ch10

CHAPTER 10

PHASE DIAGRAMS

PROBLEM SOLUTIONS

Solubility Limit

10.1 Consider the sugar–water phase diagram of Figure 10.1. (a) How much sugar will will dissolve in 1000 g of water water at 60°C? (b) If the saturated liquid solution in part (a) is cooled to 20°C, some of the sugar will precipitate precipitate out as a solid. What will be the composition composition of the saturated saturated liquid solution (in (in wt% sugar) at 20°C? (c) How much of the solid sugar will come out of solution upon cooling to 20°C? Solution (a) We are asked to determine determine how much sugar will dissolve in in 1000 g of water at 60°C. From the solubility limit curve, at 60 °C the maximum concentration concentration of sugar in the syrup is about 70 wt%. It is now possible to calculate the mass of sugar using Equation 5.6 as

C sugar (wt%) =

70 wt% =

msugar msugar

+

mwater

msugar msugar

+

1000 g

×

100

× 100

Solving for msugar yields msugar = 2330 g (b) Again using this same plot, at 20°C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. (c) The mass of sugar in this saturated solution at at 20°C ( m′ sugar ) may also be calculated using Equation 5.6 as follows:

64 wt% =

m′ sugar m′ sugar

+

1000 g

× 100

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which yields a value for m′ sugar of 1778 g. Subtracting the latter latter from the former former of these sugar concentrations concentrations yields the amount of sugar that precipitated out of the solution upon cooling m′′ sugar ; that is m′′ sugar = m sugar

− m′ sugar =

2330 g

−

1778 g = 552 g


Microstructure

10.2 Cite three variables that determine the microstructure of an alloy. Solution Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.


One-Component (or Unary) Phase Diagrams

10.3 Consider a specimen of ice that is at –15°C and 10 atm pressure. Using Figure 10.2, the pressure– pressure– temperature phase diagram for H 2O, determine the pressure to which the specimen must be raised or lowered to cause it (a) to melt, and (b) to sublime. Solution The figure below shows the pressure-temperature phase diagram for H 2O, Figure 10.2; a vertical line has been constructed at –15°C, and the location on this line at 10 atm pressure (point B (point B)) is also noted.

(a) Melting occurs, (by changing pressure) pressure) as, moving vertically vertically (upward) along this line, line, we cross the Solid-Liquid phase boundary. This occurs at approximately 1,000 atm; thus, the pressure of the specimen must must be raised from 10 to 1,000 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, temperature, we move vertically downward along this line from 10 atm until we cross the Solid-Vapor Solid-Vapor phase boundary. This intersection occurs at approximately 0.003 atm.


Binary Isomorphous Systems

10.4 Given here are the solidus and liquidus temperatures for for the copper–gold system. Construct the phase diagram for this system system and label each region.

Composition (wt% Au)

Solidus Temperature (°C)

Liquidus Temperature (°C)

0

1085

1085

20

1019

1042

40

972

996

60

934

946

80

911

911

90

928

942

95

974

984

100

1064

1064

Solution The copper-gold phase diagram is constructed below.


Interpretation of Phase Diagrams (Binary Isomorphous Systems) (Binary Eutectic Systems) (Equilibrium Diagrams Having Intermediate Phases or Compounds)

10.5 Cite the phases that are present and the phase compositions for the following alloys: (a) 25 wt% Pb–75 Pb–75 wt% Mg at 425°C (b) 55 wt% Zn–45 Zn–45 wt% Cu Cu at 600°C (c) 7.6 lbm Cu and 144.4 lbm Zn at 600°C (d) 4.2 mol Cu and 1.1 mol Ag at 900°C Solution (a) For an alloy composed of 25 wt% Pb-75 wt% wt% Mg and at 425°C, from Figure 10.20, only the

α phase is

present; its composition is is 25 wt% Pb-75 wt% wt% Mg.

(b) For an alloy composed composed of 55 wt% Zn-45 Zn-45 wt% Cu and at 600°C, from Figure 10.19, β and γ phases are present, and

C β = 51 wt% Zn-49 wt% Cu C γ = 58 wt% Zn-42 wt% Cu (c) For an alloy composed composed of 7.6 lbm Cu and 144.4 lbm Zn and at 600°C, we must first determine the Cu and Zn concentrations (using Equation 5.6), as

C Cu =

C Zn =

7.6 lb m 7.6 lbm

+

144.4 lb m

144.4 lbm 7.6 lb m

+

144.4 lbm

×

100 = 5.0 wt%

×

100 = 95.0 wt%

From Figure 10.19, only the L the L phase phase is present; its composition is 95.0 wt% Zn-5.0 Zn-5.0 wt% Cu.

(d) For an alloy composed composed of 4.2 mol Cu and 1.1 mol Ag and at 900°C, it is necessary to determine the Cu and Ag concentrations in weight percent. However, we must first compute compute the masses of Cu and Ag (in grams) grams) using a rearranged form of Equation 5.7:


′ u = nmCu ACu = (4.2 mo mol)(63.55 g/ g /mol) = 266.9 g mC ′ g = nmAg AAg = (1.1 mol)(107.87 g/mol) = 11 118.7 g mA Now, using Equation 5.6, concentrations of Cu and Ag are determined as follows:

C Cu =

C Ag =

266. 266.9 9g 266.9 g

+

118.7 g

118. 118.7 7g 266.9 g

+

118.7 g

× 100 = 69.2 wt%

× 100 = 30.8 wt%

From Figure 10.7, α and liquid phases are present; and C α = 8 wt% Ag-92 w% Cu C L = 45 wt% Ag-55 wt% Cu


10.6 Is it possible to have a copper–silver alloy that, at equilibrium, consists of an α phase of composition 4 wt% Ag–96 wt% Cu, and also a β phase of composition 95 wt% wt% Ag–5 wt% Cu? If so, what will will be the approximate temperature of the alloy? If this is not possible, explain why. Solution It is possible to possible to have a Cu-Ag alloy, which at equilibrium consists of an 96 wt% Cu and a

β phase

α phase of composition 4 wt% Ag-

of composition 95 wt% Ag-5 wt% Cu. From Figure 10.7 a horizontal tie tie can be

constructed across the α + β region at 690 °C which intersects the

α−(α + β) phase boundary at 4 wt% Ag, and also

the (α + β)– β phase boundary at 95 wt% Ag.


