Download Wind Forced Motion

12/09/11
Physical oceanography, MSCI 3001
Oceanographic Processes, MSCI 5004
Dr. Alex Sen Gupta
Equation of motion a=ΣF/ρ
What forces might cause a parcel of water
to accelerate?
[email protected]
Movement of the ocean (a)
Wind Forced Motion
= (1/ρ)x( wind - friction + rotation + tides
gravity+ buoyancy + pressure differences ….)
A sum of forces.
du 1
= ( Fg + FC + FP + Ff + ...)
dt ρ
Newton’s Laws of Motion
Vertical direction:
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)
P=F/A
Ftop
dw
1 dp
=−
+g
dt
ρ dz
In general over the ocean vertical acceleration is much smaller than g. This
means that in the vertical equation g and 1 dp must be of similar magnitude
ρ dz
P
dp
= ρg
dz
Acceleration = 1/ρ(sum of forces)
= 1/ρ(weight + Ftop - Fbottom)
Bouyancy due to
difference in
pressures
P+ΔP
Fbottom
Weight (mg)
z
dw
1 dp
=g−
dt
ρ dz
This is called the
hydrostatic equation
We can integrate this equation since density and g are essentially constant.
p
h
dp
∫0 dz dz = ∫0 ρgdz
Or simply
p = hρg
Which you are hopefully
familiar with already!
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Hydrostatic equation tells us that pressure =
weight of water above you (in this case higher
on the right)
Newton’s Laws of Motion
Horizontal direction
Acceleration = 1/ρ(sum of forces)
= 1/ρ(Fleft - Fright)
P
Barotropic and Baroclinic Motion
du
1 dp
=−
dt
ρ 0 dx
P+ ΔP
and du = − 1 ∂p
Remember, p = ρgz ,
dt
ρ ∂x
Mixed situation
Fright
Fleft
x
ρ1
Or doing the same in the y-direction:
dv
1 dp
=−
dt
ρ 0 dy
Barotropic Ocean
For a constant density ocean, we can write the pressure
gradient in an easier way.
du
1 ∂p
=−
Remember, p = ρgz , and
dt
ρ ∂x
η1
η2
d
P2=h2ρg
P1=h1ρg
h1=d+η1
So we are left with
Δx
h1=d+η1
du
∂η
= −g
dt
∂x
∂p Δp p2 − p1
=
=
∂x Δx
Δx
h ρg − h1 ρg
= 2
Δx
(d + η 2 ) ρg − (d + η1 ) ρg
=
Δx
(η 2 − η1 )
= ρg
Δx
∂p
Δη
= ρg
∂x
Δx
<
ρ2
Motion due to density differences
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
!
du 1 !
= #F
dt "
!
du 1
= #F
dt " x
Simplifying the Equations:
dv 1
= #F
dt " y
Consider the situation where there are no pressure forces
dw 1
= #F
dt " z
Acceleration = Pressure Gradient Force + Coriolis
du
1 dp
="
+ fv
dt
# dx
Horizontal Equations:
!
Acceleration = Pressure Gradient Force + Coriolis
du
1 dp
="
+ fv
dt
# dx
dv
1 dp
="
" fu
dt
# dy
Or, for a Barotropic Ocean:
dv
1 dp
="
" fu
dt
# dy
du
d#
= "g
+ fv
dt
dx
dv
d#
= "g
" fu
dt
dy
If pressure gradients are small:
du
= fv
dt
dv
= " fu
dt
!
Vertical Equation:
!
!
Pressure Gradient force = Gravitational Force
dp
= ρg
dz
Inertia currents
!
Acceleration = Pressure Gradient Force + Coriolis
Scaling arguments:
What forces are important in a bath tub?
du
1 dp
="
+ fv
dt
# dx
What kind of speeds will the water get up to? What kind of accelerations? What
surface slopes?
dv
1 dp
="
" fu
dt
# dy
Need to look at the relative sizes of the terms in the equation. But first, choose
which version of the equations are most appropriate.
du
1 dp
="
+ fv
dt
# dx
dv
1 dp
="
" fu
dt
# dy
If pressure gradients are small:
Inertia currents
!
du
= fv
dt
dv
= " fu
dt
the water flows
around in a circle
with frequency |f|.
du
d#
= "g
+ fv
dt
dx
dv
d#
= "g
" fu
dt
dy
T=2π/f
T(Sydney) = 21 hours 27 minutes
!
In a bathtub, density is pretty much constant, so conditions will be BARATROPIC
!
!
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f=1.5x10-4 x sin(lat)
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?
