Chapter 10: Phase Diagrams

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1 hapter 10: Phase Diagrams Show figures 10-1 and 10-3, and discuss the difference between a component and a phase. A component is a distinct chemical entity, such as u, Ni, NiO or MgO. A phase is a chemically and structurally homogeneous portion of the microstructure. A phase often contains more than one component. Figure 10-1 (below) illustrates a two-component (sugar, water), two-phase (sugar solution, solid sugar) system. Figures 10-3 illustrates a system with two-components (u, Ni) and two phases (liquid, solid solution). In some cases, the two different phases could have the same crystal structure, but the two materials are immiscible because of differing atomic radii, electronegativity, valence, etc. learly, solid, liquid and gas also represent different phases. Phase diagrams: Phase diagrams map the number and types of phases of phases that are present, the composition of each phase, and the microstructures that exist. For practical reasons, this course will only cover binary (two-component) phase diagrams, although of course many materials of interest contain more than two components. Type 316 stainless steel has the following composition (weight %): (0.10), Mo (2.5), Ni (11), r (18), Fe (68.4). The phase diagram of this material is 5-dimensional, and beyond the scope of this course. The u-ni and binary phase diagram (Figure 10.3) is the simplest type of binary phase diagrams for two metals. Figure 10.3 shows that u and Ni are mutually soluble at room temperature throughout the entire range of compositions. Remember from hapter 5 that we discussed the Hume-Rothery rules for deciding when two metals will be soluble, and we concluded that u and Ni form a perfect solid solution. The detailed comparison was:

2 R u nm, R Ni nm, 2.4% difference. Electronegativities of u and Ni are 1.9 and 1.8, respectively. Both have a F crystal structure. Valences are +1, +2. Our conclusion (hapter 5) that u and Ni form a complete solid solution is represented graphically by Figure 10.3 (at the right). When two metals do not form a complete solid solution, the phase diagram will become more complex. The Al 2 O 3 - r 2 O 3 phase diagram (Figure 10.23) is the simplest type of binary phase diagram for two ceramic materials. Figure is completely analogous to Figure While we did not explicitly discuss Hume-Rothery-type rules for predicting ceramic solubility, similar guidelines do exist. We expect a new criterion, the same stoichiometric ratio of anions to cations, to replace the criterion of having similar valence. From the inside cover of the text, the ionic radii of Al and r and nm. The borderline between the liquid phase and two-phase (liquid + solid) region is termed the liquidus line, while the borderline between the solid phase and two-phase (liquid + solid) region is termed the solidus line. In the two-phase region, given the temperature, we can determine the composition of each of the phases by drawing a tie line, as illustrated in figure 10.3b.

3 Example Problem: At point B (35 wt% Ni, 1250º) on figure 10.3b, what are the mass fraction liquid and the mass fraction solid? This must be solved using the lever rule, or the inverse lever rule, I have heard both terminologies used. As described in the text: W L S R + S 0 L W L The reason that some people call this the inverse lever rule is the line segment (S) chosen to calculate the mass fraction liquid is on the solid side of the phase diagram, not the liquid side. Similarly: R 0 L W R + S L W Obviously, these must add up to 1.0. Example Problem: alculate the amount of each phase present in 1 kg of a 50 wt.% Ni- 50 wt.% u alloy at a) 1400, b) 1300 and c) a) For a 50 wt.% Ni- 50 wt.% u alloy at 1400, we are in the liquid (L) region of the phase diagram. Therefore, we have 1 kg of liquid (L). b) For a 50 wt.% Ni- 50 wt.% u alloy at 1300, we are in the solid + liquid (+L) region of the phase diagram. Here we must use the lever rule to calculate the mass fraction of each phase. Unfortunately, figure 10.2 does not contain an insert enlarging the +L region of the phase diagram, so accurate reading of this figure is difficult. However, if we draw a tie line across the +L region at 1300, the endpoints are at about 45 wt.% Ni and 60 wt.% Ni. Therefore, the mass fractions are: WL L ( 1kg) ( 1kg) 0. kg

