Chapter 10 homework solutions (part 2).pdf
Chapter 10 homework solutions (part 2).pdf

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School
University of Toronto St. George
Department
Materials Science & Engineering
Course
MSE101H1
Professor
Scott Ramsay
Semester
Spring

Description
CHAPTER 10 PHASE DIAGRAMS PROBLEM SOLUTIONS Development of Microstructure in Eutectic Alloys 10.16 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. Solution Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum. 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. 10.17 Is it possible to have a magnesium–lead alloy in which the mass fractions of primary α and total α are 0.60 and 0.85, respectively, at 460°C (860°F)? Why or why not? Solution In order to make this determination we need to set up the appropriate lever rule expression for each of these quantities. From Figure 10.20 and at 460C, C = 41 wt% Mg 2b = 81 wt% Pb, and eutectic67 wt% Pb. For primary  W = C eutecticC0 = 67  C0 = 0.60 ' C C 67  41 eutectic  Solving for0C gives0C = 51.4 wt% Pb.  Now the analogous expression for total  C  C Mg 2b 0 81  C0 W  C  C = 81  41 = 0.85 Mg2Pb  which yields a value of 47 wt% Pb for0C . Therefore, since these t0o C values are different, this alloy is not  possible. 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. 10.18 For a lead–tin alloy of composition 80 wt% Sn–20 wt% Pb and at 180°C (355°F) do the following: (a) Determine the mass fractions of α and β phases. (b) Determine the mass fractions of primary β and eutectic microconstituents. (c) Determine the mass fraction of eutectic β. Solution (a) This portion of the problem asks that we determine the mass fractions of  and  phases for an 80 wt% Sn-20 wt% Pb alloy (at 180C). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the  +  phase field. From Figure 10.8 and at180C, C = 18.3 w% Sn, C = 97.8 wt% Sn, and eutectic 61.9 wt% Sn. Therefore, the two lever-rule expressions are as follows: C  C W =  0 = 97.8  80 = 0.224  C C  97.8  18.3 C0 C  80  18.3  W  C  C = 97.8  18.3 = 0.776   (b) Now it is nessary to determine the mass fractions of primary  and eutectic microconstituents for this same alloy. This requires that we utilize the lever rule and a tie line that extends from the maximum solubility of Pb in the  phase at 180C (i.e., 97.8 wt% Sn) to the eutectic composition (61.9 wt% Sn). Thus C0 C eutectic 80.0  61.9 W '= C  C = 97.8  61.9 = 0.504  eutectic C  C 97.8  80.0 W =  0 = = 0.496  e C C eutectic 97.8  61.9 (c) And, finally, we are asked to compute the mass fraction of euectic , W . This quantity is simply the  difference between the mass fractions of total  and primary  as W e= W  W '= 0.776 – 0.504 = 0.272 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. 10.21 For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 950°C (1740°F), 860°C (1580°F), 800°C (1470°F), and 600°C (1100°F). Label all phases and indicate their approximate compositions. Solution The illustration below is the Cu-Zn phase diagram (Figure 10.19). A vertical line at a composition of 52 wt% Zn-48 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement (i.e., 950C, 860C, 800C, and 600C). On the basis of the locations of the four temperature-composition points, schematic sketches of the four respective microstructures along with phase compositions are represented as follows: 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. 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. Congruent Phase Transformations Binary Eutectic Systems Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions 10.24 Figure 10.40 is the tin–gold phase diagram, for which only single-phase regions are labeled. Specify temperature–composition points at which all eutectics, eutectoids, peritectics, and congruent phase transformations occur. Also, for each, write the reaction upon cooling. Solution Below is shown the tin-gold phase diagram (Figure 10.40). There are two eutectics on this phase diagram. One exists at 10 wt% Au-90 wt% Sn and 217C. The reaction upon cooling is L   +  The other eutectic exists at 80 wt% Au-20 wt% Sn and 280C. This reaction upon cooling is  L   +   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. There are three peritectics. One exists at 30 wt% Au-70 wt% Sn and 252C. Its reaction upon cooling is as follows: L +    The second peritectic exists at 45 wt% Au-55 wt% Sn and 309C. This reaction upon cooling is  L +    The third peritectic exists at 92 wt% Au-8 wt% Sn and 490C. This reaction upon cooling is  L +    There is one congruent melting point at 62.5 wt% Au-37.5 wt% Sn and 418C. Its reaction upon cooling is  L   No eutectoids are present.  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. Ceramic Phase Diagrams 10.26 From Figure 10.24, the phase diagram for the MgO–Al O sy2te3, it may be noted that the spinel solid solution exists over a range of compositions, which means that it is nonstoichiometric at compositions other than 50 mol% MgO–50 mol% Al O . 2 3 (a) The maximum nonstoichiometry on the Al O 2r3ch side of the spinel phase field exists at about 2000°C (3630°F) corresponding to approximately 82 mol% (92 wt%) Al O . 2e3ermine the type of va
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