 Okay, so welcome everybody to this third lecture. Today I'm going to talk about a very specific application of the thermodynamics that we have learned so far. But before I do that, I'm going to demonstrate to you that this is a two-way learning experience because I learned something yesterday and I did a bit more research on that and I'll show you what I mean. So this slide here is what I used to illustrate an ideal solution. So we are plotting activity versus mole fraction and I explained to you that activity is like an effective concentration because by adding something else you might make the molecules of the solvent more agitated and therefore there will be more active than in proportion to the concentration. And I said that I found a paper in which a magnesium oxide and nickel oxide solution demonstrates ideal behavior that means the activity is proportional directly is equal to the mole fraction as opposed to a magnesium oxide and manganese oxide solution where the activity is greater than the concentration. Now during the question and answers, Professor Chakrabati, Neuropam Chakrabati, pointed out that, you know, neodymium, preseodymium also has an ideal solution. So I looked up that paper. So just before I say that, so the authors of this paper suggested that, you know, if the misfit here between the magnesium oxide and the nickel oxide is small, then you tend to get an ideal solution with good mixing. Whereas when the misfit between these two oxide molecules is large, you get non-ideal behavior. So in the case of neodymium and preseodymium, they both have a hexagonal lattice. They both have the same number of valence electrons. You know, valence electrons determine chemical activity. And their metallic radii are pretty similar as are their densities and the atomic volumes are remarkably close to each other. And they have similar electron negativity. So these guys, London and Poole, did experiments in which they made a series of preseodymium, neodymium alloys and dissolved them in liquid tin to measure the change in enthalpy. And what they found was that the change in enthalpy doesn't vary significantly between going from a pure preseodymium to pure neodymium and the alloys in between, which essentially means that the enthalpy of mixing is close to zero. So this truly is an ideal solution. And in many respects, it's consistent with the oxide story, where you know, if the misfit between the two kinds of atoms is quite small in many parameters, then you tend to get an ideal solution. And in this paper, they actually chose this system after looking at the parameters which are similar for both of these elements. So that was quite interesting. And, you know, I've already taken advantage of neuropharm suggestion and incorporated it into this lecture. So people say, you know, teaching is an exciting thing to do because you not only teach but you learn and that's absolutely true. Right. I'm going to go on to the main story today now. And I'll begin by showing you the difference between what we call a solution and a compound. So a compound has a very narrowly defined chemical composition, because when you deviate from that composition, you get a rapid rise in free energy. But it's not exact, you know, there is a certain amount of variation. So if you think about, for example, cementite, which is conventionally said to be F3C, in fact, the composition can vary, the carbon concentration can vary a little bit. And therefore we get a free energy curve which looks like this. But it doesn't vary as much as what we call a solution, which is possible over a wide range of chemical compositions. Okay. Yeah, so I'm going to talk about a solution that is produced not by melting the constituents and solidifying, but by a different technique altogether. And that is a process known as mechanical alloying, which was invented by Benjamin, who was at Inko Alloys. And his goal was to replace nickel containing thorium oxide, because the oxide particles are added in order to give creep resistance, nickel containing thorium oxide, with nickel containing yttrium oxide. Now, both thorium oxide and yttrium oxide are extremely stable at high temperatures and have a very little solution inside nickel. So even when you use the alloy at high temperatures in service, the dispersion that you introduce will work well. Now the way in which the thorium oxide is introduced into the nickel is that you add thorium to the nickel, and then you internally oxidize it. And that's a that's a messy process. And thorium can be radioactive. Okay. So I believe when I visited the BABA, BABA Atomic Research Center, they work a lot on thorium, because India has about 25% of the world's thorium. So he wanted to introduce yttrium oxide, but there's no way you can do that by adding yttrium to the melt and then solidifying, because it oxidizes extremely rapidly. I mean, that's why it's a very stable oxide. So he decided, okay, supposing I take a powder of yttrium and various elemental metals and put them into a ball mill. So a ball mill is simply a cylindrical container in which you put cast iron balls and whatever powder you want to form a solution with. So in this case, we are basically damaging the powder so intensely by these cast iron balls impacting with the powder particles and welding them