(Quasi)Equilibrium Microstructures
Previously, we asked you to provide a schematic sketch of a microstructure for an isomorphous phase diagram (Exercise 10.6.3) based on the three things we can read directly from the binary phase diagram at a specific temperature and overall composition. From those data, you could determine:
- The phases present.
- The composition of those phases.
- The weight fractions of the phases.
From this, it is clear that one could create a schematic sketch of the material's microstructure, with phases in the sketch proportional taking up an area proportional to the calculated weight fraction. This sketch is a very useful visual representation of the material's microstructure. Let's review that the one you drew for Point C, quickly, shown in Figure 10.8.1:
Figure 10.8.1 (a. )Analysis at Point C in a Cu-Ni phase diagram. (b.) A schematic microstructure.
Recall here that we 1. navigated the phase diagram and identified the phases from the phase field. 2. computed the compositions of those phases using a tie line, 3. computed the weight fractions using the lever rule, and 4. represented them as areas in a schematic sketch. Refresh yourself with Exercise 10.6.3 if this doesn't make sense to you.
Many of you may have draw different microstructures in your sketches. That's probably OK! There are many details here about the size, shape, and distribution of the microstructures that we haven't covered. This topic is called nucleation and growth and depends on kinetics (diffusion) and the thermodynamics of the self-assembly of molecules into solids during freezing. It isn't something that we cover in this class, but it is a central topic in Materials Science and Engineering. Many of you will see it again.
For our purposes, we'll limit ourselves to quasi-equilibrium cooling. This is the thermodynamic assumption that, during cooling, the structures have enough time to reach their equilibrium phases and compositions. This allows us to predict microstructures without considering the many advanced details of nucleation and growth.
So, let's step through this quasi-equilibrium cooling process for our Cu-Ni alloy, remembering that at each point on the cooling path we've giving the system enough time to form equilibrium phases at equilibrium compositions.
- Point A is entirely liquid. We can simply draw the microstructureas a single phase and label it as $(\ell_{\mathrm{s}})$. As we cool down we'll reach the liquidus at which point small crystals of $(\alpha)$-phase will form.
- By the time we've reached Point B, those crystallites have grown and will possess a phase fraction and composition that adheres to the tie line and the lever rule. You've solved these once on Section 10.6, and your sketch should match this analysis.
- As we continue to cool through the two-phase $\ell_{\mathrm{s}}-\alpha$ region to Point C we can qualitatively see that we're forming more solid phase as we approach the solidus (recall the lever rule). During this cooling process our solubility limits have changed, and we can sustain less Ni in both the liquid and the solid phase. Again, you should have done this calculation already in Section 10.6, and your sketch should match. Schematically, we show this as an increase in size in the $\alpha$-phase regions that formed earlier.
- Finally, at Point D, we've crossed over the solidus line and are in the single phase region. The $\alpha$-phase particles that formed in the $\ell_{\mathrm{s}} + \alpha$ phase field have grown to consume the liquid and coalesced. We show a likely quasi-equilibrium microstructure here as a single block of $\alpha$ phase, but indicate its polycrystallinity with lines showing where the nuclei coalesced. These are grain boundaries.
Figure 10.8.2 Schematic microstructure development during quasi equilibrium cooling.
There's a few things to ponder here, if you're interested, with regards to microstructural development. You can consider these in Exercise 10.8.1.
Quasi-equilibrium Microstructures in Binary Eutectics
Let's apply what we learned above to binary eutectics. We're going to take a number of quasi-equilibrium cooling paths, because, as we'll find, they'll yield different microstructures! We'll do the easiest first and move to the more complex ones. We'll use the famous Sn-Pb solder system (Figure 10.8.3) that we introduced in the previous section as an example.
Figure 10.8.3 A schematic Pb-Sn phase diagram.
Inclusion Formation
Let's consider the region below the maximum solubility of Sn in Pb, at about 18.3 wt% Sn and explore the results of the cooling curve based on what we know. See Figure 10.8.4 to explore the solidification of a $C_0 = 15 \text{ wt% Sn}$ alloy.
- At Point A we have just liquid.
- At Point B we have about $W_{\alpha} = 0.5$.
- At Point C we have all $\alpha$.
- At Point D we now form a new solid phase as we pass through the the solid solubility limit of Sn in the $\alpha$-phase crystal structure of lead. So, $\beta$ phase particles form. We don't show the other side of the phase diagram, so we can't be precise about the weight fraction, but we could be if we use Figure 10.8.3.
So, everything is consistent with what we know! The only thing that's slightly new is that when we pass through the solvus we gradually nucleate and grow a new, Sn-rich solid phase $\beta$ within the Pb-rich $\alpha$-phase matrix.
As we'll learn soon, these small $\beta$ phase inclusions turn out to be very important for improving mechanical properties of alloys.
