PY2N20 Material Properties and Phase Diagrams

Transcription

1 PYN0 Material Properties and Phase iagrams Lecture 7 P. Stamenov, Ph School of Physics, TC PYN0-7

2 iffusion iffusion Mass transport by atomic motion Mechanisms Gases & Liquids random (Brownian) motion Solids vacancy diffusion or interstitial diffusion gasses > liquids > solids

3 iffusion Interdiffusion: In an alloy, atoms tend to migrate from regions of high concentration to regions of low concentration. Initially After some time

4 iffusion Self-diffusion: In an elemental solid, atoms also migrate. Label some atoms C A B After some time C A B Often self-diffusion can be best characterised by radioisotope-marking, since the atoms of different isotopes are virtually identical chemically. Radiation profiles are readily traceable in macroscopic samples.

5 iffusion Mechanisms Vacancy iffusion: atoms exchange with vacancies applies to substitutional impurities atoms rate depends on: - number of vacancies - activation energy to exchange. increasing time

6 iffusion Mechanisms Interstitial diffusion smaller atoms can diffuse between atoms. More rapid than vacancy diffusion! egree of interstitialcy obvious definition

7 Processing Using iffusion Case Hardening: - iffuse carbon atoms into the host iron atoms at the surface. - Example of interstitial diffusion is a case hardened gear. Result: The presence of C atoms makes iron (steel) harder.

8 Carburising of steel Note: istances in thousands of inch. Times in hours.

9 Fe 3 C (cementite) Carburising of steel 600 d 400 T( C) L 00 g (austenite) g +L 48 C L+Fe 3 C 000 g +Fe 3 C 800 a C = T eutectoid a +Fe 3 C (Fe).4% C o, wt% C

10 Processing Using iffusion oping silicon with phosphorus for n-type regions: Process: 0.5 mm. eposit P rich layers on surface.. Heat it. silicon 3. Result: oped semiconductor regions. magnified image of a Si chip light regions: Si atoms silicon light regions: Al atoms

11 iffusion How do we quantify the amount or rate of diffusion? J Flux moles(or mass)diffusing mol surfacearea time cm s m s Measured empirically Make thin film (membrane) of known surface area Impose concentration gradient Measure how fast atoms or molecules diffuse through the membrane or kg J M At A dm dt M = mass diffused time J slope

12 Steady-State iffusion Rate of diffusion independent of time Flux proportional to concentration gradient = dc dx C C Fick s first law of diffusion C C J dc dx if x linear dc dx x x C x C x C x diffusion coefficient [] = m /s

13 Example: Chemical Protective Clothing (CPC) Methylene chloride is a common ingredient of 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? ata: diffusion coefficient in butyl rubber: = 0 x0-8 cm /s surface concentrations: C = 0.44 g/cm 3 C = 0.0 g/cm 3

14 Example (cont). Solution assuming linear conc. gradient C paint remover glove x x t b 6 C skin J - ata: dc dx C x C x = 0 x 0-8 cm /s C = 0.44 g/cm 3 C = 0.0 g/cm 3 x x = 0.04 cm J (0 x 0-8 cm (0.0 g/cm 0.44 g/cm /s) (0.04 cm) 3 3 ).6 x 0-5 g cm s

15 iffusion and Temperature iffusion coefficient increases with increasing T. o exp Q d R T o Q d R T = diffusion coefficient [m /s] = pre-exponential [m /s] = activation energy [J/mol or ev/atom] = gas constant [8.34 J/mol-K] = absolute temperature [K]

16 iffusion and Temperature has exponential dependence on T 0-8 T(C) (m /s) K/T interstitial >> substitutional C in a-fe C in g-fe Al in Al Fe in a-fe Fe in g-fe transform data ln Temp = T /T

17 Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are (300ºC) = 7.8 x 0 - m /s Q d = 4.5 kj/mol What is the diffusion coefficient at 350ºC? ln ln ln T T R Q d T 0 0 ln ln and ln ln T R Q R Q d d Example

18 Example (cont.) 573K 63K 8.34 J/mol-K 4,500 J/mol /s) exp m (7.8 x 0 exp T T R Q d = 5.7 x 0 - m /s

19 Non-steady State iffusion The concentration of diffusing species is a function of both time and position C = C(x,t) In this case Fick s Second Law is used Fick s Second Law C t C x

20 Non-steady State iffusion Copper diffuses into a bar of aluminum. Surface conc., C s of Cu atoms bar pre-existing concentration, C o of copper atoms B.C. at t = 0, C = C o for 0 x at t > 0, C = C S for x = 0 (const. surf. conc.) C = C o for x =

21 Solution: C x,t C C C s o o erf x t C(x,t) = Conc. at point x at time t erf (z) = Gaussian error function C S C(x,t) z e y 0 dy C o ( x) 6 Characteristic diffusion time necessary to penetrate an infinite slab of thickness Δx, for about 0.5% concentration threshold.

22 Non-steady State iffusion Sample Problem: An FCC iron-carbon alloy initially containing 0.0 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration constant at.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 equation C( x, t) C C C s o o erf x t

23 Example Solution (cont.): C( x,t ) C C C s o o erf x t t = 49.5 h C x = 0.35 wt% C o = 0.0 wt% x = 4 x 0-3 m C s =.0 wt% C( x, t) C C C s o o erf x t erf( z) erf(z) = 0.85

24 Example (cont.) We must now determine from table the value of z for which the error function is z erf(z) z z z 0.93 Now solve for z x t x 4z t x 4z t (4 x 0 (4)(0.93) 3 m) (49.5 h) h 3600s.6 x 0 m /s

25 To solve for the temperature at which has above value, we use a rearranged form of o exp Q d R T T Qd R( ln ln) for diffusion of C in FCC Fe o =.3 x 0-5 m /s Q d = 48,000 J/mol o T (8.34 J/mol- K)(ln.3x0 48,000 J/mol 5 m /s ln.6x0 m /s) T = 300 K = 07 C