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e45_hw_7

# e45_hw_7 - E45 HW 07 Due References J.F Shackelford...

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E45 HW 07 Due October 22, 2010 References: J.F. Shackelford, Introduction to Materials Science for Engineers, 6 th Edition, Prentice Hall, New Jersey (2005) Problem 1 10.3 Although Section 10.1 concentrates on crystal nucleation and growth from a liquid, similar kinetics laws apply to solid-state transformations. For example, Eq. 10.1 can be used to describe the rate of β precipitation upon cooling supersaturated α phase in 10 % wt Sn-90 wt % Pb alloy. Given precipitation rates of 3.77 x 10 3 s -1 and 1.40 x 10 3 s -1 at 20 °C and 0 °C, respectively, calculate the activation energy for this process. SOLUTION Ġ = C e (-Q/RT) (Source: Shackelford, Eq. 10.1, p. 359) R = 8.314 J/(mol K) T = 273 K, 293 K Ġ 293K (growth rate at 20 °C) = 3.77 x 10 3 s -1 Ġ 273K (growth rate at 0 °C) = 1.40 x 10 3 s -1 We can find Q by writing separate eqns for 20 °C and 0 °C and setting them equal to each other. Ġ 293K 3.77 x 10 3 s -1 C e -Q / [8.314J/(mol K) * 293K] e -Q / (2436 J/mol) 2.69 Ġ 273K 1.40 x 10 3 s -1 C e -Q / [8.314J/(mol K) * 273K] e -Q / (2269.72 J/mol) ln (2.69) = ln e * Q [ 1/(2269.72 J/mol) - 1/(2436 J/mol) ] Q = 0.98 / [3.01 x 10 -5 J -1 mol] = 32560 J/mol = 32.6 kJ/mol Problem 2 10.13 a. A eutectoid steel is cooled at a steady rate from 727 to 200 °C in exactly 1 day. Superimpose this

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