Chapter5 hw

# Chapter5 hw - Chapter 5 Solution 3 r —.E"kT a In...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

Chapter5 hw - Chapter 5 Solution 3 r —.E"kT a In...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online