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Class 6 CHE 333 Phase Diagrams Continued Prov08

Class 6 CHE 333

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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 Presentation

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Page 1: Class 6 CHE 333

Class 6 CHE 333

Phase Diagrams Continued

Prov08

Page 2: Class 6 CHE 333

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

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

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

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

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

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

Microstructures and Composition

Page 9: Class 6 CHE 333

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

Lead Tin Microstructures15%Pb 85%Sn

Equiaxed Single Phase Grain Structure

Page 11: Class 6 CHE 333

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?