Chapter 8 - Phase Diagram PART2
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Transcript of Chapter 8 - Phase Diagram PART2
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Pb-Sn Cooling Curve
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Pb-Sn Eutectic Reaction
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8 - 4
• a = A-rich solid solution (B solute, A host)
• b = B-rich solid solution (A solute; B host)
Definition:
• Three single-phase regions:
a solid phase
b − solid phase
L − liquid phase
• Three 2-phase regions:
(a + L); (b + L); (a + b)
Binary Eutectic Alloy System – Generic Diagram
L (eutectic) a b
Composition
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Foundations of Materials Science and Engineering, 5th Edn. Smith and Hashemi
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
Binary Eutectic Alloy System
• In binary eutectic alloy systems, components have limited solid
solubility in each other. Example: Pb-Sn alloy, Cu-Ag alloy.
• Regions designated as alpha phase (a ) and beta phase ( b ) at each
ends are called terminal solid solutions because they appear at the
ends of the phase diagram.
• Eutectic composition freezes at a lower temperature than all other
alloy compositions.
• Eutectic temperature is the lowest temperature at which the liquid
phase can exist, when cooled slowly.
• Eutectic composition and eutectic temperature define the eutectic
point.
• When liquid of eutectic composition is slowly cooled to eutectic
temperature, the single liquid phase transforms into two solid phases.
5
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• Eutectic reaction is called invariant reaction because it
occurs under equilibrium conditions at a specifictemperature and alloy composition that can not be varied.
• Eutectic reaction has zero degree of freedom (F = 0)
F + P = C + 1 = F + 3 = 2 + 1.
• At eutectic point, three phases are in equilibrium, i.e. one
liquid phase (L) is in equilibrium with two solid phases
(a and b)
• Eutectic isotherm is the horizontal solidus line at TE. There is
no liquid below TE.• General eutectic reaction:
The Eutectic Reaction
Liquid
solid solution + β solid solutionEutectic temp.
Cooling
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Invariant or eutectic point
Eutectic isotherm
Cu-Ag equilibrium phase diagram. This diagram is characterized by the limited solid
solubility of each terminal phase (a and b). The eutectic invariant reaction at 71.9% Ag
and 779°C is the most important feature of this system. At eutectic point, a (8.0% Ag),
b (91.2% Ag) and liquid (71.9% Ag) can coexist.
Binary Eutectic Alloy System: Cu-Ag
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Pb-Sn equilibrium phase diagram. Eutectic composition is at 61.9% Sn and
183°
C. At eutectic point, a (18.3% Sn), b (97.5% Sn) and liquid (61.9% Sn) cancoexist.
Binary Eutectic Alloy System: Pb-Sn
Solvus is the phase boundary below isothermal liquid + proeutectic solid
phase. Also phase boundary separating terminal solid solution from two-phase
region of solid solutions. Solvus shows the maximum solid solubility of
solute in a solid solution.
Alpha phase is Pb-rich solid
solution; at 183°C amaximum of 19.2 wt % Sn can
be dissolved in it.
Beta phase is Sn-rich solid
solution; at 183°
C amaximum of 2.5 wt % Pb can
be dissolved in it.
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9
Pb-Sn system
Lb
a b
200
T(°
C)
C, wt% Sn
20 60 80 1000
300
100
L
ab
L + a
183°C
40
TE
18.3
a: 18.3 wt%Sn
97.5
b: 97.5 wt% Sn
CE = 61.9
L: 61.9% Sn
Slow Cooling At Eutectic Composition
• Liquid at 300°C.
• At TE liquid solidifies by
eutectic reaction into two
solid phases a (19.2% Sn)
and b (97.5% Sn)
• Further cooling from TE to
room temperature, solid
solubility of solute in a and b
phases decreases, as
indicated by solvus line.
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display
Slow Cooling of 60% Pb – 40% Sn Alloy
• Liquid at 3000C.
• At about 2450C first solidforms – proeutectic or
primary a, contains 12%
Sn.
• Slightly above 183°
Ccomposition of alpha
follows solidus and
composition of Sn varies
from 40% to 61.9%.
• At eutectic temperature,all the remaining liquid
solidifies.
• Further cooling lowers alpha Sn content and beta Pb.
10
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Primary (i.e. proeutectic ) phase is formed in the (a + L) region, while
the eutectic structure that includes layers of a and b phases is formed uponcrossing the eutectic isotherm.
• Growth of a solid phase
occurs above TE
• Primary or proeutectic a
growth stops at TE
• Below TE, all remaining
liquid transforms into
eutectic structure byeutectic reaction
Result:
Slow Cooling At Hypoeutectic Composition (cont.)
