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CHAPTER 9 PHASE DIAGRAMS PROBLEM SOLUTIONSSolubility Limit 9.1 Consider the sugar–water phase diagram of Figure 9.1. (a) How much sugar will dissolve in 1500 g water at 90°C (194°F)? (b) If the saturated liquid solution in part (a) is cooled to 20 °C (68°F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (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 how much sugar will dissolve in 1000 g of water at 90°C. From the solubility limit curve in Figure 9.1, at 90°C the maximum concentration of sugar in the syrup is about 77 wt%. It is now possible to calculate the mass of sugar using Equation 4.3 as Χσυγαρ (ωτ%) = µ συγαρ µ συγαρ+ µ ωατερ msugar × 100 77 wt% = msugar + 1500 g ´ 100 Solving for msugar yields msugar = 5022 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 20°C (m' sugar ) may also be calculated using Equation 4.3 as follows: m' sugar m' sugar + 1500 g 64 wt% = ´ 100 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. which yields a value for µ∋συγαρ of 2667 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that precipitated out of the solution upon cooling µ∀συγαρ that is ; µ∀sugar = msugar - mÕ sugar = 5022 g - 2667 g = 2355 g Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9.2 At 500°C (930°F), 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 500°C corresponds to the position of the β–(α + β) 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 α–(α + β) phase boundary at this temperature, or about 1.5 wt% Ag. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3 Cite three variables that determine the microstructure of an alloy.Microstructure 9. and (3) the heat treatment of the alloy. Solution Three variables that determine the microstructure of an alloy are (1) the alloying elements present. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. . (2) the concentrations of these alloying elements. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. pressure.Phase Equilibria 9. and composition. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. .4 What thermodynamic condition must be met for a state of equilibrium to exist? Solution In order for a system to exist in a state of equilibrium the free energy must be a minimum for some specified combination of temperature. moving vertically (upward) at this temperature. and (b) to sublime. thus.One-Component (or Unary) Phase Diagrams 9.0023 atm. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. and the location on this line at 1 atm pressure (point B) is also noted. we cross the Ice-Liquid phase boundary. (by changing pressure) as. This occurs at approximately 570 atm. Figure 10. This intersection occurs at approximately 0. determine the pressure to which the specimen must be raised or lowered to cause it (a) to melt. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Solution The figure below shows the pressure-temperature phase diagram for H2O. (a) Melting occurs. a vertical line has been constructed at -10°C. we move vertically downward from 1 atm until we cross the Ice-Vapor phase boundary.5 Consider a specimen of ice that is at 210°C and 1 atm pressure. (b) In order to determine the pressure at which sublimation occurs at this temperature.2. the pressure of the specimen must be raised from 1 to 570 atm. the pressure– temperature phase diagram for H2O. Using Figure 9.2. . a horizontal line has been constructed across this diagram at a pressure of 0. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. .01 atm may be determined by moving horizontally across the pressure-temperature diagram at this pressure. which is approximately 1°C. determine (a) the melting temperature for ice.6 At a pressure of 0.01 atm. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.01 atm. The melting and boiling temperatures for ice at a pressure of 0. which is shown below. and (b) the boiling temperature for water. On the other hand.01 atm may be determined from the pressure-temperature diagram for this system. the boiling temperature is at the intersection of the horizontal line with the Liquid-Vapor phase boundary--approximately 16°C.2. Figure 10.9. Solution The melting temperature for ice and the boiling temperature for water at a pressure of 0. The temperature corresponding to the intersection of the Ice-Liquid phase boundary is the melting temperature. Binary Isomorphous Systems 9. Solidus Temperature (°C) 938 1005 1065 1123 1178 1232 1282 1326 1359 1390 1414 Liquidus Temperature (°C) 938 1147 1226 1278 1315 1346 1367 1385 1397 1408 1414 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.7 Given here are the solidus and liquidus temperatures for the germanium-silicon system. Composition (wt% Si) 0 10 20 30 40 50 60 70 80 90 100 Solution The germanium-silicon phase diagram is constructed below. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Construct the phase diagram for this system and label each region. . 19) that pertains to this problem is shown below.5 mol Sn and 0. which is the composition of the η phase. . the phase compositions are as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Similarly.12 kg Zn and 1.3 mol Cu at 1250 °C (2280°F) (h) 4.8 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) 2.2 mol Ni and 4. Thus. the point labeled “A” represents the 90 wt% Zn-10 wt% Cu composition at 400°C.5 lbm Mg at 400°C (750°F) (g) 8.45 mol Pb at 200°C (390°F) Solution This problem asks that we cite the phase or phases present for several alloys at specified temperatures.Interpretation of Phase Diagrams 9. the tie-line intersection with the ε + η−η phase boundary occurs at 97 wt% Zn. A tie line has been constructed at 400°C. its intersection with the ε−ε + η phase boundary is at 87 wt% Zn.88 kg Cu at 500°C (930°F) (f) 37 lbm Pb and 6. which corresponds to the composition of the ε phase. (a) That portion of the Cu-Zn phase diagram (Figure 9. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. point A lies within the ε and η phase field. As may be noted. the tie-line intersection with the α + β−β phase boundary occurs at 97 wt% Sn. Thus. point B lies within the α + β phase field.Cε = 87 wt% Zn-13 wt% Cu Cη = 97 wt% Zn-3 wt% Cu (b) That portion of the Pb-Sn phase diagram (Figure 9. its intersection with the α−α+ β phase boundary is at 16 wt% Sn. . the point labeled “B” represents the 75 wt% Sn-25 wt% Pb composition at 175°C. the phase compositions are as follows: Cα = 16 wt% Sn-84 wt% Pb Cβ = 97 wt% Sn-3 wt% Pb (c) The Ag-Cu phase diagram (Figure 9.8) that pertains to this problem is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. A tie line has been constructed at 175°C. Similarly. which is the composition of the β phase. the point labeled “C” represents the 55 wt% Ag45 wt% Cu composition at 900°C. As may be noted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. which corresponds to the composition of the α phase.7) is shown below. 20) is shown below.As may be noted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. its composition is 55 wt% Ag-45 wt% Cu. point C lies within the Liquid phase field. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Therefore. (d) The Mg-Pb phase diagram (Figure 9. . only the liquid phase is present. the point labeled “D” represents the 30 wt% Pb-70 wt% Mg composition at 425°C. 12 kg ´ 100 = 53 wt% 2. Similarly.88 kg CCu = That portion of the Cu-Zn phase diagram (Figure 9. the phase compositions are as follows: Cβ = 49 wt% Zn-51 wt% Cu Cγ = 58 wt% Zn-42 wt% Cu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.88 kg ´ 100 = 47 wt% 2. point E lies within the β + γ phase field. point D lies within the α phase field.12 kg Zn and 1.88 kg 1. as C Zn = 2.12 kg + 1.As may be noted.19) that pertains to this problem is shown below. its intersection with the β−β + γ phase boundary is at 49 wt% Zn. A tie line has been constructed at 500°C.12 kg + 1. (e) For an alloy composed of 2. . its composition is 30 wt % Pb-70 wt% Mg. As may be noted. the tie-line intersection with the β + γ−γ phase boundary occurs at 58 wt% Zn. the point labeled “E” represents the 53 wt% Zn-47 wt% Cu composition at 500°C.88 kg Cu and at 500°C. Thus. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. we must first determine the Zn and Cu concentrations. only the α phase is present. which corresponds to the composition of the β phase. which is the composition of the γ phase. Therefore. 5 lb m CMg = That portion of the Mg-Pb phase diagram (Figure 9. point F lies within the L + Mg2Pb phase field. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.2 mol)(58.(f) For an alloy composed of 37 lbm Pb and 6. . which is the composition of the liquid phase. the point labeled “F” represents the 85 wt% Pb-15 wt% Mg composition at 400°C. Furthermore. as CPb = 37 lb m ´ 100 = 85 wt% 37 lb m + 6. A tie line has been constructed at 400°C.3 mol Cu and at 1250°C.5 lbm Mg and at 400°C. the phase compositions are as follows: C Mg 2Pb = 81 wt% Pb-19 wt% Mg CL = 93 wt% Pb-7 wt% Mg (g) For an alloy composed of 8. it is first necessary to determine the Ni and Cu concentrations. which we will do in wt% as follows: n' = n m Ni ANi = (8. Thus.3 g Ni Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.5 lb m 6. As may be noted.5 lb m ´ 100 = 15 wt% 37 lb m + 6. the tie line intersection with the L + Mg2Pb-L phase boundary is at 93 wt% Pb. we must first determine the Pb and Mg concentrations.2 mol Ni and 4. which corresponds to the composition of Mg2Pb.69 g/mol) = 481. it intersects the vertical line at 81 wt% Pb.20) that pertains to this problem is shown below. Therefore. .3 mol)(63.8 wt% 481.' nCu = n m Cu ACu = (4. