chapter 9: phase diagrams
DESCRIPTION
Chapter 9: Phase Diagrams. ISSUES TO ADDRESS. • When we combine two elements... what is the resulting equilibrium state?. • In particular, if we specify... -- the composition (e.g., wt% Cu - wt% Ni), and -- the temperature ( T ). then.. . - PowerPoint PPT PresentationTRANSCRIPT
ISSUES TO ADDRESS...• When we combine two elements... what is the resulting equilibrium state?
• In particular, if we specify... -- the composition (e.g., wt% Cu - wt% Ni), and -- the temperature (T )
then... How many phases form? What is the composition of each phase? What is the amount of each phase? What is the structure of each phase
Chapter 9: Phase Diagrams
Phase BPhase A
Nickel atomCopper atom
Phase BehaviorLaughlin p. 67
• Components: The elements or compounds which are present in the alloy (e.g., Al and Cu)
• Phases: The physically and chemically distinct material regions that form (e.g., α and β).
Aluminum-Copper Alloy
Components and Phases
β
(darker phase)
α (lighter phase)
1µm
Optical Mettalography
Solubility Limit
Question: What is the solubility limit for sugar in water at 20ºC?
65
• Solubility Limit: Maximum concentration for which only a single phase solution exists.
Sugar/Water Phase Diagram
Tem
per
atu
re (
ºC)
0 20 40 60 80 100
Composition (wt% sugar)
L
(syrup)
Solubility Limit L
(liquid)
+ S
(solid sugar)20
40
60
80
100
• Solution – solid, liquid, or gas solutions, single phase• Mixture – more than one phase
Problem 9.2: At 170°C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in Pb?The lead-tin phase diagram is shown in the Animated Figure 9.8.
Isomorphous Binary Phase Diagram
• System is:-- binary 2 components: Cu and Ni.
-- isomorphous i.e., complete solubility of one component in another
Cu-Ni phase diagram
wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(º
C)
L (liquid)
α
solid solution
L + αliquidus
solid
us
Criteria for Solid Solubility: Hume–Rothery rules
CrystalStructure
electroneg r (nm)
Ni FCC 1.9 0.1246
Cu FCC 1.8 0.1278
• Same crystal structure• Similar electronegativities • Similar atomic radii
• Ni and Cu are totally soluble in one another for all proportions.
Structure
BCC
FCC
Periodic Table
Cu-Ni phase diagram
wt% Ni20 40 60 80 1000
1000
1100
1200
1300
1400
1500
1600
T(º
C)
L
α L + α
Determination of phases present
• If we know T and Co, then we know which phases are present.
• Examples:A(1100ºC, 60 wt% Ni): 1 phase: α
B(1250ºC, 35 wt% Ni): 2 phases: L + α B
(12
50ºC
,35)
A(1100ºC,60)
Determination of phase compositions: Lever Rule
What fraction of each phase? Think of the tie line as a lever
(teeter-totter)
ML Mα
R S
M x S ML x R
L
L
LL
LL CC
CC
SR
RW
CC
CC
SR
S
MM
MW
00
wt% Ni
20
1200
1300
T(ºC)
L
α
30 40 50
L + α B
TB
tie line
C0CL Cα
SR
Tie line –– also sometimes called an isotherm
wt% Ni20
1200
1300
30 40 50110 0
L
α
T(ºC)
35C0
L: 35wt%Ni
C0 = 35 wt% Ni alloy
Slow Cooling of a Cu-Ni Alloy
4635
4332
a: 43 wt% Ni
L: 32 wt% Ni
α: 46 wt% NiL: 35 wt% Ni
L: 24 wt% Ni
a: 36 wt% Ni
24 36
Last a to solidify:< 35 wt% Ni
Cored vs Equilibrium Structures
Uniform Ca: 35 wt% Ni
First α to solidify:46 wt% Ni
wt% Ni20
1200
1300
30 40 501100
L
α
T(ºC)
35C0
4635
4332
24 36
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Tensile strength (TS) -- Ductility (%EL)
Ten
sile
Str
eng
th (
MP
a)
Composition, wt% NiCu Ni0 20 40 60 80 100
200
300
400
TS for pure Ni
TS for pure Cu
Elo
ng
atio
n (
%E
L)
Composition, wt% NiCu Ni0 20 40 60 80 10020
30
40
50
60
%EL for pure Ni
%EL for pure Cu
2 components=low melting