and fracturing them and welding them and fracturing them, that eventually it should make a solid solution. Okay. The yttrium oxide won't go into solution, but the rest of the material will. So what do you do with the powder after you made it? Well, you put it into a can, right? So that means the powder is in perhaps a mild steel can, and you extrude it hot to basically consolidate the powder into a solid object. And this is showing you extrusion of stainless steel. So you can make a solid object from powder by extrusion, which is basically a compaction of all the powders and removing. You almost get to 100% density. And in the case of this particular alloy, which has a large amount of chromium for oxidation resistance and also aluminum for oxidation resistance and titanium in the form of oxides and yttrium in the form of oxides. After the mechanical alloying process, this is the structure that you're left with. There are extremely fine grains of very, very large dislocation density. So it's going to be an extremely hard extruded material, even though it's hot extruded. And you can see there are myriads of these small yttrium oxide particles dispersed throughout the microstructure. So there is a very large driving force for recrystallization. And you can do the same thing. So this is a commercial alloy called MA956. And in this case, it's ferritic. This is a nickel-based alloy, where we have the chromium and aluminum again for oxidation resistance, titanium, tungsten, and the yttrium particles, which you can see more clearly here. And again, this is a very fine structure and there's a large driving force for recrystallization. Now, when you extrude this material, there's a tendency for the particles to align along the extrusion direction. So when you get recrystallization, the grains are not going to be isotropic. And that can be an advantage if you're thinking about creep resistance. You want to grow the grains along the stress axis because that reduces normal grain boundaries where creep damage can happen. And we want large grain structures from the recrystallization process. Now, I'll show you some images. But before I do that, supposing that I wanted to create a tube out of this material, which would have a grain structure, which is helical, because I want to also get a good hoop properties as well as longitudinal properties. Then there is equipment which would twist the tube. And then when you recrystallize this in a temperature gradient, you get these very large grains doing a helical path around the tube. So this is where we were working on producing these tubes for biological incinerators. And you can even make the grains grow circumferential around the tube. So here, for example, the grains are going circumferentially around the tube. And just to show you how bizarre the microstructures can be, if you take the as extruded material and you bend it, then obviously along the neutral axis, there's a zero stress, but you've deformed the surrounding regions to a greater extent. And then when you recrystallize, you can see the effect of the different levels of plasticity due to the bending. And this is the neutral axis here. So the material is very good in the sense that you can produce grain structures which are very desirable for specific purposes. It's an incredibly expensive material because you're producing large quantities by a mechanical alloying process, which is a very, very noisy process. But currently, there are many projects on this material because of the fusion reactor. Now, if you have oxide particles dispersed inside your material, then they act as nuclei for helium bubbles because in a fusion reactor, you know, the neutron flux is much greater than in a fission reactor. So the helium produced by transmutation inside the steel, if you don't have lots and lots of nucleation sites, then the material would swell. If you are blowing up a balloon, it's much harder to blow it up at the beginning than when it's already large because the pressure inside the balloon scales with one upon its radius. Similarly, a pore inside the steel can hold a lot more gas when it's small than when it's large. So if you have lots and lots of nuclei for bubbles, then you hold a lot more gas in very small cavities. And therefore, the material doesn't swell during irradiation. Okay. Now, this is not the only mechanical alloying process that is important. You can get mechanical alloying in other contexts. So this is a bearing here. Okay. And these are the rolling elements of the bearings. This is a section cut from the bearing. And, you know, the load loading on this inner ring here is basically like this, that you have a rolling element on the ring, and it exerts a contact pressure at any point during the rotation of the bearing. And, you know, these are extremely hard materials, something like 60 Rockwell hardness or something like 650 wicker's hardness, if you like. So there will be some cracks which form. But remember, the load that we are applying is like a compression and a torsion. Okay. And that's why we don't require a huge toughness in these materials. I'm showing you here a crack in a very large bearing. It's a wind turbine bearing. And you see these white regions here. Basically, you've got a crack and the faces of the crack beat against each