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phasediagram

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just keep some basic in mind, its give u enough information about this topic.

just keep some basic in mind, its give u enough information about this topic.

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  • 1. PHASE DIAGRAMS Prof. H. K. Khaira Professor Deptt. of MSME M.A.N.I.T., Bhopal
  • 2. Contents • • • • 1. Phase diagram 2. Phase Rule 3. Lever Rule Microstructure Development
  • 3. Phase Diagrams
  • 4. Phase • Phase is a homogenous, physically distinct and mechanically separable portion of the material with a given chemical composition and structure.
  • 5. Phase diagrams Properties of a materials depend on 1. Number of phases present 2. Type of phases present 3. Amount of phases present and 4. Form of the phases present The Properties can be changed by altering these quantities. In order to make these changes, it is essential to know the conditions under which these quantities exist and the conditions under which a change in phase will occur.
  • 6. Phase diagrams • The best method to record the data related to phase changes in many alloy systems is in the form of phase diagrams, also known as equilibrium diagrams or constitutional diagrams.
  • 7. Phase diagrams • In order to specify completely the state of a system in equilibrium, it is necessary to specify three independent variables. • These variables, which are externally controllable, are temperature, pressure and composition. • Phase diagram is the graphical presentation of the phases present in a system under different conditions of pressure, temperature and composition.
  • 8. Phase diagrams • In metallurgical systems, the pressure is usually taken as atmospheric pressure. Thus the phase diagram shows the phases present at different compositions and temperatures.
  • 9. Phase diagrams • With pressure assumed to be constant at atmospheric pressure, the equilibrium diagram indicates the structural changes due to variation of temperature and composition. • Phase diagrams show the phases present under equilibrium conditions, that is, under conditions in which there will be no change with time. • Equilibrium conditions may be approached by extremely slow heating and cooling, so that if a phase change is to occur, sufficient time is allowed.
  • 10. Phase diagrams The phase diagram in Figure displays an alloy of two metals which forms a solid solution at all relative concentrations of the two species
  • 11. Phase diagrams • Phase diagrams are usually plotted with composition on X axis and the temperature on Y axis,.
  • 12. Types of Equilibrium Diagrams • 1. Complete solid solution type • 2. Eutectic type • 3. Peritectic type
  • 13. Complete solid solution type
  • 14. Complete solid solution type • The phase diagram in which both the constituents are soluble in each other in solid state at all compositions, is known as complete solid solution type phase diagram.
  • 15. Complete solid solution Diagram The phase diagram in Figure displays an alloy of two metals which forms a solid solution at all relative concentrations of the two species
  • 16. Nickel-Copper Phase Diagram
  • 17. Germanium-Silicon Phase Diagram
  • 18. Eutectic Type
  • 19. Eutectic Type • It is a phase diagram containing eutectic reaction. • Eutectic reaction is : L = α (s) + β (s) In this reaction, a liquid phase (L) decomposes into a mixture of two solid phases (α and β) on cooling.
  • 20. Eutectic Diagram
  • 21. Tin-Lead Phase Diagram
  • 22. Gold-Germanium Phase Diagram
  • 23. Copper-Silver Phase Diagram
  • 24. Peritectic Diagram • It is a phase diagram containing peritectic reaction. • peritectic reaction : L + α (s) = β (s) • In peritectic reaction, a liquid (L) and a solid (α) transform in to another solid (β) on cooling.
  • 25. Phase diagram for Fe–C system (dotted lines represent iron-graphite equilibrium).
  • 26. Application of Phase Diagrams • Phase diagram gives us – – – – – – – – Overall Composition Solidus Liquidus Limits of Solid Solubility Chemical Composition of Phases at any temperature Amount of Phases at any temperature Invariant Reactions Development of Microstructure
  • 27. Overall Composition • Concentration: Relative amounts of each constituent • It is the horizontal axis in all binary phase diagrams • The scale can be in weight %, atomic % or mole %
  • 28. Solidus and Solidus • Solidus – Temperature up to which alloy is completely solid – Temperature at which liquefaction begins • Liquidus – Temperature up to which alloy is completely liquid – Temperature at which solidification begins
  • 29. Chemical Composition of Phases • It is the chemical composition of each phase in the system • In a system having more than one phase, each phase will have a unique chemical composition which will be different from each other, and will also be different from the overall composition • Not to be confused with overall composition
  • 30. Invariant Reactions • • • • • Eutectic: L = α (s) + β (s); e.g., Pb-Sn Peritectic: α (s) + L = β (s); e.g., Pb-In Monotectic: L1 = α (s) + L2; e.g., Cu-Pb Syntectic: L1 + L2 = α (s); e.g., Na-Zn Metatectic: β (s) + α (s) = L1 e.g., U-Mn
  • 31. Cooling Curve • A cooling curve is a graphical plot of the changes in temperature with time for a material over the entire temperature range through which it cools.
