Lecture 12 - Ch 5 pt 2

Lecture 12 - Ch 5 pt 2 - ANNOUNCEMENTS#1 Reading Chapter 5...

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Unformatted text preview: ANNOUNCEMENTS #1 Reading: Chapter 5, all sections Recitation Problems: #s 5.3, 5.7, 5.13, 5.16, 5.22, 5.25 This topic will be included in Midterm I Midterm I: 8:00-8:50 am, Tuesday May 4, Main Auditorium Accommodations: LeBow 348; 8:00-9:15 (1.5X) or 8:00-9:40 am (2X) Be there or…;-) Help Session: Thursday, April 29, 4-6 pm, Hill Seminar Room, Room 240, LeBow Chapter 5 - 12 ANNOUNCEMENTS #2 Midterm Coverage: Material from Chapters 1, 2, 3, 4 & 5 Exam Format: 3 Questions (100 Pts.): o 1 short answer, multi-part…choose/circle the correct answer(s); write in a few words or numbers; show brief calculation… o 2 longer answer, 2-3 part…calculation + explanation…similar to the recitation problems Formula Sheet & Periodic Table provided Work smartly & efficiently…don’t write essays, show full working… Chapter 5 - 13 Diffusion - 2 Cases • Steady-State Diffusion: – Concentration profile IS NOT a function of time • Non-steady State (or Transient) Diffusion: – Concentration profile IS a function of time Chapter 5 - 14 Diffusion - Consider the Following S-S diffusion across thin plate Now let’s plot C vs. distance x Slope = concentration gradient: Conc. grad. = dC/dx f04_05_pg113 Chapter 5 - 15 Steady-State Diffusion Rate of diffusion is independent of time dC Flux J proportional to concentration gradient = dx Fick’s First Law of diffusion C1 C1 C2 x1 if linear dC dx x C2 x2 dC J= D dx D diffusion coefficient (proportionality constant) C C2 C1 = x x2 x1 Chapter 5 - 16 A Few Words About Units SINCE: dC J= D dx If the units of Flux J are (something - e.g. mol or kg)/m2s… Then the units of concentration C HAVE to be (something)/m3 and vice versa! We do not care what that “something” is: Cats, albatrosses, people, atoms, moles, kg…all fine! The units of D are ALWAYS, and always will be, m2/s Chapter 5 - 17 Example: Chemical Protective Clothing (CPC) • Methylene chloride is a common ingredient in paint removers. Besides being an irritant, it also may be absorbed through the 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/s – surface concentrations: C1 = 0.44 g/cm3 C2 = 0.02 g/cm3 Chapter 5 - 18 Example (cont). • Solution – assuming a linear conc. gradient glove C1 dC J =-D dx 2 tb = paint remover 6D skin Data: D = 110 x 10-8 cm2/s C1 = 0.44 g/cm3 C2 = 0.02 g/cm3 x2 – x1 = 0.04 cm C2 x1 x2 J = (110 x 10 -8 C2 C1 D x2 x1 (0.02 g/cm3 0.44 g/cm3 ) g cm /s) = 1.16 x 10 -5 (0.04 cm) cm2s 2 Chapter 5 - 19 Diffusion and Temperature Implicit in the equation for D: Diffusion coefficient increases with increasing T D = Do exp - Qd RT D = diffusion coefficient (m2/s) D0 = pre-exponential term (m2/s) Qd = activation energy (J/mol or eV/atom) R = the gas constant (8.314 J/mol-K) T = absolute temperature (K) Chapter 5 - 20 Activation Energy for Diffusion Qd is the height of the energy barrier atoms need to scale to make a jump… and thus diffuse Qd X Chapter 5 - 21 Diffusion and Temperature D has an exponential dependence on T: - log D = log D0 Qd 1 2.3R T - plot log D vs. 1/T 10-8 300 600 1000 1500 - Straight line, slope = -Qd/2.3R; intercept = log D0 T(°C) Dinterstitial >> Dsubstitutional C in -Fe (BCC) Al in Al Fe in -Fe C in -Fe Fe in -Fe D (m2/s) 10-14 10-20 0.5 1.0 1.5 1000 K/T Adapted 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.) Chapter 5 - 22 Example: At 300 ºC the diffusion coefficient D and activation energy Qd for Cu in Si are: D(300ºC) = 7.8 x 10-11 m2/s Qd = 41.5 kJ/mol What is the diffusion coefficient at 350 ºC? transform data D ln D Temp = T lnD2 = lnD0 1/T Qd 1 R T2 lnD2 lnD1 = ln and lnD1 = lnD0 Qd 1 D2 = D1 R T2 1 T1 Qd 1 R T1 Chapter 5 - 23 Example (cont.) D2 = D1 exp Qd 1 R T2 1 T1 T1 = 273 + 300 = 573 K T2 = 273 + 350 = 623 K D2 = (7.8 x 10 11 41,500 J/mol 1 m /s) exp 8.314 J/mol - K 623 K 2 1 573 K D2 = 15.7 x 10-11 m2/s Chapter 5 - 24 Nonsteady-State Diffusion Most practical cases…nonsteady-state diffusion: Flux and conc. gradient changing with time Implies net accumulation or depletion of diffusing species Chapter 5 - 25 Non-Steady State Diffusion • Concentration C of diffusing species is a function of both time and position, i.e. C = C(x,t) • Fick’s first