Binary-Eutectic Systems
• 3 single phase regions (L, α, β)
Ex.: Cu-Ag system
L (liquid)
α L + α L+β β
α + β
wt% Ag20 40 60 80 1000
200
1200
400
600
800
1000
CE
TE 8.0 71.9 91.2779ºC
Ag) wt%1.29( Ag) wt%.08( Ag) wt%9.71( Lcooling
heating
Eutectic Phase Reaction:
Cu-Ag system
• For alloys for which C0 < 2 wt% Sn• Result: at room temperature -- polycrystalline with grains of a phase having
composition C0
Microstructural Development in Eutectic Systems I
0
L+ α200
T(ºC)
wt% Sn10
2
20C0
300
100
L
α
30
α +β
400
(room T solubility limit)
TE
Pb-Sn
αL
L: C0 wt% Sn
α: C0 wt% Sn
• For alloys for which 2 wt% Sn < C0 < 18.3 wt% Sn• Result: at temperatures in α + β range -- polycrystalline with a grains and small β-phase particles
Microstructural Development in Eutectic Systems II
L + α
200
T(ºC)
C, wt% Sn10
18.3
200C0
300
100
L
α
30
α + β
400
(sol. limit at TE)
TE
2(sol. limit at Troom)
Lα
L: C0 wt% Sn
αβ
a: C0 wt% Sn
• For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of α and β phases.
Microstructural Development in Eutectic Systems III
160 μm
Micrograph of Pb-Sn eutectic microstructure
200
T(ºC)
wt% Sn
20 60 80 1000
300
100
L
α β
L+ α
183ºC
40
TE
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn
CE61.9
L: C0 wt% Sn
Lamellar Eutectic Structure
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn• Result: α phase particles and a eutectic microconstituent
Microstructural Development in Eutectic Systems IV
18.3 61.9
SR
97.8
SR
primary αeutectic α
eutectic β
WL = (1- Wα) = 0.50
Cα = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SWα = = 0.50
• Just above TE :
• Just below TE :C
α = 18.3 wt% Sn
Cβ = 97.8 wt% Sn
SR + S
Wα = = 0.73
Wβ = 0.27
200
T(ºC)
wt% Sn
20 60 80 1000
300
100
L
αL+
α
40
TE
L: C0 wt% Sn
Lα
L+αL+β
α + β
200
C, wt% Sn20 60 80 1000
300
100
L
α β
TE
40
System)
Hypoeutectic & Hypereutectic
160 mm
eutectic micro-constituent
hypereutectic: (illustration only)
β
ββ
ββ
β
175 mm
α
α
α
αα
α
hypoeutectic: C0 = 50 wt% Sn
T(ºC)
61.9eutectic
eutectic: C0 = 61.9 wt% Sn
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound exists as a line - not an area – because stoichiometry (i.e. composition of a compound) is fixed.
• Eutectoid –all solid phases
S2 S1+S3
ϒ α+ Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
intermetallic compound - cementite
cool
heat
Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L S1+S3 (For Pb-Sn, 183ºC, 61.9 wt% Sn) cool
heat
cool
heat
• Peritectic - liquid and one solid phase transform to a second solid phase S1 + L S2
δ + L ϒ (For Fe-C, 1493ºC, 0.16 wt% C) • Peritectoid – all solid phases
S1 + S2 S3
Eutectoid & Peritectic
Eutectoid transformation δ γ + ε
Peritectic transformation γ + L δ
Cu-Zn Phase diagram
Iron-Carbon (Fe-C) Phase Diagram
- Eutectoid (B):
γ → α +Fe3C
- Eutectic (A): L → γ + Fe3C
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ+Fe3C
α+Fe3C
α+γ
δ
(Fe) wt% C
1148ºC
T(ºC)
α 727ºC = Teutectoid
4.30
Result: Pearlite = alternating layers of α and Fe3C phases
120 mm0.76
B
γ γγγ
A L+Fe3C
Fe3C (cementite-hard)α (ferrite-soft)
Hypereutectoid Steel1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ +Fe3C
α +Fe3C
L+Fe3C
δ
(Fe) C, wt%C
1148ºC
T(ºC)
a
0.76
C0
Fe3C
γ
γγ
γγγ γ
proeutectoid Fe3C
60 μm
pearlite
pearlite
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
(Fe) C, wt% C
1148ºC
T(ºC)
α727ºC
C0
0.76
Hypoeutectoid Steel
proeutectoid ferritepearlite
100 μmγ γγγ
α
pearlite
γγ γ
γα
For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:
a) The compositions of Fe3C and ferrite (α).
