Chapter 4
Wind Forced Motions
Topics We Will Cover
1.
2.
3.
4.
5.
6.
Wind Stress
Ekman Layer Depth
Ekman Transport
Wind driven Ocean Circulation
Storm Surge, Downwelling
Upwelling
Reading:
Ch8 Cushman Roisin
Ch 10 Marshall
http://winds.jpl.nasa.gov/imagesAnim/animations.cfm
1
Dispersion Relationship
For a 2D wave travelling in the x direction:
d 2η
d 2η
− gh 2 = 0
2
dt
dx
Assumptions:
•No Coriolis
•No friction
•Constant density
•Shallow water/
motion of water
particles x-dir only
η = Acos(kx − ωt)
Try a wave like
solution
ω2 = hgk 2
Dispersion relation
2
Dispersion Relationship
For a 2D wave travelling in the x direction:
If we relax the shallow water approximation, so water
particles can move in x and z direction:
ω = gk tanh(kh)
2
Dispersion relation
But should be able to get back to shallow water dispersion relation for h<<λ …
3
Phase Velocity
ω2 = gk
h>>λ
ω2 = gk 2 h
h<<λ
Deep water waves: h > λ /2:
c=
g
=
k
gλ
2π
Shallow water waves: h < λ/20:
c = gh
ω
c=
k
Bigger wavelength =
faster waves, thus
DISPERSIVE
Shallow water =
Slower waves
4
3.2 Refraction, Diffraction and Shoaling
Wave Refraction:
Wave crests tend to become parallel to
the shore as the wave moves inshore…
Because c=√gh, c decreases as h
decreases
C j = gh j > C i = ghi
Ci
Cj
5
‘Clean’ Waves: Deep water waves
• The longer wavelength waves reach shore quicker than
the shorter wavelength waves,
• Because speed depends on their wave length.
• Waves that have traveled a long distance are ‘sorted’
into wave packets, creating ‘clean’ swell.
6
Phase Speed and Group Speed
Superposition of two
sinusoidal travelling
waves
Shows the difference
between:
•
•
Distance travelled by a crest
Phase Speed (Cp) of
the wave crests &
Group Speed (Cg) of
the wave envelope.
Distance travelled by the group
Gill1981
7
Group Velocity
• Group Velocity is the velocity at which a group of waves
travels, & the velocity that wave energy propagates:
dω
cg ≡
dk
Remember the
Phase Velocity
is given by
• Using the approximations for the dispersion
relation it can be shown:
• Deep Water Group Velocity
ω 2 = gk
h > λ/4
• Shallow-water Group Velocity
ω 2 = gk 2 h h < λ / 11
ω
c=
k
g
c
=
cg =
2ω 2
c g = gh
8
Tide generating
forces: Sum of
centrifugal and
gravitational forces
Centrifugal force:
Centrifugal
Centrifugal
forces are are
the same
everywhere …
Gravitational force:
Gravitational
force is
greatest
closest to the
sun/moon
Gravitational
• Spring tides occur when M2, and S2 reinforce.
• Neap tides occur when M2, and S2 oppose. Successive
springs occur about every 14 days.
• Mixed tides occur when the relative magnitude of K1 and 01
are sufficiently large that they interfere with the semi-diurnal
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tides.
Effect of rotation and boundaries
Co-range lines (lines of constant tidal range) run around amphidromic points in
quasi-circular fashion.
Co-phase lines (lines of constant phase, or lines which connect all places where
high water occurs at the same time) emanate from amphidromic points like
spokes of a wheel.
11
Types of Vorticity
Two types of rotation
1. Planetary Vorticity (rotating because the planet
is rotating)
f = 2 Ω sin ( φ )
2. Relative Vorticity (spinning on its own axis)
dv du
ζ=
−
dx dy
3. Total or Potential Vorticity
f +ζ
Q=
h
12
Change in Relative Vorticity
Column Stretching
Production of relative vorticity by changes in the height of a
fluid column.
As the vertical fluid column moves from left to right:
•vertical stretching reduces the moment of inertia of the
column,
•causing it to spin faster.
We are not considering the effect of planetary vorticity here.
(Stewart)
13
Relative Vorticity
Calculation
A cold core eddy (diameter 100km) formed at 30S, 200E, in 1km
depth water has:
the north end is 1.2 m/s,
the south end -1.1 m/s,
the west side 1.1 m/s,
the east side -1.5 m/s.
1. What is the relative, planetary and absolute vorticity of this eddy? Is it
cyclonic or anticyclonic?
2. The eddy moves east to 220E, what happens to the eddys
relative/planetary vorticity? Why?
3. The eddy moves south to 40S, what happens to the eddys
relative/planetary vorticity?
