Biomedical Engineering Reference
In-Depth Information
The basic pore fluid flow problem described by (
6.8
) and (
6.12
) may be
simplified by specifying the type of material symmetry. If the material has
orthotropic symmetry, employing a representation from Table 4.3 and the principal
coordinate system for orthotropic symmetry, (
6.8
) and (
6.12
) become
2
p
2
p
2
p
@
p
K
11
t
@
K
22
t
@
K
33
t
@
t
¼
x
1
þ
x
2
þ
in
O
;
(6.13)
x
3
@
@
@
@
c
1
m
n
1
K
11
@
p
n
2
K
22
@
p
n
3
K
33
@
p
x
;
þ
þ
þ
c
2
p
ð
t
Þ
@
x
1
@
x
2
@
x
3
x
x
x
ðx
;
x
@
¼
f
t
Þ;
O
;
(6.14)
and, if the material is isotropic, it follows from a representation in Table 4.3 that
(
6.13
) reduces to (
6.10
) and (
6.14
) may be specialized as follows:
x
;
x
;
x
;
x
@
ð
c
1
=mÞ
Kn
r
p
ð
t
Þþ
c
2
p
ð
t
Þ¼
f
ð
t
Þ;
O
(6.15)
The coordinate system may be rescaled so that the differential equation and
boundary conditions are those of distorted heat conduction objects with isotropic
material symmetry. To accomplish this the coordinates
x
1
,
x
2,
and
x
3
are rescaled by
r
x
1
;
k
K
11
r
x
2
;
k
K
22
r
x
3
;
k
K
33
x
¼
y
¼
z
¼
where
(6.16)
k
3
¼
K
11
K
22
K
33
:
Then the differential equation (
6.10
) for an isotropic medium applies in the
distorted or stretched O
, and the boundary conditions (
6.14
) are
x
þ
x
þ
p
kK
11
p
kK
22
p
kK
33
c
1
m
@
p
@
p
@
p
x
;
n
1
n
2
n
3
þ
c
2
p
ð
t
Þ
@
x
@
y
@
z
x
x
;
x
¼
f
ð
t
Þ;
distorted
@
O
:
(6.17)
In this restatement of the anisotropic problem one trades a slightly more com-
plicated differential equation (
6.8
) for a simpler one (
6.10
) and obtains the slightly
more complicated boundary condition above.
Example 6.2.1
A layer of thickness
L
of a rigid porous material is between two fluid reservoirs both
containing the same fluid at the same pressure
p
o
, as illustrated in Fig. 1.8. Let
x
1
be
a coordinate that transverses the perpendicular distance between two layers; one
reservoir is located at
x
1
¼
L
and the opening to the other reservoir is located at
x
1
¼
0, although the fluid level in the second reservoir is below
x
1
¼
0 to maintain
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