Solutions to the tutorial

1. (a)

   $\sqrt{Dt}$

   The growth is parabloic.

2. (c)

   increases and then decreases and hence peaks at intermediate undercooling.

   See Fig. 24

3. (a)

   $\Delta X_0$

   See. Eq. 57.

4. (b)

   the mobility of coherent boundary is lower than that of incoherent boundary

5. Consider Eq. 55
   

   $$v = \frac{dx}{dt} = \frac{D (\Delta X_0)^2}{2 x ( X_e^{\beta}-X_e^{\alpha})^2}.$$ (16)

   
   

   
   

   $$2 x dx = \frac{D (\Delta X_0)^2}{ (X_e^{\beta}-X_e^{\alpha})^2} dt.$$ (17)

   
   

   
   

   $$x^2 = \frac{D (\Delta X_0)^2}{ (X_e^{\beta}-X_e^{\alpha})^2} t.$$ (18)

   
   

   
   

   $$x = \frac{(\Delta X_0)}{ (X_e^{\beta}-X_e^{\alpha})} \sqrt{Dt}.$$ (19)

   
   

   This is Eq. 56. To obtain Eq. 57, let us start with the above equation and differentiate it with respect to time:
   

   $$x = \frac{(\Delta X_0)}{ (X_e^{\beta}-X_e^{\alpha})} \sqrt{D} t^{\frac{1}{2}}.$$ (20)

   
   

   
   

   $$v = \frac{dx}{dt} = \frac{(\Delta X_0)}{ (X_e^{\beta}-X_e^{\alpha})} \sqrt{D} \frac{1}{2 t^{\frac{1}{2}}}.$$ (21)

   
   

   Thus, we obtain the Eq. 57:
   

   $$v = \frac{dx}{dt} = \frac{(\Delta X_0)}{ 2 (X_e^{\beta}-X_e^{\alpha})} \left( \sqrt{\frac{D}{t}} \right).$$ (22)

   
   

6. We know that the increase in length is given by
   

   $$x = \frac{(\Delta X_0)}{ (X_e^{\beta}-X_e^{\alpha})} \sqrt{Dt}.$$ (23)

   
   
   Here, $\sqrt{Dt}$ is known as the diffusion distance. This distance is temperature dependent because D is temperature depdendent. Hence, let us calculate the diffusion distance at T2 and T3. Given: D$_0$ = 1.2 x 10$^{-2}$m$^2$/sec and Q = 150 kJ/mol. Also, $D = D_0 \exp{(-Q/RT)}$ where R is the universal gas constant.

   Hence, D (T2 = 650K) = $1.2 \times 10^{-2} \exp{- 150000/(8.314*650)}$ = 1.06 x 10$^{-14}$ and, D(T3= 600K) = 1.05 x 10$^{-15}$. Thus, the diffusion distance at 650 K is nearly 14 microns while it is about 4 microns at 600 K.

   $\Delta X_0$ is 0.15 (that is, 0.25-0.1) 650 K and 0.25 (that is, 0.25-0) at 600 K. Further, $X_e^{\beta}-X_e^{\alpha}$ is 0.8 (that is, 0.9-0.1) at 650 K, and is 1.0 (that is, 1.-0.) at 600 K.

   Hence, at 650 K, the growth is about 2.6 microns while at 600 K, the growth is about 1.1 microns.

7. No; we have derived Eq. 57 by assuming certain $\frac{dc}{dx}$. However, when two precipitates are close, their diffusion fields overlap. Because of this, the $\frac{dc}{dx}$ decreases much more rapidly then when there is only one precipitate. So, in the case where the diffusion fields overlap, the growth rate decelerates much more rapidly and hence the growth rate deviates from that given by Eq. 57.

8. In the case of substitutional solute, the grain boundary diffusion is relatively faster than bulk diffusion. So, from the bulk, B atoms diffuse to the grain boundary from which they diffuse along the precipitate-matrix boundary to form a lens shaped precipitate. Such rapid lengthening and thicknening of the precipitate, however, is not important in the case of interstitial solutions in which both the grain boundary and bulk diffusivities are comparable.

   Figure 31: Short circuit diffusion and the resultant precipitate morphology.

   [scale=0.4]Figures/GBAllotriomorphSolution.pdf