Geoscience Reference
In-Depth Information
if using characteristic wave scales which were kept constant in the experiment of
Babanin &
Haus
(
2009
), we suggest that
b
3
.
=
b
1
k
ω
300 for intermittent turbulence, and, as discussed above, for mean-over-
the-period dissipation rates it has to be divided by at least a factor of 10. Therefore, at this
upper margin, we will assume
In
(7.76)
,
b
=
3
b
=
b
1
k
ω
=
30
.
(7.83)
For the waves of
f
5Hz used by
Babanin & Haus
(
2009
) in their experiment that
leads us to the dimensionless proportionality coefficient
=
1
.
b
1
=
0
.
004
,
(7.84)
that is
3
a
0
=
004
ku
orb
.
dis
=
b
1
k
ω
0
.
(7.85)
Now, the turbulent kinetic energy dissipation of narrow-banded swell with wavenumber
k
per unit of time per unit of surface
(5.67)
is
b
1
k
∞
0
b
1
ku
0
∞
0
b
1
3
dz
3
u
0
.
D
a
=
u
(
z
)
=
exp
(
−
3
kz
)
dz
=
(7.86)
Here, the orbital velocity through the water column is
u
orb
(
z
)
=
u
(
z
)
=
u
0
exp
(
−
kz
)
(7.87)
and
u
0
is the orbital velocity of the surface water particles.
D
a
is dissipation of wave energy
E
(2.22)
, normalised by the water density, per unit of
time:
gH
2
D
a
=
∂(
)
8
∂(
)
c
g
∂(
)
gE
1
gE
=
=
=
c
g
D
x
.
(7.88)
∂
∂
∂
t
t
x
Dissipation per unit of propagation distance
x
, therefore, is
1
c
g
b
1
3
2
k
2
3
b
1
k
2
3
b
1
gk
2
a
0
.
u
0
=
2
a
0
=
D
x
=
D
a
=
ω
(7.89)
ω
(7.88)
and
(7.89)
lead us to differential equation
2
g
2
∂(
a
0
(
x
)
)
2
3
b
1
gk
2
a
0
(
3
=
x
)
,
(7.90)
∂
x
that is
2
∂(
a
0
(
x
)
)
4
3
b
1
k
2
a
0
(
3
3
=
x
)
=
Ba
0
(
x
)
.
(7.91)
∂
x
The solution yields
4
B
2
x
−
2
9
9
64
10
6
k
−
4
x
−
2
2
b
1
k
4
x
−
2
a
0
(
)
=
=
=
.
x
(7.92)
4
·
Search WWH ::
Custom Search