10.7 A 40 wt% Ni–60 Ni–60 wt% Cu alloy alloy is slowly cooled from 1400°C (2550°F) to 1150°C (2190°F). (a) At what temperature temperature does the first solid phase form? (b) What is the composition composition of this solid phase? (c) At what temperature temperature does the liquid solidify? solidify? (d) What is the composition composition of this last remaining remaining liquid phase? Solution Shown below is the Cu-Ni phase diagram (Figure 10.3a) and a vertical line constructed at a composition of 40 wt% Ni-60 wt% Cu.

(a)

Upon cooling a 40 wt% Ni-60 Ni-60 wt% Cu alloy form form 1400°C, the first solid phase forms at the

temperature at which this vertical line intersects the L the L –(α + L + L)) phase boundary--i.e., at about 1280 °C. (b) The composition of this solid solid phase corresponds to the intersection intersection with the L the L –(α + L + L)) phase boundary, of a tie line constructed across the

+ L phase phase α + L

region at 1320°C--i.e., C α = 53 wt% Ni-47 wt% Cu.

(c) Complete solidification solidification of the alloy occurs at the intersection intersection of this same same vertical line at 40 wt% Ni with the ( α + L)– L)– α phase boundary--i.e., at about 1230 °C.


(d) The composition of the last liquid liquid phase remaining prior to complete complete solidification solidification corresponds to the intersection with the L the L – (α + L) L) boundary, of the tie line constructed across the

L phase region at 1270°C--i.e., α + L phase

C is about 29 wt% Ni-71 wt% Cu. L


10.8 Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and temperatures given in Problem 10.5. Solution (a) For an alloy composed of 25 wt% Pb-75 Pb-75 wt% Mg and at 425°C, only the α phase is present; therefore W α = 1.0.

(b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600 °C, compositions of the β and γ phases are C β = 51 wt% Zn-49 wt% Cu C γ = 58 wt% Zn-42 wt% Cu And, since the composition of the alloy, C 0 = 55 wt% Zn-45 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent zinc

W β =

W γ =

Cγ Cγ C0 Cγ

− −

C β

− −

C β 55 − 51 = = 0.57 C β 58 − 51

C 0

=

58 − 55 58

− 51

= 0.43

(c) For an alloy composed of 7.6 lb m Cu and 144.4 lb m Zn (95.0 wt% Zn-5.0 wt% Cu) and at 600 °C, only the liquid phase phase is present; therefore, W = 1.0 L

(d) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag (30.8 wt% Ag-69.2 wt% Cu) and at 900 °C, compositions of the α and liquid phases are

C α = 8 wt% Ag-92 w% Cu C L = 45 wt% Ag-55 wt% Cu And, since the composition of the alloy, C 0 = 30.8 wt% Ag-69.2 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent silver

W α =

C L C L

− −

C 0 C α

=

45

−

45

30.8

−

8

= 0.38


W L =

C0 C L

− −

C α 30.8 − 8 = = 0.62 C α 45 − 8


10.9 A magnesium–lead alloy of mass 9.5 kg consists of a solid α phase that has a composition just slightly below the the solubility limit limit at 300°C. (a) What mass of lead is in the alloy? (b) If the alloy is heated to 400°C, 400°C, how much more lead may be dissolved dissolved in the α phase without exceeding the solubility limit of this phase? Solution (a) This portion of the problem asks that we calculate, for a Pb-Mg alloy, alloy, the mass of lead in 9.5 kg of the the solid α phase at 300 °C just below the solubility limit. limit. From the Figure, the solubility solubility limit for the α phase at 300°C corresponds to the position (composition) of the

α-α +

Mg2Pb phase boundary at this temperature, which is about

17 wt% Pb. Therefore, the mass of Pb in the alloy is just (0.17)(9.5 (0.17)(9.5 kg) = 1.615 kg (b) At 400°C, the solubility limit of the α phase increases to to approximately 32 wt% Pb. In order to determine the additional amount of Pb that may be added (m ( mPb ), we utilize a modified form of Equation 5.6 as '

C Pb = 32 wt% =

1.61 1.615 5 kg 9.5 kg

+ mPb’ × 100 + mPb’

Solving for mPb' yields mPb' = 2.096 kg.


10.10 A 40 wt% Pb–60 wt% Mg alloy is heated to a temperature within within the α + liquid-phase region. If the mass fraction of each phase is 0.5, then estimate: (a) The temperature of the alloy (b) The compositions of the two phases Solution (a) We are given that that the mass mass fractions of α and liquid phases are both 0.5 for a 40 wt% Pb-60 wt% Mg alloy and are asked to estimate the temperature temperature of the alloy. Using the appropriate phase diagram, diagram, Figure 10.20, by trial and error with a ruler, a tie line within the

α

+ L L phase region that is divided in half for an alloy of this

composition exists at about 540°C. (b) We are now asked to determine determine the compositions compositions of the the two phases. This is accomplished accomplished by noting the intersections of this tie line with with both the solidus and liquidus lines. From these intersections, intersections, C α = 26 wt% Pb, and C = 54 wt% Pb. L


10.11 For alloys of two hypothetical metals A and B, there exist an α , A-rich phase and a β , B-rich phase. From the mass fractions of both phases for two different alloys (given below), which are at the same temperature, determine the composition of the phase boundary (or solubility limit) for (a) α and (b) β (b) β phases phases at this temperature.

