1 / 11

Class 6 CHE 333

Class 6 CHE 333. Phase Diagrams Continued. Prov08. 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

holly
Download Presentation

Class 6 CHE 333

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Class 6 CHE 333 Phase Diagrams Continued Prov08

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

  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

  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. Homework • Using the Cu-Ni phase diagram, for a 50-50 Cu-Ni what are the compositions of the phases at 1400, 1300 and 1200 C and what phases would be present. • Using the Pb- Sn phase diagram,what are the compositions for points “a,c and e” on the diagram?

More Related