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Unformatted text preview: 5.9 This problems asks for us to compute the diffusion ﬂux of nitrogen gas through a 15min thick plate
of iron at 300°C when the pressures on the two sides are 0.10 and 5.0 MPa. Ultimater we will employ Equatior.
5.3 to solve this problem. However, it ﬁrst becomes necessary to determine the concentration of hydrogen at eacl’. face using Equation 5.11. At the low pressure (or B) side CNIB] = (4.90 x 10'3)«.t0.10 MPa mil—WA
~ _ (s31 J.'inolK)(3OO — 273 K) 5 a? x 107 “1% Whereas, for the high pressure (or A) sicle :— 31600 J mol _ —3
C _ (4'90 X 10 M50 MPB‘ exprtsai J.:"molK)(300 — 273 K) N (A) 4.03 x 105 “1% We now convert concentrations in weight percent to mass of nitrogen per unit volume of solid. At face B there are 537" x 10'? g (or 5.77 x 10—10 kg) ofhydrogen in 100 g ofFe. which is Virtually pure iron. From the density of iroi.‘
(TS? gr’crn3), the 1.xoluine iron in 100 g (VB) is just EB —&= 12.?cm3 =12? x 10'5 in 7.3? gt'cm3 3 Therefore, the concentration of hydrogen at the 3 face in kilograms of N per cubic meter of alloy [ C; (3)] is just a? 40 a —
= —5“' ‘10 ke: 4.54x10'3 kg."1n3 1.2.? >< 105 m3 At the A face the volume of iron in 100 g (VA) will also be 1.27" x 10—5 ms. and  may _9
2 M 2 321 x 104 ragsn? 1.27 X10_51113 Thus. the concentration gradient is just the difference between these concentrations of nitrogen divided by the thickness of the iron membrane; that is f: 113
£ 2 CNtB] Circa;
30: IB — IA —5 3_~. —4 3
= 4.54x10 kgm 3.212110 kgin =_0'184 kghﬁ 1.5 x 10—3 m At this time it becomes necessary to calculate the value of the difﬁision coefﬁcient at 300°C using Equation 5.8.
Thus.
0 D = D _;n’;
Oexpt RT} \ = (3.0 X 10—? 1n2 exp:——I “also IiiIncl \ (8.31J.'1nol—K)(3UD — 2?} K); = 3.40 x 1014 1113:; And, fmally, the diffusion flux is computed using Equation 5.3 by taking the negative product of this diffusion coefficient and the concentration gradient. as AC
it J=—D = — (3.40 x 1014 1112.53) (— 0.134 leg"1114) = 6.25 x 1015 Lagm? 4. 5. ll we are asked to compute rhe carburizing (i.e., difﬁlsion) time required for a speciﬁc nonsteady—srate difﬁiaion situation. It is ﬁrst necessary to use Equation 5.5: —'I 0: 1 —erf L;
cs — CO .200r wherein. CI = 0.30, CD = 0.10. C5 = 0.90, andx = 4 min = 4 x 10—3 1n.Thus, M = __
q.— Co 0.90 — 0.10 0'30 _ 0'10 — 0.2500 = 1 — Eff: '7‘
900:; 01' . l
erf ‘ 2'05 = 1 — 0.2500 = 0.?500 B}? linear interpolation using data from Table 5.1 g erf:
0.80 0.?431 2 0.?500
0.85 0.??0? 2 — 0.800 _ 0.?500 — 0.?421
0.850 — 0.800 0.??‘0? — 0.7421 From which Now. from Table 5.2. at 1100°C (137'3 K) 148,000 J mol D = (2.3 =< 10'5 1:13.55) exp —f.
