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Materials of Engineering

Engineering 45. Materials of Engineering. Lecture 8 - Diffusion Review & Thermal Properties. Carlos Casillas, PE 16-Jan-12 Licensed Chemical Engineer Spring 2012 CCasillas@ChabotCollege.edu Hayward, CA. Recruitment. Click to edit Master text styles Second level Third level

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Materials of Engineering

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  1. Engineering 45 Materials of Engineering Lecture 8 - Diffusion Review & Thermal Properties Carlos Casillas, PE 16-Jan-12 Licensed Chemical Engineer Spring 2012 CCasillas@ChabotCollege.eduHayward, CA

  2. Recruitment • Click to edit Master text styles • Second level • Third level • Fourth level • Fifth level

  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 Defining conductivity  (a material property) Solution: Fick’s 2nd Law Ohm’s Law • Heat Flux – • (by phonons) N = # of phonons with avg. energy  Defining thermal conductivity  (a material property) Solution: Fick’s 2nd 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 • applies to substitutional impurities • atoms exchange with vacancies • rate depends on (1) number of vacancies; (2) activation energy to exchange. Number (or concentration*) of Vacancies at T • kBT gives eV E is an activation energy for a particular process (in J/mol, cal/mol, eV/atom).

  6. Where can we use Fick’s Law? • 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. 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. CO2 in air has D~16 mm2/s, and, in water, D~ 0.0016 mm2/s

  7. Gas Diffusion in Polymers

  8. Polymer Membrane Structure

  9. Modeling rate of diffusion: flux diffused mass M Jslope time • 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 A = Area of flow

  10. Steady-state Diffusion: J ~ gradient of C • Concentration Profile, C(x): [kg/m3] Adapted from Fig. 6.2(c) • Fick's First Law: D is a constant • The steeper the concentration profile, the greater the flux

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

  12. C1 C1 C2 x1 x2 C2 x Steady-State Diffusion Rate of diffusion independent of time • J ~ Fick’s first law of diffusion D diffusion coefficient

  13. glove C1 paint remover skin C2 x1 x2 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 cm2/s • surface concentrations: • Diffusion distance: C1= 0.44 g/cm3 C2= 0.02 g/cm3 x2 – x1 = 0.04 cm

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

  15. Simple Diffusion expt • 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.

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

  17. C s 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. B.C. at t = 0, C = Co for 0  x   at t > 0, C = CS for x = 0 (fixed surface conc.) C = Co for x =  Adapted from Fig. 6.5, Callister & Rethwisch 3e.

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

  19. z erf(z) Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation. 0.90 0.7970 z 0.8125 0.95 0.8209 z= 0.93 So,  erf(z) = 0.8125 Now solve for D 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 after49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, what temperature was treatment done? Solution

  20. From Table 6.2, for diffusion of C in FCC Fe • Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol T = 1300 K = 1027°C Solution (cont.): • To solve for the temperature at which D has the above value, we use a rearranged form of Equation (6.9a); D=D0 exp(-Qd/RT) 

  21. Example:Processing • 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? Key point 1: C(x,t500C) = C(x,t600C). Key point 2: Both cases have the same Co and Cs. • Result: Dt should be held constant. Note D(T) are T dependent Values of D are provided. • Answer:

  22. transform data ln D D Temp = T 1/T Example: At 300ºC the diffusion coefficient and activation energy 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 350ºC? 22

  23. Example (cont.) T1 = 273 + 300 = 573K T2 = 273 + 350 = 623K D2 = 15.7 x 10-11 m2/s 23

  24. VMSE: Student Companion SiteDiffusion Computations & Data Plots 24

  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 25

  26. Solution (cont.): t = 49.5 h x = 4 x 10-3 m Cx = 0.35 wt% Cs = 1.0 wt% Co = 0.20 wt%  erf(z) = 0.8125 26

  27. Solution (cont.): z erf(z) 0.90 0.7970 z 0.8125 0.95 0.8209 Now solve for D 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 z= 0.93 27

  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 Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol  T = 1300 K = 1027ºC Solution (cont.): 28

  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 (tb), 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 cm2/s 29

