)

L + L+

C0,wt% Ag 20 40 60 80 100 0

200

1200 T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2 779°C



Eutectic TransitionEutectic Transition At the Eutectic

Composition there is NO “Mushy Phase”

Cu-Ag system

L (liquid)

L + L+

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

At CE the alloy when heated “flashes” to Liquid

Eutectic transition Liquid and α&β

L(CE) (CE) + (CE)



Ex: Pb-Sn Binary Eutectic SysEx: Pb-Sn Binary Eutectic Sys For 40wt%Sn-60wt

%Pb Alloy at 150 °C Find• Phases Present

• Phase Compositions

• Phase Fractions

At C0 = 40 wt% Sn @ 150C the Phases• •

L + L+

200

T(°C)

18.3

Co, wt% Sn 20 40 60 80 100 0

Co

300

100

L (liquid)

183°C

61.9 97.8 150

Pb-Sn Phase Diagram



Ex: Pb-Sn Eutectic Sys cont.Ex: Pb-Sn Eutectic Sys cont. Phase Composition

• Need to Cast Left for , and Right for

• C = 11 wt% Sn

• C = 99 wt% Sn

L+L+

+

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

183°C 61.9 97.8

Pb-Sn system

150

40Co

11C

99C

SR

For Phase Fractions Use LEVER Rule

W=C - CO

C - C

=99 - 4099 - 11

= 5988

= 67 wt%

SR+S

= W =CO - C

C - C=R

R+S

=2988

= 33 wt%=40 - 1199 - 11



Ex: Pb-Sn Eutectic Sys cont.Ex: Pb-Sn Eutectic Sys cont. For 40wt%Sn-60wt

%Pb at 200 °C

L+

+

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

L+

183°C

Pb-Sn system

40Co

46CL

17C

220SR

Phases Present → L + α

Phase Compositions• Co = 40 wt% Sn

• Cα = 17 wt% Sn

• CL = 46 wt% Sn



Ex: Pb-Sn Eutectic Sys cont.Ex: Pb-Sn Eutectic Sys cont. For 40wt%Sn-60wt

%Pb at 200 °C

L+

+

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

L+

183°C

Pb-Sn system

40Co

46CL

17C

220SR

The Relative Amounts of L & α by Lever-Law

W =CL - CO

CL - C=

46 - 40

46 - 17

= 6

29= 21 wt%

WL =CO - C

CL - C=

23

29= 79 wt%



1-Phase Cooling MicroStructure-11-Phase Cooling MicroStructure-1 Consider C0 1 Wt%

Sn Cooled from 350C• First Liquid at C0

• Then L+ – The -Particles will Have

“Cored” Structure with Very Slight Composition gradient

• Lastly @C0

– Grains Grow Out from Particles formed in the L+ Phase-Field

Pb-SnPhase

Diagram

L + 200

T(°C)

Co, wt% Sn10

2

200Co

300

100

L

30

L: Cowt%Sn

L

: Cowt%Sn

+

400

(room T solubility limit)

TE(Pb-Sn)



1-Phase Cooling MicroStructure-21-Phase Cooling MicroStructure-2 Consider 2wt%Sn < C0 <

18.3wt%Sn Cooled from 325C• First Liquid at C0

• Then L+– The -Particles with

“Cored” Structure with Significant Comp gradient

• Next Grains at Net-C0

• Lastly +(Precipitate)– Particles Could have

Cored Structure

Pb-SnPhase

Diagram

: Cowt%SnL +

200

T(°C)

Co, wt% Sn10

18.3

200Co

300

100

L

30

L: Cowt%Sn

+

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

L



1-Phase Cooling MicroStructure-31-Phase Cooling MicroStructure-3 C0 = CE,

Cooled From TE

• First Liquid at CE

• Then + in Solid State at TE–T

In the Solid, Phases Form Compositions at the ENDS of the Eutectic IsoTherm

Pb-Sn Phase Diagram

L + 200

T(°C)

Co, wt% Sn

20 400

300

100

L

60

L: Cowt%Sn

+

TE

: 18.3wt%Sn

080 100

L +

CE18.3 97.861.9

183°C

: 97.8wt%Sn

• = 18.3 wt% Sn

• = 97.8 wt% Sn



1-Phase Cooling MicroStructure-41-Phase Cooling MicroStructure-4 Eutectic Cooling

