3 This is called a DIFFUSION COUPLE a sketch of Cu – Ni here • Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.InitiallyAfter some timeAdapted 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 More rapid than vacancy diffusion Diffusion MechanismsInterstitial diffusion – smaller atoms can diffuse between atoms.this is essentially mechanism in amorphous materials such as polymer membranesMore rapid than vacancy diffusionAdapted from Fig. 5.3 (b), Callister 7e.
6 Processing Using Diffusion Case Hardening:Diffuse carbon atomsinto the host iron atomsat the surface.Example of interstitialdiffusion to produce a surface (case) hardened gear.Adapted from chapter-opening photograph, Chapter 5, Callister 7e. (Courtesy ofSurface 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 DiffusionHow do we quantify the amount or rate of diffusion? – we define a “mass flux” valueFlux is Commonly Measured empiricallyMake thin film (membrane) of known surface areaImpose concentration gradient (high conc. On 1 side low on the other)Measure how fast atoms or molecules diffuse through the membraneM =massdiffusedtimeJ slope
8 Simplest Case: Steady-State Diffusion The Rate of diffusion is independent of time Flux is proportional to concentration gradient =C1C2xx1x2This model is captured as Fick’s first law of diffusionD diffusion coefficient which is a function of diffusing species and temperatureFor 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 = 110 x10-8 cm2/ssurface concentrations:C1 = 0.44 g/cm3C2 = 0.02 g/cm3
10 Example (cont). Solution – assuming linear conc. gradient Data: glove paintremoverskinC2D = 110 x 10-8 cm2/sC2 = 0.02 g/cm3C1 = 0.44 g/cm3x2 – x1 = 0.04 cmData:x1x2
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 breakIf 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 = 110 x10-8 cm2/s
13 Example (cont). Solution – assuming linear conc. gradient gloveC1C2skinpaintremoverx1x2Equation 22.24D = 110 x 10-8 cm2/sTime required for breakthrough ca. 4 min
14 Diffusion and Temperature • Diffusion coefficient increases with increasing TD=Doexpæèçöø÷-QdRT= pre-exponential [m2/s]= diffusion coefficient [m2/s]= activation energy [J/mol or eV/atom]= gas constant [8.314 J/mol-K]= absolute temperature [K]
15 Diffusion and Temperature D has exponential dependence on T1000 K/TD (m2/s)C in a-FeC in g-FeAl in AlFe in a-FeFe in g-Fe0.51.01.510-2010-1410-8T(C)15001000600300So Note:Dinterstitial>> DsubstitutionalC in a-FeC in g-FeAl in AlFe in a-FeFe in g-FeAdapted 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 What is the diffusion coefficient at 350ºC? Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si areD(300ºC) = 7.8 x m2/sQd = 41.5 kJ/molWhat is the diffusion coefficient at 350ºC?transform dataDTemp = Tln D1/T
18 Example (cont.) T1 = 273 + 300 = 573 K T2 = 273 + 350 = 623 K D2 = 15.7 x m2/s
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 usedFick’s Second LawIn the non-steady state the concentration profile develops with time.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 whenThe 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 Non-steady State Diffusion • Copper diffuses into a bar of aluminum.pre-existing conc., Co of copper atomsSurface conc.,Cof Cu atomsbarsCsAdapted 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 5.1CSC(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 Solution (cont.): t = 49.5 h x = 4 x 10-3 m Cx = 0.35 wt% Cs = 1.0 wt% Co = 0.20 wt% erf(z) =
26 Solution (cont.):We must now determine from Table 5.1 the value of z for which the error function is An interpolation is necessary as followszerf(z)zNow By LINEAR Interpolation:z = 0.93Now solve for D
28 Solution (cont.): from Table 5.2, for diffusion of C in FCC Fe To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a):from Table 5.2, for diffusion of C in FCC FeDo = 2.3 x 10-5 m2/s Qd = 148,000 J/molT = 1300 K = 1027°C
29 Following Up: In industry one may wish to speed up this process This can be accomplished by increasingTemperature of the processSurface concentration of the diffusing speciesIf 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 (1027C)