Chapter5 hw

Chapter5 hw - Chapter 5, Solution 3 . . . . . r -...

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Unformatted text preview: Chapter 5, Solution 3 . . . . . r - —.E..-"kT a) In general. the equlllbrnun number of vaca11c1es is NV : 1V (.6 ‘ . For copper. 31-11)Cu _ (6.02><1023 atomsfat. 111ass)(8.96><106 gating) at. mass Cu (63.54 gfat. mass) Substituting and assuming EV = 1.00 eV' at 1123 K. nu =(849x1028 atoms/1113) exp 15'00 EN (8.62x10' eVr‘K)(1123 K) N — — 8.49 ><1028 ato1ns..-"1113 = 2.77 X 1024 vacancies/m3 b) The vacancy fraction at 1073 K is. ' *1 .00 V _ H" * exp 5 .e i e 1081 i 2.02 X 10'5 vacancies/atom 1V (8.62X10’ eVi-"KXIO73 K) Chapter 5, Problem 13 Consider the gas carburizing ofa gear of 1018 steel (0.18 wt 9/0) at 922°C (1200913). Calculate the time necessary t< increase the carbon content to 0.35 wt 93 at 0.40 111111 below the surface of the gear. Assume the carbon content at t] surface to be 1.15 wt % and that the nominal carbon content of the steel gear before carburizing is 0.18 wt 94'). D (C in yiron) at 922°C = 1.28 x 10'11 Inlfs. Chapter 5, Solution 13 The time required for this diffusion process is calculated using Fick‘s second law. C—C. . ‘ l=erf r CS —C‘o 2J1? where: CS 21.15% CO 20.18% Cl. 2 0.35% x = 040111111 = 4X104 111 D =1.28><10_11 1112 s 922°C S_ s —4 Substituting 1'1“ 0'31 _erf 4X10 111q “"‘0-18 2 (1.28x10'11111‘..-"s)r as 0.824? = erf [ y y '90] = erf: J? 1 Interpolating from Table 5.3. 0.8247 — 0.8209 _ :r —0.95 — x 20.959 0.8427 7 0.8209 1.0 i 0.95 el'f z z Thus, ss 0.8209 0.95 - = " " '90 = 0.959 0.8247 .1 - J; 0-8427 1-0 123397.? s= 56.6mm. Chapter 5, Problem 16 A gear made of 1020 steel (0.20 wt % C‘) is to be gas-carburized at 92 7'5'C (1700"F). Calculate the carbon content at 0.040 in. below the surface of the gear after a 70-hour carburizing time. Assume the carbon content at the surface of the gear is 1.15 “1%. D (C in yiron) at 92T'Z'C.‘ = 1.28 X 10'11 1112.55. Chapter 5, Solution 16 D(C1'11}.'1ro11)at92T°C = 1.28 x 10 '11 1112.15. Given: CS 21.00% CO 20.20% Cl. 2 1’ I: 4 h = 14,400 3 x = 009111111 = 9.0x104 111 D9350 =1.28><10_11 111215 q-q_ 115—01. _el_f 1.02x10—3 1n ("s—Co 1-15—020 2 (1.28x10'111112..-"'s)(25.2005) 1.15—(3. = e1f(0.89?98) 0.95 Interpolating from Table 5.3. 089798—085 . —0.7707 91-1.: 5 = Y x=0.?959 0.90—0.85 (17970—071707 0.??0? 0.85 I _ A 1- 0.89T9g Thus. 0.7959 — e11 (0.89798) 0.?9?0 090 Substituting. 1.15 — C I = 0.7959 C1. = 0.394 wt % 0.95 Chapter 5, Problem 21 If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100'Z'C‘ for 6 11. what is the depth below the surface at which the concentration is 101‘5 atoms.--"cm3 if the surface concentration is l 013 atoms-“c 1113‘? _ 1! 2. . . . . . . . D = 2 X 10 1‘ cm .-'s for ahuninunl difhtsrng 1n slhcon at 1100C. Chapter 5, Solution 21 Given: CS =1018 attomst’cm3 CK =1016 atoms-fun3 C0 = 0.0 r260 h=2.16x10“ s 91100.? =..0><10‘12 cm2 CS—Q. _1013—1015 3* C—C 1018—0 _erf I 42 2= 4 '5 '0 2 (2.0X10 cm ts)(2.16><10 s) 0.99 : erf ]— e1‘f: 4.157x10‘4 [ntetpolating from Table 5.3. 0.9900 _ 0-9891 x1 _1-8 2 X1 21.824 0.9928 —0.9891 1.9 — l .8 erfz z Substituting. 0.9891 1.8 0-9900 3' : 21.824 =—‘T “4 0.9928 1.9 4.157X10 x = 7".58x10—4 cm Chapter 5, Problem 22 Phosphorus is diffused into a thick slice of silicon with no previous phosphorus in it at a temperature of 1100'5'C‘. If the surface concentration of the phosphorus is l x 1018 atonisfcm3 and its concentration at l ,(1111 is l X 1015 at01115.-"'c1113. how long must the diffusion time be? D = 3.0 X 10'1" cnfis for P diffusing in Si at 1100'5'C. Chapter 5, Solution 22 Given: CS 21018 atonisr'cni3 (TX 21015 atomsr'cm3 ('0 = 0.0 . _ . —4 _ —13 2 I: , x210 ‘U111— l.0><10 c111 40110050 —3.0><10 c111 s C‘s—Cl. _1013—1015 _erf 10—4 cm (Ts—Co 1018—0 2 (3.0x10'13 mugs)! 0.999 = e1‘f[91'28? J: erf: J? Interpolating from Table 5.3. 09990 _ 09981 _ x _ 22 r _ 2 3S 09993—09981 2.4—2.2 erfz z Thus, 2f I:[91.28?]z :[91.28?]2 0.9993 2.4 I 2.35 I: 1508 s = 25.1 min. Chapter 5, Problem 25 Calculate the diffusivity D in square meters per second for the diffusion of nickel in PC C iron at 1100'Z'C. Use values ofDo = T? X 10" cmzx's; Q = 280 kanlol: R = 8.314 J.-"'(11101- K). Chapter 5, Solution 25 The diffusivity of the nickel into FCC iron at 133 K is: -280._000 Jr'nlol D :1) 9‘9“? =(7.7><10‘5 1112 s) exp . [8.314 J.-"(11101‘ K)](13?3 K) 0 = (?.?x10_5 1n2..*"s)(e'24'53) = 1.71 x 10’” mli's ...
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Chapter5 hw - Chapter 5, Solution 3 . . . . . r -...

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