10_3

Materials Science and Engineering: An Introduction

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229 The two equations are thus ln ln [ ] 1 1 - 0.2 = ln k + n ln(12.6 s) ln ln [ ] 1 1 - 0.8 = ln k + n ln(28.2 s) Solving these two expressions simultaneously for n and k yields n = 2.453 and k = 4.46 x 10 -4 . Now it becomes necessary to solve for the value of t at which y = 0.95. Algebraic manipulation of Equation (10.1) leads to an expression in which t is the dependent parameter as t = [ ] - ln(1 - y) k 1/n = - ln(1 - 0.95) 4.64 x 10 -4 1/2.453 = 35.7 s 10.6 For this problem, we are given, for the recrystallization of steel, two values of y and two values of the corresponding times, and are asked to determine the fraction recrystallized after a total
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Unformatted text preview: time of 22.8 min. The first thing necessary is to set up two expressions of the form of Equation (10.1), and then to solve simultaneously for the values of n and k . Rearrangement of Equation (10.1) and taking natural logarithms twice, leads to ln           ln [ ] 1 1 - y = ln k + n ln t The two equations are thus ln           ln [ ] 1 1 - 0.2 = ln k + n ln(13.1 min) ln           ln [ ] 1 1 - 0.7 = ln k + n ln(29.1 min)...
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This document was uploaded on 05/04/2010.

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