Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals f Diffusin Diffusin: Transprt in a slid, liquid, r gas driven by a cncentratin gradient (r, in the case f mass transprt, a chemical ptential gradient). Diffusin des nt necessarily apply t transprt f mass. Cnsider, fr example, thermal diffusin, r transprt f phnns heat quanta. Example: Tw chambers, each cntaining a different gas, separated by a remvable barrier; when the barrier is pulled away, interdiffusin ccurs N 2 O 2 Initially, a plt f cncentratin f each species wuld lk like the fllwing: 1
After a time, the cncentratin prfiles wuld lk like this: And after a lng time, the cncentratin f the tw gasses wuld be everywhere the same: We see essentially the same prcess in slids, as seen in the case f Cu and Ni: (nte, diffusin in the gaseus state des nt require an activatin energy as it des in slids) 2
The prcess f substitutinal diffusin requires the presence f vacancies (Vacancies give the atms a place t mve) 3
Diffusin tends t equalize cncentratin gradients. Nte the presence f vacancies in the material 4
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Recall, the fractin f vacancies is temperature dependent: Nv N = e Qv kt diffusin is a temperature-dependent prcess In the example f a Cu-Ni diffusin cuple, diffusin f Ni int Cu ccurs at a faster rate than Cu in Ni. Why is this? T m (Ni) = 1451 C, T m (Cu) = 1083 C larger fractin f vacancies in Cu at a given temperature than in Ni. Mathematically, we wuld write N N v Cu N N v Ni (We say that Cu is at a higher hmlgus temperature than Ni) Diffusin tends t equalize cmpsitinal gradients (if all ther factrs are equal - smetimes kinetics d nt favr diffusinal equilibratin. That is, sme diffusinal prcesses ccur s slwly as t be imperceptible.) Factrs that affect slid state diffusin: Diffusin ccurs at a higher rate at higher temperatures atms have mre energy t vercme barrier t diffusin greater prbability f activatin ver the energy barriers with smaller atms smaller atms can squeeze in between hst atms mre easily in lwer melting pint hst material lwer T m weaker bnds (easier t push apart) in lwer packing density hst material easier t migrate with fewer bnds t expand in grain bundaries mre disrdered than bulk material (lwer bnd density) 6
Steady state diffusin is described in terms f a flux and a cncentratin gradient: (steady state means the cncentratin gradient des nt change with time, r that the flux int a unit area is equal t the flux leaving the area n accumulatin r lss) Flux = number (r mass) f atms passing thrugh an area per unit time Units: atms cm 2 sec r grams cm 2 sec ( culd als be mles, Kg; the main thing is t just be cnsistent) 7
Flux is usually dented by the letter J flux cncentratin gradient C J x -r Fick s first law in wrds and mathematically dc J = D dx where the cnstant f prprtinality (D) is called the diffusin cefficient units f the diffusin cefficient: length 2 /time (cm 2 /s r m 2 /s) the diffusin cefficient is a thermally activated quantity: D = D e Qd RT Diffusin cefficient is a measure f the mbility f atms D = D e -r- qd kt where Q d and q d are activatin energies fr diffusin per mle per atm 8
Since D increases expnentially with temperature, diffusin rates increase with temperature: 9
Typical values fr preexpnential (D ) and activatin energy: (after Kittel, Slid State Physics 5 th ed.) nte: D=D exp{-q/k B T} Hst crystal D q d (ev/atm) Atm (cm 2 /s) Cu Cu 0.20 2.04 Cu Zn 0.34 1.98 Ag Ag 0.40 1.91 Ag Au 0.26 1.98 Ag Cu 1.2 2.00 Ag Pb 0.22 1.65 U U 0.002 1.20 Si Al 8.0 3.47 Si Ga 3.6 3.51 Si In 16.0 3.90 Diffusin cefficient (temperature) 2 primary mechanisms that affect flux: Slpe f cncentratin gradient D and Q d (r q d ) can be fund by pltting ln(d) vs. T 1 Then, Slpe = -Q/R (r q/k B ) Intercept = ln(d ) : Diffusin cefficient (m^/2) Reciprcal temperature (1000/K) 10
We usually dn t have steady state; mre ften, the cncentratin versus psitin curve changes with time (nn-steady state) In this case, we must use anther relatinship Fick s secnd law: (we are interested in the time rate f change f the cncentratin ) (# atms) ( J J ) 1 Aδt C = =, simplifying, we have vlume Aδx δ 2 δc δt = J x dc d D 2 dc dj dx d C - r - = = D, assuming D is independent f psitin! 2 dt dx dx dx - Fick s 2 nd 2 dc d C law is: = D 2 dt dx Which describes diffusin in cases where the cncentratin prfile changes with time Think f as the curvature f the cncentratin prfile 11