4 W L L ( 1kg) ( 1kg) 0. kg 0 Note that when determining the mass fraction of the liquid (solid) phase, the numerator contains the composition of the solid (liquid) phase. This seems backwards to some people, and they refer to the lever rule as the inverse lever rule to remind themselves of this. To ensure that you get the correct answer, you should always look at the problem visually and determine which mass fraction is larger. Then you can check the answer. In addition, the two mass fractions (0.67 and 0.33) must add to 1. c) For a 50 wt.% Ni- 50 wt.% u alloy at 1200, we are in the solid () region of the phase diagram. Therefore, we have 1 kg of solid (). Show figures 10.4 and What happens during cooling from high temperature (liquid) to room temperature (solid)? As soon as the liquidus line is crossed, the solid solution phase begins to precipitate from the liquid. What are the composition of the precipitating solid and the remaining liquid? The composition of the solid and liquid at every step along the way can be determined by drawing a tie line at that temperature. Eventually, the temperature drops below the solidus line, and precipitation of the solid is complete. No further significant change occurs in the microstructure upon cooling to room temperature.

5 Show Figure 10.6, which illustrates strain hardening for both pure u and pure Ni, when small amounts of the other are added. Eutectic phase diagrams: What happens when two materials are not completely soluble, the more typical situation? In the simplest case, a binary eutectic phase diagram is observed. Show figures 10.7 and 10.8.

6 For Ag and u, we can apply the Hume-Rothery rules as follows: R u nm, R Ag nm, 12.5% difference. Electronegativities of u and Ag are both 1.9. Both have a F crystal structure. Valences are +1, +2. Although u and Ag seem to follow the Hume-Rothery rules, Figure 10.7 shows that they do not form a complete solid solution. The critical difference is that their atomic radii are not that similar. This shows that the Hume-Rothery rules are really guidelines, not rules. We can also apply the Hume-Rothery rules to Sn and Pb as follows: R Sn nm, R Pb nm, 15.9% difference. Electronegativities of Sn and Pb are both 1.8. Sn has a BT crystal structure, while Pb has a F crystal structure Valences of both are +2, +4. Since they have different crystal structures, the relative insolubility of Sn and Pb is expected, and is shown graphically in Figure The eutectic reaction is described by L + β, occurring at the eutectic composition. Any phase transformation during which a liquid transforms simultaneously to

7 two solid phases is termed a eutectic reaction. The binary eutectic phase diagram has two liquidus lines separating L from (L + ) and (L + β) and several solidus lines separating from ( + L), β from (β + L), ( + β) from ( + L), and ( + β) from (β + L). Note that the borderline between the phase and the ( + β) and ( + L) regions defines the T-dependent solubility limit of Sn in Pb. Show figure 10.8, the Pb-Sn phase diagram. Does the eutectic reaction only occur when you happen to be cooling an alloy of exactly the eutectic composition (61.9 wt.% Sn)? No. When the Sn composition is between 19 and 97.5 wt.% Sn, liquid of the eutectic composition exists when the alloy crosses the horizontal line during cooling from high temperature (liquid) to room temperature. Explain this in detail with two examples. Show figures 10.9 through 10.11, which illustrate the microstructures that exist at various compositions and temperatures. When a liquid of eutectic composition solidifies, a eutectic microstructure is formed with alternating layers of and β phases. These thin layers form because the solidification reactions occurs all at once, rather than gradually as in the case of complete solid solubility (u-ni for example). Diffusion is not rapid enough to completely separate the and β phases that form. Note that Pb-Sn materials were widely used as solders, since the melting temperature (183 ) of the eutectic is considerably below that of pure Sn (232.0 ) and pure Pb (327.5 ). Due to environmental concerns about Pb, they are being replaced by other solder materials, such as Ag-Sn. This is an extremely good example of using phase diagrams to engineer the properties of materials!! All soldier materials have phase diagrams that look like Figure Show Figures through and explain the development of the microstructure of Sn-Pb alloys upon cooling.

8 The Figure on the left side illustrates phase changes for a Sn composition below the solubility limit in Pb, at all temperatures. The Figure on the right side illustrates the phase changes for a Sn composition below the solubility limit in the phase at elevated temperature, but above that solubility limit at room temperature.

9 Gradual cooling at the eutectic composition (Figure above) results in simultaneous conversion from 100% liquid to 100% solid. Since phase separation is instantaneous, but solid diffusion is slow, the striped eutectic microstructure in the Figure below is formed.

