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EX 1: Pb-Sn Eutectic System

EX 1: Pb-Sn Eutectic System. T (ºC). 300. L (liquid). a. L. +. a. b. b. L. +. 200. 183ºC. 18.3. 61.9. 97.8. C  - C 0. 150. S. R. S. =. W. =. . R + S. C  - C . 100. a. +. b. 99 - 40. 59. =. =. = 0.67. 99 - 11. 88. 100. 0. 11. 20. 60. 80. 99.

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EX 1: Pb-Sn Eutectic System

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  1. EX 1: Pb-Sn Eutectic System T(ºC) 300 L (liquid) a L + a b b L + 200 183ºC 18.3 61.9 97.8 C- C0 150 S R S = W =  R+S C- C 100 a + b 99 - 40 59 = = = 0.67 99 - 11 88 100 0 11 20 60 80 99 40 C0- C R W C C C0 = =  C, wt% Sn C - C R+S 40 - 11 29 = = 0.33 = 99 - 11 88 • For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine: -- the phases present Pb-Sn system Answer: a + b -- the phase compositions Answer: Ca = 11 wt% Sn Cb = 99 wt% Sn -- the relative amount of each phase Answer: Adapted from Fig. 9.8, Callister & Rethwisch 8e.

  2. EX 2: Pb-Sn Eutectic System T(ºC) CL - C0 46 - 40 = W = a CL - C 46 - 17 300 L (liquid) 6 a L + = = 0.21 29 220 a b b R L + S 200 183ºC 100 a + b 100 17 46 0 20 40 60 80 C CL C0 C, wt% Sn C0 - C 23 = WL = = 0.79 CL - C 29 • For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine: -- the phases present: Pb-Sn system Answer: a + L -- the phase compositions Answer: Ca = 17 wt% Sn CL = 46 wt% Sn -- the relative amount of each phase Answer: Adapted from Fig. 9.8, Callister & Rethwisch 8e.

  3. Microstructural Developments in Eutectic Systems I T(ºC) L: C0wt% Sn 400 L a L 300 L a + a 200 (Pb-Sn a: C0wt% Sn TE System) 100 b + a 0 10 20 30 C , wt% Sn C0 2 (room T solubility limit) • For alloys for which C0 < 2 wt% Sn (compositions ranging between pure component and the max sol at room temp) • Result: at room temperature -- polycrystalline with grains of a phase having composition C0 Adapted from Fig. 9.11, Callister & Rethwisch 8e.

  4. Microstructural Developments in Eutectic Systems II L: C0 wt% Sn T(ºC) 400 L L 300 a L + a a: C0wt% Sn a 200 TE a b 100 b + a Pb-Sn system 0 10 20 30 C , wt% Sn C0 2 (sol. limit at T ) 18.3 room (sol. limit at TE) • For alloys for which 2 wt% Sn < C0 < 18.3 wt% Sn Room tem solubility and max solid solubility at Te • Result: at temperatures in a + b range -- polycrystalline with agrains and small b-phase particles Adapted from Fig. 9.12, Callister & Rethwisch 8e.

  5. Microstructural Developments in Eutectic Systems III Micrograph of Pb-Sn T(ºC) eutectic L: C0 wt% Sn microstructure 300 L Pb-Sn system a L + a b L 200 183ºC TE 100 160m a : 97.8 wt% Sn Adapted from Fig. 9.14, Callister & Rethwisch 8e. : 18.3 wt%Sn 0 20 40 60 80 100 97.8 18.3 CE C, wt% Sn 61.9 • For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of a and b phases. Adapted from Fig. 9.13, Callister & Rethwisch 8e.

  6. Hypoeutectic & Hypereutectic hypoeutectic: C0 = 50 wt% Sn hypereutectic: (illustration only) (Figs. 9.14 and 9.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) a b a b a a b b a b a b 175 mm Adapted from Fig. 9.17, Callister & Rethwisch 8e. Adapted from Fig. 9.17, Callister & Rethwisch 8e. (Illustration only) 300 L T(ºC) Adapted from Fig. 9.8, Callister & Rethwisch 8e. (Fig. 10.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) a L + a b b L + (Pb-Sn 200 TE System) a + b 100 C, wt% Sn 0 20 40 60 80 100 eutectic 61.9 eutectic: C0=61.9wt% Sn 160 mm eutectic micro-constituent Adapted from Fig. 9.14, Callister & Rethwisch 8e.

