chapter 8 - phase diagram part2

7/18/2019 Chapter 8 - Phase Diagram PART2

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Pb-Sn Cooling Curve



Pb-Sn Eutectic Reaction



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



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



• 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



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.



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



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.)



12

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



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



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



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



16

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




17

• 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




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



780°

C

25 wt% ~ 65 wt%

PROBLEM 1

7.9



(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


CL = Co = 25 wt % Ag

(iii) Amount of each phase:

WL = 100 wt %,

Wa = 0

(iv) Microstructure: All liquid

L



(i) Phases present: 2 phases: a & b


a = 7.9 wt % Ag

b = 91.2 wt % Ag


(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


L = 71.9 wt % Ag

a = 7.9 wt % Ag


(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



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



(i) Phases present: 2 phases: L & b


L = 71.9 wt % Ag

b = 91.2 wt % Ag


(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


L = 79 wt % Ag; b = 92 wt % Ag


(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%




(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%


(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



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.



(i) Phases present: 2 phases: a & L

(ii) Composition of each phase:

a = 18.3 wt % Sn; L = 61.9 wt % Sn


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:


(ii) Composition of each phase:a = 18.3 wt % Sn; b = 97.8 wt % Sn


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%

chapter 8 - phase diagram part2

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