other. Every time a rolling element goes over that surface. And that beating is exactly what mixes up the microstructure completely. So if you have any carbides there that dissolve and you produce the so-called white etching matter, which is effectively a mechanically alloyed system. Okay. So you can get the same phenomenon in many different contexts. Now, I've asked you to assume that we have formed a solution by this process. Okay. But how do we know that mechanical processing leads to a true solution? What are the techniques that we can use to determine that? Well, the obvious method is to use an instrument called an atom probe field ion microscope. And it looks like this. You basically have a very sharp tip of your material. Apply a very large electrical field. And then you have an imaging gas which ionizes when it touches an atom. And therefore it spreads out in the electrical field giving you a magnification of something like 10 million. So here is an image from the mechanically alloyed material where each dot is actually an individual atom. Okay. So we are looking at a very high magnification. Now you also see this long object here, which is basically a time of flight mass spectrometer. So we can pull out an atom and measure its time of flight. And from that we can actually get chemical information per atom. Okay. So I'm now going to show you what the chemical map looks like for this material M956, which is mechanically alloyed. And of course it contains many elements, but I'll just show you ion and chromium for the sake of gravity. So here we are. We are looking at 50 ion clusters, although it can be one at a time. But if I showed you a plot of just one atom at a time, then it would go from 100 to 0, 100 to 0, 100 to 0. Okay. So I'm showing you 50 ion clusters. And this is what the distribution of ion atoms looks like. And this is what the distribution of chromium atoms looks like. So I want you to think now, does this represent a homogeneous solution? Okay. You've got all these peaks and troughs. Does this represent a homogeneous solution? Now, the obvious answer might be no. But I think we cannot think like that because when we have a large number of atoms and we pick out a few, you're likely, even for a random distribution of atoms, to get these variations like this. Okay. So we have to take account of the fact that we're looking at individual atoms. So the way to do this is to compare this distribution of concentrations with a binomial distribution for 50 ion clusters. And here you are. These are the comparisons for ion, chromium, and aluminium. The black bars here are the atom probe data. And the white bars represent binomial, what you would expect from a binomial distribution. And you can see that there's a fairly good match in all three cases between the calculated distribution on the basis of a random solution and the measurement. And this is a quite important demonstration because, you know, ion and chromium belong to a system, ion chromium alloys belong to a system where there's a tendency for the clustering of atoms. Okay. So chromium atoms would tend to segregate into chromium rich regions and that can lead to a phenomenon known as 475 degrees C embrittlement in the long term. But here, by mechanical alloying, you have actually produced an absolutely true solution. And just to remind you from the previous lecture, the probability of finding an A bond is the concentration of A times the concentration of A because this is the probability of finding one atom and this is a probability of finding another atom. And similarly for B, XB times XB. And the probability of AB bonds is two times the chance of finding an A atom and a chance of finding a B atom assuming that we have a random distribution. So now I can compare from my atom probe data, which is sequential, and show you that, you know, the measured quantities and the calculated quantities are so incredibly close to each other, the aluminium and calculated and measured. Okay. So we can truly conclude that this material has formed a solid solution. Okay. Simply by using mechanical force. Right. Let me just have a drink of water. So it's good in a series of lectures to remind you of what we've done before. So this is a big chunk of A and equally big chunk of B in the proportion 1 minus X and X. And this is the free energy of pure A and the free energy of pure B. And we are looking at a composition here. And the free energy of this mechanical mixture, where, you know, the atoms are not really feeling the presence of the B atoms is simply a weighted mean of the pure terms here. Okay. So we are just taking an average of A and an average of B. And I explained to you that when we take this and we form a solid solution, there is a reduction in free energy because for one thing, you know, the configuration entropy favors a random mixture. Okay. So the free energy of a solution will be different from that of a mechanical mixture. Now, this is all very well when we have put, you know, chromium inside molten iron and there's a liquid and then we form a solid alloy. But here we're not doing that. We are actually starting with large chunks of A and B, breaking them down into smaller and smaller and smaller chunks, and then obtaining a solution. So we need to divide this process up into a whole