for most of the basin question what causes the strong narrow currents on the west side of the ocean...

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β v = 1 ρH τ y x τ x y For most of the basin Question What causes the strong narrow currents on the west side of the ocean basin? The westward Intensification

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€ βv=1ρH∂τy∂x−∂τx∂y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟For most of the basin

Question

What causes the strong narrow currents on the west side of the ocean basin?

The westward Intensification

Stommel’s Model

€ −fv=−g∂η∂x+τy()ρH−Ku€

fu=−g∂η∂y−Kv€ ∂u∂x+∂v∂y=0

Rectangular ocean of constant depth

Surface stress is zonal and varies with latitude onlySteady ocean state

Simple friction term as a drag to current

Vorticity balance: Sverdrup balance +friction

€ βv=−1ρH∂τ∂y−K∂v∂x−∂u∂y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=−1ρH∂τ∂y−Kς

Flow patterns in this ocean for three conditions:(1) non-rotating ocean (f=0)(2) f-plane approximation (f=constant)(3) β-plane approximation (f=fo+βy)

€ −βv−1ρH∂τ∂y−Kς=0€

−1ρH∂τ∂y−Kς=0f-plane β-plane

€ −1ρH∂τ∂y−Kς=0Wind stress () + friction () =0

Negative vorticity generation Positive vorticity generation

Westerly winds in north, easterly winds in south

Ekman effect drives the water to the center,

Increase sea level generates anticyclonic geostrophic currents

Internal friction (or bottom Ekman layer) generate downslope cross-isobaric flow, which balance the wind-driven Ekman transport

The β effect

Generate negative vorticity Generate positive vorticity

€ −1ρH∂τ∂y−βv−Kς=0

In the west, water flows northward

Wind stress () + Planetary vorticity () + Friction () = 0

In the east, water flows southward

Wind stress () + Planetary vorticity () + Friction () = 0

Friction (W) > Friction (E)

Non-rotation Ocean, f=0

€ Ku=−g∂η∂x+τy()ρH

€ Kv=−g∂η∂y€

−1ρH∂τ∂y−Kς=0

ψψττρ

ψβψψψ 4222 1, ∇+∇−∂∂−

∂∂=

∂∂+∇+∇

∂∂

⎟⎟⎟

⎜⎜⎜

⎛⎟⎠⎞⎜

⎝⎛

Hxy Ar

yxDxJ

t

Quasi-geostrophic vorticity equation

where

4

4

22

4

4

44 2

yyxx ∂∂+

∂∂∂+

∂∂=∇ ψψψψ

( ) ( ) ( )xyyx

J∂

∇∂

∂−

∇∂

∂=∇

ψψψψψψ

222,€ ∇2ψ=∂2ψ∂x2+∂2ψ∂y2

Boundary conditions on a solid boundary L

(1) No penetration through the wall (used for the case of no horizontal diffusion)

( )0== constψ

(2) No slip at the wall

( )0== constψ

€ ∂ψ∂n=0

along the boundary L

along the boundary L

n is the unit vector perpendicular to the boundary L

Non-dimensionalize Quasi-Geostrophic Vorticity Equation

ψψττρ

ψβψψψ 4222 1, ∇+∇−∂∂−

∂∂=

∂∂+∇+∇

∂∂

⎟⎟⎟

⎜⎜⎜

⎛⎟⎠⎞⎜

⎝⎛

Hxy Ar

yxDxJ

t

Define non-dimensional variables based on independent scales L and τo

€ x=L′ x € y=L′ y € t=T′ t € ψ=Ψ′ ψ € τ=τo′ τ

The variables with primes, as well as their derivatives, have no unit and generally have magnitude in the order of 1. e.g.,

€ ′ ψ ~O1()€ ∂′ ψ ∂′ x ~O1()

€ u=−∂ψ∂x€ U

′ u =−ΨL∂′ ψ ∂′ x € U=

ΨL€ Ψ=UL€ T=LU

Note that U has not been decided yet.

€ ∂ψ∂x=ΨL∂′ ψ ∂′ x =ULL∂′ ψ ∂′ x =U∂′ ψ ∂′ x

€ ∂τy∂x−∂τx∂y=τoL∂′ τ y∂′ x −∂′ τ x∂′ y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟€ ∇2ψ=∂2ψ∂x2+∂2ψ∂y2=ULL2∂2′ ψ ∂′ x 2+∂2′ ψ ∂′ y 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=UL∇2′ ψ

€ Jψ,∇2ψ()=∂ψ∂x∂∇2ψ()∂y−∂ψ∂y∂∇2ψ()∂x=U2L2∂′ ψ ∂′ x ∂∇2′ ψ ()∂′ y −∂′ ψ ∂′ y ∂∇2′ ψ ()∂′ x ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟=U2L2J′ ψ ,∇2′ ψ ()

€ ∇4ψ=∂4ψ∂x4+2∂4ψ∂x2∂y2+∂4ψ∂y4=UL3∂4′ ψ ∂′ x 4+2∂4′ ψ ∂′ x 2∂′ y 2+∂4′ ψ ∂′ y 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=UL3∇4′ ψ

€ U2L2∂∂′ t ∇2′ ψ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+U2L2J′ ψ ,∇2′ ψ ()+βU∂′ ψ ∂′ x =τoρDL∂′ τ y∂′ x −∂′ τ x∂′ y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−rUL∇2′ ψ +AHUL3∇4′ ψ

€ UβL2∂∂′ t ∇2′ ψ +J′ ψ ,∇2′ ψ () ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+∂′ ψ ∂′ x =τoρβUDL∂′ τ y∂′ x −∂′ τ x∂′ y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−rβL∇2′ ψ +AHβL3∇4′ ψ

Non-dmensional vorticity equation

Define the following non-dimensional parameters

2

2 ⎟⎟⎟

⎜⎜⎜

==LL

U Iδ

βε , βδ UI = , nonlinearity.

