[email protected] engr-45_lec-21_phasrdia-1.ppt 1 bruce mayer, pe engineering-45: materials...
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[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt1
Bruce Mayer, PE Engineering-45: Materials of Engineering
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 45
PhasePhaseDiagrams Diagrams
(1)(1)
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt2
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt3
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt9
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt11
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt12
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
– Cu-Ni PhaseDiagram
wt% Ni20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
TA A
DTD
TBB
tie line
L +
433532C0CL C
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Bruce Mayer, PE Engineering-45: Materials of Engineering
Phase Dia.’s: Phase Comp. cont. Phase Dia.’s: Phase Comp. cont.
Example: C0 = 35 wt% Ni
• 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
– Cu-Ni PhaseDiagram
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
Phase Dia.’s: Phase Wt FractionsPhase Dia.’s: Phase Wt Fractions
Example: C0 = 35 wt% Ni
• 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)
– Cu-Ni PhaseDiagram
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
Phase Dia.’s: Wt Fractions cont. Phase Dia.’s: Wt Fractions cont.
Example: C0 = 35 wt% Ni
• At TB:
– BOTH and L
• Calc W,B & WL,B Using the INVERSE LEVER RULE
– Cu-Ni PhaseDiagram
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt19
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt20
Bruce Mayer, PE Engineering-45: Materials of Engineering
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.
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
[email protected] • ENGR-45_Lec-21_PhasrDia-1.ppt27
Bruce Mayer, PE Engineering-45: Materials of Engineering
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