1. Subtitle or main author’s name Materials of Engineering Lecture 8 - Diffusion Review & Thermal Properties Engineering 45 Carlos Casillas, PE 16-Jan-12 Licensed Chemical Engineer Spring 2012 CCasillas@ChabotCollege.edu Hayward, CA

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3. Other Types of Diffusion (Beside Mass or Atomic) Flux is general concept: e.g. charges, phonons,.. Charge Flux – e = electric chg. N = net # e- cross A N = # of phonons with avg. energy  Heat Flux – (by phonons) Defining thermal conductivity  (a material property) Solution: Fick’s 2nd Law Defining conductivity  (a material property) Solution: Fick’s 2nd Law Ohm’s Law Or w/ Thermal Diffusivity:

4. Diffusion- Steady and Non-Steady State Diffusion - Mass transport by atomic motion Mechanisms Gases & Liquids – random (Brownian) motion Solids – vacancy diffusion or interstitial diffusion

5. Substitution-diffusion: vacancies and interstitials k B T gives eV Number (or concentration*) of Vacancies at T  E is an activation energy for a particular process (in J/mol, cal/mol, eV/atom). applies to substitutional impurities atoms exchange with vacancies rate depends on (1) number of vacancies; (2) activation energy to exchange.

6. Example The total membrane surface area in the lungs (alveoli) may be on the order of 100 square meters and have a thickness of less than a millionth of a meter, so it is a very effective gas-exchange interface. Fick's law is commonly used to model transport processes in food processing, chemical protective clothing, biopolymers, synthetic polymer membranes (fluid sepaations), pharmaceuticals (controlled-release), porous soils & solids, catalysts, nuclear isotope enrichment, vapor & liquid thin-film coating & doping process, etc. Where can we use Fick’s Law? CO 2 in air has D~16 mm 2 /s, and, in water, D~ 0.0016 mm 2 /s

7. Gas Diffusion in Polymers

8. Polymer Membrane Structure

9. Flux: Directional Quantity Flux can be measured for: - vacancies - host (A) atoms - impurity (B) atoms Empirically determined: – Make thin membrane of known surface area – Impose concentration gradient – Measure how fast atoms or molecules diffuse through the membrane Modeling rate of diffusion: flux A = Area of flow J  slope diffused mass M time

10. Concentration Profile, C(x): [kg/m 3 ] Fick's First Law: D is a constant The steeper the concentration profile, the greater the flux Adapted from Fig. 6.2(c) Steady-state Diffusion: J ~ gradient of C

11. Steady State: concentration profile not changing with time. Apply Fick's First Law: Result: the slope, dC/dx, must be constant (i.e., slope doesn't vary with position) If J x ) left = J x ) right, then Steady-State Diffusion

12. Fick’s first law of diffusion C1C1 C2C2 x C1C1 C2C2 x1x1 x2x2 D  diffusion coefficient Rate of diffusion independent of timeJ ~

13. Example: Chemical Protective Clothing Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using, 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: –D in butyl rubber: D = 110 x10 -8 cm 2 /s –surface concentrations: –Diffusion distance: C 2 = 0.02 g/cm 3 C 1 = 0.44 g/cm 3 x 2 – x 1 = 0.04 cm glove C1C1 C2C2 skin paint remover x1x1 x2x2

14. Concentration profile, C(x), changes w/ time. To conserve matter: Fick's First Law: Governing Eqn. Fick’s 2 nd Law: Non-Steady-State Diffusion

15. Glass tube filled with water. At time t = 0, add some drops of ink to one end of the tube. Measure the diffusion distance, x, over some time. Compare the results with theory. Simple Diffusion expt

16. Concentration profile, C(x), changes w/ time. Non-Steady-State Diffusion: another look Rate of accumulation C(x) Using Fick’s Law: If D is constant: Fick’s 2nd Law

17. Non-Steady-State Diffusion: C = c(x,t) concentration of diffusing species is a function of both t and position x Copper diffuses into a bar of aluminum. C s Adapted from Fig. 6.5, Callister & Rethwisch 3e. B.C. at t = 0, C = C o for 0  x   at t > 0, C = C S for x = 0 (fixed surface conc.) C = C o for x = 

18. Cu diffuses into a bar of Al. Solution: "error function” Values found in Table 5.1 Non-Steady-State Diffusion CSCS CoCo C(x,t)C(x,t)

19. Example: Non-Steady-State Diffusion FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface C content 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, what temperature was treatment done?  erf(z) = 0.8125 zerf(z) 0.900.7970 z0.8125 0.950.8209 Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation. z  0.93 So, Solution Now solve for D

20. To solve for the temperature at which D has the above value, we use a rearranged form of Equation (6.9a); D=D 0 exp(-Q d /RT)  Solution (cont.): T = 1300 K = 1027°C From Table 6.2, for diffusion of C in FCC Fe D o = 2.3 x 10 -5 m 2 /s Q d = 148,000 J/mol

