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Solutions to Homework # 2
Problem 1
a)
Calculate
Dt
and compare with the sample size to see if we should use shorttime or longtime
solution.
Since
Dt
cm
cm
=× × =
<
<
−
11
0
1
0
0 0
0
1
1
6
.
we should apply the shorttime solution. One of the two surfaces at x=0 is schematiclly shown
in Figure.
x
C
o
C
s
= 0
c
x
=0
The initial and boundary conditions are given by:
cx
c
g cm
ct
ctc
(,)
/
01
1
0
00
0
53
0
==×
=
∞=
−
The general error function solution is
cxt
A Ber
f
x
Dt
(
)
=+
2
Using the above boundary conditions, the constant A and B are given by:
A
= 0
B
=
c
o
Therefore, the concentration profile after 100s is:
cer
f
x
Dt
erf
x
Dt
(
)
(
)
==
×
−
0
5
2
0
2
b)
The surface flux is given by
JD
c
x
c
D
t
g
cm
s
xx
−
−
−
=−
=− ×
×
×
×
0
5
6
10
2
0
0
100
564
10
∂
∂π
π
.(
.
)
c)
The total amount of hydrogen diffused out of the nickel at 100s is given by
mJ
d
t
c
D
t
dt
c
Dt
x
== =
=
∫∫
22
4
0
0
100
0
0
100
0
ππ
=×
×
×
−
−
−
41
0
10
100
226
5
6
7
2
g
cm
)
d)
This is a longtime problem. Since
1
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c
−
−
−
==
<
0
6
5
10
10
01
08
..
we may apply the first term of series solution.
c
c
Dt
h
−
=−=
0
2
2
2
8
π
exp(
)
.
Solve the above equation, we get t
t
h
D
s
=−
×
×
=×
−
2
2
2
26
2
5
8
1
10
80
212
10
ln( .
)
ln
.
So, after
s, the average concentration of hydrogen will reduce to 1
10
5
.
×
10
63
×
−
gc
m
/
Problem 2
a)
The failure of the detector means that the diffusion distance of elements reaches the magnitude
of the sample size, i.e.
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 Spring '02
 CHEN

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