10 In the Figure above, the Sn composition is 40 wt%, higher than the solubility limit of Sn in the phase, but less than the eutectic composition. The final microstructure is shown in the figure below, which illustrated the primary surrounded by stripes of the eutectic microstructure.

11 Example Problem: alculate the amount of each phase present in 1 kg of a 50 wt.% Pb- 50 wt.% Sn alloy at a) 300, b) 200 and c) 100. Also calculate the amount of each microstructure. a) For a 50 wt.% Pb- 50 wt.% Sn alloy at 300, we are in the liquid (L) region of the phase diagram. Therefore, we have 1 kg of liquid (L). The liquid is also the microstructure. b) For a 50 wt.% Pb- 50 wt.% Sn alloy at 200, we are in the solid + liquid (+L) region of the phase diagram. Here we must use the lever rule to calculate the mass fraction of each phase. If we draw a tie line across the +L region at 200, the endpoints are at about 17 wt.% Sn and 54 wt.% Sn. Therefore, the mass fractions are: WL L ( 1kg) ( 1kg) 0. kg W L L ( 1kg) ( 1kg) 0. kg Again, note that the numerator of each quantity contains the composition of the opposite

12 phase, but this must be true in order to obtain mostly liquid (L) phase. Looking at the tie line, the composition of interest (50 wt.% Sn) is much closer to the liquid (L) than to the solid () phase, so m L >> m. In addition, the two mass fractions (0.89 and 0.11) sum to 1. ompare the equations given above for m L and m to those employed in the previous problem for m L and m. They are not the same!! In the current problem, the liquid is on the right-hand side of the phase, but in the previous problem, the liquid is on the left-hand side of the phase. If you memorize formulas instead of understanding the use of the lever rule, you will not be able to do this type of problem correctly!! Note that the two microstructures that exist here are the same as the two phases, liquid and solid solution. c) For a 50 wt.% Pb- 50 wt.% Sn alloy at 100, we are in the + β region of the phase diagram. Drawing a tie line across this region of the phase diagram, the endpoints are at about 98 and 5 wt.% Sn. Therefore, the mass fractions are: W W β W β ( 1kg) ( 1kg) 0. kg Wβ β ( 1kg) ( 1kg) 0. kg Here the microstructure differs from the phase, and this is determined by what happens just above the eutectic temperature, 183º. First, note that we on the left side of the eutectic composition, 61.9 wt% Sn, so the primary phase is. Remember, the primary phase is the one that precipitates out prior to the eutectic reaction. The remaining liquid is transformed into the eutectic microstructure. Therefore: (100 ) W (183 + ) ( 1kg) ( 1kg) 0. kg L 0 primary 27 L (100 ) W (183 + ) ( 1kg) ( 1kg) 0. kg 0 eutectic L 73 L The distinction between the mass (mass fraction) of each phase and the mass (mass fraction) of each microstructure is an extremely important one. About 1/3 of the class will get this wrong on the next exam. Example Problem The microstructure of a u-ag alloy at 775º consists of primary and eutectic microstructures. If the mass fractions of these two micro-constituents are 0.73 and 0.27, respectively, determine the

13 alloy composition. First, you need to realize that primary refers to the phase that precipitates out prior to the eutectic reaction. Thus, the weight fraction primary is equal to the weight fraction just above the eutectic temperature, 779º. Similarly, the weight fraction eutectic microstructure is equal to the weight fraction L just above this temperature. W primary ( 775 ) W (779 + ) eut eut 0 W primary ( 775 ) wt% Show figures 10.19, and (u-zn, Mg-Pb and Ni-Ti). These are much messier phase diagrams, but every region in each phase diagram contains either one phase or two phases. In the two-phase regions, the mass fraction of each phase can be obtained from the (inverse) lever rule, just as we have done above for more complex phase diagrams.

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15 Some of these phase diagrams contain intermetallic compounds, which behave as though they are a chemically distinct component. These show up as something like a vertical line in a phase diagram. Examples include Mg 2 Pb in figure and the vertical line at 44.9 wt% Ti in figure An additional reaction is also introduced, the peritectic reaction δ + L γ ε, where a solid and a liquid phase react to form a different solid phase. Example Problem: onsider 1 kg of brass with a composition of 35 wt.% Zn- 65 wt.% u. This phase diagram is shown in figure a) Upon cooling, at which temperature does the first solid appear? Drawing a vertical line at 35 wt.% Zn, 930.

