MEC281 : MATERIALS SCIENCE

CHAPTER 2 :PHASE DIAGRAM

Rasdi bin DeramanJabatan Kejuruteraan Mekanikal

UiTM Pulau Pinang

OBJECTIVE After completing this chapter, students should be able

to :

• Understand how phase diagrams describe the equilibrium state of binary alloys materials.

• Describe the characteristics of binary isomorphous and binary eutectic alloys

• Use the lever rule technique to determine phase compositions and phase amounts at any point on a binary phase diagram.

• Understand the relation between cooling curve and phase diagram.

SOLIDIFICATION OF PURE METALS & ALLOYS 

The solidification of metals and their alloys is an important industrial process.

Not only do structure alloys start with the casting of ingots for processing into reinforcing bars or structural shape, but when a metal is welded a small portion of metal near the weld melts and resolidifies.

PHASE DIAGRAM

After pouring into the mold, the molten metals/alloys cool and solidify.

Solidification is the most important phase transformation because most of metals/alloys undergo this transformation before becoming useful products.

Solidification involve liquid-solid phase transformation, e.g : casting process.

The solidification process differs depending on whether the metal is a pure element or an alloy.

In general the solidification of metals or alloys can be divided into the following steps:

i) The formation of nuclei in the melt (nucleation)

ii) The growth of nuclei into crystal

iii) The formation of a grain structure

Liquid

Nuclei

Liquid

Grain

Grain boundaries

Irregular grain

a) b) c)

Solidification Process

Formation of the grain zones At the walls of the mold, the molten metals/alloys cools rapidly. As a result, the casting develops small equiaxed grains due to the large temp. gradient at the surface.

As cooling continues, the grains grow in the direction opposite to the heat flow where the grains grow inwardly as needles is known as columnar grains.

For pure metals, columnar grains will stop to grow until complete solidification has occurred.

For alloys, further cooling of the molten alloy will cause formation of large equiaxed grains, showing segregation of alloying components in the center of casting due to the imbalance in composition between the metal that has solidified and the remaining molten metal.

Figure : Schematic illustration of two cast structures of metals solidified in a square mold: (a) pure metals; (b) alloys

b)a)

Small equiaxed

zone

Columnar grains

Large equiaxed

zone

Most metals are combined to form alloy in order to impart specific characteristic.

An alloy is a combination of two or more elements (added impurity atoms), at least one of which is a metal.

The addition of impurity atoms to a metal will result in the formation of a solid solution which composed of solvent (host atoms) and solute (alloying elements).

E.g : Steel/cast iron (iron base alloys), bronze/brass (copper base alloys), Al alloys, Ni base alloys, Mg base alloys, Ti alloys.

METALLIC SOLID SOLUTION

Two types of solid solutions:i. Interstitial Solid Solution ii. Substitutional Solid Solution

Most engineering metals are combined with other metals or nonmetals to provide increased strength, higher corrosion resistance or the desired properties.

The simplest type of alloy is a mixture of two solution which is consists of two or more elements atomically dispersed in a single phase structure.

TYPES OF SOLID SOLUTION

Examples of atoms that can form Interstitial Solid Solutions due to their small size are hydrogen, carbon, nitrogen, and oxygen.

A solid solution formed in which solute atoms can enter the holes in the solvent-atom lattice. The solute atoms fit into the spaces between the solvent or parent atoms.

Interstitial Solid Solutions

The lattice distortion creates a strain along the slip planes and grains of the material, which results in the increase of strength and hardness because it makes atom-by-atom slip along those planes more difficult.

i) Substitutional solid solution ii) Interstitial solid solution

Figure : This is a schematic illustrating how the lattice is strained by the addition of substitutional and interstitial solute. Notice the strain in the lattice that the solute atoms

cause.

A solid solution in which solute atoms of one elements can replace those of solvent atoms of another element. It can be formed by two elements solute atoms can substitute for parent solvent atoms in a crystal lattice. The crystal structure of the solvent is unchanged, but the lattice may be distorted by the present of the solute atoms.

For example, in a Cu-Ni solid solution the Cu atoms can replace the Ni atoms in the solid solution crystal lattice.

