HW-05%20Solutions - #1 9.49 In this problem we are given...

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9.49 In this problem we are given values of W α and W Fe 3 C (0.86 and 0.14, respectively) for an iron- carbon alloy and then are asked to specify the proeutectoid phase. Employment of the lever rule for total α leads to W α = 0.86 = C Fe 3 C C 0 C Fe 3 C C α = 6.70 C 0 6.70 0.022 Now, solving for C 0 , the alloy composition, leads to C 0 = 0.96 wt% C. Therefore, the proeutectoid phase is Fe 3 C since C 0 is greater than 0.76 wt% C.
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9.52 The mass fractions of proeutectoid ferrite and pearlite that form in a 0.35 wt% C iron-carbon alloy are considered in this problem. From Equation 9.20 W p = C 0 ' 0.022 0.74 = 0.35 0.022 0.74 = 0.44 And, from Equation 9.21 (for proeutectoid ferrite) W α ' = 0.76 C 0 ' 0.74 = 0.76 0.35 0.74 = 0.56
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9.57 This problem asks that we compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy. This requires that we utilize Equation 9.23 with = 2.14 wt% C, the maximum solubility of carbon in austenite. Thus, C 1 ' W Fe 3 C' = C 1 ' 0.76 5.94 = 2.14 0.76 5.94 = 0.232
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10.22 Below is shown an isothermal transformation diagram for a 1.13 wt% C iron-carbon alloy, with time-temperature paths that will produce (a) 6.2% proeutectoid cementite and 93.8% coarse pearlite; (b) 50% fine pearlite and 50% bainite; (c) 100% martensite; and (d) 100% tempered martensite.
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This note was uploaded on 09/15/2010 for the course MATSCIE 250 taught by Professor Yalisove during the Fall '08 term at University of Michigan.

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HW-05%20Solutions - #1 9.49 In this problem we are given...

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