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SAL BARRIGA: Hi, I'm Sal.

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00:00:23,310 --> 00:00:25,840
Today, we're going to
solve Problem #5 of

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00:00:25,840 --> 00:00:28,250
Exam 3 of Fall 2009.

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00:00:28,250 --> 00:00:30,250
Now before you start this
problem, there are a couple of

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00:00:30,250 --> 00:00:32,360
things that you need to know
in order for you to fully

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00:00:32,360 --> 00:00:35,680
understand the science
behind it.

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00:00:35,680 --> 00:00:40,210
And so what you need to know
is Fick's first law.

15
00:00:40,210 --> 00:00:43,030
Not just knowledge of it,
but how it works.

16
00:00:43,030 --> 00:00:44,850
Also, Fick's second law.

17
00:00:44,850 --> 00:00:47,440
And not necessarily the
equation form, but the

18
00:00:47,440 --> 00:00:50,110
solution to Fick's second law,
which is a function of

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00:00:50,110 --> 00:00:54,060
concentrations, and which
we will refer to

20
00:00:54,060 --> 00:00:55,150
to solve the problem.

21
00:00:55,150 --> 00:01:00,240
And also know popular
dislocations, how they help

22
00:01:00,240 --> 00:01:01,930
your material and whatnot.

23
00:01:01,930 --> 00:01:04,010
So the problem reads
as follows.

24
00:01:04,010 --> 00:01:07,300
Specimens of steel are being
surface hardened by the

25
00:01:07,300 --> 00:01:08,940
introduction of carbon.

26
00:01:08,940 --> 00:01:11,630
The concentration of carbon at
the free surface of the steel

27
00:01:11,630 --> 00:01:14,480
is kept constant at c sub s.

28
00:01:14,480 --> 00:01:17,460
Chemical analysis of a number
of specimens indicate that

29
00:01:17,460 --> 00:01:18,430
after a time--

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00:01:18,430 --> 00:01:19,640
t sub 1--

31
00:01:19,640 --> 00:01:23,920
the carbon concentration has the
value of c star at a depth

32
00:01:23,920 --> 00:01:25,710
from the surface-- x1--

33
00:01:25,710 --> 00:01:28,900
shown below, which is
shown on your graph.

34
00:01:28,900 --> 00:01:30,450
So this is given to you.

35
00:01:30,450 --> 00:01:33,090
So this pretty much is your
concentration profile as a

36
00:01:33,090 --> 00:01:35,110
function of position.

37
00:01:35,110 --> 00:01:38,080
And it's telling you
what's happening.

38
00:01:38,080 --> 00:01:42,470
So a good thing to do is draw
a picture of the science

39
00:01:42,470 --> 00:01:44,730
that's happening here.

40
00:01:44,730 --> 00:01:45,830
And they tell you, first of
all, I'm going to draw it

41
00:01:45,830 --> 00:01:48,710
underneath here that you
have a rod of steel.

42
00:01:51,490 --> 00:01:55,980
And so this is steel.

43
00:01:55,980 --> 00:01:59,410
And I have a certain fixed
concentration here at the

44
00:01:59,410 --> 00:02:00,320
surface of carbon.

45
00:02:00,320 --> 00:02:08,140
So I have some type of a carbon
flux in my surface.

46
00:02:08,140 --> 00:02:10,740
And this is a visual image
of how it looks.

47
00:02:10,740 --> 00:02:16,102
And what the problem says is
that you have a certain

48
00:02:16,102 --> 00:02:19,170
concentration on your surface,
which is given by c sub s.

49
00:02:19,170 --> 00:02:22,610
And then it starts telling you
that chemical analysis was

50
00:02:22,610 --> 00:02:25,590
able to go ahead and measure
this concentration at two

51
00:02:25,590 --> 00:02:29,400
different values inside your
position at different times.