Alloy Composition

Fraction α Phase

Fraction β Fraction β Phase

60 wt% A–40 wt% B

0.56

0.44

30 wt% A–70 wt% B

0.12

0.88

Solution The problem is to solve for compositions at the phase boundaries for both α and β phases (i.e., C α and C β). We may set up two independent lever rule expressions, one for each composition, in terms of C α and C β as follows:

W α1 = 0.56 =

Cβ Cβ

− −

W α 2 = 0.12 =

Cβ Cβ

− C02 − Cα

C01 Cα

=

C β Cβ

− −

Cα

=

C β Cβ

− 30 − Cα

60

In these expressions, compositions compositions are given in wt% wt% of A. Solving for C α and C β from these equations, yield C α = 90 (or 90 wt% A-10 wt% B) C β = 21.8 (or 21.8 wt% A-78.2 wt% B)


10.12 Is it possible to have a copper–silver alloy of composition 20 wt% Ag–80 wt% Cu that, at equilibrium, consists of α and liquid phases having mass fractions W α = 0.80 and W L = 0.20? If so, what will be the the approximate temperature of the alloy? If such an alloy is not possible, explain why. Solution Yes, it is possible to have a Cu-Ag alloy of composition 20 wt% Ag-80 wt% Cu which consists of mass fractions W α = 0.80 and W L = 0.20. Using the appropriate phase diagram, diagram, Figure 10.7, by trial and error with a ruler, the tie-line segments within the

+ L phase phase region are proportioned such that α + L

W α = 0.8

=

C L C L

− −

C 0 C α

for C 0 = 20 wt% Ag. This occurs at about 800°C.


10.13 For 5.7 kg of a magnesium–lead alloy of composition 50 wt% Pb–50 wt% Mg, is it possible, at equilibrium, to have α and Mg 2 Pb phases with respective masses of 5.13 and 0.57 kg? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, then explain why. Solution It is not possible to possible to have a 50 wt% Pb-50 wt% Mg alloy that has masses of 5.13 kg and 0.57 kg for the

α

and Mg2Pb phases, respectively. In order to demonstrate this, it is first necessary to determine the mass fraction fraction of each phase as:

W α =

mα 5.13 kg = = 0.90 mα + mMg 2Pb 5.13 kg + 0.57 kg

W Mg 2Pb = 1.00 .00

−

0.90 0.90 = 0.10 0.10

Now, if we apply the lever rule expression for W α

W α =

CMg 2 Pb CMg 2Pb

− −

C 0 C α

Since the Mg2Pb phase exists only at 81 wt% Pb, and C 0 = 50 wt% Pb

W α = 0.90 0.90 =

81 81

− 50 − C α

Solving for C α from this expression yields C α = 46.6 wt% Pb. From Figure Figure 10.20, the maximum concentration of Pb in the α phase in the α + Mg2Pb phase field is about about 42 wt% Pb. Therefore, this alloy alloy is not possible. possible.


10.14 Determine the relative amounts (in terms of volume fractions) of the phases for the alloys and temperatures given in Problems 10.5a and b. Given here are the approximate densities densities of the various metals at the the alloy temperatures:

Metal

Temperature (°C)

Density 3 (g/cm )

Cu

600

8.68

Mg

425

1.68

Pb

425

10.96

Zn

600

6.67

Solution This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problems 10.5a and b. This is accomplished by using the technique illustrated illustrated in Example Problem Problem 10.3, and also the results of Problems 10.5 and 10.8. (a) This is a Pb-Mg alloy at 425°C, wherein only the α phase is present. Therefore, V α = 1.0.

(b) This is a Zn-Cu alloy at 600°C, wherein

C β = 51 wt% Zn-49 wt% Cu C γ = 58 wt% Zn-42 wt% Cu W β = 0.43 W γ = 0.57

ρZn = 6.67 g/cm3 ρCu = 8.68 g/cm3 Using these data it is first necessary to compute the densities of the β and γ phases using Equation 5.13a. Thus

ρβ

=

100 C Zn (β)

ρZn

+

C Cu (β)

ρCu


100

=

51

+

6.67 g / cm cm3

ργ

=

= 7.52 g/cm3

49 8.68 g / cm3

100 C Zn (γ )

ρZn

+

C Cu ( γ)

ρCu

100

=

58 6.67 g / cm cm3

+

= 7.39 g/cm3

42 8.68 g / cm3

Now we may determine determine the V β and V γ values using Equation 10.6. Thus, W β V β = Wβ

ρβ

ρβ +

W γ

ργ

0.43 =

7.5 7.52 g /cm3 0.43 0.57 7.52 g / cm3

+

= 0.43

7.39 g / cm cm3

W γ V γ = Wβ

ρβ

ργ +

W γ

ργ

0.57 =

7.3 7.39 g /cm3 0.43 0.57 7.52 g / cm3

+

= 0.57

7.39 g / cm cm3


Mechanical Properties of Isomorphous Alloys

10.15 It is desirable to produce a copper–nickel alloy that has a minimum noncold-worked noncold-worked tensile strength of 380 MPa (55,000 psi) and a ductility of at least 45%EL. Is such an alloy possible? If so, what must be its composition? If this is not possible, then explain why. why. Solution From Figure 10.6a, a tensile strength greater than 380 MPa is possible for compositions between about 32 and 90 wt% Ni.

