Engineering 45
Phase
Diagrams (1)
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Learning 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
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Learning Goals.2 – Phase Dia.
• Cont: Determine Structure
– HOW MANY phases Result
– The COMPOSITION of each phase
– Relative QUANTITY of each phase
Phase A
Phase B
Nickel atom
Copper atom
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Definitions – 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)
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
The Solid Solubility Limit
• Add Sugar (Solute)
to Water (Solvent)
Engineering-45: Materials of Engineering
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Solubility
Limit
80
L
(liquid)
60
L
40
(liquid solution
i.e., syrup)
20
+
S
(solid
sugar)
20
40
60
80
100
Co =Composition (wt% sugar)
Pure
Sugar
0
Pure
Water
 Example:
Water-Sugar
100
Temperature (°C)
 Solubility Limit 
Max Concentration
of Solute that will
actually DISSOLVE
in a Solvent to form
a SOLUTION
Sucrose/Water Phase Diagram
– Initially ALL the Sugar Dissolves
– But after a Certain Amount,
SOLID Sugar Starts to Collect
on the bottom of the Vessel
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
The Solid Solubility Limit cont.
– 20 °C
– 80 °C
 For 20 °C
• Cast Right from 20C
– Find Solid Sugar in
Vessel at C0 = 63 wt%
(liquid)
60
L
40
(liquid solution
i.e., syrup)
20
0
6
+
S
(solid
sugar)
75
20
40
60 63 80
100
Co =Composition (wt% sugar)
• For 80C, Again Cast Rt
– Find Solid Sugar in
Vessel at C0 = 75 wt%
• INcreased Temp INcreases Sol-Sol Limit
Engineering-45: Materials of Engineering
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Pure
Sugar
• At What wt% Sugar
does the Sugar NO
Longer Dissolve for
Solubility
Limit
80
Pure
Water
 Sol-Sol Quantitative
Example
Temperature (°C)
100
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
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., a and b below)
• Aluminum
Copper
Alloy
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Effect of T & Composition (C0)
 Changing T can change No. of phases: path A to B.
 Changing C0 can change No. of phases: path B to D
B(100,70) D(100,90)
1 phase
Temperature (°C)
100
L
80
(liquid)
60
L
liquid
solution
(
i.e., syrup)
40
+
S
(solid
sugar)
• Water
Sugar
System
A(70, 20)
20
0
2 phases
2 phases
20
40
60 70 80
100
Co= Composition (wt% sugar)Bruce Mayer, PE
Engineering-45: Materials of Engineering
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0
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Equilibria
 Consider the Cu-Ni Alloy System
Ni
Cu
Crystal
Structure electroneg r (nm)
FCC
1.9
0.1246
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
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Diagrams
 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:
1600
T(°C)
1500
L (liquid)
• 3 phase fields:
1400
L
L+ a
a
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
Engineering-45: Materials of Engineering
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L(liquid)
a(FCC solid soln)
60
80
– The Cu-Ni Phase
Diagram
100
wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Dia.’s: Phase No.s & Types
 Rule-1: Given T & C0 (for P = 1 atm) then Find
• NUMBER & TYPES of Phases Present
• Pt-A (1100C,
60wt-%)
– 1 Phase → a;
the FCC Solid
Solution
• Pt-B (1250,35)
1600
L (liquid)
1500
B(1250,35)
 Examples
T(°C)
1400
1300
1200
– 2 Phases → L+a 1100
1000
Engineering-45: Materials of Engineering
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a
– Cu-Ni
Phase
(FCC solid
Diagram
solution)
A(1100,60)
0
20
40
60
80
100
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
wt% Ni
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
T(°C)
TA
1300
TB
1200
TD
20
Engineering-45: Materials of Engineering
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A
tie line
L (liquid)
B
D
– Cu-Ni
Phase
a
(solid) Diagram
3032 35 40 43
CL C0
50
Ca wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Dia.’s: Phase Comp. cont.
 Example:
C0 = 35 wt% Ni
• At TD:
– Only Solid (a-FCC)
– Ca = C0 =
35 wt% Ni
• At TB:
– BOTH a and L
– Ca = Csolidus
 43 wt% Ni
– CL = Cliquidus
 32 wt% Ni
Engineering-45: Materials of Engineering
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T(°C)
TA
1300
TB
1200
TD
20
A
tie line
L (liquid)
B
D
a
(solid)
3032 35 40 43
CL C0
– Cu-Ni
Phase
Diagram
50
Ca wt% Ni
 Note the Use of the
IsoThermal “Tie Line”
at TB to Find CL & Ca
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Dia.’s: Phase Wt Fractions
 Rule-3: Given T & C0 (for P = 1 atm) then Find
• The AMOUNT of EACH Phase in Wt-Fraction
 Example:
C0 = 35 wt% Ni
• At TA:
– Only Liquid
– WL = 1.00 & Wa =
0.00 (wt Frac’s)
• At TD:
– Only Solid
– WL = 0.00 &
Wa = 1.00 (Frac’s)
Engineering-45: Materials of Engineering
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T(°C)
TA
1300
TB
1200
TD
20
A
tie line
L (liquid)
B
D
– Cu-Ni
Phase
a
(solid) Diagram
3032 35 40 43
CL C0
50
Ca wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Phase Dia.’s: Wt Fractions cont.