52
00:02:29,400 --> 00:02:33,470
So if I continue reading the
problem, it says that chemical

53
00:02:33,470 --> 00:02:35,930
analysis of a number of
specimens indicate that after

54
00:02:35,930 --> 00:02:37,190
a time-- t1--

55
00:02:37,190 --> 00:02:39,510
the carbon concentration
has a value of c star.

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00:02:39,510 --> 00:02:42,100
So this one right here--

57
00:02:42,100 --> 00:02:43,190
this is x1--

58
00:02:43,190 --> 00:02:49,490
so this has a value
of x1 at t1.

59
00:02:52,330 --> 00:02:57,800
And it also tells you that after
a certain time-- t2,

60
00:02:57,800 --> 00:03:00,410
which is equal to 2 times t1--

61
00:03:00,410 --> 00:03:03,750
the carbon concentration has a
value of c star, so the same

62
00:03:03,750 --> 00:03:06,090
value at a depth of x2.

63
00:03:06,090 --> 00:03:12,560
So now, this thing says that at
a depth of x2, at a time of

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00:03:12,560 --> 00:03:17,910
t2, we measure the same
concentration, c sub star in

65
00:03:17,910 --> 00:03:18,420
the specimen.

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00:03:18,420 --> 00:03:26,080
And it tells you that x2 equals
2x1, and it also tells

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00:03:26,080 --> 00:03:32,460
us that t2 equals 2t1.

68
00:03:32,460 --> 00:03:35,100
Now they tell us that for a
reason, because it's probably

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00:03:35,100 --> 00:03:40,190
expected for us to use this to
solve the problem, so don't

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00:03:40,190 --> 00:03:42,270
forget that.

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00:03:42,270 --> 00:03:45,300
So the question is, under these
conditions, would you

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00:03:45,300 --> 00:03:49,310
describe the rate of
carburization as one, steady

73
00:03:49,310 --> 00:03:50,770
state diffusion.

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00:03:50,770 --> 00:03:52,850
Two, transient state
diffusion.

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00:03:52,850 --> 00:03:57,140
Or three, not governed by any
diffusion, i.e., rate limited

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00:03:57,140 --> 00:03:58,270
by some other process.

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00:03:58,270 --> 00:04:00,290
Now you have to justify
your choice.

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00:04:00,290 --> 00:04:02,680
So this is pretty much a
problem of process of

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00:04:02,680 --> 00:04:03,920
elimination.

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00:04:03,920 --> 00:04:06,065
And for part a, which is--

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00:04:06,065 --> 00:04:08,120
I'll label this part a--

82
00:04:08,120 --> 00:04:09,610
it wants us to do
three things.

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00:04:09,610 --> 00:04:14,380
So the first thing, it asks
us if it's steady state.

84
00:04:14,380 --> 00:04:20,220
So the first thing to ask
yourself is, well, what's

85
00:04:20,220 --> 00:04:21,180
steady state?

86
00:04:21,180 --> 00:04:24,680
And the definition of steady
state is that whatever

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00:04:24,680 --> 00:04:28,080
phenomenon is happening doesn't
vary with time.

88
00:04:28,080 --> 00:04:30,810
But the problem already tells
you that it varies with time.

89
00:04:30,810 --> 00:04:33,200
It tells you that at a certain
time you measure this and at

90
00:04:33,200 --> 00:04:36,140
another time you measure this,
and you have this flux coming

91
00:04:36,140 --> 00:04:38,210
in, and this is what's
happening.

92
00:04:38,210 --> 00:04:41,510
So from just reading the
problem, I can already say

93
00:04:41,510 --> 00:04:45,840
that this process can't be
steady state, because your

94
00:04:45,840 --> 00:04:48,460
concentration profile varies
as a function of time.

95
00:04:48,460 --> 00:05:04,580
So for one, I would write, can't
be steady state due to

96
00:05:04,580 --> 00:05:11,930
the concentration being
a function of time.

97
00:05:11,930 --> 00:05:15,140
So one is out, so we
can scratch that.

98
00:05:15,140 --> 00:05:17,730
It's not a steady state.

99
00:05:17,730 --> 00:05:19,180
Now two.

