PHASE DIAGRAM · PDF file 2013. 11. 27. · L: 35wt%Ni α: 46wt%Ni C L: 35wt%Ni...
date post
25-Jan-2021Category
Documents
view
0download
0
Embed Size (px)
Transcript of PHASE DIAGRAM · PDF file 2013. 11. 27. · L: 35wt%Ni α: 46wt%Ni C L: 35wt%Ni...
BY
PHASE DIAGRAM
BY
ENGR. MS TAYYABA BANO
Cu-Ni System
• 2 phases: L (liquid) α (FCC solid solution)
• 3 phase fields: 1400
1500
1600 T(°C)
L (liquid)
• 3 phase fields:
L
L + α α
wt% Ni20 40 60 80 1000 1000
1100
1200
1300
1400
α (FCC solid solution)
L +
α liq ui du s
so lid us
• Rule 1: If we know T and Co, then we know:
the # and types of phases present# and types of phases present# and types of phases present# and types of phases present.
• Examples:
1500
1600 T(°C)
L (liquid)
liq ui du s
(1 2 5 0 ,3 5 ) Cu-Ni
phase
A(1100, 60): 1 phase: α
PHASE DIAGRAMS: # and types of phases
wt% Ni20 40 60 80 1000 1000
1100
1200
1300
1400
α (FCC solid solution)
L +
α
liq ui du s
so lid us
A(1100,60) B (1 2 5 0 ,3 5 )
phase
diagram B(1250, 35): 2 phases: L + α
Rule 2: If we know T and Co, then we
know: the composition of each phase.
• Examples:
1300
T(°C)
L (liquid) liq uid
us
sol idu
s
TA A
TB B
tie line
L + α
Cu-Ni
system
At TA:
Only Liquid (L) C = C ( = 35wt% Ni)
Co = 35wt%Ni
PHASE DIAGRAMS: composition of phases
wt% Ni
20
1200
α (solid)L +
α
sol idu
s
30 40 50
D TD
TB B
433532 CoCL Cα
CL = Co ( = 35wt% Ni)
At TB:
Both α and L CL = Cliquidus ( = 32wt% Ni here)
Cα = Csolidus ( = 43wt% Ni here)
At TD:
Only Solid (α) Cα = Co ( = 35wt% Ni)
Rule 3: If we know T and Co, then we know:
the amount of each phase (given in wt%). Cu-Ni
system• Examples:
At TA: Only Liquid (L)
WL = 100wt%, Wα = 0 At TD: Only Solid (α)
Co = 35wt%Ni
1300
T(°C)
L (liquid) liq uid
us
sol idu
s
TA A
TB B
tie line
L + α
PHASE DIAGRAMS: weight fractions of phases
At TB: Both α and L
At TD: Only Solid (α) WL = 0, Wα = 100wt%
WL = S
R +S
Wα = R
R +S
%73 3243
35430 wt CC
CC
L
= − −=
− −=
α
α
= 27wt%
wt% Ni
20
1200 α
(solid) L +
α
sol idu
s
30 40 50
D TD
TB B
433532 CoCL Cα
R S
%27wt CC
CC
L
o = − −=
α
α
• Sum of weight fractions:
• Conservation of mass (Ni):
• Combine above equations:
WL + Wα = 1
Co = WLCL + WαCα
= R R +S
Wα = Co −CL Cα −CL
= S R+ S
WL = Cα −Co Cα −CL
THE LEVER RULE: A PROOF
R S Cα −CLR SCα CL
• A geometric interpretation:
Co
R S
WαWL
CL Cα moment equilibrium:
1−Wα solving gives Lever Rule
WLR = WαS
• Phase diagram:• Phase diagram:• Phase diagram:• Phase diagram:
CuCuCuCu----Ni system.Ni system.Ni system.Ni system.
• System is:• System is:• System is:• System is:
--------binarybinarybinarybinary i.e.i.e.i.e.i.e., 2 components:, 2 components:, 2 components:, 2 components: Cu and Ni.Cu and Ni.Cu and Ni.Cu and Ni.
--------isomorphousisomorphousisomorphousisomorphous
1300
L (liquid)
L +
α
T(°C)
A
D
B
L: 35wt%Ni
α: 46wt%Ni C
L: 35wt%Ni
46
43 32
24
35
L: 32wt%Ni
Cu-Ni
system
COOLING IN A Cu-Ni BINARY
--------isomorphousisomorphousisomorphousisomorphous i.e., completei.e., completei.e., completei.e., complete
solubility of onesolubility of onesolubility of onesolubility of one
component incomponent incomponent incomponent in
another; another; another; another; αααα phasephasephasephase field extends fromfield extends fromfield extends fromfield extends from
0 to 100wt% Ni.0 to 100wt% Ni.0 to 100wt% Ni.0 to 100wt% Ni.
wt% Ni 20
1200
30 40 50 1100
α (solid)
L +
α D
35 Co
E
24 36
α: 43wt%Ni
L: 32wt%Ni
L: 24wt%Ni
α: 36wt%Ni
• Consider
Co = 35wt%Ni.