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.3 g + 273.3 g CNi = 481. point G lies within the α phase field.5 mol)(118.2 wt% Cu. its composition is 63. which we will do in weight percent as follows: nÕ = n mSn ASn = (4.45 mol Pb and at 200°C.8 wt% Ni-36.3a) is shown below.2 wt% 481.2 g Sn Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. it is first necessary to determine the Sn and Pb concentrations.8 wt% Ni-36. (h) For an alloy composed of 4. only the α phase is present.5 mol Sn and 0.55 g/mol) = 273.3 g CCu = The Cu-Ni phase diagram (Figure 9.71 g/mol) = 534. As may be noted.3 g ´ 100 = 36.3 g 273.3 g ´ 100 = 63.3 g + 273. the point labeled “G” represents the 63.2 wt% Cu composition at 1250°C. 9 wt% Pb composition at 200°C.1 wt% Sn-14. point H lies within the β + L phase field.2 g CSn = 534. the tie-line intersection with the β + L−β phase boundary occurs at 97. which is the composition of the β phase.5 wt% Pb CL = 74 wt% Sn-26 wt% Pb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. which corresponds to the composition of the L phase. Thus.2 g ´ 100 = 14. its intersection with the L−β + L phase boundary is at 74 wt% Sn. As may be noted.8) that pertains to this problem is shown below. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. the point labeled “H” represents the 85. A tie line has been constructed at 200°C.45 mol)(207. the phase compositions are as follows: Cβ = 97.2 g CPb = That portion of the Pb-Sn phase diagram (Figure 9.2 g ´ 100 = 85.2 g 93.2 g + 93. .5 wt% Sn-2.5 wt% Sn.' nPb = n m Pb APb = (0. Similarly.2 g/mol) = 93.2 g + 93.9 wt% 534.1 wt% 534. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. whereas at 37 wt% Ni the (L + α)-α phase boundary is at about 1230°C.9 Is it possible to have a copper–nickel alloy that. consists of a liquid phase of composition 20 wt % Ni-80 wt% Cu and an α phase of composition 37 wt% Ni-63 wt% Cu.3a. At 20 wt% Ni. consists of a liquid phase of composition 20 wt% Ni–80 wt% Cu and also an α phase of composition 37 wt% Ni–63 wt% Cu? If so. . which at equilibrium. explain why. the L-(α + L) phase boundary is at about 1200°C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. a single tie line does not exist within the α + L region that intersects the phase boundaries at the given compositions.9. at equilibrium. Solution It is not possible to have a Cu-Ni alloy. what will be the approximate temperature of the alloy? If this is not possible. From Figure 9. 19 a single tie line does not exist within the ε + L region which intersects the phase boundaries at the given compositions. Solution It is not possible to have a Cu-Zn alloy. consists of an ε phase of composition 80 wt% Zn-20 wt% Cu. which at equilibrium consists of an ε phase of composition 80 wt% Zn-20 wt% Cu and also a liquid phase of composition 95 wt% Zn-5 wt% Cu.10 Is it possible to have a copper-zinc alloy that.9. whereas at 95 wt% Zn the (ε + L)-L phase boundary is at about 490°C. what will be the approximate temperature of the alloy? If this is not possible. the ε-(ε + L) phase boundary is at about 575°C. and also a liquid phase of composition 95 wt% Zn-5 wt% Cu? If so. . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. From Figure 9. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. At 80 wt% Zn. at equilibrium. explain why. e. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (a) Upon heating from 1300°C..11 A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300°C (2370°F). (b) The composition of this liquid phase corresponds to the intersection with the (α + L)-L phase boundary.3a) and a vertical line constructed at a composition of 70 wt% Ni-30 wt% Cu. the first liquid phase forms at the temperature at which this vertical line intersects the α-(α + L) phase boundary--i.9.. (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. 59 wt% Ni.e. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. of a tie line constructed across the α + L phase region at 1345°C--i. . about 1345°C. about 1380°C. about 79 wt% Ni.e.(c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt% Ni with the (α + L)-L phase boundary--i.. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (d) The composition of the last solid remaining prior to complete melting corresponds to the intersection with α-(α + L) phase boundary. of the tie line constructed across the α + L phase region at 1380°C--i.e. . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. about 465°C.12 A 50 wt% Pb-50 wt% Mg alloy is slowly cooled from 700°C (1290°F) to 400°C (750°F). the first solid phase forms at the temperature at which a vertical line at this composition intersects the L-(α + L) phase boundary--i.. 21 wt% Pb-79 wt% Mg.e.e..9. . (b) The composition of this solid phase corresponds to the intersection with the α-(α + L) phase boundary. of a tie line constructed across the α + L phase region at 560°C--i. (a) Upon cooling from 700°C. (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Pb with the eutectic isotherm--i.e.20) and a vertical line constructed at a composition of 50 wt% Pb-50 wt% Mg. (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? (c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? Solution Shown below is the Mg-Pb phase diagram (Figure 9. about 560°C. e. .(d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the eutectic composition--i.. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. about 67 wt% Pb-33 wt% Mg. 19 (the Cu-Zn phase diagram). for an alloy of composition 74 wt% Zn-26 wt% Cu. From Figure 9. 750°C. CL = 77 wt% Zn-23 wt% Cu At 680°C. Cγ = 67 wt% Zn-33 wt% Cu. δ and liquid phases are present. 680°C. γ and liquid phases are present. cite the phases present and their compositions at the following temperatures: 850°C. Cγ = 69 wt% Zn-31 wt% Cu. the δ phase is present. Cε = 78 wt% Zn-22 wt% Cu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Cδ = 74 wt% Zn-26 wt% Cu At 500°C. . and 500°C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. CL = 82 wt% Zn-18 wt% Cu At 600°C. CL = 74 wt% Zn-26 wt% Cu At 750°C. which is shown below with a vertical line constructed at the specified composition: At 850°C. γ and ε phases are present. Solution This problem asks us to determine the phases present and their concentrations at several temperatures. 600°C. Cδ = 73 wt% Zn-27 wt% Cu.13 For an alloy of composition 74 wt% Zn-26 wt% Cu.9. a liquid phase is present. C0 C h . .87 Ωη = C 0 . as represented in the portion of the Cu-Zn phase diagram shown below (at point A). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.14 Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and temperatures given in Problem 9.87 = = 0. Solution This problem asks that we determine the phase mass fractions for the alloys and temperatures in Problem 9. as determined from the tie line are Cε = 87 wt% Zn-13 wt% Cu Cη = 97 wt% Zn-3 wt% Cu Inasmuch as the composition of the alloy C0 = 90 wt% Zn. Furthermore.Ce 97 .Ce = 97 . (a) From Problem 9.Ce 90 .70 97 .30 C h . the compositions of the phases.87 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.9. ε and η phases are present for a 90 wt% Zn-10 wt% Cu alloy at 400°C.8a.8. application of the appropriate lever rule expressions (for compositions in weight percent zinc) leads to Ωε = Ch .90 = 0.8. .75 = 0. Furthermore.8c.C 0 Cb .27 97 .73 Cb .0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. WL = 1. application of the appropriate lever rule expressions (for compositions in weight percent tin) leads to Ωα = Cb . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. as determined from the tie line are Cα = 16 wt% Sn-84 wt% Pb Cβ = 97 wt% Sn-3 wt% Pb Inasmuch as the composition of the alloy C0 = 75 wt% Sn.Ca 75 . as may be noted in the Ag-Cu phase diagram shown below (at point C)—i.16 = = 0.(b) From Problem 9.Ca = 97 . α and β phases are present for a 75 wt% Sn-25 wt% Pb alloy at 175°C.e.8b.. just the liquid phase is present for a 55 wt% Ag-45 wt% Cu alloy at 900°C. as represented in the portion of the Pb-Sn phase diagram shown below (at point B). the compositions of the phases.16 Ωβ = C 0 .16 (c) From Problem 9.Ca 97 . as may be noted in the Mg-Pb phase diagram shown below (at point D)—i. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. just the α phase is present for a 30 wt% Pb-70 wt% Mg alloy at 425°C. Wα = 1.e. .(d) From Problem 9..8d.0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. . the compositions of the phases.44 58 .49 Ωγ = = (f) From Problem 9. as determined from the tie line are Cβ = 49 wt% Zn-51 wt% Cu Cγ = 58 wt% Zn-42 wt% Cu Inasmuch as the composition of the alloy C0 = 53 wt% Zn and application of the appropriate lever rule expressions (for compositions in weight percent zinc) leads to Ωβ = Cg .49 = 0.Cb C 0 .8e.8f.(e) From Problem 9.Cb = 58 .88 kg Cu (i.49 53 .C0 C g .. This is represented in the portion of the Pb-Mg phase diagram shown below (at point F).53 = 0.Cb C g . This is represented in the portion of the Cu-Zn phase diagram shown below (at point E). L and Mg2Pb phases are present for an alloy composed of 37 lbm Pb and 6. of composition 53 wt% Zn-47 wt% Cu) at 500°C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Furthermore.56 58 .5 lbm Mg (85 wt% Pb-15 wt% Mg) at 400°C.12 kg Zn and 1. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. β and γ phases are present for an alloy composed of 2.e. CMg 2Pb C L .67 C L .8g..e. as determined from the tie line are C = 81 wt% Pb-19 wt% Mg Mg 2Pb CL = 93 wt% Pb-7 wt% Mg Inasmuch as the composition of the alloy C0 = 85 wt% Pb and application of the appropriate lever rule expressions (for compositions in weight percent lead) leads to ΩMg 2Pb = CL . .CMg 2Pb 93 . Wα = 1..Furthermore.CMg 2Pb = (g) From Problem 9. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.C0 93 .81 85 . just the α phase is present (i.33 93 .81 ΩL = C 0 .2 wt% Cu) at 1250°C.0) for an alloy composed of 8.e.85 = = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.81 = 0. 63.8 wt% Ni-36. the compositions of the phases.2 mol Ni and 4. such may be noted (as point G) in the Cu-Ni phase diagram shown below.3 mol Cu (i. 1 wt% Sn-14.5 mol Sn and 0.9 wt% Pb ) and at 200°C. Furthermore.(h) From Problem 9. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. . the compositions of the phases. as determined from the tie line are Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. This is represented in the portion of the Pb-Sn phase diagram shown below (at point H). β and L phases are present for an alloy composed of 4.8h.45 mol Pb (85. C L 97. application of the appropriate lever rule expressions (for compositions in weight percent lead) leads to Ωβ = C0 .CL 85. .C L 97.5 wt% Pb CL = 74 wt% Sn-26 wt% Pb Inasmuch as the composition of the alloy C0 = 85.5 .74 = = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.1 wt% Sn.5 .1 = 0.5 wt% Sn-2.5 .Cβ = 97.1 .74 Cb .85.47 Cb .53 97.C 0 Cb .74 ΩΛ = = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. At 250°C and within the α + L phase region C = 14 wt% Sn-86 wt% Pb α CL = 34 wt% Sn-66 wt% Pb Let C0 be the new alloy composition to give Wα = WL = 0.15 kg Then. Then. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.8). by adding Sn to the alloy. on the Pb-Sn phase diagram (Figure 9. The alloy is to be melted to the extent that 50% of the specimen is liquid.C0 34 .276 kg. This occurs at approximately 295°C (560°F). The amount of Sn in the original alloy is (0. let mSn be the mass of Sn added to the alloy to achieve this new composition. through the α + L region until the tie-line lengths on both sides of the given composition are the same.5 = CL . using a modified form of Equation 4.14 And solving for C0 gives 24 wt% Sn.Ca 34 . (a) To what temperature must the specimen be heated? (b) How much tin must be added to the 1. at this temperature it is entirely an α-phase solid solution (Figure 9. the remainder being the α phase.5-kg specimen of a 90 wt% Pb–10 wt% Sn alloy is heated to 250°C (480°F).15 kg + m ù Sn ´ 100 = 24 ê ú ë 1.5-kg specimen at 250°C to achieve this state? Solution (a) Probably the easiest way to solve this part of the problem is by trial and error--that is. (b) We can also produce a 50% liquid solution at 250°C.5 kg) = 0. .15 A 1. This may be accomplished either by heating the alloy or changing its composition while holding the temperature constant. moving vertically at the given composition. Ωα = 0.9.8).3 é 0.5. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Now. solving for mSn (the mass of tin to be added). yields mSn = 0.5 kg + mSn û And.C 0 = C L .10)(1. 16 A magnesium-lead alloy of mass 5. In order to determine the additional amount of Pb that may be added (mPb).5 kg consists of a solid α phase that has a composition that is just slightly below the solubility limit at 200°C (390°F). for a Pb-Mg alloy. the solubility limit for the α phase at 200°C corresponds to the position (composition) of the α-α + Mg2Pb phase boundary at this temperature. From Figure 9. we utilize a modified form of Equation 4. the mass of lead in 5.28 kg. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.28 kg + mPb ´ 100 5.5 kg of the solid α phase at 200°C just below the solubility limit. (b) At 350°C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. . how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase? Solution (a) This portion of the problem asks that we calculate.9. which is about 5 wt % Pb.3 as ΧPb = 25 wt% = 0. the solubility limit of the α phase increases to approximately 25 wt% Pb.05)(5.20.5 kg + mPb Solving for mPb yields mPb = 1.5 kg) = 0. (a) What mass of lead is in the alloy? (b) If the alloy is heated to 350°C (660°F).46 kg. the mass of Pb in the alloy is just (0. Therefore. (c) The mass fractions of the two phases are determined using the lever rule. which corresponds to about 95 wt% Ag. as β Ωβ = C0 .1 and 9. determine: (a) The temperature of the alloy (b) The composition of the β 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 β and liquid phases are present with the liquid phase of composition 85 wt% Ag.9.CL 90 .CL 95 .C 0 Cb .7 that intersects the liquidus line at 85 wt% Ag. this is possible at about 850°C.17 A 90 wt% Ag-10 wt% Cu alloy is heated to a temperature within the β + liquid phase region. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.85 ΩΛ = = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.2 with C0 = 90 wt% Ag. and C = 95 wt% Ag.50 Cb . If the composition of the liquid phase is 85 wt% Ag.90 = 0. CL = 85 wt% Ag.85 = = 0.C L 95 .50 95 . we need to construct a tie line across the β + L phase region of Figure 9. .85 Cb . (b) The composition of the β phase at this temperature is determined from the intersection of this same tie line with solidus line. Equations 9. Using the appropriate phase diagram. a tie line within the α + L phase region that is divided in half for an alloy of this composition exists at about 230°C.5 for a 30 wt% Sn-70 wt% Pb alloy and asked to estimate the temperature of the alloy. If the mass fraction of each phase is 0. and CL = 43 wt% Sn. .18 A 30 wt% Sn-70 wt% Pb alloy is heated to a temperature within the α + liquid phase region.8. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Cα = 15 wt% Sn. estimate: (a) The temperature of the alloy (b) The compositions of the two phases Solution (a) We are given that the mass fractions of α and liquid phases are both 0. (b) We are now asked to determine the compositions of the two phases. This is accomplished by noting the intersections of this tie line with both the solidus and liquidus lines. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.9. Figure 9.5. by trial and error with a ruler. From these intersections. 86 Solution The problem is to solve for compositions at the phase boundaries for both α and β phases (i.C 01 Cb .57 = Cb .14 Fraction β Phase 0. compositions are given in wt% of A. yield C = 90 (or 90 wt% A-10 wt% B) α C = 20. Cα and Cβ). From the mass fractions of both phases for two different alloys provided in the table below. determine the composition of the phase boundary (or solubility limit) for both α and β phases at this temperature. (which are at the same temperature). A-rich phase and a β.9.60 Cb .8 wt% B) β Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.e. .30 Cb .Ca Cb . there exist an α. in terms of Cα and Cβ as follows: Ωα1 = 0..19 For alloys of two hypothetical metals A and B. one for each composition.57 0.C 02 Cb .Ca = Cb .Ca Ωα2 = 0.Ca Cb . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We may set up two independent lever rule expressions.2 wt% A-79.14 = = In these expressions.2 (or 20. Alloy Composition 60 wt% A–40 wt% B 30 wt% A–70 wt% B Fraction α Phase 0. B-rich phase. Solving for Cα and Cβ from these equations.43 0. 5 = Cb .C 0 Cb . we are asked to determine the composition of the β phase given that C0 = 55 (or 55 wt% B-45 wt% A) C = 90 (or 90 wt% B-10 wt% A) β Wα = Wβ = 0. If the composition of the β phase is 90 wt% B–10 wt% A.9.55 90 . .5 for both α and β phases. what is the composition of the α phase? Solution For this problem.Ca And solving for Cα Cα = 20 (or 20 wt% B-80 wt% A) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.5 If we set up the lever rule for Wα Ωα = 0.20 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.Ca = 90 . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 50 = 0.50 100 . upon cooling through the α + β phase region will remain approximately constant at about 0. which.60 and Wβ = 0. Using the appropriate phase diagram. at equilibrium. Ωα = Cb Cb - C0 Ca = 91.50 91. . Cα = 0 wt% Ag and Cβ = 100 wt% Ag. employment of Equations 9.1 and 9. Solution It is not possible to have a Cu-Ag alloy of composition 50 wt% Ag-50 wt% Cu which consists of mass fractions Wα = 0. At just below the eutectic. what will be the approximate temperature of the alloy? If such an alloy is not possible.40? If so.00 .60 and Wβ = 0.00 . and. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.0.40 as called for in the problem.5.2 .60 and Wβ = 0. Figure 9.2 . Thus. explain why.21 Is it possible to have a copper-silver alloy of composition 50 wt% Ag-50 wt% Cu.2 yields Wa = Cb . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. consists of α and β phases having mass fractions Wα = 0.50 Furthermore. thus. Cα = 8. and will never have values of Wα = 0.9. the mass fractions of the α and β phases.50.2 let us determine Wα and Wβ at just below the eutectic temperature and also at room temperature.C a = 100 .1 and 9.8 Ωβ =1.C 0 Cb .2 wt% Ag.40.0 wt% Ag and Cβ = 91.50 = 0.50 = 0. using Equations 9.7. at room temperature. Wβ = 0.0 And.Wa = 1. 81 kg for the α and Mg2Pb phases.20 kg of a magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg.66 = 81 .Ca Solving for C from this expression yields C = 3.81 kg ΩMg 2Pb = 1.7 wt% Pb. to have α and Mg 2Pb phases having respective masses of 7. if we apply the lever rule expression for W α Ωα = CMg2Pb .66 = 0. explain why. it is first necessary to determine the mass fraction of each phase as follows: Ωα = ma 7. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.81 kg? If so. . In order to demonstrate this.39 kg + 3.39 kg and 3.39 kg and 3. The position along the α−(α + Mg2Pb) phase boundary α α of Figure 9.20 corresponding to this composition is approximately 190°C.00 . at equilibrium.Ca Since the Mg2Pb phase exists only at 81 wt% Pb.39 kg = = 0.9.34 Now. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Solution Yes. it is possible to have a 30 wt% Pb-70 wt% Mg alloy which has masses of 7.30 81 .66 ma + mMg 2Pb 7.C 0 CMg 2Pb . respectively.0. and C0 = 30 wt% Pb Ωα = 0. is it possible.22 For 11. what will be the approximate temperature of the alloy? If such an alloy is not possible. S4a) Ωβ = (9. .S2b) Substitution of these expressions into Equation 9. is defined by Equation 9. upon rearrangement yield Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Furthermore. which is used to convert from phase weight fraction to phase volume fraction.S2a) ϖ = β (9.S1 leads to ςα = µβ µα + ρα ρβ µα ρα (9. Now. respectively.S4b) Which.7a. Volume fraction of phase α. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.5 as α ςα = ϖ α ϖ + ϖ α β (9.e.S1) where v and v are the volumes of the respective phases in the alloy. which may be used to convert mass fraction to volume fraction. the density of each phase is equal α β to the ratio of its mass and volume.6a and 9. Solution This portion of the problem asks that we derive Equation 9.6a. and vice versa.23 Derive Equations 9..9.S3) in which m's and ρ's denote masses and densities. or upon rearrangement ϖ = α µα ρα µβ ρβ (9. the mass fractions of the α and β phases (i. V . W and W ) are defined in terms of the phase masses as β α Ωα = µα µα + µβ µβ µα + µβ (9. µ α = Ωα (ma + mb ) µ β = Ωβ (ma + mb ) Incorporation of these relationships into Equation 9.S3 leads to (9.S5a) (9.S5b) Ωα (ma + mb ) ςα = ra Wa (ma + mb ) ra + Wb (ma + mb ) rb ςα = Ωβ Ωα + ρα ρβ Ωα ρα (9.S6) which is the desired equation. For this portion of the problem we are asked to derive Equation 9.7a, which is used to convert from phase volume fraction to mass fraction. Mass fraction of the α phase is defined as Ωα = µα µα + µβ (9.S7) From Equations 9.S2a and 9.S2b µ α = ϖ ρα α µ β = ϖ ρβ β Substitution of these expressions into Equation 9.S7 yields (9.S8a) (9.S8b) Ωα = ϖ ρα α ϖ ρα + ϖ ρβ α β (9.S9) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. From Equation 9.5 and its equivalent for V the following may be written: β ϖ = ς α (va + vb ) α ϖ = ς β (va + vb ) β Substitution of Equations 9.S10a and 9.S10b into Equation 9.S9 yields (9.S10a) (9.S10b) Ωα = ς α (va + vb )r a Va (va + vb )r a + Vb (va + vb )r b ς α ρα Ωα = ς α ρα + ς β ρβ (9.S11) which is the desired expression. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9.24 Determine the relative amounts (in terms of volume fractions) of the phases for the alloys and temperatures given in Problem 9.8a, b, and c. Below are given the approximate densities of the various metals at the alloy temperatures: Metal Ag Cu Cu Pb Sn Zn Temperature (°C) 900 400 900 175 175 400 Density (g/cm3) 9.97 8.77 8.56 11.20 7.22 6.83 Solution This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problems 9.8a, b, and c. This is accomplished by using the technique illustrated in Example Problem 9.3, and also the results of Problems 9.8 and 9.14. (a) This is a Cu-Zn alloy at 400°C, wherein Cε = 87 wt% Zn-13 wt% Cu Cη = 97 wt% Zn-3 wt% Cu Wε = 0.70 Wη = 0.30 ρCu = 8.77 g/cm3 ρZn = 6.83 g/cm3 Using these data it is first necessary to compute the densities of the ε and η phases using Equation 4.10a. Thus ρε = 100 C Zn(e) r Zn + CCu(e) r Cu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 03 g/cm 6.= 100 = 7.83 g/cm3 8.70 ςη = Wh rh Wh We + re rh = 0.83 g/cm3 8. wherein Cα = 16 wt% Sn-84 wt% Pb Cβ = 97 wt% Sn-3 wt% Pb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.30 + 3 7.88 g/cm 3 = 0.6.70 7.70 0.03 g/cm 6.88 g/cm3 = 100 97 3 + 6.30 + 3 7.77 g/cm 3 ρη = 100 C Zn(h) r Zn + CCu(h ) r Cu = 6.03 g/cm 3 0. We re We Wh + re rh ςε = = 0.88 g/cm 3 = 0. .88 g/cm 3 0.30 (b) This is a Pb-Sn alloy at 175°C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.03 g/cm3 87 13 + 6. Thus.70 0.77 g/cm 3 Now we may determine the Vε and Vη values using Equation 9.30 6. 73 ρSn = 7.20 g/cm3 Using this data it is first necessary to compute the densities of the α and β phases. Thus ρα = 100 CSn(a ) r Sn + CPb(a ) r Pb = 100 = 10.27 10.20 g/cm 3 ρβ = 100 CSn(b ) r Sn + CPb(b ) r Pb = 100 = 7.73 + 10.30 g/cm3 97 3 + 7. Wa ra Wb Wa + ra rb ςα = 0. Thus.Wα = 0.22 g/cm3 11.27 Wβ = 0. .22 g/cm3 11. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.6.20 g/cm 3 Now we may determine the Vα and Vβ values using Equation 9.29 g/cm 3 = = 0.21 0.27 0.29 g/cm3 7.29 g/cm3 16 84 + 7.30 g/cm 3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.22 g/cm3 ρPb = 11. 0.30 g/cm 3 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .79 (c) This is a Ag-Cu alloy at 900°C. Therefore. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.27 0.73 + 10.30 g/cm 3 = 0.73 7. wherein only the liquid phase is present. VL = 1.Wb ςβ = rb Wb Wa + ra rb = 0.29 g/cm3 7. (b) Cite one undesirable consequence of coring.Development of Microstructure in Isomorphous Alloys 9. with higher concentrations of the component having the lower melting temperature at the grain boundaries. Solution (a) Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys. this melting results in a loss in mechanical integrity of the alloy. It occurs. during solidification. . upon heating.25 (a) Briefly describe the phenomenon of coring and why it occurs. the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram. as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (b) One undesirable consequence of a cored structure is that. Solution From Figure 9. according to Figure 9.750 psi) and a ductility of at least 48%EL. . On the other hand. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.Mechanical Properties of Isomorphous Alloys 9. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.750 psi) is possible for compositions between about 22.6a. a tensile strength greater than 350 MPa (50.26 It is desirable to produce a copper-nickel alloy that has a minimum noncold-worked tensile strength of 350 MPa (50.6b. Is such an alloy possible? If so. what must be its composition? If this is not possible. Therefore. the stipulated criteria are met only at a composition of 98 wt% Ni. ductilities greater than 48%EL exist for compositions less than about 8 wt% and greater than about 98 wt% Ni. then explain why.5 and 98 wt% Ni. 6 wt% Pb is about 360°C (680°F). We can write a lever-rule expression for the mass fraction of the α phase as Ωα = 0. Determine the approximate temperature from which the alloy was quenched. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. which is independent of 2 temperature (Figure 9. This microstructure is found to consist of the α phase and Mg2Pb.20) has a value of 25. . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and CMg Pb is 81 wt% Pb-19 wt% Mg.65 and 0.C 0 CMg 2Pb .65 = CMg 2Pb .Binary Eutectic Systems 9. Solution We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% Mg alloy was quenched.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.Ca 0. given the mass fractions of α and Mg2Pb phases. 81 .35.6 wt% Pb α The temperature at which the α–(α + Mg2Pb) phase boundary (Figure 9.Ca The value of C0 is stated as 45 wt% Pb-55 wt% Mg.65 = which yields C = 25.20).45 81 . thus. having respective mass fractions of 0. 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. upon solidification. Solution Upon solidification. an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases.28 Briefly explain why. .Development of Microstructure in Eutectic Alloys 9. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. and with this layered configuration the diffusion path length for the atoms is a minimum. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .29 What is the difference between a phase and a microconstituent? Solution A “phase” is a homogeneous portion of the system having uniform physical and chemical characteristics. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. whereas a “microconstituent” is an identifiable element of the microstructure (that may consist of more than one phase). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.9. 925. For primary β ΩβÕ = C 0 .9 = 0.0 wt% Ag. at 775°C (1425 °F)? Why or why not? Solution This problem asks if it is possible to have a Cu-Ag alloy for which the mass fractions of primary β and total β are 0.2 .0 = 0 = 0.2 .9.68 91. and Ceutectic = 71.925.Ca 91. Therefore.9 wt% Ag.8. From Figure 9.30 Is it possible to have a copper-silver alloy in which the mass fractions of primary β and total β are 0.71.925 Cb .7 and at 775°C. In order to make this determination we need to set up the appropriate lever rule expression for each of these quantities. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. .C eutectic Cb Ê C eutectic - = C 0 .68 and 0.2 wt% Ag.71.0 And this value of C0 is 85 wt% Ag. since these two C0 values are the same (85 wt% Ag). respectively. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. respectively at 775°C.Ca C . Now the analogous expression for total β Ωβ = C 0 . Cα = 8.68 and 0. Cβ = 91.8.9 Solving for C0 gives C0 = 85 wt% Ag. this alloy is possible. C 0 = 0.70 kg Ωα = Next it is necessary to set up the appropriate lever rule expression for each of these quantities.2 wt% Pb. respectively. since these two C0 values are different. is it possible to have the masses of primary α and total α of 4.896 81 .70 kg 6.C 0 = 0. respectively in 6. at 460°C (860°F)? Why or why not? Solution This problem asks if it is possible to have a Mg-Pb alloy for which the masses of primary α and total α are 4. Ωα' = 4. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.23 kg and 6.41 And solving for C0 gives C0 = 50.2 wt% Pb. Cα = 41 wt% Pb.20 and at 460°C.896 6.9. . From Figure 9.631 66 .31 For 6. Now the analogous expression for total α Ωα = CMg 2Pb .23 kg = 0. Thus. CMg Pb = 81 wt% Pb.00 kg.631 6.C 0 CMg 2Pb . and Ceutectic = 66 wt% Pb 2 For primary α Ωα' = C eutectic C eutectic - C0 Ca = 66 .00 kg = 0. In order to make this determination we first need to convert these masses to mass fractions.70 kg total of the alloy at 460°C. Therefore.23 kg and 6.00 kg. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.70 kg of a magnesium-lead alloy.41 And this value of C0 is 45. this alloy is not possible.Ca = 81 . 0 (b) Now it is necessary to determine the mass fractions of primary α and eutectic microconstituents for this same alloy.Ca 25 . .9 . 8.8.Ca 25 .062 α Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.0 wt% Ag. Thus Ωα' = C eutectic Ê C 0 71.0 wt% Ag) to the eutectic composition (71. Therefore.9 wt% Ag).796 – 0. Weα. Cα = 8. we are asked to compute the mass fraction of eutectic α. finally. Solution (a) This portion of the problem asks that we determine the mass fractions of α and β phases for an 25 wt% Ag-75 wt% Cu alloy (at 775°C). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.32 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.8.C 0 Cb . and Ceutectic = 71.25 = = 0. In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the α + β phase field.9 .8.e. (b) Determine the mass fractions of primary α and eutectic microconstituents. the two lever-rule expressions are as follows: Ωα = Cb . This requires us to utilize the lever rule and a tie line that extends from the maximum solubility of Ag in the α phase at 775°C (i.Ca 91.796 91.2 .7 and at 775°C.734 C eutectic Ê Ca 71.204 Cb .25 = 0.2 wt% Ag.0 Ωe = = (c) And.266 71.0 = 0.8.2 .0 = = 0. This quantity is simply the difference between the mass fractions of total α and primary α as We = Wα – Wα' = 0. From Figure 9. Cβ = 91.9 .8.9.9 wt% Sn. (c) Determine the mass fraction of eutectic α..0 C 0 .Ca = 91.0 Ωβ = C 0 .2 .734 = 0.Ca C eutectic .8. determine the composition of the alloy.61. C0 is between 61.9 and 97.8 .9. then we know that the alloy composition.8 wt% Sn (Figure 9. Applying the appropriate lever rule expression for Wβ' Ωβ' = C 0 . . If the mass fractions of these two microconstituents are 0.9 wt% Sn.8).8 wt% Sn and Ceutectic = 61.57 Cb Ê C eutectic 97.57 and 0. respectively.C eutectic C .9 = 0 = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.61.43. Furthermore. Solution Since there is a primary β microconstituent present.9 and solving for C0 yields C0 = 82. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.4 wt% Sn. this figure also indicates that Cβ = 97.33 The microstructure of a lead-tin alloy at 180 °C (355°F) consists of primary β and eutectic structures. 34 Consider the hypothetical eutectic phase diagram for metals A and B.693 92. Let us write lever rule expressions for Wα' and Wα Ωα = Cb Ê C 0 Cb .6 .6 .693.6 wt% B. and (3) the composition of the β phase at the eutectic temperature is 92.4 wt% A. respectively.9.356 and Wα = 0.6 wt% B-7. . (2) the eutectic composition is 47 wt% B-53 wt% A. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Figure 9.6 wt% B at the eutectic temperature. Cβ = 92.C a 47 . from this we are asked to determine the composition of the alloy. Solution We are given a hypothetical eutectic phase diagram for which Ceutectic = 47 wt% B. Solving them for C0 gives C0 = 32.356 and 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.8. Determine the composition of an alloy that will yield primary α and total α mass fractions of 0.Ca = 92.C 0 = 0. we have two simultaneous equations with C0 and Cα as unknowns. and also that Wα' = 0. Assume that (1) α and β phases exist at the A and B extremities of the phase diagram.356 47 .C a Ωα' = C eutectic Ê C 0 C eutectic Ê Ca - = Thus.693. which is similar to that for the lead-tin system.C 0 = 0. respectively. and. . 270°C. On the basis of the locations of the four temperature-composition points. A vertical line at a composition of 85 wt% Pb-15 wt% Mg has been drawn. Label all phases and indicate their approximate compositions..9. in addition. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.35 For an 85 wt% Pb-15 wt% Mg alloy.e. 500°C. and 200°C). 500°C (930°F). make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 600 °C (1110 °F). 600°C. and 200°C (390°F).20). horizontal arrows at the four temperatures called for in the problem statement (i. Solution The illustration below is the Mg-Pb phase diagram (Figure 9. schematic sketches of the four respective microstructures along with phase compositions are represented as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 270°C (520°F). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. e. make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1000°C (1830°F). in addition. . schematic sketches of the four respective microstructures along with phase compositions are represented as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 1000°C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 760°C (1400°F). 600°C. 760°C. 600°C (1110°F).9. On the basis of the locations of the four temperature-composition points. horizontal arrows at the four temperatures called for in the problem statement (i. and 400°C (750°F). and 400°C).. Solution The illustration below is the Cu-Zn phase diagram (Figure 9. A vertical line at a composition of 68 wt% Zn-32 wt% Cu has been drawn.36 For a 68 wt% Zn-32 wt% Cu alloy. Label all phases and indicate their approximate compositions. and.19). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. . . and 700°C (1290°F). 900°C (1650°F).9. 950°C (1740°F). Solution The illustration below is the Cu-Zn phase diagram (Figure 9. and 700°C).e. schematic sketches of the four respective microstructures along with phase compositions are represented as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 1100°C. 950°C. in addition. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1100°C (2010°F).37 For a 30 wt% Zn-70 wt% Cu alloy. On the basis of the locations of the four temperature-composition points.19). A vertical line at a composition of 30 wt% Zn-70 wt% Cu has been drawn. Label all phases and indicate their approximate compositions. . 900°C. horizontal arrows at the four temperatures called for in the problem statement (i. and. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (2) the densities of the α and β phases as well as the eutectic structure are 11. Below is shown a square grid network onto which is superimposed outlines of the primary α phase areas. . respectively.7 g/cm3. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.17 and the Pb–Sn phase diagram (Figure 9.38 On the basis of the photomicrograph (i. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.9. 7. Make the following assumptions: (1) the area fraction of each phase and microconstituent in the photomicrograph is equal to its volume fraction. and 8.e..17: Primary α and eutectic microconstituents are present in the photomicrograph. the relative amounts of the microconstituents) for the lead–tin alloy shown in Figure 9.2.17. Figure 9.7 g/cm3.3. respectively. and it is given that their densities are 11. and then compare this estimate with the composition given in the figure legend of Figure 9. and (3) this photomicrograph represents the equilibrium microstructure at 180°C (355°F). estimate the composition of the alloy. Solution Below is shown the micrograph of the Pb-Sn alloy.2 and 8.8). Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.9 . it is necessary to utilize Equations 9. the area fraction of primary α is 104/644 = 0.The area fraction of this primary α phase may be determined by counting squares. Solving for C0 yields C0 = 53. To do this.197 Ωευτεχτιχ= ς ευτεχτιχ ευτεχτιχ ρ ς α∏ α∏+ ς ευτεχτιχ ευτεχτιχ ρ ρ = (0.3 wt% Sn.16)(11. approximately 104 lie within the primary α phase particles.7 g /cm3 ) = 0.16. Accordingly Ωα∋ = 0. This value is in good agreement with the actual composition—viz.84)(8.16)(11.2 g /cm 3) (0. 50 wt% Sn.84)(8.197 = 61.84)(8.7 g /cm 3 ) (0.16)(11.3 wt% Sn).9 .9 wt% Sn) to the α–(α + β) phase boundary at 180°C (about 18. and of these. Thus.7a and 9.C 0 61.7b as follows: Ωα∋ = ς α∋ ρα∋ ς α∋ρα∋ + ς ευτεχτιχ ευτεχτιχ ρ = (0.2 g /cm3 ) + (0.7 g /cm3 ) = 0.803 From Figure 9.3 wherein C0 is the alloy composition (in wt% Sn). . we want to use the lever rule and a tie-line that extends from the eutectic composition (61.18. There are a total of 644 squares. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. which is also assumed to be the volume fraction.8.2 g /cm3 ) + (0. We now want to convert the volume fractions into mass fractions in order to employ the lever rule to the PbSn phase diagram. (Hint: you may want to consult Sections 9.39 The room-temperature tensile strengths of pure lead and pure tin are 16. at room temperature (20°C). (a) Make a schematic graph of the room-temperature tensile strength versus composition for all compositions between pure lead and pure tin. for compositions between about 1. (c) Upon consultation of the Pb-Sn phase diagram (Figure 9.10 and 9.8) we note that. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus.24 in Problem 9. as well as Equation 9. . (Figure 9. both α and β phase will coexist.8). at room temperature. there will a solidsolution strengthening effect on both ends of the phase diagram—strength increases slightly with additions of Sn to Pb [in the α phase region (left-hand side)] and with additions of Pb to Sn [in the β phase region (right-hand side)]. such a graph is shown below. tensile strength will depend Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. about 1.5 MPa. (c) Explain the shapes of these two curves. for compositions within this range.5 wt% Sn and 99 wt% Sn. which is in agreement with tensile strength values provided in the problem statement.5 wt% of Sn is soluble in Pb (within the α-phase region at the left extremity of the phase diagram). 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 150°C. these effects are noted in the above figure. Furthermore. Similarly.9.11. only about 1 wt% of Pb is soluble in Sn (within the β-phase region at the left extremity).) (b) On this same graph schematically plot tensile strength versus composition at 150°C. This figure also shows that the tensile strength of pure lead is greater than pure tin. respectively.8 MPa and 14. as well as any differences between them. In addition.64. the compositional ranges over which solid-solution strengthening occurs increase somewhat from the room-temperature ranges. and for compositions over which both α and β phases coexist. according to Figure 9. although there is some disparity between the densities of Pb and Sn (11.24 for the elastic modulus (Problem 9.5 to 10 wt% Sn.(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 9.6. In addition.14). strength will decrease approximately linearly with increasing Sn content. Furthermore. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Furthermore. Also. respectively. these effects are also noted on the 150°C curve above. and the subscripts represent the alloy/phases.64). That is.6). the curve will be shifted to significantly lower tensile strengths inasmuch as tensile strength diminishes with increasing temperature (Section 6. Thus.35 versus 7. mass fractions of the α and β phases change linearly with changing composition (according to the lever rule). weight and volume fractions of the α and β phases will also be similar (see Equation 9. at 150°C. .8. it would be expected that the tensile strength of lead will be greater than that of tin. solubility limits for both α and β phases increase—for the α phase from 1. for this problem (TS) alloy @ (TS)a Va + (TS)b Vb in which TS and V denote tensile strength and volume fraction. Figure 6. and for the β phase from 1 to about 2 wt% Pb.27 g/cm3). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. At 150°C. 916)(39. AB = 1. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. consider the AB intermetallic compound. from which an identification may be made.92 g/mol). its composition in atomic percent is 50 at% A-50 at% B.3 wt% B. since we know the atomic weight of potassium (39. If the compositions for AB and AB 2 are 34.6a.10 g/mol. then.6a may be written in the form: ' ΧB = CB AA ´ 100 C A AB + CB AA where AA and AB are the atomic weights for elements A and B.7 wt% B)(AA ) ´ 100 (34. Probably the easiest way to solve this problem is to first compute the ratio of the atomic weights of these two elements using Equation 4. it is possible to determine the atomic weight of element B. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and then asks us to identify element B if element A is potassium.3 wt% A)(AB ) + (65. For this AB compound.Equilibrium Diagrams Having Intermediate Phases or Compounds 9. the atomic weight of element B is just AB = (1.7 wt% A–79.10 g/mol) = 74.10 g/mol.6) we note the element that has an atomic weight closest to this value is arsenic (74. inasmuch as it contains the same numbers of A and B atoms. Solution This problem gives us the compositions in weight percent for the two intermetallic compounds AB and AB 2. First of all. Equation 4. and the two intermetallic compounds are KAs and KAs2. per inside the front cover). .40 Two intermetallic compounds.3 wt% A–65.916 AA Since potassium is element A and it has an atomic weight of 39. AB and AB 2. and element A is potassium. element B is arsenic. exist for elements A and B.7 wt% B)(AA ) 50 at% B = Now.92 g/mol Upon consultation of the period table of the elements (Figure 2. and CA and CB are their compositions in weight percent. solving this expression yields. Therefore. and making the appropriate substitutions in the above equation leads to (65. identify element B. respectively.7 wt% B and 20. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. For incongruent there will be compositional alterations of the phases. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.41 What is the principal difference between congruent and incongruent phase transformations? Solution The principal difference between congruent and incongruent phase transformations is that for congruent no compositional changes occur with any of the phases that are involved in the transformation. .Congruent Phase Transformations Eutectoid and Peritectic Reactions 9. for which only single-phase regions are labeled. Solution Below is shown the aluminum-neodymium phase diagram (Figure 9. This reaction upon cooling is Λ® AlNd 3 + Nd There are four peritectics. The reaction upon cooling is L ® Al + Al11Nd 3 The other eutectic exists at about 97 wt% Nd-3 wt% Al and 635°C. peritectics. Its reaction upon cooling is as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Specify temperature-composition points at which all eutectics. There are two eutectics on this phase diagram. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.36 is the aluminum-neodymium phase diagram. One exists at 12 wt% Nd-88 wt% Al and 632°C. write the reaction upon cooling. One exists at 59 wt% Nd-41 wt% Al and 1235°C.36). .9. Also. and congruent phase transformations occur. for each. eutectoids.42 Figure 9. This reaction upon cooling is Λ + AlNd 2 ® AlNd 3 There is one congruent melting point at about 73 wt% Nd-27 wt% Al and 1460°C. This reaction upon cooling is Λ + Al 2 Nd ® AlNd The third peritectic exists at 91 wt% Nd-9 wt% Al and 795°C. This reaction upon cooling is Λ + AlNd ® AlNd 2 The fourth peritectic exists at 94 wt% Nd-6 wt% Al and 675°C. .Λ + Al 2 Nd ® Al11Nd 3 The second peritectic exists at 84 wt% Nd-16 wt% Al and 940°C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Its reaction upon cooling is Λ® Al 2 Nd No eutectoids are present. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9. Its reaction upon cooling is Λ ® Ti2Cu + TiCu There is one eutectoid for this system. It exists at about 7. Specify all temperature-composition points at which eutectics. for each. There is one eutectic on this phase diagram. . and congruent phase transformations occur.5 wt% Cu-92. write the reaction upon cooling.37 is a portion of the titanium-copper phase diagram for which only single-phase regions are labeled. peritectics.43 Figure 9.37). Solution Below is shown the titanium-copper phase diagram (Figure 9. This reaction upon cooling is β ® a + Ti2Cu There is one peritectic on this phase diagram. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.5 wt% Ti and 790°C. which exists at about 51 wt% Cu-49 wt% Ti and 960°C. It exists at about 40 wt% Cu-60 wt% Ti and 1005°C. Also. eutectoids. The reaction upon cooling is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 5 wt% Ti and 982°C.5 wt% Cu-42.β + L ® Ti2Cu There is a single congruent melting point that exists at about 57. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. . The reaction upon cooling is Λ ® TiCu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. the solubility of A in B is 8 wt% A.44 Construct the hypothetical phase diagram for metals A and B between temperatures of 600 °C and 1000°C given the following information: ● ● ● ● ● ● ● ● ● ● ● ● The melting temperature of metal A is 940°C. The intermetallic compound AB2 exists at 67 wt% B–33 wt% A. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. At 600°C. Solution Below is shown the phase diagram for these two A and B metals. The maximum solubility of A in B is 12 wt% A. The intermetallic compound AB exists at 51 wt% B–49 wt% A. One eutectic occurs at 700°C and 75 wt% B–25 wt% A. The melting temperature of metal B is 830°C. which occurs at 700°C. A second eutectic occurs at 730°C and 60 wt% B–40 wt% A.9. One congruent melting point occurs at 780°C and 51 wt% B–49 wt% A. . The solubility of B in A is negligible at all temperatures. A third eutectic occurs at 755°C and 40 wt% B–60 wt% A. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. A second congruent melting point occurs at 755°C and 67 wt% B–33 wt% A. temperature and pressure. the phase rule now becomes P+F=1+2=3 Or F=3–P where P is the number of phases present at equilibrium. and C on the pressure-temperature diagram for H2O. and C.45 In Figure 9. the number of externally controllable variables that need be specified to completely define the system. is 2--viz. . the number of noncompositional variables. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. B.38. whereas N. Gibbs phase rule in general form is P+F=C+N For this system. specify the number of degrees of freedom at each of the points—that is. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. which is shown below. the number of components. is 1. B. Solution We are asked to specify the value of F for Gibbs phase rule at points A. Apply the Gibbs phase rule at points A. Figure 9.38 is shown the pressure–temperature phase diagram for H 2O. C. that is.The Gibbs Phase Rule 9. Thus. point A is an invariant point (in this case a triple point). At point B on the figure. P = 1). three phases are present (viz. hence F=3–P=3–2=1 Or that we need to specify the value of either temperature or pressure. thus. ice III. which determines the value of the other parameter (pressure or temperature). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. the number of degrees of freedom is zero since F=3–P=3–3=0 Thus.e. finally.At point A. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .. only a single (vapor) phase is present (i. And. two phases are in equilibrium (P = 2). ice I. at point C which is on the phase boundary between liquid and ice I phases. and liquid) and P = 3. and we have no choice in the selection of externally controllable variables in order to define the system. or F=3–P=3–1=2 which means that specification of both temperature and pressure are necessary to define the system. 70 . for cementite ΩΦε Χ = 3 Χ0 − Χα 0. since CFe C = 6. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.70 . C0 = 0. Solution This problem asks that we compute the mass fractions of α ferrite and cementite in pearlite.76 . .022 wt% C 3 Ωα = 6.0.70 .022 3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.0.46 Compute the mass fractions of α ferrite and cementite in pearlite. and Cα = 0.70 wt% C. The lever-rule expression for ferrite is Ωα = ΧΦε Χ − Χ0 ΧΦε Χ − Χα 3 3 and.The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram Development of Microstructure in Iron-Carbon Alloys 9.0.11 ΧΦε Χ − Χα 6.022 Similarly.022 = = 0.89 6.76 wt% C.0.76 = 0. 022 wt% C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. a “hypereutectoid” steel has a carbon content greater than the eutectoid. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (b) For a hypoeutectoid steel. on the other hand.9. What will be the carbon concentration in each? Solution (a) A “hypoeutectoid” steel has a carbon concentration less than the eutectoid. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. both eutectoid and proeutectoid ferrite exist.47 (a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel. The carbon concentration for both ferrites is 0. the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. . Explain the difference between them. 94 = CFe3C .48 What is the carbon concentration of an iron–carbon alloy for which the fraction of total ferrite is 0.94. .94? Solution This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.C0 6. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.C 0 ' 6.12) yields ' CFe 3C .42 wt% C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.022 Ωα = 0. Application of the lever rule (of the form of Equation 9.70 .70 .Ca = ' and solving for Χ0 ' Χ0 = 0.0.9. solving for C0. the proeutectoid phase is α-ferrite since C0 is less than 0.56 wt% C. leads to C0 = 0.49 What is the proeutectoid phase for an iron–carbon alloy in which the mass fractions of total ferrite and total cementite are 0.0.C 0 CFe 3C . respectively) for an iron-carbon alloy α 3 and then are asked to specify the proeutectoid phase.70 . Employment of the lever rule for total α leads to Ωα = 0.76 wt% C. Therefore.Ca = 6.08.92 and 0.08.70 .92 and 0. the alloy composition. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.92 = CFe 3C .C 0 6.9. respectively? Why? Solution In this problem we are given values of W and WFe C (0.022 Now. . Likewise.15 wt% C is greater than the eutectoid composition (0. Solution (a) The proeutectoid phase will be Fe3C since 1. . Applying Equation 9. in which Χ ' = 1.76 wt % C).Ca 6.9.70 . Similarly.0.1.83 kg of total ferrite.022 And the mass of total cementite that forms is (0.70 . the microstructure would appear as: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (d) Schematically.0 kg) = 0.94 5.0.70 .23 C1' .0.76 = = 0.70 .15 .022 which.17)(1.76 1.07 kg of the total 1. Application of the appropriate lever rule expression yields Ωα = CFe 3C .C 0 CFe 3C .1.0 kg mass. when multiplied by the total mass of the alloy (1.76 6. ΩFe 3C = C 0 .0 kg of austenite containing 1.0.50 Consider 1.17 kg. (b) For this portion of the problem.0.70 .07 5.C 1 6.93 6.15 wt% C 1 ' 6.022 = = 0.Ca 1. (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. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.93 kg.0 kg). we are asked to determine how much total ferrite and cementite form.22.17 CFe 3C . for total cementite.70 . (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. gives 0.Ca = 6.15 = = 0.0.15 .94 ΩFe 3C' = which is equivalent to 0. from Equation 9.15 wt% C.0.15 = 0.70 .83 6.76 Ωp = which corresponds to a mass of 0. cooled to below 727°C (1341°F). . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 27 kg of total ferrite. . Ωα' = 0.65 .C 0 CFe 3C .65 = 0.12 kg of pearlite.65 wt% C. ΩFe 3C = C 0 .09)(2.76 wt% C.15 0.20 ' C 0 .9.91)(2.0.70 .85 0.5 kg) = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.022 0. Using Equation 9. there are (0.022 = = 0.0.022 Or (0. Similarly.5 kg) = 0. cooled to below 727°C (1341°F).Ca 0.5 kg) = 2.76 .09 CFe 3C . application of the appropriate lever rule expression yields Ωα = CFe 3C .022 = = 0.0. For ferrite.38 kg of proeutectoid ferrite.15)(2.70 .0. from Equation 9.65 wt% C is less than 0. we are asked to determine how much total ferrite and cementite form. (b) For this portion of the problem.21.74 Ωp = This corresponds to (0.85)(2.70 . (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) Ferrite is the proeutectoid phase since 0.65 .0. the microstructure would appear as: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.74 Or.51 Consider 2. (c) Now consider the amounts of pearlite and proeutectoid ferrite.74 0.5 kg) = 2.65 = 0.022 which corresponds to (0. Also.91 6.0. for total cementite.0.Ca = 6.23 kg of total cementite form. (d) Schematically.Ca 6.5 kg of austenite containing 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. . 25 wt% C.74 Ωα' = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.9.74 0.0.022 = = 0.74 0.25 wt% C iron-carbon alloy are considered in this problem.52 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy containing 0.69 0.25 .76 . Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.C 0 0.0.20 ' C 0 . .21 (for proeutectoid ferrite) ' 0.022 0.25 = = 0.74 Ωp = And.31 0.76 . From Equation 9. from Equation 9.0. respectively. the mass fractions of these two microconstituents are 0.20 ' C 0 .022 0.714 = which yields Χ ' = 0. From Equation 9.74 Determine the concentration of Ω p = 0.9. 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.53 The microstructure of an iron–carbon alloy consists of proeutectoid ferrite and pearlite. Solution This problem asks that we determine the carbon concentration in an iron-carbon alloy. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .55 wt% C.286 and 0.0.714. carbon in this alloy. given the mass fractions of proeutectoid ferrite and pearlite. 70 .82 wt% C.C 0 6. α 3 and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid. Therefore.Ca = 6.9. respectively. .88 = CFe 3C .76 wt% C.12.022 Now.88 and 0. respectively).C 0 CFe 3C .12. Is this a hypoeutectoid or hypereutectoid alloy? Why? Solution In this problem we are given values of W and WFe C for an iron-carbon alloy (0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. the alloy is hypereutectoid since C0 is greater than 0.70 .0.54 The mass fractions of total ferrite and total cementite in an iron-carbon alloy are 0.88 and 0. solving for C0. leads to C0 = 0. Employment of the lever rule for total α leads to Ωα = 0. the alloy composition. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 61 wt% C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 0 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Equation 9.80. leads to Ωp = ' C 0 . Solution We are asked in this problem to determine the concentration of carbon in an alloy for which Ωα∋ = 0.20.0. respectively.20 and 0. Determine the concentration of carbon in this alloy.74 Solving for Χ ' yields Χ ' = 0.20 and Wp = 0. . the mass fractions of these microconstituents are 0. If we let Χ ' equal the carbon concentration in the alloy. employment of the appropriate lever rule 0 expression.80.9.80 0.55 The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite.022 = 0. 76 .98 kg of proeutectoid ferrite forms. it first becomes necessary to compute the amount of total ferrite using the lever rule applied entirely across the α + Fe C phase field. (0. or 1.022 = = 0.40 = = 0.0 kg) = 1.94)(2.0. (b) In order to determine the amount of eutectoid ferrite. the amount of eutectoid ferrite is just the difference between total and proeutectoid ferrites.0 kg) = 0.88 kg – 0.057)(2.9. Thus. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.98 kg = 0.C0 CFe 3C . leads to Õ C 0 .022 ΩFe 3C = which amounts to (0.0. (a) How many kilograms of proeutectoid ferrite form? (b) How many kilograms of eutectoid ferrite form? (c) How many kilograms of cementite form? Solution In this problem we are asked to consider 2. ' 0.49 0.0.0.4 wt% C alloy that is cooled to a temperature below the eutectoid. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.057 CFe 3C .49)(2.74 0.0 kg of a 99. as 3 Ωα = Õ CFe 3C .94 6.Ca 0.76 .21 must be used in computing the amount of proeutectoid ferrite that forms.40 = 0.0 kg) = 0.0.C 0 0.88 kg. Now.70 .40 .4 wt% C alloy that is cooled to a temperature just below the eutectoid.0 kg of a 99.90 kg (c) With regard to the amount of cementite that forms.56 Consider 2.022 which corresponds to (0. (a) Equation 9.70 .70 .6 wt% Fe–0. again application of the lever rule across the entirety of the α + Fe3C phase field.Ca = 6.114 kg cementite in the alloy.74 Ωα' = Or.6 wt% Fe-0.Ca 6. . 76 2. Thus.9. This requires that we utilize Equation 9. ∋ Χ1 − 0.94 5.57 Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron– carbon alloy. