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Unformatted text preview: 9.52 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an ironcarbon alloy containing 0.25 wt% C. Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0.25 wt% C ironcarbon alloy are considered in this problem. From Equation 9.20 W p = C ' ! 0.022 0.74 = 0.25 ! 0.022 0.74 = 0.31 And, from Equation 9.21 (for proeutectoid ferrite) W ! ' = 0.76 " C ' 0.74 = 0.76 " 0.25 0.74 = 0.69 9.54 The mass fractions of total ferrite and total cementite in an ironcarbon alloy are 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why? Solution In this problem we are given values of W and W Fe 3 C for an ironcarbon alloy (0.88 and 0.12, respectively), and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid. Employment of the lever rule for total leads to W ! = 0.88 = C Fe 3 C " C C Fe 3 C " C ! = 6.70 " C 6.70 " 0.022 Now, solving for C , the alloy composition, leads to C = 0.82 wt% C. Therefore, the alloy is hypereutectoid since C is greater than 0.76 wt% C. 9.54 The mass fractions of total ferrite and total cementite in an ironcarbon alloy are 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why? Solution In this problem we are given values of W and W Fe 3 C for an ironcarbon alloy (0.88 and 0.12, respectively), and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid. Employment of the lever rule for total leads to W ! = 0.88 = C Fe 3 C " C C Fe 3 C " C ! = 6.70 " C 6.70 " 0.022 Now, solving for C , the alloy composition, leads to C = 0.82 wt% C. Therefore, the alloy is hypereutectoid since C is greater than 0.76 wt% C. 9.55 The microstructure of an ironcarbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these microconstituents are 0.20 and 0.80, respectively. Determine the concentration of carbon in this alloy....
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This document was uploaded on 11/08/2011.
 Spring '09

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