# Phase Diagrams Continued

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Phase Diagrams Continued
Class 5 Phase Diagrams Continued

1085C 420C

Substitutional Solid Solubilty
How much one element will dissolve in another is determined by the Hume Rothery rules Atomic radii should be within 15% of each other Crystal structure should be the same for each element for good solubility Electronegativities should be similar. The valences of the atoms should be similar. Good solubility Cu –Ni, Cu-Au; Cu r=0.128A, Ni r=0.125, Au r=0.144 Crystal structure Ti- HCP, Al - FCC Poor soluility Na-Cl Na electronegativity 0.9, Cl 3.0 Valences – Zn 2+, Cu 1+ Only indicate solubility from these rules. DOES NOT APPLY TO THE ELEMENTS H,C,O,N,B THESE FORM INTERSTITIAL SOLID SOLUTIONS.

Iron Carbon Phase Diagram
Liquid 1538C 3367 SUBLIMES Fe - a then g then d 912C 1394C g +L Steels Eutectoid S1 -> S2 + S3 -> a + F e3C Peritectic S1+L -> S2 d + L -> g Fe3C- cementite A compound g + Fe3C a + Fe3C

Stainless Steel Phase Diagram
Ternary phase diagram for stainless steels. In this case an isothermal section at a constant temperature is used.

Lever Arm Rule Determine the AMOUNTS of each phase use the Inverse Lever Arm Rule. Amount of solid = wa-wl/ws-wl Amount of liquid = ws-wa/ws-wl Amount of solid at 1300C is therefore / = 8/13 = 0.615= 61.5% Amount of liquid at 1300C is therefore / = 5/13 = = 38.5%

Change Average Composition 50%
Determine the AMOUNTS of each phase use the Inverse Lever Arm Rule. Amount of solid = wa-wl/ws-wl Amount of liquid = ws-wa/ws-wl Amount of solid at 1300C is therefore / = 5/13 = 0.385= 38.5% Amount of liquid at 1300C is therefore / = 8/13 = = 61.5%

Microstructures and Composition
Pro-eutectic phase is formed before the eutectic reaction, eg a in the a+L region of phase diagram.

38.1%Pb 61.9% Sn 90 %Pb 10%Sn 70% Pb 30%Sn 50%Pb 50%Sn