0:32

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

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

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

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

So, we have then determine the single phase liquid,

the single phase alpha, and the single phase beta.

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

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

1:32

Now, when we go below that in variant reaction,

we have two two-phase regions.

One of them contains liquid and alpha.

That's on the left.

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

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

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

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

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

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

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

That is, liquid plus beta going into alpha.

And that defines for us, the peritectic.

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

Remember we did this with respect to the isomorphous system.

We did it with respect to the eutactic system.

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

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

We have, essentially, two liquidus boundaries.

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

much like we talked about with eutectic.

And the other with the B ridge diagram.

And so, that's our two liquidus boundaries.

Again, remember what the liquidus boundary does.

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

3:13

Now, when we look at the other phase boundaries,

we find the solidus phase boundary.

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

which is an equilibrium with a single solid phase alpha.

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

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

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

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

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

those are the solvus boundaries.

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

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

So this is the solvus boundary.

4:05

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

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

would allow you to make calculations that would tell you and

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

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

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

the calculations that are associated with alloy three.

4:37

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

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

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

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

5:00

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

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

and the associated composition of the solid phase.

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

which gives composition.

Now, what we can do is take that alloy and

we can cool it down to a lower temperature.

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

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

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

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

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

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

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

6:05

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

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

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

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

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

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

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

is at a much, much lower temperature.

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

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

by coupling peritectic reactions across the diagram.

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

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

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

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

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

7:20

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

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

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

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

zinc, the copper is face centered cubic.

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

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

elements have effectively the same atomic radius.

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

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

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

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

because the FCC phase and the HCP phase are different.

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

the highest temperature to the lowest temperature.

Those two melting temperatures are coupled by

looking at five different peritectic reactions.

So the peritectic system,

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

8:47

You can still analyze these systems precisely the way

we analyze the eutectic or the isomorphous.

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

the composition of the phase are the same.

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

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

will be determined by the tie line that ties together

those two phases in that particular two-phase field.

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

tell precisely what the compositions are at the peritectic temperature.

Thank you.