phase dia.pdf

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LEARNING SEQUENCE -PHASES AND PHASE DIAGRAMS

INTRODUCTION

When a lake freezes the water changes from a liquid to a solid. This process is called a PHASETRANSFORMATION. Initially we had a liquid phase, and after the freezing we then have a solidphase.

The study of the possible ways in which various materials can exist by themselves or in contact withothers, as a function of temperature, pressure, and time, is a very important part of Materials Science.The reason is that we have to be able to tell what the stability of materials will be in all sorts ofenvironments, since the mechanical performance of these materials depends very much on this. Anobvious example is that we cannot use structural materials above their melting point. Other lessobvious changes can, however, happen inside materials that may be just as important from the pointof view of strength. These change in phase, called phase transformations, can be recorded indiagrams called PHASE DIAGRAMS. The phase diagram is a type of map that allows us to predictwhat will happen when we change the temperature or the overall composition of the material.

Suppose we have a closed imaginary box that contains the materials under study. Whatever is in thatbox we will call the SYSTEM. In this learning sequence we will consider only systems that are inequilibrium; that is we consider only equilibrium phase relations. This is important, since non-equilibrium phase relations can be quite different from equilibrium ones for the same system.

Definitions:

PHASE: to say exactly what a phase is, is not a trivial matter. It is easy to become confused aboutwhether something is or is not a phase. In our definition a phase must have the followingcharacteristics:

(1) must be a physically distinct region

(2) must be microscopically homogeneous (that is it must have one particularstructure)

(3) Its properties must change continuously* with temperature, pressure, andcomposition (Note: many mixtures maintain the same structure for different ratiosof the components; in that case they are still of the same phase).

COMPONENT: the number of components of a system is the minimum number of independent

substances from which the system with all its phases can be constructed.

I. ONE COMPONENT SYSTEMS

EXAMPLE I: H2O

H2O is a ONE COMPONENT system that can exist in 3 different PHASES: liquid, solid and gas,

depending on pressure or temperature. This is shown in the diagram below.

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Pressure vs Temperature Diagram for Water

Question: Derive an experiment to make the measurements necessary toconstruct this phase diagram.

Answer: One possible way would be to have a transparent cylinder with atightly fitting movable piston. We have to start out with nothing but H

2O in

the cylinder (e.g., no air). The external force on the cylinder will tell uswhat the pressure in the system is. A thermometer could tell us thetemperature. Changing the pressure and the temperature, and watching whathappens inside would give the points on the diagram. The experimentwould have to be done very slowly, to be sure that we are as close to

equilibrium as possible.

Question: Suppose the pressure P and the temperature T are as follows,what phases would be present in out cylinder:

Answer: Phases present

1 solid (ice) only - no vapor - the movable piston would sit against the ice,with no space for water vapor

Condition P (mm Hg) T ( C) Phasespresent?

1 600 -20

2 700 40

3 900 160

4 100 0

5 1E-4 -60

6 760 100

7 760 102

8 760 95

9 Triple point Triple point




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2 liquid water only

3 vapor only

4 solid (ice) + liquid

5 vapor only

6 liquid + vapor

7 vapor only

8 liquid only

9 liquid + solid + vapor: this situation can only exist at one particularpressure and temperature. It can be used as a standard.

EXAMPLE II Iron

The equilibrium diagram of iron is shown below. Fe can have five different phases: vapor, liquid,

and the solid phases (BCC), (FCC), and(BCC).

Figure III 2

Question: Can and exist together in equilibrium?

Answer: No, there is no point or line where they are adjacent on thediagram.

Question: Suppose we had iron vapor at a pressure of 10-8 atm and 1450 C. Then, suppose we increased the pressure to 1 atm (T = constant). Then (p= constant), we cool down to room temperature. What would happen?

Answer: First, at around 10-5 atm -Fe, a solid phase with the BCC

structure, would form. Then, upon cooling at p = 1 atm, at 1400 C Fe

Fe, and at 910 C Fe Fe.




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The phase transformations that occur in the solid iron as it is heated or cooled are calledALLOTROPIC transformations (from Greek allo = other, tropic = manner, way). So, the

transformations of pure iron at 910 C is an allotropic transformation.

Many substances exhibit such allotropic phase transformations in the solid state.