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .76 = = 0.23 with Χ1 = 2. Solution This problem asks that we compute the maximum mass fraction of proeutectoid cementite possible for a ∋ hypereutectoid iron-carbon alloy.0.14 .232 5. the maximum solubility of carbon in austenite.14 wt% C.94 ΩΦε Χ∋ = 3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 05 wt% C.76 = 0. if both alloy composition values are equal.0.23 as ' C1 . Now for WFe 3C¢ we utilize Equation 9.Ca = 6.9. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 1 1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.049 5. The expression for the mass fraction of total ferrite is Ωα = CFe 3C .05 wt% C.846 and WFe C¢ = 0.70 .846 and 0. And.70 .94 ΩFe 3C' = This expression leads to Χ ' = 1.58 Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0. In α 3 order to make this determination.0. it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition. this alloy is possible.C 0 CFe 3C . then such an alloy is possible. then to solve for the alloy composition of each.049. since C0 = Χ ' .049. respectively? Why or why not? Solution This problem asks if it is possible to have an iron-carbon alloy for which W = 0.846 6.C 0 = 0. .022 Solving for this C0 yields C0 = 1. 74 Ωp = This expression leads to Χ '0 = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. respectively? Why or why not? Solution This problem asks if it is possible to have an iron-carbon alloy for which WFe 3C = 0. Since C0 and Χ '0 are different.9. .Ca 6.20 for Wp as C '0 . then to solve for the alloy composition of each.417 0.039 and Wp = 0. then such an alloy is possible.33 wt% C.417.039 and 0.022 = 0 = 0. this alloy is hypoeutectoid since C0 is less than the eutectoid composition (0. In order to make this determination.28 wt% C.70 .022 = 0. if both alloy composition values are equal. it is necessary to use Equation 9.417.0.59 Is it possible to have an iron-carbon alloy for which the mass fractions of total cementite and pearlite are 0.022 Solving for this C0 yields C0 = 0. Thus.039 CFe 3C .0.Ca C . Therefore. The expression for the mass fraction of total cementite is ΩFe 3C = C 0 . this alloy is not possible. it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition.76 wt% ).0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.446 0. To calculate the mass fraction of total ferrite.939 –0.9.43 = = 0.939 6.76 .60 Compute the mass fraction of eutectoid ferrite in an iron-carbon alloy that contains 0. the mass fraction of eutectoid ferrite Wα'' is just Wα'' = Wα – Wα' = 0. finally.493 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.022 Now.43 wt% C.70 . Solution In order to solve this problem it is necessary to compute mass fractions of total and proeutectoid ferrites. it is necessary to use the lever rule and a tie line that extends across the entire α + Fe3C phase field as Ωα = CFe 3C .76 .0.0.C 0 CFe 3C .0.74 Ωα' = And. and then to subtract the latter from the former. for the mass fraction of proeutectoid ferrite we use Equation 9.70 .Ca = 6.43 = 0.446 = 0.21 as ' 0.74 0.C 0 0. . 0. what is its composition? If this is not possible. it is possible to determine the alloy composition. that is WFe C'' = WFe C – WFe C' 3 3 3 Now. Yes. and if so.104.0. For the first. For the second possibility.022 5.61 The mass fraction of eutectoid cementite in an iron-carbon alloy is 0. . solving for C0 yields C0 = 1. Thus.022 And.9. Thus.0. ΩΦε Χ∀ = 3 Χ0 − Χα Χ − 0.76 − 0 ΧΦε Χ − Χα 5.Ca C . explain why. is it possible to determine the composition of the alloy? If so. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. it is necessary to set up a lever rule expression wherein the mass fraction of total cementite is 0. the alloy composition.72 wt% C.12) and WFe C' (Equation 9. it is possible to write expressions for WFe C (of the form of Equation 9.70 . Solution This problem asks whether or not it is possible to determine the composition of an iron-carbon alloy for which the mass fraction of eutectoid cementite is 0. For this case the mass fraction of eutectoid cementite (WFe C'') is just the 3 difference between total cementite and proeutectoid cementite mass fractions.022 C 0 .94 And.104 6.76 = 0.0.104.23) in terms 3 3 of C0. there are two possible answers. we have a hypoeutectoid alloy wherein all of the cementite is eutectoid cementite. to calculate the composition. ΩFe 3C = C 0 .11 wt% C.70 . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.0. in fact.104 CFe 3C . and.104.94 3 = C 0 .Ca 6. Therefore.022 = 0 = 0. solving for C0 yields C0 = 0. the eutectoid cementite exists in addition to proeutectoid cementite. On the basis of this information. in fact. is it possible to determine the composition of the alloy? If so. Therefore.12) and W ' (Equation 9. it is possible to determine the alloy composition. Thus. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.C 0 0. Yes.74 = 6. Solution On the basis of this information. and.82. it is possible to write expressions for W (of the form of Equation 9.70 wt% C. α α the alloy composition.74 And.9. explain why.022 And.C 0 = 0. Ωα = CFe 3C .C 0 0. it is necessary to set up a lever rule expression wherein the mass fraction of total ferrite is 0.0.76 .70 .82 6. For the first.82 6.22 wt% C. Ω α∀ = ΧΦε Χ − Χ0 ΧΦε Χ − Χα 3 3 − 0. and if so.21) in terms of C0. solving for C0 yields C0 = 1.70 . to calculate the composition.82.C 0 = 0.70 .0.Ca = 6. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.C 0 CFe 3C . we have a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite. possible.76 . Thus. there are two possible answers. what is its composition? If this is not This problem asks whether or not it is possible to determine the composition of an iron-carbon alloy for which the mass fraction of eutectoid ferrite is 0. that is W '' = W – W ' α α α Now.62 The mass fraction of eutectoid ferrite in an iron-carbon alloy is 0. For the second possibility. For this case the mass fraction of eutectoid ferrite (W '') is just the difference α between total ferrite and proeutectoid ferrite mass fractions. . solving for C0 yields C0 = 0.70 .022 0. the eutectoid ferrite exists in addition to proeutectoid ferrite.82. A vertical line at a composition of 5 wt% C-95 wt % Fe has been drawn. .24). On the basis of the locations of the three temperature-composition points. 1175°C. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. schematic sketches of the respective microstructures along with phase compositions are represented as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.9. Label the phases and indicate their compositions (approximate). horizontal arrows at the three temperatures called for in the problem statement (i.63 For an iron-carbon alloy of composition 5 wt% C-95 wt% Fe. Solution Below is shown the Fe-Fe3C phase diagram (Figure 9.e. in addition. make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1175 °C (2150°F). and 700°C (1290°F).. 1145°C (2095 °F). and 700°C). 1145°C. and. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9.64 Often, the properties of multiphase alloys may be approximated by the relationship E (alloy) = EαVα + EβVβ (9.24) 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. Employ the relationship above to determine the approximate Brinell hardness of a 99.80 wt% Fe–0.20 wt% C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volume fractions may be approximated by mass fractions. Solution This problem asks that we determine the approximate Brinell hardness of a 99.80 wt% Fe-0.20 wt% C alloy, using a relationship similar to Equation 9.24. First, we compute the mass fractions of pearlite and proeutectoid ferrite using Equations 9.20 and 9.21, as C '0 - 0.022 0.20 - 0.022 = = 0.24 0.74 0.74 0.76 - C '0 0.76 - 0.20 = = 0.76 0.74 0.74 Ωp = Ωα' = Now, we compute the Brinell hardness of the alloy using a modified form of Equation 9.24 as HBalloy = HBa ' Wa ' + HB pW p = (80)(0.76) + (280)(0.24) = 128 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. The Influence of Other Alloying Elements 9.65 A steel alloy contains 97.5 wt% Fe, 2.0 wt% Mo, and 0.5 wt% C. (a) What is the eutectoid temperature of this alloy? (b) What is the eutectoid composition? (c) What is the proeutectoid phase? Assume that there are no changes in the positions of other phase boundaries with the addition of Mo. Solution (a) From Figure 9.34, the eutectoid temperature for 2.0 wt% Mo is approximately 850°C. (b) From Figure 9.35, the eutectoid composition is approximately 0.22 wt% C. (c) Since the carbon concentration of the alloy (0.5 wt%) is greater than the eutectoid (0.22 wt% C), cementite is the proeutectoid phase. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and 0.0 0.20 ' C0 .2 wt%) is less than the eutectoid (0.62 Likewise. Modifying Equation 9. Since the carbon concentration in the alloy (0. the eutectoid composition is approximately 0.32 0.0 0.62 wt% C).62 .66 A steel alloy is known to contain 93.0 wt% Ni.62 Wp = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. .9.62 wt% C.34. the proeutectoid phase is ferrite.0 0.0 wt% Ni is approximately 650°C (1200°F).68 0.20 = = 0. 6.62 .62 .2 wt% C. (b) From Figure 9.20 = = 0.8 wt% Fe. (c) Assume that the α–(α + Fe3C) phase boundary is at a negligible carbon concentration. the eutectoid temperature for 6. Solution (a) From Figure 9. (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 amounts of the proeutectoid phase and pearlite.0. using a modified Equation 9.62 . Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.35.21 leads to Ωα' = 0.C '0 0.