0:32

Up at the elevated temperatures, what we have is the liquid phase.

Â Now, look at the diagram, on the left hand side, we have TA,

Â it has a lower melting temperature than on the right hand side which is

Â the melting temperature of pure component B, that is T beta.

Â So, we have then determine the single phase liquid,

Â the single phase alpha, and the single phase beta.

Â Now, when we start putting in the two-phase fields, that is,

Â the two-phase fields are bound by two single phases.

Â 1:32

Now, when we go below that in variant reaction,

Â we have two two-phase regions.

Â One of them contains liquid and alpha.

Â That's on the left.

Â And we have, on the right, solid alpha plus solid beta.

Â Now that we have looked at the diagram, we know that flat

Â line represents the invariant reaction where we have three phases in equilibrium.

Â We have the liquid composition given to the left, the alpha composition

Â given in the middle, and the beta composition on the right hand side.

Â And remember, what we're looking at as we cool down below the liquidus temperature,

Â what we're going to be looking at are two phases going into a single phase.

Â That is, liquid plus beta going into alpha.

Â And that defines for us, the peritectic.

Â Now, what we need to do is let's look at all of the phase boundaries.

Â Remember we did this with respect to the isomorphous system.

Â We did it with respect to the eutactic system.

Â So let's do the same thing here, but this time with the peritectic.

Â So, the first thing we're going to take a look at are the liquidus boundaries.

Â We have, essentially, two liquidus boundaries.

Â One of them is for the A ridge side of the diagram,

Â much like we talked about with eutectic.

Â And the other with the B ridge diagram.

Â And so, that's our two liquidus boundaries.

Â Again, remember what the liquidus boundary does.

Â It separates a single phase liquid from a liquid plus solid two-phase region.

Â 3:13

Now, when we look at the other phase boundaries,

Â we find the solidus phase boundary.

Â On the left, what we have is liquid plus alpha, a two-phase region,

Â which is an equilibrium with a single solid phase alpha.

Â So that's what we remember as a solidus boundary,

Â that is a liquid plus solid separating a single phase solid.

Â We have one on the left, which is the alpha solidus,

Â we have one on the right, which is the beta solidus.

Â Now, what we are interested in is the third set of boundaries and

Â those are the solvus boundaries.

Â That is, they tell us how the solubility of the various components

Â change with respect to temperature for the alpha phase and the beta phase.

Â So this is the solvus boundary.

Â 4:05

Now, what I have illustrated here is the same diagram, but

Â this time what I've done is to include actual compositions that

Â would allow you to make calculations that would tell you and

Â have you use the lever rule to calculate the fractions of the various

Â phases that are present in alloy one, alloy two, and alloy three.

Â But what we're going to do here is we're going to do

Â the calculations that are associated with alloy three.

Â 4:37

So in this particular case, what we're going to do is we're going to cool alloy

Â three down until it reaches the liquidious boundary, and at the liquidious boundary,

Â as soon as it forms at this higher temperature, what we're going to see is

Â the formation of the solid phase that begins to form which is called beta.

Â 5:00

And as a result of that, we can calculate from the phase diagram

Â what the composition of the liquid is, which is essentially .6,

Â and the associated composition of the solid phase.

Â And you can determine that by dropping a straight line down to the X axis,

Â which gives composition.

Â Now, what we can do is take that alloy and

Â we can cool it down to a lower temperature.

Â And when we cool this down to a lower temperature, now, what we're going to look

Â at are the phase boundaries that are associated with this new temperature.

Â So again, we have a liquid phase and a solid phase, and what we'll do is to

Â calculate the fraction of the liquid phase and the fraction of the solid phase.

Â Now, what you should do is to take that diagram and

Â make some more calculations just to make sure that you really understand

Â the process of determining the fractions of phases inside of a two-phase field.

Â 6:05

Now, what I'd like to do is to turn our attention to a system,

Â the system that I have up on the board is the copper-zinc system.

Â And I'm illustrating this, because it's a very interesting system in two regards.

Â First of all, the copper-zinc system is the basis for brasses and bronzes.

Â So here what we're looking at is copper, which has a high melting point,

Â all the way up on the left hand side of the diagram.

Â Alternatively, zinc, which lies all the way on the right hand side of the diagram,

Â is at a much, much lower temperature.

Â So now, it turns out that what we need to do is to bring in

Â a sequence of reactions that will actually permit the temperature to drop

Â by coupling peritectic reactions across the diagram.

Â And this is what's illustrated in this particular system.

Â But what I'd like to do is to then focus on all of these different peritectics.

Â The red circles show you where each of the peritectic reactions are occurring.

Â Remember, a peritectic is a liquid and solid going into a single faced solid,

Â and that's what each one of these illustrates on the diagram.

Â 7:20

The other thing that is important is that whole region of solid solution

Â on the left hand diagram is referred to as alpha brass.

Â So we see a very large solubility of zinc and the copper, but it's a curious thing.

Â Because what we know is that brass being made up of copper and

Â zinc, the copper is face centered cubic.

Â But the zinc, on the other hand, is hexagonal close pack.

Â Now, what's interesting about it is each of those two

Â elements have effectively the same atomic radius.

Â But what's interesting, of course, is that the crystal structure is different.

Â So, because of the similarity and their location to one another on the phase

Â diagram, what we're able to do is to have a reasonably extensive solubility.

Â However, we cannot have a single phase that goes all the way across the boundary,

Â because the FCC phase and the HCP phase are different.

Â So as a matter of fact, what we're actually looking at is going from

Â the highest temperature to the lowest temperature.

Â Those two melting temperatures are coupled by

Â looking at five different peritectic reactions.

Â So the peritectic system,

Â a little bit more complicated than what we described with respect to the eutectic.

Â 8:47

You can still analyze these systems precisely the way

Â we analyze the eutectic or the isomorphous.

Â Remember that in a single phase field, the composition of the alloy and

Â the composition of the phase are the same.

Â When we get into a two-phase field, the fractions of the phases

Â are determined by the lever rule and the particular composition of the phases

Â will be determined by the tie line that ties together

Â those two phases in that particular two-phase field.

Â And of course, along the three-phase line, the invariant line, we can only

Â tell precisely what the compositions are at the peritectic temperature.

Â Thank you.

Â