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Chapter 6 Diffusion in Solids

Chapter 6 Diffusion in Solids. In general. Diffusion is a process by which: -- a matter is transported through another matter. OR -- atoms are moved through solids. Occurs at an elevated temperature , where atoms tend to move from high conc. region

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Chapter 6 Diffusion in Solids

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  1. Chapter 6 Diffusion in Solids

  2. In general • Diffusion is a process by which: • -- a matter is transported through • another matter. OR • -- atoms are moved through solids. • Occurs at an elevated temperature, where • atoms tend to move from high conc. region • to low conc. region. • . • Atoms diffuse in solids if: • 1. vacancies or other crystal defects are • present. • 2. there is enough activation energy. • - Atoms move into the vacancies present. • - Normally, at higher temperature; • -- more vacancies are created. • -- diffusion rate is higher. solid Diffusion process… diffusing species diffusing species heat solid Diffusion in solids Example of diffusion… 1. Movement of smoke particles in air: -- very fast. 2. Movement of dye in water: -- relatively slow. 3. Movement of atom in solid (solid state reactions): -- very restricted movement due to bonding. Diffusion in solids type (classification) mechanism (method) state (condition) industrial application inter-diffusion @ impurity diffusion steady state diffusion vacancy @ substitutional non-steady state diffusion interstitial self-diffusion case hardening doping silicon Example: If atom ‘A’ has sufficient activation energy, it moves into the vacancy. Activation energy of diffusion Activation energy to form a vacancy Activation energy to move a atom = +

  3. C C D A A D B B Diffusion in solids type (classification) - atoms migrate itself in solid. -- change its position. • diffusion occurs in 2 different solids. • atoms migrate between 2 solids. After some time Initially After some time Initially self-diffusion inter-diffusion • atoms move from one interstitial • site to another. • the diffusing atoms that move • must be much smaller than the • host atom. • i.e: carbon interstitially diffuses • into α-iron (BCC) @ -iron (FCC). Example: Cu-Ni alloy 1. Join 2 metals (as diffusion couple). 2. Heat (below Tm) & cool. 3. After heat-treated, the couple: -- contains alloyed diffusion region. -- occurs diffusion process. interstitial diffusion Interstitial diffusion more rapid than vacancy diffusion • - atoms exchange with vacancies. • applies to substitutional impurities • atoms. • - Rate of diffusion depends on: • -- number of vacancies (vacancy • concentration). • . -- activation energy to exchange • (frequency of jumping). Mechanism (method) vacancy diffusion

  4. rate of diffusion  concentration gradient C1 dC dx  J dC dx = -D J C1 C2 glove C1 x1 x2 paint remover skin C2 C2 C2 – C1 x2 – x1 dC dx = -D = -D J x1 x2 M (0.02 g/cm3 – 0.44 g/cm3) (0.04 cm – 0) = -(110x10-8 cm2/s) J Jslope x t Diffusion in solids state (condition) Steady state diffusion • rate of diffusion independent of time • (@ does not change with time). • concentration of diffusing species • depends with position. Example: Chemical Protective Clothing (CPC) Rate of diffusion @ flux, J 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? Given: D in butyl rubber = 110 x10-8 cm2/s surface concentrations: How do we quantify the amount or rate of diffusion? J = moles (or mass) diffusing surface area x time C1= 0.44 g/cm3 C2= 0.02 g/cm3 Answer: mol cm2s @ kg m2s unit: assuming linear conc. gradient • Measured empirically: • make thin film (membrane) of known surface area. • impose concentration gradient. • measure how fast atoms or molecules diffuse through the membrane. • apply Fick’s 1st law D diffusion coefficient C surface concentration x position J = 1.16 x 10-5 g/cm2s

  5. Cs time = t2 time= t1 C(x,t) time = to Co x Position, x M Jslope t Diffusion in solids state (condition) Non-steady state diffusion • occur for most practical diffusion • situations. • rate of diffusion & concentration • gradient vary with time & position. • concentration of diffusing species is • a function of both time and position. • C = C(x,t) • - apply Fick’s 2nd law - boundary conditions (B. C) Rate of diffusion @ flux, J How do we quantify the amount or rate of diffusion? J = moles (or mass) diffusing surface area x time at t = 0, C = Co for 0  x   at t > 0, C = CS for x = 0 (constant surface conc.) C = C(x,t) for x = xt = x2 C = Co for x =  Example: mol cm2s @ kg m2s unit: Copper diffuses into a bar of aluminum. before diffusion Al bar Co CS • Measured empirically: • make thin film (membrane) of known surface area. • impose concentration gradient. • measure how fast atoms or molecules diffuse through the membrane. - B. C are related to equation: during diffusion C(x,t) after some time x1 x2 where; Co = pre-existing concentration CS = surface concentration erf (z) = error function Note: erf(z) values are given in Table 6.1 C(x,t) = concentration at depth x after time t