II. TWO COMPONENT SYSTEMS OR BINARY SYSTEMS

In two component or binary systems a new variable has entered: the ratio of the two components, inother words, the COMPOSITION. The presence of a second component can have a very profoundinfluence on the material; e.g., in many cases it may cause an "unmixing" or phase separation in thesolid material. These phase separations, or splitting up of the solid into different solid phases, cantake place because even in solid materials the atoms can move around (diffusion).

Suppose now that we dissolve salt in water. Then the ions Na+ and Cl- are floating around randomlyin between the water molecules. If we don't put too much salt in the water we have a solution, and nolittle crystal salt can possible be extracted from the solution by just pouring it through a fine sieve.

The salt solution is then a single liquid phase. If we put more and more salt in the water, eventuallywe cannot get any more salt in solution and all additional salt will just sit there at the bottom. Whenno more salt can be dissolved we have reached the SOLUBILITY LIMIT of salt in water. Puttingmore salt in over and above this solubility limit will just cause the accumulation of solid salt at thebottom of the jar. When that happens we have two phases present: a salt solution in equilibrium withthe solid phase of salt.

Question: Suppose we have two salt crystals at the bottom of this jar inequilibrium with the salt solution. How many phases are present?

Answer: Two: salt solution + solid NaCl.

The above situation can be plotted in a diagram. Suppose we put down the point at which the secondphase (solid NaCl) forms in the NaCl-H2O system (i.e., the solubility limit) as a function of

temperature, T, we would get something as shown below:

Fig. III 3

A similar situation can be found in many binary solid systems. First we will extend the idea ofsolutions to solution of a solid in a solid, a SOLID SOLUTION, and then we will discuss TWOPHASE SOLIDS

(1) Solid Solutions




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When we mix two different metals together, say by first melting them in one pot and stirring themwell, and then letting them solidify, we can get a solution of one solid in the other. The mixture ofthe solids will be a solid solution when:

1 The crystal lattice is the same everywhere in the solid.

2 The two metal atoms occupy the atom sites randomly.

Actually, we would have to examine the solid to see if it complies with all the requirements for beinga single phase (see DEFINITIONS). As in the case of salt and water, we could go so high intemperature that the mixture starts to melt. Then we can have the situation that we have a liquidmetal mixture in equilibrium with solid metal. It's important to note that (just as in the case of saltand water) the liquid phase does NOT have the same composition as the solid. Suppose we depictedall this in a diagram in which we plotted what is present as a function of the ratio of the metals, andtemperature, in an equilibrium situation. This plot would be a PHASE DIAGRAM. Below is anexample of a phase diagram of a binary system of A and B exhibiting complete solid solubility:

Fig. III.4

Some definitions:

LIQUIDUS: gives the temperature of the liquid phase at which, for a given overall composition, thefirst solid forms.

SOLIDUS: gives the temperature of the solid phase at which, for a given overall composition, thefirst liquid forms.

As you can see, this diagram tells us that at a given temperature, T3 the liquid can have at most~35% B. If we try to put in more B, we get a solid phase with a composition, c", of ~85% B. Notethat we have some A of the liquid phase that has to go into c". So, somehow the relative amounts ofsolid c" and liquid c' will be determined by the overall composition of the mixture of A and B.Suppose that the overall ratio of A and B is say: c (~ 50 /50)

weight fraction of solid = Xs = (c-c')/(c"-c')

weight fraction of liquid = XL = (c"-c)/(c"-c')

Question: Prove this. HINT: Use a mass balance.




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Answer: XL = fraction of liquid

Xs = fraction of solid

XL

+ Xs

= XT=1

Start with M grams of total material. Then XL

M is the mass of material in the liquid phase and XsM

is the mass of material in the solid phase. Now the total mass of B is the overall weight fraction of Btimes M or cM. The mass of B in the liquid is c'X

LM and the mass of B in the solid is c"X

sM so that:

cM = c"XsM + c' (1 Xs) M

or

Similarly,

This kind of procedure is very general and follows directly from a mass balance. It is the famousLEVER RULE. If we call the line c'-c" the TIE LINE then we can state the lever rule as follows:

LEVER RULE: The relative fraction of material present in a particular phase is equal to theratio of the tie line portion which lies between the overall composition and the phase boundarycurve away from the phase of interest, to the total tie line length.

