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Ekman FlowEkman Flow
September 27, 2006September 27, 2006
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22
Remember from last time…Remember from last time…
2
2
2
22
2
22
2
22
1
1
yx
z
vAvAfu
y
p
dt
dv
z
uAuAfv
x
p
dt
du
H
zHH
zHH
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33
fvdt
duu
fudt
dvv
0)(2
1 22 uvuvfvudt
d 0
22
dt
dc
vuc
Define c (current speed) as:
u and v change, but c stays constant: Coriolis force does no work!
002
12
2
dt
dc
td
dc
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Flow is in a circle: Inertial or Centripetal force = Coriolis force
fcr
c
2where
f
cr
Inertial radius
If
c ~ 0.1 m/s
f ~ 10-4
then
r ~ 1 km
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55
Inertial Period is given by T where:
fT f
2
If
f ~ 10-4
then
T ~ 6.28x104sec ~ 17.4 hrs
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66
r
fc
fc
fc
fc
cc
cc
c2/r
c2/r
c2/r
c2/r
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77
Latitude () Ti (hr) D (km)
for V = 20 cm/s
90° 11.97 2.7
35° 20.87 4.8
10° 68.93 15.8
Table 9.1 in Stewart
Inertial Oscillations
Note: V is equivalent to c from previous slides, D is equal to the diameter
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88
Figure 9.1 in Stewart
Inertial currents in the North Pacific in October 1987
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99
Ekman flowEkman flow
Fridtjof Nansen noticed that wind tended to blow Fridtjof Nansen noticed that wind tended to blow ice at an angle of 20ice at an angle of 20°-40° to the right of the wind °-40° to the right of the wind in the Articin the Artic
Nansen hired Ekman (Bjerknes graduate Nansen hired Ekman (Bjerknes graduate student) to study the influence of the Earth’s student) to study the influence of the Earth’s rotation on wind-driven currentsrotation on wind-driven currents
Ekman presented the results in his thesis and Ekman presented the results in his thesis and later expanded the study to include the influence later expanded the study to include the influence of continents and differences of density of water of continents and differences of density of water (Ekman, 1905)(Ekman, 1905)
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1010
Figure 9.2 in Stewart
Balances of forces acting on an iceberg on a rotating earth
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1111
So again…So again…
2
2
2
22
2
22
2
22
1
1
yx
z
vAvAfu
y
p
dt
dv
z
uAuAfv
x
p
dt
du
H
zHH
zHH
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1212
So again…So again…
2
2
2
22
2
22
2
22
1
1
yx
z
vAvAfu
y
p
dt
dv
z
uAuAfv
x
p
dt
du
H
zHH
zHH
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1313
Balance in the surface boundary layer is between vertical Balance in the surface boundary layer is between vertical friction and Coriolis – all other terms are neglectedfriction and Coriolis – all other terms are neglected
0
0
2
2
2
2
z
vAfu
z
uAfv
z
z
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1414
friction
u
c.f.
45°
At the surface (z=0)
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1515
If we assume the wind is blowing in the x-direction only, we can show:
4sin
4cos
)(
)(
D
ze
fAv
D
ze
fAu
Dz
z
x
Dz
z
x
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1616
Wind StressWind Stress Frictional force acting on the surface skinFrictional force acting on the surface skin
2WcDa ρa: density of air
cD: Drag coefficient – depends on atmospheric conditions, may depend on wind speed itself
W: usually measured at “standard Anemometer height” ~ 10m above the sea surface
2
3
3
/5.01.0
/1
1021
mNt
mkg
c
a
D
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1717
Ekman Depth (thickness of Ekman layer)Ekman Depth (thickness of Ekman layer)
f
AD z
2
For Mid-latitudes:
Av = 10
ρ = 103
f = 10-4
Plug these into the D equation and:
4520031010
1023
43
D meters
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1818
Figure 9.3 in Stewart
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1919
U10(m/s)Latitude
15° 45°
540m 30m
10 90m 50m
20 180m 110m
Typical Ekman Depths
Table 9.3 in Stewart
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2020
4sin
4cos
)(
)(
z
x
z
x
fAv
fAu
At z=0
y
x
v
u45°u
)(x
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2121
4sin
4cos
)(
)(
efA
v
efA
u
z
x
z
x
At z=-D
y,v
x,u
v
u -π/4
u
)(x
-π
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2222
Ekman NumberEkman Number
The depth of the Ekman layer is closely related to the The depth of the Ekman layer is closely related to the depth at which frictional force is equal to the Coriolis depth at which frictional force is equal to the Coriolis force in the momentum equation force in the momentum equation
The ratio of the forces is known as Ekman depthThe ratio of the forces is known as Ekman depth
2fD
AE zz
Solving for d:
z
z
fE
AD
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2323
Ekman TransportEkman Transport
fvdzM
fudzM
xyE
yxE
)(0)(
)(0)(
Transport in the Ekman Layer is 90° to the right of the wind stress in Northern Hemisphere
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2424
Ekman PumpingEkman Pumping
)()0(
0
0 00
0
Dwwy
M
x
M
dzz
wvdz
yudz
x
dzz
w
y
v
x
u
yxEE
D DD
D
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2525
By definition, the vertical velocity at the sea surface w(0)=0, and the vertical velocity at the base of the layer wE (-D) is due to divergence of the Ekman flow:
)(
)(
DwM
Dwy
M
x
M
EEH
E
EE yx
vector mass transport due to Ekman flow
horizontal divergence operator
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2626
If we use the Ekman mass transports in we can relate Ekman pumping to the wind stress.
yxcurl
fcurlDw
fyfxDw
xy
E
xyE
)(
1)(
wind stress
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2727
ME Ek
pile up
of water
wE wE
Hi P
Lo P Lo P
anticyclonic
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2828
Hydrostatic EquilibriumHydrostatic Equilibrium
L 106m f 10-4s-1
U 10-1m/s g 10 m/s2
H 103m 103kg/m3
Typical Scales:
with these we can “scale” the equations of motion
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2929
sU
LT
PagzP
smsmL
UHW
L
U
H
W
y
v
x
U
z
W
7
7313
46
31
10
10101010
/10/10
1010;
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3030
The momentum equation for vertical velocity is therefore:
10101010101010
cos21
514111111
2
gfUH
P
L
W
L
UW
L
UW
T
W
guz
p
z
ww
y
wv
x
wu
t
w
and the only important balance in the vertical is hydrostatic:
gz
p
Correct to 1:106
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3131
The momentum equation for horizontal velocity in the x direction is
558888 101010101010
1
fvx
P
z
uw
y
uv
x
uu
t
u
The Coriolis balances the pressure gradient, known as the geostrophic balance
gz
pfu
y
pfv
x
p
1
;1
;1