  • 32. Cooling Curve for Pure Metals
  • 33. Cooling Curve • This is by far the most widely used experimental method. • It relies on the information obtained from the cooling diagrams. • In this method, alloys with different compositions are melted and then the temperature of the mixture is measured at a certain time interval while cooling back to room temperature.
  • 34. Cooling Curve • A cooling curve for each mixture is constructed and the initial and final phase change temperatures are determined. Then these temperatures are used for the construction of the phase diagrams
  • 35. Cooling Curve • Under equilibrium conditions, all metals exhibit a definite melting or freezing point. If a cooling curve is plotted for a pure metal. It will show a horizontal line at the melting or freezing temperature.
  • 36. Cooling curve for the solidification of a pure metal.
  • 37. Cooling curve for a pure metal showing possible undercooling.
  • 38. Cooling curve for a solid solution.
  • 39. Phase Diagram of alloy A+B
  • 40. Series of cooling curves for different alloys in a completely soluble system. The dotted lines indicate the form of the phase diagram
  • 41. Cooling Curves
  • 42. The Phase Rule
  • 43. The Phase Rule Degree of freedom or Variance (f): the number of intensive variables that can be changed independently without disturbing the number of phases in equilibrium. The phase rule: a general relation among the variance f, the number of components c and the number of phases p at equilibrium for a system of any composition. f=c–p+2 43
  • 44. Chapter 7 Physical Chemistry The Phase Rule (a) (b) The difference between (a) a single-phase solution, in which the composition is uniform on a microscopic scale, and (b) a dispersion, in which regions of one component are embedded in a matrix of a second component. 45
  • 45. Chapter 7 Physical Chemistry The Phase Rule Phase: a state of matter that is uniform throughout in chemical composition and physical state. (Gibbs) Number of phase (p): Gas or gaseous mixture – single phase Liquid – one, two and three phases two totally miscible liquids – single phase a slurry of ice and water – two phases Solid – a crystal is a single phase an alloy of two metals – two phases (immiscible) - one phase (miscible) 46
  • 46. Simple Example
  • 47. Chapter 7 Physical Chemistry H2O phase diagram: P — T Region (s, l, g): D P / 10 5 Pa 218 atm C f=2, one phase Y I solid Line (OA, AD, AC): liquid S f=1, two phases in equilibrium 1 atm R 0.00611 gas A Point (A): O 0.0024 0.01 T3 Tf 99.974 Tb t/℃ 374.2 Tc f=0, three phases in equilibrium 51
  • 48. Chapter 7 Physical Chemistry One-component phase equilibrium For a one-component system (pure water) f=1-p+2=3-p,(C=1) f ≥0, p ≥1, 3≥p≥1 p=1,f=2 p=2,f=1 p=3,f=0 52
  • 49. Lever Rule
  • 50. Lever Rule • The compositions of the two coexisting phases at a point in a two phase region is given by the points of intersection of the tie line from that point with the boundaries of the respective phases. • The relative amounts of two coexisting phases at a point are INVERSELY proportional to the distances of the point from intersection points of the tie line from the point with the phase boundaries.” 54
  • 51. Translating This Statement • “two coexisting phases” Means you are in a 2 phase region. Pick an arbitrary point C. • There are two phases at point C. • These phases are A an B • Hence Phases A and B will be in equilibrium at point C • L A+L B+L C P Q A+B A B