2.2.2 Substitutional Solid Solutions

This configuration can be tend to occur when the atoms do not differ greatly in size and must be quite similar :

• Less than about 15% difference in atomic radii

• The same crystal structure

• Similar electronegativity (the ability of the atom to

attract an electron)

• The same valence.

2.3 SOLUTION

When two components combined they can either remain separate entities or combine to form a single phase which is referred to as a solution. Degree to which the two components mix is referred to as the solubility.

Solutions can also occur in the solid state and with the same range of solubility as described for liquids.

Examples: Cu and Ni are completely soluble

Zinc has limited solubility in copper

Pb insoluble in copper

Examples:

Alcohol and water – completely solubleHot choc – powder mix soluble in water but limited extent

Oil and vinegar – insoluble liquids can be temporarily mixed

A phase diagram is a graphical representation of the phases present, composition, and temperature at equilibrium of the structure of an alloy or ceramics in which the phases are stable. Phase Diagram most important tools available to materials scientist and engineers.

2.4.1 Binary Phase Diagrams

A binary phase diagrams consist of two elements in

one alloy. There are three type of phase diagrams:

i. Complete Solid Solution

ii. No Solid Solution

iii. Limited Solid Solution

2.4              THE PHASE DIAGRAMS

2.4.2 Binary Phase Diagrams for Complete Solid Solution

This phase diagram shows an isomorphous* system in which only one solid phase forms. *Isomorphicity means having same crystal structure or complete solid solublity for all composition.

The alloy is solid for all composition at temperature below Tm,A

The alloy is liquid for all composition at temperature above Tm,B

liquidus line

Solidus line

Tm,A

Tm,B

The elements are dissimilar that their solubility in other is nearly can negligible.

Liquidus line

Solidus line

Temperature

Composition

Eutectic temperature

Eutectic point

L

A + B

A + L B + L

A B

2.4.3 Binary Phase Diagrams for No Solid Solution

The figure below shows element A and B cannot dissolves with each other. The elements are only completely soluble in the liquid state but completely insoluble in the solid state.

The elements are dissimilar that their solubility in other is nearly can negligible.

Liquidus line

Solidus line

Temperature

Composition

Eutectic temperature

Eutectic point

Liquid, L

+ L + L

A B

2.4.4 Binary Phase Diagrams for Limited Solid Solution

Solidus line

Solvus lineSolvus

line+

The figure below shows element A and B cannot dissolves with each other. The elements are only completely soluble in the liquid state but completely insoluble in the solid state.

2.5 THE GIBBS PHASE RULE

Phase diagram describe thermodynamic equilibrium condition. Number of phases in equilibrium limited by thermodynamics.

This is described by Gibbs phase rule:

p + f = c + np = number of phases present

f = number of degrees of freedom

(state variables : Temperature, Pressure

and composition)

c = no. of components

n = no. of noncompositional variables

(n= 1 for alloy/ceramic, n= 2 for condensate)

ExamplesIce-water-steam : 3 distinct phases

gas

solidliquid

Temperature

Pre

ssu

re

solidliquid

Temperature

Pre

ssu

re

One phase – two degrees of freedom

P1

P2

T1 T2 T3

solidliquid

Temperature

Pre

ssu

re

Two phases – one degree of freedom

T1 T2

P1P2

solidliquid

Temperature

Pre

ssu

re

Three phases – zero degree of freedom

P1

T1

When the degree of freedom is equal to zero, it is called invariant point.

gas

Phase rule for binary system , c=2 (A and B)

F = 3 - P

A single phase , P=1 , F=2

- composition and temperature can be changed

independently

Two-phase mixture F=1

- at a given temp, the composition of the liquid and solid

is fixed , given by the liquidus and solidus line

2.6 INTERPRETATION OF PHASE DIAGRAM BY USING LEVER RULES METHOD

A phase diagram gives the following information:

• The phase present (solids and liquids or combinations)

• The composition (chemical makeup) of these phases • The amount of each phase

2.6.1 The phase present

One just locates the temperature – composition point

on the diagram and notes the phase(s) with which the

corresponding phase field is labeled.

2.6.2 The composition

i. Draw the tie line across the two-phase region at the temp. of the alloy.

ii. The intersections of the tie line & phase boundaries on either side are noted.

iii. Draw the perpendicular line from these intersections to the horizontal composition axis,

2.6.3 Determination of Phase Amount

Temperature = 1250 ºC Composition = 35 wt%Ni-65 wt % NiPhases present = + Liquid

Mass fraction of , W andMass fraction of Liquid, WL

L

oL

L

CC

CCW

SR

SW

The relative amounts of the phases present at equilibrium

can be determined by using Lever Rule method.

