Cu-Ag phase diagram Brazil s map

Transcription

1 Phase Diagrams [9] Cu-Ag phase diagram Brazil s map 1>

2 Some important definitions Component - chemically recognizable species (Fe and C in carbon steel, H2O and NaCl in salted water). A binary alloy contains two components, a ternary alloy three, etc. Phase a portion of a system that has uniform physical and chemical characteristics. Two distinct phases in a system have distinct physical or chemical characteristics (e.g. water and ice) and are separated from each other by definite phase boundaries. A phase may contain one or more components. A single-phase system is called homogeneous, systems with two or more phases are mixtures or heterogeneous systems. Solvent - host or major component in solution. Solute - minor component Solubility Limit of a component in a phase is the maximum amount of the component that can be dissolved in it (e.g. alcohol has unlimited solubility in water, sugar has a limited solubility, oil is insoluble). The same concepts apply to solid phases: Cu and Ni are mutually soluble in any amount (unlimited solid solubility), while C has a limited solubility in Fe. 2>

3 Some important definitions Solubility Limit: after Callister & Rethwisch (2014) 3>

4 Some important definitions Microstructure: The properties of an alloy depend not only on proportions of the phases but also on how they are arranged structurally at the microscopic level. Thus, the microstructure is specified by the number of phases, their proportions, and their arrangement in space. matrix graphite This is an alloy of Fe with 4 wt.% C. There are several phases. The long gray regions are flakes of graphite. The matrix is a fine mixture of BCC Fe and Fe 3 C compound. 4>

5 Some important definitions Equilibrium thermodynamic state where a system is stable at constant temperature, pressure and composition, not changing with time. 5>

6 Gibb s Phase Rule A tool to define the number of phases and/or degrees of phase changes that can be found in a system at equilibrium P + F = C + N P: # of phases present F: degrees of freedom (temp., pressure, composition) C: components or compounds N: noncompositional variables (usually 1) For the Cu-Ag 1 atm for a single phase P: N=1 (temperature), C = 2 (Cu-Ag), P= 1 (a, b, L) F = = 2 This means that to characterize the alloy within a single phase field, 2 parameters must be given: temperature and composition. If 2 phases coexist, for example, +L, β+l, +β, then according to GPR, we have 1 degree of freedom: F = = 1. So, if we have temperature or composition, then we can completely define the system. If 3 phases exist (for a binary system), there are 0 degrees of freedom. This means the composition and temperature are fixed. This condition is met for a eutectic system by the eutectic isotherm. 6>

7 Unary phase diagram water pure iron 7>

8 Binary Systems Binary Isomorphous Systems: An alloy system that contains two components that attain complete liquid and solid solubility of the components, e.g. Cu and Ni alloy. It is the simplest binary system Crystal Structure Electronegativity r (nm) Ni FCC Cu FCC Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume Rothery rules) suggesting high mutual solubility. 8>

9 Cu-Ni Isomorphous phase diagram Information: Phases, Compositions, Weight fractions. Lever Rule 9>

10 Determination of phase(s) present Rule 1: If we know T and C o, then we know: - which phase(s) is (are) present. Examples: A (1100ºC, 60 wt% Ni): 1 phase: B (1250ºC, 35 wt% Ni): 2 phases: L + T(ºC) L (liquid) B (1250ºC,35) L + liquidus solidus (FCC solid solution) A(1100ºC,60) Cu-Ni phase diagram wt% Ni 10>

11 Determination of phase compositions Rule 2: If we know T and C 0, then we can determine: - the composition of each phase. Examples: Consider C 0 = 35 wt% Ni At T A = 1320ºC: Only Liquid (L) present C L = C 0 ( = 35 wt% Ni) At T D = 1190ºC: Only Solid () present C = C 0 ( = 35 wt% Ni) At T B = 1250ºC: Both and L present C L = Cliquidus ( = 32 wt% Ni) C = Csolidus ( = 43 wt% Ni) T(ºC) T A 1300 T B 1200 T D 20 L (liquid) L + Cu-Ni system A B D tie line liquidus L + solidus (solid) CL C0 C wt% Ni 11>

