EXAMPLE PROBLEM II-3-1

FIND:

Wave height *H *and angle θ at water depths of 200, 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, 16, 14, 12, 10, 8,

6, and 4 m for deepwater wave angles of 0o, 15o, and 45o.

GIVEN:

A wave 1 m high and 15-sec period in 500 m of water, with a plane, sloping beach.

SOLUTION:

Routine solutions for a plane beach can be obtained using the ACES wave transformation code, by direct

calculation, or graphically using Figure II-3-6.

Table II-3-1 provides the results obtained by directly using the ACES code. On a personal computer with a

486-level microprocessor, the results may be obtained in seconds.

For a wave with a depth of 10 m and an initial wave angle of 45 deg, wave height and angle are calculated as

follows:

Since the deepwater wave length of a 15-sec wave is

and since 500 m is greater than *L*0/2, the given initial wave is a deepwater condition. The wave length of the wave

in 10 m must be estimated from

2π*d*

tanh

2π

and is 144 m (see Problem II-1-1).

The shoaling coefficient *K*s can be estimated from

1

2

In deep water *C*g0 for a 15-sec wave is

1

1

23.4

( 1.56 *T *) '

' 11.7 m / s

2

2

2

The group velocity is given by

1

4 π *d*/*L*

2π*d*

tanh

1%

2

sinh (4 π *d*/*L*)

2π

Substitution of *d *= 10 m, *L *= 144 m, *T *= 15 sec, and *g *= 9.8 m/sec2 yields 9.05 m/s.

1

11.70

2

' 1.14

9.05

Solution for *K*r involves

1

1 & sin2θ0

4

1 & sin2θ1

(Continued)

Estimation of Nearshore Waves

II-3-13

Integrated Publishing, Inc. |