4. The water then moves into a region where depth increases to 1.5km.
What are the possible changes to the eddy?
Schmitz (1996)
15
Wind Stress
• Wind blowing over the ocean produces a force per unit
area
τ xτ y
• Given by
(τ x ,τ y ) ≈ ρ air cD (u , v )
2
2
• where CD is a drag co-efficient, typically 0.002
• ρair is 1.3 kg/m3
• u,v is the speed of the wind (x,y compnent) at a height of
10m above sealevel
• Units are Pa or N/m2
16
Wind stress
U10
•
•
Analogous to a wind blowing over the ocean.
It is for this reason that we generally measure wind speed at a
height of 10m above the ocean.
Cushman Roisin Ch8
17
Is the Ocean speed important?
(τ x ,τ y ) ≈ ρ air cD (u , v )
2
2
CD 0.002
ρair is 1.3 kg/m3
u is the speed at ~10 m, but the important number
is actually the relative wind speed, between ocean
and atmosphere.
If u=10m/s calculate the wind stress
What is the % error, if the ocean speed id 1m/s in
the same direction as the wind?
Cushman Roisin Ch8
18
Wind drag
(τ x ,τ y ) ≈ ρ air cD (u , v )
2
2
The drag coefficient is a measure of friction
between wind and ocean, but this depends on the
roughness of the surface
CD = (0.44
+ 0.063 U10 )/1000
Annual Mean Surface Wind Stress
CI = 0.05 N/m2
Marshall Ch10
20
Geostrophic Balance
In the equation of motion, acceleration is 1/100
smaller than the Coriolis Force and Pressure Force
Over long
timescales
ocean is in
“Steady State”
(no acceleration)
Thus the balance is
between Pressure
and Coriolis Forces
-Geostrophic
Balance
OR
∂η
= fv
∂x
∂η
g
= − fu
∂y
g
For a barotropic ocean
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.
du 1
a=
= ∑F
dt m
τ force per m2
To get F/m need to divide by (ρh)
F
τ
=
m ρh
Check that the units are correct
Ekman Velocity
Consider the steady (accn=0) equations of motion for a barotropic
ocean of depth h
In the absence of wind with weak (zero) bottom stress the equations of
motion would be the geostrophic equations (Ch 2).
du
dη
= −g
+ fv + friction
dt
dx
dη
dv
= −g
− fu + friction
dy
dt
If we add surface wind forcing they become
dη τ x
− fv = − g
+
dx ρh
dη τ y
fu = − g
+
dy ρh
• We split the velocity field into the sum of the geostrophic
and wind driven components i.e.
u = u g + ue
v = v g + ve
• Where ug, vg denote geostrophic velocities such that
dη
− fv g = − g
dx
dη
fu g = − g
dy
25
• We are then left with a wind-driven component
of velocity (ue, ve )
• i.e the remainder of velocity must balance the
wind forcing
τx
− fv e =
ρh
τy
fu e =
ρh
These equations give the velocity averaged over
the full ocean depth. In fact the direct effects of
wind only felt over Ekman layer depth (He)
τx = 0 τy > 0
Steady State
• Assume initially
• Initially there is no motion
• Column accelerates,
• As the water column moves faster,
–
–
–
–
•
Coriolis force increases (fv)
Column is deflected to the right (NH)
This continues until Ve vanishes and
A steady state is reached.
Coriolis and Wind forcing balance
27
The Ekman Depth (He)
The effects of the wind stress on the ocean extend through the surface
of the ocean to a depth He
He =
2K
| f |
K is the vertical eddy diffusivity(viscosity) which indicates vertical
mixing.
Typically K~2 x 10-2 m2/s
so that with |f|=10-4 s-1, He ~20m
In regions of high wind driven
turbulence, K increases (K~0.5 m2/s)
so that He can reach ~100m
Marshall Ch10
The wind stress
diminishes to
zero with depth
28
Depth Averaged Ekman Layer
•
•
•
Direct effects of wind only felt over Ekman
layer depth (He)
What then is the velocity in the Ekman layer?