Size of the pressure force:
dη
= 0.1m / 1m = 0.1
dx
dη
g
= 10(0.1) = 1ms − 2
dx
Size of the Coriolis force:
fu = 7 x10 −5 ×1 = 7 x10 −5 ms − 2
du
d#
= "g
+ fv
dt
dx
dv
d#
= "g
" fu
dt
dy
So Coriolis<<Pressure, so we can
neglect rotation effects
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
du
1
="
dt
#
Scaling Analysis:
dv
1
T~10 days = 8.64 x 105 s ~ 106 s
="
dt
#
u,v ~ U ~ 1cms-1 - 1ms-1
dp
+ fv
dx
dp
" fu
dy
f~ 10-4 s-1
du
dη
≈ −g
dt
dx
!
Acceleration in the bathtub is driven by pressure
differences (due to changes in surface slopes)
Acceleration << Coriolis
!
Pressure forced motion – Geostrophic Transport
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
du
1 dp
="
+ fv
dt
# dx
1 dp
= fv
" dx
dv
1 dp
="
" fu
dt
# dy
1 dp
= # fu
" dy
dη
dx
dη
fu = − g
dy
fv = g
Or if conditions are baratropic (constant density)
!
!
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? L H 4
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Pressure forced motion – Geostrophic Current
Geostrophic Currents
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 = Coriolis Force SH Example So, no net force on the water (looking down on the ocean) So keeps on going L Pressure Force High pressure
H Low pressure
horizontal pressure
gradient force
Coriolis Force Extends down
to the deep
ocean
Water flow Geostrophic Eddy
1 dp
= fv
! dx
(Northern Hemisphere)
Which direction is the
Geostrophic wind? (f <0 SH) y
Geostrophic Current
•  Current that is produced when Pressure Force and
Coriolis Force balance
•  In the NH the final geostrophic current will flow at 90
degrees to the right of the pressure force
•  In the SH the final geostrophic current will flow at 90
degrees to the left of the pressure force
x
PG
CF
V
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1
!
1
!
dp
= fv
dx
dp
= ! fu
dy
dη
dx
dη
fu = − g
dy
fv = g
Like this the equations say:
Pressure force is balancing
Coriolis
Like this the equations say:
If we know the pressure force (or surface
slope), we can calculate how fast the water
is moving
1 dp
! f dx
1 dp
u=!
! f dy
v=
g d!
f dx
g d!
u=!
f dy
v=
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Effects of Friction
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
So far we have neglected friction.
A simple model for the frictional force at the sea floor in the x and y
direction is:
− ru
− rv
and
h
h
h is the depth and r is the dissipation
constant
Baroclinic
velocity changes with depth
CORRECTION
Rayleigh frictional dissipation, r is a coefficient (r ~ 10-7ms-1)
du
d#
ru
= "g
+ fv "
dt
dx
h
dv
d#
rv
= "g
" fu "
dt
dy
h
Hence the equations of motion become :
ρ1
<
ρ2
Motion due to density differences
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!
Thermal Wind Balance
Thermal Wind Balance
For geostrophic conditions:
The vertical structure of u and v is related to the horizontal
density gradients
dv
g d"
=
dz "f dx
du
g d"
=#
dz
"f dy
Baroclinic
velocity changes with depth
NO CORIOLIS
Baroclinic
velocity changes with depth
WITH CORIOLIS
THERMAL WIND
What does this mean…
d!
dx dv
Then the north-south water velocity will change with depth
dz
If there are density changed in the east-west direction
!
ρ1
<
ρ2
ρ1
<
ρ2
Motion due to density differences
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Use thermal wind to calculate the east-west water speed at 45 S?
Use thermal wind to calculate the east-west water speed at 45 S?
S Pole
1000km
Equator
dv g d !
=
dz ! f dx
du
g d!
=!
dz
! f dy
1027
2000
m
1027.5
Depth of no motion
d ! 1027 !1027.5
=
dy
1000, 000
=-5x10-7 kg/m4
Given that u & v = 0 at 2000m (depth of no motion)
Given that u & v = 0 at 2000m (depth of no motion).
du
10
=!
(!5 "10 !7 )
dz
1000(!10 !4 )
=-0.00005 m/s/s
Use thermal wind to calculate the east-west water speed at 45 S?