On the other hand, according to Figure 10.6b, 10.6b, ductilities greater than 45%EL 45%EL exist for

compositions less than about 13 wt% and greater than about 94 wt% wt% Ni. Therefore, such an alloy is not possible inasmuch, that in order to meet the stipulated criteria: For a TS > TS > 380 MPa

32 wt% < C < 90 wt% Ni

For %EL > 45%

C < 13 wt% or C > 94 wt% Ni Ni


Development of Microstructure in Eutectic Alloys

10.16 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. Solution Upon solidification, an alloy of eutectic composition forms a microstructure consisting consisting of alternating layers of the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.


10.17 Is it possible to have a magnesium–lead alloy in which the mass fractions of primary α and total α are 0.60 and 0.85, respectively, respectively, at 460°C? Why or why not? Solution In order to make this determination, we need to set up the appropriate lever rule expression for each of these quantities. From Figure 10.20 and at 460°C, C α = 41 wt% Pb, C Mg 2Pb = 81 wt% Pb, and C eutectic = 67 wt% Pb. For primary α

W α' =

Ceutectic − C0 Ceutectic − C α

=

− C 0 67 − 41

67

= 0.60

Solving for C 0 gives C 0 = 51.4 wt% Pb. Now the analogous expression for total α

W α =

CMg 2Pb CMg 2Pb

− C 0 − C α

=

81 − C 0 81

−

41

= 0.85

which yields a value of 47 wt% Pb for C 0. Therefore, since these two C 0 values are different, this alloy is not possible. possible.


10.18 For a lead–tin alloy of composition 78.0 wt% Sn–22 wt% Pb and at 180°C do the following: (a) Determine the mass fraction of α phase. (b) Determine the mass fraction of β of β phase. phase. (c) Determine the mass mass fraction of primary β primary β microconstituent. microconstituent. (d) Determine the mass fraction fraction of eutectic microconstituent. (e) Determine the mass mass fraction of eutectic eutectic β β . Solution (a & b) This portion of the problem problem asks that we determine the mass mass fractions of

α and β phases for an

78 wt% Sn-22 wt% Pb alloy (at 180 °C). In order to do this, it is necessary to to employ the lever rule using a tie line line From Figure Figure and at 180°C, C α = 18.3 wt% Sn, C β = 97.8 wt% Sn, and C eutectic = 61.9 wt% Sn. Therefore, the two lever-rule expressions expressions are as follows: that extends entirely across the

α + β phase field.

(a)

W α =

(b)

W β =

Cβ Cβ

− −

C α

C0

− −

C α 78 − 18.3 = = 0.751 C α 97.8 − 18.3

Cβ

C 0

=

97.8 97.8

− 78 − 18.3

= 0.249

(c & d) Now it is necessary to determine the mass mass fractions of primary β and eutectic microconstituents for this same alloy. This requires that we utilize utilize the lever rule and a tie line that extends from the maximum maximum solubility solubility of Pb in the β phase at 180°C (i.e., 97.8 wt% Sn) to the eutectic composition composition (61.9 wt% Sn). Thus

(c)

W β ' =

(d)

W e =

C0 − C eutectic Cβ − C eutectic Cβ Cβ

− C 0

− C eutectic

=

=

− 61.9 97.8 − 61.9 78

97.8 97.8

− 78 − 61.9

= 0.448

= 0.552

(e) And, finally, we are asked to compute compute the mass fraction fraction of eutectic β, W eβ. This quantity quantity is simply simply the difference between the mass fractions of total β and primary β as W eβ = W β – W β' = 0.751 – 0.448 = 0.303


10.19 Consider the hypothetical eutectic phase diagram for metals A and B, which is similar to that for the lead–tin system. system. Assume that: that: (1) α and β and β phases phases exist at the A and B extremities of the phase diagram, respectively; (2) the eutectic composition composition is 36 wt% A–64 wt% B; and (3) the composition of the α phase at the eutectic temperature is 88 wt% A–12 wt% B. Determine the composition composition of an alloy that will yield primary primary β and and total β mass fractions of 0.376 and 0.761, respectively. Solution We are given a hypothetical eutectic phase diagram for which C eutectic = 64 wt% B, C α = 12 wt% B at the eutectic temperature, and also that W β' = 0.376 and W β = 0.761; from this this we are are asked to to determine determine the composition of the alloy. alloy. Let us write lever rule expressions for W β' and W β

W β =

W β' =

C0 − Cα C = 0 Cβ − C α Cβ

C0 Cβ

− Ceutectic − C eutectic

=

− 12 − 12

= 0.761

C 0 − 64 Cβ − 64

= 0.376

Thus, we have two simultaneous equations with C 0 and C β as unknowns. Solving them for C 0 gives C 0 = 76.1 wt% B.


10.20 For a 64 wt% Zn–36 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 900°C, 820°C, 750°C, and 600°C. Label all phases and indicate indicate their approximate compositions. compositions. Solution The illustration below is the Cu-Zn Cu-Zn phase diagram (Figure 10.19). A vertical line at a composition composition of 64 wt% Zn-36 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement statement (i.e., 900°C, 820°C, 750°C, and 600°C).

On the basis of the locations of the four temperature-composition points, schematic sketches of the four respective microstructures along with phase compositions are represented as follows:



10.21 For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow slow cooling at the following following temperatures: 950°C, 860°C, 800°C, and 600°C. Label all phases and indicate indicate their approximate compositions. compositions. Solution The illustration below is the Cu-Zn Cu-Zn phase diagram (Figure 10.19). A vertical line at a composition composition of 52 wt% Zn-48 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement statement (i.e., 950°C, 860°C, 800°C, and 600°C).

