EM 1110-2-1100 (Part II)
30 Apr 02
Example Problem II-6-2 (Concluded)
Bay tide range is 0.96 m and is not symmetric about mean sea level, due to river flow.
Maximum channel velocities are determined from Figure II-6-27 (maximum ebb velocity)
)
Umaxe ' 0.80
Using Equation II-6-5, and substituting UNmaxe for VNm and Umaxe for Vm , and solving for Umaxe
)
Umaxe 2π aoAb
0.80 (2) (3.14) (0.65) (1.90) 107
' 2.08 m/s
Umaxe '
'
Aavg T
666 (12.42) (3600)
From Figure II-6-28 (maximum flood velocity)
)
Umaxf ' 0.50
and in a manner similar to solving for ebb velocity, flood velocity is determined by
)
Umaxf 2π ao Ab
0.50 (2) (3.14) (0.65) (1.90) 107
' 1.30 m/s
Umaxf '
'
Aavg T
666 (12.42) (3600)
Maximum discharges are determined from
Qmaxe ' Umaxf Ac ' 2.08 (666) ' 1385 m 3/s
Qmaxf ' Umaxf Ac ' 1.30 (666) ' 866 m 3/s
River flow increases maximum ebb discharge and decreases maximum flood discharge, as one would expect.
The values are nearly balanced around the no-river-discharge value of 1,145 m3/s of Example 1.
Using Figure II-6-29, the value of ε (phase lag of bay tide relative to ocean tide) is determined to be 19 deg,
indicating high water occurs earlier than for the non-river-inflow condition of Example 1.
Hydrodynamics of Tidal Inlets
II-6-35
Previous Page