Example: Based on the diagram below, determine the phase present,

the chemical composition and relative phase amount at the composition and temperature 53 wt% Ni and 1300 ºC respectively.

Solution:

Example: Based on the diagram below, determine the phase present,

the chemical composition and relative phase amount at the compo. and temp. 30 wt% Sn and 200 ºC respectively.

Solution:

Example: Based on the diagram below, determine the phase

present, the chemical composition and relative phase amount at the eutectic point and slightly below eutectic temperature.

Solution:

CE

Eutectic temp. (TE) : No liquid below TE

and form 2 different solid phases.

Eutectic temp. (TE) a.k.a. triple point.

Eutectic concentration (CE) :

Min. melting temp. composition (CE)Figure : Copper-silver phase diagram

Gradual solidification of the 50%A – 50%B composition

Microstructural development during slow cooling

Figure : Pb-Sn system

Cooling curve – useful tools for studying the solidification process.

In solidification of alloyed metals, the temp. of phase changes will be different for every composition (because alloy consists of 2 or more metals).

The temp. & comp. of phase changes can be graphed so that all possible combination of two pure metals are represented.

This will produce a graph known as PHASE DIAGRAM which shows the relationship among temp., comp. & phases present in alloy system.

COOLING CURVE DIAGRAM

• For pure metal, the cooling curves show horizontal thermal arrest at their freezes points.

• Different composition will give different cooling curves.

• The slope changes at L1-S1 etc., are correspond to the liquidus and solidus.

L1

S1Freezing zone

COOLING CURVE FOR ISOMORPHOUS BINARY

DIAGRAM

1

equilibrium phase diagram

A binary isomorphous cooling curves

Liquid + Solid

Liquid

Solid

A B C D

COOLING CURVE FOR EUTECTIC BINARY DIAGRAM

Solid

Liquid

2.8.1 Eutectice Transformations

L + cooling

The eutectic transformation occurs at 183oC at a comp. of 61% Sn.

+

In the eutectic reaction the single liquid phase transforms simultaneously into the two solid phases form at the eutectic temperature. The eutectic reaction can be written as,

2.8 VARIOUS TYPE OF ALLOY TRANSFORMATIONS FOR BINARY PHASE DIAGRAMS

1517oC

2.8.2 Peritectic alloy transformations

In the peritectic reaction a liquid phase reacts with a solid phase to form a new and different solid phase. The peritectic reaction can be written as, Liquid, L + For example, L (5.4% Ni) + (4.0% Ni) (4.3% Ni)

cooling

2.8.3 Monotectic transformation

L1 + L2

The monotectic reaction occurs where at a specific temp. the liquid solution of composition L1 transforms into a solid

-phase and a second liquid of composition L2 which is

different from L1. The equation reads:

cooling

The monotectic transformation occurs at 955oC at a comp.of 36% Pb. From 955oC down, pockets of lead-rich melt are entrapped in the solid Cu alpha-phase.

The eutectoid reaction is similar to the eutectic reaction in that one phase decompose into two phases upon cooling. A eutectoid reaction where the single solid phase transforms into solid phases having different compositions from the phase.The eutectic reaction can be written as,  

2.8.4 Eutectoid transformation

cooling

A in peritectoid reaction the two solid phases transform into the single solid phase. The peritectoid reaction can be written as,  

2.8.5 Peritectoid transformation

cooling

2.8.6 Monotectoid transformation

21cooling

The monotectoid reaction occurs where at a specific temp. the solid solution of composition transforms into the two solid phases . The equation reads:

1

The transformation of a Eutectic, Peritectic and Monotectic is consider as the transformation form liquid to solid phase. While the transformation of a Eutectoid, Peritectoid and Monotectoid is consider as the transformation form solid to solid phase.

2.9 INVARIANT REACTIONS

· Invariant reactions are found in phase diagrams of many binary alloys. An alloy may have only one invariant reaction ( e.g. Pb-Sn) or have many invariant reactions depending on concentration and temperature (see engineering alloys).