series of stages. So here, for example, is what I mean. We start with the separate A and B, put them together, and then we severely deform them together until, you know, breaks up into ever smaller particles, ever smaller particles, and eventually a solution. So we need to calculate now what is the configurational energy, configurational entropy change in going from here to here, here to here, and here to here. No longer dealing with atoms. Okay. We're dealing with particles. So if you remember the number of configurations when we had an atomic mixture was the total number of atoms factorial divided by the atoms of B factorial and the atoms of A factorial. All we have to do is to express these terms in terms of particles rather than atoms. So here is the term where we take the concentration of A and divide by the number of atoms in the particle of A. So that reduces this number quite dramatically, right, because, you know, you're likely to have millions if not billions of atoms in a chunk of material. And similarly, the concentration of B divided by the number of atoms in a particle of B. And that gives us the total number of particles here, which is like this term. At the bottom, we have the total number of particles of A factorial, right, and the total number of particles of B factorial here. So we've done a small modification of our calculation of the number of part number of configurations, because we no longer have atoms, but we have particles which consist of so many atoms per particle. Okay, so nothing complicated here. Now I need to take the logarithm of this, and, you know, we use Sterling's approximation to do that. And we come up with an equation for configurational entropy, which looks like this. And it's a longer equation here, okay. But if I set MA and MB to be a single atom, right, so there's only one atom per particle, then it reduces to our standard equation there, okay. Okay, so we've got the configurational entropy, and we can work out the evolution of the configurational entropy as the particles go from large chunks to smaller and smaller chunks. So the next question to ask is when do the A atoms and B atoms actually feel at each other's presence? That means that when they start to behave somewhat like a solution. Of course, when they're single atom particles, that's different, but is there a particular size of a particle at which, you know, the presence of other particles becomes important? Now, just to put this into context, you know, a typical driving force for a transformation might be something like 10 joules per mole, right? So we need to work out at what point will these particles feel a mixing free energy of 10 joules per mole, assuming, you know, that it's an ideal solution. And here I have a plot, okay, assuming it's an ideal solution, this is the atoms per particle, and this is 10 joules per mole. So the particles begin to behave differently when their size is approximately a thousand atoms, okay. About that, it really doesn't matter. You can still treat it as a mechanical mixture. Now, of course, this may not be an ideal solution. So we need to think about enthalpy. And just to remind you, we had the enthalpy of mixing as the concentration of A times the concentration of B, coordination number, avogados number, and omega, which takes account of the change in energy when we break an air bond, a BB bond, and create two AB bonds, okay. However, once again, you see this is the equation for the final solution. It doesn't represent the different stages in the evolution of the solution, because only the atoms at the interfaces between the two different kinds of particles actually see each other. These atoms don't care about the presence of the black atoms on this side. Okay, so we've got to scale this function by the number of atoms that lie within that interface region. So we have to multiply it by the thickness of that interface, and two because it's on both sides. And Sv is the amount of surface, that means the amount of interface per unit volume that you've created. And this term actually is the chemical component of interfacial energy, this term here. You know, if you've got different atoms on each side, so even if this was absolutely perfectly coherent, you would have an interfacial energy, which corresponds to this. And this is the amount of interface we have per unit volume. So this is quite interesting now. So we've got an enthalpy of mixing, but it only applies to the atoms at the boundary, because the other atoms are separated, and therefore they don't have a chemical interaction with the other species. Now, this is the chemical component, and I said to you that, you know, if this is perfectly coherent, that means there is no structure in the interface, but that's very, very unlikely. There will be some sort of a defect structure in the interface. So we also need to take account of the structural component of interfacial energy. And that is much simpler. You have the amount of surface per unit volume multiplied by the interfacial energy per unit area. So that gives you the energy of the interfaces. And this is just a molar volume. We immediately face a problem. Okay. Okay. What this is saying is that as our particles become finer and finer, this term increases indefinitely, because, you know, the amount of