Forms Lamellar Structure

160 µm

Micrograph of Pb-SnEutectic

MicroStructure

• Composition Relaxation by Atomic Diffusion

18.3-SnLEAD Rich

Dumps Sn

97.8-SnTIN Rich

61.9 Sn

DumpsPb



1-Phase Cooling MicroStructure-51-Phase Cooling MicroStructure-5 18.3wt%Sn < C0 <

61.9wt%Sn • C,max < C0 < CE

Result → -Crystals + eutectic-microstructure

Calc Compositions and Phase-Fractions by Lever Rules• Ref Levers:

Pb-Sn Phase Diagram

L + 200

T(°C)

Co, wt% Sn

20 400

300

100

L

60

L: Cowt%Sn

+

TE

080 100

L +

Co18.3 61.9

L

L

primary

97.8

QR

PP

eutectic eutectic

P Q R



1-Phase Cooling MicroStructure-61-Phase Cooling MicroStructure-6Pb-Sn Phase Diagram

L + 200

T(°C)

Co, wt% Sn

20 400

300

100

L

60

L: Cowt%Sn

+

TE

080 100

L +

Co18.3 61.9

L

L

primary

97.8

QR

PP

eutectic eutectic

P Q R

Just BELOW TE

Just ABOVE TE in L+ Field

WL = (1 -W) = 50 wt%

C = 18.3wt%Sn

CL = 61.9wt%SnQ

R + QW = = 50 wt%

C = 18.3wt%Sn

C = 97.8wt%SnR

P + RW = =73wt%

W = 27wt%



[[HYPOHYPO//HYPERHYPER]-eutectic]-eutecticT(°C)

L + 200

Co, wt% Sn20 400

300

100

L

60

+

TE

080 100

L +

18.361.9

97.8

C0hypoeutectic

C0hypereutectic

eutectic

hypereutectic: (illustration only)

160 µm

eutectic: C0= 61.9wt%Sn

175 µm

hypoeutectic: C0= 50wt%Sn

eutectic micro-constituent

Pb-SnPhase

Diagram

HYPOeutectic →BELOW Eutecticcomposition• Yields Island-like

-regions with Lamellar Eutectic Structure

HYPEReutectic →ABOVE EutecticComposition• Yields

lsland-like-regions withLamellar Eutectic Structure



InterMetallic CompoundsInterMetallic Compounds

Mg2Pb

An Intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact by a chemical reaction between the pure constituents



Eutectoid & PeritecticEutectoid & Peritectic Eutectic liquid in equilibrium with two solids

• L α + β

Eutectoid solid phase in equilibrium with two OTHER solid phases• S1 S2 + S3

• Example of Iron-Carbon @ 727 °C: α + Fe3C

Peritectic liquid + solid1 solid 2• L + S1 S2

• Example of Iron-Carbon @ 1493 °C: L + δ



Eutectoid & PeritecticEutectoid & Peritectic

Cu-Zn Phase diagram

Eutectoid transition +

Peritectic transition + L



Iron-Carbon Phase DiagramIron-Carbon Phase Diagram

Fe3C

(cem

enti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe3C+

L+Fe3C

(Fe) Co, wt% C0.77 4.30

727°C = Teutectoid

1148°C

T(°C)

A

B

SR

R S

Fe3C(cementite-hard)(ferrite-soft)

Ceute

ctoid

Fe-C Phase Diagram Two Significant (C0,T)

points on the Fe-Fe3C Phase Diagram

1. Eutectic Point-A

– L + Fe3C

2. Eutectoid Point-B– + Fe3C

Result: PEARLITE = Alternating Layers of and Fe3C Phase

120 µm



HYPOHYPOeutectoid Steeleutectoid SteelFe-C Phase Diagram

Cool From Solid Austenite @1460C

1. @1000C → Grains of -Only

2. @~800C → Tiny Islands of (ferrite) Form Along -Grain Boundaries

3. @727+ °C → + Ferrite in proportions as given by Lever Rule

Co

Fe3C

(ce

menti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

727°C

1148°C

T(°C)

