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Chapter 6 Diffusion in Solids

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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 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. 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 self-diffusioninter-diffusion @ impurity diffusion vacancy @ substitutional interstitialsteady state diffusion non-steady state diffusion case hardeningdoping 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 =+ diffusing species solid diffusing species heat Diffusion process…

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Diffusion in solids type (classification) self-diffusion Mechanism (method) A B C D After some time Initially After some time Initially interstitial 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). 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). - atoms migrate itself in solid. -- change its position. - diffusion occurs in 2 different solids. - atoms migrate between 2 solids. Example: Cu-Ni alloy 1. Join 2 metals (as diffusion couple). 2. Heat (below T m ) & cool. 3. After heat-treated, the couple: -- contains alloyed diffusion region. -- occurs diffusion process. inter-diffusion Interstitial diffusion more rapid than vacancy diffusion

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Diffusion in solids state (condition) Steady state diffusion How do we quantify the amount or rate of diffusion? Measured empirically: 1.make thin film (membrane) of known surface area. 2.impose concentration gradient. 3.measure how fast atoms or molecules diffuse through the membrane. J = moles (or mass) diffusing surface area x time mol cm 2 s kg m 2 s unit: @ Rate of diffusion @ flux, J M t J slope - rate of diffusion independent of time (@ does not change with time). - concentration of diffusing species depends with position. C1C1 C2C2 x C1C1 C2C2 x1x1 x2x2 concentration gradient rate of diffusion J dC dx - apply Fick’s 1 st law J dC dx = -D D diffusion coefficient C surface concentration x position 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? Given: D in butyl rubber = 110 x10 -8 cm 2 /s surface concentrations: C 2 = 0.02 g/cm 3 C 1 = 0.44 g/cm 3 glove C1C1 C2C2 skin paint remover x1x1 x2x2 Answer: assuming linear conc. gradient J = 1.16 x 10 -5 g/cm 2 s J dC dx = -D C 2 – C 1 x 2 – x 1 = -D (0.02 g/cm 3 – 0.44 g/cm 3 ) (0.04 cm – 0) = -(110x10 -8 cm 2 /s) J

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Diffusion in solids state (condition) Non-steady state diffusion How do we quantify the amount or rate of diffusion? Measured empirically: 1.make thin film (membrane) of known surface area. 2.impose concentration gradient. 3.measure how fast atoms or molecules diffuse through the membrane. J = moles (or mass) diffusing surface area x time mol cm 2 s kg m 2 s unit: @ Rate of diffusion @ flux, J M t J slope - 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 2 nd law Copper diffuses into a bar of aluminum. Example: before diffusion after some time during diffusion Al bar CoCo CSCS C (x,t) - boundary conditions (B. C) at t = 0, C = C o for 0 x at t > 0, C = C S for x = 0 (constant surface conc.) C = C (x,t) for x = x t = x 2 C = C o for x = C o = pre-existing concentration C S = surface concentration C (x,t) = concentration at depth x after time t where; - B. C are related to equation: x1x1 x2x2 erf (z) = error function Note: erf(z) values are given in Table 6.1 Position, x CoCo C (x,t) CsCs time = t 2 time= t 1 time = t o x

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Diffusion in solids state (condition) Non-steady state diffusion Table 6.1: Tabulation of Error Function Values z erf (z)z z 000.550.56331.30.9340 0.0250.02820.600.60391.40.9523 0.050.05640.650.64201.50.9661 0.100.11250.700.67781.60.9763 0.150.16800.750.71121.70.9838 0.200.22270.800.74211.80.9891 0.250.27630.850.77071.90.9928 0.300.32860.900.79702.00.9953 0.350.37940.950.82092.20.9981 0.400.42841.00.84272.40.9993 0.450.47551.10.88022.60.9998 0.500.52051.20.91032.80.9999

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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 950 o C. 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 m 2 /s. Answer: x = 0.5 mm = 5 x 10 -4 m C (x,t) = 0.80 wt% C C s = 1.20 wt% C C o = 0.25 wt% C D = 1.6 x 10 -11 m 2 /s Given; erf (z) = 0.4210 zerf (z) 0.35 z 0.40 0.3794 0.4210 0.4284 C (x,t) – C o = 1 – erf (z) C s – C o 0.80 – 0.25 = 1 – erf (z) 1.20 – 0.25 z – 0.35 = 0.4210 – 0.3794 0.40 – 0.35 0.4284 – 0.3794 z = 0.392 use table 6.1 to find z value An interpolation is necessary as follows x = z 2 (Dt) 1/2 Therefore, 5 x 10 -4 m = 0.392 2 [(1.6 x 10 -11 m 2 /s)(t)] 1/2 t = 25400 s = 7.1 h

