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

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.

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

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

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

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

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

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

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

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)

The Hypoeutectic Alloy Pro eutectic The Hypoeutectic Alloy

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?

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) = 45.3 % Wb = 100 R / (R+S) = 100 (61.9 – 19.0) / (97.5 – 19.0) = 54.7 %

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) = 86.0 % Wb = 100 R / (R+S) = 100 (30.0 – 19.0) / (97.5 – 19.0) = 13.0 %

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) = 74.4 % WE = 100 R / (R+S) = 100 (30.0 – 19.0) / (61.9 – 19.0) = 23.6 %

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

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

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