Environmental Engineering Reference
In-Depth Information
The following relationships apply for a primary wave (see also Section 2.5.3):
H
2
cos k
x
v
t
ð
Þ
z
ðÞ¼
x
;
u ¼
@F
ð
x
;
z
;
t
Þ
H
2
cosh k
½
ð
z þ d
Þ
¼ v
cos k x v t
ð
Þ
@
x
sinh k d
ð
Þ
_
u ¼
@
u
t
¼
@
2
F
H
2
cosh k
½
ð
z þ d
Þ
2
t
¼ v
sin k x v t
ð
Þ
@
@
x
@
sinh k d
ð
Þ
Substituting in the Morison formula results in
dz
¼
c
M
r
p
D
2
dF
x
ðÞ
dz
¼
dF
Ix
ðÞ
dz
þ
dF
Dx
ðÞ
H
2
cosh k
½
ð
z
þ
d
Þ
2
v
sin k
x
v
t
ð
Þ
4
sinh k
d
ð
Þ
2
1
2
r
D
H
2
cosh k
½
ð
z
þ
d
Þ
þ
c
D
v
cos k
x
v
t
ð
Þ
cos k
x
v
t
j
ð
Þ
j
sinh k
d
ð
Þ
If we are dealing with a cylinder diameter that is narrow in hydrodynamic terms, that is
the diameter of the cylinder is relatively small in relation to the wavelength, then the
orbital velocity and acceleration hardly change over the width of the cylinder. Using the
values in the position of the cylinder axis (x
¼
0) is then permissible. Therefore, for
x
¼
0 it follows that
dz
¼
c
M
r
p
D
2
dF
x
ðÞ
dz
¼
dF
Ix
ðÞ
dz
þ
dF
Dx
ðÞ
H
2
cosh k
½
ð
z
þ
d
Þ
2
v
sin
ð
v
t
Þ
4
sinh k
d
ð
Þ
2
1
2
r
D
H
2
cosh k
½
ð
z
þ
d
Þ
þ
c
D
v
cos
ð
v
t
Þ
j
cos
ð
v
t
Þ
j
sinh k
d
ð
Þ
Example of application:
Parameters: H
¼
10m,
l¼
150m, D
¼
2.0m, d
¼
30 m
0419 m
1
It follows from the dispersion equation that
k
¼
2
p=l ¼
2
p=
150
¼
0
:
0
:
5
0
:
5
591 s
1
v ¼½
g
k
tanh
ð
k
d
Þ
¼½
9
:
81
0
:
0419
tanh
ð
0
:
0419
30
Þ
¼
0
:
T ¼ 2 p=v ¼ 2 p=
0
:
591 ¼ 10
:
63 s
10
2
cosh 0
½
:
0419
ð
0 þ 30
Þ
ux¼ 0
ð
;
z ¼ 0
;
t ¼ 0
Þ ¼ 0
:
591
1
:
0 ¼ 3
:
48 m
=
s
sinh 0
ð
:
0419 30
Þ
¼ 5
3 10
6
35 10
6
Re z ¼ 0
ð
Þ¼
uz¼ 0
ð
Þ D
=n ¼ 3
:
48 2
:
0
=
1
:
:
N
KC
z ¼ 0
ð
Þ¼
uz¼ 0
ð
Þ T
=
D ¼ 3
:
48 10
:
63
=
2
:
0 ¼ 18
:
5 ð<
30Þ
