10 Askeland Chap

10 Dispersion Strengthening and Eutectic Phase Diagrams 10–22

A hypothetical phase diagram diagram is shown in Figure 10–32. (a) Are any intermetallic intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. (b) Identify the solid solutions present in the system. Is either material  A or  B allotropic? Explain. (c) Identify the three-phase reactions by writing down the temperature, the reaction in equation form, the composition of  each phase in the reaction, and the name of the reaction.

Solution: (a) u  non-stoichiometric intermetallic compound. (b) a, h, g, and b; material  B is allotropic, existing in three different forms at different temperatures (c) 1100° 1100°C:

g   L

900° 900°C:  L1 680° 680°C:  L 600° 600°C: 300° 300°C:

10–23

S

S

S

b;

L2  a;

a  b;

ab b

S

S

u;

u  h;

peri perittect ectic; ic;

 L: 82%  B g: 97%  B

monotectic;  L1: 28%  B  L 2: 50%  B eutectic; peri perite tect ctoi oid; d; eutectoid;

b: 90%  B a: 5%  B

 L: 60%  B a: 5%  B

b: 90%  B

a: 5%  B b: 80%  B

u: 37%  B

b: 90%  B u: 40%  B

h: 95%  B

The Cu–Zn phase diagram is shown in Figure 10–33. (a) Are any intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. (b) Identify the solid solutions present in the system. (c) Identify the three-phase reactions by writing down the temperature, the reaction in equation form, and the name of the reaction.

111

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The Science and Engineering of M aterials

Solution: (a) b, b, g, d, e: all nonstoichiometric. (b) a, u

10–24

(c) 900°C:

a   L

S

b;

peritectic

830°C:

b   L

S

g;

peritectic

700°C:

g   L

S

d;

peritectic

600°C:

d   L

S

550°C:

d

420°C:

e   L

250°C:

b¿

S

S

peritectic

g  e;

eutectoid

S

u;

peritectic

a  g;

eutectoid

A portion of the Al–Cu phase diagram is shown in Figure 10 –34. (a) Determine the formula for the u compound. (b) Identify the three-phase reaction by writing down the temperature, the reaction in equation form, the composition of each phase in the reaction, and the name of the reaction. 54 g  63.54 g/mol

Solution: (a) u at 54% Cu; (b) 548°C;  L

10–25

e;

S

54  63.54  46  26.981 a  u;



33 at% Cu;

CuAl2

eutectic;  L: 33.2% Cu, a: 5.65% Cu,

u: 52.5% Cu.

The Al–Li phase diagram is shown in Figure 10 –35. (a) Are any intermetallic compounds present? If so, identify them and determine whether they are stoichiometric or nonstoichiometric. Determine the formula for each compound. (b) Identify the three-phase reactions by writing down the temperature, the reaction in equation form, the composition of each phase in the reaction, and the name of the reaction.

Solution: (a) b is non-stoichiometric @ 21 wt% Li: at% Li 

21 g  6.94 g/mol 21  6.94  79  26.981



100%  50 at% Li ∴ AlLi



100%  66.7% Li ∴ AlLi2

g, is stoichiometric @ 34 wt% Li: at% Li  (b) 600°C:  L

34 g  6.94 g/mol 34  6.94  66  26.981 S

ab

eutectic

 L: 9.9% Li

a: 4% Li 510°C: b   L

S

g

peritectic

b: 20.4% Li

b: 25% Li  L: 47% Li

170°C:  L

10–26

S

g  a 1 Li 2 

eutectic

 L: 98% Li g: 34% Li

g: 34% Li

a 1 Li 2 : 99% Li

An intermetallic compound is found for 10 wt% Si in the Cu –Si phase diagram. Determine the formula for the compound.

Solution:

at% Si 

10 g  28.08 g/mol 10  28.08  90  63.54



0.20

or

SiCu4

CHAPTER 10 10–27

Dispersion Strengthening and Eutect ic Phase Diagrams

Using the phase rule, predict and explain how many solid phases will form in an eutectic reaction in a ternary (three-component) phase diagram, assuming that the pressure is fixed.

