class 6 che 333
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/1.jpg)
Class 6 CHE 333
Phase Diagrams Continued
Prov08
![Page 2: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/2.jpg)
Copper Zinc Phase Diagram
Peritectic reaction Liquid + solid 1 -> solid 2. Eutectoid reaction Solid 1 -> solid 2 + solid 3Maximum solid solubilty of zinc in copper is 38% at around 450C
1085C
420C
![Page 3: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/3.jpg)
Substitutional Solid SolubiltyHow much one element will dissolve in another is determined by the Hume Rothery rules
1. Atomic radii should be within 15% of each other
2 Crystal structure should be the same for each element for good solubility
3 Electronegativities should be similar.
4 The valences of the atoms should be similar.
Good solubility Cu –Ni, Cu-Au; Cu r=0.128A, Ni r=0.125, Au r=0.144Crystal structure Ti- HCP, Al - FCCPoor soluility Na-Cl Na electronegativity 0.9, Cl 3.0Valences – Zn 2+, Cu 1+Only indicate solubility from these rules.
DOES NOT APPLY TO THE ELEMENTS H,C,O,N,B THESE FORM INTERSTITIALSOLID SOLUTIONS.
![Page 4: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/4.jpg)
Iron Carbon Phase Diagram
SteelsEutectoidS1 -> S2 + S3-> + F e3C
PeritecticS1+L -> S2 + L ->
Fe3C- cementiteA compound
1538C
3367 SUBLIMES
Fe - then then C 1394C
![Page 5: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/5.jpg)
Stainless Steel Phase Diagram
Ternary phase diagram for stainless steels. In this case an isothermal section at a constanttemperature is used.
![Page 6: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/6.jpg)
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%
![Page 7: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/7.jpg)
Change Average Composition 50%
Determine the AMOUNTS of each phase use the Inverse Lever Arm Rule.Amount of solid = wa-wl/ws-wlAmount 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%
![Page 8: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/8.jpg)
Microstructures and Composition
![Page 9: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/9.jpg)
Lead Tin Microstructures90 %Pb 10%Sn
70% Pb 30%Sn
38.1%Pb 61.9% Sn
50%Pb 50%Sn
![Page 10: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/10.jpg)
Lead Tin Microstructures15%Pb 85%Sn
Equiaxed Single Phase Grain Structure
![Page 11: Class 6 CHE 333](https://reader035.vdocuments.us/reader035/viewer/2022062314/56814691550346895db3aed9/html5/thumbnails/11.jpg)
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?