Home-work #9_solutions

9.2 At 500C (930F), what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?

Solution

(a) From Figure 9.7, the maximum solubility of Cu in Ag at 500C corresponds to the position of

the –( + ) phase boundary at this temperature, or to about 2 wt% Cu.

(b) From this same figure, the maximum solubility of Ag in Cu corresponds to the position of the

–( + ) phase boundary at this temperature, or about 1.5 wt% Ag.

9.5 Consider a specimen of ice that is at 210C and 1 atm pressure. Using Figure 9.2, the pressure–

temperature phase diagram for H2O, 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 H2O, Figure 10.2; a vertical

line has been constructed at -10C, and the location on this line at 1 atm pressure (point B) is also noted.

(a) Melting occurs, (by changing pressure) as, moving vertically (upward) at this temperature, we

cross the Ice-Liquid phase boundary. This occurs at approximately 570 atm; thus, the pressure of the

specimen must be raised from 1 to 570 atm.

(b) In order to determine the pressure at which sublimation occurs at this temperature, we move

vertically downward from 1 atm until we cross the Ice-Vapor phase boundary. This intersection occurs at

approximately 0.0023 atm.

9.8 Cite the phases that are present and the phase compositions for the following alloys:

(a) 90 wt% Zn-10 wt% Cu at 400C (750F)

(b) 75 wt% Sn-25 wt% Pb at 175C (345F)

(c) 55 wt% Ag-45 wt% Cu at 900C (1650F)

(d) 30 wt% Pb-70 wt% Mg at 425C (795F)

(e) 2.12 kg Zn and 1.88 kg Cu at 500C (930F)

(f) 37 lbm Pb and 6.5 lbm Mg at 400C (750F)

(g) 8.2 mol Ni and 4.3 mol Cu at 1250C (2280F)

(h) 4.5 mol Sn and 0.45 mol Pb at 200C (390F)

Solution

This problem asks that we cite the phase or phases present for several alloys at specified

temperatures.

(a) That portion of the Cu-Zn phase diagram (Figure 9.19) that pertains to this problem is shown

below; the point labeled ―A‖ represents the 90 wt% Zn-10 wt% Cu composition at 400C.

As may be noted, point A lies within the and phase field. A tie line has been constructed at 400C; its

intersection with the + phase boundary is at 87 wt% Zn, which corresponds to the composition of the

phase. Similarly, the tie-line intersection with the + phase boundary occurs at 97 wt% Zn, which is

the composition of the phase. Thus, the phase compositions are as follows:

C = 87 wt% Zn-13 wt% Cu

C = 97 wt% Zn-3 wt% Cu

(b) That portion of the Pb-Sn phase diagram (Figure 9.8) that pertains to this problem is shown

below; the point labeled ―B‖ represents the 75 wt% Sn-25 wt% Pb composition at 175C.

As may be noted, point B lies within the + phase field. A tie line has been constructed at 175C; its

intersection with the + phase boundary is at 16 wt% Sn, which corresponds to the composition of the

phase. Similarly, the tie-line intersection with the + phase boundary occurs at 97 wt% Sn, which

is the composition of the phase. Thus, the phase compositions are as follows:

C = 16 wt% Sn-84 wt% Pb

C = 97 wt% Sn-3 wt% Pb

(c) The Ag-Cu phase diagram (Figure 9.7) is shown below; the point labeled ―C‖ represents the

55 wt% Ag-45 wt% Cu composition at 900C.

As may be noted, point C lies within the Liquid phase field. Therefore, only the liquid phase is present; its

composition is 55 wt% Ag-45 wt% Cu.

(d) The Mg-Pb phase diagram (Figure 9.20) is shown below; the point labeled ―D‖ represents the

30 wt% Pb-70 wt% Mg composition at 425C.

As may be noted, point D lies within the phase field. Therefore, only the phase is present; its

composition is 30 wt% Pb-70 wt% Mg.

(e) For an alloy composed of 2.12 kg Zn and 1.88 kg Cu and at 500C, we must first determine

the Zn and Cu concentrations, as

CZn 2.12 kg

2.12 kg 1.88 kg 100 53 wt%

CCu 1.88 kg

2.12 kg 1.88 kg 100 47 wt%

That portion of the Cu-Zn phase diagram (Figure 9.19) that pertains to this problem is shown below; the

point labeled ―E‖ represents the 53 wt% Zn-47 wt% Cu composition at 500C.

