Diffusion – And Its Role In Material Property Control

1. Diffusion – And Its Role In Material Property Control R. Lindeke, Ph. D. Engr 2110

2. IN SOLIDS DIFFUSION IN LIQUIDS IN GASES Carburization Surface coating

3. Diffusion • Interdiffusion: In an alloy, atoms tend to migrate from regions of high conc. to regions of low conc. Initially After some time Adapted from Figs. 5.1 and 5.2, Callister 7e. This is called a DIFFUSION COUPLE a sketch of Cu – Ni here

4. Diffusion Mechanisms Vacancy Diffusion: • atoms exchange with vacancies • applies to the atoms of substitutional impurities • rate depends on: -- number of vacancies -- activation energy to exchange – a function of temp. and size effects. increasing elapsed time

5. Diffusion Mechanisms • Interstitial diffusion – smaller atoms can diffuse between atoms. More rapid than vacancy diffusion Adapted from Fig. 5.3 (b), Callister 7e.

6. Processing Using Diffusion • Case Hardening: • Diffuse carbon atoms • into the host iron atoms • at the surface. • Example of interstitial • diffusion to produce a surface (case) hardened gear. Adapted from chapter-opening photograph, Chapter 5, Callister 7e. (Courtesy of Surface Division, Midland-Ross.) The carbon atoms (interstitially) diffuse from a carbon rich atmosphere into the steel thru the surface. Result: The presence of C atoms makes the iron (steel) surface harder.

7. M = mass diffused Jslope time Diffusion • How do we quantify the amount or rate of diffusion? – we define a “mass flux” value • Flux is Commonly Measured empirically • Make thin film (membrane) of known surface area • Impose concentration gradient (high conc. On 1 side low on the other) • Measure how fast atoms or molecules diffuse through the membrane

8. C1 C1 C2 x1 x2 C2 x Simplest Case: Steady-State Diffusion The Rate of diffusion is independent of time •  Flux is proportional to concentration gradient = This model is captured as Fick’s first law of diffusion D diffusion coefficient which is a function of diffusing species and temperature For steady state diffusion concentration gradient = dC/dx is linear

9. F.F.L. Example: Chemical Protective Clothing (CPC) • Methylene chloride is a common ingredient in paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn. • If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the glove? • Data: • diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s • surface concentrations: C1= 0.44 g/cm3 C2= 0.02 g/cm3

10. Data: D = 110x10-8 cm2/s C1 = 0.44 g/cm3 C2 = 0.02 g/cm3 x2 – x1 = 0.04 cm Example (cont). • Solution – assuming linear conc. gradient glove C1 paint remover skin C2 x1 x2

11. What happens to a Worker? • If a person is in contact with the irritant and more than about 0.5 gm of the irritant is deposited on their skin they need to take a wash break • If 25 cm2 of glove is in the paint thinner can, How Long will it take before they must take a wash break?

12. Another Example: Chemical Protective Clothing (CPC) • If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand? • Data (from Table 22.5) • diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s

13. glove C1 paint remover skin C2 x1 x2 D = 110x10-8 cm2/s Example (cont). • Solution – assuming linear conc. gradient Equation 22.24 Time required for breakthrough ca. 4 min

14. æ ö ç ÷ = D Do exp è ø Qd D = diffusion coefficient [m2/s] - Do R T = pre-exponential [m2/s] Qd = activation energy [J/mol or eV/atom] R = gas constant [8.314 J/mol-K] T = absolute temperature [K] Diffusion and Temperature • Diffusion coefficient increases with increasing T

15. T(C) 1500 1000 600 300 10-8 D (m2/s) C in g-Fe C in a-Fe D >> D interstitial substitutional C in a-Fe Al in Al C in g-Fe Fe in a-Fe 10-14 Fe in g-Fe Fe in a-Fe Fe in g-Fe Al in Al 10-20 1000K/T 0.5 1.0 1.5 Diffusion and Temperature D has exponential dependence on T So Note: Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

17. transform data ln D D Temp = T 1/T Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are D(300ºC) = 7.8 x 10-11 m2/s Qd = 41.5 kJ/mol What is the diffusion coefficient at 350ºC?

18. T1 = 273 + 300 = 573K T2 = 273 + 350 = 623K D2 = 15.7 x 10-11 m2/s Example (cont.)

19. Non-steady State Diffusion • If the concentration of diffusing species is a function of both time and position that is C = C(x,t) • In this case Fick’s Second Law is used Fick’s Second Law Concentration (C) in terms of time and position can be obtained by solving above equation with knowledge of boundary conditions The solution depends on the specific case we are treating

20. One practically important solution is for a semi-infinite solid in which the surface concentration is held constant. Frequently source of the diffusing species is a gas phase, which is maintained at a constant pressure value. A bar of length l is considered to be semi-infinite when • The following assumptions are implied for a good solution: • Before diffusion, any of the diffusing solute atoms in the solid are uniformly distributed with concentration of C0. • The value of x (position in the solid) at the surface is zero and increases with distance into the solid. • The time is taken to be zero the instant before the diffusion process begins.

21. • Copper diffuses into a bar of aluminum. Surface conc., bar C C of Cu atoms s s pre-existing conc., Co of copper atoms Non-steady State Diffusion Adapted from Fig. 5.5, Callister 7e. Notice: the concentration decreases at increasing x (from surface) while it increases at a given x as time increases! Boundary Conditions: at t = 0, C = Co for 0  x   at t > 0, C = CS for x = 0 (const. surf. conc.) C = Co for x = 

22. Solution: C(x,t) = Conc. at point x at time t erf (z) = error function erf(z) values are given in Table 5.1 CS C(x,t) Co

23. Non-steady State Diffusion • Sample Problem: An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration (C0 ) constant at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out. • Solution: use Eqn. 5.5

24. Notice that the solution requires the use of the erf function which was developed to model conduction along a semi-infinite rod as we saw earlier

25.  erf(z) = 0.8125 Solution (cont.): • t = 49.5 h x = 4 x 10-3 m • Cx = 0.35 wt% Cs = 1.0 wt% • Co = 0.20 wt%

26. z erf(z) 0.90 0.7970 z 0.8125 0.95 0.8209 Now solve for D Solution (cont.): We must now determine from Table 5.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows Now By LINEAR Interpolation: z= 0.93

28. from Table 5.2, for diffusion of C in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol  T = 1300 K = 1027°C Solution (cont.): • To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a):

29. Following Up: • In industry one may wish to speed up this process • This can be accomplished by increasing • Temperature of the process • Surface concentration of the diffusing species • If we choose to increase the temperature, determine how long it will take to reach the same concentration at the same depth as in the previous study?

30. Diffusion time calculation: • X and concentration are equal therefore: • D*t = constant for non-steady state diffusion! • D1300 = 2.6x10-11m2/s (1027C)

31. Summary Diffusion FASTER for... • open crystal structures • materials w/secondary bonding • smaller diffusing atoms • lower density materials Diffusion SLOWER for... • close-packed structures • materials w/covalent bonding • larger diffusing atoms • higher density materials