Material Properties and Phase Diagrams

Transcription

1 PY2M20 Material Properties and Phase Diagrams ecture 5 P. Stamenov, PhD School of Physics, TCD PY2M20-5

2 Phase Diagrams - Introduction How much can be done with pure elemental compounds? How many combinations of elements could be imagined? 2 100??? How many of these combinations will have the structure (crystallographic, nanoscale, microscale, etc. ) of the end members? How is mixing them going to affect the resulting mechanical, electronic and other physical and chemical properties? Can the properties of the end members be improved on?

3 Phase Diagrams Why? When we combine two or more constituents (elements)... what equilibrium state do we get? In particular, if we specify... c, p, T, but also H, E (all are intensive thermodynamic parameters) then... - composition (e.g., wt% Cu - wt% Ni), and - temperature t (T) 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 Does this phase segregation really occur for Cu 1-x Ni x?

4 Phase Equilibria: Solubility imit Introduction Solutions solid solutions, single phase Mixtures more than one phase Solubility imit: Max concentration for which only a single phase solution occurs. Question: What is the solubility limit at 20 C? Answer: 65 wt% sugar. If Co < 65 wt% sugar: syrup If Co > 65 wt% sugar: syrup + sugar. ( C) Tempe erature e ter Pure Wat Sucrose/Water Phase Diagram Solubility imit (liquid solution i.e., syrup) (liquid) id) + S (solid sugar) Co=Composition Composition (wt% sugar) e gar Pure Sug

5 Components and Phases Components: The elements or compounds which h are present in the mixture (e.g., Al and Cu) Phases: The physically and chemically distinct material regions that result (e.g., and ). Aluminum- Copper Alloy (lighter phase) (darker phase)

6 Effect of T & Composition (C o ) Changing T can change # of phases: path A to B. Changing C o can change # of phases: path B to D. water- sugar system Tempe erature e ( C) 100 B (100 C,70) 1 phase (liquid solution i.e., syrup) (liquid) + S (solid sugar) A (20 C,70) 2 phases D (100 C,90) 2 phases C o =Composition (wt% sugar)

7 Phase Equilibria Simple solution system (e.g., Ni-Cu solution) Crystal Structure Electroneg. 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. Ni and Cu are totally miscible in all proportions. Hence, the answer to the earlier question is is No

8 Phase Diagrams Indicate phases as function of T, C o, and P. For this course: -binary systems: just 2 components. -independent variables: T and C o (P = 1 atm is almost always a used). Phase Diagram for Cu-Ni system T( C) (liquid) 1200 (FCC solid 1100 solution) phases: (liquid) (FCC solid solution) 3 phase fields: + wt% Ni

9 Phase Diagrams: Number and types of phases Rule 1: If we know T and C o, then we know: - the number and types of phases present. Examples: A(1100 C, 60): 1 phase: B(1250 C, 35): 2 phases: + T( C) (liquid) B ( C,35) (FCC solid solution) Cu-Ni phase diagram 1100 A(1100 C C,60) wt% Ni

10 Composition of phases Rule 2: If we know T and C o, then we know: - the composition of each phase. Examples: Co = 35 wt% Ni At TA = 1320 C: Only iquid () C = Co ( = 35 wt% Ni) TB At TD = 1190 C: Only Solid ( ) C = Co (=35wt%Ni) At TB = 1250 C: T( C) Cu-Ni system A TA tie line 1300 (liquid) 1200 TD Both and C = Cliquidus ( = 32 wt% Ni here) C = Csolidus ( = 43 wt% Ni here) C o C B D (solid) C wt% Ni

11 Weight fractions of phases Rule 3: If we know T and C o, then we know: - the amount of each phase (given in wt%), via the so-called: centre of gravity principle or the lever rule Examples: T( C) Cu-Ni system Co = 35 wt% Ni At TA: A Only iquid () W = 100 wt%, W = 0 At TD: Only Solid ( ) At T B: W W = 0, W = 100 wt% Both and S R +S W R R +S = 27 wt% 73 wt% T( C) TA 1300 (liquid) B TB R S 1200 TD A tie line D C o 43 C C (solid) wt% Ni

