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MSE 5321 Homework 07 & Solution - University of Texas, Arlington | Phase Transformation (MSE 5321)

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MSE 5321 Homework 07 & Solution - University of Texas, Arlington Phase Transformation (MSE 5321) Homework #7 1. (a) Show that surface melting is to be expected below  Tm in gold (1336 K) ... given   SL = 0.132,  LV =1.128,   SV =1.400 Jm-2 (b) Given that the latent heat of fusion of gold is 1.2  109 Jm-3 estimate whether sensible liquid layer thicknesses are feasible at measurably lower temperatures than  Tm - - - - - - - - - Calculate the homogeneous nucleation rate in liquid copper at undercooling of 180, 200 and 220 K, using the following data: - - - - - - - Show that Equation 4.23 applies to homogeneous nucleation and heterogeneous nucleation on a flat mould wall. - - - - - - - Show that Equation 4.16 follows from 4.15 using the following relationships for a spherical cap - - - - - - - - - Thus the sum of the solid-liquid and liquid-vapor interfacial free energies is less than the solid-vapor free energy, and there is no increase of free energy in the early stages of melting. Therefore, it would be expected that a thin layer of liquid should form on the surface below the melting point, because the difference in free energies could be used to convert solid into liquid. =- - - - - - -- But from Equation 4.19 and 4.17 we have 3 2 2 16 (2 cos )(1 cos ) 3 4 SL V G G           Which is identical to that obtained using     G V G   12 V7. G V G A A A het s V SL SL SM SM SM ML                      3 2 3 2 3 2 3 2 3 2 cos 1 cos 3 2 cos 1 cos ( cos ) 3 2 cos 1 cos cos 3 2 cos 1 cos cos 3 2 3 het V SL SL SM SM SM ML V SL SL SM SM SM SM SL V SL SL SM SM SM SM SM SL V SL SL SM SL r G G A A A r G A A A r G A A A A r G A A  r                                                                           2 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 3 3 2 cos 1 cos 2 (1 cos ) sin cos 2 cos 1 cos 2 2 cos (1 cos ) cos 3 2 cos 1 cos 2 2 cos cos cos 3 2 cos 1 cos 2 3 V SL SL V SL SL SL V SL SL SL SL V G r r r G r r r r G r r r r r G r                                                                          2 2 2 3 3 2 2 2 3 2 2 3 cos cos 2 cos 1 cos 2 cos 1 cos 3 2 cos 1 cos 3 SL SL SL V SL V SL r r r G r r G r                                        So,     (2cos )(1 cos ) / 4 [Show More]

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