16 b) What is the first solid phase to appear, and what is its composition? On the same vertical line, the sample crosses from the L phase to the two-phase + L region. Thus, the first solid that forms is the u-rich solid solution, which probably has a F structure. Drawing a tie line at 930, the composition of the first to precipitate is about 25 wt.% Zn. c) At which temperature will the alloy completely solidify? From the same vertical line, complete solidification occurs at about 900. d) Over which temperature range will the microstructure be completely in the phase? The alloy is 100% phase from 670 down to about 200, where this alloy enters the + β two-phase region. Show a series of binary ceramic phase diagrams(al 2 O 3 - r 2 O 3, MgO- Al 2 O 3, ZrO 2 -ao), and discuss the analogies to binary metal phase diagrams:

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18 Eutectoid phase diagrams: Show figure Obviously, the Fe-Fe 3 phase diagram is of enormous commercial importance, since the primary ingredients of most steels are Fe and. The addition of small amounts of ( wt.%) increases the ductility of Fe tremendously and prevents brittle failure. Figure is not a true phase diagram, since Fe 3 is metastable, gradually degrading to form Fe and (graphite). However, this reaction is nearly infinitely slow at room temperature, so common practice is to treat Fe 3 as though it were a stable compound. Note that pure Fe has a room-temperature B phase (ferrite) and a higher temperature F γ phase (austenite). This is not uncommon, particularly among ceramic materials. The Fe-Fe 3 phase diagram shows a eutectic reaction, but it also shows a eutectoid reaction, during which γ + Fe 3. Any phase transformation during which a solid phase transforms simultaneously to two other solid phases is referred to as a eutectoid reaction. Figures through illustrate the microstructures that are present at different compositions and temperatures. The transformations associated with the eutectoid reaction are similar to those associated with the eutectic reaction. The main difference is that for the eutectoid reaction, the initial reactant is a solid, whereas for the eutectic reaction, the initial reactant is a liquid. This explains the difference between Figure (Fe 3 precipitates along the grain boundaries of γ) and Figure ( precipitates as solid chunks out of liquid L).

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20 Example Problem onsider 2.5 kg of austenite containing 0.65 wt%, cooled to below 727º. a) What is the pro-eutectoid phase? The pro-eutectoid phase forms prior to the eutectoid reaction, so that is the (ferrite) phase. b) How many kg each form of ferrite and cementite? This involves using the inverse lever rule: W Fe Fe ( 2.5kg) ) ( 2.5kg) 2.3 kg W ( 2.5kg) ) ( 2.5kg) 0. kg 0 alpha Fe 2 3 Fe c) How many kg each form of pearlite and the pro-eutectoid phase? Remember, the pro-eutectoid precipitates above 727º, while the remaining γ is

21 W transformed at that temperature into pearlite. Therefore: (727 ) W (727 + ) ( 2.5kg) ) ( 2.5kg) 0. kg eutectoid 0 pro 4 eutectoid W pearlite (727 ) Wγ (727 + ) eutectoid ( 2.5kg) ) ( 2.5kg) 2. kg Once again, a large percentage of the class will confuse the answers to parts b and c. Example Problem: alculate the amount of the proeutectoid phase present in 1 kg of steel containing 1.20 wt.%. The proeutectoid phase refers to the phase that precipitates out of solid solution prior to the eutectoid reaction. When the content is less than 0.77 wt.%, the proeutectoid phase is ferrite. When the content is greater than 0.77 wt.%, the proeutectoid phase is Fe 3. From section and figure 10.26, this problem is really asking what is the amount of Fe 3 that

22 exists just above the eutectoid temperature (727 +). This can be determined using the lever rule: W ( 1kg) ( 1kg) 0. kg γ eut Fe 07 3 Fe eut You should realize that other alloying elements beyond are commonly used in many types of steel, especially stainless steels. Show figures and 10.39, which illustrate the effect of some alloying elements on the eutectoid temperature and eutectoid composition of steel.

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