12 Determination of phase weight fractions Rule 3: If we know T and C 0, then can determine: - the weight fraction of each phase. Examples: Consider C 0 = 35 wt% Ni At T A : Only Liquid (L) present At T D : At T B : W L = W L = 1.00, W = 0 Only Solid ( ) present W L = 0, W = 1.00 Both and L present S R +S = = 0.73 T(ºC) T A 1300 T B 1200 T D 20 L (liquid) L + Cu-Ni system A B R S D 3235 tie line liquidus L + solidus (solid) 43 C CLC0 wt% Ni W = R R + S = >

13 The Lever Rule: A Proof Sum of weight fractions: W L + W = 1 Conservation of mass: Combine above equations: W L = C C o C CL = S R + S A geometric interpretation: CL Co C R S C o = W L C L + W C W = C o C L C C L moment equilibrium: W L R = W S = R R + S WL W 1 W solving gives Lever Rule 13>

14 Development of the microstructure Solidification in the solid + liquid phase occurs gradually upon cooling from the liquidus line. The equilibrium composition of the solid and the liquid change gradually during cooling (as can be determined by the tie-line method.) Nuclei of the solid phase form and they grow to consume all the liquid at the solidus line. 14>

15 Development of the microstructure When cooling is too fast for atoms to diffuse and produce equilibrium conditions, nonequilibrium concentrations are produced. The nonuniform composition produced by nonequilibrium solidification is known as segregation. Microsegregation can cause hot shortness which is the melting of the material below the melting point of the equilibrium solidus. Homogenization, which involves heating the material just below the non-equilibrium solidus and holding it there for a few hours, reduces the microsegregation by enabling diffusion to bring the composition back to equilibrium. 15>

16 Cored vs Equilibrium Structures C changes as we solidify. Cu-Ni case: Slow rate of cooling: Equilibrium structure First to solidify has C = 46 wt% Ni. Last to solidify has C = 35 wt% Ni. Uniform C : 35 wt% Ni Fast rate of cooling: Cored structure First to solidify: 46 wt% Ni Last to solidify: < 35 wt% Ni 16>

17 Mechanical Properties of the Cu-Ni System Effect of solid solution strengthening on: 17>

18 Three-phase Reactions in Phase Diagrams 18>

19 Eutectoid & Peritectic Cu-Zn Phase diagram: Peritectic transition γ + L δ Eutectoid transition δ γ + ε 19>

20 Intermetallic Compounds Mg 2 Pb Note: an intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact. 20>

21 Microstructures in Eutectic Systems Pb-Sn system cooling L( 61.9 wt% Sn) (18.3 wt% Sn) + β (97.8 wt% Sn) heating 21>

22 Lamellar Eutectic Structure A 2-phase microstructure resulting from the solidification of a liquid having the eutectic composition where the phases exist as a lamellae that alternate with one another. Formation of eutectic layered microstructure in the Pb-Sn system during solidification at the eutectic composition. Compositions of and β phases are very different. Solidification involves redistribution of Pb and Sn atoms by atomic diffusion. Pb rich Sn rich 22>

23 Hypoeutectic & Hypereutectic Compositions 300 T( C) 200 T E 100 L+ L + β L+β β (Pb-Sn System) hypoeutectic: C o = 50 wt% Sn eutectic 61.9 C o, wt% Sn hypereutectic: (illustration only) eutectic: C o = 61.9 wt% Sn β β β β β β 175 μm 160 μm eutectic micro-constituent 23>

24 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 Max. C solubility in γ iron = 2.11 wt% γ γ γ γ 1148 C R S L A γ+fe 3 C 727 C = Teutectoid +Fe 3 C L+Fe 3 C S Fe 3 C (cementite) 120 μm Result: Pearlite = alternating layers of and Fe 3 C phases (Fe) 0.76 C eutectoid 4.30 Fe 3 C (cementite-hard) (ferrite-soft) C o, wt% C 24>

25 Iron-Carbon (Fe-C) Phase Diagram γ γ γ γ γ γ γ γ γ γ γ γ w =s/(r+s) wγ =(1- w) 1600 δ T( C) γ γ+l (austenite) r s R S Hypoeutectoid Steel 727 C 1148 C L γ +Fe 3 C +Fe 3 C L+Fe 3 C Fe 3 C (cementite) (Fe-C System) w =S/(R+S) w Fe3 C =(1-w) (Fe) C 0 pearlite w pearlite = wγ 0.76 pearlite C o, wt% C 100 μm Hypoeutectoid steel proeutectoid ferrite 25>

