typical mean dynamic balances in estuaries along-estuary component 1. barotropic pressure gradient...

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Typical Mean Dynamic Balances in Estuaries Along-Estuary Component . Barotropic pressure gradient vs. friction 2 2 z u A x g z z z y x z u A z y u A y x u A x dz x g x g fv z u w y u v x u u t u Steady state, linear motion, no rotation, homogeneous fluid, friction in the vertical only (A z is a constant) S x S A g z u z 2 2 The balance can then be rewritten as: z S x x

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Typical Mean Dynamic Balances in Estuaries

Along-Estuary Component

1. Barotropic pressure gradient vs. friction

2

2

z

uA

xg z

z

zyx zu

Azy

uA

yxu

Ax

dzx

gx

gfvzu

wyu

vxu

utu

Steady state, linear motion, no rotation, homogeneous fluid, friction in the vertical only (Az is a constant)

Sx

SAg

z

u

z

2

2

The balance can then be rewritten as:

z

Sx

x

SAg

z

u

z

2

2Let’s solve this differential equation; integrating once:

1cSzAg

zu

z

Integrating again:

212

2czcSz

Ag

uz

This is the solution, but we need two boundary conditions:

0.,.,0,0@

zu

eizu

Az z This makes c1 = 0

z

b

z

bbz A

ucAz

uzu

AHz2

;,@

z

b

z Auc

SHAg

zu

Hz2

,@

HzcgSH

uSHcg

ubb

@2

Substituting in the solution at z = -H: 2

2

2c

AgSH

cgSH

zb

zb A

gSHcgSH

c2

2

2

z

Sx

x

212

2czcSz

Ag

uz

zb AgSH

cgSH

cc2

02

21

bzz cgSH

AgSH

zAgS

u 22

22

bz c

gSHHz

AgS

u 22

2

General Solution

from boundary conditions

Particular solution, which can be re-arranged:

2. Pressure gradient vs. vertical mixing

2

21

z

uA

xp

z

)( zgp

)( z

zx

gx

gxp

expanding the pressure gradient:

We can write: Gx

Ix

;

The momentum balance then becomes:

zxA

gxA

g

z

u

zz

2

2

zAgG

AgI

z

u

zz

2

2

O.D.E. with general solutionobtained from integrating twice:

12

2cz

AgG

zAgI

zu

zz

2132

62)( czcz

AgG

zAgI

zuzz

2132

62)( czcz

AgG

zAgI

zuzz

General solution:

c1and c2 are determined with boundary conditions:

z

z Acz

zu

A

1gives0)1

2

32

620gives0)2 c

AH

AgGH

AgIH

Hzuzzz

zzz AH

AgGH

AgIH

c

62

32

2

This gives the solution:

Hz

AH

H

zA

gGH

H

zA

gIHzu

zzz11

61

2)( 3

33

2

22

Third degree polynomial proportional to depth and inversely proportional to friction.

Requires knowledge of I, G, and wind stress.

12

2cz

AgG

zAgI

zu

zz

Hz

AH

H

zA

gGH

H

zA

gIHzu

zzz11

61

2)( 3

33

2

22

We can express I in terms of River Discharge R, G ,and wind stress if we restrict thesolution to:

0

)(H

Rzzui.e., the river transport per unit width provides the water added to the system.

Integrating u(z) and making it equal to R, we obtain:

zzz A

H

A

gGH

A

gIHR

2

1

24

3

3

243

Which makes:gHgH

RAGHI z

233

83

3

Note that the effects of G and R are in the same direction, i.e., increase I.The wind stress tends to oppose I.

Hz

AH

H

zA

gGH

H

zA

gIHzu

zzz11

61

2)( 3

33

2

22

gHgH

RAGHI z

233

83

3 Substituting into:

We get:

2

2

2

2

3

3

2

23

131441

123

181948

)(H

zHz

AH

H

zHR

H

z

H

zA

gGHzu

zz

inducedWind

2

2

inducedRiver

2

2

inducedDensity

3

3

2

23

131441

123

181948

)(

H

zHz

AH

H

zHR

H

z

H

zA

gGHzu

zz

Density-induced: sensitive to H and Az; third degree polynomial - two inflection points

River induced: sensitive to H; parabolic profile

Wind-induced: sensitive to H (dubious) and Az; parabolic profile

inducedWind

2

2

inducedRiver

2

2

inducedDensity

3

3

2

23

131441

123

181948

)(

H

zHz

AH

H

zHR

H

z

H

zA

gGHzu

zz

0.0125 Pa

If we take no bottom stress at z = -H (instead of u(-H) = 0):0zu

2

2

3

33

64124

)(H

z

H

zA

gGHzu

z

smAI

H

z

H

zAHIg

zu

z

z

246

2

2

3

32

1025.1;102

64112

)(

0z

zx

gx

gz

uA

zz

xg

xg z

Momentum balance, steady with friction (tidally averaged)

Geyer et al (2000) approximations: H

UUC

HzETBB

UT is the tidal current amplitude UE is the magnitude of the estuarine circulation.

40 10771

01

.;s

H

UUC

aH

x

sg ETB

o

where ao is an O (1) constant related to the vertical variability of stress.