On the basis of the locations of the four temperature-composition points, schematic sketches of the four respective microstructures along with phase compositions are represented as follows:



10.22 The room-temperature tensile strengths of pure copper and pure silver are 209 MPa and 125 MPa, respectively. (a)

Make a schematic graph of the room-temperature room-temperature tensile tensile strength versus versus composition for for all

compositions between pure copper and pure silver. silver. (Hint: You may want to consult Sections 10.10 and 10.11, as well as Equation 10.24 in Problem 10.36.) (b) On this same graph, schematically plot tensile tensile strength versus composition at 600°C. (c) Explain the shapes of these these two curves, as well as any differences differences between them. Solution The (a) and (b) portions of the problem ask that we make schematic plots on the same graph for the tensile strength versus composition for copper-silver alloys at both room temperature and 600 °C; such a graph is shown below.

(c) Upon consultation of the Cu-Ag phase diagram (Figure 10.7) we note that, at room temperature, temperature, silver is virtually insoluble in copper (i.e., there is no

diagram); α-phase region at the left extremity of the phase diagram);

may be said the solubility of copper in silver and for the

β phase.

Thus, only the

the same

α and β phase will exist for all

compositions at room temperature; temperature; in other words, there will be no solid-solution strengthening strengthening effects at room temperature. All other things being equal, the tensile tensile strength will depend (approximately) (approximately) on the tensile strengths of each of the α and β phases as well as their phase fractions in a manner described by Equation 10.24 for the elastic modulus (Problem 10.36). That is, for this problem


(TS )alloy

≅

(TS )αVα + (TS )βVβ

in which TS and V denote tensile strength and volume fraction, respectively, and the subscripts represent the alloy/phases. Also, mass mass fractions of the α and β phases change linearly with changing composition (according to the lever rule). Furthermore, inasmuch inasmuch as the densities of both Cu and Ag are similar, weight and volume fractions of the

similar (see Equation Equation 10.6). α and β phases will also be similar

In summary, the the previous discussion explains the

linear dependence of the room temperature tensile strength on composition as represented in the above plot given that the TS of of pure copper is greater than for pure silver (as stipulated in the problem statement). At 600°C, the curve will be shifted to significantly lower tensile strengths inasmuch as tensile strength diminishes with increasing increasing temperature (Section (Section 7.6, Figure 7.14). In addition, according to Figure 10.7, about 4 wt% of silver will dissolve in copper (i.e., in the

α phase), and about 4 wt% of copper will dissolve in silver (i.e., in

the β phase). Therefore, solid-solution solid-solution strengthening will occur occur over these compositions ranges, ranges, as noted in the graph shown above. Furthermore, between 4% Ag and 96% Ag, the curve will be approximately approximately linear for the same reasons noted in the previous paragraph.


Equilibrium Diagrams Having Intermediate Phases or Compounds

10.23 Two intermetallic compounds, A 3 B and AB3 , exist for elements A and B. If the compositions for A3 B and AB3 are 91.0 wt% A–9.0 wt% B and 53.0 wt% A–47.0 wt% B, respectively, and element A is zirconium, identify element B. Solution Probably the easiest way to solve this problem is to first compute the ratio of the atomic weights of these two elements using Equation 5.9a; then, since we know the atomic weight of zirconium (91.22 g/mol), it is possible to determine the atomic weight of element B, from which an identification may be made. First of all, consider the A 3B intermetallic intermetallic compound; inasmuch as it contains three times times the number of A atoms than and B atoms, its composition in atomic atomic percent is 75 at% A-25 at% B. Equation 5.9a may be written in the form:

′ = 25 at% = C B

CB AA CA AB

+

CB AA

× 100

where A where AA and A and AB are the atomic weights for elements A and B, and C A and C B are their compositions in weight percent. For this A3B compound, and making the appropriate substitutions in the above equation leads to

25 at % B =

(9.0 wt % B)( AA ) (91.0 wt % A )( AB )

+

(9.0 wt % B) B )( AA )

× 100

Now, solving this expression expression yields, AB = 0.297 A 0.297 AA

Since zirconium is element A and it has an atomic weight of 91.22 g/mol, the atomic weight of element B is just AB = (0.297)(91.22 g/mol) = 27.09 g/mol

Upon consultation of the period table of the elements (Figure 2.6) we note the element that has an atomic weight closest to this value is aluminum aluminum (26.98 g/mol). Therefore, element element B is aluminum, and the two two intermetallic intermetallic compounds are Zr 3Al and ZrAl 3.


Congruent Phase Transformations Binary Eutectic Systems Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions

10.24 Figure 10.40 is the tin–gold phase diagram, for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write write the reaction upon cooling. cooling. Solution Below is shown the tin-gold phase diagram (Figure 10.40).

There are two eutectics eutectics on this phase diagram. One exists at 10 wt% Au-90 wt% Sn and 217°C. The reaction upon cooling is

L

→ α +β

The other eutectic exists at 80 wt% Au-20 wt% Sn and 280°C. This reaction reaction upon cooling is

L

→ δ +ζ


There are three peritectics. peritectics. One exists at 30 wt% wt% Au-70 wt% Sn and 252 °C. Its reaction reaction upon cooling cooling is as follows: L + γ

→ β

The second peritectic exists at 45 wt% Au-55 wt% Sn and 309 °C. This reaction reaction upon cooling cooling is L + δ

→ γ

The third peritectic exists at 92 wt% Au-8 wt% Sn and 490 °C. This reaction reaction upon cooling cooling is L + η

→ ζ

There is one congruent melting point at 62.5 wt% Au-37.5 wt% Sn and 418 °C. Its reaction upon cooling is

L

→ δ

No eutectoids are present. present.


10.25 Construct the hypothetical phase diagram for metals A and B between room temperature 20°C and 700°C given the following information: ●

The melting temperature of metal A is 480°C.

●

The maximum solubility of B in A is 4 wt% B, which occurs at 420°C.

●

The solubility of B in A at room temperature is 0 wt% B.