  6. Diffusion in solids state (condition) Non-steady state diffusion Table 6.1: Tabulation of Error Function Values

  7. C(x,t) – Co = 1 – erf (z) Cs – Co 5 x 10-4 m = 0.392 2 [(1.6 x 10-11 m2/s)(t)]1/2 Diffusion in solids state (condition) Non-steady state diffusion Example: Consider one such alloy that initially has a uniform carbon concentration of 0.25 wt% and is to be treated at 950oC. If the concentration of carbon at the surface is suddenly brought to & maintained at 1.20 wt%, how long will it take to achieve a carbon content of 0.80 wt% at a position 0.5 mm below the surface? The D for carbon in iron at this temperature is 1.6 x 10-11 m2/s. Answer: Given; An interpolation is necessary as follows • x = 0.5 mm = 5 x 10-4 m • C(x,t) = 0.80 wt% C • Cs = 1.20 wt% C • Co = 0.25 wt% C • D = 1.6 x 10-11 m2/s z – 0.35 = 0.4210 – 0.3794 0.40 – 0.35 0.4284 – 0.3794 z = 0.392 Therefore, x = z 2 (Dt)1/2 0.80 – 0.25 = 1 – erf (z) t = 25400 s = 7.1 h 1.20 – 0.25  erf (z) = 0.4210 use table 6.1 to find z value

  8. 1500 1000 T(C) 600 300 10-8 D (m2/s) C in a-Fe C in g-Fe C in a-Fe Al in Al C in g-Fe Fe in a-Fe Qd D = diffusion coefficient [m2/s] 10-14 Fe in g-Fe - Fe in a-Fe Do = pre-exponential [m2/s] Fe in g-Fe R T Al in Al Qd = activation energy [J/mol or eV/atom] R = gas constant [8.314 J/mol-K] T = absolute temperature [K] ln D = ln Do – Qd 10-20 R (T) 1000 K/T 0.5 1.0 1.5 Diffusion in solids Factors that influence diffusion Temperature Diffusing species Smaller diffusing species, diffusion faster. • diffusion coefficient, D increases with increasing T. • apply the formula: D vs T for several metals… æ ö ç = D Do exp è ø - rearrange; D >> D interstitial substitutional The above formula can be used for steady state & non-steady state diffusion Note: Diffusion data values are given in Table 6.2

  9. Diffusion in solids Factors that influence diffusion Temperature Diffusing species Table 6.2: Tabulation of diffusion data

  10. T1 = 273 + 300 = 573 K transform data T2 = 273 + 350 = 623 K ln D D Qd - R T Temp = T 1/T D2 = 15.7 x 10-11 m2/s Diffusion in solids Factors that influence diffusion Temperature Diffusing species Example 1: At 300ºC the diffusion coefficient and activation energy for Cu in Si are 7.8 x 10-11 m2/s and 41.5 kJ/mol. What is the diffusion coefficient at 350ºC? Answer: æ ö ç = D Do exp è ø

  11. z erf(z) 0.90 0.7970 z 0.8125 0.95 0.8209 z= 0.93  from Table 6.2, for diffusion of Carbon in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol T = 1300 K = 1027°C Diffusion in solids Factors that influence diffusion Temperature Diffusing species Example 2: (for non-steady state diffusion application) 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 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. Answer: Given; • t = 49.5 h • x = 4 x 10-3 m • C(x,t) = 0.35 wt% • Cs = 1.0 wt% • Co = 0.20 wt% Now solve for D  erf(z) = 0.8125 We must now determine from Table 6.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows

  12. glove C1 paint remover skin C2 x1 x2 Diffusion in solids Factors that influence diffusion Temperature Diffusing species Example 3: 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 breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand? Given; D in butyl rubber = 110 x10-8 cm2/s Answer: Breakthrough time = tb assuming linear conc. gradient  Time required for breakthrough ca. 4 min

  13. silicon silicon Diffusion in solids • sliding and rotating parts needs to have hard • surfaces. • - these parts are usually machined with low • carbon steel as they are easy to machine. • - their surface is then hardened by carburizing. • steel parts are placed at elevated temperature • (927oC) in an atmosphere of hydrocarbon gas • (CH4). • carbon diffuses into iron (steel) surface and fills • interstitial space to make it harder. Case hardening Diffusing carbon atoms Low carbon steel part Industrial application • Doping silicon with phosphorus, P for n-type semiconductors. • impurities (P) are made to diffuse into silicon wafer to change • its electrical characteristics. • - used in integrated circuits. • silicon wafer is exposed to vapor of impurity at 1100oC in a • quartz tube furnace. • the concentration of impurity at any point depends on • depth and time of exposure. Doping silicon wafer Doping process… 3. Result: Doped semiconductor regions. 1. Deposit P rich layers on surface. 2. heat

  14. 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

  15. End of Chapter 6

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