EXAMPLE PROBLEM II-6-4

Find:

Determine the wave height in an inlet channel that is 5.0 m (16.4 ft) deep.

Given:

A 5.0-sec, 1.0-m (3.28-ft) wave is entering an inlet having a 1.0 m/s (3.3 ft/s) ebb current. The angle

between current and wave orthogonal is 180 deg.

Solution:

Ω = (2π/T) (dT/g)1/2

θ = angle between horizontal velocity vector and horizontal wave vector = 180o

Therefore:

1/2

2 (3.14)

5

Ω'

' 0.90

5.0

9.8

1.0 (cos 180)

' &0.14

1/2

( 9.8 @ (5) )

From Figure II-6-34, *R*H = 1.35

Wave height *H*A, originally 1 m, is modified by the current to a wave height of:

II-6-42

Hydrodynamics of Tidal Inlets

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