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Chapter9 hw

# Chapter9 hw - Chapter 9 Problem 12 A 0.65 percent C...

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Unformatted text preview: Chapter 9, Problem 12 A 0.65 percent C hypoeutectoid plain-carbon steel is slowly cooled from about 950°C to a temperature just slightly above 725°C. Calculate the weight percent austenite and weight percent proeutectoicl fe1rite in this steel. Chapter 9, Solution 12 0’ 7 0.02% C 0.65% c 0.80% c The weight percent austenite is calculated from the ratio of the segment of the tie line to the left of the 0.65 percent C‘ to the entire length of the tie line. -: _ Wt % austenite = 100% 2 Ex 100% = 80.8% 0.80—0.02 0.78 The weight percent proeutectoid fen‘ite is calculated from the ratio of the segment of the tie line to the right of the 0.65 percent C.‘ to the entire length of the tie line. . . . — . I . 5 Wt % proeutectord ferrite = MXIOOQ” = g><100% = 19.2% 0.80—0.02 0.78 Chapter 9, Problem 13 A 0.25 percent C hypoeutectoid plain-carbon steel is slowly cooled from 950°C to a temperature just slightly below TEB'Z'C. (0) Calculate the weight percent proeutectoid ferrite in the steel. (1)) Calculate the weight percent eutectoid ferrite and weight percent eutectoid cementite in the steel. Chapter 9, Solution 13 (a) The weight percent proeutectoid ferrite just below “93°C will be the same as that just above T23°C‘. It is therefore calculated based upon a tie line similar to that shown in P912. 0.80—0.25 0 55 Wt % proeutectoid ferrite = —><100% = '—>< 100% = 70.5% 0.80—0.02 0.78 (b) The weight percent total cementite and total FesCI Of ferrite are calculated based on the tie line A A shown to the right. f ‘r H“ o: l_’—l Fe 3C . 0.25 —0.02 0. 1' ‘ , “No.1” . 70/ Wt % total cententite : —><'100%: ﬁlbé‘l’o 2'3." (3% 6'6“ ’0 C 6.67—0.02 6.65 . 6.67—0.25 6.42 , Wt % total fen'ite = —x 100% = x100% = 96. 5% 6.67—0.02 6.65 The eutectoid ferrite is equal to the difference between the total ferrite and the proeutectoid fen‘ite: Wt % eutectoid ferrite = 96.5% - 70.5% = 26.0% Since the steel contains less than 0.8 percent carbon. no proeutectoid celnentite was formed during cooling. Thus. W' t % eutectoid celnentite = Wt % total cementite = 3.46% Chapter 9, Problem 14 A plain-carbon steel contains 93 wt % ferrite and 7 wt % Fe3C. What is its average carbon content in weight percent? Chapter 9, Solution 14 The average weight percent carbon is calculated based upon the tie line shown below. F€3C of 0.93 2—,) f A VF A K 6.6?—0.0;. a I I PCQC x =6.67—0.93(6.67—0.02) I l _ 0.02% C .T 9/3 C 6.670%) C x = 0.49% C Chapter 9, Problem 15 A plain-carbon steel contains 45 wt ‘34: proeutectoid ferrite. \Vhat is its average carbon content in weight percent? Chapter 9, Solution 15 0-45 : 0.80—x 0.80—0.02 x=0.80—0.45(0.80—0.02) x: 0.45% C 0.02% C x 9’6 C 0.30% p Chapter 9, Problem 16 A plain-carbon steel contains 5.9 wt % hypoeutectoid ferrite. What is its average carbon content? Chapter 9, Solution 16 The carbon content is related to the eutectoid ferrite by considering the proeutectoid ferrite and the total ferrite present after the eutectoid reaction. Wt % eutectoid ferrite = Wt % total ferrite — \Vt % proeutectoid ferrite FejC + y o: i Feat? V ‘ﬁ total Fe3C total or 0.02% C 0.80% C 6.6?le C 6.6773: 0.807x 6.67—0.02—080—002 6.67 x 0.80 x + 6.65 6.65 0.?8 0.?8 0.059=1.003—0.150x—l.026+l.282x 0.082 21.132x x = 0.0?2 01' 0.072% 0.059 : 0.059 — Chapter 9, Problem 25 A 1.10 percent C hypereutectoid plain-carbon steel is slowly cooled from 900°C to a temperature just slightly below TZB'Z'C. (a) Calculate the weight percent proeutectoid ceinentite present in the steel. (5) Calculate the weight percent eutectoid cementite and the weight percent eutectoid ferrite present in the steel. Chapter 9, Solution 25 l Fe3C or eutectoid or 0.02% C 0.80% C 6.67% C (a) The weight percent proeutectoid cententite will be: . . 1.70 — 0.80 0.9 . W’t % proeutectord cementite : —><1009/ : — x100°/o : 15.3% 6.67—0.80 5.87 (b) As depicted by the tie lines above. the eutectoid reaction will produce: Wt % eutectoid ferrite = ﬂ><100% = = 74.7% 6.67—0.02 6.65 Wt % eutectoid ceinentite = Wt % total cementite — Wt % proeutectoid cementite [ 1 .70 — 0.02 6.6? —0.02 = 9.96% X'1009/E):|—15.39/b Chapter 9, Problem 44 What types of microsttuct'ures are produced by tempering a plain-carbon steel with more than 0.2 percent carbon in the temperature ranges (a) 20—250c'C. (2)) 250—3 50°C. and (c) 400—60099? Chapter 9, Solution 44 (a) Between 30-2 00°C. very small epsilon (s) carbide precipitate forms. (b) Between BOO-300°C. cementite (1:63C ) precipitates which is rodlike in shape. (c) Between 400-700°C.‘. the rodlike carbides coalesce to form spherical particles. Chapter 9, Problem 52 Thin pieces of 0.3 mm thick hot-rolled strips of 1080 steel are heat-treated in the following ways. Use the IT diagram of Fig. 9.23 and other knowledge to determine the 111icrostructure of the steel samples after each heat trea ment. ((1) eat 1 h at 860°C; water-quench. (3)) eat 1 h at 860°C; water-quench: reheat 1 11 at 350°C. What is the name of this heat treatment'.7 (c) - eat 1 h at 860°C: quench in molten salt bath at 700°C and hold 2 11: water quench. (d) ’ eat 1 h at 860°C: quench in molten salt bath at 260°C and hold 1 min: air-cool. What is the name of this heat treatment? - eat 1 h at 860°C: quench in molten salt bath at 350°C: hold 1 h: air-cool. What is the name of this heat treatment? (f) Heat 1 11 at 860°C: water-quench; reheat l h at 700°C. Chapter 9, Solution 52 'C l basin-I! Gin“! ‘(.I. - A21 .' mm) W I-Too Ivan-no- “mm! 93L, ’5': .3- . i- EHI! 2!! 1200— l_ 1!: 3| — 600 "00 ISOTHERMAL TRANSFORMﬁTID‘N DIAGRAM 0F I000 — EUTEGTDiD CARBON STEEL c — 0.891 M4 29'; mun-mo nl w'f. [null 59: 4-3 W um um ‘lsoo 500 —' 400 700"“ . mu m 300 500 400 300 200 I00 TmE - SECONDS ® Figure 9.23 (a) nurtensite (b) tempered martensite, quenching and tempering process ((2) coarse pearlite (d) martensite, marquenching process (e) baini‘re= austempering (ﬂ spheroidite ...
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Chapter9 hw - Chapter 9 Problem 12 A 0.65 percent C...

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