EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-3-1 (Concluded)
In deep water, θ is 45 deg. From Equation II-3-9,
C1 sin θ0
sin θ '
C0
In deep water C0 = 1.56T = 23.4 m/s. In 10 m of water, C1 = L1/T = 144 m/15s = 9.60 m/s.
9.6 sin (450 )
9.6 (7.07)
' 0.29
sin θ '
'
23.4
23.4
1
1
1
1 & sin2θ0
1 & (0.707)2
0.50
4
4
4
Kr '
' 0.86
'
'
0.91
2
1 & (0.29)2
1 & sin θ
Therefore: Hi = H0 Ks Kr = 1(1.14) (0.86) = 0.98 m.
The angle of approach is arc sin (sinθ) = 16.80. Thus, the l-m, 15-sec wave has changed 2 percent in height by
28.2 deg in angle of approach.
The largest differences caused by refraction and shoaling will be seen at the shallowest depths. From Table II-3-1
at the 4-m depth, the wave height for a 45-deg initial angle is 1.18 m compared to 1.39 m for a wave with initial
angle of 0 deg. If the initial angle had been 70 deg, Kr Ks would be about 0.8.
Table II-3-1
Example Problem II-3-1 Refraction and Shoaling Results
θo = 0o
θo = 15o
θo = 45o
Depth
θ
H
θ
H
θ
H
500
0
1.00
15.0
1.00
45.0
1.00
400
0
1.00
15.0
1.00
45.0
1.00
300
0
1.00
15.0
1.00
45.0
1.00
200
0
1.00
15.0
1.00
45.0
1.00
100
0
0.94
14.3
0.94
42.4
0.92
90
0
0.93
14.0
0.93
41.2
0.91
80
0
0.93
13.7
0.92
30.4
0.89
70
0
0.92
13.2
0.91
38.9
0.88
60
0
0.91
12.7
0.91
37.0
0.86
50
0
0.91
12.0
0.91
34.5
0.85
40
0
0.92
11.1
0.92
31.8
0.84
30
0
0.95
9.9
0.94
28.1
0.85
20
0
1.00
8.4
0.99
23.4
0.88
18
0
1.02
8.0
1.01
22.3
0.89
16
0
1.04
7.8
1.03
21.1
0.91
14
0
1.07
7.1
1.05
19.8
0.92
12
0
1.10
6.6
1.08
18.4
0.95
10
0
1.14
6.1
1.12
16.8
0.98
8
0
1.19
5.5
1.17
15.1
1.02
6
0
1.27
4.8
1.25
13.15
1.08
4
0
1.39
3.9
1.37
10.8
1.18
II-3-14
Estimation of Nearshore Waves
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