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In eutectic reaction, both a and b phases are formed simultaneously,
resulting in a microstructure called eutectic structure. In Pb-Sn alloy,eutectic reaction involves re-distribution of Pb and Sn atoms by atomic
diffusion.
In binary eutectic reaction, the two solid phases (a + b ) can have various
structures as shown above. The most common is lamellar eutectic
structure of alternating layers of and phases.
Various Eutectic Structures
160m
Micrograph of Pb-Sn
eutectic microstructurelamellar
rodlike
globular
acicular
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13
L+aL+b
a + b
200
Co, wt% Sn20 60 80 1000
300
100
L
a bTE
40
(Pb-Sn System)
Eutectic structure
hypereutectic: Co > 61.9wt% Sn
b
bb
bb
b
a
a
a
aaa
hypoeutectic: Co< 61.9wt% Sn
T(°C)
eutectic
eutectic: Co =61.9wt% Sn
Microstructure: Eutectic, Hypoeutectic and Hypereutectic
of Pb-Sn Alloy
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8 - 14
• Growth of b solid phase occurs above TE
• Growth of this primary b or proeutectic b stops at TE
• Below TE, all remaining liquid transforms into eutectic structure.
Microstructure Evolution of Pb-Sn Alloy:
Slow Cooling At Hypereutectic Composition
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15
L + a L+b
a + b
200
T(°C)
18.3
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
a 183°C
61.9 97.5
b
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: a + b--compositions of phases:
CO = 40 wt% Sn
--relative amount of each phase:
150
40Co
11Ca
99Cb
SR
Ca = 11 wt% Sn
Cb = 99 wt% Sn
Wa=Cb - CO
Cb - Ca
=99 - 40
99 - 11 =59
88 = 67 wt%
SR+S
=
Wb =CO - Ca
Cb - Ca
=R
R+S
=29
88
= 33 wt%=40 - 11
99 - 11
Phase Diagram Analysis : Pb-Sn Eutectic Alloy
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L+b
a + b
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
a b
L + a
183°C
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...
--the phases present: a + L--compositions of phases:
CO = 40 wt% Sn
--relative amount of of each phase:
Wa =CL - CO
CL - Ca
=46 - 40
46 - 17
= 629
= 21 wt%
WL =CO - Ca
CL - Ca
=23
29= 79 wt%
40Co
46CL
17Ca
220 SR
Ca = 17 wt% SnCL = 46 wt% Sn
Phase Diagram Analysis : Pb-Sn Eutectic Alloy
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• For a 40 wt% Sn-60 wt% Pb alloy slow-cooled to just below TE
• Result: a crystals and eutectic microstructure
18.3 61.9
SR
97.8
SR
primary a
eutectic a
eutectic b
WL = (1-Wa) = 50 wt%
Ca = 18.3 wt% Sn
CL = 61.9 wt% Sn
SR + SWa= = 50 wt%
• Just above TE :
• Just below TE :
Ca = 18.3 wt% Sn
Cb = 97.5 wt% SnS
R + SWa= = 73 wt%
Wb = 27 wt%
L+b200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
ab
L+a
40
a+b
TE
L: Co wt% Sn LaL
a
Phase Diagram Analysis : Pb-Sn Eutectic Alloy
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8 - 18
SAMPLE PROBLEM 1
For a 75 wt.% Cu – 25 wt.% Ag alloy at the following temperatures: (a) 1000oC,
(b) 800oC, (c) 780oC +DT, and (d) 780oC – DT, state the:
i)phase(s) present.
ii)chemical composition of the phases.iii)amount of each phase.
iv)Sketch the microstructure.