12 The ever Rule Tie line connects the phases in equilibrium with each other - essentially an isotherm T( C) 1300 (liquid) tie line TB B 1200 (solid) R S 20 30C C o C wt% Ni How much of each phase? Think of it as a lever M M M R S S M R W M M M S R S C C C C W R R S C C 0 0 C C

13 Cooling in the Cu-Ni Binary System 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 100 wt% Ni. Consider C o = 35 wt%ni. T( C) 1300 : 35 wt% Ni : 46 wt% Ni 1200 (liquid) A B C D (solid) E : 35wt%Ni Co Cu-Ni system : 32 wt% Ni : 43 wt% Ni : 24 wt% Ni : 36 wt% Ni wt% Ni

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

15 Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: - Tensile strength (TS) - Ductility (%E,%AR) Ten nsile Str rength (MPa) TS for pure Cu Cu Ni TS for pure Ni Composition, wt% Ni Elong gation (% %E) 60 %E for pure Cu 50 %E for pure Ni Cu Ni Composition, wt% Ni - Maximum as a function of Co - Minimum as a function of Co

16 Binary Eutectic Systems ευτηκτικός - from Greek easiest to melt

17 Binary-Eutectic Systems Cu-Ag system 3 single phase regions (, ) imited solubility: ) : mostly Cu : mostly Ag T E : No liquid below T E C E : Composition with min. melting T E Eutectic transition T( C) TE (C E ) (C E ) + (C E ) + (liquid) Cu-Ag system C C E C E C o, wt% Ag

18 Pb-Sn Eutectic System (1) For a 40 wt% Sn-60 wt% Pb alloy at 150 C, find... W= - the phases present: - compositions of phases: C O =40wt%Sn C = 11 wt% Sn C = 99 wt% Sn - the relative amount of each phase: = T( C) C 18.3 (liquid) S C - C O 150 R R+S = C - C = 59 = 67 wt% 88 R C W = O - C = R+S C -C Ca ste e = = 29 = 33 wt% Pb-Sn system S C Co C Adapted from Fig. 9.8, Callister 7e. C, wt% Sn

19 Pb-Sn Eutectic System (2) For a 40 wt% Sn-60 wt% Pb alloy at 220 C, find... - the phases present: + Pb-Sn - compositions of phases: T( C) C O =40wt%Sn C = 17 wt% Sn C = 46 wt% Sn - the relative amount of each phase: W = C -C O C -C = (liquid) system 200 R S C = 6 29 = 21 wt% W = C O -C = 23 C -C 29 = 79 wt% C o C C C, wt% Sn

20 Microstructures in Eutectic ti Systems: I C o < 2 wt% Sn Result: - at extreme ends - polycrystal of grains i.e., only one solid phase. T( C) 400 : C o wt% Sn T E : C o wt% Sn + (Pb-Sn System) C o 2 (room T solubility limit) C o, wt% Sn

21 Microstructures in Eutectic ti Systems: II 2 wt% Sn < C o < 18.3 wt% Sn Result: Initially liquid + then alone finally two phases polycrystal fine -phase inclusions T( C) T E : C o wt% Sn : C o wt% Sn Pb-Sn system C o (sol. limit at Troom) ) 18.3 (sol. limit at T E ) C o, wt% Sn

22 Microstructures in Eutectic ti Systems: III C o = C E Result: Eutectic microstructure (lamellar structure) - alternating layers (lamellae) of and crystals. Pb-Sn system T( C) C T E Micrograph of Pb-Sn eutectic : C o wt% Sn microstructure t 100 : 97.8 wt% Sn : 18.3 wt%sn 160m Adapted from Fig. 9.14, Callister 7e C E C, wt% Sn

23 amellar Eutectic Structure Other possible eutectic structures are: rod-like, globular and acicular.

24 Microstructures in Eutectic ti Systems: IV 18.3 wt% Sn < C o < 61.9 wt% Sn Result: crystals and an eutectic microstructure Pb-Sn system T( C) : C o wt% Sn R S + T E R S Just above T E : C = 18.3 wt% Sn C = 61.9 wt% Sn S W= = 50 wt% R + S W = (1- W) = 50 wt% Just below T E : C = 18.3 wt% Sn primary C = 97.8 wt% Sn eutectic eutectic S W= = 73 wt% R + S W = 27 wt% C o, wt% Sn 0 40