[email protected] engr-45_lec-21_phasrdia-1.ppt 1 bruce mayer, pe engineering-45: materials...

[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt1

Bruce Mayer, PE Engineering-45: Materials of Engineering

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 45

PhasePhaseDiagrams Diagrams

(1)(1)



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)

– a pressure (P)almost ALWAYS assume ROOM Pressure

• 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



Definitions – Phase SystemsDefinitions – Phase Systems

Component Pure Constituent of a Compound• Typcially an ATOM, but can also be a

Molecular Unit

Solvent/Solute• Solvent Majority Component in a Mixture

• Solute Minority Component in a Mixture

System Possible Alloys Formed by Specific Components (e.g. C-Fe Sys)



The Solid Solubility LimitThe Solid Solubility Limit Solubility Limit

Max Concentration of Solute that will actually DISSOLVE in a Solvent to form a SOLUTION

Example: Water-Sugar• Add Sugar (Solute)

to Water (Solvent)

– Initially ALL the Sugar Dissolves

– But after a Certain Amount, SOLID Sugar Starts to Collect on the bottom of the Vessel

Sucrose/Water Phase Diagram

Pu

re

Su

gar

Tem

per

atu

re (

°C)0 20 40 60 80 100Co =Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid)

+ S

(solid sugar)20

40

60

80

100

Pu

re

Wat

er



The Solid Solubility Limit cont.The Solid Solubility Limit cont. Sol-Sol Quantitative

Example• At What wt% Sugar

does the Sugar NO Longer Dissolve for– 20 °C

– 80 °C

For 20 °C• Cast Right from 20C

– Find Solid Sugar in Vessel at C0 = 63 wt%

• For 80C, Again Cast Rt– Find Solid Sugar in

Vessel at C0 = 75 wt%

• INcreased Temp INcreases Sol-Sol Limit

Su

gar

Pu

re

Tem

per

atu

re (

°C)

0 20 40 60 80 100Co =Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid)

+ S

(solid sugar)20

40

60

80

100

Pu

re

Wat

er

63

75



Components & Phases Components & Phases Components The elements or compounds

which are mixed initially (e.g., Al and Cu) Phases The PHYSICALLY and

CHEMICALLY DISTINCT material regions that result from mixing (e.g., and below)

• AluminumCopperAlloy

(darker phase)

(lighter phase)



Effect of T & Composition (CEffect of T & Composition (C00) )

• WaterSugarSystem

70 80 1006040200

Tem

pera

ture

(°C

)

Co = Composition (wt% sugar)

L (liquid solution i.e., syrup)

A(70,20) 2 phases

B(100,70) 1 phase

20

100

D(100,90) 2 phases

40

60

80

0

L (liquid)

+ S

(solid sugar)

Changing T can change No. of phases: path A to B.

Changing C0 can change No. of phases: path B to D



Phase EquilibriaPhase Equilibria Consider the Cu-Ni Alloy System

CrystalStructure electroneg r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (c.f. Hume – Rothery rules) suggesting high mutual solubility.

Copper and Nickel are, in fact, totally miscible in all Proportions



Phase Diagrams Phase Diagrams

– The Cu-Ni Phase Diagram

Describes Phase Formation as a Function of T, C0, P

This Course Considers• binary systems: 2 components

• independent variables: T & C0 (P = 1atm in all Cases)

• 2 phases: L (liquid) (FCC solid soln)

• 3 phase fields: L

L+

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600

T(°

C)

L (liquid)

(FCC solid solution)

L +

liquidus

solid

us



Phase Dia.’s: Phase No.s & Types Phase Dia.’s: Phase No.s & Types Rule-1: Given T & C0 (for P = 1 atm) then Find

• NUMBER & TYPES of Phases Present

Examples• Pt-A (1100C,

60wt-%)– 1 Phase → ;

the FCC Solid Solution

• Pt-B (1250,35)

– 2 Phases → L+ wt% Ni20 40 60 80 1000

1000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solid

us

A(1100,60)B(

1250,3

5)

– Cu-Ni PhaseDiagram



Phase Dia.’s: Phase Composition Phase Dia.’s: Phase Composition Rule-2: Given T & C0 (for P = 1 atm) then Find

• The COMPOSITION (wt% or at%) for EACH Phase

Example: C0 = 35 wt% Ni

• At TA:

– Only Liquid

– CL = CO = 35 wt% Ni


wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

TA A

DTD

TBB

tie line

L +

433532C0CL C



Phase Dia.’s: Phase Comp. cont. Phase Dia.’s: Phase Comp. cont.


• At TD:

– Only Solid (-FCC)

– C = C0 = 35 wt% Ni

• At TB:

– BOTH and L

– C = Csolidus

43 wt% Ni

– CL = Cliquidus

32 wt% Ni


wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

TA A

DTD

TBB

tie line

L +

433532C0CL C

Note the Use of the IsoThermal “Tie Line” at TB to Find CL & C



Phase Dia.’s: Phase Wt FractionsPhase Dia.’s: Phase Wt Fractions


• At TA:

– Only Liquid

– WL = 1.00 & W = 0.00 (wt Frac’s)

• At TD:

– Only Solid

– WL = 0.00 & W = 1.00 (Frac’s)


wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

TA A

DTD

TBB

tie line

L +

433532C0CL C

Rule-3: Given T & C0 (for P = 1 atm) then Find

• The AMOUNT of EACH Phase in Wt-Fraction



Phase Dia.’s: Wt Fractions cont. Phase Dia.’s: Wt Fractions cont.


