MSCI3001 wk5 S2-2011-dynamics4web.science.unsw.edu.au/~alexg/course_msci3001/... · 29/08/11 2 g dz...

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Physical oceanography, MSCI 3001 Oceanographic Processes, MSCI 5004

Dr. Alex Sen Gupta [email protected]

Ocean Dynamics

Newton’s Laws of Motion An object will continue to move in a straight line and at a constant speed unless acted on by a NET force The change in the velocity (speed and/or direction) of an object (e.g. a bit of water) is proportional to the force and inversely proportional to the mass of the object.

zyx FdtdwF

dtdvF

dtduSo

dtduaonAccelerati

volumeunitperforconsiderwillhersOceanograp

FV

aso

VmbutFm1a

Σ=Σ=Σ=

=

Σ=

=Σ=

ρρρ

ρ

ρ

111,

.

1,

u,x: west to east v,y: south to north w,z: up to down

Movement of the ocean (a)

= (1/ρ)x( wind - friction + rotation + tides

gravity+ buoyancy + pressure differences ….)

A sum of forces.

Equation of motion a=ΣF/ρ

What forces might cause a parcel of water to accelerate?

...)(1 ++++= fPCg FFFFdtdu

ρ

Newton’s Laws of Motion Vertical direction:

P

P+ΔP

What are the forces acting in the up-down direction? The boxes weight is acting downwards (mg) The pressure at the top of the box is also trying to force the box downwards. But the pressure at the bottom of the box is trying to force it upwards (the difference in the pressure forces is just the buoyancy force discussed for Archimedes)

dzdpg

dtdw

ρ1

−=Weight (mg)

P=F/A

Ftop

Fbottom

Acceleration = 1/ρ(sum of forces) = 1/ρ(weight + Ftop - Fbottom)

z

Bouyancy due to difference in pressures

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gdzdp

dtdw

+−=ρ1

Newton’s Laws of Motion Vertical direction

But we can normally make this even simpler. We can us SCALE ANALYSIS to see the size of the terms. e.g. Suppose A = B + C If we know that A=0.00001 and B=10, then to a good approximation we could simplify this equation to

CB −≈

Vertical acceleration Buoyancy or pressure force

Weight

gdzdp

dtdw

+−=ρ1

In general over the ocean vertical acceleration is much smaller than g. This means that in the vertical equation g and must be of similar magnitude

dzdp

ρ1

gdzdp

ρ= This is called the hydrostatic equation

We can integrate this equation since density and g are essentially constant.

∫∫ =hp

gdzdzdzdp

00

ρ Or simply ghp ρ=Which you are hopefully familiar with already!

Newton’s Laws of Motion Horizontal direction

P P+ ΔP

dxdp

dtdu

0

1ρ

−=

dydp

dtdv

0

1ρ

−=Or doing the same in the y-direction:

x

Fleft Fright

Acceleration = 1/ρ(sum of forces) = 1/ρ(Fleft - Fright)

Hydrostatic equation tells us that pressure = weight of water above you (in this case higher on the right) Barotropic and Baroclinic Motion

Remember, p = ρgz , and

ρ0=1027 ρ1=1026 ρ2=1028 kgm-3 <

xp

dtdu

∂

∂−=ρ1

5cm 5cm

1cm 2cm

2cm

2cm

2cm

2cm

2cm

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Barotropic and Baroclinic Motion Remember, p = ρgz , and

ρ0 ρ1 ρ2 <

xp

dtdu

∂

∂−=ρ1

Barotropic – Velocity is constant with depth Baroclinic –

velocity changes with depth

Motion due to surface slopes Motion due to density differences

BAROTROPIC BAROCLINIC

Barotropic Currents move the whole water column. Can be induced by horizontal changes in elevation, which produce a pressure gradient throughout the whole water column.

Baroclinic Currents vary over depth. This can be induced by a horizontal density gradient. This effect is known as the thermal wind balance (more later).