  • 52. Tie Lines The line from A to B through C is a tie line. A tie line is, • An isothermal line • That connects two equilibrium phases
  • 53. Translating This Statement The tie line from point C intersect boundaries of phase A and B at 0% A and 0% B respectively. Hence the composition of phase A will be 0% B or pure A. Similarly, the composition of B will be pure B or 0% A. L A+L B+L C A+B A The distances are AC and AB B
  • 54. Translating This Statement L A+L “The intercepts of the tie line with the phase boundaries A and B are PA and CQ respectively. B+L C P Q A+B A B
  • 55. Translating This Statement The intercepts of the tie line with the phase boundaries A and B are CA and CB respectively. L A+L B+L C A+B A B
  • 56. Translating This Statement • “The amounts . . . Are inversely proportional” Means AC / AB is the fraction of B L B= And, CB / AB is the fraction of A A+L B+L A= C A+B A B
  • 57. A Sample Calculation • Draw a horizontal tie line through the point. • Identify the phases. • Measure its length L (liquid) C • Measure the length of each side D AD = 1 cm A+L B+L AC = .75 cm CD = .25 cm • Calculate the amounts A & L A+B A (solid) B A .25cm 1cm 25% L .75 1 75%
  • 58. A more complex system . . . • Draw a tie line. • Identify the phases. • Measure the line lengths. • Calculate the amounts of each phase present. A = Li2O-B2O3 B = Li2O-2B2O3 C A CB = .562 cm AC = .188 cm AB = .75 cm B A .562 .75 75% B .188 .75 25%
  • 59. One Last Note. C If a point is in a single phase region (including a solid solution), NO tie line is used. There is 100% of that phase.
  • 60. Summary • The lever rule is used to calculate the relative percents of each phase when 2 or more phases are present. • The first step in lever rule calculation is to draw a tie line through the composition. • Next one measures the lengths of the tie line, and the distance from the composition to each phase. • The relative concentration of a phase is proportional to the distance from the other phase to the composition, divided by the length of the tie line. (Opposite length / total length)
  • 61. 65
  • 62. Cu-Ni Phase Diagram: T vs. c (wt% or at%) • Tell us about phases as function of T, Co, P. • For this course: --binary systems: just 2 components. --independent variables: T and Co (P = 1atm is always used). -- isomorphous i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni. T(°C) 1600 1500 L (liquid) Contains 2 Phases 1. L 2. There are 3 phase fields 1. L 2. L + 3. 1400 1300 1200 (FCC solid 1100 1000 solution) 0 20 40 60 80 100 wt% Ni 66
  • 63. Phase Diagrams: # and types of phases • Rule 1: If we know T and Co, then we know: --the # and types of phases present. A(1100 C, 60 wt% Ni): 1 phase: B(1250 C, 35 wt% Ni): 2 phases: L + 1500 L (liquid) 1400 1300 (FCC solid solution) 1200 A(1100 C,60) 1100 1000 Cu-Ni phase diagram B (1250 C,35) • Examples: T(°C) 1600 0 20 40 60 80 100 wt% Ni 67
  • 64. Phase Diagrams: composition of phases • Rule 2: If we know T and Co, then we know: --the composition of each phase. • Examples: Consider C0 = 35 wt% Ni At TA = 1320 C: Only Liquid (L) present CL = C0 ( = 35 wt% Ni) At TD = 1190 C: Only Solid ( ) present C = C0 ( = 35 wt% Ni) At TB = 1250 C: Both and L present CL = C liquidus ( = 32 wt% Ni) C = C solidus ( = 43 wt% Ni) Cu-Ni system T(°C) A TA 1300 TB 1200 TD 20 tie line L (liquid) B D 30 35 32 C L C0 (solid) 4043 C 50 wt% Ni 68