As may be noted, point E lies within the + phase field. A tie line has been constructed at 500C; its

intersection with the + phase boundary is at 49 wt% Zn, which corresponds to the composition of the

phase. Similarly, the tie-line intersection with the + phase boundary occurs at 58 wt% Zn, which is

the composition of the phase. Thus, the phase compositions are as follows:

C = 49 wt% Zn-51 wt% Cu

C = 58 wt% Zn-42 wt% Cu

(f) For an alloy composed of 37 lbm Pb and 6.5 lbm Mg and at 400C, we must first determine the

Pb and Mg concentrations, as

CPb 37 lbm

37 lbm 6.5 lbm

100 85 wt%

CMg 6.5 lbm

37 lbm 6.5 lbm

100 15 wt%

That portion of the Mg-Pb phase diagram (Figure 9.20) that pertains to this problem is shown below; the

point labeled ―F‖ represents the 85 wt% Pb-15 wt% Mg composition at 400C.

As may be noted, point F lies within the L + Mg2Pb phase field. A tie line has been constructed at 400C;

it intersects the vertical line at 81 wt% Pb, which corresponds to the composition of Mg2Pb. Furthermore,

the tie line intersection with the L + Mg2Pb-L phase boundary is at 93 wt% Pb, which is the composition of

the liquid phase. Thus, the phase compositions are as follows:

CMg2Pb

= 81 wt% Pb-19 wt% Mg

CL = 93 wt% Pb-7 wt% Mg

(g) For an alloy composed of 8.2 mol Ni and 4.3 mol Cu and at 1250C, it is first necessary to

determine the Ni and Cu concentrations, which we will do in wt% as follows:

nNi' nmNi

ANi (8.2 mol)(58.69 g/mol)= 481.3 g

nCu' nmCu

ACu (4.3 mol)(63.55 g/mol)= 273.3 g

CNi 481.3 g

481.3 g + 273.3 g 100 63.8 wt%

CCu 273.3 g

481.3 g + 273.3 g 100 36.2 wt%

The Cu-Ni phase diagram (Figure 9.3a) is shown below; the point labeled ―G‖ represents the 63.8 wt% Ni-

36.2 wt% Cu composition at 1250C.

As may be noted, point G lies within the phase field. Therefore, only the phase is present; its

composition is 63.8 wt% Ni-36.2 wt% Cu.

(h) For an alloy composed of 4.5 mol Sn and 0.45 mol Pb and at 200C, it is first necessary to

determine the Sn and Pb concentrations, which we will do in weight percent as follows:

nSn’ nmSn

ASn (4.5 mol)(118.71 g/mol)= 534.2 g

nPb' nmPb

APb (0.45 mol)(207.2 g/mol)= 93.2 g

CSn 534.2 g

534.2 g + 93.2 g 100 85.1 wt%

CPb 93.2 g

534.2 g + 93.2 g 10014.9 wt%

That portion of the Pb-Sn phase diagram (Figure 9.8) that pertains to this problem is shown below; the

point labeled ―H‖ represents the 85.1 wt% Sn-14.9 wt% Pb composition at 200C.

As may be noted, point H lies within the + L phase field. A tie line has been constructed at 200C; its

intersection with the L + Lphase boundary is at 74 wt% Sn, which corresponds to the composition of

the L phase. Similarly, the tie-line intersection with the + Lphase boundary occurs at 97.5 wt% Sn,

which is the composition of the phase. Thus, the phase compositions are as follows:

C = 97.5 wt% Sn-2.5 wt% Pb

CL = 74 wt% Sn-26 wt% Pb

9.13 For an alloy of composition 74 wt% Zn-26 wt% Cu, cite the phases present and their compositions at

the following temperatures: 850C, 750C, 680C, 600C, and 500C.

Solution

This problem asks us to determine the phases present and their concentrations at several

temperatures, for an alloy of composition 74 wt% Zn-26 wt% Cu. From Figure 9.19 (the Cu-Zn phase

diagram), which is shown below with a vertical line constructed at the specified composition:

At 850C, a liquid phase is present; CL = 74 wt% Zn-26 wt% Cu

At 750C, and liquid phases are present; C = 67 wt% Zn-33 wt% Cu; CL = 77 wt% Zn-23 wt%

Cu

At 680C, and liquid phases are present; C = 73 wt% Zn-27 wt% Cu; CL = 82 wt% Zn-18 wt%

Cu

At 600C, the phase is present; C = 74 wt% Zn-26 wt% Cu

At 500C, and phases are present; C = 69 wt% Zn-31 wt% Cu; C = 78 wt% Zn-22 wt% Cu

9.17 A 90 wt% Ag-10 wt% Cu alloy is heated to a temperature within the + liquid phase region. If the

composition of the liquid phase is 85 wt% Ag, determine:

(a) The temperature of the alloy

(b) The composition of the phase

(c) The mass fractions of both phases

Solution

(a) In order to determine the temperature of a 90 wt% Ag-10 wt% Cu alloy for which and liquid

phases are present with the liquid phase of composition 85 wt% Ag, we need to construct a tie line across

the + L phase region of Figure 9.7 that intersects the liquidus line at 85 wt% Ag; this is possible at about

850C.

