This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 surfacetovolume 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....
View
Full
Document
This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.
 Spring '11
 tsakal

Click to edit the document details