Solution:

F   C   P  1

At the eutectic, F   0, C   3 Therefore,  L S a  b  g 10–30

113

03P1

or

P4

and 3 solid phases form.

Consider a Pb–15% Sn alloy. During solidification, determine (a) the composition of the first solid to form, (b) the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy, (c) the amounts and compositions of  each phase at 260°C, (d) the amounts and compositions of each phase at 183°C, and (e) the amounts and compositions of each phase at 25 °C.

Solution: (a) 8% Sn (b) liquidus  290°C, solvus  170°C,

solidus  240°C, freezing range  50°C

(c) L: 30% Sn a: 12% Sn; 15  12 % L   100%  17% 30  12 (d) a: 15% Sn

100% a

(e) a: 2% Pb b: 100% Sn 100  15 %a   100  87% 100  2 10–31

%a  83%

%b  13%

Consider an Al–12% Mg alloy (Figure 10 –36). During solidification, determine (a) the composition of the first solid to form, (b) the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy, (c) the amounts and compositions of each phase at 525 °C, (d) the amounts and compositions of each phase at 450°C, and (e) the amounts and compositions of each phase at 25°C.

Solution: (a) 2.5% Mg (b) liquidus  600°C, solvus  400°C,

solidus  470°C, freezing range  130°C

(c) L: 26% Mg a: 7% Mg; 26  12 %a   100%  74% 26  7 (d) a: 12% Mg

% L  26%

100% a

(e) a: 1% Mg b: 34% Mg 34  12 %a   100%  67% 34  1

%b  33%

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The Science and Engineering of M aterials 10–32

Instructor’s Solution Manual

Consider a Pb–35% Sn alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 184 °C, (d) the amounts and compositions of each phase at 182°C, (e) the amounts and compositions of each microconstituent at 182°C, and (f) the amounts and compositions of each phase at 25 °C.

Solution: (a) hypoeutectic

(b) 14% Sn

(c) a: 19% Sn  L: 61.9% Sn 61.9  35  100%  63% %a  61.9  19 (d) a: 19% Sn b: 97.5% Sn 97.5  35 %a   100%  80% 97.5  19 (e) primary a: 19% Sn eutectic: 61.9% Sn

%b  20%

%primary a  63% %eutectic  37%

(f) a: 2% Sn b: 100% Sn 100  35  100%  66% %a  100  2 10–33

% L  37%

%b  34%

Consider a Pb–70% Sn alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 184 °C, (d) the amounts and compositions of each phase at 182°C, (e) the amounts and compositions of each microconstituent at 182°C, and (f) the amounts and compositions of each phase at 25 °C.

Solution: (a) hypereutectic

(b) 98% Sn

(c) b: 97.5% Sn  L: 61.9% Sn 70  61.9  100%  22.8% %b  97.5  61.9 (d) a: 19% Sn b: 97.5% Sn 97.5  70 %a   100%  35% 97.5  19 (e) primary b: 97.5% Sn eutectic: 61.9% Sn

%b  65%

%primary b  22.8% %eutectic  77.2%

b: 100% Sn (f) a: 2% Sn 100  70  100%  30% %a  100  2 10–34

% L  77.2%

%b  70%

Calculate the total % b and the % eutectic microconstituent at room temperature for the following lead-tin alloys: 10% Sn, 20% Sn, 50% Sn, 60% Sn, 80% Sn, and 95% Sn. Using Figure 10–22, plot the strength of the alloys versus the % b and the % eutectic and explain your graphs.

