Modeling Diffusion: Flux

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1 Modeling Diffusion: Flux Flux (#/area/time): J = 1 A dm dt Directional Quantity y Jy kg atoms m 2 or s m 2 s Jx Jz x z Flux can be measured for: --vacancies and interstitials --host (A) atoms --impurity (B) atoms x-direction Unit area A through which atoms move.

2 Concentration profile, C(x), changes w/ time. Non Steady State Diffusion To conserve matter: J(right) J(left) = dc dx dt dj = dc dx Governing Eqn.: dt J(left) equate dc dt = D d2 C dx 2 dx Fick's First Law: J = D dc or dx dj dx = D d2 C dx 2 J(right) Concentration, C, in the box (if D does not vary with x)

3 Ex: Non Steady State Diffusion Copper diffuses into a bar of aluminum. Surface conc., Cs of Cu atoms Cs C(x,t) bar pre-existing conc., Co of copper atoms Co to t 1 t 2 t Adapted from 3 Fig. 5.5, Callister 6e. General solution: position, x C(x,t) C o C s C o "error function" Values calibrated in Table 5.1, Callister 6e. = x 1 erf 2 Dt

4 Diffusion and Temperature x i Dt i Make simple estimates for diffusion lengths x i 1 sec 100 sec (1.667 min) 10,000 sec (2.7 hrs) 100,000 sec (1.15 days) 1,000,000 sec (11.6 days)

5 Summary: Structure & Diffusion Diffusion FASTER for... Diffusion SLOWER for... open crystal structures lower melting T materials materials w/secondary bonding smaller diffusing atoms cations lower density materials close-packed structures higher melting T materials materials w/covalent bonding larger diffusing atoms anions higher density materials

6 Example of Diffusion Studies: Using Secondary Ion Mass Spectroscopy Incident sputter ions e.g. O 2+, Cs +, Ar +, Ga + Breaks bonds Sputters sample Measure: Secondary ions Measure secondary ions in mass spectrometer

7 Natural GaSb Example: SIMS Study of Self-Diffusion 69 Ga 121 Sb 71 Ga 123 Sb As-grown Annealed 105 min Natural isotope abundance Annealed 18 days Ga: 69 Ga (60.1%) and 71 Ga (39.9%) Sb: 121 Sb (57.4%) and 123 Sb (42.6%) Which diffuses faster: Ga or Sb? Why?

8 Example: SIMS Study of Self-Diffusion Solve diffusion equation dc dt = D d2 C dx 2 Fit expt. C(t,x) to solution Determine D D(T) activation energy

9 Chapter 9: Phase Diagrams ISSUES TO ADDRESS... When we combine two elements or compounds... what equilibrium state do we get? In particular, if we specify... --a composition (e.g., wt%cu - wt%ni), and --a temperature (T) then... How many phases do we get? What is the composition of each phase? How much of each phase do we get? Phase A Phase B Nickel atom Copper atom

10 Solubility Limit: Max concentration for which only a solution occurs. Ex: Phase Diagram: Water-Sugar System Question: What is the solubility limit at 20 C? Answer: 65 wt% sugar. The Solubility Limit If Co < 65 wt% sugar: sugar If Co > 65 wt% sugar: syrup + sugar. Solubility limit increases with T: e.g., if T = 100 C, solubility limit = 80wt% sugar. Temperature ( C) Pure Water Solubility Limit L (liquid solution i.e., syrup) L (liquid) + S (solid sugar) Co=Composition (wt% sugar) Adapted from Fig. 9.1, Callister 6e. Pure Sugar

11 Components and Phases Components: The elements or compounds which are mixed initially (e.g., Al and Cu) Phases: The physically and chemically distinct material regions that result (e.g., and β). Aluminum- Copper Alloy β (lighter phase) (darker phase) Adapted from Fig. 9.0, Callister 3e.