 Example:
C0 = 35 wt% Ni
• At TB:
– BOTH a and L
• Calc Wa,B & WL,B
Using the
INVERSE LEVER
RULE
WL 
TA
1300
A
tie line
L (liquid)
B
TB
S
1200
TD
20
43  35
S

 73wt %
+
R S 43  32
Engineering-45: Materials of Engineering
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T(°C)
R
D
a
(solid)
3032 35 40 43
CL C0
– Cu-Ni
Phase
Diagram
50
Ca wt% Ni
R
= 27wt%

Wa
R +S
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Lever Rule Proof
 Sum of weight fractions: WL + Wa  1
 Conservation of mass (Ni): C0  Wa Ca + WLCL
 Combine These Two Equations for WL & Wα
 A Geometric Interpretation
moment equilibrium:
WLR  WaS
Balance massXdist
at Tip-Pt
Engineering-45: Materials of Engineering
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1 Wa
solving gives Lever Rule
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Cooling Cu-Ni Binary Phase-Sys
 Phase Diagram forT(°C) L (liquid)
Cu-Ni System →
1300
 System
L: 35wt%Ni
35
Characteristics: a:46wt%Ni
32
• BINARY → 2
components: Cu & Ni
• ISOMORPHOUS →1200
Complete Solubility of
one Component in
Another
– At least One Solid 1100
Phase-Field Extends
20
from 0 to 100 wt% Ni
Engineering-45: Materials of Engineering
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L: 35wt%Ni
A
B
C
46
43
D36
24
L: 32wt%Ni
a: 43wt%Ni
E
L: 24wt%Ni
a:36wt%Ni
a
(solid)
30
35
C0
40
50
wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ex: Cu-Ni Binary Cooling
 Consider 35 wt%
Ni Cooled: 1300
°C → Rm-Temp
 Pt-A
• 1.00 Liquid
• 35 wt% Ni
T(°C) L (liquid)
1300
L: 35wt%Ni
a:46wt%Ni
18
35
1200
1100
20
B
C
46
43
D36
24
• Tiny Amount of solida in Liq. Suspension
Engineering-45: Materials of Engineering
A
32
 Pt-B on Liquidus
– Liq → 35 wt% Ni
– a → 46 wt% Ni
L: 35wt%Ni
L: 32wt%Ni
a: 43wt%Ni
E
L: 24wt%Ni
a:36wt%Ni
a
(solid)
30
35
C0
40
50
wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ex: Cu-Ni Binary Cooling cont.
 Pt-C in 2-Ph Region
• (43-35)/(43-32) =
0.727 Liquid
– Liq → 32 wt% Ni
– a → 43 wt% Ni
T(°C) L (liquid)
1300
L: 35wt%Ni
a:46wt%Ni
• Small Liq Pockets in
Solid Suspension
• 1.00 a, @ C0
Engineering-45: Materials of Engineering
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35
B
C
46
43
D36
24
1200
L: 32wt%Ni
a: 43wt%Ni
E
L: 24wt%Ni
a:36wt%Ni
a
(solid)
– Liq → 24 wt% Ni
– a → 36 wt% Ni
 Pt E
A
32
 Pt-D on Solidus
L: 35wt%Ni
1100
20
30
35
C0
40
50
wt% Ni
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
NonEquilibrium 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.
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
NonEquil Cool → Cored Structure
 Ca Changes Composition Upon Cooling
• First a to solidify has Ca = 46 wt%Ni
• Last a to solidify has Ca = 35 wt%Ni
 Fast Cool Rate →
Cored structure
 Slow Cool Rate →
Equil. Structure
First a to solidfy:
46wt%Ni
Last a to solidfy:
< 35wt%Ni
Engineering-45: Materials of Engineering
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Uniform Ca
35wt%Ni Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Mech Props → Cu-Ni System
 Recall Solid-Solution Strengthening
400
TS for
pure Ni
300
TS for pure Cu
200
0 20 40 60 80 100
Cu
Ni
Composition, wt%Ni
• Max As Fcn of C0
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• Ductility (%EL,%AR)
Elongation (%EL)
Tensile Strength (MPa)
• Tensile Strength, TS
60
%ELfor pure Cu
%EL for
pure Ni
50
40
30
20
0 20
Cu
40 60 80 100
Ni
Composition, wt%Ni
• Min as Fcn of C0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
WhiteBoard 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
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ice is Nice – Problem 9.5
 Given Ice-I at −15C & 10atm → Find
MELTING and SUBLIMATION PRESSURES
• Note Typo in
Book
• Temperature
needs to be
–15 °C for
this to work
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ice is Nice P9.5a – Melt Temp
 At −15C Cast UPward to the Solid-LIQUID
Phase Boundary
1000
Engineering-45: Materials of Engineering
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• Find that Ice-I,
when held at
−15C, MELTS
at about 1000
atm (~15000
psi, ~100 Mpa)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ice is Nice P9.5b – Sublime Temp
 At −15C Cast DOWNward to the Solid-VAPOR
Phase Boundary
• Find that
Ice-I, when
held at −15C,
VAPORIZES
at about
0.003 atm
(~0.0002 psi,
~20 Pa)
0.003
Engineering-45: Materials of Engineering
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt
Ice is Nice P9.6  P = 0.1 Atm
 At 0.1 Atm Cast RIGHTward to intercept the
Sol-Liq and Liq-Vap Phase-Boundaries
• Ice-I MELTS
at 2 °C
• Water BOILS
at 75 °C
2.0
Engineering-45: Materials of Engineering
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75
– i.e., the
VAPOR
PRESSURE
of Water at
75 °C is
10% of Atm
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-45_Lec-21_PhasrDia-1.ppt