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.
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)
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
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)
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
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
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?
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
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
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) 100 80 60 40 20 Pure Water Solubility Limit L (liquid solution i.e., syrup) L (liquid) + S (solid sugar) 0 20 40 6065 80 100 Co=Composition (wt% sugar) Adapted from Fig. 9.1, Callister 6e. Pure Sugar
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.
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) 100 80 60 40 20 0 0 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 20 40 60 70 80 100 Co=Composition (wt% sugar)
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) 1600 1500 1400 1300 1200 1100 1000 0 L (liquid) liquidus L + (FCC solid solution) solidus 20 40 60 80 100 2 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
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) 1600 1500 1400 1300 1200 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). 1100 1000 0 A(1100,60) 20 40 60 80 100 wt% Ni
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 1300 1200 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 30 40 50 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.)
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 = 43 35 43 32 = 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 30 40 50 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.)
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
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 1200 1100 20 L (liquid) 24 35 32 L + L + (solid) Adapted from Fig. 9.3, Callister 6e. A B C D E 35 Co 36 L: 35wt%Ni 43 46 30 40 50 L: 32wt%Ni : 43wt%Ni L: 24wt%Ni : 36wt%Ni wt% Ni Cu-Ni system
Cored vs. Equilibrium Phases
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
Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: --Tensile strength (TS) --Ductility (%EL,%AR) Tensile Strength (MPa) 400 300 TS for pure Cu 200 0 20 40 60 80 100 Cu Ni TS for pure Ni Composition, wt%ni Elongation (%EL) 20 0 20 40 60 80 100 Cu Ni Composition, wt%ni Adapted from Fig. 9.5(a), Callister 6e. Adapted from Fig. 9.5(b), Callister 6e. 60 50 40 30 %EL for pure Cu %EL for pure Ni --Peak as a function of Co --Min. as a function of Co
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) 1200 1000 TE 800 600 400 200 0 Has a special composition with a min. melting T. L + L (liquid) + β L+β β 779 C 8.0 71.9 91.2 20 40 60 CE 80 100 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
Phase Diagrams: Labeling the Lines and Fields
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 1 + 1 = 2 Change T and C independently in phase field 2 phase field: F = 2 2 + 1 = 1 C depends on T not independent 3 phase point: F = 2 3 +1 = 0 C and T defined only at one point (Eutectic point) (no degrees of freedom)
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: 300 200 150 100 L + 18.3 L (liquid) + β Pb-Sn system L+β 183 C 61.9 97.8 β 0 20 40 60 80 100 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.)
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 = 59 88 = 67wt % W β = 29 88 = 33 wt % 300 200 150 100 0 L + 18.3 R L (liquid) + β Pb-Sn system L+β 183 C 61.9 97.8 S 11 20 40 60 80 99100 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
Microstructures in Eutectic Systems (1) C o < 2wt%Sn Result: --polycrystal of grains. T( C) 400 300 L: Cowt%Sn L L 200 TE : Cowt%Sn L + (Pb-Sn System) 100 + β Adapted from Fig. 9.9, Callister 6e. 0 10 20 Co 2 (room T solubility limit) 30 Co, wt% Sn
Microstructures in Eutectic Systems (2) 2wt%Sn < C o < 18.3wt%Sn Result: -- polycrystal with fine β crystals. 400 300 200 TE T( C) L L + L: Cowt%Sn L : Cowt%Sn β 100 + β Pb-Sn system Adapted from Fig. 9.10, Callister 6e. 0 10 20 30 2 Co (sol. limit at Troom) 18.3 (sol. limit at TE) Co, wt% Sn
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 100 0 0 Adapted from Fig. 9.11, Callister 6e. + β β: 97.8wt%Sn : 18.3wt%Sn 20 40 60 80 100 18.3 CE 97.8 61.9 Co, wt% Sn 160µm Adapted from Fig. 9.12, Callister 6e. (Fig. 9.12 from Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
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 20 40 60 18.3 Co 61.9 L S L + β Co, wt% Sn β L primary eutectic eutectic β 80 100 97.8 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%
Hypoeutectic & Hypereutectic T( C) 300 L (Figs. 9.12 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 100 0 0 L + 20 40 18.3 hypoeutectic: Co=50wt%Sn 175µm + β 60 61.9 L + β Co Co hypoeutectic hypereutectic eutectic 160µm β 80 100 97.8 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)
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
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
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)
Eutectoid and Peritectic Reactions Eutectoid δ γ+ ε Pertectic δ + L ε Section of Cu-Zn Phase Diagram
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
Iron-Carbon (Fe-C) Phase Diagram 2 important points -Eutectic (A): L γ + Fe 3 C -Eutectoid (B): γ +Fe 3 C 1600 δ 1400 1200 1000 800 600 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. 9.24 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 400 0 1 2 3 4 5 6 6.7 (Fe) 0.77 4.30 Co, wt% C Fe3C (cementite-hard) (ferrite-soft) Ceutectoid Adapted from Fig. 9.21,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
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) 1400 1200 1000 800 T( C) 600 γ γ+l (austenite) r s R S 400 0 1 2 3 4 5 6 6.7 Co 0.77 pearlite w pearlite = wγ 727 C Adapted from Fig. 9.27,Callister 6e. (Fig. 9.27 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. 9.21 and 9.26,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)
Fe3C γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ 1400 1200 1000 800 T( C) Hypereutectoid Steel γ γ+l (austenite) R w Fe3C =r/(r+s) 600 wγ =(1-w Fe3C ) +Fe3C 400 0 1 2 3 4 5 6 6.7 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. 9.30 copyright 1971 by United States Steel Corporation.) Fe3C (cementite) Co, wt% C (Fe-C System) 60µm Hypereutectoid steel Adapted from Figs. 9.21 and 9.29,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)
Alloying Steel with more Elements TEutectoid ( C) T eutectoid changes: C eutectoid changes: 1200 Ti Mo 1000 800 600 Ni Si W Cr Mn 0 4 8 12 wt. % of alloying elements Ceutectoid (wt%c) 0.8 0.6 Ni Cr 0.4 Si Mn 0.2 W Ti Mo 0 0 4 8 12 wt. % of alloying elements Adapted from Fig. 9.31,Callister 6e. (Fig. 9.31 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. 9.32 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)