Barotropic and Baroclinic currents

10

Barotropic and Baroclinic Motion Remember, p = ρgz , and

ρ1 ρ2 <

xp

dtdu

∂

∂−=ρ1

Mixed situation

Motion due to density differences

Barotropic Ocean

xg

xp

xg

xgdgd

xghghxpp

xp

xp

Δ

Δ=

∂

∂Δ

−=

Δ

+−+=

Δ

−=

Δ

−=

Δ

Δ=

∂

∂

ηρ

ηηρ

ρηρη

ρρ

)(

)()(

12

12

12

12

xp

dtdu

∂

∂−=ρ1

For a constant density ocean, we can write the pressure gradient in an easier way. Remember, p = ρgz , and

Δx

P1=h1ρg h1=d+η1

P2=h2ρg h1=d+η1

d

η1 η2

So we are left with x

gdtdu

∂

∂−=

η

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...1++

∂

∂−= forceCoriolis

xp

dtdu

ρ

The Coriolis Force F

m1a Σ=

Coriolis hand out

The Coriolis Force

The strength of the Coriolis force varies with latitude It is proportional to the Coriolis parameter f=2Ωsin(Φ), where Φ is latitude And Ω is the angular velocity (in radians per second) It is maximum at the poles, zero at the equator and changes sign from NH to SH

Coriolis Force - Summary •  The Acceleration due to the Coriolis force is

fv (x direction) and

-fu (y direction)

i.e the Coriolis Force x the velocity…

•  It only acts if water/air is moving (i.e a secondary force)

•  Acts at right-angles to the direction of motion

•  causes water/air to move to the right in the northern hemisphere

•  causes water/air to move to the left in the southern hemisphere

E.g. Foucault’s Pendulum …

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The Equations of Motion

Horizontal Equations:

Acceleration = Pressure Gradient Force + Coriolis

Vertical Equation:

Pressure Gradient force = Gravitational Force gdzdp

ρ=!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fuOr, for a Barotropic Ocean:

!

dudt

= "gd#dx

+ fv

dvdt

= "gd#dy

" fu

!

d! u dt

=1"#! F

!

dudt

=1"#Fx

dvdt

=1"#Fy

dwdt

=1"#Fz

!

dudt

= fv

dvdt

= " fuIf the only force acting on a water parcel is Coriolis, the Navier Stokes equations can be simplified to:

What does this mean?

x

y v > 0 v < 0

u > 0 u < 0

!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

!

dudt

= fv

dvdt

= " fu

If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to:


x

y v > 0 v < 0

u > 0 u < 0

Northern Hemisphere: f > 0

!

dudt

= fv

dvdt

= " fu


x

y v > 0 v < 0

u > 0 u < 0

Northern Hemisphere: f > 0

If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to:

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!

dudt

= fv

dvdt

= " fu


x

y v > 0 v < 0

u > 0 u < 0

Southern Hemisphere: f < 0

If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to: Scaling arguments:

These are the equations you need when there are both pressure and Coriolis forces in play. But this is not always going to be the case. E.g. •  If there are no surface slopes or horizontal density differences then there will

be no pressure force (i.e. left with du/dt=fv etc.). •  If we are on the equator there will be no Coriolis force

!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fug

dzdp

ρ=

Scaling arguments:

Consider the situation where there are no pressure forces


!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

If pressure gradients are small:

!

dudt

= fv

dvdt

= " fuInertia currents


!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

If pressure gradients are small:

!

dudt

= fv

dvdt

= " fu

Inertia currents

the water flows around in a circle with frequency |f|.

T=2π/f

T(Sydney) = 21 hours 27 minutes

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!

dudt

= "g d#dx

+ fv

dvdt

= "g d#dy

" fu

Scaling arguments: What forces are important in a bath tub? What kind of speeds will the water get up to? What kind of accelerations? What surface slopes? Need to look at the relative sizes of the terms in the equation. But first, choose which version of the equations are most appropriate.

!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

In a bathtub, density is pretty much constant, so conditions will be BARATROPIC

!

dudt

= "g d#dx

+ fv

dvdt

= "g d#dy

" fu

Scaling arguments: What forces are important in a bath tub? What kind of speeds will the water get up to? What kind of accelerations? What surface slopes?

255

2

1071107

1)1.0(10

1.01/1.0

−−−

−

=×=

==

==

msxxfu

msdxdg

mmdxd

η

η

Size of the pressure force:

Size of the Coriolis force:

So Coriolis<<Pressure, so we can neglect rotation effects

dxdg

dtdu η

−≈

Acceleration in the bathtub is driven by pressure differences (due to changes in surface slopes)

Scaling arguments: 1. If there are no surface slopes or horizontal density differences then there will be no pressure gradient force (i.e. left with du/dt=fv, dv/dt=-fu, inertia currents):

2. If you are sitting in your bathtub, you are in a barotropic environment AND the Coriolis force can be neglected:

!

dudt

= "gd#dx

dvdt

= "gd#dy

!

dudt

= fv

dvdt

= " fu

Scaling Analysis:

T~10 days = 8.64 x 105 s ~ 106 s

u,v ~ U ~ 1cms-1 - 1ms-1

f~ 10-4 s-1

But the ocean is not a bathtub….

