0:20

What we mean by the word eutectic, it actually comes from the Greek, and

Â it means low melting point material.

Â And as you look at the red arrows that are coming in,

Â what you're seeing is, the temperature of the liquidus

Â boundary is beginning to reduce as the composition of the alloy changes.

Â And, ultimately, it moves all the way down to a point on that

Â horizontal line called the eutectic point.

Â 0:59

So if we look at all the phase fields we have on the left side alpha,

Â we have on the right side beta, and we have a single phase liquid phase and

Â then we have, below, a two phase region of alpha plus beta.

Â 1:19

Now, if we look at the invariant reaction,

Â we see that the invariant reaction is referred to as the eutectic isotherm.

Â If you recall, what we mean by the term invariant reaction,

Â it means that the number of degrees of freedom we have is equal to 0.

Â Now if we go back and we think about a pure substance like aluminum, for

Â example, aluminum melts at 660 degrees C.

Â And at that temperature, at constant pressure,

Â two phases are in equilibrium with one another, a solid phase and a liquid phase.

Â According to the phase rule,

Â what we have at that fixed pressure are two phases that are in equilibrium.

Â When we refer to the reaction as being invariant,

Â what it means is that we know that there will be two phases,

Â a liquid phase and a solid phase, that are in equilibrium.

Â But we cannot tell you how much liquid or

Â how much solid there is at any point in time.

Â And it obviously varies and goes back and fourth between liquid and solid.

Â Exactly the same thing happens along the Eutectic isotherm.

Â In other words we have three phases in equilibrium, thus giving us, for

Â a two component system with pressure fixed, it gives us 0 degrees of freedom.

Â So what that means then is,

Â we have the alpha phase, we have the liquid phase and the beta phase.

Â All three of those phases are in equilibrium.

Â But regardless of the overall composition of the alloy,

Â we can't tell which of the three phases we have with respect to their content.

Â We can only say that all three are going to be present.

Â 3:15

Certainly the first set of phase boundaries that we've seen previously

Â are the boundaries we refer to as the liquidus.

Â Again, just like we have in the other phase diagrams in terms

Â of the isomorphous diagrams, we said basically that these liquidus boundaries

Â separate the single phase liquid field from the two phase liquid plus solid.

Â Sometimes when we're referring to these phase diagrams,

Â we might talk about the alpha branch of the liquidus,

Â meaning all of those alloys that are to the left of that eutectic point.

Â And sometimes we'll say the beta branch of the liquidus,

Â meaning we're looking at the alloys that are to the right of the Eutectic,

Â and those are rich in the amount of B that we have in the system.

Â 4:07

Again, when we identify the solidus and once again,

Â that solidus is the boundary that separates the single phase solid and

Â the two phase liquid plus solid phase field.

Â And when we're looking at these boundaries here, again we have

Â the alpha branch of the solidus and we have the beta branch of the solidus.

Â 4:26

The last boundary that we have that we want to include which is

Â an additional type of boundary, and it's referred to as the solvus boundary.

Â The solvus boundary essentially tells us how the solubility of

Â one of the components in the other changes as a function of temperature.

Â So when we're looking to the left, what we see is, as the temperature goes up,

Â the amount of B in the alpha phase progressively increases.

Â And likewise the amount of A in the beta phase progressively increases

Â with increasing temperature.

Â 5:02

Now what you'll see periodically through these various

Â lessons that we're going to be describing, I have a full Eutectic Phase Diagram.

Â And on the diagram, I've labeled various points, and

Â we'll be able to use those points to make some calculations.

Â We're not going to be looking at all the different alloys, alloys 1 through 5, but

Â what you should do is, on your own, go through, do cooling curves so

Â that you can calculate what the compositions of the phases are and

Â how much of each of the phases.

Â Now what we're looking at here are three regions of two phase equilibrium.

Â Regardless of how complicated the diagram may appear, when we're doing our analysis,

Â we do exactly as we had done in the previous lessons

Â where we were describing the isomorphous diagrams.

Â So we focus on regions of two phase equilibrium.

Â So in that region of two phase equilibrium,

Â we can determine what the compositions of the phases are, and we can also

Â then determine the fractions of the phases that we have at a given temperature for

Â a given overall alloy composition.

Â 6:15

When we go to three phase equilibrium, we have three phases in equilibrium.

Â We know what their compositions are,

Â those compositions are identified as those points on the Eutectic isotherm.

Â However, we cannot tell how much we have.

Â So when we look at a diagram like the Eutectic diagram,

Â we have single phase fields.

Â And in the single phase field, we know what the composition of the alloy is,

Â and therefore we know what the composition of the phase is.

Â When we get into a two phase region,

Â not only can we tell what the compositions of the two phases are, but we can also

Â determine how much of the two phases that we have in that two phase field.

Â In a three phase region, where we're looking at the straight line,

Â we can only tell what the compositions of the phases are.

Â 7:04

So what I've done then is I've looked at Alloy number 3 from

Â the previous diagram and I have chosen that particular alloy and

Â I have come down through higher temperatures.

Â So I was in a liquid phase field and

Â came down to the two phase field as we dropped down in temperature.

Â Now when we hit the first point on the diagram,

Â which represents our liquidus temperature, at that point we have essentially all

Â liquid and the composition of the liquid is equal to the composition of the alloy.

Â We have effectively no solid present, but

Â we do know what the composition of that solid phase would be

Â in that really small, inperceptible amount of solid that we have present.

Â So, consequently, we know, using the tie line,

Â what the compositions of the phases that are present.

Â 8:03

Now when we drop down into the two phase field below the liquidus temperature,

Â we have the fractions of the phases, the compositions of the phases and

Â we can solve everything at temperature T5, knowing that,

Â our overall alloy composition is equal to a value of 0.5.

Â So we continue to do that all the way through the table so

Â that we can determine the fractions of the phases as a function of temperature.

Â So those are the fractions that we're going to be looking for,

Â given the fact that we know what the composition of the alloy and

Â the composition of the phases that we are studying.

Â 8:47

Now, for example, let's take a quick look at Alloy number 4.

Â Alloy number 4 happens to be the Eutectic composition.

Â And when we go through the eutectic composition,

Â what we find is above the Eutectic isotherm, we have 100% liquid.

Â Below the Eutectic isotherm, we have a mixture of two phases, alpha phase and

Â beta phase.

Â 9:10

And what we can do is, we can actually make a calculation

Â of what the fraction of those phases are as a function of temperature.

Â So above the temperature T6, we have only one phase present.

Â That's 100% liquid.

Â When we reach the point at T6, what we begin to

Â see is we have liquid and two solids in equilibrium with one another, but

Â the amount of solid we have is imperceptibly small.

Â But as we drop below T6, what happens is both alpha phase and

Â beta phase increase and the amount of liquid we have now goes to 0.

Â So below T6, we have no liquid phase, we only have alpha plus n beta.

Â Now the thing that we also know is, as a check of our answers,

Â we know that at any particular temperature, the fraction of alpha and

Â the fraction of beta must equal 1.

Â In subsequent lessons, we're going to be looking at analyzing

Â these diagrams from the point of view of the phases that are present,

Â the compositions of the phases, and, in addition to that,

Â how the microstructure actually evolves as we go through a cooling process.

Â Thank you.

Â