Fe-Fe 3 C phase diagram is given on the last page of the exam. Multiple choices (2.5 points each):

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1 Materials Science and Engineering Department MSE 200, Exam #3 ID number First letter of your last name: Name: No notes, books, or information stored in calculator memories may be used. Cheating will be punished severely. All of your work must be written on these pages and turned in. Mark your answer on this paper first, and then copy onto the answer sheet at the end of the test. Use #2 pencil to mark the answer sheet. Fe-Fe 3 C phase diagram is given on the last page of the exam. Multiple choices (2.5 points each): 1. A phase is defined as a matter with A. distinct composition B. distinct structure C. distinct structure and composition D. all of above 2. See the phase diagram of water. On the liquid/solid boundary line, the freedom is A. 1 B. 2 C. 3 D See the phase diagram of water. On the liquid/vapor boundary line, the number of phase is A. 0 B. 1 C. 2 D represents A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 1

2 5. As shown in the binary phase diagram, an alloy with a composition C 0 = 48% is cooled down from liquid to form solid. At the temperature T, C s =26%, C L = 72%, the fraction of the solid phase is A. 48% B. 52% C. 50% Answer: According to the level rule, the fraction of the solid phase is Therefore, choose B., 6. A peritectoid can be expressed as A. L α + β B. α β + γ C. L + α β D. α + β γ 7. In the phase diagram on the right, the three phase transition marked as 1 is A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 2

3 8. In the phase diagram on the right, the three phase transition marked as 2 is A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 9. In the phase diagram on the right, at the three-phase transition the freedom is A. 0 B. 1 C. 2 D In the phase diagram on the right, the number of component is A. 0 B. 1 C. 2 D In the phase diagram on the right, the eutectic point is A. 20% B. 80% C. 50% 3

4 12. In the phase diagram on the right, an alloy with 30%B is cooled from liquid, calculate weight fraction of α phase at just above the eutectic temperature A. 2/3 B. 1/3 C. 1/2 D. 2/5 Answer: The alloy cooled from liquid has a 30% weight fraction of B, so when it cools down, it gives rise to proeutectic phase first, Basing on the graph, above eutectic temperature, the solid weight fraction Cs=20%, liquid phase weight fraction Cl=50%. Co=30%. According to lever rule, the weight fraction of phase is X α = C l C 0 = C l C s Therefore, choose A. 13. In the phase diagram on the right, an alloy with 50%B is cooled from liquid, calculate weight fraction of α phase at slightly above the eutectic temperature A. 38% B. 48% C. 26% D. 0% 50% 30% 50% 20% = 2 3, Answer: The alloy cooled from liquid has a 50% weight fraction of B, so it gives rise to eutectic, which means that before the liquid hits the eutectic temperature and turns into phase, it will remains liquid. (The eutectic can be written as: ) Therefore, there are no phase at slight above the eutectic temperature, choose D. and 14. In the phase diagram on the right, an alloy with 40%B is cooled from liquid, calculate weight fraction of α phase in the eutectic structure with respect to the total alloy weight at slightly below the eutectic temperature A. 38% B. 48% C. 32% D. 0% 4

5 Answer: The alloy cooled from liquid has a 40% weight fraction of B, so when it cools down, it goes through two process: 1 st, it goes hypoeutectic, and will gives rise to proeutectic phase first, 2 nd, the remaining liquid phase will go through eutectic, forming and. We calculated them separately: At the point just before eutectic, the solid weight fraction Cs=20%, liquid phase weight fraction Cl=50%. Co=40%. According to lever rule, the weight fraction of remaining liquid phase is, During eutectic, it reacts as, Basing on the graph, below eutectic temperature the phase weight fraction =20%, phase weight fraction =78%. Co=50%. According to lever rule, the second time phase weight fraction is, is what we need because it is in the eutectic structure. Therefore, weight fraction of phase in the eutectic structure with respect to the total alloy weight at slightly below the eutectic temperature is,choose C 15. An alloy has a composition of 40%B. The microstructure after slow cooling from liquid to room temperature can be described by A). Fig. (a) B). Fig. (b) C). Fig. (c) Fig. (a) Fig. (b) Fig. (c) 16. In the Pb-Sn phase diagram on the right, an alloy contains 64 wt% proeutectic α and 36 wt% of eutectic α+β at 180 o C ΔT, find the average composition of this alloy A. 34.6% Sn B % Sn C. 64 % Sn D. 27.3% Sn Answer: Since the alloy contains proeutectic, so the average composition of this alloy must vary between 19.2%( ) and 61.9%( ). 5

