Handout 1 - Cooling Transformations

7/28/2019 Handout 1 - Cooling Transformations

1/15

Material Science

1

Lecture 1Cooling Curves

Most substances can exist as a solid, liquid and gas. For example water, if it is sufficienthot it is steam and if it is sufficiently cold it is ice. Between these two temperatures it is

obviously a liquid. If water is brought to above its boiling temperature and then allowedslowly to cool, the change in temperature against time can be plotted and a cooling curvecan be constructed. This is shown in figure 1.

Figure 1Cooling curve of pure water

It can be seen that where a change of state occurs i.e. a change in phase, there is a short

pause in the cooling process. This pause is referred to as an arrest point and it is theresult of the water giving up latent heat energy. Latent heat is the heat energy required to

produce a change in state in a substance at a constant temperature. Thus a physical

change in state during cooling or heating is always accompanied by an arrest point in the

cooling or heating curve.

The cooling curve of water is typical of all pure substances and applies equally to pure

metals. Alloys, however, consist of two or more components and to understand theirbehaviour we will consider a solution of water and salt.

Figure 2Cooling curve for salt solution

It can be seen that the salt-water solution has two arrest points and that both of these arebelow the freezing point of pure water. As cooling continues below 0C droplets of pure

water separate out from solution and immediately turn into ice particles. This occurs at

the upper arrest point, which is usually not too well defined. This process of separationcontinues as the temperature falls therefore, more and more water separates out causing

the concentration of the salt water to increase. When the lower arrest point is reached,


2/15

Material Science

2

even the concentrated salt solution freezes and no liquid phase is left. The solid formed

consists of a mixture of fine crystals of pure water and fine crystals of salt.

If the experiment is repeated several times using stronger and weaker solutions a family

of cooling curves can be plotted.

Figure 3Cooling curves for salt solutions of different concentrations

Four important observations can be made:

Since the amount of salt which can be held in solution with water varies withtemperature, water separates out as ice crystals between the arrest points of hypo-eutectic

solutions as the temperature falls, and salt crystals separate out between the arrest points

of hyper-eutectic solutions as the temperature falls, therefore the remaining solution isalways of a constant concentration and therefore, has a constant arrest point temperature.

This concentration is the eutectic composition. The fact that excess water or salt is

rejected from solution means that the eutectic balance or equilibrium is always

ultimately achieved. When a family of cooling curves is plotted in this way it is referredto as aphase equilibrium diagram.

Summary Questions Lecture 1

1. What is latent heat?2. Differentiate between asaturatedand asuper saturatedsolution.3. Explain why impure substances have a less defined freezing/melting point.4. Why is salt spread onto the roads in icy conditions?5. What would be the ideal salt concentration for reducing the risk of ice?


3/15

Material Science

3

Lecture 2Phases in solidified alloys

Until now weve understood the term phase to describe whether a material is a solid,liquid or gaseous state. Let us now consider it in more detail and define it as:

A portion of a system, which is of uniform composition and texture throughout,and, which is separated from the other phases by clearly defined surfaces.

Thus for the salt solution considered last lecture there are four possible phases:

Each of these four phases is uniform composition and is separated from adjacent phases

by definite boundaries. It is necessary to distinguish between crystals (or grains) and

phases. The ice phase may appear as separate lumps with each lump containing manysingle-phase water crystals and similarly the single-phase crystalline sodium chloride

may appear as lumps of salt containing many separate crystals.

Extending this argument to metal alloys, when a liquid solution of two metals (a binaryalloy) solidifies, one of the following conditions will occur.

Simple eutectic type - Metalsthat are soluble in the liquidstate, become totally insoluble in

the solid state and they separate

out as grains of two pure metals.In practice this is rare since there

is usually some solubility of one

metal in another.

Solid solution type - If thealloying elements have similar

chemical properties, and theiratoms are of similar size theywill form a solid solution, that is,

they remain totally soluble in the

solid state.