(b) The composition of the phase at this temperature is determined from the intersection of this

same tie line with solidus line, which corresponds to about 95 wt% Ag.

(c) The mass fractions of the two phases are determined using the lever rule, Equations 9.1 and

9.2 with C0 = 90 wt% Ag, CL = 85 wt% Ag, and C = 95 wt% Ag, as

W =C0 CL

C CL

=90 85

95 85= 0.50

WL =C C0

C CL

=95 90

95 85= 0.50

9.27 A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature from an elevated

temperature in such a way that the high-temperature microstructure is preserved. This microstructure is

found to consist of the α phase and Mg2Pb, having respective mass fractions of 0.65 and 0.35. Determine

the approximate temperature from which the alloy was quenched.

Solution

We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% Mg

alloy was quenched, given the mass fractions of and Mg2Pb phases. We can write a lever-rule expression

for the mass fraction of the phase as

W = 0.65 =CMg2Pb C0

CMg2Pb C

The value of C0 is stated as 45 wt% Pb-55 wt% Mg, and CMg2Pb is 81 wt% Pb-19 wt% Mg, which is

independent of temperature (Figure 9.20); thus,

0.65 =81 45

81 C

which yields

C = 25.6 wt% Pb

The temperature at which the –( + Mg2Pb) phase boundary (Figure 9.20) has a value of 25.6 wt% Pb is

about 360C (680F).

9.33 The microstructure of a lead-tin alloy at 180C (355F) consists of primary β and eutectic structures.

If the mass fractions of these two microconstituents are 0.57 and 0.43, respectively, determine the

composition of the alloy.

Solution

Since there is a primary microconstituent present, then we know that the alloy composition, C0

is between 61.9 and 97.8 wt% Sn (Figure 9.8). Furthermore, this figure also indicates that C = 97.8 wt%

Sn and Ceutectic = 61.9 wt% Sn. Applying the appropriate lever rule expression for W'

W' =C0 Ceutectic

CÊ Ceutectic

=C0 61.9

97.8 61.9= 0.57

and solving for C0 yields C0 = 82.4 wt% Sn.

9.37 For a 30 wt% Zn-70 wt% Cu alloy, make schematic sketches of the microstructure that would be

observed for conditions of very slow cooling at the following temperatures: 1100C (2010F), 950C

(1740F), 900C (1650F), and 700C (1290F). Label all phases and indicate their approximate

compositions.

Solution

The illustration below is the Cu-Zn phase diagram (Figure 9.19). A vertical line at a composition

of 30 wt% Zn-70 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures

called for in the problem statement (i.e., 1100C, 950C, 900C, and 700C).

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:

9.39 The room-temperature tensile strengths of pure lead and pure tin are 16.8 MPa and 14.5 MPa,

respectively.

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

compositions between pure lead and pure tin. (Hint: you may want to consult Sections 9.10 and 9.11, as

well as Equation 9.24 in Problem 9.64.)

(b) On this same graph schematically plot tensile strength versus composition at 150°C.

(c) Explain the shapes of these two curves, as well as any 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 lead-tin alloys at both room temperature and 150C; such a

graph is shown below.

(c) Upon consultation of the Pb-Sn phase diagram (Figure 9.8) we note that, at room temperature

(20C), about 1.5 wt% of Sn is soluble in Pb (within the -phase region at the left extremity of the phase

diagram). Similarly, only about 1 wt% of Pb is soluble in Sn (within the -phase region at the left

extremity). Thus, there will a solid-solution strengthening effect on both ends of the phase diagram—

strength increases slightly with additions of Sn to Pb [in the phase region (left-hand side)] and with

additions of Pb to Sn [in the phase region (right-hand side)]; these effects are noted in the above figure.

This figure also shows that the tensile strength of pure lead is greater than pure tin, which is in agreement

with tensile strength values provided in the problem statement.

In addition, at room temperature, for compositions between about 1.5 wt% Sn and 99 wt% Sn,

both and phase will coexist, (Figure 9.8), Furthermore, for compositions within this range, tensile

strength will depend (approximately) on the tensile strengths of each of the and phases as well as their

phase fractions in a manner described by Equation 9.24 for the elastic modulus (Problem 9.64). That is, for

this problem

(TS)alloy (TS)V + (TS)V

in which TS and V denote tensile strength and volume fraction, respectively, and the subscripts represent

the alloy/phases. Also, mass fractions of the and phases change linearly with changing composition

(according to the lever rule). Furthermore, although there is some disparity between the densities of Pb and

Sn (11.35 versus 7.27 g/cm3), weight and volume fractions of the and phases will also be similar (see

Equation 9.6).

