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Material Science I Phase diagrams II 1 Material Science I Introduction ( Binary Phase diagrams) • A different kind of phase diagram which is very common is where the temperature and composition are varied when the pressure is constant • We will focus on binary- alloys • Binary phase diagrams are helpful in predicting phase transformations and the resulting microstructure • Therefore, binary phase diagrams are maps representing the relationships between temperature and compositions as well as quantities of phases at equilibrium 2 Material Science I Objectives • Determine for a given binary phase diagram • Phase(s) present • Composition of the phase(s) • Mass fraction of the phase(s) • Locate the temperature and compositions of all eutectic, eutectoid, peritectic and congruent phase transforms • Be able to write reactions for all the transforms for either heating or cooling 3 Material Science I Binary Phase diagrams (Isomorphous systems) • The easiest type of binary phase diagrams to understand and interpret is a copper-nickel system • The system is said to be isomorphous because they (components) are completely soluble in liquid and solid forms 4 Material Science I Binary Phase diagrams (Isomorphous systems) NOTE: In the phase diagrams, for metallic alloys, solid solutions are commonly designated by Greek letters (, , etc) Consider the copper – nickel phase diagram given 5 Material Science I Binary Phase diagrams (Isomorphous systems) In the phase diagrams, there are three different phase regions or fields () an alpha field () a liquid field ( + ) two-phase field Each region is defined by the phase or phases that exist over the range of temperatures and compositions The phases are delimited by phase boundaries 6 Material Science I Binary Phase diagrams (Isomorphous systems) The liquid is a homogeneous solution composed of nickel and copper phase is a substitutional solid solution with both copper and nickel having an FCC crystal structure At temperatures below 1080℃, copper and nickel are mutually soluble in each other in the solid state at all compositions 7 Why? Material Science I Interpretation of Phase diagrams In a binary system with known composition and temperature at equilibrium, at least three kinds of information are available Phase(s) present Composition of the phases present Percentages or fractions of the phases We now show how to determine these three pieces of information in a nickel-copper system 8 Material Science I Interpretation of Phase diagrams Phases Present In order to determine the phase(s) present, one needs to know the temperature and composition at a point on the phase diagram e.g. 60wt% − 40% at 1100℃ 9 Material Science I Interpretation of Phase diagrams How to determine the Phase compositions Locate the temperature –composition point with concentration in terms of concentration of the components For a single phase, the composition is just the same as the overall composition of an alloy. To determine the phase composition for an alloy with composition and temperature located in a two-phase region is not easy but can be determined using a tie line A tie line, can be thought of as a horizontal lines with same temperature (isotherm) They are drawn so as to terminate at phase boundary lines on either side 10 Material Science I Interpretation of Phase diagrams How to determine the Phase compositions Using the tie line we can compute the equilibrium concentrations of the two phase (only) using the following procedure: Draw a tie line across the two – phase region at the temperature of the alloy Note the intersection points on the phase boundaries Drop the perpendiculars from intersection points to composition axis where each respective phase is read 11 Material Science I Interpretation of Phase diagrams How to determine the Phase compositions Example: Determine the phase composition for copper-nickel alloy at 35% and 65% alloy at 1250℃ 12 Material Science I Interpretation of Phase diagrams How to determine the Phases compositions Solution The problem is to determine the composition in % of both and for both the and phases 13 Material Science I Interpretation of Phase diagrams How to determine the Phase compositions Solution Liquidus boundary is by the composition axis at 31.5% Ni and 68.5% Cu at Solidus tie line intersection , 42.5% − 57.5% 14 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts It is important to determine the relative amounts as fractions or percentage of the phases present at equilibrium In a single phase, only one phase is present therefore the fraction is 1.0 or 100% as a percentage If the composition and temperature position is located within a two – phase region, one needs to think Here again the tie line in conjunction with a procedure called lever rule must be applied. 15 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts In the lever rule, the following procedure is followed i. Draw a tie line across the two-phase region at the temperature of the alloy ii. Overall alloy composition is located an the tie line iii. To compute the fraction of one phase, take the length of the tie line from the overall alloy composition to the phase boundary for the other phase and divide by the total tie length. iv. Use the above steps to determine the fraction of the other phase v. Multiply the phase fractions by 100 to get the phase percentages 16 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts How do you determine the tie line segment lengths? i. Either by direct measurement from phase diagram using a linear scale graduated in millimeters or ii. By subtracting compositions as taken from the composition axis 17 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts Consider the diagram below, we want to compute the fraction of each of the and liquidus phase 18 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts Locate the overall alloy composition along the tie line and denoted by and the mass fraction and for respective phases From the lever rule = " #$" = %& '%( %& '%) using direct measurement by subtraction using compositions 