surface per unit volume scales with one upon the particle size. So the amount of surface will increase indefinitely as the particle size begins to get smaller. If that is the case, then the cost here is enormous. And there is no way that you can form a solid solution. That is what this equation is predicting. So there's something missing in the theory, because we've already demonstrated that we have a true solid solution. And I now want you to think completely differently. You've probably done nucleation theory. And nucleation is necessary because when you have a small particle and an interface around it, the amount of surface per unit volume is large. And it is overwhelms the chemical driving force. Now, nucleation is easiest if the interfacial energy is small. In other words, particles that are able to form a coherent interface will nucleate at the expense of particles that are not able to form a coherent interface, even though there might be metastable phases. A classic example is the aluminum copper system, where you get these GP zones and then the theta prime and so on. And finally, the stable theta, which has an incoherent interface with the aluminum. So at nucleation, you go from a coherent to a semi-coherent and finally to an incoherent interface when the particle is large enough. What we've got to do is to think in the opposite sense that when we start with large particles in the mechanical alloying process, they are going to be incoherent, because you can't sustain coherency over large length scales, unless it's a perfectly coherent interface, which is very unlikely. Now, as we make the particles smaller, they will gain coherency. This is exactly the opposite of what happens in a nucleation sequence, where you lose coherency as the particle grows. So as this grows smaller, we will gain coherency until it eventually becomes fully coherent. After all, you cannot have an interface around a single atom. So it makes complete common sense. So in that equation, the interfacial energy cannot be constant. It's got to decrease as a function of particle size. Otherwise, it's impossible to form a solution. So now we have a complete picture of the structural and chemical components of interfacial energy contributing to the enthalpy of mixing. And we have the configurational entropy as a function of particle size as well. And both both enthalpy terms and entropy terms are a function of particle size. So I'll show you a graph now of the free energy of mixing as a function of particle size. So here we are starting with the large particle. And as we make the particles smaller and smaller, the free energy actually rises. You can see this is zero. So all this is positive. And then it decreases in this manner. Now, if you think about this, this is exactly the opposite of the nucleation barrier. Here, the free energy is rising because the interfacial energy is not decreasing sufficiently as the particles become smaller. And eventually, when it starts, you start to get some sort of coherency, you begin to gain from the fact that the atoms can be more intimately mixed. So this is exactly the opposite of a nucleation barrier. It will be hard during the mechanical processing to go below this sort of a size for the parameters that I've used. Now, there's one more assumption here. And that is that we've got an enthalpy of mixing, which means that unlike atoms attract. There's a minus sign here. This is the regular solution constant. Now, supposing that the atoms don't like to be next to unlike atoms, then the plot looks somewhat different. At fairly large particle sizes, we still have that barrier that I talked about, because the interfacial energy is not decreasing fast enough as the particle size becomes bigger. But then we have a second barrier when we get to the point where the particles feel each other's presence. And that is because the clusters of atoms have to be disrupted by the mechanical alloying process. And until those are disrupted, you don't get this decrease in pre-energy. So there are actually two barriers to the formation of an ion chrome mechanically alloyed solid solution. So these are quite interesting results, but you need to think completely opposite of nucleation theory. Okay. So we had this as our entropy of mixing and this as the free energy of mixing. I'm now talking about a solution that is already formed and you get a free energy curve which looks like this. Now, if I differentiate this equation with respect to x, then you'll find that the slope at x equals 0 has to be minus infinity and the slope at x equals 1 plus infinity. So you'll find many textbooks. We say, look, you need to draw your free energy curves so that at the pure element axes, they have infinite slope but a finite intercept, which corresponds to mu note, the free energy of the pure component. Again, this is actually missing something, right? Infinite quantities make one feel uncomfortable. And what we have to do is not treat composition as a continuous variable. It is not a continuous variable. The smallest change corresponds to one atom being added. So in fact, you know, you don't have infinite slope. I'm exaggerating here because I can't do the illustration for one atom, but you will have a finite slope at both ends because the smallest change in concentration that