100
00:05:19,180 --> 00:05:22,060
This is where it gets a little
bit more complicated.

101
00:05:22,060 --> 00:05:24,780
Two is a transient
state diffusion.

102
00:05:24,780 --> 00:05:29,000
Well, how do we know whether
or not this is

103
00:05:29,000 --> 00:05:29,780
governed with time?

104
00:05:29,780 --> 00:05:34,190
Well, you would argue, because
this is not steady state, then

105
00:05:34,190 --> 00:05:35,920
it has to be time dependent.

106
00:05:35,920 --> 00:05:42,330
But, there are certain rules
that you need to know to be

107
00:05:42,330 --> 00:05:44,040
able to make that
justification.

108
00:05:44,040 --> 00:05:48,080
First of all, if it's time
dependent, then whatever your

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00:05:48,080 --> 00:05:51,140
concentration and whatever your
position in your time is,

110
00:05:51,140 --> 00:05:55,460
has to fit into your solution
to Fick's law.

111
00:05:55,460 --> 00:05:59,810
So if I look at my solution to
Fick's law right here, I know

112
00:05:59,810 --> 00:06:03,300
that on the left side of
my equation, I have the

113
00:06:03,300 --> 00:06:05,440
concentration at any
point in x--

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00:06:05,440 --> 00:06:07,170
that's what c stands for--

115
00:06:07,170 --> 00:06:09,060
minus the concentration
at my surface,

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00:06:09,060 --> 00:06:12,560
which is given as cs--

117
00:06:12,560 --> 00:06:16,660
sorry, this is supposed
to be sub s--

118
00:06:16,660 --> 00:06:18,580
divided by the initial
concentration.

119
00:06:18,580 --> 00:06:21,300
So c sub naught is if there was
already some carbon in my

120
00:06:21,300 --> 00:06:25,480
steel, minus, again, the
surface concentration.

121
00:06:25,480 --> 00:06:27,590
This equals to an
error function--

122
00:06:27,590 --> 00:06:30,140
which is the solution to
Fick's second law--

123
00:06:30,140 --> 00:06:33,880
and this error function has
the values of position,

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00:06:33,880 --> 00:06:35,840
diffusion coefficient,
and time.

125
00:06:35,840 --> 00:06:43,180
So in order to verify if it's a
transient state, we're going

126
00:06:43,180 --> 00:06:45,800
to go ahead and take that
equation and break it down.

127
00:06:45,800 --> 00:06:49,500
So if I assume that there's no
carbon, I can go ahead and

128
00:06:49,500 --> 00:06:52,690
write this down.

129
00:06:52,690 --> 00:06:57,360
Assume c0 is 0, just to make
the equation easy.

130
00:06:57,360 --> 00:07:00,410
I can, because there was no
information given about that.

131
00:07:00,410 --> 00:07:02,390
I can go ahead and--

132
00:07:02,390 --> 00:07:04,100
because we're trying
to figure out if

133
00:07:04,100 --> 00:07:06,830
it's a transient state--

134
00:07:06,830 --> 00:07:07,770
write this down.

135
00:07:07,770 --> 00:07:13,340
So c minus c sub s over
negative c sub s.

136
00:07:13,340 --> 00:07:18,380
This equals to my error
function, which is x divided

137
00:07:18,380 --> 00:07:19,630
by 2 squared of Dt.

138
00:07:22,050 --> 00:07:28,220
And I know that at x1
and time-- t1--

139
00:07:28,220 --> 00:07:32,240
I have this value of my
concentration, c sub star.

140
00:07:32,240 --> 00:07:36,800
So I'm going to go ahead
and plug that in

141
00:07:36,800 --> 00:07:38,030
here into this equation.

142
00:07:38,030 --> 00:07:40,530
So I know that my concentration
is c sub star.

143
00:07:40,530 --> 00:07:48,600
So that's what we measured, c
sub star minus c sub s over

144
00:07:48,600 --> 00:07:50,680
negative c sub s.