• Invariant reactions are always determined in a neighborhood of the triple point that pins the invariant reaction. According to Gibbs phase rule there is zero degrees of freedom at these reaction points.

When the degree of freedom is equal to zero, the reaction is called invariant reaction.

EXAMPLEFind the peritectic and eutectoid reactions in the Cu-Zn system?

• TUTORIAL • Bismuth (melting point 271°C) and cadmium

(melting point 321°C) are assumed to be completely soluble in the liquid state and completely insoluble in solid state. They form a eutectic at 144°C containing 40 % cadmium.

• Draw the equilibrium diagram to scale on a piece of graph paper labeling all points, lines and areas.

• For an alloy containing 70% cadmium,• Determine the temperature of initial and final

solidification.• Determine the chemical composition and relative

amounts of phases present at a temperature of 200°C.

• Sketch the microstructure at room temperature.• Draw the cooling curve.

OCT 2004The cooling curves obtained when two pure metals and several of their alloys were cooled from the liquid phase are shown in figure below.

a) Draw the phase diagram for the alloy system to scale on a piece of graph paper and label all points, lines and areas.

b) For an alloy containing 40 wt%A, determine the liquidus temperature, solidus temperature and freezing range of the alloy.

c) Consider an alloy containing 10 wt%A. During solidification, determine composition of the first solid to form.

d) For an alloy containing 70wt% A, determine the chemical composition and relative amount of phases present at 600°C.

MAR 2002Table 1.0 shows temperatures of the beginning and ending of the solidification for Germanium-Silicon system. Construct a phase diagram to scale on a piece of graph paper labeling all points, lines and areas.

iii. Describe the solidification process of a 40wt%Siand determine its chemical compositions and relative amounts of phases present at a temperature of 1250°C.

iv. Draw the cooling curve for the 40wt%Si alloy.

v. Describe the solidification process.

For an alloy containing 40wt%Ge which is cooled from liquid phase to room temperature, determine: i. the composition of the first liquid solidified. ii. the composition of the last liquid solidified.

MAR 2004A hypothetical A-B alloy of composition 55wt%B - 45wt%A at some temperature is found to consist of relative amount fraction of 0.5 for both a and (3 phases. If the composition of the p phase is 90wt%B - 10wt%A, what is the composition of the a phase?

OCT 2003The Bismuth (Bi) and Cadmium (Cd) alloy system is generally known as the system with no single phase of solid solution. It forms eutectic at the composition (CE) and temperature (TE) of 40wt%Cd and 114°C respectively. Based on the information given,answer the following questions :i) Draw the phase diagram on the graph paper and label all points, lines and areas.Given that the melting point for Bismuth and Cadmium are 280°C and 310°C respectively. (Assume all lines are straight).

ii) With the aid of neat sketches, distinguish the microstructure at room temperature for the eutectic alloy from an alloy containing 80wt%Cd.

Eg 1: For an alloy containing 75 wt% aluminium, determine the relative amount:a) The temperature just slightly below 577C.b) The temperature just slightly above 577C. SOLUTION:

54

Iron Carbide Phase Diagram (Fe- Fe3C)

The Iron-Iron Carbide Phase Diagram

55

Iron Carbide Phase Diagram (Fe- Fe3C)

Alpha, α Ferrite: - --Very low

solubility of carbon. - BCC structure. - Max solid

solubility of 0.022 wt% C at 727°C and the solubility of carbon in ferrite decreases to 0.008 wt%C at 0°C.

56

Iron Carbide Phase Diagram (Fe- Fe3C)

Austenite, : - Interstitial solid solution of C in γ

iron. - FCC structure.- Solubility of C is

2.11% at 1148°C and 0.77% at 727°C.

57

Iron Carbide Phase Diagram (Fe- Fe3C)

Delta, iron- BCC structure - Max. %C : 0.09

at 1495 ℃Cementite, Fe3C : -

Intermetallic compound. - 6.67% C and 93.33% Fe. - Orthorombic

structure - Brittle

58

• Allotropic changes / Polymorphic transformation: Species with the same chemical composition, but different crystal structures

• MICROSTRUCTURECHANGES

59

1) Peritectic reaction.

2) Eutectic reaction.

3) Eutectoid reaction.