●

One eutectic occurs at 420°C and 18 wt% B–82 wt% A.

●

A second eutectic occurs at 475°C and 42 wt% B–58 wt% A.

●

The intermetallic compound AB exists at a composition of 30 wt% B–70 wt% A, and melts

congruently at 525°C. ●

The melting temperature of metal B is 600°C.

●

The maximum solubility of A in B is 13 wt% A, which occurs at 475°C.

●

The solubility of A in B at room temperature is 3 wt% A. Solution

Below is shown the phase diagram for these two A and B metals.


Ceramic Phase Diagrams 10.26 From Figure 10.24, the phase diagram for the MgO–Al 2O3 system, it may be noted that the spinel solid solution exists over a range of compositions, which means that it is nonstoichiometric at compositions other than 50 mol% MgO–50 mol% Al 2O3. (a) The maximum nonstoichiometry on the Al 2O3-rich side of the spinel phase field exists at about 2000°C corresponding to approximately 82 mol% (92 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. (b) The maximum nonstoichiometry on the MgO-rich side of the spinel phase field exists at about 2000°C corresponding to approximately 39 mol% (62 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. Solution (a) For this portion of the problem, we are to determine the the type of vacancy defect that is produced on the Al2O3-rich side of the spinel phase field (Figure 10.24) and the percentage of these vacancies at the maximum nonstoichiometry (82 mol% Al2O3). On the alumina-rich side of this this phase field, there is an excess of Al3+ ions, which means that some of the Al 3+ ions substitute for Mg2+ ions. In order to maintain charge charge neutrality, Mg2+ 3+

vacancies are formed, and for every Mg 2+ vacancy formed, two Al ions substitute for three Mg 2+ ions. Now, we will calculate the percentage of Mg2+ vacancies that exist at 82 mol% Al2O3. Let us arbitrarily arbitrarily choose as our basis 50 MgO-Al2O3 units of the stoichiometric material, which consists of 50 Mg 2+ ions and 100 Al3+ ions. Furthermore, let let us designate the number of Mg2+ vacancies as x as x,, which means that 2 x Al x Al 3+ ions have been added and 3 x Mg2+ ions have been removed (two of which are filled with Al 3+ ions). Using our 50 MgOAl2O3 unit basis, the number of moles of Al 2O3 in the nonstoichiometric material is (100 + 2 x)/2; x)/2; similarly the number of moles of MgO is (50 – 3 x). x). Thus, the expression for the the mol% of Al2O3 is just

mol% Al2O3 =

100 + 2 x ⎡ ⎢ 2 ⎢ 100 + 2 x ⎢ + (50 − ⎢⎣ 2

⎤ ⎥ ⎥ × 3 x ) ⎥ ⎥⎦

100

If we solve for x for x when when the mol% of Al2O3 = 82, then x then x = = 12.1. Thus, adding 2 x or x or (2)(12.1) = 24.2 Al 3+ ions to the original material consisting of 100 Al 3+ and 50 Mg2+ ions will produce 12.1 Mg2+ vacancies. Therefore, the percentage of vacancies is just

% vacancies =

12.1 100

+

50

× 100

= 8.1%


(b) Now, we are asked to make the same determinations determinations for the MgO-rich side of the spinel phase field, for 39 mol% Al 2O3. In this this case, Mg2+ ions are substituting for Al 3+ ions. Since the Mg2+ ion has a lower charge than the Al3+ ion, in order to maintain charge neutrality, negative charges must be eliminated, which may be accomplished by introducing O2– vacancies. For every 2 Mg2+ ions that substitute for 2 Al 3+ ions, one O2 – vacancy is formed. Now, we will calculate the percentage of O2– vacancies that exist at 39 mol% Al 2O3. Let us arbitrarily choose as our basis 50 MgO-Al 2O3 units of the stoichiometric material which consists of 50 Mg 2+ ions 100 Al 3+ ions. Furthermore, let let us designate the number of O2– vacancies as y, y, which means that 2 y Mg2+ ions have been added and 2 y Al y Al3+ ions have been removed. Using our 50 MgO-Al2O3 unit basis, the number of moles of Al 2O3 in the nonstoichiometric material is (100 – 2 y 2 y)/2; )/2; similarly the the number of moles of MgO is (50 + 2 y). y). Thus, the expression for the mol% of Al 2O3 is just

mol% Al2O3 =

⎡ ⎢ ⎢ 100 − ⎢ 2 ⎣⎢

100

−

2 y

2 2 y

+

(50

⎤ ⎥ ⎥ × 100 + 2 y) ⎥ ⎦⎥

If we solve for y y when the mol% of Al2O3 = 39, then y y = 7.91. Thus, 7.91 7.91 O2– vacancies are produced in the original material that had 200 O2– ions. Therefore, the percentage percentage of vacancies is just

% vacancies =

7.91 200

× 100

= 3.96%


The Gibbs Phase Rule

10.27 In Figure 10.41 is shown the pressure–temperature phase diagram for H 2O. Apply the the Gibbs phase rule at points A, B, and C; that is, specify the number of degrees of freedom at each of the points—that is, the number of externally controllable variables that need be specified to completely define the system. Solution We are asked to specify the value of F for Gibbs phase rule at points A, B, and C on the pressuretemperature diagram for H2O, Figure 10.42, which is shown below.

