1 / 38

Diffusion Chapter 8

Diffusion Chapter 8. Selection of steel for gears. Wear resistance. 1410. 1150. 910. 725. 0.8. 0.02. Fraction of cementitite by LEVER rule:. More carbon. More cementite. More wear resistance. Cementite is hard as well as brittle. Hardness or strength is desirable.

verdi
Download Presentation

Diffusion Chapter 8

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. Diffusion Chapter 8

  2. Selection of steel for gears Wear resistance

  3. 1410 1150 910 725 0.8 0.02 Fraction of cementitite by LEVER rule: More carbon More cementite More wear resistance

  4. Cementite is hard as well as brittle Hardness or strength is desirable. But brittleness is not. Silica Glass is also hard and brittle

  5. Selection of steel for gears High Carbon Steel: Good wear resistance but Btittle Low Carbon steel: Good ductility but poor wear resistance

  6. Wear resistance is required only at the surface High C steel on the surface Mild steel inside Q: How do you achieve this? Ans: By case carburization

  7. Case carburization Pack a mild steel gear in carbon and heat at a high temperature in the austenite phase field for some time. Carbon will enter into the mild steel to give a high-carbon wear resistant surface layer called case. How do carbon enter into solid steel? At what temperature and how long should we do the carburization?

  8. The process why which Carbon enters into solid steel during Case carburization is an example of DIFFUSION Diffusion is relative movement of atoms inside a solid We can find appropriate time and tempearture for case carburization by solution of Fick’s second law

  9. Si substrate Si substrate How do we create an n-p junction in silicon chip? Ans: by DIFFUSION Deposit p type element Deposit n type element Heat Si substrate

  10. Different kinds of flows in material Heat: flow of thermal energy Electric current: flow of electric charge Diffusion: flow of matter

  11. Heat: flow of thermal energy q: heat flux (J m-2 s-1) Gradient? Temperature gradient Fourier’s law of heat conduction (1811) Joseph Fourier (1768-1830) Thermal conductivity

  12. Electric current: flow of charge j : charge flux (C m-2 s-1), current density Gradient? Electric potential gradient, electric field E Ohm’s law of electrical conduction (1827) Georg Simon Ohm (1787-1854) electrical conductivity

  13. Diffusion: flow of mass j : mass flux (kg m-2 s-1, moles m-2 s-1) Gradient? concentration gradient, kg m-4 Fick’s first law of diffusion1855 1829-1901 D: Diffusivity, m2 s-1

  14. Temperature dependence of Diffusivity D0 = preexponential factor Empirical constants Q = activation energy

  15. ln D -28 -30 -32 -34 1/T 0.00305 0.00315 0.00325 Self-Diffusion in Amorphous Se (Problem 8.3) T (ºC) D (m2s-1) 35 7.7 x 10-1640 2.4 x 10-1546 3.2 x 10-1456 3.2 x 10-13 D0 = 2 x 1027 m2 s-1 Q = 250 kJ mol-1

  16. c c+c j + j j x x+x Mass in at x: min = At j Mass out at x+ x: mout = At (j + j) Mass accumulation between x and x+  x m = min-mout = At ( j – j - j ) = -At j

  17. c c+c j + j j x x+x m = -At j Change in concentration in a volume V = Ax and time interval  t : Average rate of change of concentration between x and x +  x in time interval t:

  18. Instantaneous change in concentration at a time t, at a point x: Fick’s 2nd Law

  19. Using Fick’s First Law If D is independent of x Fick’s 2nd law

  20. Solution to Fick’s 2nd law: Solution depends on the boundary condition. A and B : constants depending on the boundary conditions erf (z) : Gaussian error function

  21. exp (-2) 1 0.8 0.6 0.4 0.2  -3 -2 -1 1 2 3 0 The Gaussian Error Function Hatched area  (2/) = erf (z) erf (0) = 0, erf (-z) = - erf (z), erf (+ ) = +1, erf (- ) = -1 z