  • 65. Phase Diagrams: wt. fraction of phases • Rule 3: If we know T and C0, then can determine: -- the weight fraction of each phase. • Examples: Consider C0 = 35 wt% Ni T(°C) At TA : Only Liquid (L) present WL = 1.00, W = 0 At TD : Only Solid ( ) present WL = 0, W = 1.00 At TB : Both and L present WL W S R +S R R +S 43 35 43 32 Cu-Ni system 0.73 A TA 1300 TB 1200 TD 20 tie line L (liquid) B R S D 30 35 32 CL C0 (solid) 40 43 C 50 wt% Ni = 0.27 W = wt. fraction of phase out of whole. 69
  • 66. The Lever Rule • Sum of weight fractions: W W L • Conservation of mass (Ni): 1 C W C W C o L L • Combine above equations: T(°C) tie line 1300 WLR W S B TB (solid) 1200 R 20 moment equilibrium: L (liquid) 30C L S C0 40 C 1 W solving gives Lever Rule 50 wt% Ni 70
  • 67. The Lever Rule: an interpretation • Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm T(°C) What fraction of each phase? tie line 1300 L (liquid) B TB (solid) 1200 R 20 M ML 30C L R S C0 40 C 50 M xS Think of tie line as a lever S ML x R wt% Ni WL ML ML M S R S C C C0 CL W R R S C0 CL C CL 71
  • 68. Phase and Microstructure (equilibrium) Example: Cooling in Cu-Ni Binary • Consider microstuctural changes that accompany the cooling of a C0 = 35 wt% Ni alloy T(°C) L (liquid) 130 0 L: 35 wt% Ni : 46 wt% Ni • From liquid, solid phase nucleates. A 35 32 • From solid, other phases can nucleate. • Like ice, many grains of solid form. L: 35wt%Ni B C 46 43 D 24 L: 32 wt% Ni 36 120 0 : 43 wt% Ni E L: 24 wt% Ni • wt% of SOLUTE given by line dropped from boundaries : 36 wt% Ni • Fraction of PHASES present given by the “lever rule”. • Microstructure different depending on cool slowly or quench. (solid) 110 0 20 30 35 C0 40 50 wt% Ni 72
  • 69. Binary-Eutectic Systems has a special composition with a minimum melting T. • Ex: Cu-Ag 3 single-phase regions (L, , ) • Limited solubility : mostly Cu : mostly Ag • TE: no liquid below TE. • cE: composition for min. melting T (Eutectic). Cu-Ag system T(°C) 1200 L (liquid) 1000 L+ TE 800 779°C 8.0 L+ 71.9 91.2 600 400 200 0 20 40 100 60 CE 80 C , wt% Ag Eutectic: direct from liquid to 2-phase solid upon cooling: L cooling L(71.9 wt% Ag) (8.0 wt% Ag) (91.2 wt% Ag) + heating 73
  • 70. Example 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220 C, determine: -- phases present: +L T(°C) -- phase compositions C = 17 wt% Sn CL = 46 wt% Sn -- relative amt of phases 300 L (liquid) L+ 220 200 CL - C0 46 - 40 = W = = 0.21 CL - C 46 - 17 100 C0 - C 23 = = 0.79 WL = CL - C 29 R S L+ 183 C + 0 17 20 C 40 46 60 80 C0 CL C, wt% Sn 100 74
  • 71. Example 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220 C, determine: -- phases present: +L T(°C) -- phase compositions C = 17 wt% Sn CL = 46 wt% Sn -- relative amt of phases 300 L (liquid) L+ 220 200 CL - C0 46 - 40 = W = = 0.21 CL - C 46 - 17 100 C0 - C 23 = = 0.79 WL = CL - C 29 R S L+ 183 C + 0 17 20 C 40 46 60 80 C0 CL C, wt% Sn 100 75
  • 72. Solder for electronics Example 1: Lead-Tin (Pb-Sn) Eutectic Diagram • For a 40wt%Sn-60wt%Pb alloy at 150oC, determine... --phases present: + --compositions of phases: C = 11 wt% Sn T(°C) 300 L (liquid) C = 99 wt% Sn -- relative amount Use the of each phase: “Lever Rule” 200 C - C0 S = W = R+S C -C 99 - 40 99 - 11 W = R = R+S 40 - 11 = 99 - 11 = 59 = 0.67 88 C0 - C C -C 150 100 L+ L+ 183°C 18.3 61.9 R 97.8 S + = = 29 = 0.33 88 0 11 20 C 40 C0 60 80 C, wt% Sn 99100 C Adapted from Fig. 10.8, Callister & Rethwisch 3e. 76
