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Class 5

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

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Class 5

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  1. Class 5 Phase Diagrams Continued

  2. 1085C 420C

  3. Substitutional Solid Solubilty How much one element will dissolve inanother 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.

  4. 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

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

  6. 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 53-45 / 58-45 = 8/13 = 0.615= 61.5% Amount of liquid at 1300C is therefore 58-53 / 58-45 = 5/13 = 0.385 = 38.5%

  7. 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 50-45 / 58-45 = 5/13 = 0.385= 38.5% Amount of liquid at 1300C is therefore 58-50 / 58-45 = 8/13 = 0.615 = 61.5%

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

  9. Lead Tin Microstructures 38.1%Pb 61.9% Sn 90 %Pb 10%Sn 70% Pb 30%Sn 50%Pb 50%Sn

  10. Lead Tin Microstructures 15%Pb 85%Sn Equiaxed Single Phase Grain Structure

  11. Homeworks. • From the Pb­Sn phase diagram, for a 70% Pb, 30%Sn alloy, at what temperature does solid first form.  At the eutectic temperature, how much pro­eutectic phase is present and how much eutectic phase?  What are the compositions at 100oC? • For a steel containing 0.4% C:- a) At what temperature does austenite start to transform on cooling from 950oC?  b) How much proeutectoid phase is present at the eutectoid temperature  c) What is its composition? d) Upon quenching the steel from 950oC to ­50oC what phases will be present and what will be the composition?

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