780°C
7.9
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780°
C
25 wt% ~ 65 wt%
PROBLEM 1
7.9
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(i) Phases present: 2 phases: L & a
(ii) Composition of phases:
L = 65 wt % Ag;
a = 7.9 wt % Ag
(iii) Amount of each phase
(iv) Microstructure: proeutectic a + L
WL
257.9
657.9
x 100% 30.0%
Wa
6525
657.9
x 100% 70.0%
SOLUTION: PROBLEM 1
(a) At 1000oC: (b) At 800oC:
Proeutectic a
L
(i) Phases present: 1 phase: L
(ii) Composition of phases:
CL = Co = 25 wt % Ag
(iii) Amount of each phase:
WL = 100 wt %,
Wa = 0
(iv) Microstructure: All liquid
L
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(i) Phases present: 2 phases: a & b
(ii) Composition of phases:
a = 7.9 wt % Ag
b = 91.2 wt % Ag
(iii) Amount of each phase
(iv) Microstructure:
Wb 91.271.991.27.9
x 100% 20.5%
Wa 91.225
91.27.9
x 100% 79.5%
SOLUTION: PROBLEM 1 (continue)
(c) At 780oC + T: (d) At 780oC - T:
Proeutectic a
(i) Phases present: 2 phases: L & a
(ii) Composition of phases:
L = 71.9 wt % Ag
a = 7.9 wt % Ag
(iii) Amount of each phase:
(iv) Microstructure:
Wa 71.925
71.97.9
x 100% 73.3%
WL
257.9
71.97.9
x 100% 26.7%
Proeutectic a
L
eutectic a
eutectic b
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8 - 22
SAMPLE PROBLEM 2
If 750g of 80 wt.% Ag – 20 wt.% Cu alloy is slowly cooled from 1000oC to just below
780°C (i.e. cooled to 780oC – DT), determine:
i)How many grams of liquid and proeutectic beta are present at 800°C?
ii)How many grams of liquid and proeutectic beta are present at 780°C + ΔT?iii)How many grams of alpha are present in the eutectic structure at 780°C − ΔT?
iv)How many grams of beta are present in the eutectic structure at 780°C – ΔT?
780°C
7.9
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(i) Phases present: 2 phases: L & b
(ii) Composition of phases:
L = 71.9 wt % Ag
b = 91.2 wt % Ag
(iii) Amount of each phase
(iv) Mass of each phase:
L = (750g)(0.58) = 435g
b = (750g)(0.42) = 315g
Wb 8071.991.271.9
x 100% 42.0%
WL 91.280
91.27.9
x 100% 58%
SOLUTION: PROBLEM 2
(a) At 800oC: (b) At 780oC + T:
(i) Phases present: 2 phases: L & b
(ii) Composition of phases:
L = 79 wt % Ag; b = 92 wt % Ag
(iii) Amount of each phase:
(iv) Mass of each phase:
L = (750g)(0.923) = 692.3gb = (750g)(0.077) = 57.7g
WL
9280
9279
x 100% 92.3%
Wb 80799279
x 100% 7.7%
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(i) Phases present: 2 phases: a & b
(ii) Composition of a phase:
a = 7.9 wt % Ag
(iii) Amount of a phase:
% total a = % eutectic a
(iv) Mass of a phase:
a = (750g)(0.1345) = 100.8g
Wb
807.9
91.27.9
x 100% 86.55%
SOLUTION: PROBLEM 2 (continue)
(c) At 780oC - T: (d) At 780oC -
T:
Wa
91.280
91.27.9
x 100% 13.45%
% eutecticb % Total b - % proeutectic b = 86.55% - 42%
= 44.55%
(i) Phases present: 2 phases: a & b
(ii) Composition of b phase:
b = 91.2 wt % Ag
(iii) Amount of b phase
Note this amount is the total b, which
includes both proeutectic b and eutectic b
(iv) Mass of eutectic b:
Eutectic b = (750g)(0.446) = 334.5g
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SAMPLE PROBLEM 3
An alloy of 75 wt.% Pb – 25 wt.% Sn alloy is slow-cooled from 300oC to 27°C.
Determine:
a) Whether this alloy is hypoeutectic or hypereutectic. Hypoeutectic
b) The composition of the first solid to form. 12% Sn
c) The amount and composition of each phase present at 183°
C + ΔTd) The amount and composition of each phase present at 183°C – ΔT
e) The amount of each phase present at room temperature. Assume solid solubility
of Sn in Pb is 1% at room temperature.
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(i) Phases present: 2 phases: a & L
(ii) Composition of each phase:
a = 18.3 wt % Sn; L = 61.9 wt % Sn
(iii) Amount of each phase
Wa 61.925
61.918.3
x 100% 84.6%
WL
2518.3
61.918.3
x 100% 15.4%
SOLUTION: PROBLEM 3
(c) At 183oC + T: (d) At 183oC -
T:
(i) Phases present: 2 phases: a & b
(ii) Composition of each phase:a = 18.3 wt % Sn; b = 97.8 wt % Sn
(iii) Amount of each phase
Wa 97.825
97.818.3
x 100% 91.6%
Wb 2518.397.818.3
x 100% 8.4%
(amount of total a phase)
(amount of total b phase)(e) At room temperature:
Wa 100251001
x 100% 75.8%
Wb 2511001
x 100% 24.2%