4/15

Material Science

4

Combination type - they remainjust partially soluble in the solid

state

Intermetallic Compound - In asituation where atoms with

sufficiently different chemicalproperties attract each other the

two metals react together to form

an intermetallic compound.Upon cooling a single phase

consisting of many grains of thesame composition will be

present. In general, intermetalliccompounds tend to be hard and

brittle and are thus less useful for

engineering alloys than the toughand ductile solid solutions. The

exceptions being some bearing

metals where they form hard,

wear resistant pads with a lowcoefficient of friction, set in a

matrix of tough, ductile solidsolution.

The individual grains found in any of these phases may vary considerably in size. Some

are large enough to see with the unaided eye, whilst others are so small that a high-powered microscope is required. Note that although the phases found in alloys, as

described above, are formed from two metals, they may equally be formed between a

metal and a non-metal. For example, austenite is a solid solution of carbon in iron,

whilst cementiteis the compound iron carbide. Well consider these in a lot more detailin future lectures.


1. How manyphases are present in the solid form of salt water?2. Which of the four solid compositions (as discussed above) would best describethe solid form of salt water?

3. State two metals that are insoluble in each other in their molten form.4. Define asolid solution.5. What factors determines which solid composition an alloy will exhibit?


5/15

Material Science

5

Lecture 3Phase Equilibrium Diagrams (eutectic type)

In the next couple of lectures we will continue to consider binary alloys and the threemain ways in which they solidify, namely; eutectic, solid solution and combination type.

Eutectic typeFigure 1 shows a eutectic-type phase equilibrium diagram and it can be seen that it is

very similar to the diagram produced for a sodium chloride and water solution. That is,

total solubility of the salt in water in the liquid state and total insoluble in the solid state.

In figure 1 the two components are referred to as metal A and metal B. although they aremutually soluble in the liquid state, both components retain their identities of crystal A

and crystal B in the solid state.

Figure 1 - Phase equilibrium diagram (eutectic type)

This type of equilibrium diagram gets its name from the fact that at one particular

composition (E), the temperature at which solidification commences is a minimum forthe alloying elements present. With this composition the liquidus and thesolidus coincide

at the same temperature, i.e. crystals of A and B form instantaneously at the same

temperature. In practice, few metal alloys form simple eutectic type equilibriumdiagrams.

The diagram can be used to further categorise the alloys into two groups; hypo-eutectic

and hyper-eutectic. Hypo-eutectic simply means that it is to the left of the eutectic pointand hyper-eutectic is to the right.

Consider 25%A, 75%B alloy. Firstly, this is a hypo-eutectic alloy as it is to the left of the

eutectic point. Let us consider what happens as this alloy cools. To begin with it is amolten mixture of A and B. Solidification commences as it cools at the liquidus line. The

first crystals to precipitate out are crystals of B (as this is in excess). As crystals of B

fall out of solution the proportion of A in the remaining solution increases. At thesolidus line the concentration of the remaining solution is the same as the eutectic

composition and on reaching the eutectic temperature crystals of A and B precipitate out

simultaneously to form lamellar eutectic crystals (see figure 3). Therefore, the solid alloyis made up of two phases; B phase and a A+B phase (eutectic phase).


6/15

Material Science

6

ExampleAlloys of cadmium and bismuth are an example of the simple eutectic type. The phase

diagram is shown below.

Figure 2Cadmium-bismuth phase equilibrium diagram

Figure 3Lamellar structure of eutectic composition

Figure 4Solid compositions a)20% Cd, 80% Bi; b)40% Cd, 60% Bi; c)80% Cd, 20% Bi

Summary Questions Lecture 3Referring to figure 1;

1. What is the name of the line joining the points where solidification begins?2. What is the line joining the points where solidification is complete?3. What is name given to the point where the liquidus and thesolidus coincide?4. What is the temperature it occurs at called?5. What is the composition at this point called?6. State the range of both hypo and hyper-eutectic alloys.