CHAPTER 10

Dispersion Strengthening and Eutect ic Phase Diagrams

Solution:

%b 10% Sn 20% Sn 50% Sn 60% Sn 80% Sn 95% Sn

10  2 99  2 20  2 99  2 50  2 99  2 60  2 99  2 80  2 99  2 95  2 99  2

%eutectic



8.2%



18.6%



49.5%



59.8%



80.4%



95.9%

0% 20  19 61.9  19 50  19 61.9  19 60  19 61.9  19 97.5  80 97.5  61.9 97.5  95 97.5  61.9



2.3%



72.3%



95.6%



49.2%



7.0%

8000

8000

   )    i   s   p    (    h    t   g   n   e   r    t   s   e    l    i   s   n   e    T

115

   )    i   s   p    (    h    t   g   n   e   r    t   s   e    l    i   s   n   e    T

7000

6000

5000

7000

 o  H y p

  r   p e   y   H

6000

5000

4000 20

40

60 %

10–35

80

100

20

40

60

80

% eutectic

β

Consider an Al–4% Si alloy. (See Figure 10 –23.) Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 578 °C, (d) the amounts and compositions of each phase at 576 °C, the amounts and compositions of each microconstituent at 576°C, and (e) the amounts and compositions of each phase at 25°C.

Solution: (a) hypoeutectic (b) 1% Si (c) a: 1.65% Si  L: 12.6% Si 12.6  4 %a   78.5% 12.6  1.65 (d) a: 1.65% Si b: 99.83% Si 99.83  4  97.6% %a  99.83  1.65

% L  21.5%

%b  2.4%

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The Science and Engineering of M aterials

Instructor’s Solution Manual %primary a  78.5% %eutectic  21.5%

(e) primary a: 1.65% Si eutectic: 12.6% Si a: 0% Si 10–36

%a 

b: 100% Si

100  4 100  0



96%

%b  4%

Consider a Al–25% Si alloy. (See Figure 10 –23.) Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 578 °C, (d) the amounts and compositions of each phase at 576 °C, (e) the amounts and compositions of each microconstituent at 576 °C, and (f) the amounts and compositions of  each phase at 25°C.

Solution: (a) hypereutectic (b) 100% Si (c) b: 99.83% Si  L: 12.6% Si 99.83  25 % L   85.8% 99.83  12.6

%b  14.2%

(d) a: 1.65% Si b: 99.83% Si 99.83  25 %a   76.2% 99.83  1.65

%primary b  14.2% %eutectic  85.8%

(e) primary b: 99.83% Si eutectic: 12.6% Si (f) a: 0% Si 10–37

100  0



75%

%b  25%

%a  45 

98.0   x 98.0  5



or  x  56.15% Sn

100

Hypoeutectic

%a  85 

100   x 100  1



100

or  x  15.85% Si

Hypereutectic

A Pb–Sn alloy contains 23% primary a and 77% eutectic microconstituent. Determine the composition of the alloy.

Solution: 10–40

100  25

An Al–Si alloy contains 85% a and 15% b at 500°C. Determine the composition of  the alloy. Is the alloy hypoeutectic or hypereutectic?

Solution: 10–39

%a 

b: 100% Si

A Pb–Sn alloy contains 45% a and 55% b at 100°C. Determine the composition of  the alloy. Is the alloy hypoeutectic or hypereutectic?

Solution: 10–38

%b  23.8%

%primary a  23 

61.9   x 61.9  19



100

or  x  52% Sn

An Al–Si alloy contains 15% primary b and 85% eutectic microconstituent. Determine the composition of the alloy.

CHAPTER 10

Solution: 10–41

Dispersion Strengthening and Eutect ic Phase Diagrams

%eutectic  85 

100   x 100  12.6



100

117

or  x  25.71% Si

Determine the maximum solubility for the following cases. (a) lithium in aluminum (Figure 10 –35), (b) aluminum in magnesium (Figure 10 –37), (c) copper in zinc (Figure 10–33), and (d) carbon in g-iron (Figure 10–38)

Solution: (a) 4% Li dissolves in aluminum (b) 12.7% Al dissolves in magnesium (c) 3% Cu dissolves in zinc (d) 2.11% C dissolves in g-iron 10–42

Determine the maximum solubility for the following cases. (a) (b) (c) (d)

magnesium in aluminum (Figure 10–36), zinc in copper (Figure 10–33), beryllium in copper (Figure 10–33), and Al2O3 in MgO (Figure 10–39)

Solution: (a) 14.9% Mg dissolves in aluminum (b) 40% Zn dissolves in copper (c) 2.5% Be dissolves in copper (d) 18% Al2O3 dissolves in MgO 10–43

Observation of a microstructure shows that there is 28% eutectic and 72% primary b in an Al–Li alloy (Figure 10 –35). (a) Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic. (b) How much a and b are in the eutectic microconstituent?