You should remember the lever rule and know how to apply it in all equilibrium phase diagrams. Byphase boundary curve we mean lines like the solidus and the liquidus. The lines are bounding singlephase regions (or single phase fields as they are called) in the phase diagram -hence the name.

Question: Below is given the phase diagram of the binary system MgO-NiO. What are thecompositions of the phases in equilibrium when the overall composition of the solid is 40% MgO(and 60% NiO) at 2300 C. Same question for 20% MgO, and 80% MgO.

Figure III 5




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Answer: Composition in % MgO

40 liquid with 25% MgO + solid with 49% MgO

20 single phase liquid solution

80 single phase solid solution

Question: Shown is the Cu-Ni equilibrium phase diagram.

Fig III 6

Answer: From lever rule and diagram we have 53.5% liquid withcomposition ~77% Cu. The mass of liquid phase = 5 lbs x .535 = 2.67 lbs.The rest must be solid at a composition ~60% Cu.




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2) Unmixing in the Solid State

Most binary systems are not as simple as the ones we just looked at. They do not form solid solutionsfor all ratios of the two components. We then have two phases coexisting in equilibrium in the solidstate. Suppose for example we would cool the mixture of salt and brine to a low enough temperature,then eventually we would get ice and salt. That would be a two phase solid. For mixtures of metals anumber of conditions have to be satisfied to possibly have solid solutions. If these conditions are notfulfilled the solid will "unmix" or PHASE SEPARATE. The conditions for solid solubility of 2components are

1) The crystal structure of each component must be the same.

2) The lattice parameter of those components must not differ by more than 15%.

3) The elements of the components should not form chemical compounds.

More often than not, all these conditions are not met, and the solid in equilibriumactually consists of two solid phases.

A typical case is Cu-Ag (see next page).

To understand this binary phase diagram it is best to follow an alloy as it is cooled down from themelt. Three parts of the phase diagram look somewhat familiar:

1) the liquid phase field.

2) the solid solutionrich in silver (shaded).

3) the solid solution rich in copper (shaded).

Further, other parts of the phase diagram may also look familiar: the two regions where+ liquid

and + liquid coexist, above 779 C. They are like parts of the diagrams that we saw earlier. Then,below 779 C something happens in the solid; a solid of say 24% Cu will, below 779 C in

equilibrium exist as blocks of phase and will both contain as much copper and as much silver asthey can.

Question: Suppose we have an alloy that has an overall Cu content of 20%. What does the solidconsist of at 778 C?

Fig. III 7




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Answer: Blocks of with 8.8% Cu and blocks ofwith 92% Cu.

A binary system that has phase relations as the example of Cu-Ag is called a EUTECTIC SYSTEM.A eutectic phase transformation is one in which

liquid solid phase A + solid phase B

The relative amounts ofandare, however, the equilibrium ones.

as occurs in Ag-Cu at 779 C. The liquid in this case contains 28.1% Cu: the eutectic composition.

Question: Given an alloy with overall Cu concentration = 20 weight %,what are the phases in equilibrium at 779.2 C. Same question for an alloywith 60% Cu overall. What weight fraction (weight percentage) of eachphase is present?

Answer: a) solid solution (8.8% Cu) + liquid of eutectic composition (28.1%Cu) using the lever rule

X solid = 0.42 or 42 %

b) Solid solution (92% Cu) and liquid of eutectic composition (28.1% Cu) usingthe lever rule again

X solid = 0.50 or 50%

Upon going from 779.1 to 778 C it is that liquid of eutectic composition that goes through a eutecticphase transformation. Again, the lever rule applies. Below is the phase diagram of the binary systemof lead with tin.

Figure III 8




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a) What is the eutectic temperature? What is the eutectic composition?

b) Suppose an alloy contains 25% tin overall, what are the compositions of the phase in equilibriumat the eutectic temperature +0.5 C, and 0.5 C?

c) A Pb-Sn alloy contains 45% Sn.

i) Which phases are in equilibrium at 100 C?

ii) Apply the lever rule to find the relative amounts of each phase.

Answer: a) eutectic temperature: 183 C

eutectic composition: 61.9% Sn (this is the composition of 60/40 soft solder-lowmelting point!)

b) at 183.5 C: (19.2% Sn) + liquid (61.5% Sn)

at 182.5 C: (19.2% Sn) + (97.5% Sn)

c) wt% :Thus (i ) at 100 C we have

(4.5% Sn) + (~100% Sn)

wt% : = 42.4%

Question: a) Starting with an alloy of 15% Sn 85% Pb, heat to 182 C.What phases are present and what are their compositions?

b) Cool to 150 C. What phases, what are their compositions and whatpercentage of each is present?

c) Cool to 50 C. What phases, what are their compositions and whatpercentage of each is present?