Ekman layer velocities (Ue and Ve):
h
τy
h
Ue =
ue =
He
ρH e f
τx
h
Ve =
ve =
He
ρH e f
Cushman Roisin Ch8
29
Depth Averaged Ekman Velocities
• While driven by the wind stress, Ue, Ve are actually
perpendicular to the wind stress
wind
τy
ρh
Ekman Velocity
Ue
fu e
Northern Hemisphere, Aerial View, F >0
fve
τx
ρh
wind
Ekman Velocity
Ve
30
Question P 48
a) Find the Ekman layer depth He and the Ekman
Velocity ue and ve when a wind of stress 0.5 Pa
(0.5N/m2) blows to the north in the mid-latitudes
(NH). Assume h is 1000m, take K=0.1 m2/s and
f=10-4s-1
b) What is the Ekman layer velocity Ue and Ve ?
c) Sketch the resultant flow
31
τx
− fve =
=0
ρh
τy
0.5
fue =
=
ρ h 1000(1000)
ue =
0.5
×10000
1000(1000)
2K
He =
| f |
τy
=0.005m/s
=√(2x0.1/10-4)=44.7m
0.5
Ue =
=
= 0.11m / s
−4
ρ H e f 1000 × 44.7 ×10
Ekman Spiral
Consider the ocean in terms of multiple layers:
First layer - forces acting are wind, friction from layer below and Coriolis
1.
Wind
Friction
Coriolis
Friction
2.
Wind
Coriolis
At the next layer:
But, in this configuration forces can’t balance
Flow curves to the right due to Coriolis – and speeds up
due to wind. As flow changes direction and increases in
speed Coriolis and Friction change direction and become
greater also. Wind (an external force) remains the same.
But now the forces can balance
Wind
3.
Fri
c
Flow adjusts until a balance can exist:
tio
Fo
n
Co
rio
li
Fo
rc
s
ea
bov
e
Fr i
Layer 2 current at an
even greater angle
And so on for subsequent layers
Coriolis
Friction
Surface current at an
angle to the wind
c t io
rce
n
abo
ve
Co
ri
ol i
s
34
Summary
• On average wind driven ‘Ekman flow’ is at
90 degrees to the right (left) of the wind
direction in the NH (SH)
• Ekman flow is confined to the top few 10s
of meters (unlike geostrophic flow)
• Because of friction between layers of
water the flow actually forms a spiral that
diminishes with depth (with the average
flow perpendicular to the wind)
Divergence and convergence
• If there are differences in
Ekman
wind
wind strength (or direction)
it can lead to converging
or diverging flow
• This causes a piling up of
Convergence
water (for convergence) or
Ekman
a trough in the surface (for wind
divergence)
Divergence
• This leads to pressure
wind
gradients
Ekman
NH
Gyre – Southern Hemisphere
Consider a rectangular basin
with a wind stress as shown
The mass transports are
convergent in the centre of
the Basin causing a sealevel rise.
The pressure gradient drives a
geostrophic current
Winds
Ekman Transport
Geostrophic Current
H
H
37
Sea Surface Elevation
Jason Satellite uses radar altimetry to collect
sea surface height data of all the world's
oceans.
Images are processed to highlight the interannual signal of sea surface height.
The mean signal, seasonal signal, and the
trend have been removed.
38
Now it’s your turn ……
Use the map of the global ocean wind field.
Using the concepts that we have been discussing, draw:
1.arrows showing the Ekman transport
2.contours showing the adjusted sea-level
3.the upper ocean currents
39
EQ
40
Remember Ekman Transport is:
the movement of the surface waters as a response to
wind forcing
EQ
EKMAN TRANSPORT
41
Ekman Transport causes a rise in the sea surface
elevation… HIGH
HIGH
EQ
HIGH
HIGH
SEALEVEL ELEVATION
HIGH
HIGH
42
Because of the rise in sea surface elevation a
balance is set up between the pressure gradient
(PG) and rotation (CF)
HIGH
HIGH
EQ
HIGH
HIGH
HIGH
PG
CF
SEALEVEL ELEVATION
43
This causes Currents in the ocean… Geostrophic
currents
HIGH
EQ
HIGH
HIGH
HIGH
HIGH
PG
CF
GEOSTROPHIC FLOW
44
46
Wind Forced Ocean Circulation
•
•
Schematic diagram showing the classification of ocean gyres and
major ocean current systems and their relation to the prevailing
zonal winds.
Patterns of Ekman transport in regions of upwelling and downwelling
are also marked.
Marshall Ch10
47
48
49
Q 2b:
f = 10-4 s-1
τx = 0.1 Pa
h = 4000 m
ρ = 1000 kg m-3
He = 200 m
2K
He =
| f |
Ekman velocities:
(depth-averaged)
ve = -τx / ρhf
= (-0.1) / (1000 x 4000 x 10-4)
= 2.5 x 10-4 m s-1
Ekman transport:
(Ekman-layer velocity)
= (4000 / 200) x 2.5 x 10-4 m s-1
= 5 x 10-3 m s-1
50
Summary
• Wind Stress on the ocean surface moves
the ocean
• Effects are felt through the Ekman Layer
• Major ocean gyres are driven by the
resulting geostrophic balance.
51