Solution: (a) 28



20.4   x 20.4  9.9

(b) %aEut  10–44



100

20.4  9.9 20.4  4



or  x  17.46% Li

100%  64%

and

Hypereutectic

%bEut  36%

Write the eutectic reaction that occurs, including the compositions of the three phases in equilibrium, and calculate the amount of  a and b in the eutectic microconstituent in the Mg –Al system, (Figure 10 –36).

Solution:

 L32.3 S a12.7  g40.2

∴ %aEut 

40.2  32.3 40.2  12.7



100%  28.7%

and

%gEut  71.3%

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The Science and Engineering of M aterials 10–45

Instructor’s Solution Manual

Calculate the total amount of  a and b and the amount of each microconstituent in a Pb–50% Sn alloy at 182°C. What fraction of the total a in the alloy is contained in the eutectic microconstituent?

Solution:

atotal  aPrimary 

97.5  50 97.5  19 61.9  50 61.9  19



100%  60.5%

bTotal  39.5%



100%  27.7%

Eutectic  72.3%

ain eutectic  60.5  27.7  32.8%  f   32.8  60.5  0.54 10–46

Figure 10–40 shows a cooling curve for a Pb –Sn alloy. Determine (a) the pouring temperature, (b) the superheat, (c) the liquidus temperature, (d) the eutectic temperature, (e) the freezing range, (f) the local solidification time, (g) the total solidification time, and (h) the composition of the alloy.

Solution: (a) pouring temperature



360°C

(b) superheat  360  250  110°C (c) liquidus temperature  250°C (d) eutectic temperature  183°C (e) freezing range  250  183  67°C (f) local solidification time  600  110  490 s (g) total solidification time  600 s (h) approximately 32% Sn 10–47

Figure 10–41 shows a cooling curve for an Al –Si alloy. Determine (a) the pouring temperature, (b) the superheat, (c) the liquidus temperature, (d) the eutectic temperature, (e) the freezing range, (f) the local solidification time, (g) the total solidification time, and (h) the composition of the alloy.

Solution: (a) pouring temperature



1150°C

(b) superheat  1150  1000  150°C (c) liquidus temperature  1000°C (d) eutectic temperature  577°C (e) freezing range  1000  577  423°C (f) local solidification time  11.5  1  10.5 min (g) total solidification time  11.5 min (h) approximately 45% Si

CHAPTER 10 10–48

Dispersion Strengthening and Eutect ic Phase Diagrams

119

Draw the cooling curves, including appropriate temperatures, expected for the following Al–Si alloys. (a) Al–4% Si

(b) Al–12.6% Si

(c) Al–25% Si

(d) Al–65% Si

Solution:

780°

630° 577°

T 

577°

T 

Al – 4% Si

10–49

577°

T 

Al – 12.6% Si

t 

1200° T 

Al – 25% Si

t 

577° Al – 65% Si

t 

t 

Based on the following observations, construct a phase diagram. Element  A melts at 850°C and element  B melts at 1200°C. Element  B has a maximum solubility of 5% in element  A, and element  A has a maximum solubility of 15% in element  B. The number of degrees of freedom from the phase rule is zero when the temperature is 725°C and there is 35%  B present. At room temperature 1%  B is soluble in  A and 7%  A is soluble in  B.