145
00:07:50,680 --> 00:07:59,815
This equals to my error
function, x1 over 2 square

146
00:07:59,815 --> 00:08:01,460
root of Dt1.

147
00:08:04,680 --> 00:08:09,510
So this is for the first
time point, x1, t1.

148
00:08:09,510 --> 00:08:13,210
Now I want to go ahead
and do the same thing

149
00:08:13,210 --> 00:08:15,980
for the second point.

150
00:08:15,980 --> 00:08:19,630
And if I do it for the second
point, I know that just by

151
00:08:19,630 --> 00:08:25,330
looking at the equation, I have
the same concentration at

152
00:08:25,330 --> 00:08:29,000
x2, and I have the same
surface concentration.

153
00:08:29,000 --> 00:08:34,670
So what that tells me is that
this value is going to be the

154
00:08:34,670 --> 00:08:41,430
same for x1, t1, and x2, t2.

155
00:08:41,430 --> 00:08:46,950
So if I go ahead and plug
this into my equation.

156
00:08:46,950 --> 00:08:53,190
I can go ahead and look at this
and say, this is for t

157
00:08:53,190 --> 00:08:57,020
equals 1 for that, and
this is going to be--

158
00:08:57,020 --> 00:09:00,752
or, t equals t sub 1, sorry--
and this is t sub 2.

159
00:09:00,752 --> 00:09:05,650
I have c star minus cs
all over negative cs.

160
00:09:05,650 --> 00:09:14,280
This equals to my error function
of x2 over 2 square

161
00:09:14,280 --> 00:09:15,530
root of Dt2.

162
00:09:19,110 --> 00:09:24,190
So I know that this value is
the same as this value.

163
00:09:24,190 --> 00:09:28,850
So therefore, this should
be the same as this.

164
00:09:28,850 --> 00:09:32,020
The argument in both your
error functions.

165
00:09:32,020 --> 00:09:33,640
So we have to go ahead
and prove that.

166
00:09:33,640 --> 00:09:34,870
Well, how are we going
to do that?

167
00:09:34,870 --> 00:09:37,880
Do I know x2 as a
function of x1?

168
00:09:37,880 --> 00:09:41,680
I do, because that's
given in the data.

169
00:09:41,680 --> 00:09:52,220
And I know again, that x2 equals
2x1 and t2 equals 2t1.

170
00:09:52,220 --> 00:10:03,740
So what I want to know, does x1
over 2 root Dt1, does this

171
00:10:03,740 --> 00:10:12,590
equal to x2 over 2 root Dt2?

172
00:10:12,590 --> 00:10:15,560
Well, I'm going to go ahead and
plug in my values that I

173
00:10:15,560 --> 00:10:25,490
know, and this becomes 2x1
divided by 2 times the square

174
00:10:25,490 --> 00:10:27,530
root of D--

175
00:10:27,530 --> 00:10:30,100
and t2 happens to be 2t1.

176
00:10:33,220 --> 00:10:38,250
So those two cancel each other
out, and I end up getting that

177
00:10:38,250 --> 00:10:45,225
this is actually equal to x1
over square root of 2Dt1.

178
00:10:48,270 --> 00:10:55,400
Now this and this are
not the same.

179
00:10:55,400 --> 00:10:59,300
Because this has a square root
of 2, where the other

180
00:10:59,300 --> 00:11:02,130
one is just a 2.

181
00:11:02,130 --> 00:11:05,730
So because it did not satisfy
the solution to Fick's second

182
00:11:05,730 --> 00:11:09,940
law, we can say that it's
not case two either.

183
00:11:09,940 --> 00:11:14,770
This is not transient
state diffusion.

184
00:11:14,770 --> 00:11:17,380
So therefore, by the process of
elimination, it has to be

185
00:11:17,380 --> 00:11:19,080
governed by some other
phenomena.

186
00:11:19,080 --> 00:11:20,450
And that is the answer
to this problem.

187
00:11:20,450 --> 00:11:25,190
The fact that it's governed by
something that we don't have

188
00:11:25,190 --> 00:11:26,590
enough information to answer.

