Phy351 ch 5

Phase Diagrams

Introduction Phase: A region in a material that differs in structure and function

from other regions.

Phase diagrams:

Represents phases present in metal at different conditions

(Temperature, pressure and composition).

Indicates equilibrium solid solubility of one element in

another.

Indicates temperature range under which solidification

occurs.

Indicates temperature at which different phases start to

melt.

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Phase Diagram of Pure Substance Pure substance exist as solid, liquid and vapor.

Phases are separated by phase boundaries.

Different phases coexist at triple point.

Example : Water, Pure Iron.

3PT phase diagram

Question 1In the pure water pressure-temperature equilibrium phase

diagram, what phases are in equilibrium for the following

conditions:

a. Along the freezing line

b. Along the vaporization line

c. At the triple point

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Question 2a. How many triple points are there in the pure iron pressure-

temperature equilibrium phase diagram of below?

b. What phases are in equilibrium at each of the triple points?

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P+F = C+2

P = number of phases that coexist in a system

C = Number of components

F = Degrees of freedom (is the number of variables such

pressure, temperature or composition that can changed

independently without changing the number of phases in

equilibrium in the chosen system)

Gibbs Phase Rule

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Example 1

For pure water, at triple point, 3 phases coexist.

There is one component (water) in the system.

Therefore 3 + F = 1 + 2 F = 0.

Degrees of freedom indicate number of variables that can be

changed without changing number of phases.

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Question 3Refer to the pressure-temperature equilibrium phase diagram for

pure water below:

a. How many degrees of freedom are there at the freezing line?

b. How many degrees of freedom are there at the vaporization

line?

c. How many degrees of freedom

are there inside a single phase?

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Cooling Curves Used to determine phase transition temperature.

Temperature and time data of cooling molten metal is recorded and plotted.

Thermal arrest : heat lost = heat supplied by solidifying metal

Alloys solidify over a range of temperature (no thermal arrest)

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Pure Metal

Iron

Consider a mixture or alloy of two metals instead of pure

substances.

A mixture of two metals is called a binary alloy and constitutes a

two component.

Example:

pure substance : cooper is one component

alloy : cooper + nickel is two component

Binary Alloy

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Binary alloy Mixture of Two component

two systems system

Isomorphous system:

Two elements completely soluble in each other in liquid and solid

state.

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Solid Solution Solid substitutional atoms into an existing lattice.

In a solid solution, some atoms of the solute replace those of the

solvent in the solvent's normal lattice structure and randomly

distributed among the lattice sites.

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Solid Solution Phase diagram

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Question 4 Determine the liquidus temperature, solidus temperature, and

freezing range for the following MgO-FeO ceramic compositions:

a) MgO -25wt% FeO

(Answer: 22600C, 25900C, 22600C-25900C)

b) MgO - 45wt% FeO

c) MgO - 65wt% FeO

d) MgO -80wt% FeO

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Example: Cu-Ni solution.

Figure 8.5: The cooper-nickel phase diagram- Cooper and nickel have complete liquid solubility and complete solid solubility. - Cooper nickel solid solutions melt over a range of temperatures rather than at a fixed temperature, as is the case for pure metals.

Binary Diagrams

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Phase Diagram from Cooling

Curves Series of cooling curves at different metal composition are first

constructed.

Points of change of slope of cooling curves (thermal arrests) are

noted and phase diagram is constructed.

More the number of cooling curves, more accurate is the phase

diagram.

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The Lever Rule The Lever rule gives the WEIGHT % of phases in any two phase

regions.

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Wt fraction of solid phase

Xs = w0 – w1

ws – w1

Wt fraction of liquid phase

X1 = ws – w0

ws – w1

Question 5 A cooper-nickel alloy contains 47 wt % Cu and 53 wt % Ni and is at

13000C. Use Fig 8.5 and answer the following:

a. What is the weight percent on cooper in the liquid and solid

phases at this temperature?

(Answer:55% wt Cu , 42% wt Cu)

b. What weight percent of this alloy is liquid and what weight

percent is solid?

(Answer: 35% , 62%)

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Binary Eutectic Alloy System

In some binary alloy systems, components have limited solid

solubility.