We will conduct a scaling analysis on our equations of motion ...

to find further simplifications for motions with a period greater than ~10 days

!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

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Scaling Analysis:

T~10 days = 8.64 x 105 s ~ 106 s

u,v ~ U ~ 1cms-1 - 1ms-1

f~ 10-4 s-1



!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu


Acceleration << Coriolis

Scaling Analysis:

T~10 days = 8.64 x 105 s ~ 106 s

u,v ~ U ~ 1cms-1 - 1ms-1

f~ 10-4 s-1



!

dudt

= "1#dpdx

+ fv

dvdt

= "1#dpdy

" fu

!

1"dpdx

= fv

1"dpdy

= # fu

Geostrophic Balance Acceleration is much smaller than Coriolis and Pressure

Gradient Force (this is true almost everywhere in the ocean)

The ocean is in Geostrophic Balance (= balance between Pressure Gradient and Coriolis Forces)

Ocean is in

“Steady State”

(no acceleration)

du/dt is negligible

Pressure forced motion – Geostrophic Transport

Ingredients: (1) Pressure force acts from high pressure to low pressure (2) Coriolis always tries to push a moving object to the left

(SH) or right (NH)

Imagine that somehow water has been piled up in some area e.g. by the ac7on of winds. What happens when the wind stops?

H L

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Pressure forced motion – Geostrophic Current


(SH) or right (NH)

Pressure Force

SH Example (looking down on the ocean)

H L Water flow


H L



(SH) or right (NH)

Coriolis Force


H L



(SH) or right (NH)

Coriolis Force


H L



(SH) or right (NH)

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H L



(SH) or right (NH)


H L



(SH) or right (NH)


Coriolis Force

Water flow

Pressure Force

Pressure Force = Coriolis Force So, no net force on the water So keeps on going

H L



(SH) or right (NH)

Extends down to the deep ocean

40

Geostrophic Balance: How is it set up?

• What does the geostrophic balance mean physically?

• Suppose we have a difference in sea-level height.

• Water will want to move from the region of high pressure towards the region of low pressure.

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41

Geostrophic Balance: How is it set up? (cont)

•  As the water starts to move, the Coriolis effect (rotation) deflects the water to the right (NH) or left (SH).

•  The water keeps getting deflected until the force due to the pressure difference balances the Coriolis force.

•  This balance is called a geostrophic balance and the resulting current is referred to as a geostrophic current.

42

Geostrophic Currents

Geostrophic Eddy (Northern Hemisphere)

High pressure

Low pressure

horizontal pressure gradient force

Which direction is the Geostrophic wind? (f <0 SH)

PG

CF

V

x

y Concept Problem (NH):

In the NH: Which way does the current flow if sea level height is increasing towards the South? West? North?

dydgfu

dxdgfv

η

η

−=

=

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Concept Problem (NH):

In the NH: Which way does the current flow if sea level height is increasing towards the South?

dydgfu

dxdgfv

η

η

−=

=

y

NH: f>0 dη/dy<0 u=(-)(+)(+)(-)>0 East!

You can use the equations, or just think about the forces

P C

Geostrophy Problem 2: Which direction does the water flow around this pressure feature if it is in the Northern Hemisphere?

Geostrophy Problem 2: Which direction does the water flow around this pressure feature if it is in the Northern Hemisphere?

Which direction does the water flow around this pressure feature if it is in the Southern Hemisphere?

Geostrophy Problem 3:

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Which direction does the water flow around this pressure feature if it is in the Southern Hemisphere?

Geostrophy Problem 3: Anticyclone Cyclone

Northern Hemisphere

Southern Hemisphere

clockwise

counter-clockwise

Anticyclonic circulation: ² ALWAYS around a high pressure system.

² clockwise in the Northern Hemisphere

² counter-clockwise in the Southern Hemisphere

Cyclonic circulation: ² ALWAYS around a low pressure system.