  30. CPC Example (cont.) glove Equation from online CPC Case Study 5 at the Student Companion Site for Callister & Rethwisch 8e (www.wiley.com/ college/callister) C1 paint remover skin C2 x1 x2 D = 110x10-8 cm2/s • Solution – assuming linear conc. gradient Breakthrough time = tb Time required for breakthrough ca. 4 min 30

  31. Summary Diffusion SLOWER for... • close-packed structures • materials w/covalent bonding • larger diffusing atoms • higher density materials Diffusion FASTER for... • open crystal structures • materials w/secondary bonding • smaller diffusing atoms • lower density materials 31

  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. energy input (J/mol or J/kg) heat capacity (J/mol-K) temperature change (K) Heat Capacity (Specific Heat) • The ability of a material to absorb heat is specified by its heat capacity, C • Cp= heat capacity measured @ constant pressure • Cv= 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)

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

  35. Cv as Function of Temperature Cv  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. Cv as Function of Temp cont For Many Crystalline Solids • Where • A  Material Dependent CONSTANT • TD  Debye Temperature, K • Thermal Physics • Energy is Stored in Lattice Vibration Waves Called Phonons • Analogous to OpticalPHOTONS

  37. Atomic Vibrations Atomic vibrations are in the form of lattice waves or phonons • 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 Adapted from Fig. 19.1, Callister & Rethwisch 8e. 37

  38. Specific Heat: Comparison cp (J/kg-K) at room T Material • Polymers cp(specific heat): (J/kg-K) 1925 Polypropylene 1850 Polyethylene 1170 Polystyrene 1050 Teflon • Ceramics Magnesia (MgO) 940 Alumina (Al2O3) 775 Glass 840 • Metals Aluminum 900 Steel 486 Selected values from Table 19.1, Callister & Rethwisch 8e. Tungsten 138 Gold 128 Cp(heat capacity): (J/mol-K) • Why is cpsignificantly larger for polymers? increasing cp 38

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

  40. Coefficient of Thermal Expansion: Comparison a (10-6/C)at room T Material • Polymers Polypropylene 145-180 Polyethylene 106-198 Polystyrene 90-150 Teflon 126-216 • Metals Aluminum 23.6 Steel 12 increasing  Tungsten 4.5 Gold 14.2 • Ceramics Magnesia (MgO) 13.5 Alumina (Al2O3) 7.6 Soda-lime glass 9 Silica (cryst. SiO2) 0.4 Why do Polymers have larger  values? • Q: Why does a generally decrease with increasing bond energy? Selected values from Table 19.1, Callister & Rethwisch 8e. 40

  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 (W/m2) or (J/m2-s) Thermal conductivity (W/m-K) or(J/m-K-s) T > T T 2 1 1 x x 1 2 heat flux • Heat Flux given by Temperature Gradient (K/m) Fourier’s Law • Q: Why theNEGATIVE Sign before k?

  42. Thermal Conductivity: Comparison Material k (W/m-K) Energy Transfer • Metals Aluminum 247 By vibration of Steel 52 atoms and Tungsten 178 motion of Gold 315 electrons • Ceramics Magnesia (MgO) 38 By vibration of Alumina (Al2O3) 39 increasing k atoms Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4 • Polymers Polypropylene 0.12 By vibration/ Polyethylene 0.46-0.50 rotation of chain Polystyrene 0.13 molecules Teflon 0.25

  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) • lo  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 0 lbs 8200 lbs • E = 10.4 Mpsi = 71.7 GPa • α= 13.5 µin/in-°F = 13.5 µm/m-°F • Recall the Thermal Stress Eqn • Need E & α • Consult Matls Ref

  47. Thermal Stress Example cont Solve Thermal Stress Reln for ΔT 0 lbs 8200 lbs • Since The Bar was Originally at Room Temp • 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 T1 to T2: set equal Tension develops at surface Critical temperature difference for fracture (set s = sf) Temperature difference that can be produced by cooling: • Result: 10

  49. Thermal Shock Resistance con’t

  50. TSR – Physical Meaning The Reln • For Improved (GREATER) TSR want • σf↑  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

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