3. Sédiment Transport
89
Problem VII
Evaluate P[3 (longshore energy flux component) using energy flux method,
LEO (Littoral Environment Observation) method and compare the energy
flux and sédiment transport Q for the given data as below.
p = 1024 kg/m3, g — 9.81 m/s2, C'y = 0.01 (friction factor), X = 50 m
(distance of dye patch from shoreline), Bed slope = 1:80, angle between
shoreline geo-north (u) = 650. Therefore angle between shore normal and
geo-north = 155°.
Given data
i?°
w.r.t shore
normal
Month
Ho (m)
T (s)
i?°
w.r.t north
Measured
Vieo (m/s)
J an
1.5
9
105
-0.32
-50
Feb
1.7
9
110
-0.31
-45
Mar
1.5
9
120
-0.28
-35
Apr
1.0
9
140
-0.09
-15
May
1.2
9
140
-0.14
-15
J un
2.5
10
200
0.40
45
Jul
2.8
10
205
0.35
50
Aug
2.0
10
185
0.19
30
Sep
2.0
10
190
0.22
35
Oct
1.5
8
110
-0.38
-45
Nov
0.9
8
135
-0.12
-20
Dec
1.2
8
130
-0.18
-25
Solution
(a) Energy Flux Method
The detailed calculations for the month of January
Pis = ^H]bCgbsm2ab
89
Problem VII
Evaluate P[3 (longshore energy flux component) using energy flux method,
LEO (Littoral Environment Observation) method and compare the energy
flux and sédiment transport Q for the given data as below.
p = 1024 kg/m3, g — 9.81 m/s2, C'y = 0.01 (friction factor), X = 50 m
(distance of dye patch from shoreline), Bed slope = 1:80, angle between
shoreline geo-north (u) = 650. Therefore angle between shore normal and
geo-north = 155°.
Given data
i?°
w.r.t shore
normal
Month
Ho (m)
T (s)
i?°
w.r.t north
Measured
Vieo (m/s)
J an
1.5
9
105
-0.32
-50
Feb
1.7
9
110
-0.31
-45
Mar
1.5
9
120
-0.28
-35
Apr
1.0
9
140
-0.09
-15
May
1.2
9
140
-0.14
-15
J un
2.5
10
200
0.40
45
Jul
2.8
10
205
0.35
50
Aug
2.0
10
185
0.19
30
Sep
2.0
10
190
0.22
35
Oct
1.5
8
110
-0.38
-45
Nov
0.9
8
135
-0.12
-20
Dec
1.2
8
130
-0.18
-25
Solution
(a) Energy Flux Method
The detailed calculations for the month of January
Pis = ^H]bCgbsm2ab
