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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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Class 5

Class 5

Phase Diagrams Continued


Class 5

1085C

420C


Substitutional solid solubilty

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.


Class 5

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


Class 5

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

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%


Change average composition 50

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%


Microstructures and composition

Microstructures and Composition

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


Lead tin microstructures

Lead Tin Microstructures

38.1%Pb 61.9% Sn

90 %Pb 10%Sn

70% Pb 30%Sn

50%Pb 50%Sn


Lead tin microstructures1

Lead Tin Microstructures

15%Pb 85%Sn

Equiaxed Single Phase Grain Structure


Homeworks

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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