Nucleation and Growth in Phase Transformations Tutorial 3

1. Name the two stages involved in the formation of particles of a new phase. Briefly describe each. Answer The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleation process involves the formation of normally very small particles of the new phase(s), which are stable and capable of continued growth. The growth stage is simply the increase in size of the new phase particles. 2. (a) Rewrite the expression for the total free energy change for nucleation 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 and solve for both the critical cube edge length, a*, and ΔG*. (b) Is ΔG* 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 a3, whereas the total surface area is 6a2 (since there are six faces each of which has an area of a2). Thus, the expression for G is as follows: D G = a3D Gv + 6a g2 Differentiation of this expression with respect to a is as 2 d D G d ( a3D Gv ) d (6a g ) + = da da da = 3a2 D Gv + 12ag If we set this expression equal to zero as 3a2 D Gv + 12ag = 0 and then solve for a (= a*), which gives a* = - 4g D Gv Substitution of this expression for a in the above expression for G yields an equation for G* as g2 D G * = (a*)3D Gv + 6(a*) æ g ö 3 æ g ö 2 ç 4 ÷ D Gv +g6 çè D4 G ø÷ = è - D Gø - = 32g 3 (D Gv )2 é g3 ù é g3 ù æ 16p ö é g 3 ù ú —is greater that for a sphere—i.e., ç ê ú = (16.8)ê ú. (b) G for a cube—i.e., (32) ê v è 3 ÷ø ê (D G )2ú êë (D Gv )2úû êë (D Gv )2úû ë v û The reason for this is that surface-to-volume ratio of a cube is greater than for a sphere. 3 It is known that the kinetics of phase transformation for some alloy obeys the Avrami equation, and that the value of n in the exponential is 5.0. If, at some temperature, the fraction transformed is 0.30 after 100 min, determine the rate of transformation at this temperature. Solution This problem gives us the value of y (0.30) at some time t (100 min), and also the value of n (5.0) for the transformation of an alloy at some temperature, and then asks that we determine the rate of transformation at this same temperature. It is first necessary to calculate the value of k. We first rearrange Equation 10.17 as exp (- kt n ) = 1 - y and then take natural logarithms of both sides: - kt n = ln(1 - y ) Now solving for k gives k=- ln (1 - y ) (10.17b) tn which, using the values cited above for y, n, and t yields ln (1 - 0.30) k== 3.57 ´ 10- 11 (100 min)5 At this point we want to compute t0.5, the value of t for y = 0.5, which means that it is necessary to establish a form of Equation 10.17 in which t is the dependent variable. Rearrangement of Equation 10.17b above leads to tn = - ln (1 - y ) k And solving this expression for t leads to 1/n é ln (1 - y )ù t = êú k ë û For t0.5, this equation takes the form é 1/ n t = ëê ln (1 k- 0.5)úù û and incorporation of the value of k determined above, as well as the value of n cited in the problem statement (5.0), we compute the value of t0.5 as follows: 1/5 é ln (1 - 0.5) ù t0.5 = êú êë 3.57 ´ 10- 11 úû = 114.2 min Therefore, from Equation 10.18, the rate is just rate = 1 1 = = 8.76 ´ 10- 3 (min)-1 t0.5 114.2 min