12 Effect of T and Composition (C o ) Changing T can change # of phases: path A to B. Changing C o can change # of phases: path B to D. watersugar system Adapted from Fig. 9.1, Callister 6e. Temperature ( C) L (liquid solution i.e., syrup) B(100,70) 1 phase L (liquid) + S (solid sugar) A(70,20) 2 phases D(100,90) 2 phases Co=Composition (wt% sugar)

13 Phase Diagrams Tells us about phases as function of T, C o, P. For this course: --binary systems: just 2 components. --independent variables: T and C o (P = 1 atm is used here). Phase Diagram for Cu-Ni system T( C) L (liquid) liquidus L + (FCC solid solution) solidus phases: L (liquid) (FCC solid solution) 3 phase fields: L L + Adapted from Fig. 9.2(a), Callister 6e. (Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). wt% Ni

14 Phase Diagrams: # and Types of Phases Rule 1: If we know T and C o, then we know: --the # and types of phases present. Examples: A(1100, 60): 1 phase: B(1250, 35): 2 phases: L + T( C) L (liquid) B(1250,35) solidus L + liquidus (FCC solid solution) Cu-Ni phase diagram Adapted from Fig. 9.2(a), Callister 6e. (Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991) A(1100,60) wt% Ni

15 Phase Diagrams: Composition of Phases Rule 2: If we know T and C o, then we know: --the composition of each phase. Examples: Co = 35wt%Ni At TA: Only Liquid (L) CL = Co ( = 35wt% Ni) At TD: Only Solid () C = Co ( = 35wt% Ni) At TB: T( C) TA Both and L CL = Cliquidus ( = 32wt% Ni here) C = Csolidus ( = 43wt% Ni here) TB TD 20 L (liquid) L + A B D 3235 CL C o Cu-Ni system tie line liquidus L + solidus (solid) 43 C wt% Ni Adapted from Fig. 9.2(b), Callister 6e. (Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)

16 Phase Diagrams: Weight Fractions of Phases Rule 3: If we know T and C o, then we know: --the amount of each phase (given in wt%). Examples: Co = 35wt%Ni At TA: Only Liquid (L) WL = 100wt%, W = 0 At TD: Only Solid () WL = 0, W = 100wt% At TB: Both and L W L = W = S R + S R R + S = = 73wt % = 27wt% Lever rule (or inverse lever rule!) T( C) TA 1300 TB 1200 TD 20 L (liquid) L + R A B S D 3235 CL C o Cu-Ni system tie line liquidus L + solidus (solid) 43 C wt% Ni Adapted from Fig. 9.2(b), Callister 6e. (Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)

17 The Lever Rule: A Proof Sum of weight fractions: W L + W = 1 Conservation of mass (Ni): C o = W L C L + W C Combine above equations: W L = C C o C CL = S R + S W = C o C L C C L = R R + S A geometric interpretation: CL Co C R S moment equilibrium: W L R = W S WL W 1 W solving gives Lever Rule

18 Ex: Cooling in a Cu-Ni Binary Phase diagram: Cu-Ni system. System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100wt% Ni. Consider C o = 35wt%Ni. T( C) 1300 L: 35wt%Ni : 46wt%Ni L (liquid) L + L + (solid) Adapted from Fig. 9.3, Callister 6e. A B C D E 35 Co 36 L: 35wt%Ni L: 32wt%Ni : 43wt%Ni L: 24wt%Ni : 36wt%Ni wt% Ni Cu-Ni system

19 Cored vs. Equilibrium Phases

20 Cored vs. Equilibrium Phases C changes as we solidify. Cu-Ni case: Fast rate of cooling: Cored structure First to solidify has C = 46wt%Ni. Last to solidify has C = 35wt%Ni. Slow rate of cooling: Equilibrium structure First to solidfy: 46wt%Ni Last to solidfy: < 35wt%Ni Uniform C: 35wt%Ni