Solution:

1200 1000    )    C             °    (   e   r   u    t   a   r   e   p   m   e    T

L

b+L

800

a+L

a

b

600 a+b

400 200

A

5

20

40

60 % B 

80 85

B 

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Instructor’s Solution Manual

The Science and Engineering of M aterials 10–50

Cooling curves are obtained for a series of Cu –Ag alloys, (Figure 10–42). Use this data to produce the Cu–Ag phase diagram. The maximum solubility of Ag in Cu is 7.9% and the maximum solubility of Cu in Ag is 8.8%. The solubilities at room temperature are near zero.

Solution: 0% Ag 8% Ag 20% Ag 50% Ag 71.9% Ag 90% Ag 100% Ag

S S S S S S S

T liq

T sol

1085°C 1030°C 975°C 860°C 780°C 870°C 961°C

950°C 780°C 780°C 780°C 780°C

1100

   ) 1000    C             °    (   e   r   u    t   a   r 900   e   p   m   e    T

L

a+L

a

b+L

800

b a+b

700 Cu

29

40

60

80

Ag

% Ag

10–51

The SiO2 –Al2O3 phase diagram is included in Figure 10 –27(b). A refractory is required to contain molten metal at 1900 °C. (a) Will pure Al 2O3 be a potential candidate? Explain. (b) Will Al2O3 contaminated with 1% SiO 2 be a candidate? Explain.

Solution: (a) Yes. T m  2040°C  1900°C

No liquid will form.

(b) No. Some liquid will form. % L 

100  99 100  80



100%  5%  L

This liquid will weaken the refractory.

CHAPTER 10 10–66

10–67

10–68

Dispersion Strengthening and Eutect ic Phase Diagrams

121

Consider the ternary phase diagram shown in Figures 10 –30 and 10–31. Determine the liquidus temperature, the first solid to form, and the phases present at room temperature for the following compositions. (a) 30%  B –20% C , balance  A (c) 60%  B –10% C , balance  A

(b) 10%  B –25% C , balance  A

Solution: (a) T Liq



220°C;

b;

agb

(b) T Liq  330°C;

a;

ag

(c) T Liq  390°C;

b;

ab

Consider the ternary phase diagram shown in Figures 10 –30 and 10–31. Determine the liquidus temperature, the first solid to form, and the phases present at room temperature for the following compositions. (a) 5%  B –80% C , balance  A (c) 30%  B –35% C , balance  A

(b) 50%  B –5% C , balance  A

Solution: (a) T Liq



390°C;

g;

ag

(b) T Liq  330°C;

b;

ab

(c) T Liq  290°C;

b;

abg

Consider the liquidus plot in Figure 10 –30. (a) For a constant 20%  B, draw a graph showing how the liquidus temperature changes from 20% B –0% C, balance A to 20% B–80% C, balance A, (b) What is the composition of the ternary eutectic in this system? (c) Estimate the temperature at which the ternary eutectic reaction occurs.

Solution:

% A % B %C

T liquidus

80–20–0 70–20–10 60–20–20 50–20–30 40–20–40 30–20–50 20–20–60 10–20–70 0–20–80

390C 355°C 300°C 210°C 150°C 210°C 270°C 320°C 400°C

400

   )    C             °    ( 300   e   r   u    t   a   r   e 200   p   m   e    T 100

L

a+L g+L

B =

0

20

40 %C

60

20% 80

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Instructor’s Solution Manual

The Science and Engineering of M aterials

(b) The composition of the ternary eutectic is about 40% 20%  B –40% C , balance  A (c) The ternary eutectic temperature is about 150°C 10–69

From the liquidus plot in Figure 10 –30, prepare a graph of liquidus temperature versus percent  B for a constant ratio of materials  A and C  (that is, from pure  B to 50%  A –50% C  on the liquidus plot). Material  B melts at 600°C.

Solution:

A

B C 

50– 0–50 45–10–45 40–20–40 35–30–35 30–40–30 25–50–25 20–60–20 15–70–15 0–100–0

200°C 180°C 150°C 280°C 330°C 375°C 415°C 485°C 580°C

600 500    )    C             °    ( 400   e   r   u    t 300   a   r   e   p   m200   e    T

L

b + L

a+L

% A = % C 

100

0

20

40

60 % B 

80

100