² counter-clockwise in the Northern Hemisphere

² clockwise in the Southern Hemisphere

Geostrophy Problems: A certain ocean current has a height change of 1.1 m (increasing to the east) over its width of 100 km at 45° N. How fast is the current flowing?

52

f=1x10-4 s-1

g=10ms-2 Δη=1.1m Δx=100x1000m V=1 m/s

Summary: Geostrophy is the balance between pressure forces and Coriolis. In most of the open ocean (away from boundaries) motion will become geostrophic after a few days. Geostrophy doesn’t work over short periods of time or small distances (other forces become dominant). Geostrophy also fails in regions where friction becomes important

xg

dxdgfv

Δ

Δ≈=

ηη

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!

dudt

= "1#o

dpdx

+ fv

dvdt

= "1#o

dpdy

" fu

dpdz

= #g

Recap

Acceleration = Pressure + Coriolis Force }

Weight of water is balanced by its buoyancy (or vertical pressure force)

After a period of time flow becomes steady, so, acceleration becomes zero. This occurs when the pressure force balances the Coriolis force

!

dudt

= "1#o

dpdx

+ fv

dvdt

= "1#o

dpdy

" fu

dpdz

= #g

Recap

Acceleration = Pressure + Coriolis Force }

Weight of water is balanced by its buoyancy (or vertical pressure force)

After a period of time flow becomes steady, so, acceleration becomes zero. This occurs when the pressure force balances the Coriolis force

=0

=0

...)(1 ++++= fPCg FFFFdtdu

ρ

The last force to consider is friction. This is only important at continental boundaries, at the bottom of the ocean, and at the surface (due to wind) What will the friction term look like? We know that friction always tries to retard motion.

du/dt = ?

56

Effects of Friction So far we have neglected friction.

A simple model for the frictional force at the sea floor in the x and y direction is:

hrv

hru −− and

!

dudt

= "g d#dx

+ fv " ruh

dvdt

= "g d#dy

" fu " rvh

h is the depth and r is the dissipation constant

Hence the equations of motion become :

Rayleigh frictional dissipation, r is a coefficient (r ~ 10-7s-1)

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57 Time

U

Uo

h/r

1/3Uo ≈ Uoe-1

To examine the effects of friction, consider the simple balance:

A solution is:

So that at t=0, u=uo

hru

dtdu

−=

Velocity decreases with time because of friction Known as the e-folding time scale of frictional spin down.

hrtoeuu /−= h/r represents the time

it takes for the speed to drop to about 1/3 of its initial value

i.e no pressure forces and no coriolis. Motion is just decelerated because of friction

•  So far we have assumed that density ρ is constant (barotropic) •  Small horizontal changes in ρ can result in large vertical changes in current/wind – e.g. near fronts and eddies

Thermal Wind Balance

ρ1 ρ2 <

Baroclinic velocity changes with depth

Motion due to density differences


Thermal Wind Balance •  So far we have assumed that density ρ is constant (barotropic) •  Small horizontal changes in ρ can result in large vertical changes in current/wind – e.g. near fronts and eddies

• Starting with the geostrophic balance

• We can differentiate the equations with regard to depth (z), then substitute for dp/dz


!

v =1"f

dpdx

u = #1"f

dpdy

dpdz

= "g

!

1"dpdx

= fv

1"dpdy

= # fu

!

dvdz

=ddz

1"f

dpdx

#

$ %

&

' (

dudz

=ddz

)1"f

dpdy

#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddz

dpdx#

$ %

&

' (

dudz

= )1"f

ddz

dpdy#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddx

dpdz#

$ %

&

' (

dudz

= )1"f

ddy

dpdz#

$ %

&

' (

dpdz

= "g

!

u d"dz

<< "dudz

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!

v =1"f

dpdx

u = #1"f

dpdy

dpdz

= "g

!

dvdz

=ddz

1"f

dpdx

#

$ %

&

' (

dudz

=ddz

)1"f

dpdy

#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddz

dpdx#

$ %

&

' (

dudz

= )1"f

ddz

dpdy#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddx

dpdz#

$ %

&

' (

dudz

= )1"f

ddy

dpdz#

$ %

&

' (

dpdz

= "g

!

u d"dz

<< "dudz

!

dvdz

=1"f

ddx

"g( )

dudz

= #1"f

ddy

"g( )

dpdz

= "g





!

v =1"f

dpdx

u = #1"f

dpdy

dpdz

= "g

!

dvdz

=ddz

1"f

dpdx

#

$ %

&

' (

dudz

=ddz

)1"f

dpdy

#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddz

dpdx#

$ %

&

' (

dudz

= )1"f

ddz

dpdy#

$ %

&

' (

dpdz

= "g

!

dvdz

=1"f

ddx

dpdz#

$ %

&

' (

dudz

= )1"f

ddy

dpdz#

$ %

&

' (

dpdz

= "g

!

u d"dz

<< "dudz

!

dvdz

=1"f

ddx

"g( )

dudz

= #1"f

ddy

"g( )

dpdz

= "g

!