Gibbs phase rule in general form is

P + F + F = = C + N + N

For this system, the number of components, C , is 1, whereas N whereas N , the number of noncompositional variables, is 2--viz. temperature and pressure. pressure. Thus, the phase rule now becomes

P + F + F = = 1 + 2 = 3 Or F = = 3 – P – P

where P where P is is the number of phases present at equilibrium. At point A, only a single (liquid) phase is present (i.e., P (i.e., P = = 1), or


F = = 3 – P – P = = 3 – 1 = 2

which means that both temperature and pressure are necessary to define the system. At point B which is on the phase boundary between liquid and vapor phases, two phases are in equilibrium ( P P = = 2); hence

F = = 3 – P – P = = 3 – 2 = 1

Or that we need to specify the value of either temperature or pressure, which determines the value of the other (pressure or temperature). And, finally, at point C, three phases are present—viz. ice I, vapor, and liquid—and the number of degrees of freedom is zero since

F = = 3 – P – P = = 3 – 3 = 0

Thus, point C is an invariant point (in this case a triple point), and we have no choice in the selection of externally controllable variables in order to define the system.


The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram Development of Microstructure in Iron-Carbon Alloys

10.28 What is the carbon concentration of an iron–carbon alloy for which the fraction of total ferrite 0.95? Solution This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.95. Application of the lever rule (of the form of Equation Equation 10.12) yields

W Fe3C = 0.95 =

− C Fe C −

C Fe3C 3

C 0′ C α

=

6.7 6.70

− C 0′ − 0.022

and solving for C 0′ C 0′

= 0.356 wt% C


10.29 Consider 1.1 kg of austenite containing 0.20 wt% C, cooled to below 727°C. (a) How many kilograms of total ferrite form? form? (b) How many kilograms kilograms of cementite cementite form? (c) How many kilograms of pearlite phase form? (d) How many kilograms of proeutectoid phase form? Solution This problem asks us to consider various aspects of 1.1 kg of austenite containing 1.2 wt% C that is cooled to below the eutectoid. (a & b) For this portion of the problem, we are asked to determine how much total total ferrite and cementite form. Application of the appropriate appropriate lever rule expression expression yields

(a)

W α =

CFe3C CFe3C

− −

C 0 C α

−

6.70

=

6.70

−

1.2

0.022

= 0.824

which, when multiplied by the total mass of the alloy, gives (0.824)(1.1 kg) = 0.906 kg of total ferrite. Similarly, for total cementite,

(b)

W Fe3C =

C0

−

CFe3C

C α 1.2 = − C α 6.70

− −

0.022 0.022

= 0.176

And the mass of total cementite that forms is (0.176)(1.1 kg) = 0.194 kg. (c & d) Now we are asked to calculate how much much pearlite and the proeutectoid phase (cementite) (cementite) form. Applying Equation 10.22, in which C 1′ = 1.2 wt% C

(c)

W p =

6.70 6.70

− C 1′ − 0.76

=

6.70 6.70

− 1.2 − 0.76

= 0.926

which corresponds to a mass of (0.926)(1.1 kg) = 1.02 kg. Likewise, from from Equation 10.21

(d)

W Fe3C' =

C 1′

−

0.76

5.94

=

1.2

−

0.76

5.94

= 0.074

which is equivalent to (0.074)(1.1 kg) = 0.0815 kg of the total 1.1 k g mass.


10.30 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy containing 0.23 wt% C. Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0.23 wt% C iron-carbon alloy are considered in this problem. From Equation 10.20

W p =

C 0′

−

0.022

0.74

=

0.23

−

0.022

0.74

= 0.281

And, from Equation 10.21 (for proeutectoid ferrite)

W α' =

0.76

−

0.74

C 0′

=

0.76

−

0.23

0.74

= 0.716


10.31 The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.866 and 0.134, respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why? Solution In this problem we are given values of W α and W Fe C for an iron-carbon alloy (0.866 and 0.134, 3 respectively), and then are asked to specify whether the alloy is hypoeutectoid hypoeutectoid or hypereutectoid. Employment of the lever rule for total α leads to

W α = 0.866 =

CFe3C CFe3C

− −

C 0 C α

=

− C 0 − 0.022

6.70 6.70

Now, solving for C 0, the alloy composition, leads to C 0 = 0.92 wt% C. Therefore, the alloy alloy is hypereutectoid since C is greater than 0.76 wt% C. 0


10.32 Consider 2.9 kg of a 99.5 wt% Fe–0.5 wt% C alloy that is cooled to a temperature just below the eutectoid. (a) How many kilograms of proeutectoid ferrite ferrite form? (b) How many kilograms of eutectoid ferrite ferrite form? (c) How many kilograms of cementite form? Solution In the problem, we are asked to consider 2.9 kg of a 99.5 wt% Fe-0.5 wt% C alloy that is cooled to a temperature below the eutectoid. (a) Equation 10.21 must be used in computing the amount of proeutectoid ferrite ferrite that forms. forms. Thus,

W α' =

−

0.76

C 0′

0.74

=

0.76

−

0.50

0.74

= 0.351

Or, (0.351)(2.9 kg) = 1.02 kg of proeutectoid ferrite forms. (b) In order to determine the amount of eutectoid ferrite, it first becomes becomes necessary to compute the amount amount of total ferrite using the lever rule applied entirely across the

W α =

CFe3C CFe3C

− −

C 0′ C α

=

α + Fe3C phase field, as

6.70 6.70

− 0.50 − 0.022

= 0.928

which corresponds to (0.928)(2.9 kg) = 2.692 kg. Now, the amount of eutectoid ferrite ferrite is just the difference between total and proeutectoid proeutectoid ferrites, or

2.692 kg – 1.02 kg = 1.67 kg

(c) With regard to the amount of cementite that forms, again application application of the lever rule across the entirety of the α + Fe3C phase field, leads to

W Fe3C =

C0′

−

CFe3C

C α 0.5 − 0.022 = = 0.072 − C α 6.70 − 0.022

which amounts to (0.072)(2.9 kg) = 0.208 kg cementite in the alloy.


