27-302, Microstructure & Properties II, A.D. Rollett. Fall Homework 2, due Weds, Nov. 6 th

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1 27-32, Microstructure & Properties II, A.D. Rollett Fall 2 Homework 2, due Weds, Nov. 6 th 1. Materials Design In the first discussion, we discussed materials in preference to an engineered device that might include an interesting material. The class chose to work on Ni-Ti alloys as a class of shape-memory materials. This is very suitable for analysis of microstructure-property relationships because of the complex behaviors associated with the interplay between phase transformation and mechanical loading. The second discussion on Weds, Oct. 3 th should therefore focus on choosing an application of Nitinol alloys so that we can define the properties that are required. Once we have all the properties defined, we can work on understanding the microstructures that are possible in these materials as a preliminary step towards understanding the shape memory effect. So the assignment is to find an application of Nitinol alloys that you find interesting and make a list of the properties required for the application. 2. Required Reading The original reference for this is I.S. Servi & D. Turnbull, Acta metall., 14, 161 (166). You should read this classic paper in its entirety. The next two or three homeworks will ask you to calculate nucleation rates and transformation kinetics in the Cu-Co system that you will compare with the experimental results of Servi & Turnbull. 3. Nucleation Rate The critical free energy for nucleation depends on the interfacial free energy and the driving force. We will pick the system of Cu with a small amount of Co, which is known to exhibit homogeneous nucleation. In order to start on the problem we will find the appropriate driving force based on a heat of transformation (which was obtained from the original paper), and define the solvus temperature at any specific composition. We will assume that Co precipitates as pure Co. 3a. [2 points] Plot (linear-linear) the solvus temperature as a function of composition (weight % of Co) on a temperature (vertical) versus molar fraction (horizontal) plot over the range 5-1 C (just as in a phase diagram) based on the relationship, where X Co is the weight % of Co in Cu, and T is temperature in Kelvin: log 1 (X Co ) = / T Check your answer by comparing the calculated solvus curve against a published phase diagram for Cu-Co. Note that we are simply reproducing the solvus curve determined by Servi & Turnbull in order to use it for determining driving forces.

2 X(Co) T( C) X(Co) 3b. [2 points] Plot (linear-linear) the volumetric driving force (horizontal) as a function of [temperature] undercooling (vertical) for the following composition: 1.48 weight % Co (corresponding to one of the compositions investigated by Servi & Turnbull). Mark the equilibrium temperature for each composition on your plot (i.e. the T for which the undercooling is zero). You may use the relation on slide 11 of lecture 3, since you know the matrix composition and the equilibrium composition at any temperature. Be careful about what you mean by positive and negative free energies. Density of Copper = 82 kg/m 3 ; atomic wt. Cu = gm. Density of Copper = 82 kg/m 3 ; atomic wt. Cu = gm; atomic wt. Co = gm Ignore difference between Cu and Co: molar volume, V m = /( ) m 3 = m -3. Volumetric driving force, G V = RT ln(x /X e ) /V m. To find X simply use the formula above. A note about the plot: I ve plotted the undercooling with zero at the top and increasing down the page. This matches the diagrams in Porter & Easterling, for convenience and to make it easy to recognize the result. Note that the temperature at which the alloy composition is equal to the solubility limit is T = / (2.853 log ) = 7 C.

3 Under-cooling Driving Force.E+ 5.E+ 8 1.E+ 1.5E+ 2.E+ 2.5E+ 5 Undercooling In the plot, positive values represent a negative driving force for precipitation because we are below the solvus. The units of driving force are J.m -3. As some did, a better unit would be GPa, so that the driving force is about 2GPa at an undercooling of 3 degrees. 3c. [2 points] Plot the critical free energy, G*, (horizontal, log-scale) for nucleation versus [temperature] undercooling (vertical, linear-scale) for the same composition, assuming an interfacial free energy of.2 J.m -2. You should obtain values of the order of 1-1 J when the undercooling has increased to about 5 C.

4 Nucleation Energy, G* G* 1E-2 1E-1 1E-18 1E-17 1E-16 5 Undercooling d. [2 points] Plot the critical radius, r*, (horizontal, log-scale) versus [temperature] undercooling (vertical, linear) for the same range of temperatures. The critical radius should be in the nm range when the undercooling reaches 5 C.

5 Critical Radius I changed my mind about this chart when I came to do it. It makes more sense to leave the horizontal scale for the size linear; in this example the units are nanometers. 3e. [2 points] The pre-factor and activation energy for diffusion of Co in Cu are m 2.s -1 and Q=217.2 kj.mole -1, respectively. Using these data and estimating any quantities needed (take the jump frequency, w, as 1 13 s -1 ), plot (linear-log) [sorry, this should have been undercooling on a linear scale, and nucleation rate on a log scale!] the nucleation rate (horizontal) versus undercooling (vertical). Since you are given the activation energy for bulk diffusion in units of J.mole -1 then you should write the activation term as exp-{q/rt}. The term for the critical free energy (which gives the population of critical size embryos) is expressed in Joules per cubic meter (from part 3c above) so this activation term should also be calculated as exp-{ G*/RT}. Correction: the critical energy for nucleation is in Joules but the quantity is very small because it pertains to an energy at the atomistic scale. Therefore the appropriate thermal energy is kt, not RT, in the exponential term that scales the nucleation rate.

6 Nucleation Rate Nucleation Rate Undercooling.1 1 1E+ 1E+14 1E+1 1E+24 1E In the next homework we will estimate the number density of nuclei and then calculate the growth rate of the precipitates as a function of temperature. Armed with this information, we will calculate transformation kinetics using the KJMA equation and compare our results to the Servi & Turnbull measurements! The final step will be to extend the application of KJMA analysis in order to estimate a TTTdiagram. Once you have completed this exercise, you will have learned what lies behind the TTT diagrams for 12 and 862 that you saw in the second laboratory for