Biomedical Engineering Reference
In-Depth Information
•
Resonance.
A serious disadvantage of our formulation for discontinous material
properties is the potential occurrence of
resonance
. This happens when either of
the eigenvalues
c
vanish, creating locally critical flow with
a non-linear wave overlapping the stationary wave associated with the eigenvalue
λ
2
=
λ
1
=
u
−
c
or
λ
3
=
u
+
0. Theoretical issues regarding resonance are found in [10, 14], and refer-
ences there in. A serious difficulty is the loss of uniqueness, as shown by our
preliminary results to be reported elsewhere.
2.5 Sample solutions
We consider two test problems for a long, straight tube of length 1
m
, of constant
equilibrium cross sectional area
A
0
10
−
4
m
2
, of wall thickness
h
0
=
2
.
1124
×
=
8
.
2
×
10
−
4
m, Young's modulus
E
10
5
N/m
2
, exponent in tube law
m
1
2
and
=
3
.
0
×
=
1
2
. The coefficient
K
is taken as
Poisson ratio
ν
=
4
√
π
3
E
h
0
√
A
0
(
x
)
K
0
=
,
1050 kg/m
3
. Initial data for two Riemann problems are
given in Table 2.1. Table 2.2 shows the values of the exact solution in the Star Re-
gion for areas
A
∗
L
,
A
∗
R
and velocities
u
∗
L
,
u
∗
R
. These numbers can be useful to test
numerical methods.
while blood density is
ρ
=
Test 1: left rarefaction and right shock.
Fig. 2.3 shows profiles of vessel diameter,
velocity and Froude number at the output time
T
out
=
012 s. Also shown is the
resulting wave configuration in the
x
-
t
plane. The solution consists of a left facing
rarefaction, a stationary contact and right facing shock.
0
.
Test 2: two rarefactions.
Fig. 2.4 shows solution profiles of vessel diameter, ve-
locity and Froude number at
T
out
0075 s. See also wave configuration in the
x
-
t
plane. The solution consists of a left rarefaction, a stationary contact and a right
rarefaction.
=
0
.
Table 2.1.
Initial conditions for two test Riemann problems
Test
A
L
[
m
2
A
R
[
m
2
]
u
L
[
m
/
s
]
K
L
]
u
R
[
m
/
s
]
K
R
10
−
4
10
−
5
10
−
6
13
.
000
×
−
2
.
6575
×
50
×
K
0
3
.
000
E
−
04
6
.
1230
×
K
0
10
−
4
10
−
4
23
.
100
×
−
3
.
9967
K
0
3
.
100
×
11
.
9943
40
×
K
0
Table 2.2.
Exact solution in the Star Region for two test Riemann problems
Test
m
2
m
2
A
∗
L
[
]
u
∗
L
[
m
/
s
]
K
∗
L
A
∗
R
[
]
u
∗
R
[
m
/
s
]
K
∗
R
10
−
4
10
−
4
1
2
.
0119
×
12
.
8103
K
L
6
.
3403
×
4
.
0648
K
R
10
−
5
10
−
4
2
9
.
3917
×
0
.
9625
K
L
2
.
0784
×
0
.
4349
K
R