6 We assign the average composition of the alloy Co. Co=34.6% Therefore, choose A. 17. In a peritectoid, on cooling A. liquid reacts to form two solid phases B. a solid phase reacts to form two different solid phases C. two solid phases react to form a solid phase D. a solid phase reacts to form a liquid and a different solid phase 18. In a peritectic, on cooling A. liquid reacts to form two solid phases B. a solid phase reacts to form two different solid phases C. a solid phase and a liquid react to form a solid D. a solid phase reacts to form a liquid and a different solid phase 19. See the Al-Ni phase diagram. In area (B) the phases that are present: A). Al 3 Ni+L B). L + AlNi C). L+Al 3 Ni See the Al-Ni phase diagram. The that occurs at 854 o C is (A) Eutectic (B) peritectic (C) peritectoid 21. In the Al-Ni phase diagram, the horizontal line marked as 1 represents A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 22. In the Al-Ni phase diagram, the horizontal line marked as 2 represents A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 23. In the Al-Ni phase diagram, the horizontal line marked as 4 represents A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 24. In the Al-Ni phase diagram, the horizontal line marked as 5 represents A. an eutectic three phase B. an eutectoid three phase C. a peritectic three phase D. a peritectoid three phase 25. In the Al-Ni phase diagram, the Al 3 Ni is A. congruently melting stoichiometric compound 6

7 B. incongruently melting stoichiometric compound C. intermediate non-stoichiometric phase D. intermediate stoichiometric phase 26. In the Al-Ni phase diagram, the Al 3 Ni 2 is A. congruently melting stoichiometric compound B. incongruently melting stoichiometric compound C. intermediate non-stoichiometric phase D. intermediate stoichiometric phase 27. In the Fe-Fe 3 C phase diagram, the eutectic point is A. 0.02% C B. 0.77% C C. 2.11%C D. 4.3% C 28. In the Fe-Fe 3 C phase diagram, the maximum solubility of the carbon in austenite is A. 0.02% C B. 0.77% C C. 2.11%C D. 4.3% C 29. See the Fe-Fe 3 C phase diagram (the carbide composition is 6.67%). A steel with 0.4 wt% C is first heated to form austenite, and then cooled down slowly, calculate the weight fraction of pearlite A. 50.7% B. 49.3% C. 40.2% D. 67.1% Answer: When steel is cooled down slowly, it will give rise to eutectic to form pearlite at the temperature of 727 o C, The steel contains 0.4 wt% carbon, so it will react as Co=0.4%, =0.02%, =0.77%, according to lever rule, Therefore, choose A. 30. See the Fe-Fe 3 C phase diagram (the carbide composition is 6.67%). A steel with 1 wt% C is first heated to form austenite, and then cooled down slowly, calculate the weight fraction of carbide at the temperature slightly below 727 C A. 3.9% B. 96.1% C. 85.3% D. 14.7% 7

8 Answer: When steel is cooled down slowly, it will give rise to eutectic to form pearlite at the temperature of 727 o C, The steel contains 1 wt% carbon, so it will react as. But the pearlite is composed of Fe3C(carbide) and Ferrite, So at the end, the whole structure is composed of Ferrite and Fe3C(carbide). Co=1%, =0.02%, =6.67%, according to lever rule, Therefore, choosing D. 31. See the Fe-Fe 3 C phase diagram. At a temperature slightly higher than 727 C a steel with 1 wt% C consists of A. austenite + carbide B. ferrite +carbide C. austenite + ferrite D. none of the above 32. See the Fe-Fe 3 C phase diagram. At a temperature slightly higher than 727 C a steel with 0.4 wt% C consists of A. austenite + carbide B. ferrite +carbide C. austenite + ferrite D. none of the above 33. See the Fe-Fe 3 C phase diagram. At a temperature slightly lower than 727 C a steel with 0.4 wt% C consists of A. austenite + carbide B. ferrite +carbide C. austenite + ferrite D. none of the above 34. The crystal structure of Martensite is A. fcc B. bcc C. bct D. hcp 35. The transformation of austenite to pearlite is A. diffusionless B. diffusion controlled C. none of the above 36. The transformation of austenite to ferrite is 8

9 A. diffusionless B. diffusion controlled C. none of the above 37. The CCT diagram below, cooling curve C will result in A. martensite B. coarse pearlite C. fine pearlite D. martensite + pearlite 38. The crystal structure of austenite is A. fcc B. bcc C. bct D. hcp 39. The limitations of plain carbon steel include A. low strength B. limited hardenability C. formation of perlite D. both A and B 40. In an aged Al-4% Cu alloy, the GP2 zone is A. Segregation of Cu atoms B. Coherent precipitates C. Non-coherent precipitates D. None of the above Answers: 1B 2A 3C 4D 5B 11C 12A 13D 14C 15C 21A 22C 23C 24A 25B 21A 32C 33B 34C 35B 6D 7A 8C 9A 10C 16A 17C 18C 19C 20B 26C 27D 28C 29A 30D 36B 37D 38A 39D 40B 9

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