Referring to figure 2;7. State the eutectic composition and temperature.8. Consider alloy 80% cadmium and 20% bismuth. Is it a hypo or a hyper-eutectic

alloy?9. Describe what is happening, in terms of crystal growth, as the liquid cools from

above the liquidus temperature to below the solidus temperature.

10.What are alloys of Cadmium and Bismuth commonly used for?


7/15

Material Science

7

Lecture 4Phase Equilibrium Diagrams (solid solution and

combination types)

Solid Solution

Copper and Nickel are an example of solid solution type binary alloy, that is, not only arethey mutually soluble in the liquid state, they are also mutually soluble in the solid state.

The phase diagram for copper-nickel is shown in figure 1. Again, the line marked

liquidus joins the points where solidification commences, whilst the solidus joins thepoints where solidification is complete. This time there is no eutectic composition.

Figure 1Phase diagram for copper-nickel alloy

The following points can be made;

Pure copper

Pure nickel

At the solidus

At the liquidus

Between the liquidus and solidus

Below the solidus


8/15

Material Science

8

Combination TypeMany metals and non-metals are neither completely insoluble nor are they completely

soluble. Therefore, they form a phase diagram of the type shown in figure 2.

Figure 2Combination type phase diagram

Greek symbols are used and are defined as follows;

Tin-lead alloys are a typical example of the combination type of phase diagram.

Figure 3Combination type phase diagram for Tin and Lead


1. What are alloys of tin and lead commonly used for?Referring to figure 3;

2. Define thesolvus line3. What is the composition of the and the phase at the eutectic temperature?4. Describe the solidification of an alloy of composition 10% tin and 90% lead.5. Describe the solidification of an alloy of composition 30% tin and 70% lead.


9/15

Material Science

9

Lecture 5Plain Carbon Steels

Ferrous MetalsFerrous metals and alloys are based upon the element iron (Fe). Iron is a soft, grey metal

and rarely found in its pure state. When iron is alloyed with carbon it forms solidsolutions and the compound iron carbide. The carbon content is carried over from thesmelting process during which iron is extracted from its ore. Table 1 shows the

relationship between the amount of carbon present and the resulting ferrous metal.

Table 1Ferrous metals

Figure 1 shows the iron-carbon phase equilibrium diagram.

Figure 1Iron-carbon phase equilibrium diagram

You should recognize it as a combination type, that it, the elements are only partiallysoluble in each other in the solid state. It is different and more complex than the examplelast lecture due to the fact that iron is allotropic, that is, it can exist in more than one

form.

However, we will only be concerned with the solid phases of the diagram, conventionallyknown as the steel section. This is drawn in figure 2 and will be the focus for the next

few lectures.


10/15

Material Science

10

Figure 2Iron-carbon phase equilibrium diagram (steel section)

EutectoidThe steel section of the iron-carbon phase equilibrium diagram is very similar to the

combination type of phase diagram. In the combination type diagram there was oneeutectic composition at which both alloying elements crystallized out simultaneously at

the same temperature to form lamellar structure. However, in the steel section of the iron-

carbon phase diagram such transformations occur in the solid state and the point at which

ferrite and cementite precipitate out from the solid solution of austenite is called theeutectoidpoint. Thus, the eutectoid point is similar to the eutectic point but occur in the

solid state.

There are only three important phases in the steel portion of the equilibrium diagram:

Ferrite ( phase)

Austenite ( phase)

Cementite (iron-carbide phase)

Steels can be defined as

Those alloys of iron andcarbon in which the entire

carbon content is combinedwith the iron in solid solution

or as iron carbide (or both)and that no free carbon is

present.


11/15

Material Science

11

Eutectoid SteelSteel with the eutectoid composition (i.e. with a carbon content of 0.83%) at room

temperature consists entirely of pearlite. A wholly pearlitic structure gives steelmaximum toughness. Similarly to previous eutectic type binary alloys previously

discussed, the resulting structure is lamellar i.e. alternating layers of ferrite and cementite.

