1. Working with Phase Diagrams
 Overall composition of the alloy
 Liquidus and solidus temperatures
 Limits of solid solubility
 Chemical composition of phases at any temperature
 Amount of phases at any temperature
 Structural change with temperature

2. Isomorphous Phase Diagrams
 Phase diagrams of components that are completely soluble to each other (Type 1 phase diagrams).
Example: Copper – Nickel phase diagram.
Wt% Ni 
Both Cu and Ni can go into solution completely into each other in the solid state without forming any supersaturated solution
i.e., there is no solubility limits.

3. Phase Diagram Nomenclatures
Liquidus
Start of solidification or, end of liquification 
Temperature X Y
% Y added
Liquid
Solidus
End of solidification or, start of liquification 
Liquid +
Solid
Freezing range
Not to be confused with Solvus, which indicates the limit of solid solubility in a solid solution
Solid
Composition of alloy
 Relative amounts of X and Y
 The scale can be in weight %, atomic % or mole % 
Phase diagram with complete solubility of one component into another
(Type 1 Phase Diagram)

4. Chemical composition of Phases
 Chemical composition of each phase in the system
(indicating how much X and Y in the phase).
 For systems having more than one phase, each phase will have a unique chemical composition which will be different from each other, and will also be different from the overall composition.
 Not to be confused with overall composition of the alloy.
Relative amounts of Phases
 For system having more than one phase, relative amount is
the amount of each phase relative to overall amount of the alloy (indicating kg of each phase in overall weight of the alloy).
 Depends on temperature and composition of the alloy.
 Not to be confused with composition of phases.

5. The Cooling Curve
 Cooling curve indicates the gradual change in temperature with time during cooling of alloy X Y
% Y L S
L+ S TX
Temperature
ARREST POINTS
Temperature
Time
Freezing range
Alloy 1
Solid
Liquid
Liquid + Solid
start of solidification
end of solidification TX
Pure metal X
Alloy 1

6. Interpreting Phase Diagrams
 Predicting the nature of stable phases as a function of T, P and X
 Number and types of phases
 Composition of each phase
 Relative amount of each phase

7.  Predicting the nature of stable phases as a function of T, P and X
Independent variables:
T and X ( always P=1 atm)
one-phase
area (L)
two-phase
area (L + a)
one-phase
area (a)

8.  Number and types of phases
Phase diagram rule #1:
If we know the alloy composition, C0, then we can tell the number and type of phases present in the structure at any given temperature.
wt.%Ni
Point A: (30%Ni, 1225°C)
Phases are:
L and a
Point B: (60%Ni, 1100°C)
Phases are:
a only
Cu-Ni Phase Diagram

9.  Composition of each phase
Phase diagram rule #2: Tie Line Rule
If we know C0, then we can tell the composition of each phase at any temperature.
tie line
At TA: L phase only
CL =
35%Ni
At TB: L and a phases
CL =
Ca =
Cliquidus = 32%Ni
Csolidus = 43%Ni
At TD: a phase only
Ca =
35%Ni

10.  Relative amount of each phase
Phase diagram rule #3: Lever Rule
If we know C0, then we can tell the weight fraction of each phase at any temperature.
Consider C0 = 35 wt.%Ni
 At TA, only L phase  WL = 100 wt.%
 At TD, only a phase  Wa = 100 wt.%
 At TB, both L and a phases 
WL = = = 73 wt.%
100 (43-35)
43-32
100 S
R+S
Wa = = = 27 wt.%
100 (35-32)
43-32
100 R
R+S

11. To summarise: Finding the composition in a two phase region:
1. Locate composition and temperature in diagram.
2. In the two phase region, draw a tie line at the given temperature.
3. The liquid composition will be on the liquidus line and the solid composition will be on the solidus line.
Note the intersections with phase boundaries. Read compositions at the intersections.
Finding the amounts of phases in a two phase region:
1. Locate composition and temperature in diagram.
2. In the two phase region, draw a tie line at the given temperature.
3. Fraction of a phase is determined by taking the length of the tie line to the phase boundary for the other phase, and dividing it by the total length of tie line.

12. Problem
One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following:
1. Chemical composition of the liquid and a phase at 225ºC.
2. Weight % of liquid and a at 225ºC.

13. Problem
One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following: R S
1. Chemical composition of the liquid and a phase at 225ºC.
2. Weight % of liquid and a at 225ºC.
CL = 45 wt.% Sn Ca = 17 wt.% Sn
100 (30 – 17)
WL = = 56 wt.%
100R
R+S =
100 (45 – 30)
Wa = = 54 wt.%
100S

14. Development of Microstructures
The microstructural development depends on the overall composition and the cooling rate of the alloy
1300
1200
1100 20 30 40 50
C0=35
T (°C)
wt.% Ni
L+a L a
Equilibrium Cooling of Alloy  A
C0=35  B 34 46  C 43 32
Point A
L: 35 wt.%Ni
Point E
a: 35 wt.%Ni  D 36 24
Point D
a: 36 wt.%Ni
L: 24 wt.%Ni  E
Point C
a: 43 wt.%Ni
L: 32 wt.%Ni
Point B
a: 46 wt.%Ni
L: 34 wt.%Ni
Binary isomorphous phase diagram of Copper and Nickel

15. All microstructures at a glance
Summary of Solidification Process
Solidification begins at the liquidus line and continues upon cooling
The composition and amount of the solid and the liquid change gradually during cooling
Solid phase grows by consuming all the liquid and solidification ends at the solidus line

16. Non-Equilibrium Cooling of Alloy
 The cooling of alloy is so fast that the time required for transformation is not given.
Faster cooling rate
cored structure – a.k.a. microsegregation
Slower cooling rate
equilibrium structure
Uniform Ca
35 wt.% Ni
First a to solidify:
46 wt.% Ni
Last a to solidify:
<35 wt.% Ni
Ca composition is equals to C0 throughout
Ca changes during solidification

17. Problem with cored structure
 On re-heating, grain boundaries will melt first, as they are rich in lower melting-point constituent. This can lead to premature mechanical failure!
Lighter phase with low Al content
Darker phase with high Al content
Cored structure in Al-Cu alloy
Homogenising the cored structure
 Microsegregation or the cored structure can be eliminated by reheating the material at a higher temperature.
 Diffusion of inhomogeneous element takes place.
 Macrosegregation, however, cannot be eliminated by reheating.