[email protected] engr-45_lec-22_phasedia-2.ppt 1 bruce mayer, pe engineering-45: materials...
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[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt1
Bruce Mayer, PE Engineering-45: Materials of Engineering
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 45
PhasePhaseDiagrams Diagrams
(2)(2)
[email protected] • ENGR-45_Lec-22_PhaseDia-2.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)
• Determine Structure
[email protected] • ENGR-45_Lec-22_PhaseDia-2.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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt4
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt5
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt6
Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt7
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt9
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt10
Bruce Mayer, PE Engineering-45: Materials of Engineering
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%
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt11
Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt12
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt13
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt14
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt15
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt16
Bruce Mayer, PE Engineering-45: Materials of Engineering
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%
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt17
Bruce Mayer, PE Engineering-45: Materials of Engineering
[[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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt18
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt19
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt20
Bruce Mayer, PE Engineering-45: Materials of Engineering
Eutectoid & PeritecticEutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition +
Peritectic transition + L
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt21
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt22
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt23
Bruce Mayer, PE Engineering-45: Materials of Engineering
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 )
w =S/(R+S)wFe3C =(1-w)
wpearlite = w
pearlite
100 µm
HypoEutectoidSteel
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
r
Fe3C
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt25
Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
r
Fe3C
60
µm
HyperEutectoidSteel
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Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt27
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt28
Bruce Mayer, PE Engineering-45: Materials of Engineering
Lever Rule for Lever Rule for + Cementite + Cementite
C = 1.5-.022
Ccem = 6.7-1.5
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt29
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt30
Bruce Mayer, PE Engineering-45: Materials of Engineering
Lever Rule for Pearlite + ProEuLever Rule for Pearlite + ProEu
Cp = 6.7-1.5
CFe3C = 1.5-0.76
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt31
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt32
Bruce Mayer, PE Engineering-45: Materials of Engineering
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%
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt33
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
[email protected] • ENGR-45_Lec-22_PhaseDia-2.ppt34
Bruce Mayer, PE Engineering-45: Materials of Engineering
WhiteBoard WorkWhiteBoard Work