{0.31 J.'1nolK)[13?3 L.)_ = 5.35;;10'11 111353 Thus , 4 x 10—3 n1 (2)«,;(5.35 >< 1011 1:12 mm Solving for F}'i€1d'3 :=1.13x105 5:313 h 5.15 problem calls for an estimate of the time necessar},r to achieve a carbon concentration of 0.35 wt% at: a point 6.0 111111 ﬁom the sulface. From Equation 5.61:, 1
I
— = constant
Dr But since the temperature is constant, so also is D constant, and T
— = COﬂStﬂJJI
:
01'
.2 2
L1 = :2
F1 F2
Thus,
(2.0 mm)2 = (6.0 mm}2
15 1: t2
from which r2=135h 5.21 (a) Using Equarion 5.9a. we set up two simultaneous equations with a, and D0 as Lmknowns as follows: 111: — i
R 1mm =111DO—& ii
4 R K13; Now. solving for Qa. in terms oftemperatures T1 and T3 {14?} K and 16?} K) and D.l and D: (2.2 x 10'15 and 4.8 x :10"H 1n2."s). we get 0 _ RlnDl—lnD2
._.d—_ —i—i
s 13 Erin: x 1045) — m(4.s X 1044)]
1 _ 1
14.133: 16?3K = — (3.31 J.'mo1— K} = 315300 J."mol Now. solving for D0 from Equation 5.8 (and using the 1473 K value of D) :{L .31] '~,
D0 = D1 exp ‘ : (22 X 1045111335)?pr 31?.700 JImol‘ I
_(8.3]. Ismol—leﬂa K)_ = 3.5 x 1LT4 1112.55 (b) Using these values ofD and Q ,9 at 15??» K is just
0 0’ i? D = (3.5 K 210—4 m3."s)expl—31f$m,l
_ (8.31 .T.'molK)(15?3 KL = 1.1 3:19—14111253 6.4 we are asked to compute the inaximmn length of a cylindrical nickel specimen (before deformation) that is deformed elastically in tension. For a cylindrical specimen its d“
at at  0.15 r ' WT; AIL“nag
to =  a Z: —
a g F F 4F
E An _ {0.25 >< 1041:1100? >< 109N‘1112j (n) (10.2 x 1031103
(4)(3900 N) = 0.435 in =45 mm (18.7" in.) 6.9 This problem asks that we calculate the elongation ﬁtt of a specimen of steel the stressstrait] behavior of which is shown in Figure 6.21. First it becomes necessary to compute the stress when a load of 65,250 N is applied using Equation 6.1 as G :27 = {Fuz = I 51250 1‘ 5,, = 115030001 (1?0,000psi)
0 T0_0 8.5x10'31n'
J r} it —_ I _ 2 _. '\ Referring to Figure 6.2]. at this stress level we are in the elastic region on the stressstrait] curve, which corresponds to a strain of 0.0054. Now, utilization of Equation 6.2 to compute the yalue of if a: = 350 = (0.0054)(80 mm) = 0.43 min (0.01? in.) 6.22 This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile load of 35.000 N is applied; this
means that the stress corresponding to this load not exceed the yield strength of the material. Upon computing the SHESS G _ i: _ fr“  31000 T‘ a — 200x 1061mm; = 200 ma
0 larCI '_ 15 x10_31nl_
1. It I
t 2 .5 . 2 _.I
0f the alloys listed, the Al, Ti and steel alloys have yield strengths greater than 200 MPa. . . . . _'1 . .
Relative to the second criterion (1e, that id be less than 1.2 x 10 ‘ mm). 1t is necessary to calculate the change in diameter for these three alloys. From Equation 6.3 a
1... = _3_x 2 _d_o=_EM
a: E 0 (£0
E
Now. solving for Ad from this expression.
v o d
M = — '5'
E
For the aluminum allo},r
ﬂn' = ——(U'JJ)QUO MPaKb mm) = —l.4l 1103 mm 70 x103MPa Therefore, the Al alloy is not a candidate. For the steel alloy _ (0.23000 MPa){lS mm) = —0.40 310'2 mm
:05 11(103 MPa M: Therefore, the steel is a candidate. For the Ti alloy _(o.35)(200 Mpamﬁ mm) = 3 —1.0x10'31mn
105x10 am Mr: Hence, the titanium alloy is also a candidate. 6.30 This problem calls for the computation of ductility in both percent reduction in area and percent elongation. Percent reduction in area is computed using Equation 6.l2 as 1 a; :‘d 33 a _ rtl—U‘ — 'E fl . 2 2 . aan = I \1 ' x100 l' ‘
n —
'x2 2' in which (in and (ifare. respectively. the original and fracture cross—sectional areas. Thus, .1 2 "’128 1nle {3.13 mm"
1T — TE —
. a 2 ,e l 2 ,
9'0RA — _. E x 100 — 60%
[12.8 1ntn''
I — 2 ..
\Vltile, for percent elongation. we use Equation 6.11 as
"'t — f "1
“a; = f _ O t x 100
1'0 '_I
7' — 5 .
_ 11.1. mm 080 mm X100 _ 46% 50.30 mm ...
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 Spring '08
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 Tensile strength

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