law no longer applicable.. • Now use Fick’s Second Law: Fick’s Second Law (assuming D independent of comp’n.) 2 C C =D 2 t x Chapter 5 - 26 A Few Words About…Differential Equations • The Unreasonable Effectiveness of Mathematics in the Natural Sciences (Eugene Wigner) “It is difficult to avoid the impression that a miracle confronts us here, quite comparable in its striking nature to the miracle that the human mind can string a thousand arguments together without getting itself into contradictions, or to the two miracles of the existence of laws of nature and of the human mind’s capacity to divine them.” http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences Chapter 5 - 27 Differential Equations • You CANNOT solve a D-E, e.g. conc. as fn. (position and time) without… • BOUNDARY and INITIAL conditions… Chapter 5 - 28 Nonsteady-State Diffusion • Copper (Cu) diffuses into a bar of aluminum (Al) Surface conc., CS of Cu atoms Al bar pre-existing conc., Co of copper atoms Cs Adapted from Fig. 5.5, Callister 7e. Boundary Cond.: at t = 0, C = Co for 0 x at t > 0, C = CS for x = 0 (const. surf. conc.) C = Co for x = Chapter 5 - 29 Solution: C (x , t ) Co x = 1 erf Cs Co 2 Dt C(x,t) = Conc. at point x at time t erf(z) = Gaussian error function: CS = 2 z 0 e y2 dy C(x,t) erf(z) values are given in Table C o 5.1 for various x/2 Dt values Chapter 5 - 30 Solution to Differential Equation given Initial and Boundary conditions shown in previous slides is: C (x , t ) Co x = 1 erf Cs Co 2 Dt This is a very common solution…many diffusion problems solutions look like this… which leads to: Chapter 5 - 31 Some Implications of the Solution • Suppose it is desired to have a specific conc. of solute C1 in an alloy, then LH side becomes: C1 C0 = constant Cs C0 • If so, then the RH side of the eqn. must also be a constant, thus… x 2 Dt or = constant x2 = constant Dt and… Chapter 5 - 32 VERY IMPORTANT SLIDE x Dt Excellent “back of the envelope” calc’n. relating characteristic diffusion distance x, to time t, and diffusion coefficient D Here is what the critics said about it: Amazing equation!; Very powerful!; Never seen anything like it! And to think such a complex phenomenon like non-steady-state diffusion could be captured so easily. A must use… Chapter 5 - 33 EMVISI 1) Temperature: Q. Does higher T lead to higher or lower D? A. Higher D… 2) Diffusion Mechanism: Interstitial diffusion is much faster than vacancy diffusion 3) Activation Energy: High Qd = ???? Ans. = low D Qd determined by diffusing species and host: Small diffusing species, open structure leads to low Qd or high Qd ? which leads to low D or high D? Large diffusing species, close-packed structure leads to: low Qd or high Qd ? which leads to a low D or high D? *Even More Very Important Slide Indeed! Chapter 5 - 34 Nonsteady-State Diffusion • Example 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 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 C ( x, t ) C o x = 1 erf Cs Co 2 Dt Chapter 5 - 35 Solution: C (x , t ) Co x = 1 erf Cs Co 2 Dt C(x,t) = Conc. at point x at time t erf(z) = Gaussian error function: CS = 2 z 0 e y2 dy C(x,t) erf(z) values are given in Table C o 5.1 for various x/2 Dt values Chapter 5 - 36 Solution (Cont.): C ( x , t ) Co x = 1 erf Cs Co 2 Dt Boundary Conditions: – t = 49.5 h x = 4 x 10-3 m Cs = 1.0 wt.% – Cx = 0.35 wt.% – Co = 0.20 wt.% C( x, t ) Co 0.35 0.20 x = 1 erf = = 1 erf ( z ) Cs Co 1.0 0.20 2 Dt erf(z) = 0.8125 Chapter 5 - 37 Solution (cont.): We now determine, from Table 5.1 (p.116), the value of z for which the error function is = 0.8125. A linear interpolation is necessary, as follows: z erf(z) 0.90 z 0.95 0.7970 0.8125 0.8209 Now solving for D x2 z 0.90 0.8125 0.7970 = 0.95 0.90 0.8209 0.7970 z = 0.93 x z= 2 Dt D= x2 4 z 2t ( 4 x 10 3 m)2 1h D= = = 2.6 x 10 11 m2 /s 4 z 2t ( 4)(0.93)2 ( 49.5 h) 3600 s Chapter 5 - 38 Solution (cont.): • Now, to solve for the temperature T at which D has the above value, we use a rearranged form of Equation (5.9a): Qd T= R(lnDo lnD ) from Table 5.2 (p.119), for the diffusion of C in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol T= 148,000 J/mol (8.314 J/mol - K)(ln 2.3 x10 5 m2 /s ln 2.6 x10 11 m2 /s) T = 1300 K = 1027 °C Chapter 5 - 39 ...
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