you can make is by one atom. So that paradox of infinite slopes at the pure axis also disappears when you think about a solution as consisting of discrete atoms. So this mechanical allowing process is revealing both from a fundamental point of view and it produces useful materials which are extremely expensive, but in some circumstances, you have to have those materials. So if you have a finite material, you can use those materials. So it's not material which is for mass production and consumption, but for very specific applications where you can't do without, for example, the dispersion of the yttrium oxide particles. I'm going to show you another very interesting application of mechanical alloy and that relates to a different kind of a problem. Now most of us think of austenite as being non-magnetic. You know, if you take a magnet and you touch an austenitic stainless steel, unless it contains some ferrite, it's not likely to get attached to the magnet. But actually, you know, if you go deep into the subject, you'll find that any lump of austenite actually has two kinds of austenite within it, two electronic states. One of them has a high volume and has a high magnetic moment, something like two Bohr magnetons per ion atom, and is ferromagnetic. And the other one is a low volume and low magnetic moment and anti-ferromagnetic. And we have experimental proof for this. So these both of these states exist in the same austenite and the proportions of high volume and low volume austenite change as you alter the temperature. The high volume state is more dominant at higher temperatures. So what this means is that even though each of these forms has the same expansion coefficient as ferrite, a very low expansion coefficient, because the proportions of these two forms of austenite are changing when you heat the material, the expansion coefficient of austenite as a whole is much bigger. And this rules it out for many power plant applications, where even though austenitic steels have a good creep resistance, their thermal expansion coefficient is too large and you get thermal fatigue as a consequence. So you won't find a thick walled stainless steel in any steam generating power plant, for example, because the expansion coefficient of austenite is something like 1.8 times 10 to the minus 6 per degree centigrade, whereas that of ferrite is something like 1.15 times 10 to the minus 6, almost the same as that of nickel. So there's an overall expansion and the expansion coefficient is much greater than that of ferrite. And just by intuitive thinking, you might think, okay, austenite is a closed back structure and therefore its expansion coefficient should be smaller. But it's not because of this two state issue. Now, there are many, many experimental data which prove this and we can even preserve the low volume and the high volume forms in isolation. For example, if we take pure iron and precipitate it inside copper, it will retain the FCC structure to cryogenic temperatures and we can measure the nil temperature for the antiferromagnetism and so on. So one particular experiment is quite clever. Supposing that we took copper which has a larger lattice parameter than austenite and we formed an alloy between iron and copper, then you would effectively put the iron into a larger lattice parameter material and therefore it should have a high magnetic moment per ion atom, whereas if we make the concentration of iron too large, then it's not going to adopt the lattice parameter of the copper and therefore we'll end up with a low magnetic moment per ion atom. But iron and copper are not missable, are not at temperatures below about, I think, 800 degrees centigrade. There's almost no solubility of copper in iron and iron in copper. So how can we make these solutions? Well, another feature of mechanical alloying is when you have to force elements together, you can do that, okay, whether or not they have solubility in each other or in the liquid states. So these experiments were done where mechanically alloyed iron, copper alloys were produced at a low concentration of iron and a large concentration of iron. Sure enough at the low concentration where the iron is in a large lattice parameter material, we end up with a large magnetic moment per ion atom, whereas at large ion concentrations, the magnetic moment per ion atom is much smaller. So in today's lecture, I've demonstrated to you that only from the material that we've learned in the first two lectures, okay, nothing else. We can actually apply it to study real materials which are of practical importance both in materials engineering and from a fundamental point of view. So if someday you need to make a strange alloy, which the phase diagrams tell you should not be possible, this is a technique that you can use. And the ball mill that I showed you, you can actually get laboratory sized ball mills, you can make them yourselves, put ball bearings inside instead of cast iron balls, but industrial scale ball mills also exist to produce larger quantities of material. The only issue I suppose is the consolidation part where you would need some help with either extrusion or hot ice pressing or something which doesn't allow the atoms to move about too much during the process. Okay, so that's the end of today's lecture.