Invariant reactions in Fe – Fe3C system

L (0.53 %C) + (0.09 %C) (0.17 %C)1495 ℃

L (4.3 %C) (2.11 %C) + Fe3C (6.67 %C)1148 ℃

(0.77 %C) (0.02 %C) + Fe3C (6.67 %C)727 ℃

60

Slow Cooling of Plain Carbon Steel• Eutectoid plain carbon steel: If a sample is heated up

to 7500C and held for sufficient time, structure will become homogeneous austenite.

• Below eutectoid temperature, layers of ferrite and cementite are formed. Pearlite.

• The microstructure of eutectoid steels

61

Slow Cooling of Plain Carbon Steel (Cont..)

• Hypoeutectoid plain carbon steel: If a sample of 0.4% C is heated up to 9000C, it gets austenitized, .

• Further cooling gives rise to α and pearlite. Pearlite

62

Slow Cooling of Plain Carbon Steel (Cont..)• Hypereutectoid plain carbon steel: If a 1.2% C sample

is heated up to 9500C and held for sufficient time, it entirely gets austenitized.

• Further cooling results results in eutectoid cementite and pearlite.

63

Formation of pearlite• Two-dimensional morphology of pearlite,

apparently consisting of alternating layers of cementite and ferrite.

• Three-dimensional analogy to the morphology of pearlite, i.e. the cabbage represents a single crystal of pearlite, and the water in the bucket the single crystal of ferrite.

64

Eg 2: Sketch of microstructural changes for a steel containing 0.4 % C as it is cooled from within the austenite phase region to below eutectoid temp.

SOLUTION:

(Ans: pearlite 51%)

65

Evolution of Microstructure

Euctectoid

66

HYPOEUTECTOID STEEL

(Fe-C System)

Co

Fe3C

(ce

menti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

727°C

1148°C

T(°C)

R S

r s

w =s/(r+s)w =(1-w)

w =S/(R+S)wFe3C =(1-w)

wpearlite = w

pearlite

100m Hypoeutectoid steel

67

(Fe-C System)

Co

Fe3C

(ce

menti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

Co, wt% C0.7

7

1148°C

T(°C)

R S

s

wFe3C =r/(r+s)w =(1-wFe3C)

w =S/(R+S)wFe3C =(1-w)

wpearlite = wpearlite

60m Hypereutectoid steel

r

Fe3C

HYPEREUTECTOID STEEL

68

SUMMARY OF Fe-Fe3C P.D

A carbon steel alloy containing 0.3 wt% C when cooled just below eutectoid temperature level. Determine:

a) The relative amount of proeutectoid ferrite form.

b) The relative amount of eutectoid ferrite form.

c) Sketch the microstructure of item (a) and (b) respectively.

70

Mar 2005

With the aid of a neat sketch, describe the microstructural changes which take place when a 1.1 wt% plain carbon steel is slowly cooled from the austenite region just above the upper critical temperature line.

Sep 2001 A carbon steel alloy containing 0.5 wt% C when cooled

from the austenite region crosses the A3 line. Describe with neat sketches the microstructural changes that take place during the slow cooling of this steel from the austenite region.

71

Example 3.

A 0.25% C hypoeutectoid plain-carbon steel is slowly cooled from 950°C to a temp. just slightly below 727°C. Calculate:

(a) the weight percent proeutectoid ferrite in the steel.(b) the weight percent eutectoid ferrite and weight

percent eutectoid cementite in the steel.

A 0.90% C hypereutectoid plain-carbon steel is slowly cooled from 900°C to a temperature just slightly above727°C. Calculate the weight percent proeutectoid cementite and weight percent austenite present in the steel.Example 4

72

Example 5.

What is the Carbon concentration of an iron-carbon alloy for which the relative amount (fraction) of total ferrite is 0.94?

A 1.25% C hypereutectoid plain-carbon steel is slowly cooled from 950°C to a temp. just slightly below 727°C. Calculate:

(a) the weight percent proeutectoid cementite in the steel.

(b) the weight percent eutectoid ferrite and weight percent eutectoid cementite in the steel.

Example 4 (Due date:1/9/2008)

End of the Chapter 2

OPEN BOOK QUIZ

1. Answer Tutorial 1 Q1.2 2. Answer Tutorial 2 Q2.3

Note: ; Angstrom = m

MARK = 3/100