21. Copper diffuses into a bar of aluminum. 10 hours processed at 600 C gives desired C(x). How many hours needed to get the same C(x) at 500 C? Result: Dt should be held constant. Answer: Note D(T) are T dependent Values of D are provided. Key point 1: C(x,t 500C ) = C(x,t 600C ). Key point 2: Both cases have the same C o and C s. Example: Processing

22. 22 Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are D(300ºC) = 7.8 x 10 -11 m 2 /s Q d = 41.5 kJ/mol What is the diffusion coefficient at 350ºC? transform data D Temp = T ln D 1/T

23. 23 Example (cont.) T 1 = 273 + 300 = 573 K T 2 = 273 + 350 = 623 K D 2 = 15.7 x 10 -11 m 2 /s

24. VMSE: Student Companion Site Diffusion Computations & Data Plots 24

25. 25 Non-steady State Diffusion Sample 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

26. 26 Solution (cont.): –t = 49.5 h x = 4 x 10 -3 m –C x = 0.35 wt%C s = 1.0 wt% –C o = 0.20 wt%  erf(z) = 0.8125

27. 27 Solution (cont.): We must now determine from Table 5.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows zerf(z) 0.900.7970 z0.8125 0.950.8209 z  0.93 Now solve for D

28. 28 To solve for the temperature at which D has the above value, we use a rearranged form of Equation (5.9a); from Table 5.2, for diffusion of C in FCC Fe D o = 2.3 x 10 -5 m 2 /s Q d = 148,000 J/mol  Solution (cont.): T = 1300 K = 1027ºC

29. 29 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 breakthrough time (t b ), i.e., how long could the gloves be used before methylene chloride reaches the hand? Data –diffusion coefficient in butyl rubber: D = 110 x10 -8 cm 2 /s

30. 30 CPC Example (cont.) Time required for breakthrough ca. 4 min glove C1C1 C2C2 skin paint remover x1x1 x2x2 Solution – assuming linear conc. gradient Equation from online CPC Case Study 5 at the Student Companion Site for Callister & Rethwisch 8e (www.wiley.com/ college/callister) D = 110 x 10 -8 cm 2 /s Breakthrough time = t b

31. 31 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

32. Chapter 19 Thermal Properties Response of materials to the application of heat As a material absorbs energy in the form of heat, its temperature rises and its dimensions increase. This dimensional change is different for different materials. Also, the ability to transfer heat by conduction varies greatly between materials. How metals, ceramics, and polymers rank in hi-temp applications Important concepts: –heat capacity (C) –thermal expansion (a) –thermal conductivity (k) –thermal shock resistance (TSR)

33. Heat Capacity (Specific Heat) The ability of a material to absorb heat is specified by its heat capacity, C C p = heat capacity measured @ constant pressure C v = heat capacity measured @ constant volume Heat capacity is proportional to the amount of energy required to produce a unit of temperature change in a specified amount of material (Will measure in Lab3) The specific heat (lower case c) describes the heat capacity per unit mass (Joules/kg-K) heat capacity (J/mol-K) energy input (J/mol or J/kg) temperature change (K)

34. Materials Comparison of c p and C @ 298K Battery Insulation (Lab3 set up)

35. C v as Function of Temperature C v  The vibrational contribution to heat capacity varies as a function of temperature for a crystalline solid –Increases with Increasing T –Tends to a limiting value of 3R = 24.93 J/mol-k 3R=24.93

36. C v as Function of Temp cont For Many Crystalline Solids Where –A  Material Dependent CONSTANT –T D  Debye Temperature, K  Thermal Physics Energy is Stored in Lattice Vibration Waves Called Phonons –Analogous to Optical PHOTONS

37. 37 Atomic Vibrations Atomic vibrations are in the form of lattice waves or phonons Adapted from Fig. 19.1, Callister & Rethwisch 8e. Vibrations become coordinated & produce elastic waves that propagate through the solid Waves have specific wavelengths and energies characteristic of the material – i.e., they are quantized A single quantum of vibrational energy is a phonon Phonons are responsible for the transport of energy by thermal conduction

38. 38 increasing c p Why is c p significantly larger for polymers? Selected values from Table 19.1, Callister & Rethwisch 8e. Polymers Polypropylene Polyethylene Polystyrene Teflon c p (J/kg-K) at room T Ceramics Magnesia (MgO) Alumina (Al 2 O 3 ) Glass Metals Aluminum Steel Tungsten Gold 1925 1850 1170 1050 900 486 138 128 c p (specific heat): (J/kg-K) Material 940 775 840 Specific Heat: Comparison C p (heat capacity): (J/mol-K)