Thermal Wind Balance
For geostrophic flow (i.e. pressure is balanced by Coriolis):
S Pole
1000km
Equator
1027
2000
m
1027.5
Depth of no motion
du
= !0.00005 m/s/s
dz
This means that every meter that you go
downwards the eastward water velocity (u)
decreases by 0.00005m/s
OR
that every meter that you go upwards the
eastward water velocity (u) increases by
0.00005m/s
Moving upwards 2000m the increase in u
velocity will be 2000x0.00005=0.1m/s
But we know at 2000m u=0
So the surface velocity will be 0 + 0.1 = 0.1m/s
• 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)
dv
g d"
=
dz "f dx
du
g d"
=#
dz
"f dy
!
-  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.
Water at the surface is moving at 10cm/s to
the east (the Antarctic Circumpolar Current)
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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.
dp
= ρg
dz
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.
Summary: Ocean Dynamics
Or for a constant density (barotropic ocean):
du
1 dp
="
+ fv " ru
dt
# dx
dv
1 dp
="
" fu " rv
dt
# dy
du
d#
= "g
+ fv " ru
dt
dx
dv
d#
= "g
" fu " rv
dt
dy
dp
= ρg
dz
!
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
! at right angles to the pressure gradient.
balance coriolis. The current moves
P
1 dp
1 dp
= fv ,
= # fu
" dx
" dy
C
When there is a horizontal density gradient the velocity changes with depth.
This can be calculated using the thermal wind equations
!
dv
g d"
=
dz "f dx
,
du
g d"
=#
dz
"f dy
!
Wind Driven Motion
Background – wind forced motion
Track of the Fram as Fridtjof Nansen
attempts to drift to the North Pole in his
ship (1893-1896).
Vagn Walfrid Ekman (1874-1954),
Swedish oceanographer, developed
theory of Ekman spiral from
Nansens observations.
http://maritime.haifa.ac.il/departm/lessons/ocean/lect01.htm
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Wind forced motion – EkmanTransport
Ingredients:
(1)  Wind force acts in the direction of the wind
(2)  Coriolis always tries to push a moving object to the left
(SH) or right (NH)
Wind forced motion – EkmanTransport
Ingredients:
(1)  Wind force acts in the direction of the wind
(2)  Coriolis always tries to push a moving object to the left
(SH) or right (NH)
SH Example (looking down on the ocean) Wind Force Wind forced motion – EkmanTransport
Ingredients:
(1)  Wind force acts in the direction of the wind
(2)  Coriolis always tries to push a moving object to the left
(SH) or right (NH)
Wind Force = Coriolis Force SH Example So, no net force on the water (looking down on the ocean) So keeps on going Wind Force Coriolis Force Only affects water
in the top few 10s
of meters
Equations of Motion
du
d!
"
= !g
+ fv + x ! ru
dt
dx
#h
"
dv
d!
= !g
! fu + y ! rv
dt
dy
#h
i.e the acceleration of water towards the east depends on:
•  any east-west pressure force
•  The velocity of the water in the north-south direction (which
determines the strength of the east west Coriolis force
•  The east-west wind force
•  The velocity of water in the east-west direction (which
determines the strength of the friction)
Water flow 9
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Equations of Motion
du
d!
"
= !g
+ fv + x ! ru
dt
dx
#h
"
dv
d!
= !g
! fu + y ! rv
dt
dy
#h
For large scale, long timescale,
throughout most of the ocean the
terms that matter are Coriolis and
Pressure forces
Near the surface in the presence of
sustained wind, we also need to
worry about the wind force.
Without Coriolis
Equations of Motion
If we know the wind force
(normally called wind stress) we
can calculate the Ekman
transport
!x
Wind driven Ekman transport is
a balance between Coriolis and
!x
wind forces
fve = !
d!
dx
d!
fug = !g
dy
"x
fve = !
#h
"y
fue =
#h
fvg = g
With Coriolis
ve = !
"h
f "h
!y
ue =
f "h
!
fue = y
"h
In words: Ekman transport is proportional to the strength of the wind. Ekman flow
is at 90° to the right of the direction of the wind in the NH and 90° to the left of the
direction of the wind in the southern hemisphere. NB the Ekman transport in
these equations are the depth averaged current. In reality Ekman transport
only occurs in the top few 10s of meters (the Ekman Layer)
E.G Wind blowing to the north in the southern hemisphere
Τy>0 and f<0, so ue <0 i.e Ekman transport is to the west (90° to the left of the
direction of the wind)
Ekman Layer
- wind blowing on an ocean produces a force per unit area called a wind
stress τ.
