Hi. My name is Dr. Billy Wu and in this video, we’ll be talking about metal alloys and how we can use phase diagrams, and the lever rule, to help us understand their composition and properties. First of all, lets have a look at why metal alloys are an important class of materials. An everyday example, might be coins. Now if we think about what are the material requirements, there’s a number of key considerations. First of all, there’s a number of different coins, and we want to be able to distinguish one from another. Secondly, security. We obviously want to make these coins difficult to counterfeit. But at the same time we want these to be easy to manufacture. When in use, we also want these coins to be wear resistant And have a high corrosion resistance finally we also want them to be anti-bacterial. In terms of the last criteria, copper has natural anti-bacterial properties, which is great. However its very ductile and soft. But what we can do is alloy this with other metals to change it’s properties. The 2 euro coin, for instance, has 2 different alloys. A outer ring made of 75% copper and 25% nickel which is silver in colour, and an inner ring with 75% copper, 20% zinc and 5% nickel which has a gold colour The 1 euro coin has the opposite of this and the 50, 20 and 10 cent coins are made up of an alloy of 89% copper, 5% aluminium, 5% zinc and 1% tin. Combined this is called Nordic gold The 5, 2 and 1 cent coins on the other hand are copper plated steel, since we want to reduce the cost of these low denomination coins. If the value of the metals, were greater than the stated value of the coin, then people would just collect these and melt them down. Now, if we look at what options we have for material selection, the periodic table gives us our basic building blocks. This consists of over 100 elements, which might sound like a lot, but many of these are either scarce or would be unsuitable in most material selection problems, limiting our options. However, if we mix elements together, or alloy them, this allows us to augment the material properties and opens up many possibilities. The types of elements that we alloy together, their proportions and processing conditions have a huge impact on the eventual properties. Take for example bronze, which is an alloy of copper and tin. This is a relatively hard material that resists corrosion which makes it ideal for making statues. However, if we change tin for zinc, we can then make brass, which is softer and more malleable, making it suitable for manufacturing musical instruments. Therefore, you can see that alloying materials has a strong influence on a its properties, and phase diagrams can help us the understand what is happening. So now lets look at a few basic concepts which we’ll be discussing in more detail throughout the video. Firstly, the term alloy refers to a mixture of a metal with one or more other metals or non-metals. We term this binary when we have 2 elements, ternary when we have 3 and so on. When we talk about a component, we’re referring to an element in the alloy. A phase is a region of material which has uniform physics and chemical properties. This will be an important concept later on. In the image you see here, this is an optical micrograph of steel which is an alloy of iron and carbon. The different coloured grains you see here are different phases. The composition of a phase is the mass of each component in the alloy or phase. The constitution is the sum of the phases, the mass of each phase and the composition And finally the phase diagram is the equilibrium constitution of all combinations of temperature and composition for that particular alloy. Now given that this video is about phase diagrams it’s worth discussing in a bit more detail what a phase is. Earlier we defined this as a region of a material which has uniform physics and chemical characteristics but lets have a look at an example to help us visualise this. The plot here is the phase diagram for H2O, with the x-axis being temperature and the y-axis being pressure. Now we know at room temperature and atmospheric pressure, H2O exists in a liquid state we know as water. As we decrease temperature below zero degrees C the H2O undergoes a phase transformation from a liquid to a solid we know as ice. Clearly here the physical properties of water are very different to ice. If we then increase the temperature to above 100°C, the ice transitions to liquid water and then to gaseous steam. Again, steam has very different physical properties to ice and water but has the some composition, so this tells us that temperature is important when describing physical properties. Now when we start to mix different elements together a number of different things can happen. If the 2 elements are similar to each other, they can form what we call a solid solution. A solid solution is a single phase region in which the solute atoms have mixed with the solvent atoms to form a homogeneous composition. The term solute atoms refer to the element in the smaller proportion, and the solvent atoms refer to element in the larger proportion. If the 2 elements are very similar to each other, they can have complete solid solubility, which means that the solute atoms are completely soluble in the solvent atoms for any composition of solute and solvent atoms. A second different phase therefore does not form. In order to visualise this, lets take the example of a copper alloy. Here you can see a cartoon which represents the copper which is the solvent in this case. We then want to alloy this with nickel, which is the solute in this case, since there is a smaller proportion of this element. In the case of copper and nickel they have complete solid solubility and therefore for any amount of copper or nickel they’ll form a homogenous phase with uniform physical and chemical properties, as shown in the diagram. If we dive a bit deeper into alloys which have complete solid solubility, we can start to see the utility of the phase diagram. So, Lets again look at our nickel and copper alloy example, which exhibits complete solid solubility From our earlier example with H2O we know that temperature is a key metric for determining the phase and physical properties of a material. In the case of nickel, the melting point is 1453°C Copper on the other hand has a lower melting point of 1084°C Now if we draw the equilibrium phase diagram for the copper-nickel alloy, it will look like this. On the x-axis we have the composition in terms of weight percentage of nickel and on the y-axis we have the temperature. We can also see several lines which represent the phase boundaries between