Example: Cnsider diffusin f C int γ-fe (this is ne example f carburizatin) In carburizatin, we diffuse C atms int the surface f lw C steels, heated t ~ 1000 C (FCC structure) Carbn cncentratin (wt. %) Surface cncentratin (held cnstant) Prfile at time t1 Prfile at (later) time t2 Distance int the plate (The prfile abve is cnsistent with an initial carbn cncentratin f ~ 0. Steels, by definitin, always have a nn-zer carbn cntent, w. Slutin t Fick s 2 nd law: Emplying apprpriate bundary cnditins, Fick s 2 nd law can be slved, with the slutin written in a useful frm fr metallurgical applicatins: C C x s C C = 1 erf 2 x Dt where C x = carbn cncentratin at any pint x in the steel during diffusin, C = initial (unifrm) carbn cncentratin in the steel (at t = 0), C s = surface cncentratin f carbn during diffusin (we can cntrl this) erf{} = errr functin (a mathematical expressin: -1 < erf(x) < 1 ) While nt strictly true, fr this class we will assume erf{x} x. 12
(The assumptin that erf(x) x isn t t bad fr x < 0.7. In situatins where x > 0.7 r where erf(x) > 0.65, yu shuld really use the crrect value. The fllwing table may be f help. Remember t interplate t btain values between thse listed.) Then, the (smewhat versimplified) slutin becmes: x + s 2 Dt C(x) C ( C C ) 1 (x in cm, t in sec, D in cm^2/s, C1, C, Cx all in wt. %) Next, sme example prblems 13
I. Callister: 5.6 The purificatin f H 2 (gas) by diffusin thrugh a Pd sheet was discussed in Sectin 5.3. Cmpute the number f kilgrams f hydrgen that pass per hur thrugh a 5- mm thick sheet f Pd having an area f 0.20 m 2 at 500 C. Assume a diffusin cefficient f 1.0x10-8 m 2 /s, that the cncentratins at the high- and lw-pressure sides f the plate are 2.4 and 0.6 kg f H 2 per m 3 f Pd, and that steady state cnditins have been attained. This prblem calls fr the mass f hydrgen per hur that diffuses thrugh a Pd sheet. It first becmes necessary t emply bth Equatins (5.1a) and (5.3). Cmbining these expressins and slving fr the mass yields M = JAt = - DAt DC Dx = - (1.0 x 10-8 m 2 /s)(0.2 m 2 )(3600 s/h) 0.6-2.4 kg/m 3 5 x 10-3 m Recall, flux is mass per unit time per unit area. Thus, multiplying J by area and time will give ttal mass. = 2.6 x 10-3 kg/h II. Callister 5.12 (decarburizatin) An FCC Fe-C ally initially cntaining 0.35 wt. % C is expsed t an xygen-rich (and carbn-free) atmsphere at 1400 K (1127 C). Under these cnditins the carbn in the ally diffuses tward the surface and reacts with the xygen in the atmsphere; that is, the carbn cncentratin at the surface is maintained essentially at 0 wt. % C. (This prcess f carbn depletin is termed decarburizatin.) At what psitin will the carbn cncentratin be 0.15 wt. % after a 10-hur treatment. The value f D at 1400 K is 6.9x10-11 m 2 /s. 14
This prblem asks that we determine the psitin at which the carbn cncentratin is 0.15 wt% after a 10-h heat treatment at 1400 K when C = 0.35 wt% C. Frm Equatin (5.5) C x - C 0.15-0.35 C s - C = 0-0.35 = 0.5714 = 1 - erf x 2 Dt Thus, erf x 2 Dt = 0.4286 Since we are assuming erf(z) z, we will simply let z = 0.4: Which means that And, finally x 2 Dt = 0.4 x = 2(0.4) Dt = (0.8004) (6.9 x 10-11 m 2 /s)(3.6 x 10 4 s) = 1.3 x 10-3 m = 1.3 mm III. Callister 5. 26. At apprximately what temperature wuld a specimen f γ-fe have t be carburized fr 2 hurs t prduce the same diffusin result as at 900 C fr 15- hurs? T slve this prblem it is necessary t emply Equatin (5.7) (Dt = cnstant) which takes n the frm (Dt) 900 = (Dt) T (the hint is that the prblem stated the same diffusin result, meaning that we want the same specific carbn cncentratin in bth cases) At 900 C, and using the data frm Table 5.2 15
D 900 = (2.3 x 10-5 m 2 /s) exp 148000 J/ml - (8.31 J/ml-K)(900 + 273 K) = 5.9 x 10-12 m 2 /s Thus, (5.9 x 10-12 m 2 /s)(15 h) = D T (2 h) And D T = 4.43 x 10-11 m 2 /s Slving fr T frm Equatin (5.9a) T = - Q d R(ln D T - ln D ) 148000 J/ml = - (8.31 J/ml-K)[ ln (4.43 x 10-11 ) - ln (2.3 x 10-5 )] = 1353 K = 1080 C Next tpic: mechanical prperties f materials 16