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy

dpdz

= "g


i.e. horizontal density gradients in temperature (T) and salinity (S) can explain the change in horizontal velocity with depth (vertical profile of horizontal velocity).

The vertical structure of u and v is related to the horizontal density gradients

!

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy

For geostrophic conditions: Thermal Wind Balance

Velocity is into the page

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!

dvdz

=g"f

d"dx

!

d"dx

> 0

NH: f > 0

Velocity is into the page


Velocity is into the page and increases with depth

!

dvdz

=g"f

d"dx

!

d"dx

> 0

NH: f > 0, so,

!

dvdz

> 0

i.e. v gets more positive with increasing depth

Thermal Wind Balance: Cold Core Eddy Example


Potential Density through an Eddy near the Gulf Stream. How does the geostrophic current velocity in the eddy change with depth?

cold/saline core

x y

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27.9

x y !

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy


!

dvdz

=g"f

d"dx

#g"f

$"$x

27.9

x y !

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy


Estimate the density gradient at x=40km: ρ changes from 1026.8 to 1027.2 kgm-3 over 25 km.

!

dvdz

=g"f

d"dx

#g"f

$"$x

27.9

x y !

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy


Estimate the density gradient at x=40km: ρ changes from 1026.8 to 1027.2 kgm-3 over 25 km.

!

dvdz

=g"f

d"dx

#g"f

$"$x

!

"#"x

=1026.8 $1027.2

25km= $1.6 %10$5

dvdz

=g#f

% $1.6 %10$5

= $1.5 %10$3 s$127.9

x y !

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy

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19


This means that at z = 500m, let’s see how much v varies over 100m of depth at z=500m (Δz=100m, Δv=?):

27.9

x y

!

dvdz

=g"f

d"dx

#g"f

$"$x

!

dvdz

= "1.5 #10"3 s"1


This means that at z = 500m, let’s see how much v varies over 100m of depth at z=500m (Δz=100m, Δv=?):

!

"v = #1.5 $10#3 s#1 $100m= #0.15ms#1 = #15cms#1

27.9

x y

!

dvdz

=g"f

d"dx

#g"f

$"$x

!

dvdz

= "1.5 #10"3 s"1


This means that at z = 500m, let’s see how much v varies over 100m of depth at z=500m (Δz=100m, Δv=?): i.e the velocity shear ~ -15 cm/s per 100m depth increase. This means that v (velocity into page) is getting more negative as we get deeper

!

"v = #1.5 $10#3 s#1 $100m= #0.15ms#1 = #15cms#1

27.9

x y

!

dvdz

=g"f

d"dx

#g"f

$"$x

!

dvdz

= "1.5 #10"3 s"1


To work out the actual velocities,

we need one more piece of information: at depth (say 2000m) the velocity is 0. This is called the

depth of no motion.

27.9

x y

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•  Going down in depth, the density surfaces flatten out. •  We can assume a ‘level of no motion’ where there is no longer a

change in density. •  Hence we can calculate the change in velocity up through the

water column. •  Note that on the other side of the eddy, the density surfaces

slope the other way, so the circulation must also be in the opposite sense.