21 Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: --Tensile strength (TS) --Ductility (%EL,%AR) Tensile Strength (MPa) TS for pure Cu Cu Ni TS for pure Ni Composition, wt%ni Elongation (%EL) Cu Ni Composition, wt%ni Adapted from Fig. 9.5(a), Callister 6e. Adapted from Fig. 9.5(b), Callister 6e %EL for pure Cu %EL for pure Ni --Peak as a function of Co --Min. as a function of Co

22 Binary Eutectic Systems 2 components Ex.: Cu-Ag system 3 single phase regions (L,, β) Limited solubility: : mostly Cu β: mostly Ni TE: No liquid below TE CE: Min. melting T composition Eutectic Reaction L(CE) (C E )+ β(c βe ) T( C) TE Has a special composition with a min. melting T. L + L (liquid) + β L+β β 779 C CE Co, wt% Ag Adapted from Fig. 9.6, Callister 6e. (Fig. 9.6 adapted from Binary Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) Cu-Ag system

23 Phase Diagrams: Labeling the Lines and Fields

24 P + F = C + 2 Gibbs Phase Rule P: # of phases F: Degrees of freedom C: # of components Normally, pressure = 1 atm P + F = C + 1 or F = C - P + 1 Apply to eutectic phase diagram 1 phase field: F = = 2 Change T and C independently in phase field 2 phase field: F = = 1 C depends on T not independent 3 phase point: F = = 0 C and T defined only at one point (Eutectic point) (no degrees of freedom)

25 Ex: Pb-Sn Eutectic System (1) For a 40wt%Sn-60wt%Pb alloy at 150 C, find... --the phases present: + β T( C) --the compositions of the phases: L L (liquid) + β Pb-Sn system L+β 183 C β Co Co, wt% Sn Adapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)

26 Ex: Pb-Sn Eutectic System (2) For a 40wt%Sn-60wt%Pb alloy at 150 C, find... --the phases present: + β --the compositions of T( C) the phases: C = 11wt%Sn C β = 99wt%Sn --the relative amounts of each phase: W = = 67wt % W β = = 33 wt % L R L (liquid) + β Pb-Sn system L+β 183 C S Co Co, wt% Sn Adapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) β

27 Microstructures in Eutectic Systems (1) C o < 2wt%Sn Result: --polycrystal of grains. T( C) L: Cowt%Sn L L 200 TE : Cowt%Sn L + (Pb-Sn System) β Adapted from Fig. 9.9, Callister 6e Co 2 (room T solubility limit) 30 Co, wt% Sn

28 Microstructures in Eutectic Systems (2) 2wt%Sn < C o < 18.3wt%Sn Result: -- polycrystal with fine β crystals TE T( C) L L + L: Cowt%Sn L : Cowt%Sn β β Pb-Sn system Adapted from Fig. 9.10, Callister 6e Co (sol. limit at Troom) 18.3 (sol. limit at TE) Co, wt% Sn

29 Microstructures in Eutectic Systems (3) C o = C E Result: Eutectic microstructure --alternating layers of and β crystals. 300 Pb-Sn system 200 TE T( C) L + L 183 C L: Cowt%Sn L + β β Micrograph of Pb-Sn eutectic microstructure Adapted from Fig. 9.11, Callister 6e. + β β: 97.8wt%Sn : 18.3wt%Sn CE Co, wt% Sn 160µm Adapted from Fig. 9.12, Callister 6e. (Fig from Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

30 Microstructures in Eutectic Systems (4) 18.3wt%Sn < C o < 61.9wt%Sn Result: crystals and a eutectic microstructure 300 Pb-Sn system 200 TE 100 T( C) 0 0 Adapted from Fig. 9.14, Callister 6e. L L + R R L: Cowt%Sn + β S Co 61.9 L S L + β Co, wt% Sn β L primary eutectic eutectic β Just above TE: C = 18.3wt%Sn CL = 61.9wt%Sn S W = =50wt% R + S WL = (1-W) =50wt% Just below TE: C = 18.3wt%Sn Cβ = 97.8wt%Sn S W = =73wt% R + S Wβ = 27wt%