Ask yourself quickly - will inclusion formation also form on the Sn-rich side of Figure 10.8.3? Why or why not?

Figure 10.8.4 Solidification of a $C_0 = 15 \text{ wt% Sn}$ alloy.
Eutectic Composition and Microstructure
In Section 10.8.6 we passed through a solvus line. However, something different and a bit more complicated happens when we pass through the eutectic isotherm (from Point A to Point B in Figure 10.8.5). First, let's show you the result, and then we'll discuss why.
This microstructure looks like none that we've seen before. It has an alternating zebra-skin pattern of alternating black and white phases. The black phase is the Pb-rich $\alpha$ crystallites and the white phase is the Sn-rich $\beta$ crystals. While we can see both phases in the microstructure, we can also identify a this zebra-skin pattern, which is distinct itself, so we call it a microconstituent. Importantly zebra-skin pattern is not a phase, it is instead comprised of two phases.
Figure 10.8.5 The full Pb-Sn phase diagram. The inset shows the microstructure at Point B that evolves from quasi-equilibrium cooling at the eutectic composition $C_{\mathrm{E}}$, with layered, alternating phases of black ($\alpha$-phase) and white $\beta$-phase.
So what's happened here? Let's explore the process. There's nothing new at Point A. We have $C_0 = C_{\mathrm{E}}$ and we're completely liquid. When we pass through the eutectic point, however, something new happens. In previous examples we'd always had a gradual nucleation and growth of new crystals, where (for example) liquid is slowly converted to solid. However, as we pass through the eutectic point we suddenly move from a liquid solution phase field to a phases field with two solids, $\alpha$ and $\beta$. That means that immediately the liquid is an unstable phase, and so it will convert to the two solids, as shown in Section 10.7. Furthermore, the $\alpha$-phase can only tolerate a very small amount of Sn, on the order of 20 wt%, while the $\beta$ phase can only tolerate a very small amount of Pb, about 2 wt%.
What does this mean? It means that the moment you cross the eutectic isotherm, all liquid converts to two new solid phases $\ell_{\mathrm{s}} \rightarrow \alpha + \beta$. Additionally, while the liquid as a fully mixed solution of Sn and Pb, now we have two phases which are completely segregated in composition. Any $\alpha$ phase that forms does so by spitting out a bunch of Sn, any $\beta$ phase that forms does so by spitting out a bunch of Pb. Let's look at a schematic of this in Figure 10.8.6. Or, look at the video simulating the process in Section 10.8.9, below the figure.
In this image, $\alpha$-phase crystals are forming, and when they do so they expel Sn. The Sn is transported (trough liquid phase diffusion) through the liquid solution to a nearby $\beta$ phase, where they contribute to the growth of the $\beta$ phase. In turn, the $\beta$ phase expels Pb, which is transported to nearby $\alpha$ phase regions, and contributes to their growth. This process leads to alternating regions of $\alpha$ and $\beta$ solid forming from liquid. We call this a eutectic or lamellar microstructure.
Figure 10.8.6 The formation of eutectic microstructure from liquid solution.
Let's draw a schematic of this happening in as we pass through the eutectic point in Figure 10.8.5. Remember, we our tie lines and lever rule will still work, so we can leverage them to make an accurate sketch of the quasi-equilibrium microstructure.
- At Point A we simly have $W_{\ell_{\mathrm{s}}} = 1.0$ and $C_{\ell_{\mathrm{s}}} = 61.9 \text{wt\% Sn}$.
- At Point B all the liquid has been transitioned to two solids during cooling. We can use a tie line (like you've done before) to find that
- $C_{\alpha} \approx 18 \text{wt% Sn}$
- $C_{\beta} \approx 98 \text{wt% Sn}$
- $W_{\alpha} \approx \frac{98-62}{98-18} = 0.38$
- $W_{\beta} \approx 0.62$
Our sketches should reflect that in Figure 10.8.7. I've made the $\alpha$-phase regions about half the size of the $\beta$ phase in an attempt to reflect this.
Figure 10.8.7 Quasi-equilibrium microstructure development at the euctectic composition.
Microstructure in Hypo- and Hypereutectic Alloys
We've covered the microstructures that develop in binary eutectic alloys when you don't pass through a eutectic isotherm. We've covered what happens exactly at $C_{\mathrm{E}}$. But what happens if you cool through a composition that isn't exactly at the eutectic composition but does pass through the isotherm.
Before we cover this, one bit of terminology: we designate alloys that have compositions less than the euectic composition as hypoeuctectics. "Hypo" means "beneath" or "below" - as in a hypodermic needle: below the skin. Hypereutectics are those with composition more than the eutectic temperature. "Hyper" means "over" or "above", i.e., to hyperactive or hyperbolic.
Before we do this, let's try a brief exercise, Exercise 10.8.2.