Figure 3(a)Structure of pearlite

Summary Questions Lecture 51. Differentiate between the eutectic point and the eutectoid point.2. By referring to figure 1, state the following; the eutectic temperature, the

eutectoid temperature, the eutectic composition and the eutectoid composition.3. State the carbon content range of hypo-eutectoid steels and a hyper-eutectoid

steels

4. Is there such a thing as a hyper-eutectic steel?5. Differentiate between steel and cast iron.6. Define toughness.


12/15

Material Science

12

Lecture 6Cooling Transformations

Cooling transformation for a steel with eutectoid composition

Cooling transformation for a steel with hyper-eutectoid composition

At T1

At T2

At T3

At T4

At T1

At T2

At T3

Final structurecrystals

of pearlite surrounded by

bands of primary cementiteat crystal boundaries.

Final structurecrystals of pearlite.

Cementite


13/15

Material Science

13

Cooling transformation for a steel with hypo-eutectoid composition

At T1

At T2

At T3

At T4

At T5


1. By referring to both table 1 in lecture 5 and the steel section of the phase diagramsuggest how the proportion of ferrite, pearlite and cementite in the final

composition of a steel affects the use of the steel.

2. Explain why eutectoid steel is used for chisels and springs.

Final Structure

crystals of ferriteand crystals of

pearlite


14/15

Material Science

14

Lecture 7The Lever Rule

Composition of phasesThe phase diagram for a binary alloy showssingle-phase fields and two-phase fields (e.g.

liquid plus A). The fields are separated by phase boundaries. When a phase boundary iscrossed, a phase change starts, or finishes, or both.

When a single point lies in a single-phase region, the alloy consists of a single,

homogeneous, phase. Its composition must be that of the alloy. Therefore, we can saythat thephase composition and the alloy composition coincide in single-phase fields.

When a single point lies in a two-phase field the alloy breaks up into two phases that do

not have the same composition as the alloy (e.g. + ). The composition of each phase is

then given by constructing a tie line (the isotherm spanning the two-phase region,

terminating at the nearest phase boundary on either side). The ends of the tie line define

the composition of each phase. Any alloy that lies on the tie line breaks into the same twophases at its end. The proportions of each phase depend on the alloy composition.

Example 1Consider the alloy 60% Pb and 40% Sn (refer to fig.1)

a) Name the phases present at 250 C

b) Name the phases present at 200 C

What is the composition of phase 1?


c) Name the phases present at 150 CWhat is the composition of phase 1?



15/15

Material Science

15

Proportion of phase in two-phase alloysYou can get relative amounts of each phase in a two-phase alloy from the phase diagram.

The weight fraction of phase is W and that of phase is W .In a binary alloy,

W + W = 1

The weight fractions of each phase, W + W , are fixed by the requirement that matter is

conserved. Then the weight fraction of -phase in the alloy is

W = a/l

And the weight fraction of the -phase is

W = b/l

This is the lever rule. The easiest way to understand this result is to notice that, if the

point coincides with the left-hand end of the tie line the alloy is all -phase and if itcoincides with the right-hand end it is all -phase. Or if the point is closer to the left-hand

end -phase is the dominantphase, if the point is closer to the right-hand end -phase isthe dominant phase.

Example 2Consider a lead-tin alloy with composition 80% Pb and 20% Sn.

a) At 280 C which is the dominant phase?b) At 200 C which is the dominant phase?c) At 200 C what are the approximate composition of the phases and the

proportions by weight of each phase?

d) At 150 C what are the approximate composition of the phases and theproportions by weight of each phase?

Figure 1Tin-Lead phase equilibrium diagram

Solid

L+L+

Solid

+

Temp

Handout 1 - Cooling Transformations

Documents

Transcript of Handout 1 - Cooling Transformations