2 components has a special composition
with a min. melting T.
• 3 single phase regions
(L, α, β) • Limited solubility:
α: mostly Cu
Ex.: Cu-Ag system
L (liquid)
α L + α L+β β
1200 T(°C)
800
1000
TE 779°C
Cu-Ag
system
BINARY-EUTECTIC SYSTEMS
α: mostly Cu β: mostly Ni • TE: No liquid below TE
• CE: Min. melting T
composition
L+β β
α + β
Co, wt% Ag 20 40 60 80 100 0
200
400
600
800
CE
TE 8.0 71.9 91.2 779°C
L (liquid)
α L + α L+β β
120 0 T(°C)
800
1000
TE 8.0 71.9 91.2 779°C
α + β
Co, wt% Ag 20 40 60 80 100 0
200
400
600
CE
IRON-CARBON/ SYSTEM/ PHASE DIAGRAMPHASE DIAGRAM
• 2 important points
-Eutectic (A):
-Eutectoid (B):
L⇒ γ +Fe3C
γ ⇒ α +Fe3C
C ( c e m e n ti te )
1600
1400
1200
1000
800
L
γ (austenite)
γ+L
γ+Fe3Cα +γ
L+Fe3C
δ
727°C = Teutectoid
1148°C
T(°C)
A
B
SR
γ γ γγ
α
IRON-CARBON (Fe-C) PHASE DIAGRAM
Result: Pearlite = alternating layers of α and Fe3C phases.
120µm
F e 3 C ( c e m e n ti te )
800
600
400 0 1 2 3 4 5 6 6.7
α+Fe3C
+γ
(Fe) Co, wt% C 0.77 4.30
727°C = TeutectoidB R S
Fe3C (cementite-hard)
α (ferrite-soft)
α C e u te c to id
DEVELOPMENT OF MICROSTRUCTURESDEVELOPMENT OF MICROSTRUCTURES
� Austenite precipitates Fe3C at Eutectoid Transformation Temperature (727°C).
γγγγ αααα + Fe3C Cooling
Heating
� When cooled slowly, forms Pearlite, which is a micro-contituent made of ferrite (αααα) and Cementite (Fe3C),
Heating
HYPOEUTECTOID STEEL
(Fe-C System)
C ( c e m e n ti te )
1600
1400
1200
1000
L
γ (austenite)
γ+L
γ+Fe3C
L+Fe3C
δ
1148°C
T(°C)
γ γ γγ
γγ γ γ
α
Adapted from
Fig. 9.27,Callister 6e. (Fig. 9.27 courtesy Republic Steel Corporation.)
Co
F e 3 C ( c e m e n ti te )
800
600
400 0 1 2 3 4 5 6 6.7
α+Fe3C
Co, wt% C0 .7 7
727°C
R Sα γ γ γ
γ r s
wα =s/(r+s) wγ =(1-wα)
wα =S/(R+S) wFe3C =(1-wα)
wpearlite = wγ
α αα
α α
α pearlite
100µm Hypoeutectoid steel
HYPOHYPO--EUTECTOIDEUTECTOID
“Proeutectoid” means it formed ABOVE or BEFORE the Eutectoid Temperature!
How to calculate the relative amounts of proeutectoid
phase (α or Fe3C) and pearlite? Eutectoid ,0.76% C Eutectoid ,0.76% C
2.14
HypoEutectoid HyperEutectoid
2.14
� Fraction of pearlite:
74.0
022.0'
022.076.0
022.0' −= − −=
+ = CoCo
UT
T Wp 74.0022.076.0
= −
= +
= UT
Wp
� Fraction of Proeutectoid :
74.0
'76.0
022.076.0
'76.0 CoCo
UT
U W
−= − −=
+ =α
� Fraction of pearlite: (for Hypereutectoid alloys)
94.5
'170.6
76.070.6
'170.6 CC
XV
X Wp
−= − −=
+ =
94.576.070.6XV Wp =−
= +
=
� Fraction proeutectoid cementite: (Hypereutectoid)
76.0'176.0'1 −=−== CCVW 94.576.070.63
= −
= +
= XV
W CFe
(Fe-C System)
C ( c e m e n ti te )
1600
1400
1200
1000
L
γ (austenite)
γ+L
γ+Fe3C
L+Fe3C
δ
1148°C
T(°C)
γ γ γγ
γγ γ γ
Fe3C
HYPEREUTECTOID STEEL
Co
F e 3 C ( c e m e n ti te )
800
600
400 0 1 2 3 4 5 6 6.7
α+Fe3C
Co, wt% C0 .7 7
R Sα s
wFe3C =r/(r+s) wγ =(1-wFe3C)
wα =S/(R+S) wFe3C =(1-wα)
wpearlite =