different regions. Here there are 3 distinct regions. A single phase liquid phase region, a single phase solid solution region and a 2 phase region. Given that this is a binary alloy of 2 metals: copper and nickel, when the weight percentage of nickel is 0 this means we have 100% copper. And when we have 100% nickel we have zero percent copper. Now if we assume we have an arbitrary composition of nickel and copper where the temperature is above the melting point of nickel, we will be in a single phase liquid region, where the structure will look very homogenous. If we remind ourselves, a phase is defined as a region of a material with uniform chemical and physical properties, so at this temperature and composition everything will behave like a liquid. As we cool the alloy, the overall composition stays the same but then a solid phase starts to precipitate out of the liquid. As we continue to cool the material, we can see that eventually all of the liquid becomes a solid, where the grains which initially started to form, have grown and joined together. Where these grains join are what we call grain boundaries. So that was the situation where 2 metals have complete solid solubility. However, in many cases where the 2 metal elements are considerably different from each other, in terms of atomic size and charge, complete solid solubility is not possible. Limited solid solubility therefore means, solute atoms are not completely soluble in the solvent atoms for any composition and thus a 2nd phase forms. Now lets look at the example of aluminium and copper. These 2 elements don’t have complete solid solubility, however at low concentrations, of solute atoms, a solid solution can form as shown in the diagram. However, as the concentration of solute atoms increase, in a limited solid solubility system, a second phase can precipitate out as this is the more thermodynamically stable state. Here you can see in the diagram, that as we added more copper, a different phase has formed, where the composition and structure has changed. Now this can seem a little abstract so lets have a look at a example of a limited solid solubility system which you might come across in every day life. Here, lets take the example of adding sugar to water. The plot on the bottom shows the phase diagram for a mixture of water and sugar. The x-axis is the weight percentage of sugar, where 0% sugar is 100% water and vice versa. On the y-axis we have temperature and we can also see 2 regions. Initially, at low concentrations of sugar, this dissolves completely to form a single phase syrup. However, as we add more and more sugar, and pass the 65% solubility limit, the liquid can’t hold any more sugar and solid sugar starts to precipitate out to leave us with a 2 phase system of liquid syrup and solid sugar. Now if we return back to our binary metal alloy system, and look at silver-copper alloys, which have limited solid solubility, this is what the phase diagram looks like. This is obviously a little bit more complicated than the phase diagram for the complete solid solubility case but we can break this down. Here we have 3 single phase regions. Firstly, at low weight percentage concentrations of silver, we form a solid solution with silver as a solute which we term alpha. The naming is somewhat arbitrary, and in reality we could have called this anything else. Then, at high silver concentrations we have another solid solution region where copper is the solute, which we term the beta phase. At high temperatures we then have a pure liquid phase where everything has melted. Beyond these single phase regions we also have 3 2-phase regions. Firstly, we have a 2 phase region of alpha and liquid Then a 2 phase region of beta and liquid And finally a 2 phase region consisting of solid alpha and solid beta The lines on the phase diagram mark the boundaries between where these phases form, and each of these have a different name. The liquidus line, for instance, represents the temperature and composition at which all components are liquid. The solidus line is the boundary beyond which no liquids form And the solvus line is the solid solubility limit between the alpha and beta phase regions. We also have a region called the invariant point where the liquidus lines meet as shown by point E. Now if we focus on this phase diagram and visualise the equilibrium microstructures which form this will help us understand what is going on. If we select a low silver composition and start from a high temperature, we can see we have a pure liquid phase. Then as we decrease the temperature we go through the liquidus line into the 2 phase alpha plus liquid region. Here, the microstructure looks like a soup with bits inside. Then as we cool the material further, we pass the solidus line where all the remaining liquid has turned into solid alpha. Finally, if we continue to decrease the temperature we pass the solvus line, where beta is no longer stable in the material and starts to precipitate out into a 2nd phase. This type of microstructure with 2 phases can impact the mechanical properties of a material since it can affect how easy it is for atoms to move past each other. So that was useful for use to get a qualitative view on what is going on in the material as we cool it, however if we want to quantify the amounts of each phase in the 2-phase region we have to take a different approach. In this case, we can use something called the lever rule to achieve this. To run through this example, lets start again with a simple complete solid solubility system First of all we start by drawing a horizontal tie line at the temperature of interest in the 2 phase region. We then mark the overall alloy composition on this tie line which I’ve arbitrarily put at 50% nickel. If we want to calculate the fraction of one phase, we do this by taking the length of the tie line from the overall alloy composition, to the phase boundary of the other phase and divide this by the total tie line length. This equation is given at the bottom, for the generalised form describing the mass of liquid and mass of solid in this 2-phase region. Now if we put some numbers in to help us visualise what this means, lets assume we have a 50% concentration of nickel, which means we have 50% copper in our alloy. At this arbitrary temperature, the liquid-solid phase boundaries occur at 20% and 80% nickel. If we then take the equation for the mass fraction for solid, we can see the length of the tie line, from the overall composition to that of the phase boundary, is 0.5 minus 0.2, giving us our numerator. The length of the tie line is the composition of the solid phase