LLr S

S

δβε == βδ r

S = , bottom friction.

3

3 ⎟⎟⎟

⎜⎜⎜

==LL

A MHM

δβε

,

3βδ H

MA= , lateral friction.,

€ τoρβUDL=1If we choose

€ U=τoρβDLwe have Sverdrup relation

€ ε∂∂′ t ∇2′ ψ +J′ ψ ,∇2′ ψ () ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+∂′ ψ ∂′ x =∂′ τ y∂′ x −∂′ τ x∂′ y −εS∇2′ ψ +εM∇4′ ψ

Interior (Sverdrup) solutionIf <<1, S<<1, and M<<1, we have the interior (Sverdrup) equation:

yxxxyI

∂∂−

∂∂=

∂∂ ττψ

∫ ∂

∂−∂∂−=

⎟⎟⎟

⎜⎜⎜

⎛Ex

x

xyEI

dxyxττψ

(satistfying eastern boundary condition)

∫ ∂∂−∂

∂=⎟⎟⎟

⎜⎜⎜

⎛x

Wx

xyWI

dxyxττψ

Example:Let ( )yx πτ cos−= ,

0=yτOver a rectangular

basin (x=0,1; y=0,1)

( )yxEI ππψ sin1⎟

⎠⎞⎜

⎝⎛ −−=

( )yxWI ππψ sin−=

(satistfying western boundary condition)

.

Westward IntensificationIt is apparent that the Sverdrup balance can not satisfy the mass conservation and vorticity balance for a closed basin. Therefore, it is expected that there exists a “boundary layer” where other terms in the quasi-geostrophic vorticity is important. This layer is located near the western boundary of the basin. Within the western boundary layer (WBL),

IB ψψ ~ , for mass balance

δξ x=

In dimensional terms,

⎟⎟⎟

⎜⎜⎜

⎟⎟⎟

⎜⎜⎜

⎟⎟

⎜⎜

⎟⎟

⎜⎜

⎟⎟⎟

⎜⎜⎜

∂∂−

∂∂>>

∂∂−

∂∂=

=∂∂

xyDxyDLO

ULOOx

yx

o

yx

o

BB

ττρ

ττρδ

δβ

δψβψβ

11

~

The Sverdrup relation is broken down.

, the length of the layer δ <<L The non-dimensionalized distance is

The Stommel modelBottom Ekman friction becomes important in WBL.

( )yxS ππψψε sin2 −=

∂∂+∇ , S<<1.

0=ψ

(Since the horizontal friction is neglected, the no-slip condition can not be enforced. No-normal flow condition is used).

( )yx

I ππψ

sin−=∂

( )yxI ππψ sin1 ⎟⎠⎞⎜

⎝⎛ −=

Interior solution

at x=0, 1; y=0, 1. No-normal flow boundary condition

Let S

S

xxδεξ

*

==, we

have

( )ySyySS ππψεψεψε ξξξ sin11 −=++ −−

( ) 0sin2 ==−=+ ⎟⎠⎞

⎜⎝⎛

SSyySOy εππεψεψψ ξξξ

Re-scaling in the boundary layer:

€ ∂ψ∂x=∂ψ∂ξ∂ξ∂x=1εS∂ψ∂ξ

€ ∂2ψ∂x2=1εS∂∂ξ∂ψ∂x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=1εS2∂2ψ∂ξ2

€ ∇2ψ=∂2ψ∂x2+∂2ψ∂y2=1εS∂2ψ∂ξ2+∂2ψ∂y2

( )yxS ππψψε sin2 −=

∂∂+∇Take into

As ξ=0, ψ=0. As ξ,ψψI

The solution for 0=+ ξξξ ψψ is

( ) ( ) S

x

BeAeyxByxA εξψ−− +=+= ,,

0=ξ , 0=ψ . A=-B

( ) ⎟⎟

⎜⎜

⎛ −−= S

x

eyxA εψ 1,

, ( ) ( ) ( )yxyxyxA I ππψψ sin1,, ⎟⎠⎞⎜

⎝⎛ −==→

⎟⎟

⎜⎜

⎛ −−= S

x

Ie εψψ 1 ( Iψ can be the interior solution under different winds)

For ( )SOx ε<

( )ye S

xB ππψ ε sin1 ⎟

⎜⎜

⎛ −−=

( )yevS

x

B S

ππεε

sin−

=

For ( ) 1≤≤ xO Sε ,

( )yxI ππψ sin1 ⎟⎠⎞⎜

⎝⎛ −= ,

( )yv I ππ sin−= .

,

.

,

The dynamical balance in the Stommel model

In the interior,Dx

pfvo

x

o ρτ

ρ +∂∂−=− 1

Dypfu

o

y

o ρτ

ρ +∂∂−= 1

( )D

curlvoρ

τβ = ( )D

curldt

dfoρ

τ=

Vorticity input by wind stress curl is balanced by a change in the planetary vorticity f of a fluid column.(In the northern hemisphere, clockwise wind stress curl induces equatorward flow).

In WBL,xpfv

o ∂∂=ρ

1

rvypfu

o−

∂∂−= ρ

10=+

∂∂ vxvr β x

vrdtdf

∂∂−=

Since v>0 and is maximum at the western boundary, 0<∂∂xv

the bottom friction damps out the clockwise vorticity.

,

Question: Does this mechanism work in a eastern boundary layer?