10.33 Is it possible to have an iron–carbon alloy for which the mass fractions of total cementite and proeutectoid ferrite are 0.057 and 0.484, respectively? respectively? Solution This problem asks if it is possible to have an iron-carbon alloy for which W Fe Fe3C = 0.057 and W α' = 0.484. In order to make this determination, it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition, composition, then to solve for the alloy composition composition of each; if both alloy composition values values are equal, then such an alloy is possible. The lever-rule expression for the mass fraction of total cementite is

W Fe3C =

C0 − Cα C 0 − 0.022 = = 0.057 CFe3C − C α 6.70 − 0.022

Solving for this C 0 yields C 0 = 0.40 wt% wt% C. Now for W α' we utilize Equation 10.21 as

W α' =

0.76

− C 0′

0.74

= 0.484

This expression leads to C 0′ = 0.40 wt% wt% C. And, since C 0 and C 0′ are the same, this alloy is is possible. possible.


10.34 Compute the mass fraction of eutectoid cementite in an iron–carbon alloy that contains 0.84 wt% C. Solution This problem asks that we compute the mass fraction of eutectoid cementite in an iron-carbon alloy that contains 0.84 wt% C. In order to solve this problem it is necessary to compute mass fractions of total and proeutectoid cementites, and then to subtract the latter from the former. To calculate the mass fraction of total cementite, it is necessary to use the lever rule and a tie line that extends across the entire α + Fe3C phase field as

W Fe3C =

C0 − C α 0.84 = CFe3C − C α 6.70

− −

0.022 0.022

= 0.122

Now, for the mass mass fraction of proeutectoid proeutectoid cementite we use Equation 10.21

W Fe3C' =

C 1′ − 0.76

=

5.94

And, finally, the mass fraction of eutectoid cementite W Fe

0.84

−

0.76

5.94

= 0.013

is just

3C"

W Fe

3C''

= W Fe C – W Fe 3

3C'

= 0.122 – 0.013 = 0.109


10.35

The mass fraction of eutectoid ferrite in an iron–carbon alloy alloy is 0.71. On the basis of this

information, is it possible possible to determine the composition composition of the alloy? If so, what is its its composition? If this is not possible, explain why. Solution Yes, it is possible possible to determine the alloy composition; composition; and, in fact, there there are two possible answers. answers. For the first, the eutectoid ferrite exists in addition to proeutectoid ferrite (for a hypoeutectoid alloy). For this case the mass fraction of eutectoid ferrite (W ( W α'') is just the difference between total ferrite and proeutectoid ferrite mass fractions; that is W α'' = W α – W α' Now, it is possible to write expressions for W α (of the form of Equation 10.12) and W α' (Equation 10.21) in terms of C 0, the alloy composition. composition. Thus,

Wα" =

=

CFe3C CFe3C

− −

C 0 C α

− C0 − − 0.022

6.70 6.70

0.76

−

−

C 0

0.74

−

0.76

0.74

C 0

= 0.71

And, solving for C 0 yields C 0 = 0.61 wt% C. For the second possibility, we have a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite. Thus, it is necessary to set up a lever rule expression wherein wherein the mass fraction of total ferrite ferrite is 0.71. Therefore,

W α =

CFe3C CFe3C

− −

C 0 C α

=

− C 0 − 0.022

6.70 6.70

= 0.71

And, solving for C 0 yields C 0 = 1.96 wt% C.


10.36 Often, the properties of multiphase alloys may be approximated by the relationship

(alloy) = E = E αV α + E + E β V β E (alloy)

(10.24)

where E represents a specific property (modulus of elasticity, hardness, etc.), and V is the volume fraction. The subscripts α and β denote the existing phases or microconstituents. microconstituents. Employ this relationship relationship to determine determine the approximate Brinell hardness of a 99.8 wt% wt% Fe–0.2 wt% C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volume fractions may may be approximated by mass fractions. Solution This problem asks that we determine the approximate Brinell hardness of a 99.8 wt% Fe-0.2 wt% C alloy, using a relationship similar similar to Equation 10.24. First, we compute compute the mass fractions fractions of pearlite and proeutectoid ferrite using Equations 10.20 and 10.21, as

W p =

C 0′

W α' =

−

0.022

0.74 0.76

−

0.74

C 0′

=

=

0.20

−

0.022

0.74 0.76

−

0.20

0.74

= 0.241

= 0.757

Now, we compute the Brinell hardness of the alloy using a modified modified form of the Equation as HBalloy = HBα’Wα’ + HBpW p

= (80)(0.757) + (280)(0.241) = 128


The Influence of Other Alloying Elements

10.37 A steel alloy is is known to contain 93.65 wt% wt% Fe, 6.0 wt% wt% Mn, and 0.35 wt% C. (a) What is the approximate eutectoid temperature of this alloy? (b) What is the proeutectoid phase when this alloy is cooled to a temperature just below the eutectoid? (c) Compute the relative relative amounts of the proeutectoid phase and pearlite. pearlite. Assume that there are no alterations in the positions of other phase boundaries with the addition of Mn. Solution (a) From Figure 10.38, the eutectoid eutectoid temperature for 6.0 wt% Mn is approximately approximately 700°C. (b)

From Figure Figure 10.39, the eutectoid eutectoid composition composition is approximately 0.44 wt% C.

Since the the carbon

concentration in the alloy (0.35 wt%) is less than the eutectoid (0.44 wt% C), the proeutectoid phase is ferrite. (c) Assume that the α –(α + Fe3C) phase boundary is at a negligible negligible carbon concentration. concentration. Modifying Equation 10.21 leads to

W α' =

− C 0′ 0.44 − 0

0.44

=

0.44

−

0.35

0.44

= 0.20

Likewise, using a modified Equation 10.20

W p =

C 0′

− 0 0.44 − 0

=

0.35 0.44

= 0.80


ch10

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