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Example - 718 Ni-base superalloy

Example - 718 Ni-base superalloy. Composition: 52.5Ni 19Cr 3Mo 19Fe 0.4Al 5Nb-1Ti 0.04C . Example - 718 Ni-base superalloy. Example - 718 Ni-base superalloy. Example - 718 Ni-base superalloy. Possibilities for strengthening 718:. Example - 718 Ni-base superalloy.

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Example - 718 Ni-base superalloy

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  1. Example - 718 Ni-base superalloy Composition: 52.5Ni 19Cr 3Mo 19Fe 0.4Al 5Nb-1Ti 0.04C

  2. Example - 718 Ni-base superalloy

  3. Example - 718 Ni-base superalloy

  4. Example - 718 Ni-base superalloy Possibilities for strengthening 718:

  5. Example - 718 Ni-base superalloy

  6. Example - 718 Ni-base superalloy 52.5Ni Provides friendly matrix for alloying elements 19Cr Oxidation and corrosion resistance, solid solution strengthening 3Mo Solid solution strengthening, stabilizes chromium oxide film, carbide formation for Hi T strength 19Fe Solid solution strengthening 0.4Al Increases oxidation resistance, forms strengthening pptates 5Nb-1Ti Form strengthening pptates 0.04C Carbide formation for creep resistance

  7. Diffusion concn concn concn concn distance distance distance distance Movement of atoms (or molecules) down a concentration gradient How??

  8. Diffusion Vacancies occur naturally in materials vacancy One way - vacancy movement No. fract. of vacancies = exp (-Q/RT)

  9. Diffusion Vacancies occur naturally in materials vacancy One way - vacancy movement No. fract. of vacancies = exp (-Q/RT) For Cu @ rm T: 1/1014; @ 1080°C: 1/1600

  10. Diffusion Vacancy movement

  11. Diffusion concn concn concn concn distance distance distance distance Movement of atoms (or molecules) down a concentration gradient I lied

  12. Diffusion Movement of atoms (or molecules) over a hump Where do we get the activation energy?

  13. Diffusion Where do we get the activation energy? From energy stored as thermal vibrations # jumps/sec = A exp (-Q/RT) For Cu: # jumps/sec = 1015 exp (-29/RT) = 1.1 x 10-6 @ rm T = 2.2 x 1010 @ 1080° C

  14. Diffusion Diffusion coefficient: D(T) = Do exp (-Q/RT) flux J = D(T) dC/dx For C diffusing in Fe: D(T) = 0.21 exp (-33.8/RT) = 6.7 x 10-11 cm2/sec @ 500° C = 4 x 10-7 cm2/sec @ 1000° C

  15. Diffusion For C diffusing in Fe: D(T) = 0.21 exp (-33.8/RT) = 6.7 x 10-11 cm2/sec @ 500° C = 4 x 10-7 cm2/sec @ 1000° C Ex - carburizing a steel shaft: how long to carburize to depth of 1 mm x2 = D t D(1000° C) = 3 x 10-7 cm2/sec t = 10 hr D(1200° C) = 1.8 x 10-6 cm2/sec t = 1.5 hr

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