TIDAL INLETS

• Natural of man-made cut through barrier island• Allows for bay flushing • Provides access for maritime traffic• Normally migrate unless restrained by hard structures Puts system in an unnatural state and causes sediment to build up or be lost

http://www.cccturtle.org/images/inlet.jpg

TIDAL HYDRODYNAMICS

Caused by differences in bay and ocean water levels.

Assumptions:1) Steep sided bay so the cross-section area is not a function of water level2) The cross-sectional area of the channel changes as the water level moves up and

down3) Length and width of channel are fixed4) Mean depth in the channel is hi

5) Bay depth is much greater than mean channel depth so wave travel times are not important

6) Assume only one tidal component driven by simple cosine wave

AB

AC

Lb

ocean

bay

TIDAL HYDRODYNAMICS

tato cos)( 0 Ocean tide with amplitude a0, and frequency, σ

BBB tat cos)( Bay tide with different amplitude and phase shifted. Bay tide lags ocean tide.

D&D

TIDAL HYDRODYNAMICS

Mass Conservation: flow rate through the inlet, Q, assuming bays fills uniformly over plan area

20 B

ihUbQ

Q=velocity times width times mean depth plus the average tidal elevation

Note that the flow rate can also be found from the time derivative of the volume in the bay as

dt

dAhA

dt

dQ B

BBBB

Assume that hi>> average tidal amplitude (i.e. linearize)

cB

i UAhUbQ

20

dt

dAUA BBc

DYNAMIC EQUATION

From Navier-Stokes along channel axis

zx

P

Dt

Du

11

Assume P is hydrostatic, P=ρg(η-z)

zxg

x

uu

t

u

1

Integrate over depth

biii xgh

x

U

ht

Uh

12

2

Where U = depth-averaged velocity

And the shears are denoted by at the surface and b at the bed

0

)( dzzuU

DYNAMIC EQUATION

Define the bed shear stress using the quadratic drag law and assume no wind such that the surface shear is zero, and integrate along the channel to get

8

ULUfgh

t

ULh ILii

Where L is the channel length and the water level elevations at location L and location I are just inside the channel on the bay and ocean side and are not typically equal to the tidal elevation in the bay and ocean respectively.

g

UKenI 2

)1(2

0

Where Ken is an entrance coefficient to account for losses due to the a velocity head in the inlet; typically 0.1 to 0.3

g

UKexBL 2

)1(2

Where Kex is an exit coefficient to account for flow expansion losses; typically on the order of 1.0

DYNAMIC EQUATION

Finally substitute for the bay and tidal amplitudes and integrate across the channel section to get

0240

g

UU

R

fLKK

t

U

g

L

hexenB

Where Rh is the wetted perimeter or hydraulic radius (part where water touches for frictional drag along walls

i

ih hb

bhR

2

This equation is the one that is often used to estimate things like maximum tidal velocities, channel widths etc after some additional assumptions are made. Typically used in an engineering fashion because one has to utilize coefficients like f, the Ken and Kex and keep in mind the initial assumptions of bay geometries.

KEULEGAN METHOD

Consider the intertial term to be small then

g

UU

R

fLKK

hexenB 240

Potential energy = allocation of energy losses through various means

Solving for U

)(

4

20

0B

hexen

B sign

RfL

KK

gU

KEULEGAN METHOD

Nondimensionalize and replace U by dt

d

A

AU B

c

B From mass conservation

B

BB aa

tt

,,0

00

Leaves )( 00 BBB signKtd

d

where

hexen

B

c

RfL

KK

ga

Aa

AK

4

2 0

0

K is called the repletion coefficient (filling coefficient). Signifies degree to which a bay will fill depending on bay, inlet and tidal parameters

KEULEGAN METHOD

Increases in the numerator of K imply an increase in the relative tidal amplitude

Increases in the denominator of K (increases in friction) decreases the relative tidal amplitude. Large bays likely have a small K because the area of walls is large

Keulegan assumed that ta

sin0

00

D&D

Large bays

88.08.0

18.02.0

0

0

Ka

a

Ka

a

B

B

Small bays

KEULEGAN METHOD

Dimensionless maximum velocity can be found from conservation equation

maxmaxmax

dt

dAUAQ BBc

B

B adt

d

max

Bc

B aA

AU max

D&D

D&D

KEULEGAN METHOD

Bay tide lags ocean tide by a phase shift defined by

0

cosa

aBB

Note that as K gets large, implying a small bay, the phase shift approaches zero. As the bay gets very large (small K), the phase shift approaches but does not reach 90 degrees.

All these types of analyses can be repeated for a Linear approach rather than Keulegan’s

TIDAL PRISMThe tidal prism, Ω, is the volume of water flowing in or out during flood or ebb tide.

BBAa2

Example: For a channel that is 120 m wide and 4 m deep, determine K, aB, Umax and the tidal prism if L= 600m, a0=0.5m, AB=20x106 m2, f=0.08, Rh=3.75m and a tidal period of 12.4 hours.

47.0

4

2 0

0

hexen

B

c

RfLKK

ga

Aa

AK

From figure 13.4, 25.05.00

BB aa

a

From figure 13.5, s

mUU 15.14.0 maxmax

3626 1010102025.022 mxmxmAa BB

INLET AREA RELATED TO TIDAL PRISM

O’Brien and Jarret separately related the Ac to the tidal prism.

5102xAc For natural sandy inlets with Ac in ft2 and Ω in ft3

From tUU cosmax 4/

0

max cos2T

c tdtUA

sft

xUor

UAc 5.3104

25max

max

After substituting for Ac

Rough estimate of a max velocity for a natural inlet to be viable.

EBB TIDAL SHOALSCreated by sediment being carried out of inlet and deposited offshore

http://www.offshorecoastal.com/images/KFA/History/4-22-97.jpg

FLOOD SHOAL

EBB SHOAL

EBB TIDAL SHOALS

wavesEbb currents pushes seds offshoreWaves pushing seds onshore

Reach a dynamic equilibrium where shoals form

Note that if the inlet was closed, the shoal would move shoreward under wave action and spread out along coast

Large waves = small shoalSmall waves = large shoal

Some empirical relationships for shoal volume (Walton and Adams, 1976)

23.1aVshoal wavessmall108.13

wavesmedium105.10

waveslarge107.8

5

5

5

xa

xa

xa