Answer: a) At 182 C, a 15% Sn alloy is in the solid solution phase field. At




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equilibrium we would have homogeneous solid solution of 15% Sn in Pb.

b) On cooling from 182 to 150 C we cross thesolid solution phase boundary

(or solidus) into the + two phase region. At 150 C the alloy will consist of

phase solid solution (10% Sn) andphase solid solutions (99% Sn) by the leverrule

X 5.6% phase

X = 1 - X =94.4% phase

c) At 50 C we are further from the solidus in the 2 phase ( & ) region. At

equilibrium we would have phase solid solution consisting of 2% Sn and phase solid solution consisting of 100% Sn. By the lever rule.

X 13.3% phase

X 86.7% phase

Note that one can start at 182 C with single phase solid solution (15% Sn) and by cooling produce

more and more of a new phase. We could write the solid state reaction that occurs on cooling as

(high wt% Sn) (lower wt% Sn) +

This reaction is known as a precipitation reaction. Much as we can heat up an aqueous solution ofsalt (NaCl) to dissolve more salt, then cool the solution and observe that the solution which wassaturated at high temperature exceeds the solubility limit (Fig III.3) so that new NaCl crystalsprecipitate from the super-saturated solution, we can heat up a Pb-15% Sn alloy, dissolve all the Sn

insolid solution and then cool to a temperature where the solubility limit is exceeded so that phase precipitates from the solid solution.

The more we exceed the solubility limit of Sn in (given by the solidus curve), the higher the

percentage ofprecipitates will be in the two phase mixture. Since the solubility of Sn insolid

solution decreases with decreasing temperature we can achieve a higher fraction ofprecipitates bycooling to a lower temperature. Solid state precipitation reactions like this one are an important wayof strengthening metallic alloys.

MORE COMPLICATED PHASE DIAGRAMS

The phase relations in binary systems can be extremely complicated. We only have seen the mostsimple ones. As an example we show finally 2 complex phase diagrams - one of Cu-Zn, the other ofthe very important system Fe -C (steel). Note that there are transformations of the kind

solid solution 1 solid solution 2 + solid solution 3.

This is called EUTECTOID TRANSFORMATION.




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Fig III 9

Question: Find eutectoid transformations in both the Cu-Zn and the Fe-Cdiagrams, and write the transformation reaction.

Fig III 10

Answer: Cu-Zn: + at about 550 C.

Fe-C: + Fe3C at 723 C.

or austenite (FCC)ferrite (BCC) + cementite (complex orthorhombiccrystal structure).

Question: What phases are present, what are their compositions and whatare their compositions and what are their relative amounts for each of thefollowing points on the Fe-C phase diagram?




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a) 1%C, 1000 C

b) 1%C,724 C

c) 1% C, 722 C

d) 0.6%C, 724 C

e) 0.6%C, 722 C

Answer: a) At 1000 C, an Fe 1% C alloy is a homogeneous solid solution

of C iniron (FCC austenite)

b) At 724 C an Fe-1% C alloy is a two phase mixture ofiron(composition = 0.8%C) and Fe

3C (6.67%C). By the lever rule

XFe3C 3.4% Fe3C

X 96.6% -iron

c) At 722 C an Fe-1% C alloy is a two phase mixture of-iron (ferrite(BCC) iron composition 0.025%C) and Fe

3C (6.67% C). By the lever rule

XFe3C 14.7% Fe3C

X 85.3% -iron

d) At 724 C an Fe-0.6%C alloy is a two phase mixture of-iron (0.025%

C) and -iron (0.8%C) by the lever rule

X 25.8% -iron

X 74.2% -iron

e) At 722 C an Fe - 0.6%C alloy is a two phase mixture ofiron (0.025%C) and Fe3C (6.67%C).

X 91.4% -iron

XFe3C 8.6% Fe3C

The FeC phase diagram is so important that you should memorize the major features of thediagram. This includes the important temperatures, such as 723 C which is called an Eutectoid




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Temperature, and compositions, such as 0.8%C.


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