31 Hypoeutectic & Hypereutectic T( C) 300 L (Figs and 9.15 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) Adapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in- Chief), ASM International, Materials Park, OH, 1990.) 200 TE L hypoeutectic: Co=50wt%Sn 175µm + β L + β Co Co hypoeutectic hypereutectic eutectic 160µm β eutectic: Co=61.9wt%Sn eutectic micro-constituent Adapted from Fig. 9.15, Callister 6e. Adapted from Fig. 9.12, Callister 6e. Co, wt% Sn (Pb-Sn System) hypereutectic: (illustration only) β β β β β β Adapted from Fig. 9.15, Callister 6e. (Illustration only)

32 Eutectic Microstructure Formation Elemental partitioning -phase: Pb-rich rejects Sn β-phase: Sn-rich rejects Pb Higher cooling rate Less time for elemental redistribution Finer microstructure greater strength less ductility

33 Phase Diagrams with Intermediate Phases/Compounds Example: Cu-Zn binary phase diagram Terminal phases/solid solutions: and η Intermediate phases/solid solutions: β, β. γ, and ε

34 Phase Diagrams with Intermediate Phases/Compounds Example: Mg-Pb binary phase diagram Terminal phases/solid solutions: and β Intermediate compound: Mg 2 Pb (line compound)

35 Eutectoid and Peritectic Reactions Eutectoid δ γ+ ε Pertectic δ + L ε Section of Cu-Zn Phase Diagram

36 Congruent Phase Transformations Congruent transformations No composition change for phase transformation e.g., L γ (for 44.9% Ti in Ni-Ti system) Incongruent transformations Composition change for at least one phase for phase transformation Section of Ni-Ti Phase Diagram

37 Iron-Carbon (Fe-C) Phase Diagram 2 important points -Eutectic (A): L γ + Fe 3 C -Eutectoid (B): γ +Fe 3 C 1600 δ T( C) γ γ+l (austenite) +γ R B γ γ γ γ 1148 C R L 727 C = Teutectoid A γ+fe3c S +Fe3C L+Fe3C S Fe3C (cementite) 120µm Result: Pearlite = alternating layers of and Fe3C phases. (Adapted from Fig. 9.24, Callister 6e. (Fig from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) (Fe) Co, wt% C Fe3C (cementite-hard) (ferrite-soft) Ceutectoid Adapted from Fig. 9.21,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

38 Hypoeutectoid Steel γ γ γ γ γ γ γ γ γ γ γ γ w =s/(r+s) wγ =(1-w) 1600 δ w =S/(R+S) w Fe3C =(1-w) T( C) 600 γ γ+l (austenite) r s R S Co 0.77 pearlite w pearlite = wγ 727 C Adapted from Fig. 9.27,Callister 6e. (Fig courtesy Republic Steel Corporation.) L 1148 C γ+fe3c +Fe3C L+Fe3C Fe3C (cementite) Co, wt% C (Fe-C System) 100µm Hypoeutectoid steel Adapted from Figs and 9.26,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)

39 Fe3C γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ T( C) Hypereutectoid Steel γ γ+l (austenite) R w Fe3C =r/(r+s) 600 wγ =(1-w Fe3C ) +Fe3C w =S/(R+S) w Fe3C =(1-w) 0.77 pearlite w pearlite = wγ r Co s S L 1148 C γ+fe3c L+Fe3C Adapted from Fig. 9.30,Callister 6e. (Fig copyright 1971 by United States Steel Corporation.) Fe3C (cementite) Co, wt% C (Fe-C System) 60µm Hypereutectoid steel Adapted from Figs and 9.29,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)

40 Alloying Steel with more Elements TEutectoid ( C) T eutectoid changes: C eutectoid changes: 1200 Ti Mo Ni Si W Cr Mn wt. % of alloying elements Ceutectoid (wt%c) Ni Cr 0.4 Si Mn 0.2 W Ti Mo wt. % of alloying elements Adapted from Fig. 9.31,Callister 6e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Adapted from Fig. 9.32,Callister 6e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

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