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
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