MSCI3001 wk5 S2-2011-dynamics4web.science.unsw.edu.au/~alexg/course_msci3001/... · 29/08/11 2 g dz...
Transcript of 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
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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:
What does this mean?
x
y v > 0 v < 0
u > 0 u < 0
Northern Hemisphere: f > 0
!
dudt
= fv
dvdt
= " fu
What does this mean?
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
What does this mean?
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
Acceleration = Pressure Gradient Force + Coriolis
!
dudt
= "1#dpdx
+ fv
dvdt
= "1#dpdy
" fu
If pressure gradients are small:
!
dudt
= fv
dvdt
= " fuInertia currents
Acceleration = Pressure Gradient Force + Coriolis
!
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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But the ocean is not a bathtub….
Scaling Analysis:
T~10 days = 8.64 x 105 s ~ 106 s
u,v ~ U ~ 1cms-1 - 1ms-1
f~ 10-4 s-1
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
!
dudt
= "1#dpdx
+ fv
dvdt
= "1#dpdy
" fu
But the ocean is not a bathtub….
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
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
!
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
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)
Pressure Force
SH Example (looking down on the ocean)
H L Water flow
SH Example (looking down on the ocean)
H L
Pressure forced motion – Geostrophic Current
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)
Coriolis Force
SH Example (looking down on the ocean)
H L
Pressure forced motion – Geostrophic Current
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)
Coriolis Force
SH Example (looking down on the ocean)
H L
Pressure forced motion – Geostrophic Current
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)
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SH Example (looking down on the ocean)
H L
Pressure forced motion – Geostrophic Current
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)
SH Example (looking down on the ocean)
H L
Pressure forced motion – Geostrophic Current
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)
SH Example (looking down on the ocean)
Coriolis Force
Water flow
Pressure Force
Pressure Force = Coriolis Force So, no net force on the water So keeps on going
H L
Pressure forced motion – Geostrophic Current
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)
Extends down to the deep ocean
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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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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.
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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?
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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
• 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 • 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
Thermal Wind Balance
!
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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• 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
Thermal Wind Balance
!
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
• 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
Thermal Wind Balance
!
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
Thermal Wind Balance
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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Thermal Wind Balance
!
dvdz
=g"f
d"dx
!
d"dx
> 0
NH: f > 0
Velocity is into the page
Thermal Wind Balance
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
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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Thermal Wind Balance: Cold Core Eddy Example
27.9
x y !
dvdz
=g"f
d"dx
dudz
= #g"f
d"dy
Thermal Wind Balance: Cold Core Eddy Example
!
dvdz
=g"f
d"dx
#g"f
$"$x
27.9
x y !
dvdz
=g"f
d"dx
dudz
= #g"f
d"dy
Thermal Wind Balance: Cold Core Eddy Example
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
Thermal Wind Balance: Cold Core Eddy Example
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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Thermal Wind Balance: Cold Core Eddy Example
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
Thermal Wind Balance: Cold Core Eddy Example
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
Thermal Wind Balance: Cold Core Eddy Example
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
Thermal Wind Balance: Cold Core Eddy Example
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.
• 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
• 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+.15
V=0 V is positive into the page
100m
• 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=+.15+.15
V=0+.15
V=0 V is positive into the page
100m
100m
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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=+.3+.15
V=+.15+.15
V=0+.15
V=0 V is positive into the page
100m
100m
100m
Thermal Wind Balance: Cold Core Eddy Example
A physical explanation
NH f>0
dxdp
fv
ρ1
=
Thermal Wind Balance: Cold Core Eddy Example
A physical explanation
NH f>0
Step 1: denser in the middle so the surface will be depressed
dxdp
fv
ρ1
=
Thermal Wind Balance: Cold Core Eddy Example
NH f>0
dxdp
fv
ρ1
=
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22
Thermal Wind Balance: Cold Core Eddy Example
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
Thermal Wind Balance: Cold Core Eddy Example
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 <
Remember the fishtank experiment
Thermal Wind Balance: Cold Core Eddy Example
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
=
Thermal Wind Balance: Cold Core Eddy Example
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
Remember the fishtank experiment
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23
Thermal Wind Balance: Cold Core Eddy Example
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
=
Thermal Wind Balance: Cold Core Eddy Example
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.
Thermal Wind Balance
• 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