Example : Pb-Sn alloy.

Eutectic composition freezes at lower temperature than all other

compositions.

This lowest temperature is called eutectic temperature.

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Fig 8.12 The lead-tin equilibrium phase diagram. -This diagram is characterized by the limited solid solubility of each terminal phase (a and b).- The eutectic invariant reaction at 61.9% Sn and 1830C is the most important feature of this system.- At the eutectic point, a (19.2% Sn),b (97.5% Sn) and liquid (61.9 % Sn) can coexist.

In simple binary eutectic systems like the Pb-Sn, there is a specific

alloy composition known as the eutectic composition that freezes

at lower temperature than all other compositions.

Eutectic temperature is the low temperature which corresponds

to the lowest temperature at which the liquid phase can exist when

cooled slowly.

In the Pb-Sn system, the eutectic temperature (1830C) determine a

point on the phase diagram called the eutectic point.

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When liquid of eutectic composition is slowly cooled to the

eutectic temperature, the single liquid phase transforms

simultaneously into two solid forms (solid solution a and b). This

transmission is known as the eutectic reaction and written as:

The eutectic reaction is called an invariant reaction since it

occurs under equilibrium conditions at a specific temperature and

alloy composition that cannot be varied (according Gibbs rule,

F=0).

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Liquid α solid solution + β solid solutionEutectic temperature

Cooling

Question 6a. Make phase analyses of the equilibrium (ideal solidification of

lead-tin alloys at the following points in the lead-tin phase

diagram of Fig 8.12 (alloy 1):

i. At the eutectic composition just below 1830C (eutectic temperature). (Answer: a, 19.2% Sn in a phase, 45.5%, b, 97.5% Sn in b phase, 54.5%)

ii. The point C at 40% Sn and 2300C. (Answer: a, 15% Sn in a phase, 24.2%, liquid, 48% Sn in liquid phase, 75.8%)

iii. The point d at 40% Sn and 1830C + T. (Answer: a, 19.2% Sn in a phase, 51%, liquid, 61.9% Sn in liquid phase, 49%)

iv. The point e at 40% Sn and 1830C – T. (Answer: a, 19.2% Sn in a phase, 73%, b, 97.5% Sn in b phase, 27%)

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b. One kilogram of an alloy of 70 percent Pb and 30 percent Sn is

slowly cooled from 3000C. Refer to the lead-tin diagram of Fig.

8.12 and calculate the following:

i. The weight percent of the liquid (Answer: 64.3%)

ii. The weight percent of the liquid just above the eutectic temperature

(1830C) and the weight in kilograms of these phases. (Answer: 25.3%,

252.9g)

iii. The weight in kilograms of alpha and beta formed by the eutectic

reaction. (Answer: 86.2% wt a, 862g, 13.8% b, 138g)

c. An Pb-Sn alloy (Figure 8.12) contains 40% wt b and 60% wt a at

500C. What is the average composition of Pb and Sn in this alloy? (Answer: 25.3%, 252.9g)

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Various Eutectic Structures Structure depends on factors like minimization of free energy at α /

β interface.

Manner in which two phases nucleate and grow also affects

structures.

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Peritectic reaction:

Liquid phase REACTS with a solid phase to form a new and

different solid phase.

Binary Peritectic Alloy System

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Liquid + α β

cooling

Peritectic reaction occurs when a slowly COOLED alloy of Fe-4.3

wt% Ni passes through Peritectic temperature of 15170C in Figure

8.17.

Peritectic point is invariant.

The reaction can be written as:

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Liquid(5.4 wt% Ni) + δ (4.0 wt% Ni) γ 4.3 wt % Nicooling

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Figure 8.17:

- The peritectic region of the iron-nickel phase diagram.

- The peritectic point is located at 4.3% Ni and 15170C, which is point c.