19 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts NOTE: Composition need to be specified in terms of only one of the constituents for a binary alloy: so far we have made our computations in weight percent nickel i.e. = 35%, = 42.5%, = 31.5% * − 42.5 − 35 = = = 0.68 = + + * − 42.5 − 31.5 For the phase * , − 35 − 31.5 = = = 0.32 = + + * − 42.5 − 31.5 20 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts We can conclude by stating that the lever rule can be employed to determine the relative amounts or fractions of phases in any twophase region for a binary alloy if temperature and composition are known and if equilibrium has been established. IMPORTANT Compositions of phases are expressed in terms of % of components i.e. % , % Fractional phase amounts (e.g. mass fraction of or liquid phase) of a two –phase alloy 21 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts Home work Derive the lever rule for determining mass fraction of a two-phase region (see example 9.1 ) William D Callister Jr. 22 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts For multiphase alloys, it is often more convenient to specify relative phase amounts in terms of volume fraction rather than mass fraction. Phase volume fractions cab be examined from examination of the microstructure Similarly, properties of multiphase alloy may be estimated on the basis of volume fractions 23 Material Science I Interpretation of Phase diagrams How to determine the Phase amounts For an alloy consisting of phase - is defined as . - = ./ + . Where . and ./ denote the volumes of the respective phases in the alloy We also know that for an alloy with two phases - + -/ = 1 24 Material Science I Interpretation of Phase diagrams How to convert from mass fraction to volume fraction and vice versa Converting from mass fraction to volume fraction 0 - = 0 + / 0/ / 0/ -/ = 0 + / 0/ 25 Material Science I Interpretation of Phase diagrams How to convert from mass fraction to volume fraction and vice versa Converting from volume fraction to mass fraction - 0 = - 0 + -/ 0/ / -/ 0/ = - 0 + -/ 0/ 26 Material Science I Interpretation of Phase diagrams How to convert from mass fraction to volume fraction and vice versa Note : 0 and 0/ are the densities of respective phases and their appropriate values can be found by using the equation: 0123 0123 100 = 4 5 + 04 05 46 74 + 56 75 = 6 4 74 56 75 + 04 05 27 Material Science I Development of microstructure in Isomorphus alloys Consider the figure below We are interested in determining (uniform equilibrium cooling) the development of microstructure that occurs for isomorphous alloys during solidification Solidification occurs very slowly and continuously maintained The overall composition remains unchanged during cooling even though there is a redistribution of copper and nickel between phases and is along the dashed solid line 28 Material Science I Development of microstructure in Isomorphus alloys Non - equilibrium cooling (self study) William D callister Jr (2009), Material Science and Engineering, an Introduction, 7th Edition, chapter 9 29 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS 30 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS For the copper-silver system shown in the phase diagram called eutectic phase diagram, there are a number of features which are important and worth noting 31 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS The three single phase regions are found i.e. , and liquid phase is a solid solution rich in copper and has silver as the solute component (an FCC crystal structure) phase is a solid solution rich in silver with copper as the solute ( FCC crystal structure) and are pure copper and pure silver respectively Solubility of these solid phases is limited since at any temperature below line BEG only a limited concentration of silver will dissolve in copper (for the phase) Similarly, for copper in silver ( for phase) 32 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS Solubility limit for the phase corresponds to the boundary line CBA between the ⁄( + ) and ⁄( + ) phase regions The solubility increases with temperature to a maximum ( 8.0%7DE779℃) at point B and decreases back to zero at the melting temperature of pure copper at point A (1085℃) The maximum solubility of copper in the phase point G (8.8%) also occurs at 779℃ Line BEG represents the lowest temperature at which a liquid phase may exist for any copper-silver alloy that is at equilibrium 33 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS Note: There are three two-phase regions for the copper-silver system, i.e. + , + and α + compositions and relative amounts for the phases may be determined using tie lines and the lever rule Point E is called the invariant point designated by H (71.6%) and temperature IH (779℃) 34 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS At the invariant point H (71.6%), as it changes temperature in passing through IH , the reaction is expressed as This equation is called eutectic reaction in which H and IH represent the eutectic composition and temperatures respectively H and /H are the respective compositions of the and phases at IH For this reaction 35 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS The horizontal solidus line at IH is called eutectic isotherm Note: The eutectic reaction upon cooling is similar to solidification and is always two-phases , where as for pure component only a single phase forms Phase diagrams similar to the one discussed here are termed eutectic phase diagrams and components which exhibit this behaviour comprise a eutectic system 36 Material Science I BINARY EUTECTIC (EASILY MELTED) SYSTEMS When constructing a binary phase diagram, it is important to understand that one or at most two phases may be in equilibrium within a phase field For a eutectic system, three phases (, and ) may be in equilibrium but only at a point along the eutectic isotherm The other general rule is that a single-phase regions are always separated from each other by a two phase region that consists of two single phases that is supports JKLMKNO(PQELRSM9.2ETU9.3) 37