boundary minus that of the liquid phase boundary. This acts as our denominator. If we take the ratio of these 2 we can see that the mass fraction of the solid phase is 50%. We can then repeat this process for the liquid phase which should also be 1 minus the solid mass fraction. Now lets say for example, we increase the amount of nickel in our alloy to 80%. When we then put in the numbers into the equation we can see that we have 100% solid and therefore zero percent liquid. This makes sense, since we are now that the solidus line where the entire alloy should be in a solid phase. Conversely, if we reduce the amount of nickel to 20%, we can see that the mass fraction of solid becomes zero and therefore we have 100% liquid. This again makes sense since we are now at the liquidus line composition, highlights another important factor, that alloying metals allows us to control the melting point of alloy. So hopefully that example of the lever rule helps you to understand how to use it, however it’s also useful to know how this is derived, so that we don’t just take this at face value. To do this, we firstly take a mass balance of our 2 phase alloy which consists of a solid phase and liquid phase. Since there are only 2 phases, the sum of the 2 must equal that of the total mass of the alloy. Next, if we assume an arbitrary alloy composition of X% nickel, we can say that the mass of nickel, over the mass of the alloy, gives us our mass fraction of nickel. If we rearrange this equation to make the mass of nickel the subject, and then substitute in the equation for total mass, we get to a description of the mass of nickel in the system. Now, if we create an independent mass balance for nickel what we can say is that the total mass of nickel can be described as the mass fraction of nickel in the liquid, plus the mass fraction of nickel in the solid since these are the only 2 places nickel could be. It’s important to remember at this point that nickel generally exists in both the solid phase and the liquid phase. We now have 2 independent equations which describe the mass of nickel and if we equate the 2 together and rearrange them slightly we come to this equation. Finally, if we say that in the 2 phase region the mass of liquid and the mass of solid equals one, we can then solve the equations for the mass fraction of liquid and mass of solid. So, now that we’ve done an example of how to use the lever rule for a complete solid solubility system and also understood where this comes from, lets look at our final example where we apply this to a limited solid solubility system such as silver-copper. Lets say for example, we have an alloy consisting of 40 weight percent silver and 60 weight percent copper at 600°C. The first question might be what phases are present? This is a relatively straight forward, as we find the composition and temperature and then look at what region we are in. In this case, we can see that this is in the solid alpha plus beta 2-phase region. Now whilst we know at 40 weight percent silver and 600°C that we have 2 phases, what we don’t know is what their compositions are? To answer this, we can use the lever rule. Firstly we draw a tie line at 600°C between the alpha and beta region. Then if we read the composition of alpha at the solvus line, we can see that we have a composition of 4 percent silver and therefore 96% copper. Since at this point we only have alpha, we can say that this is the composition of alpha. We can then repeat this process for the solvus line between the alpha plus beta region and the pure beta region, where this tells us that beta has a composition of 97% silver and therefore 3% copper. Ok, so we now that that we have 2 phases, and what the composition of each of these phases are. The only missing element of this is now calculating what are the mass and volume fractions of each of these phases. To do this lets first introduce some definitions. C1 is the overall alloy composition, w alpha is the mass fraction of alpha and w beta is the mass fraction of beta. We can then use the lever rule equation to calculate the relative amounts of alpha and beta which are 61% and 39% respectively. This looks slightly different to our early implementation, however in actual fact this is the same since we’ve just replaced liquid and solid, with alpha and beta. Great, so we now know the mass fractions. The final step is to calculate the volume fractions. This requires us to revisit the amount of each element in each phase and take an average of their densities. Here, we define C-Ag-alpha as the concentration of silver in alpha, and C-Cu-alpha as the concentration of copper in alpha. The density of silver is taken as 10.49 g/cm3 and the density of copper is 8.96 g/cm3. Since we know the concentrations of each component and their densities now, we can calculate the overall density of each phase by using the equation below. If we then put the numbers into the equation we can see that the density of alpha is 9.01 g/cm3 and that of beta is 10.44 g/cm3. This makes sense since alpha is copper rich which has a lower density than silver, and beta has more silver. Also, it’s worth doing a quick check to see if the resulting density is between that of the 2 base metals. Finally, now that we have the densities and mass fractions, we can calculate the volume of each component. If we take the ratio of the volume of a single phase and the combined volume this then gives us the volume fractions of each phase which are 0.64 and 0.36 for alpha and beta respectively. So, to summarise. Hopefully, you've seen that alloys are used in a range of applications but changing their composition can drastically change their physical and chemical properties. Phase diagrams are a tool which can help us to understand what sorts of structures they should form at different compositions. Where a phase a region of a material with uniform chemical and physical properties. In cases where the alloying elements are similar to each other, they can have complete solid solubility, where any proportion of solute atoms are soluble in the solvent atoms. However, in many cases where we are using relatively dissimilar metals, we have the case of limited solid solubility where the solute atoms aren’t completely soluble in the solvent atoms and another phase forms. Through using the phase diagrams we could then see how different temperatures and compositions gave rise to different microstructures. And finally we applied the lever rule to help use quantify these changes. So hopefully this video was a useful introduction to metal alloys, phase diagrams and the lever rule, and thank you for listening.