Example Problem II-6-1 (Concluded)

Therefore, from Equation II-6-5

)

0.66 (2) (3.14) (0.65) (1.9) 107

' 1.72 m/s (5.64 ft/s)

666 (12.42) (3,600)

Since *a*b/as = 0.78, *a*b = 0.78 (0.65) = 0.51 m (1.67 ft) and the bay tidal range is 0.51 (2) or 1.02 m (3.35 ft).

The tidal prism is

2 *a*b Ab ' 2 (0.51) (1.90) (107) ' 1.94 (107) m3

(6.27 108 ft 3)

If the average depth of the bay is 6.0 m and the distance to the farthest point in the bay is 6.0 km, the time *t** it

will take for the tide wave to propagate to that point is

6000

' 782 sec or 0.22 hr

'

9.8 (6.0)

Since this time is significantly less than 12.42 hr, the assumption that the bay surface remains horizontal is

quite satisfactory.

Hydrodynamics of Tidal Inlets

II-6-25

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