-  effects of a wind stress on the ocean surface are transmitted down
through the water column by the action of
•  turbulent eddies that are themselves generated by the wind
•  breaking waves
•  boundary shear stresses
Depth
- depth to which the effects of wind are felt is called
the Ekman layer thickness or Ekman depth (He), where
Net flow of water
is 90° to
direction of wind
He =
2K
| f |
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Ekman Layer – cont‘d
- depth to which the effects of wind are felt is called
the Ekman layer thickness or Ekman depth (He), where
He =
2K
| f |
-  Coriolis force and wind balance depth-averaged Ekman
velocities (ue and ve)
τx
-  But u and v in these equations
- fve = ρh
are depth averaged velocitties
and we know that flow is actually
τy
onlu in the surface Ekman layer
fue =
ρh
depth (He)
- where K is the eddy diffusivity and
- K ≃ WL where W and L denote a characteristic eddy
velocity and size
So need to scale up to get the
Ekman later velocities
He
Ekman layer velocities (Ue and Ve):
Typically K ≃ 2 × 10−2 m2 s−1, so that with |f| = 10−4 s−1,
He ≃ 20 m.
However in regions of high wind-driven turbulence, K can be up to 0.5
m2 s−1, so that He can reach ≃ 100 m.
h
Ue =
τy
h
ue =
He
ρH e f
Ve =
τx
h
ve = −
He
ρH e f
Exercise:
Find the Ekman layer depth He and the Ekman layer velocities Ue and
Ve when a wind of strength 0.5 Pascals (0.5 N/m2) blows towards
the north in the mid-latitudes of the NH.
Take: vertical diffusivity K = 0.1m2/s and f = 10-4 s-1 and h = 1000m
He =
2K
| f |
Ue =
τy
h
ue =
He
ρH e f
Ve =
τx
h
ve = −
He
ρH e f
Schematic of movement in column of water in the Ekman layer.
Which hemisphere is this example from?
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Upwelling &
Downwelling
E.g. Northern
Hemisphere
Schematic diagrams of winddriven upwelling in SH.
http://oceanmotion.org/html/background/upwelling-and-downwelling.htm
TEMP
Winds
SST in California
Current under the
influence of a northerly
alongshore wind.
NITRATE
Temperature and nitrate
distributions off Pt. Sur,
California, from satellite and
shipboard operations. From
Traganza et al. (1983).
Mann & Lazier (1996)
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Full effect of alongshore wind …
SH
Wind
Upwe
lling
Geostrophic current
Ekman current
1.  Wind drives offshore
Ekman Current in first
few 10’s of meters.
2.  In boundary layer,
offshore flow is
compensated by
upwelling
3.  Because there is a
surface slope, there
must also be a
pressure force (towards
the coast)
4.  This drive a geostrophic
current that stretches
through the full depth of
the water column
5.  Except at very bottom
(where friction is
important) and there is
an onshore current
Storm Surge
Storm Surge
Idealised model of a
storm surge.
Which hemisphere?
Storm Surge
A storm surge is an offshore rise of water associated with a low pressure weather
system, typically tropical cyclones and strong extratropical cyclones.
When rotation effects
are not important –
For example, in semienclosed bays
5) The effect of waves, while directly powered by the wind, is distinct from a
storm's wind-powered currents. Powerful wind whips up large, strong waves in
the direction of its movement
6) The rainfall effect is experienced predominantly in estuaries. Hurricanes may
dump as much as 300 mm of rainfall in 24 hours over large areas, and higher
rainfall densities in localized areas
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Exercise:
Exercise:
Q 4)
Q 5)
A low-pressure system sits off the NSW coast as shown below.
The southerly winds exert a stress τy = 1 N/m2 over a period of 5 days.
Assuming an Ekman layer depth He = 100m over the shelf region,
calculate the onshore Ekman layer velocity Ue.
Calculate how far and in which direction the front at A will move
during the 5 day period.
Comment on the likely sealevel change and resulting geostrophic
adjustment.
The wind stress τ and wind speed U are related by
τ = CD ρa U2
where ρa = 1.23 kg/m3 (air density)
and
CD = 10-3 (0.61 + 0.063U)
is a drag coefficient for U between 6m/s and 22m/s.
Given U = 8m/s calculate τ.
The units of τ are kg/(ms2) = 1 N/m2 = 1 Pa.
A northerly wind speed of U = 10m/s blows off the west coast
of
Africa during 9-13 March. Draw an idealised sketch of the
resulting
Ekman transport, sealevel change, and geostrophic
adjustment to this wind field. Discuss this diagram in relation to the
observations
of upwelling shown below.
Q 6a)
u<0
Density
Eastward velocity (u)
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Q 7)
Discuss the density sections drawn below. The sections
correspond to waters off the NSW coast.
Why might the density field slope upward in the initial conditions?
What can you say about summertime swimming conditions in Sydney?
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