  7. Intermetallic Compounds Adapted from Fig. 9.20, Callister & Rethwisch 8e. Mg2Pb Note: intermetallic compound exists as a line on the diagram - not an area - because of stoichiometry (i.e. composition of a compound is a fixed value). – 2- melting Tem. 3- solubility of each element in other

  8. cool cool cool heat heat heat • Eutectoid – one solid phase transforms to two other solid phases S2S1+S3  + Fe3C (For Fe-C, 727ºC, 0.76 wt% C) intermetallic compound - cementite • Peritectic - liquid and one solid phase transform to a second solid phase S1 + LS2  + L (For Fe-C, 1493ºC, 0.16 wt% C) Eutectic, Eutectoid, & Peritectic • Eutectic - liquid transforms to two solid phases L +  (For Pb-Sn, 183ºC, 61.9 wt% Sn)

  9. Iron Carbon Phase diagram

  10. Comments • 1- BCC, FCC, BCC • 2- % is max 6.70 then pure graphite • Fe3C cementite • Carbon max solubility in Ferrite and austenite and at what temp • fe3C exist where (ranges of Tem)? And it is hard and brittle and not equilibrium • Eutictoid a = pearlite • Proeutictoid a + eutectoid a • a (proeutictiod + eutectoid)

  11. T(ºC) 1600 d - Eutectic (A): L 1400 Þ g + L Fe3C g +L g A 1200 1148ºC - Eutectoid (B): (austenite) g Þ a + Fe3C g g 1000 g +Fe3C g g a Fe3C (cementite) + 800 B g a 727ºC = T eutectoid 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 4.30 0.76 C, wt% C (Fe) 120 mm Result: Pearlite = Fe3C (cementite-hard) alternating layers of a (ferrite-soft) a and Fe3C phases (Adapted from Fig. 9.27, Callister & Rethwisch 8e.) Adapted from Fig. 9.24,Callister & Rethwisch 8e. Iron-Carbon (Fe-C) Phase Diagram • 2 important points L+Fe3C

  12. Ferrous alloys : based on C content 1- Iron < .008 wt% C : ferrite phase at room Tepm 2- Steel : .008 – 2.14 wt% C : Fe3C + a 3- Cast iron : 2.14 -6.70 wt%

  13. T(ºC) 1600 d L 1400 (Fe-C g +L g g g System) 1200 L+Fe3C 1148ºC (austenite) g g g 1000 g g +Fe3C g g Fe3C (cementite) a g g 800 727ºC a a a g g 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 a C, wt% C (Fe) C0 0.76 pearlite Hypoeutectoid 100 mm steel pearlite proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e. Hypoeutectoid Steel Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

  14. T(ºC) 1600 d L 1400 (Fe-C g +L g System) 1200 L+Fe3C 1148ºC (austenite) 1000 g +Fe3C Wa’= S/(R+S) Wa= s/(r+s) Fe3C (cementite) a r s g g 800 Wg=(1 - Wa) W=(1 – Wa’) 727ºC a a a g g R S 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 a C, wt% C (Fe) C0 0.76 Wpearlite=Wg pearlite Hypoeutectoid 100 mm steel Fe3C pearlite proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e. Hypoeutectoid Steel Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

  15. T(ºC) 1600 d L 1400 (Fe-C g +L g g g System) 1200 L+Fe3C 1148ºC g g (austenite) g 1000 g g +Fe3C g g Fe3C (cementite) Fe3C 800 g g a g g 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 C, wt%C (Fe) pearlite Hypereutectoid 60mm steel pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e. Hypereutectoid Steel Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) C0 0.76

  16. T(ºC) 1600 d L 1400 (Fe-C g +L g Fe3C System) 1200 L+Fe3C 1148ºC (austenite) 1000 g +Fe3C Wg=x/(v + x) Fe3C (cementite) x v 800 g W=(1-Wg) g Fe3C a g g V X 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 C, wt%C (Fe) pearlite Wpearlite=Wg Wa= X/(V+X) Hypereutectoid 60mm W=(1 - Wa) steel Fe3C’ pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e. Hypereutectoid Steel Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) C0 0.76

  17. Hypo and hyper – lever rule

  18. Example Problem For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following: • The compositions of Fe3C and ferrite (). • The amount of cementite (in grams) that forms in 100 g of steel. • The amounts of pearlite and proeutectoid ferrite () in the 100 g.

  19. 1600 d L 1400 T(ºC) g +L g 1200 L+Fe3C 1148ºC (austenite) 1000 g +Fe3C Fe3C (cementite) 800 727ºC Amount of Fe3C in 100 g = (100 g)WFe3C = (100 g)(0.057) = 5.7 g R S 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 C0 C C, wt% C CFe C 3 Solution to Example Problem a) Using the RS tie line just below the eutectoid Ca = 0.022 wt% CCFe3C = 6.70 wt% C • Using the lever rule with the tie line shown

  20. 1600 d L 1400 T(ºC) g +L g 1200 L+Fe3C 1148ºC (austenite) 1000 g +Fe3C Fe3C (cementite) 800 727ºC V X Amount of pearlite in 100 g = (100 g)Wpearlite = (100 g)(0.512) = 51.2 g 600 a +Fe3C 400 C0 0 1 2 3 4 5 6 6.7 C C C, wt% C Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and realizing that C0 = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C

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