V=0





V=0+.15

V=0 V is positive into the page

100m





V=+.15+.15

V=0+.15


100m

100m

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V=+.3+.15

V=+.15+.15

V=0+.15


100m

100m

100m


A physical explanation

NH f>0

dxdp

fv

ρ1

=


A physical explanation

NH f>0

Step 1: denser in the middle so the surface will be depressed

dxdp

fv

ρ1

=


NH f>0

dxdp

fv

ρ1

=

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22


NH f>0

Step 2: Start by figuring out the pressure force due to the surface slope: Pressure increases moving away from centre

P

Opposite on the other side of the eddy

Remember the fishtank experiment


NH f>0

Step 3: Forget the surface pressure part. What happens at greater depth? Density is higher at the centre than further to the right. So the pressure force, just due to density would be in the positive x-direction (on the right hand side of the gyre). The density gradient causes a pressure force that increases with depth Opposite happens on the left side of the gyre

ρ1 ρ2 <



NH f>0

Step 3: Keep repeating this down the water column, until there is no density gradient (i.e. at the bottom of the cold core eddy) But a smaller pressure force means a smaller geostropic velocity (going into the page) with depth

dxdp

fv

ρ1

=


NH f>0

Step 3: So if we add up the forces due to the surface slope and due to the density gradient we get a pressure force that decreases with depth


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NH f>0

Step 3: Keep repeating this down the water column, until there is no density gradient (i.e. at the bottom of the cold core eddy) But a smaller pressure force means a smaller geostropic velocity (going into the page) with depth

dxdp

fv

ρ1

=


NH f>0

P C

Finally we need to use the geostrophic relationship: If we have a pressure gradient in the x direction (as we do), it will create a geostrophic velocity in the y direction, proportional in strength to the pressure gradient.

C P

dxdp

fv

ρ1

=

91

•  Example: •  Derive the rotation of these cold and warm

water eddies, in the SH, using the thermal wind balance.

Example: Derive the rotation of these cold and warm water

eddies, in the SH, using the thermal wind balance.

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Given that u & v = 0 at 2000m (depth of no motion) estimate the surface current.


• Geostrophic flow in the presence of horizontal density gradients

• Horizontal density gradients (T,S) can explain vertical velocity changes

• To know absolute velocity we need extra information (i.e. we need to know absolute velocity at some depth)

!

dvdz

=g"f

d"dx

dudz

= #g"f

d"dy

For geostrophic flow (i.e. pressure is balanced by Coriolis):

-  Can figure out surface velocity from surface heights. -  Often we assume that at a certain depth (e.g. 2000m) velocities are zero – this is called “the depth of no motion”. -  Once we know the velocity at the surface or the depth of no motion we can calculate velocity at all other depths using the thermal wind equation.

Summary: Ocean Dynamics Most of the motion in the ocean can be understood in terms of Newton’s Law that

the acceleration of a parcel of water (how fast its velocity changes with time – du/dt) is related to the sum of forces acting on that parcel of water.

We can split the forces, velocities and accelerations into south-north (y,v), west-east (x,u) and up-down (z,w) components. In the vertical direction the acceleration is related to the difference between the water weight and the bouyancy (or pressure) force. When there is a vertical density gradient this leads to oscillations (Brunt Väisälä frequency N). The density gradient tries to inhibit vertical motion (and mixing). These vertical accelerations are generally very weak, so we get the hydrostatic equation.

If the hydrostatic equation is integrated over depth, it just says that the pressure at a point just equals the weight of water above that point. Acceleration in the horizontal can be driven by a number of different forces: (1) The pressure gradient force. This exists whenever there is a surface slope and/

or a horizontal density gradient. (2) Coriolis (because we live on a rotating planet). It is very weak, so we only feel its

effect over long times (> few days) and large distances (> 10s of km). Coriolis only affects moving fluids, deflecting to the right in the NH and to the left in SH.

(3) Friction. Also only acts on moving water. Always acts to slow down motion. Important at the boundaries of the ocean.

gdzdp

ρ=

Summary: Ocean Dynamics

gdzdp

ρ=

!

dudt

= "1#dpdx

+ fv " ru

dvdt

= "1#dpdy

" fu " rv

!

dudt

= "gd#dx

+ fv " ru

dvdt

= "gd#dy

" fu " rv

Or for a constant density (barotropic ocean):

Over much of the ocean, the flow is steady (i.e. du/dt=dv/dt=0) and friction is negligible, so we are left with the geostrophic balance i.e. pressure gradient forces balance coriolis. The current moves at right angles to the pressure gradient.

!

1"dpdx

= fv , 1"dpdy

= # fu

When there is a horizontal density gradient the velocity changes with depth. This can be calculated using the thermal wind equations

!

dvdz

=g"f

d"dx

, dudz

= #g"f

d"dy

P C

MSCI3001 wk5 S2-2011-dynamics4web.science.unsw.edu.au/~alexg/course_msci3001/... · 29/08/11 2 g dz...

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