1. Eutectic Type Phase Diagrams
MME 293: Lecture 07
Eutectic Type Phase Diagrams
Department of MME
BUET, Dhaka

2. Today’s Topics  Binary eutectic systems with partial miscibility
 Nomenclatures and invariant reaction
 Determining composition and relative amounts
 Development of microstructures
Reference:
1. WD Callister, Jr. Materials Science and Engineering: An Introduction, 5th Ed., Ch. 9.
2. SH Avner. Introduction to physical metallurgy, 2nd Ed., Ch. 6.

3. Binary Eutectic Systems with Partial Solubility
 Both metals are only partially soluble to each other in the solid state.
 Solids are formed in the form of solutions.
 The solvus line indicates the limit of solid solubility.
Lead - Tin Binary Phase Diagram

4. Lead - Tin Binary Phase Diagram
The Eutectic Point E
 where three phases are in equilibrium.
 through this point, the horizontal isotherm line passes through and the eutectic reaction occurs
 Point E is designated by the composition, CE and the temperature, TE of the alloy.
Melting point of eutectic alloy is lower than that of the components
(eutectic = “easy to melt”)
Lead - Tin Binary Phase Diagram

5. The Eutectic Reaction
 Eutectic reaction is the transition between liquid and mixture of two solid phases, S1 and S2, at eutectic concentration CE.
L (CE) S1 (CS1E) + S2 (CS2E)
cooling
heating
eutectic mixture
 For Pb-Sn system, the eutectic reaction may be written as:
L 61.9 Sn a19 Sn + b97.5Sn
cooling
heating

6. Lead – Tin Binary Phase Diagram
Eutectic alloy
 Short-freezing-range alloy
 Behaves like pure metal
 No “mushy” zone
Hypo / hypereutectic alloy
 Long-freezing-range alloys
 Wide mushy zone
Hypoeutectic Alloys
Hypereutectic
EUTECTIC ALLOY
Lead – Tin Binary Phase Diagram

7. Microstructural Development
In the case of lead-rich alloy (0-2 wt. % of tin)
solidification proceeds in the same manner as for isomorphous alloys (e.g. Cu-Ni) that we discussed earlier.
L  L + a  a
Alloys with no Eutectic and forming Unsaturated Solid Solutions

8. Alloys with no Eutectic and forming Supersaturated Solid Solutions
At compositions between the minimum and the maximum solid solubility limit, β phase nucleates as the α solid solubility is exceeded upon crossing the solvus line.
L  L + a
 a  a + b
Precipitates in Al-Si alloy
Alloys with no Eutectic and forming Supersaturated Solid Solutions

9. Micrograph of Pb-Sn eutectic, containing alternate layers of Pb-rich a phase (dark layers), and Sn-rich b phase (light layers)
alternate layers of a and b crystals
The Eutectic Alloy

10. Microstructural Development in Partially Soluble Systems
How does eutectic microstructure form?
 Compositions of α and β phases are very different. The eutectic reaction involves redistribution of Pb and Sn atoms by atomic diffusion. This simultaneous formation of α and β phases result in a layered (lamellar) microstructure that is called the eutectic structure.
Micrograph of Pb-Sn eutectic, containing alternate layers of Pb-rich a phase (dark layers), and Sn-rich b phase (light layers)

11. The Hypoeutectic Alloy
Pro eutectic
The Hypoeutectic Alloy

12. Problem For the Pb-Sn alloy system,
(a) What is the compositions and relative amounts of the phases that constitute the eutectic microstructure?
(b) For an alloy containing 30 % Sn,
(i) What are the relative amounts of a and b phases present at the eutectic temperature?
(ii) What are the relative amounts of the eutectic and the proeutectic a phase present at the same temperature?

13. Solutions  S R For the Pb-Sn alloy system,
(a) What is the compositions and relative amounts of the phases that constitute the eutectic microstructure? R S 
Eutectic structure consists of a and b phases
Ca = 19.0 wt% Sn
Cb = 97.5 wt% Sn
Wa = 100 S / (R+S) = 100 (97.5 – 61.9) / (97.5 – 19.0) = %
Wb = 100 R / (R+S) = 100 (61.9 – 19.0) / (97.5 – 19.0) = %

14. Solutions  R S (b) For an alloy containing 30 % Sn,
(i) What are the relative amounts of a and b phases present at the eutectic temperature? R S 
Wa = 100 S / (R+S)
= 100 (97.5 – 30.0) / (97.5 – 19.0)
= %
Wb = 100 R / (R+S)
= 100 (30.0 – 19.0) / (97.5 – 19.0)
= %

15. Solutions  R S (b) For an alloy containing 30 % Sn,
(ii) What are the relative amounts of the eutectic and the proeutectic a phase present at the same temperature? R S 
Wa = 100 S / (R+S)
= 100 (61.9 – 30.0) / (61.9 – 19.0)
= %
WE = 100 R / (R+S)
= 100 (30.0 – 19.0) / (61.9 – 19.0)
= %

16. The Cooling Curves L Temperature L + a a a + b Time
Alloys with no eutectic

17. L
L + (a+b)
(a + b)
Temperature
Time
Eutectic alloy

18. L
L + a
Temperature
Time
(a+b) + a
Hypo / hyper eutectic alloy