  22. Mistakes in the textbook in the boxed values erf (z) z TABLE 8.1

  23. Carburisation of steel c (wt% C) cs Surface concentration Concentration profile after carburization for time t at a temperature T Carburising atmosphere Steel initial concentration c0 x Distance in steel from surface Boundary conditions: 1. c=c0 at x>0 , t=0 2. c=cs at x=0 , t>0

  24. Carburisation of steel (contd.) Boundary conditions: 1. c(x,t) = cs at x = 0, t > 0 2. c(x,t) = c0 at x > 0, t = 0 B.C. 1  cs = A – B erf(0) = A A = cs B = cs – c0 B.C. 2  c0 = A – B erf(+) = A-B

  25. Case carburization of steel Problem 8.4 Initial concentration c0 = 0.2 wt% C Surface concentration cs = 1.4 wt % C Temperature = 900 ºC = 1173 K Desired concentration c = 1.0 wt% C at x = 0.2 mm At 900 ºC the equilibrium phase of steel is austenite () Diffusivity data for C in austenite: D0 = 0.7 x 10-4m2s-1 Q = 157 kJ mol-1 = 7.13688 x 10-12 m2s-1

  26. wt% C 1.4 10000 s 1.2 1.0 0.8 1000 s 0.6 0.4 100 s 0.2 0.1 0.2 0.3 0.4 0.5 Distance in steel from surface, mm Carburization of steels

  27. z erf(z) 0.30 0.3286 0.35 0.3794 0.305 0.3333

  28. t = 15062 s = 4 h 11 min Ans This is reasonable. If not, change D by changing T

  29. Atomic Mechanism of Diffusion How does C enter into solid steel?  is INTERSTITIAL solid solution of C in FCC Fe C occupies octahedral voids in FCC Fe Maximum solubility of C in austenite () is 2.14 wt% at 1150 ºC. Thus 9 out of 91 OH voids are occupied. 90 % of OH voids are empty

  30. C atoms can jump from one interstitial site to another vacant interstitial site. This is interstitial diffusion. For OH voids the void size is 0.414 R but the window through which C atoms can jump outside is only 0.155 R. Thus to jump out of an interstitial OH site the C atoms will have to displace neighbouring Fe atoms. This will increase the energy of the system

  31. Potential energy Hm OH void OH void Only a fraction of these attempts A carbon atom can jump to a neighbouring site if it has sufficient energy Hm. It can gain this energy only through random thermal vibration. If thermal vibration frequency is  then it makes  attempts per second. will have an energy Hm and will be successful.

  32. 1 2 No. of successful jumps per second from plane 1 to plane 2, n1->2 = A  c1  exp(-Hm/RT) p No. of successful jumps per second from plane 2 to plane 1, n2->1 = A  c2  exp(-Hm/RT) p  C1 C2 Net jumps per second from plane 1 to plane 2 n = n1->2 - n2->1 = A  (c1-c2)  exp(-Hm/RT) p Flux:

  33. Fick’s 1st Law An atom making a successful jump may remain in plane 1, go to the back plane or jump to forward plane 2. Thus only a fraction p of successful jumps are from plane 1 to plane 2. This factor has been omitted in the textbook.

  34. Initially After some time Wt % Ni Ni 100% 100% 0 0 x Adapted from Figs. 5.1 and 5.2, Callister 6e.

  35. Mechanism of substitutional diffusion How is diffusion taking place in a substitutional solid solution ? Vacancy mechanism of substitutional diffusion

  36. However, only a very small fraction of the order of 10-4 to 10-30 are vacant. A jump can only be successful if the neighbouring site is vacant. Probability of finding a vacant site = fraction of vacant site

  37. Sbstitutional diffusion is usually slower than interstitial diffusion due to difficulty of finding a vacant site.

  38. Other Diffusion Paths Lattice diffusion Grain boundary diffusion Surface diffusion Experimentally Qsurface < Qgrain boundary < Qlattice

More Related