Hyper/Hypoeutectic Microstructure Development
Alright -let's give it a try, and then you'll try yourself in Exercise 10.8.3.
Figure 10.8.8 Hypereutectic microstructural development in Pb-Sn alloys.
We'll do the hypereutectic path along Point D to Point F at 85 wt% Sn shown in Figure 10.8.8. We'll step through this explicitly, but understand that 90% of the work is in Point F.
- At Point D, we have a fully liquid solution $W_{\ell_{\mathrm{s}}} = 1.0$ with a composition of $C_{\ell_{\mathrm{s}}} = 85 \text{ wt\% Sn}$. The drawing in D is simple.
- At Point E we are in the $\ell_{\mathrm{s}} + \beta$ phase field, directly above the eutectic temperature. Using a tie line, we find that $C_{\ell_{\mathrm{s}}} \approx 61.9 \text{ wt\% Sn}$ and $C_{\beta} \approx 98 \text{ wt\% Sn}$. WE can compute the weight fractions with this information:
$$W_{\beta} = \frac{C_{0}-C_{\ell_{\mathrm{s}}}}{C_{\beta}-C_{\ell_{\mathrm{s}}}} = \frac{85-61.9}{98-61.5} = 0.64$$
and so
$$W_{\ell_{{\mathrm{s}}}} = 1- W_{\beta} = 0.36$$
This is reflected, more or less, in the drawing for E.
- Now as we pass from Point E through the isotherm to Point F (immediately below the isotherm) all the liquid is thermdynamically unstable and will convert to a eutectic structure of $\alpha$ and $\beta$, just like it did when we were at the eutectic composition $C_{\mathrm{E}}$! In other words, $W_{\ell_{\mathrm{s}}} \rightarrow W_{\mathrm{E}}$. Remember this.
The $\beta$ phase, on the other hand is stable both above an below the isotherm, so we expect nothing to change with respect to the that phase. There's no thermodynamic driving force to do so! It is already stable. However, to denote that this $\beta$ phase formed above the eutectic, we change the notation for this proeutectic or primary $\beta$ phase to $\beta^{\prime}$ at Point F. Importantly nothing has changed about $\beta$. Not the phase, not the microstructure. We just change the notation.
Knowing this, let's analyze our new structure.
- Since $W_{\beta} \rightarrow W_{\beta^{\prime}}$, we can simply draw the $\beta$ phase from Point E and relabel it as the primary $\beta$ phase.
- Since $W_{\ell_{\mathrm{s}}} \rightarrow W_{\mathrm{E}}$, we know the liquid will all convert to the lamellar eutectic microconstituent.
- We can still use a tie line to compute the compositions of the $\alpha$ and $\beta$ phase. Note: there are only two phases, still, even though the microstructure is complicated. We find $C_{\alpha} \approx 18 \text{ wt\% Sn}$ and $C_{\beta} \approx 98 \text{ wt\% Sn}$.
- We can still use a the lever rules to simply compute the weight fractions for the total $\alpha$ and $\beta$. Let's do it:
$$W_{\beta} = \frac{C_{0}-C_{\alpha}}{C_{\beta}-C_{\alpha}} = \frac{85-18}{98-18} = 0.84$$
and so
$$W_{\alpha} = 1- W_{\beta} = 0.16$$
I've written these values in the solution and reflected them in my drawing. Observe the following:
- $W_{\beta}$ in Point E is the same as $W_{\beta^{\prime}}$ in Point F because nothing changed for that phase as we passed through the isotherm. We just relabeled it. So the drawing is clear there.
- $\ell_{\mathrm{s}}$ converted to eutectic.
How did I choose how thick to draw the ratio of the $\alpha_{\mathrm{E}}$ and $\beta_{\mathrm{E}}$ lamellae? Let's state a few facts and see if we can work it out:
- The total weight fraction for all of the $\beta$ phase is solved through the lever rule. This $\beta$ phase is all of the $\beta$ phase in the microstructure, meaning $W_{\beta} = W_{\beta_{\mathrm{E}}} + W_{\beta^{\prime}}$.
- We know that $W_{\mathrm{\beta}} = 0.84$ and $W_{\beta^{\prime}} = 0.64$. So, $W_{\beta_{\mathrm{E}}} = 0.20$.
- All of the $\alpha$ phase is in the euctectic microconstituent, so $W_{\alpha} = W_{\alpha_{\mathrm{E}}}$, and $W_{\alpha} = W_{\alpha_{\mathrm{E}}} = 0.16$.
- This means that the ratio $W_{\beta_{\mathrm{E}}}/W_{\alpha_{\mathrm{E}}} = 0.2/0.16 = 1.25$. The green stripes should be about 25% thicker than the pink ones.
Nice. Now you try in Exercise 10.8.3.
Figure 10.8.9 Hypoeutectic microstructure analysis.