26 Iron-Carbon (Fe-C) Phase Diagram Fe 3 C γ γ γ γ γ γ γ γ γ γ γ γ 1600 δ T( C) γ γ+l (austenite) R Hypereutectoid Steel 1148 C 600 w Fe3 C =r/(r +s) +Fe 3 C wγ =(1-w Fe3 C) C o pearlite (Fe) C o, wt%c w pearlite = wγ w =S/(R+S) w Fe3 C =(1-w ) 0.76 r s pearlite S L γ +Fe 3 C L+Fe 3 C Fe 3 C (cementite) (Fe-C System) 60 μm Hypereutectoid steel proeutectoid Fe 3 C 26>

27 Application of the Lever Rule (I) Example: For a AISI 1040 carbon steel (0.40 wt% C) at a temperature just below the eutectoid, determine the following: a) composition of cementite (Fe 3 C) and ferrite () b) the amount of carbide (cementite) in grams that forms per 100 g of steel c) the amount of pearlite and proeutectoid ferrite () 27>

28 Application of the Lever Rule (I) Solution: a) composition of Fe 3 C and ferrite () b) the amount of carbide (cementite) in grams that forms per 100 g of steel Fe3C Fe C Fe 3 3 = = C C = o C + CFe C C = 5.7 g 94.3 g x 100 = x g T( C) 1600 δ C O = 0.40 wt% C C = wt% C C Fe C = 6.70 wt% C 3 γ γ+l (austenite) R 727 C L 1148 C γ +Fe 3 C S +Fe 3 C L+Fe 3 C Fe 3 C (cementite) C o, wt% C C C O C Fe C 3 28>

29 Application of the Lever Rule (I) Solution (cont.): c) the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of γ just above T E C o = 0.40 wt% C C = wt% C C pearlite = C γ = 0.76 wt% C γ γ+ = C o C C γ C x 100 = 51.2 g pearlite = 51.2 g proeutectoid = 48.8 g T( C) 1600 δ γ γ+l (austenite) R S 727 C L 1148 C γ +Fe 3 C +Fe 3 C L+Fe 3 C Fe 3 C (cementite) C C γ C o, wt% C C O 29>

30 Application of the Lever Rule (II) Pb-Sn system For Pb-Sn system at 150 C, calculate relative amounts of each phase by (a) Mass fraction (b) Volume fraction 30>

31 Application of the Lever Rule (II) Solution (a): Given: ρ :11.2 gm/cm 3 ρ :7.3 gm/cm β 3 W C = C β β C C 1 = 99 40= C C 1 W = 40 11= C C β = =0.33 β 31>

32 Application of the Lever Rule (II) (b) Solution Volume Fraction v ( ) 67gm = 11.2 gm/cm3 = 5.98cm 3 v ( β ) 33gm = 7.3gm/cm 3 = 4.52cm 3 v V = = 5.98 = 0.57 v + v V β = = 0.43 β 32>

33 Ternary Phase Diagrams Many alloy systems are based on three or more elements. To describe the changes in structure with temperature, a threedimensional phase diagram is required, which is somewhat complicated. In the plot below the shaded area is the temperature at which freezing begins. Hypothetical ternary phase diagrams where binary phase diagrams are present at the three faces. A 33>

34 Ternary Phase Diagrams The isothermal plot of a ternary phase diagram is usually constructed using an equilateral triangle with the composition of each pure element (or compound) at each corner of the triangle to predict the phases at a particular temperature. 20A-20B-60C An isothermal plot at room temperature for the hypothetical ternary phase diagram. 34>

35 References CALLISTER JR, W. D. AND RETHWISCH, D. G. Materials Science and Engineering: An Introduction, 9th edition. John Wiley & Sons, Inc. 2014, 988p. ISBN: ASHBY, M. and JONES, D. R. H. Engineering Materials 1: An Introduction to Properties, Applications and Design. 4th Edition. Elsevier Ltd. 2012, 472p. ISBN ASKELAND, D. AND FULAY, P. Essentials of Materials Science & Engineering, 2nd Edition. Cengage Learning. 2009, 604p. ISBN SMALLMAN, R. E. and NGAN, A.H.W. Physical Metallurgy and Advanced Materials, 7th Edition. Elsevier Ltd. 2007, 650p. ISBN Nota de aula preparada pelo Prof. Juno Gallego para a disciplina Ciência dos Materiais de Engenharia Permitida a impressão e divulgação. 35