EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-6-1 (Concluded)
Therefore, from Equation II-6-5
)
Vm 2π as Ab
Vm '
Ac T
0.66 (2) (3.14) (0.65) (1.9) 107
Vm '
' 1.72 m/s (5.64 ft/s)
666 (12.42) (3,600)
Qm ' Vm Ac ' (1.72) (666) ' 1145 m 3/s (40,430 ft 3/s)
Since ab/as = 0.78, ab = 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 ab 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
Lb
6000
' 782 sec or 0.22 hr
t( '
'
gdb
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.
Figure II-6-23.
Results from Example Problem II-6-1
Hydrodynamics of Tidal Inlets
II-6-25
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