At 150C, the curve will be shifted to significantly lower tensile strengths inasmuch as tensile

strength diminishes with increasing temperature (Section 6.6, Figure 6.14). In addition, according to Figure

9.8, solubility limits for both and phases increase—for the phase from 1.5 to 10 wt% Sn, and for the

phase from 1 to about 2 wt% Pb. Thus, the compositional ranges over which solid-solution strengthening

occurs increase somewhat from the room-temperature ranges; these effects are also noted on the 150C

curve above. Furthermore, at 150C, it would be expected that the tensile strength of lead will be greater

than that of tin; and for compositions over which both and phases coexist, strength will decrease

approximately linearly with increasing Sn content.

9.42 Figure 9.36 is the aluminum-neodymium 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 the reaction upon cooling.

Solution

Below is shown the aluminum-neodymium phase diagram (Figure 9.36).

There are two eutectics on this phase diagram. One exists at 12 wt% Nd-88 wt% Al and 632C.

The reaction upon cooling is

LAl Al11Nd3

The other eutectic exists at about 97 wt% Nd-3 wt% Al and 635C. This reaction upon cooling is

L AlNd3 + Nd

There are four peritectics. One exists at 59 wt% Nd-41 wt% Al and 1235C. Its reaction upon

cooling is as follows:

L + Al2Nd Al11Nd3

The second peritectic exists at 84 wt% Nd-16 wt% Al and 940C. This reaction upon cooling is

L + Al2Nd AlNd

The third peritectic exists at 91 wt% Nd-9 wt% Al and 795C. This reaction upon cooling is

L + AlNd AlNd2

The fourth peritectic exists at 94 wt% Nd-6 wt% Al and 675C. This reaction upon cooling is

L + AlNd2 AlNd3

There is one congruent melting point at about 73 wt% Nd-27 wt% Al and 1460C. Its reaction

upon cooling is

L Al2Nd

No eutectoids are present.

9.43 Figure 9.37 is a portion of the titanium-copper phase diagram for which only single-phase

regions are labeled. Specify all temperature-composition points at which eutectics, eutectoids, peritectics,

and congruent phase transformations occur. Also, for each, write the reaction upon cooling.

Solution

Below is shown the titanium-copper phase diagram (Figure 9.37).

There is one eutectic on this phase diagram, which exists at about 51 wt% Cu-49 wt% Ti and

960C. Its reaction upon cooling is

L Ti2Cu + TiCu

There is one eutectoid for this system. It exists at about 7.5 wt% Cu-92.5 wt% Ti and 790C.

This reaction upon cooling is

+ Ti2Cu

There is one peritectic on this phase diagram. It exists at about 40 wt% Cu-60 wt% Ti and

1005C. The reaction upon cooling is

L Ti2Cu

There is a single congruent melting point that exists at about 57.5 wt% Cu-42.5 wt% Ti and

982C. The reaction upon cooling is

L TiCu

.50 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F).

(a) What is the proeutectoid phase?

(b) How many kilograms each of total ferrite and cementite form?

(c) How many kilograms each of pearlite and the proeutectoid phase form?

(d) Schematically sketch and label the resulting microstructure.

Solution

(a) The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid

composition (0.76 wt% C).

(b) For this portion of the problem, we are asked to determine how much total ferrite and

cementite form. Application of the appropriate lever rule expression yields

W =CFe3C C0

CFe3C C

=6.70 1.15

6.70 0.022= 0.83

which, when multiplied by the total mass of the alloy (1.0 kg), gives 0.83 kg of total ferrite.

Similarly, for total cementite,

WFe3C =C0 C

CFe3C C

=1.15 0.022

6.70 0.022= 0.17

And the mass of total cementite that forms is (0.17)(1.0 kg) = 0.17 kg.

(c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite)

form. Applying Equation 9.22, in which

C1' = 1.15 wt% C

Wp =6.70 C 1

'

6.70 0.76=

6.70 1.15

6.70 0.76= 0.93

which corresponds to a mass of 0.93 kg. Likewise, from Equation 9.23

WFe3C' =C1

' 0.76

5.94=

1.15 0.76

5.94= 0.07

which is equivalent to 0.07 kg of the total 1.0 kg mass.

(d) Schematically, the microstructure would appear as: