ME330_HW_Chapter 9-1_solutions

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ME330 Recommended Homework Problems Chapter 9 Part 1 Problems will not be collected or graded. Use them to prepare for the exams. 9.10 Cite the phases that are present and the phase compositions for the following alloys: (a) 90 wt% Zn-10 wt% Cu at 400 ° C (750 ° F) (b) 75 wt% Sn-25 wt% Pb at 175 ° C (345 ° F) (c) 55 wt% Ag-45 wt% Cu at 900 ° C (1650 ° F) (d) 30 wt% Pb-70 wt% Mg at 425 ° C (795 ° F) (e) 2.12 kg Zn and 1.88 kg Cu at 500 ° C (930 ° F) (f) 37 lb m Pb and 6.5 lb m Mg at 400 ° C (750 ° F) (g) 8.2 mol Ni and 4.3 mol Cu at 1250 ° C (2280 ° F) (h) 4.5 mol Sn and 0.45 mol Pb at 200 ° C (390 ° F) Solution (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 400 ° C.

As may be noted, point A lies within the e + h phase field. A tie line has been constructed at 400 ° C; its intersection with the e- ( e + h ) phase boundary is at 87 wt% Zn, which corresponds to the composition of the e phase. Similarly, the tie-line intersection with the ( e + h ) -h phase boundary occurs at 97 wt% Zn, which is the composition of the h phase. Thus, the phase compositions are as follows: C e = 87 wt% Zn-13 wt% Cu C h = 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 175 ° C. As may be noted, point B lies within the a + b phase field. A tie line has been constructed at 175 ° C; its intersection with the a- ( a + b ) phase boundary is at 16 wt% Sn, which corresponds to the composition of the a phase. Similarly, the tie-line intersection with the ( a + b ) -b phase boundary occurs at 97 wt% Sn, which is the composition of the b phase. Thus, the phase compositions are as follows: C a = 16 wt% Sn-84 wt% Pb C b = 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 900 ° C.

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 425 ° C.

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As may be noted, point D lies within the a phase field. Therefore, only the a 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 500 ° C, we must first determine the Zn and Cu concentrations using Equation 4.3a, as 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 500 ° C. C Zn = 2.12 kg 2.12 kg + 1.88 kg × 100 = 53 wt% C Cu = 1.88 kg 2.12 kg + 1.88 kg × 100 = 47 wt%

As may be noted, point E lies within the b + g phase field. A tie line has been constructed at 500 ° C; its intersection with the b- ( b + g ) phase boundary is at 49 wt% Zn, which corresponds to the composition of the b phase. Similarly, the tie-line intersection with the ( b + g ) -g phase boundary occurs at 58 wt% Zn, which is the composition of the g phase. Thus, the phase compositions are as follows: C b = 49 wt% Zn-51 wt% Cu C g = 58 wt% Zn-42 wt% Cu (f) For an alloy composed of 37 lb m Pb and 6.5 lb m Mg and at 400 ° C, we must first determine the Pb and Mg concentrations using Equation 4.3a, as 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 400 ° C. C Pb = 37 lb m 37 lb m + 6.5 lb m × 100 = 85 wt% C Mg = 6.5 lb m 37 lb m + 6.5 lb m × 100 = 15 wt%

As may be noted, point F lies within the L + Mg 2 Pb phase field. A tie line has been constructed at 400 ° C; it intersects the vertical line at 81 wt% Pb, which corresponds to the composition of Mg 2 Pb. Furthermore, the tie line intersection with the ( L + Mg 2 Pb)- L phase boundary is at 93 wt% Pb, which is the composition of the liquid phase. Thus, the phase compositions are as follows: C Mg 2 Pb = 81 wt% Pb-19 wt% Mg C L = 93 wt% Pb-7 wt% Mg (g) For an alloy composed of 8.2 mol Ni and 4.3 mol Cu and at 1250 ° C, it is first necessary to determine the Ni and Cu concentrations in wt%; however, before this is possible it is necessary to compute the mass (in grams) of Ni and Cu using a rearranged form of Equation 4.4 as follows: We may now calculate the composition in weight percent of both Ni and Cu using Equation 4.3a: ( ′ m Ni and ′ m Cu ) ′ m Ni = n m Ni A Ni = (8.2 mol)(58.69 g/mol) = 481.3 g ′ m Cu = n m Cu A Cu = (4.3 mol)(63.55 g/mol) = 273.3 g C Ni = ′ m Ni ′ m Ni + ′ m Cu × 100 = 481.3 g 481.3 g + 273.3 g × 100 = 63.8 wt% C Cu = 273.3 g 481.3 g + 273.3 g × 100 = 36.2 wt%

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The Cu-Ni phase diagram (Figure 9.3 a ) is shown below; the point labeled “G” represents the 63.8 wt% Ni-36.2 wt% Cu composition at 1250 ° C. As may be noted, point G lies within the a phase field. Therefore, only the a 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 200 ° C, it is first necessary to determine the Ni and Cu concentrations in wt%; however, before this is possible it is necessary to compute the mass (in grams) of Sn and Pb using a rearranged form of Equation 4.4 as follows: We may now calculate the composition in weight percent of both Sn and Pb using Equation 4.3a: ( ′ m Sn and ′ m Pb ) ′ m Sn = n m Sn A Sn = (4.5 mol)(118.71 g/mol) = 534.2 g ′ m Pb = n m Pb A Pb = (0.45 mol)(207.2 g/mol) = 93.2 g C Sn = ′ m Sn ′ m Sn + ′ m Pb × 100 = 534.2 g 534.2 g + 93.2 g × 100 = 85.1 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 200 ° C. As may be noted, point H lies within the b + L phase field. A tie line has been constructed at 200 ° C; its intersection with the L - ( b + L ) phase boundary is at 74 wt% Sn, which corresponds to the composition of the L phase. Similarly, the tie-line intersection with the ( b + L ) -b phase boundary occurs at 97.5 wt% Sn, which is the composition of the b phase. Thus, the phase compositions are as follows: C b = 97.5 wt% Sn-2.5 wt% Pb C L = 74 wt% Sn-26 wt% Pb C Pb = 93.2 g 534.2 g + 93.2 g × 100 = 14.9 wt%

9.13 A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300 ° C (2370 ° F). (a) At what temperature does the first liquid phase form? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting? Solution Shown below is the Cu-Ni phase diagram (Figure 9.3 a ) and a vertical line constructed at a composition of 70 wt% Ni-30 wt% Cu. (a) Upon heating from 1300 ° C, the first liquid phase forms at the temperature at which this vertical line intersects the a -( a + L ) phase boundary--i.e., about 1345 ° C. (b) The composition of this liquid phase corresponds to the intersection with the ( a + L )- L phase boundary, of a tie line constructed across the a + L phase region at 1345 ° C--i.e., 59 wt% Ni. (c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt% Ni with the ( a + L )- L phase boundary--i.e., about 1380 ° C.

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(d) The composition of the last solid remaining prior to complete melting corresponds to the intersection with a -( a + L ) phase boundary, of the tie line constructed across the a + L phase region at 1380 ° C--i.e., about 79 wt% Ni. 9.15 For an alloy of composition 74 wt% Zn-26 wt% Cu, cite the phases present and their compositions at the following temperatures: 850 ° C, 750 ° C, 680 ° C, 600 ° C, and 500 ° C. 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 850 ° C, a liquid phase is present; C L = 74 wt% Zn-26 wt% Cu At 750 ° C, g and liquid phases are present; C g = 67 wt% Zn-33 wt% Cu; C L = 77 wt% Zn-23 wt% Cu At 680 ° C, d and liquid phases are present; C d = 73 wt% Zn-27 wt% Cu; C L = 82 wt% Zn-18 wt% Cu At 600 ° C, the d phase is present; C d = 74 wt% Zn-26 wt% Cu At 500 ° C, g and e phases are present; C g = 69 wt% Zn-31 wt% Cu; C e = 78 wt% Zn-22 wt% Cu

9.16 Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and temperatures given in Problem 9.10. Solution This problem asks that we determine the phase mass fractions for the alloys and temperatures in Problem 9.10. (a) From Problem 9.10a, e and h phases are present for a 90 wt% Zn-10 wt% Cu alloy at 400 ° C, as represented in the portion of the Cu-Zn phase diagram shown below (at point A). Furthermore, the compositions of the phases, as determined from the tie line are C e = 87 wt% Zn-13 wt% Cu C h = 97 wt% Zn-3 wt% Cu Inasmuch as the composition of the alloy C 0 = 90 wt% Zn, application of the appropriate lever rule expressions (for compositions in weight percent zinc) leads to W ε = C η − C 0 C η − C ε = 97 − 90 97 − 87 = 0.70 W η = C 0 − C ε C η − C ε = 90 − 87 97 − 87 = 0.30

(b) From Problem 9.10b, a and b phases are present for a 75 wt% Sn-25 wt% Pb alloy at 175 ° C, as represented in the portion of the Pb-Sn phase diagram shown below (at point B). Furthermore, the compositions of the phases, as determined from the tie line are C a = 16 wt% Sn-84 wt% Pb C b = 97 wt% Sn-3 wt% Pb Inasmuch as the composition of the alloy C 0 = 75 wt% Sn, application of the appropriate lever rule expressions (for compositions in weight percent tin) leads to (c) From Problem 9.10c, just the liquid phase is present for a 55 wt% Ag-45 wt% Cu alloy at 900 ° C, as may be noted in the Ag-Cu phase diagram shown below (at point C)—i.e., W L = 1.0. W α = C β − C 0 C β − C α = 97 − 75 97 − 16 = 0.27 W β = C 0 − C α C β − C α = 75 − 16 97 − 16 = 0.73

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(d) From Problem 9.10d, just the a phase is present for a 30 wt% Pb-70 wt% Mg alloy at 425 ° C, as may be noted in the Mg-Pb phase diagram shown below (at point D)—i.e., W a = 1.0.

(e) From Problem 9.10e, b and g phases are present for an alloy composed of 2.12 kg Zn and 1.88 kg Cu (i.e., of composition 53 wt% Zn-47 wt% Cu) at 500 ° C. This is represented in the portion of the Cu-Zn phase diagram shown below (at point E). Furthermore, the compositions of the phases, as determined from the tie line are C b = 49 wt% Zn-51 wt% Cu C g = 58 wt% Zn-42 wt% Cu Inasmuch as the composition of the alloy C 0 = 53 wt% Zn and application of the appropriate lever rule expressions (for compositions in weight percent zinc) leads to (f) From Problem 9.10f, L and Mg 2 Pb phases are present for an alloy composed of 37 lb m Pb and 6.5 lb m Mg (85 wt% Pb-15 wt% Mg) at 400 ° C. This is represented in the portion of the Pb-Mg phase diagram shown below (at point F). W β = C γ − C 0 C γ − C β = 58 − 53 58 − 49 = 0.56 W γ = C 0 − C β C γ − C β = 53 − 49 58 − 49 = 0.44

Furthermore, the compositions of the phases, as determined from the tie line are C Mg 2 Pb = 81 wt% Pb-19 wt% Mg C L = 93 wt% Pb-7 wt% Mg Inasmuch as the composition of the alloy C 0 = 85 wt% Pb and application of the appropriate lever rule expressions (for compositions in weight percent lead) leads to (g) From Problem 9.10g, just the a phase is present (i.e., W a = 1.0) for an alloy composed of 8.2 mol Ni and 4.3 mol Cu (i.e., 63.8 wt% Ni-36.2 wt% Cu) at 1250 ° C; such may be noted (as point G) in the Cu-Ni phase diagram shown below. W Mg 2 Pb = C L − C 0 C L − C Mg 2 Pb = 93 − 85 93 − 81 = 0.67 W L = C 0 − C Mg 2 Pb C L − C Mg 2 Pb = 85 − 81 93 − 81 = 0.33

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(h) From Problem 9.10h, b and L phases are present for an alloy composed of 4.5 mol Sn and 0.45 mol Pb (85.1 wt% Sn-14.9 wt% Pb) and at 200 ° C. This is represented in the portion of the Pb-Sn phase diagram shown below (at point H). Furthermore, the compositions of the phases, as determined from the tie line are C b = 97.5 wt% Sn-2.5 wt% Pb

C L = 74 wt% Sn-26 wt% Pb Inasmuch as the composition of the alloy C 0 = 85.1 wt% Sn, application of the appropriate lever rule expressions (for compositions in weight percent lead) leads to 9.19 A 90 wt% Ag-10 wt% Cu alloy is heated to a temperature within the b + 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 b phase I 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 b and liquid phases are present with the liquid phase of composition 85 wt% Ag, we need to construct a tie line across the b + L phase region of Figure 9.7 that intersects the liquidus line at 85 wt% Ag; this is possible at about 850 ° C. (b) The composition of the b phase at this temperature is determined from the intersection of this same tie line with solidus line, which corresponds to about 95 wt% Ag. I The mass fractions of the two phases are determined using the lever rule, Equations 9.1 and 9.2 with C 0 = 90 wt% Ag, C L = 85 wt% Ag, and C b = 95 wt% Ag, as W β = C 0 − C L C β − C L = 85.1 − 74 97.5 − 74 = 0.47 W L = C β − C 0 C β − C L = 97.5 − 85.1 97.5 − 74 = 0.53 W β = C 0 − C L C β − C L = 90 − 85 95 − 85 = 0.50 W L = C β − C 0 C β − C L = 95 − 90 95 − 85 = 0.50

9.22 A hypothetical A–B alloy of composition 55 wt% B–45 wt% A at some temperature is found to consist of mass fractions of 0.5 for both α and β phases. If the composition of the β phase is 90 wt% B–10 wt% A, what is the composition of the α phase? Solution For this problem, we are asked to determine the composition of the b phase given that C 0 = 55 (or 55 wt% B-45 wt% A) C b = 90 (or 90 wt% B-10 wt% A) W a = W b ° = 0.5 If we set up the lever rule for W a And solving for C a C a = 20 (or 20 wt% B-80 wt% A) 9.28 It is desirable to produce a copper-nickel alloy that has a minimum non-cold-worked tensile strength of 350 MPa (50,750 psi) and a ductility of at least 48%EL. Is such an alloy possible? If so, what must be its composition? If this is not possible, then explain why. Solution From Figure 9.6 a , a tensile strength greater than 350 MPa (50,750 psi) is possible for compositions between about 22.5 and 98 wt% Ni. On the other hand, according to Figure 9.6 b , ductilities greater than 48%EL exist for compositions less than about 8 wt% and greater than about 98 wt% Ni. Therefore, the stipulated criteria are met only at a composition of 98 wt% Ni. 9.30 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. W α = 0.5 = C β − C 0 C β − C α = 90 − 55 90 − C α

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Answer Upon solidification, an alloy of eutectic composition forms a microstructure 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. 9.34 For a copper-silver alloy of composition 25 wt% Ag-75 wt% Cu and at 775 ° C (1425 ° F) do the following: (a) Determine the mass fractions of α and β phases. (b) Determine the mass fractions of primary α and eutectic microconstituents. (c) Determine the mass fraction of eutectic α. Solution (a) This portion of the problem asks that we determine the mass fractions of a and b phases for an 25 wt% Ag-75 wt% Cu alloy (at 775 ° C). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the a + b phase field. From Figure 9.7 and at 775 ° C, C a = 8.0 wt% Ag, C b = 91.2 wt% Ag, and C eutectic = 71.9 wt% Sn. Therefore, the two lever-rule expressions are as follows: (b) Now it is necessary to determine the mass fractions of primary a and eutectic microconstituents for this same alloy. This requires us to utilize the lever rule and a tie line that extends from the maximum solubility of Ag in the a phase at 775 ° C (i.e., 8.0 wt% Ag) to the eutectic composition (71.9 wt% Ag). Thus W α = C β − C 0 C β − C α = 91.2 − 25 91.2 − 8.0 = 0.796 W β = C 0 − C α C β − C α = 25 − 8.0 91.2 − 8.0 = 0.204 W ′ α = C eutectic − C 0 C eutectic − C α = 71.9 − 25 71.9 − 8.0 = 0.734 W e = C 0 − C α C eutectic − C α = 25 − 8.0 71.9 − 8.0 = 0.266

(c) And, finally, we are asked to compute the mass fraction of eutectic a , W e a . This quantity is simply the difference between the mass fractions of total a and primary a as W e α = W α – W ′ α = 0.796 – 0.734 = 0.062

9.35 The microstructure of a lead-tin alloy at 180 ° C (355 ° F) 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 b microconstituent present, then we know that the alloy composition, C 0 is between 61.9 and 97.8 wt% Sn (Figure 9.8). Furthermore, this figure also indicates that C b = 97.8 wt% Sn and C eutectic = 61.9 wt% Sn. Applying the appropriate lever rule expression for and solving for C 0 yields C 0 = 82.4 wt% Sn. 9.37 For an 85 wt% Pb-15 wt% Mg alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 600 ° C (1110 ° F), 500 ° C (930 ° F), 270 ° C (520 ° F), and 200 ° C (390 ° F). Label all phases and indicate their approximate compositions. Solution The illustration below is the Mg-Pb phase diagram (Figure 9.20). A vertical line at a composition of 85 wt% Pb-15 wt% Mg has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement (i.e., 600 ° C, 500 ° C, 270 ° C, and 200 ° C). W ′ β W ′ β = C 0 − C eutectic C β − C eutectic = C 0 − 61.9 97.8 − 61.9 = 0.57

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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:

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