b) The amount of cementite (in grams) that forms in 100 g of steel.
c) The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.
Solution to Example Problem
WFe3C R
R S
C0 C
CFe3C C
0.40 0.0226.70 0.022
0.057
b) Wight Fraction of cementite
a) The compositions of Fe3C and ferrite (α).RS tie line just below the eutectoidCα = 0.022 wt% C
CFe3C = 6.70 wt% C
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
C , wt% C
1148ºC
T(ºC)
727ºC
C0
R S
CFe C3Cα
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
pearlite
The amounts of pearlite in the 100 g.
c) Using the VX tie line just above the eutectoid and realizing thatC0 = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 0.76 wt% C
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ(austenite)
γ+L
γ + Fe3C
+ Fe3C
L+Fe3C
δ
C, wt% C
1148ºC
T(ºC)
727ºC
C0
V X
CγCα
Wpearlite V
V X
C0 C
C C
0.40 0.0220.76 0.022
0.512
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g
α
pearlite
Alloying with Other Elements
• Teutectoid changes:
TE
ute
cto
id (
ºC)
wt. % of alloying elements
Ti
Ni
Mo SiW
Cr
Mn
• Ceutectoid changes:
wt. % of alloying elements
Ce
ute
cto
id (
wt%
C)
Ni
Ti
Cr
SiMn
WMo
VMSE: Interactive Phase Diagrams
Change alloy composition
Microstructure, phase compositions, and phase fractions respond interactively
• Phase diagrams are useful tools to determine:-- the number and types of phases present,-- the composition of each phase,-- and the weight fraction of each phase
given the temperature and composition of the system.
• The microstructure of an alloy depends on -- its composition, and -- whether or not cooling rate allows for maintenance of equilibrium.
• Important phase diagram phase transformations include eutectic, eutectoid, and peritectic.
Summary
70 80 1006040200
Tem
pera
ture
(ºC
)
C = Composition (wt% sugar)
L (liquid solution
i.e., syrup)
20
100
40
60
80
0
L (liquid)
+ S
(solid sugar)
Effect of Temperature & Composition
• Altering T can change # of phases: path A to B.• Altering C can change # of phases: path B to D.
D (100ºC,C = 90)
2 phases B (100ºC,C = 70)
1 phase
A (20ºC,C = 70)
2 phases
L+αL+β
α + β
200
T(ºC)
18.3
wt% Sn20 60 80 1000
300
100
L (liquid)
α 183ºC 61.9 97.8
β
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC
Pb-Snsystem
EX 1: Pb-Sn Eutectic System
-- the relative amount of each phase
150
40C0
11Cα
99Cβ
SR
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
Wα
=Cβ - C0
Cβ - Cα
=99 - 4099 - 11
= 5988
= 0.67
SR+S
=
Wβ
=C0 - Cα
Cβ - Cα
=RR+S
=2988
= 0.33=40 - 1199 - 11
Ca = 17 wt% Sn
L+β
a + b
200
T(ºC)
wt% Sn20 60 80 1000
300
100
L
α β
L+ α
183ºC
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC
EX 2: Pb-Sn Eutectic System
the relative amount of each phase
Wa =CL - C0
CL - Cα
=46 - 40
46 - 17
= 6
29= 0.21
WL =C0 - Cα
CL - Cα
=23
29= 0.79
40C0
46CL
17Cα
220SR
CL = 46 wt% Sn