• At TB:

– BOTH and L

• Calc W,B & WL,B Using the INVERSE LEVER RULE


wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

TA A

DTD

TBB

tie line

L +

433532C0CL C

43 3543 32

73wt%WL S

R S= 27wt%W

RR S

S

R



Lever Rule ProofLever Rule Proof Sum of weight fractions: 1 WWL

Conservation of mass (Ni): LLCWCWC 0

Combine These Two Equations for WL & Wα

RR S

W Co CLC CL

SR S

WLC Co

C CL

A Geometric Interpretation Co

R S

WWL

CL C

Balance massXdist at Tip-Pt

moment equilibrium:

1 Wsolving gives Lever Rule

WLR WS



Cooling Cu-Ni Binary Phase-SysCooling Cu-Ni Binary Phase-Sys Phase Diagram for

Cu-Ni System →

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

D

B

35C0

L: 35wt%Ni

: 46wt%Ni

C

E

L: 35wt%Ni

464332

24

35

36: 43wt%Ni

L: 32wt%Ni

L: 24wt%Ni

: 36wt%Ni

System Characteristics:• BINARY → 2

components: Cu & Ni

• ISOMORPHOUS → Complete Solubility of one Component in Another

– At least One Solid Phase-Field Extends from 0 to 100 wt% Ni



Ex: Cu-Ni Binary CoolingEx: Cu-Ni Binary Cooling Consider 35 wt%

Ni Cooled: 1300 °C → Rm-Temp

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

D

B

35C0

L: 35wt%Ni

: 46wt%Ni

C

E

L: 35wt%Ni

464332

24

35

36: 43wt%Ni

L: 32wt%Ni

L: 24wt%Ni

: 36wt%Ni

Pt-A• 1.00 Liquid

• 35 wt% Ni

Pt-B on Liquidus• Tiny Amount of solid-

in Liq. Suspension– Liq → 35 wt% Ni

– → 46 wt% Ni



Ex: Cu-Ni Binary Cooling cont.Ex: Cu-Ni Binary Cooling cont.

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

D

B

35C0

L: 35wt%Ni

: 46wt%Ni

C

E

L: 35wt%Ni

464332

24

35

36: 43wt%Ni

L: 32wt%Ni

L: 24wt%Ni

: 36wt%Ni

Pt-C in 2-Ph Region• (43-35)/(43-32) =

0.727 Liquid– Liq → 32 wt% Ni

– → 43 wt% Ni

Pt-D on Solidus• Small Liq Pockets in

Solid Suspension– Liq → 24 wt% Ni

– → 36 wt% Ni

Pt E• 1.00 , @ C0



NonEquilibrium CoolingNonEquilibrium Cooling Phases Diagrams are Constructed Under the

Assumption of ThermoDynamic Equilibrium• i.e., All Phases have Formed Sufficiently Slowly to

allow for HOMOGENOUS (same) Concentrations WITHIN ALL Phases

In the Previous Example The Solid STARTS at 46 wt%-Ni (pt-B) and ENDS at 35 wt%-Ni (Pt-E)• Thus Solid particles that WERE 46Ni Had to

CHANGE to 35Ni by SOLID STATE DIFFUSION

But Solid-State Diffusion Proceeds Slowly• Rapid Cooling Can result in NonUniform Comp.



NonEquil Cool → Cored StructureNonEquil Cool → Cored Structure C Changes Composition Upon Cooling

• First to solidify has C = 46 wt%Ni

• Last to solidify has C = 35 wt%Ni Fast Cool Rate →

Cored structure Slow Cool Rate →

Equil. Structure

Uniform C

35wt%Ni

to solidfy:

First to solidfy: 46wt%Ni

Last < 35wt%Ni



Mech Props → Cu-Ni SystemMech Props → Cu-Ni System Recall Solid-Solution Strengthening

• Tensile Strength, TS • Ductility (%EL,%AR)

Ten

sile

Str

en

gth

(M

Pa)

Composition, wt%NiCu Ni0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu

Elo

ng

ati

on

(%

EL)

Composition, wt%NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

• Max As Fcn of C0 • Min as Fcn of C0



WhiteBoard PPT WorkWhiteBoard PPT Work

Problems 9.[5,6]• The Affect of PRESSURE

on Phase Diagrams

• Water Ice, Has at Least TEN, yes 10, Distinct Structural Phases– Phases form in Response to

the PRESSURE Above The Ice



Ice is Nice – Problem 9.5Ice is Nice – Problem 9.5

Starting Point

• Note Typo in Book

• Temperature needs to be –15 °C for this to work

Given Ice-I at −15C & 10atm → Find MELTING and SUBLIMATION PRESSURES



Ice is Nice P9.5a – Melt TempIce is Nice P9.5a – Melt Temp

At −15C Cast UPward to the Solid-LIQUID Phase Boundary

• Find that Ice-I, when held at −15C, MELTS at about 1000 atm (~15000 psi, ~100 Mpa)

1000



Ice is Nice P9.5b – Sublime TempIce is Nice P9.5b – Sublime Temp

At −15C Cast DOWNward to the Solid-VAPOR Phase Boundary

0.003

• Find that Ice-I, when held at −15C, VAPORIZES at about 0.003 atm (~0.0002 psi, ~20 Pa)



Ice is Nice P9.6 Ice is Nice P9.6 P = 0.1 Atm P = 0.1 Atm At 0.1 Atm Cast RIGHTward to intercept the Sol-Liq and

Liq-Vap Phase-Boundaries

2.0

• Ice-I MELTS at 2 °C

• Water BOILS at 75 °C– i.e., the

VAPOR PRESSURE of Water at 75 °C is10% of Atm75

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