189
00:11:26,590 --> 00:11:29,390
We don't have enough data
points to answer.

190
00:11:29,390 --> 00:11:32,580
So that's part a,
which is good.

191
00:11:32,580 --> 00:11:41,350
Now part b asks, if the steel
specimens in part a were

192
00:11:41,350 --> 00:11:44,480
replaced with steel specimens
of identical composition but

193
00:11:44,480 --> 00:11:47,900
with a dislocation density 10
times greater than that of the

194
00:11:47,900 --> 00:11:50,260
steel in part a, how
would the rate of

195
00:11:50,260 --> 00:11:51,760
uptake of carbon change?

196
00:11:51,760 --> 00:11:53,650
Explain.

197
00:11:53,650 --> 00:11:56,340
Again, this requires you
to know properties of

198
00:11:56,340 --> 00:11:57,620
dislocations, and what

199
00:11:57,620 --> 00:11:59,640
dislocations do to your materials.

200
00:11:59,640 --> 00:12:00,890
Your metals, for example.

201
00:12:04,420 --> 00:12:05,610
What is a dislocation?

202
00:12:05,610 --> 00:12:09,430
Well, a dislocation is--
if you have a bunch

203
00:12:09,430 --> 00:12:12,050
of a plane of atoms--

204
00:12:12,050 --> 00:12:15,230
what a line dislocation is, or
an edge dislocation is, this

205
00:12:15,230 --> 00:12:18,110
line defect such that
it just terminates.

206
00:12:18,110 --> 00:12:20,010
So you have a plane of atoms
that just comes to a point and

207
00:12:20,010 --> 00:12:21,090
then it stops.

208
00:12:21,090 --> 00:12:23,200
And what do you create here?

209
00:12:23,200 --> 00:12:27,240
Well it creates void, and void
facilitates diffusion.

210
00:12:27,240 --> 00:12:28,490
It increases--

211
00:12:30,540 --> 00:12:33,970
in our case, we're talking
about carbon

212
00:12:33,970 --> 00:12:36,000
diffusion into steel--

213
00:12:36,000 --> 00:12:41,430
so I would argue that the more
dislocations that you have,

214
00:12:41,430 --> 00:12:45,500
the greater the diffusion
should be.

215
00:12:45,500 --> 00:12:47,660
Because you have more
void-- more spaces--

216
00:12:47,660 --> 00:12:51,540
for your carbon atom to
jump into and from.

217
00:12:51,540 --> 00:12:56,260
So for part b, just by arguing
the fact that dislocations

218
00:12:56,260 --> 00:13:00,090
represent defects in
your crystal--

219
00:13:00,090 --> 00:13:03,900
specifically a defect that
can create void--

220
00:13:03,900 --> 00:13:09,050
then with that argument, the
fact that more void means

221
00:13:09,050 --> 00:13:14,030
higher diffusion, I would
conclude that--

222
00:13:14,030 --> 00:13:15,190
for this part--

223
00:13:15,190 --> 00:13:39,340
that my rate of diffusion of
carbon should increase when

224
00:13:39,340 --> 00:13:50,390
using steel of a greater
dislocation density.

225
00:13:58,220 --> 00:14:02,025
So the rate of carbon uptake
or diffusion of carbon into

226
00:14:02,025 --> 00:14:05,620
your steel increases
as you increase

227
00:14:05,620 --> 00:14:07,920
your dislocation density.

228
00:14:07,920 --> 00:14:12,600
And with that, you're to be able
to solve a lot of these

229
00:14:12,600 --> 00:14:15,110
problems. And these are
common in 3.091.

230
00:14:15,110 --> 00:14:19,150
And again, knowledge of Fick's
first and second law, how to

231
00:14:19,150 --> 00:14:23,910
use the solutions to Fick's
first, second law, and knowing

232
00:14:23,910 --> 00:14:26,150
what a dislocation is, that will
help you a lot in solving

233
00:14:26,150 --> 00:14:27,550
these problems.