R S

r s

w =s/(r+s)w =(1-w )

w =S/(R+S)wFe3C =(1-w)

wpearlite = w

pearlite



HYPOHYPOeutectoid Steel conteutectoid Steel contFe-C Phase Diagram

4. @727- °C → ProEutectoid-FERRITE and Pearlite in proportions asgiven by the Lever Rule

Co

Fe3C

(ce

menti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

727°C

1148°C

T(°C)

R S

r s

w =s/(r+s)w =(1-w )


wpearlite = w

pearlite

100 µm

HypoEutectoidSteel



HYPERHYPEReutectoid Steeleutectoid SteelFe-C Phase Diagram

Cool From Solid Austenite @1000C

1. @1000C → Grains of -Only

2. @~860C → Tiny Islands of Cementite Form Along -Grain Boundaries

3. @727+ °C → + Cementite in proportions as given by Lever Rule

Co

Fe3C (

cem

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

1148°C

T(°C)

R S

s

wFe3C =r/(r+s)w =(1-wFe3C)


wpearlite = wpearlite

r

Fe3C



HYPERHYPEReutectoid Steel cont.eutectoid Steel cont.Fe-C Phase Diagram

4. @727– °C → ProEutectoid-CEMENTITE and Pearlite in proportions asgiven by the Lever Rule

Co

Fe3C (

cem

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

1148°C

T(°C)

R S

s

wFe3C =r/(r+s)w =(1-wFe3C)


wpearlite = wpearlite

r

Fe3C

60

µm

HyperEutectoidSteel



WhiteBoard PPT WorkWhiteBoard PPT Work

Given: 1.8 kg of 1.5 wt%-C Austenite is Cooled 1050C → 725C

• Find1. ProEutectoid Phase

2. TOTAL kg’s of Ferrite & Cementite

3. kg of MicroConstituents Pearlite & ProEu-Phase

Starti

ng Poin

t



HyperEutectoid SteelHyperEutectoid Steel PROeutectoid Phase

The FINAL, or Persistent, Phase that is Present ABOVE the Euectoid Temperature

1. In this Case ProEutectoid Phase is: Fe3C, a.k.a. CEMENTITE



Lever Rule for Lever Rule for + Cementite + Cementite

C = 1.5-.022

Ccem = 6.7-1.5



From Lever Rule From Lever Rule +Fe +Fe33C C

Total Lever Length Ctot

678.6022.07.6 totC

Since 1.5 wt%C is Closer to the Terminous, Then Expect more

221.01

779.0678.6

5.17.6

WW

W

cemen

Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg

2. Thus

kgkgM

kgkgM

cemen 398.08.1221.0

402.18.1779.0



Lever Rule for Pearlite + ProEuLever Rule for Pearlite + ProEu

Cp = 6.7-1.5

CFe3C = 1.5-0.76



From Lever Rule for Pearlite From Lever Rule for Pearlite Total Lever Length

Ctot

94.576.07.6 totC

Now All present at 727+ °C converts to PEARLITE• Since 1.5wt%C is

closer to -line than Fe3C, expect More PEARLITE than ProEutectoid Fe3C

Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg

kgkgM

kgkgM

cemenoEu

p

225.08.1125.0

575.18.1875.0

Pr

125.01

875.094.5

5.17.6

ProEuCemen

pearl

pearl

WW

W



Cementite MicroConstituentsCementite MicroConstituents

Quantities of the two forms of Cementite• ProEutectoid Cementite = 0.225 kg

• Total Cementite = 0.398 kg

Thus Lamellar (Pearlite) Cementite = kgkgkgMMM cemenTotalcemenpearl 173.0225.0398.0

cemenProEu

Then the Cementite-Form Fractions• ProEutectoid = 225/398 = 56.5%

• Pearlitic Cementite = 173/398 =43.5%



Pearlite Mass BalancePearlite Mass Balance

Note at the ALL the α-Iron is in the MicroForm of PEARLITE• Recall Mα = 1.402 kg

So Total Pearlite by Phase Addition = kgkgkgMMM cemenpearlp 575.1402.1173.0

This is the SAME value for Mp as that Calculated by an independent LEVER Rule Calculation kgM p 8.1

94.5

5.17.6



WhiteBoard WorkWhiteBoard Work

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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