SolutionsHW8

SolutionsHW8 - HW 8 Problem 10.2 a. To Find: (a) The...

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HW 8 Problem 10.2 a. To Find: (a) The expression for total free energy change for a cubic nucleus, critical cube edge length, a*, and G* (b) Compare G* for a cube with that for a sphere b. Given: Nucleus is cubic; edge length of the cubic nucleus = ‘a’ units c. Assumptions: 1. Nucleation is homogeneous/no impurities are present in the material. 2. We are asked to compare the G* for a cubic nucleus with the G* for spherical nucleus of the same material . (This allows us to use the same values of G v and for both cases.) d. Solution: (a) Step 1: Volume of a cubic nucleus of edge length ‘a’ = a 3 Total surface area a cubic nucleus of edge length ‘a’ = 6 a 2 (since there are six faces each of which has an area of a 2 ). Step 2: Equation 10.1 for the case of the spherical nucleus of radius ‘r’ can be re-written as: G = [(Volume of spherical nucleus (of radius r)) * G v ] + [(Surface area of spherical nucleus (of radius r))* ] Analogously, for the case of the cubic nucleus of edge length ‘a’ : G = [(Volume of cubic nucleus (of edge length a)) * G v ] + [(Surface area of cubic nucleus (of edge length a))* ] G = a 3 G v 6 a 2 …(1)
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Step 3: Differentiating this expression with respect to a : da a d da G a d da G d v ) ( ) ( 2 3 6 = 3 a 2 G v 12 a Step 4: To calculate a*, we need to set this expression equal to zero. 0 = da G d => 3 a 2 G v 12 a   0 Solving for a (= a*) : a * = 4 G v …(2) Step 5: Substituting the value of a* from equation (2) for ‘a’ in equation (1) gives the expression for G * : G * = ( a *) 3 G v 6 ( a* ) 2 4 G v 3 G v 6   4 G v 2 * G 32 3 ( G v ) 2 …(3) (b) From equation 10.3, for a spherical nucleus of radius ‘r’ : * G 2 3 ) 3( 16 v G …(4) Comparing (3) and (4), G * is greater for a cube than for a sphere. This is because the surface-to-volume ratio for a cube is greater than that for a sphere. (a) ; ; G * (b) G * cubic nucleus > G * spherical nucleus since the surface-to-volume ratio for a cube is greater than that for a sphere.
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Problem 10.4 a. To Find: (a) r*and G* for Fe. (b) Number of atoms in a nucleus of critical size. b. Given: Heat of fusion, H f = -1.85 X 10 9 J/m 3 = 0.204 J/m 2 T= 295 0 C = 295 K (please see ‘Note on T’ on Pages 3/ 4 to see why this is so) Lattice parameter of pure Fe at melting temperature,’a’ = 0.292 nm c. Assumptions: (i) Homogeneous nucleation (ii) Spherical nucleus (iii) No other element is present d. Solution: (a) Calculation of r*: From equation 10.6 : r * 2 T m H f 1 T m T …(1) This expression may be re-written as T
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This note was uploaded on 12/08/2011 for the course MSE 2090 taught by Professor Leoindzhiglei during the Fall '10 term at UVA.

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SolutionsHW8 - HW 8 Problem 10.2 a. To Find: (a) The...

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