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- diffusion coefficient, D increases with increasing T. - apply the formula: = pre-exponential [m 2 /s] = diffusion coefficient [m 2 /s] = activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K] = absolute temperature [K] D DoDo QdQd R T D DoDo exp QdQd RT Diffusion in solids Factors that influence diffusion Temperature D interstitial >> D substitutional C in -Fe C in -Fe Al in Al Fe in -Fe Fe in -Fe 1000 K/T D (m 2 /s) C in a-Fe C in g-Fe Al in Al Fe in a-Fe Fe in g-Fe 0.51.01.5 10 -20 10 -14 10 -8 T( C) 15001000 600 300 D vs T for several metals… Diffusing species - rearrange; ln D = ln D o – Q d R (T) Note: Diffusion data values are given in Table 6.2 Smaller diffusing species, diffusion faster. The above formula can be used for steady state & non- steady state diffusion

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Diffusion in solids Factors that influence diffusion TemperatureDiffusing species Diffusing Species Host Metal D o (m 2 /s) Activation Energy Q d Calculated Values kJ/moleV/atomT ( o C)D (m 2 /s) Fe α-Fe (BCC) 2.8 x 10 -4 2512.605003.0 x 10 -21 9001.8 x 10 -15 Fe -Fe (FCC) 5.0 x 10 -5 2842.949001.1 x 10 -17 11007.8 x 10 -16 Cα-Fe6.2 x 10 -7 800.835002.4 x 10 -12 9001.7 x 10 -10 C -Fe 2.3 x 10 -5 1481.539005.9 x 10 -12 11005.3 x 10 -11 Cu 7.8 x 10 -5 2112.195004.2 x 10 -19 ZnCu2.4 x 10 -5 1891.965004.0 x 10 -18 Al 2.3 x 10 -4 1441.495004.2 x 10 -14 CuAl6.5 x 10 -5 1361.415004.1 x 10 -14 MgAl1.2 x 10 -4 1311.355001.9 x 10 -13 CuNi2.7 x 10 -5 2562.655001.3 x 10 -22 Table 6.2: Tabulation of diffusion data

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Example 1: At 300ºC the diffusion coefficient and activation energy for Cu in Si are 7.8 x 10 -11 m 2 /s and 41.5 kJ/mol. What is the diffusion coefficient at 350ºC? transform data D Temp = T ln D 1/T D DoDo exp QdQd RT Diffusion in solids Factors that influence diffusion TemperatureDiffusing species Answer: T 1 = 273 + 300 = 573 K T 2 = 273 + 350 = 623 K D 2 = 15.7 x 10 -11 m 2 /s

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Diffusion in solids 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: Factors that influence diffusion TemperatureDiffusing species t = 49.5 h x = 4 x 10 -3 m C (x,t) = 0.35 wt% C s = 1.0 wt% C o = 0.20 wt% Given; 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 zerf(z) 0.90 0.7970 z 0.8125 0.95 0.8209 z 0.93 Now solve for D from Table 6.2, for diffusion of Carbon in FCC Fe D o = 2.3 x 10 -5 m 2 /s Q d = 148,000 J/mol T = 1300 K = 1027°C

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Diffusion in solids 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 (t b ), i.e., how long could the gloves be used before methylene chloride reaches the hand? Answer: Factors that influence diffusion TemperatureDiffusing species Given; D in butyl rubber = 110 x10 -8 cm 2 /s assuming linear conc. gradient glove C1C1 C2C2 skin paint remover x1x1 x2x2 Breakthrough time = t b Time required for breakthrough ca. 4 min

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Diffusion in solids Industrial application Case hardening Doping silicon wafer - 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 (927 o C) in an atmosphere of hydrocarbon gas (CH 4 ). - carbon diffuses into iron (steel) surface and fills interstitial space to make it harder. Low carbon steel part Diffusing carbon atoms - 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 1100 o C in a quartz tube furnace. - the concentration of impurity at any point depends on depth and time of exposure. 1. Deposit P rich layers on surface. silicon 2. heat Doping process… 3. Result: Doped semiconductor regions.

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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 Summary

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End of Chapter 6

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