HW7 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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Unformatted text preview: 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 07 Question 10.2: (a) Rewrite the expression for the total free energy change for nucleation (Equation 10.1) for the case of a cubic nucleus of edge length a (instead of a sphere of radius r). Now differentiate this expression with respect to a (per Equation 10.2) and solve for both the critical cube edge length, a*, and also G*. (b) Is G* greater for a cube or a sphere? Why? Solution: (a) This problem first asks that we rewrite the expression for the total free energy change for nucleation (analogous to Equation 10.1) for the case of a cubic nucleus of edge length a . The volume of such a cubic radius is a 3 , whereas the total surface area is 6 a 2 (since there are six faces each of which has an area of a 2 ). Thus, the expression for G is as follows: G = a 3 G v + 6 a 2 Differentiation of this expression with respect to a is as d G da = d ( a 3 G v ) da + d ( 6 a 2 ) da = 3 a 2 G v + 12 a If we set this expression equal to zero as 3 a 2 G v + 12 a = 0 and then solve for a (= a *), we have a * = 4 G v Substitution of this expression for a in the above expression for G yields an equation for G * as G * = ( a *) 3 G v + 6( a* ) 2 = 4 G v 3 G v + 6 4 G v 2 = 32 3 ( G v ) 2 (b) G v for a cubei.e., (32) 3 ( G v ) 2 is greater that for a spherei.e., 16 3 3 ( G v ) 2 = (16.8) 3 ( G v ) 2 . The reason for this is that surface-to-volume ratio of a cube is greater than for a sphere. Question 10.3: If copper (which has a melting point of 1085C) homogeneously nucleates at 849C, calculate the critical radius given values of 1.77 10 9 J/m 3 and 0.200 J/m 2 , respectively, for the latent heat of fusion and the surface free energy. Solution: This problem states that copper homogeneously nucleates at 849 C, and that we are to calculate the critical radius given the latent heat of fusion (1.77 10 9 J/m 3 ) and the surface free energy (0.200 J/m 2 ). Solution to this problem requires the utilization of Equation 10.6 as r * = 2 T m H f 1 T m T = (2) ( 0.200 J /m 2 ) (1085 + 273 K) 1.77 10 9 J /m 3 1 1085 C 849 C = 1.30 10 9 m = 1.30 nm Question 10.4: (a) For the solidification of iron, calculate the critical radius r* and the activation free energy G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are 1.85 10 9 J/m 3 and 0.204 J/m 2 , respectively. Use the supercooling value found in Table 10.1....
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This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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HW7 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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