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Unformatted text preview: EMSE 201 – Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 1 of 4 Case School of Engineering Solution to Homework #7 Explain how answers were obtained. Give appropriate units for all numerical answers. 1) (6 points) C&R problem 10.18. Solution (modified from the publisher’s solution manual; used with permission) a) This portion of the problem asks that we determine the mass fractions of α and β phases for an 80 wt% Sn-20 wt% Pb alloy (at 180 ° C). Use the lever rule, placing a tie line entirely across the α + β phase field. In problems like this one, unless a temperature is specified, set the tie line at the eutectic temperature. That tie line gives the compositions that the α and β phases will have when they first form during the eutectic solidification reaction. In actual instances of cooling to lower temperatures, equilibration of the compositions of the solid phases (to follow the solvus curves) would require solid-state diffusion, a relatively slow process (and one that gets even slower as temperature drops). Under normal cooling rates, it can be assumed that the compositions of the solids stay “locked in” at the values they had at the eutectic temperature. From Figure 10.8 and at 183 ° C, C α = 18.3 wt% Sn, C β = 97.8 wt% Sn, and C eutectic = 61.9 wt% Sn. Therefore, the two lever-rule expressions are as follows: W α = C β − C C β − C α = 97.8 − 80 97.8 − 18.3 = 0.224 (1 pt) W β = C − C α C β − C α = 80 − 18.3 97.8 − 18.3 = 0.776 (1 pt) b) Now determine the mass fractions of primary β and eutectic microconstituents for this same alloy. Utilize the lever rule and a tie line that extends from the maximum solubility of Pb in the β phase at 183 ° C (i.e., 97.8 wt% Sn) to the eutectic composition (61.9 wt% Sn). Thus W β ’ = C − C eutectic C β − C eutectic = 80.0 − 61.9 97.8 − 61.9 = 0.504 (1 pt) W e =...
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This document was uploaded on 11/08/2011.
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