  • 73. Example 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220 C, determine: -- phases present: +L T(°C) -- phase compositions C = 17 wt% Sn CL = 46 wt% Sn -- relative amt of phases 300 L (liquid) L+ 220 200 CL - C0 46 - 40 = W = = 0.21 CL - C 46 - 17 100 C0 - C 23 = = 0.79 WL = CL - C 29 R S L+ 183 C + 0 17 20 C 40 46 60 80 C0 CL C, wt% Sn 100 77
  • 74. Microstructure “below” Eutectic (hypoeutectic) • For alloys for 18.3wt%Sn < Co < 61.9wt%Sn • Result: a crystals and a eutectic microstructure T(°C) L: C0 wt% Sn L 300 L C = 18.3 wt% Sn CL = 61.9 wt% Sn W = S = 0.50 R+S WL = (1- W ) = 0.50 L Pb-Sn system L+ 200 R TE R 100 L+ S + S • Just below TE : primary eutectic eutectic 0 20 18.3 Adapted from Fig. 10.16, Callister & Rethwisch 3e. • Just above TE : 40 60 61.9 80 C, wt% Sn 100 97.8 C = 18.3 wt% Sn C = 97.8 wt% Sn W = S = 0.73 R+S W = 0.27 78
  • 75. Solder: Lead-Tin (Pb-Sn) microstructure L For 50 wt% Pb alloy: • Lead-rich phase (dark) • Lamellar eutectic structure of Sn-rich phase (light). L+ + * Why is Liquid-phase ~62.9wt%Sn and -phase ~16.3wt%Sn at 180 C? * For fraction of total phase (both eutectic and primary), use the Lever Rule. 79
  • 76. Hypoeutectic & Hypereutectic Adapted from Fig. 10.8, Callister & Rethwisch 3e. (Figs. 10.14 and 10.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, Amer ican Society for Metals, Materials Park, OH, 1985.) 80
  • 77. Example Problem Steel For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following: a) The compositions of Fe3C and ferrite ( ). b) The amount of cementite (in grams) that forms in 100 g of steel. c) The amounts of pearlite and proeutectoid ferrite ( ) in the 100 g. 81
  • 78. Solution to Problem a) Use RS tie line just below Eutectoid b) Use lever rule with the tie line shown WFe 3C R R S 1600 T(°C) 1200 C0 C CFe 3C C 0.40 0.022 6.70 0.022 L 1400 +L 1000 + Fe3C 800 727°C R 0.057 S + Fe3C 600 400 0 Amount of Fe3C in 100 g L+Fe3C 1148°C (austenite) Fe3C (cementite) C = 0.022 wt% C CFe3C = 6.70 wt% C C C0 1 2 3 4 C, wt% C 5 6 6.7 CFe 3C = (100 g)WFe3C = (100 g)(0.057) = 5.7 g 82
  • 79. Solution to Problem c) Using the VX tie line just above the eutectoid and realizing that C0 = 0.40 wt% C C = 0.022 wt% C Cpearlite = C = 0.76 wt% C V X T(°C) C0 C C C 0.40 0.022 0.76 0.022 L 1400 1200 +L L+Fe3C 1148°C (austenite) 1000 + Fe3C 0.512 800 727°C VX Amount of pearlite in 100 g = (100 g)Wpearlite = (100 g)(0.512) = 51.2 g 600 400 0 + Fe3C 1 C C0 C 2 3 4 5 6 C, wt% C 83 Fe C (cementite) Wpearlite V 1600 6.7
  • 80. Example Problem • One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following: • a) Weight % of liquid and α at 250ºC • b) Chemical composition of the liquid and α at 250ºC • c) Weight % of the liquid and α just above the eutectic temperature • d) Chemical composition of the liquid and α at just above the eutectic temperature 90
  • 81. Summary Lever Rule • Lever Rule is useful to determine: - the composition of each phase, - and the wt% of each phase • 92
  • 82. Microstructure Development • The microstructure developed depends on the overall composition and the cooling rate
  • 83. Composition dependence of microstructure
  • 84. Composition dependence of microstructure
  • 85. Composition dependence of microstructure
  • 86. Composition dependence of microstructure
  • 87. Composition dependence of microstructure
  • 88. Phase diagram for Pb–Sn system. Alloy 1: 63Sn–37Pb, Alloy 2: 70Pb–30Sn, Alloy 3: 70Sn–30Pb.

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