Question 71. Make phase analyses at the

following points in the platinum-silver equilibrium phase diagram given:

a. The point at 42.2 % Ag and 14000C

b. The point at 42.2 % Ag and 11860C + T

c. The point at 42.2 % Ag and 11860C - T

d. The point at 60 % Ag and 11500C

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PHASES PRESENT Liquid Alpha

COMPOSITION 55% Ag 7%Ag

AMOUNT OF 42.4 –7 55-42.4

PHASES 55 – 7 55 - 7

= 74% = 26%30

Solution 1 (a) :

2. Consider an Fe -4.2 wt % Ni alloy (Fig 8.17) that is slowly cooled

from 15500C to 14500C. What weight percent of the alloy

solidifies by the peritectic reaction? (Answer: 66.7%)

3. Consider an Fe-5.0 wt % Ni alloy (Fig 8.17) that is slowly cooled

from 15500C to a4500C. What weight percent of the alloy

solidifies by the peritectic reaction? (Answer: 36.4%)

4. Determine the weight percent and composition in weight

percent of each phase present in an Fe-4.2 wt % Ni alloy (Fig

8.17) at 15170C + T. (Answer: 14.3% wt in liquid, 85.7% wt in

d)

5. Determine the composition in weight percent of the alloy in the

Fe-Ni system (Fig 8.17) that will produce a structure of 40 wt % d

and 60 wt % a just below the peritectic temperature. (Answer:

4.18% Ni , 95.82% Fe)31

Binary Monotectic Systems

Monotectic Reaction:

Liquid phase TRANSFORMS into solid phase and another liquid.

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L1 α + β

cooling

Two liquids are immiscible.

Example:-

Copper – Lead (system at 9550C and 36% Pb)

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Figure 8.24:

- The cooper-lead phase diagram.

- The most important feature of this diagram is the monotectic invariant reaction at 36% Pb and

9550C,.

- At the monotectic point a (100%Cu), L1 (87%Pb) can coexist. Note that cooper and lead are

essentially insoluble in each other.

Question 81. In the cooper-lead (Cu-Pb) system (Figure 8.24 ) for an alloy of

Cu-10 wt % Pb, determine the amounts and compositions of the

phases present at

a. 10000C (Answer: 47.7% wt a , 100% Cu, 0% Pb in a phase, 52.6% wt L1, 81% Cu, 19% Pb in

L1 phase)

b. 9550C + T (Answer: 72.2% wt a , 100% Cu, 0% Pb in a phase, 27.8% wt L1, 64% Cu, 36% Pb in

L1 phase)

c. 9550C – T (Answer: 88.5% wt a , 100% Cu, 0% Pb in a phase, 11.5% wt L2, 13% Cu, 97% Pb in

L2 phase)

d. 2000C (Answer: 90% wt a , 99.995% Cu, 0.005% Pb in a phase, 10% wt b, 0.007%

Cu, 99.993% Pb in b phase)

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2. For an alloy of Cu-70 wt % Pb (Figure 8.24), determine the

amounts and compositions in weight percent of the phases

present at

a. 9550C + T (Answer: 66.7% wt L2 , 64% Cu, 36% Pb in L1 phase, 33.3% wt L1 , 13% Cu, 87% Pb

in L2 phase)

b. 9550C – T (Answer: 19.5% wt a , 100% Cu, 0% Pb in a phase, 80.5% wt L2 , 13% Cu, 87% Pb

in L2 phase)

c. 2000C (Answer: 30% wt a , 99.995% Cu, 0.005% Pb in a phase, 70% wt b , 0.007%

Cu, 99.993% Pb in b phase)

3. What is the average composition (weight percent) of a Cu-Pb

alloy that contains 30 wt % L1 and 70 wt % a at 9550C + T ? (Answer: 10.8% Pb. 89.2% Cu)

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Invariant Reaction

Eutectoid Reaction

Eutectoid reaction is similar with eutectic reaction which two solid

phases are formed from one phase on cooling.

However, in the eutectoid reaction, the decomposing phase is

solid.

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Peritectoid Reaction

Peritectoid reaction two solid phase reacts to form new solid

phase.

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References

A.G. Guy (1972) Introduction to Material Science, McGraw Hill.

J.F. Shackelford (2000). Introduction to Material Science for Engineers, (5th Edition), Prentice Hall.

W.F. Smith (1996). Priciple to Material Science and Engineering, (3rd Edition), McGraw Hill.

W.D. Callister Jr. (1997) Material Science and Engineering: An Introduction, (4th Edition) John Wiley.

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