Also considering that the surface slope scales in proportion to the baroclinic pressure gradient, the momentum eqn becomes

solving for UE

TB

2

oE UC

Hxs

gaU

where ao0.3 based on Hudson data. (Its value depends on whether the layer average or the maximum is considered).

-UE

UE

TB

2

oE UC

Hxs

gaU

ao0.3 on Hudson

(2000, JPO, 30, 2035)

What drives Gravitational Circulation?

Pressure Gradient

x

zRiver

Ocean

isopycnals

isobars?

3. Advection vs. Pressure Gradient

2

Take:

Upper layer homogeneous and mobile

Lower layer immobile

Consider inertial effects

Ignore friction

@ lower layer there is no horizontal pressure gradient

x

h1

1

2

2

1 u1

U2 = 0z

interface

surface

At interface: 111 ghP

At lower layer: zhgghP 112112

dx

zdg

dx

hdg

dx

dP 2

2112 0

112

2 1h

dxd

dxd

The interface slope is of opposite sign to the surface slope

2

The basic balance of forces is

xP

dxdu

xu

u

1

1

211

11

21

02

xP

Over a volume enclosing the upper layer:

dzxP

uhdxd

2

1

12111

But 2111 ,, zPP

x

h1

1

2

2

1 u1

U2 = 0z

interface

surface

Using Leibnitz rule for differentiation under an integral (RHS of last equation; seeOfficer ((1976), p. 103) we get:

0'21

12

111

hguh

dxd

This is the momentum balance integrated over the entire upper layer (i.e., energy balance)The quantity inside the brackets (kinetic and potential energy) must remain constant

2

112

111 '21

hgdxd

uhdxd

2

12

g'g

0'21

12

111

hguh

dxd

Defining transport per unit width:

111 huq

0'21 2

11

21

1

hg

hq

dxd

Total energy (Kinetic plus Potential energy) remains constant along the system

If the density ρ1 and the g’ do not change much along the system, we can estimate the changes in h1 as a function of q1 (i.e. how upper layer depth changes with flow)

x

h1

1

2

2

1 u1

U2 = 0z

interface

surface

constant'21 2

11

21 hghq

121

1

1

1

'

2

hgu

udqdh

0'2

1

11

1

121

21

1

1 dqdh

hgdqdh

h

qhqDifferentiating with

Respect to ‘q1’

21

21

1

12

1

1

1

1 2

'

2

cu

u

hgu

udqdh

Subcritical flow causes 011 dqdh and supercritical flow causes 011 dqdh

This results from a flow slowing down as it moves to deeper regions or accelerating as it moves to shallower waters or through constrictions

x

z

h1 supercritical flow

x

z

h1 subcritical flow

Apparent PARADOX!?

Pressure gradient vs. friction: bz c

gSHHz

AgS

u 22

2i.e., u proportional to H 2

This can also be seen from scaling the balance:

22

2

2

HAgS

UAgS

H

U

AgS

z

u

zz

z

If we include non-linear terms: 02

2

gSz

uA

xu

u z

Which may be scaled as: 02

2

gSH

UALU z 0

22

2

22

gSLH

LA

H

LAU zz

i.e., U proportional to H -2 !!! Ahaa! If L is very large, we go back to u proportional to H 2

Physically, this tells us that when L is small enough the non-linear terms are relevant to the dynamics and the strongest flow will develop over the shallowest areas (fjords). When frictional effects are more important than inertia, then the strongest flow appears over the deepest areas (coastal plain estuaries)!!!

This competition inertia vs. friction to balance the pressure gradient can be exploredwith a non-dimensional number:

ULAHU

H

UALU

zz

22

2

2

LA

UH

z

2 When this ratio > 1, inertia dominates

When the ratio < 1, friction dominates

LCH

HUC

LU

bb

2

2

Alternatively:

bb

CLH

LCH

1 if

When H / L > Cb, inertia dominates

When H / L < Cb, friction dominates

4. Surface Pressure + Advection vs. Interfacial Friction

x

h1

1

2

2

1

u1

u2

z

interface

surface

h2

Similar situation as before (advection vs. presure gradient) but with interfacial friction (fi).

Flow in the lower layer but interface remains.

There is frictional drag between the two layers. The drag slows down the upper layer and drives a weak flow in the lower layer.

In the upper layer, over a volume enclosing the layer: 211

12111 ;

2

1

uCffdzxP

uhdxd

bii

The momentum balance becomes(Officer, 1976; pp. 106-107):

friction

linterfacia

and layer ofthinning by

balanced

2

gradientpressuresfce to dueforce driving

121

31

bCdxd

dxd

qgh

211

211

2111

211 2

1uC

dxd

hghghudxd

b

The solution has a parabolic shape

Boundary conditions:

u1 = u2 at the interfaceu2 =0 at the bottom, z = -H

0

2

2

h

H

u

because interface touches the bottom

In the lower layer, the balance is:2

221

z

uA

xp

z

x

h1

1

2

2

1

u1

u2

z

interface

surface

h2

u

z Salt-wedge

Typical Mean Momentum Balances

Barotropic pressure gradient vs. Friction Rivers, Homogeneous Estuaries

Total pressure gradient vs. Friction Partially Mixed Estuaries

Total pressure gradient vs. Advection Fjords

Total pressure gradient vs. Advection + Interfacial Friction Salt Wedge