39. Thermal Expansion Concept  Materials Change Size When Heated T init T final L L init Coefficient of Thermal Expansion   due to Asymmetry of PE InterAtomic Distance Trough T↑  E↑ r i is at the Statistical Avg of the Trough Width Bond energy Bond length (r) increasing T T 1 r(T 5 ) 1 ) T 5 Bond-energy vs bond-length curve is “asymmetric”

40. 40 Coefficient of Thermal Expansion: Comparison Q: Why does  generally decrease with increasing bond energy? Polypropylene145-180 Polyethylene106-198 Polystyrene90-150 Teflon126-216 Polymers Ceramics Magnesia (MgO)13.5 Alumina (Al 2 O 3 )7.6 Soda-lime glass9 Silica (cryst. SiO 2 )0.4 Metals Aluminum23.6 Steel12 Tungsten4.5 Gold14.2  (10 -6 /  C) at room T Material Selected values from Table 19.1, Callister & Rethwisch 8e. Why do Polymers have larger  values? increasing 

41. Thermal Conductivity Concept  Ability of a solid to transport heat from high to low temperature regions  atomic vibrations and free electrons Consider a Cold←Hot Bar  Heat Flux given by T 2 > T 1 T 1 x 1 x 2 heat flux Thermal conductivity (W/m-K) or (J/m-K-s) Heat flux (W/m 2 ) or (J/m2-s) Q: Why the NEGATIVE Sign before k? Temperature Gradient (K/m) Fourier’s Law

42. Thermal Conductivity: Comparison increasing k Polymers Polypropylene0.12 Polyethylene0.46-0.50 Polystyrene0.13 Teflon0.25 By vibration/ rotation of chain molecules Ceramics Magnesia (MgO)38 Alumina (Al 2 O 3 )39 Soda-lime glass1.7 Silica (cryst. SiO 2 )1.4 By vibration of atoms Metals Aluminum247 Steel52 Tungsten178 Gold315 By vibration of atoms and motion of electrons k (W/m-K)Energy TransferMaterial

43. Thermal Stresses As Noted Previously a Material’s Tendency to Expand/Contract is Characterized by α If a Heated/Cooled Material is Restrained to its Original Shape, then Thermal Stresses will Develop within the material For a Solid Material Where –   Stress (Pa or typically MPa) –E  Modulus of Elasticity; a.k.a., Young’s Modulus (GPa) –  l  Change in Length due to the Application of a force (m) –l o  Original, Unloaded Length (m)

44. Thermal Stresses con’t

45. Thermal Stresses cont. From Before  Sub  l/l into Young’s Modulus Eqn To Determine the Thermal Stress Relation  Eample: a 1” Round 7075-T6 Al (5.6Zn, 2.5Mg, 1.6Cu, 0.23Cr wt%’s) Bar Must be Compressed by a 8200 lb force when restrained and Heated from Room Temp (295K) Find The Avg Temperature for the Bar

46. Thermal Stress Example Find Stress  Recall the Thermal Stress Eqn  E = 10.4 Mpsi = 71.7 GPa  α = 13.5 µin/in-°F = 13.5 µm/m-°F 8200 lbs 0 lbs  Need E & α Consult Matls Ref

47. Thermal Stress Example cont Solve Thermal Stress Reln for ΔT  Since The Bar was Originally at Room Temp 8200 lbs 0 lbs Heating to Hot-Coffee Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)

48. Thermal Shock Resistance Occurs due to: uneven heating/cooling. Ex: Assume top thin layer is rapidly cooled from T 1 to T 2 : Tension develops at surface Temperature difference that can be produced by cooling: set equal Critical temperature difference for fracture (set  =  f ) 10 Result:

49. Thermal Shock Resistance con’t

50. TSR – Physical Meaning The Reln  For Improved (GREATER) TSR want σ f ↑  Material can withstand higher thermally-generated stress before fracture k↑  Hi-Conductivity results in SMALLER Temperature Gradients; i.e., lower ΔT E↓  More FLEXIBLE Material so the thermal stress from a given thermal strain will be reduced (σ = Eε) α↓  Better Dimensional Stability; i.e., fewer restraining forces developed

51. 51 Application: Space Shuttle Orbiter Silica tiles (400-1260ºC) : -- large scale application-- microstructure: Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J. Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The Shuttle Orbiter Thermal Protection System", Ceramic Bulletin, No. 11, Nov. 1981, p. 1189.) Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.) Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace Ceramics Systems, Sunnyvale, CA.) Thermal Protection System reinf C-C (1650ºC) Re-entry T Distribution silica tiles (400-1260ºC) nylon felt, silicon rubber coating (400ºC) ~90% porosity! Silica fibers bonded to one another during heat treatment. 100  m Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.)

52. WhiteBoard Work Problem 19.5 – Debye Temperature Charles Kittel, “Introduction to Solid State Physics”, 6e, John Wiley & Sons, 1986. pg-110