Biomedical Engineering Reference
In-Depth Information
vanishingly small makes the approximation in Equations
4.11 and 4.13 exact and we can assume that the contribution to the concentration at
a point,
y
Making
r
,
δ θ
and
δ φ
=(
r
,
0
,
0
)
, from the volume
X
is as if from a point source at the point:
=(
,
θ
,
φ
)
x
. Inspecting Equation 4.7 we see that the concentration depends on the
time after synthesis and the distance between
x
and
y
,
r
−
. Substituting this
and our strength/second term,
S
X
, from Equation 4.13 into Equation 4.7 we obtain
the concentration of NO at
y
due to
x
at a time
t
after synthesis:
x
y
ex p
−
e
−
λ
r
2
sin
2
Q
drd
d
x
−
y
t
C
P
(
x
−
y
,
t
)=
(4.14)
8
(
π
Dt
)
3
/
2
4
Dt
Now, to get the concentration at
y
due to the whole sphere we must sum up the
contributions from all the points
x
=(
r
,
θ
,
φ
)
inside the sphere,
M
, (i.e., 0
≤
r
≤
a
;0
≤
θ
≤
π
and 0
≤
φ
≤
2
) as shown below:
)=
∑
x
∈
M
C
P
(
x
−
y
,
t
)
C
S
(
a
,
r
,
t
(4.15)
exp
−
(4.16)
a
π
2π
r
2
sin
2
Q
x
−
y
=
8
(
π
Dt
)
3
/
2
4
Dt
0
0
0
e
−
λ
t
drd
·
d
(4.17)
Using:
2
r
2
r
2
x
−
y
=
+
−
2
rr
cos
(4.18)
and noting that there is radial symmetry so that the concentration at any point
z
=
(
r
,
θ
,
φ
)
at a distance of
r
from the origin is equal to the concentration at
y
=(
r
,
0
,
0
)
,
we therefore obtain:
t
1
2
erf
a
erf
a
+
−
r
2
√
Dt
r
2
√
Dt
e
−
λ
C
S
(
a
,
r
,
t
)=
Q
+
Dt
π
exp
(
exp
(
2
2
1
r
a
−
r
)
a
+
r
)
−
−
(4.19)
4
Dt
4
Dt
where:
x
exp
−
u
2
du
2
√
π
erf
(
x
)=
(4.20)
0
for the concentration at a distance
r
from the centre of a solid sphere of radius
a
at
a time
t
after synthesis. This leads naturally to the solution for a hollow sphere of
inner radius
a
and outer radius
b
:
C
H
(
a
,
b
,
r
,
t
)=
C
S
(
b
,
r
,
t
)
−
C
S
(
a
,
r
,
t
)
(4.21)
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