Download Ocean Dynamics

Physical Oceanography, MSCI 3001
Oceanographic Processes, MSCI 5004
Dr. Katrin Meissner
[email protected]
Week 5
Ocean Dynamics
Transport of Volume, Heat & Salt
Flux: Amount of heat, salt or volume that flows through a unit
area per unit time (e.g. in Jm-2s-1 (for heat), kgm-2s-1 (for salt)
or m3m-2s-1=ms-1 (for volume)).
Volume transport: The volume of moving water measured
between two points of reference and expressed in cubic
meters per second or in Sverdrup (Sv).
1 Sv = 1 million cubic meters per second
= 106 m3/s
Harald Sverdrup
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The amount of salt in a parcel of water moving
through the ocean remains relatively constant
(unless the parcel is in contact with the
surface, then river discharge, evaporation,
precipitation, sea ice, will change it!)
 Salinity is conservative.
The flux of salt into the cube includes an
advective contribution (ρsu) and a diffusive
contribution (-Kρds/dx)
Diffusive flux (turbulent and molecular) of salt
has a negative sign because it is directed in
the opposite direction to the salinity gradient
Advective Flux
Diffusive Flux
The conversation equation for salt per unit
volume is given by
But don’t panic...
In most instances the advective flux will dominate, so the diffusive
part can be ignored. We are also usually interested in 1D flow
(e.g. flow through a channel or across a section)
Volume transport of an ocean
current with velocity ‘u’ through
an area A is given by:
Volume transport = u . A = u.H.L
u
A
Volume
swept out
2
Transport of Volume, Heat & Salt
If we know the heat (or salt) per m3 we can convert
volume flux into heat (or salt) flux.
Heat per m3 = ρcpT
Salt per m3 = ρS
So, multiplying by uA
Q=ρcpTuA
Sf=ρSuA
ρ is density, cp ~ 4000 J/Ckg, salinity is in absolute units, 35kg/
1000kg
UNITS heat flux – Watts (1 Watt = 1 J/s). Salt flux - kg/s
Total Heat and Salt Fluxes
VT = u⋅ L⋅ H
QT = ρc p T⋅ u⋅ L⋅ H
m3/s
S fT = ρS⋅ u⋅ L⋅ H
kg/s
J/s
T, S
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3
Example
Calculate the total volume, heat and salt fluxes
a)  Past Tasmania
b)  Through the Drake Passage.
Assume the ocean currents persist over the upper 1000m only.
c)  Why might the volume flux be different?
The Equations of Motion
•  Dynamics = how things move and why things move.
• The fundamental equations describing this motion is Newton’s second law:


i.e. F = ma
€
or

F = force
m = mass

a = acceleration
 1 
a = ΣF
m
€
ρ = ρ0(1-αT+βS)
4
Forces
Slope force – component of
weight acting down the slope
Friction force – only
acts on moving car
The car will accelerate du/dt>0 until
the ‘slope’ force is balanced by friction. At that point the net force will
be zero and the car will continue at a steady speed. Just because
the net force is 0 doesn’t mean that the car won’t be moving.
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.
 1 
a = ΣF but m = ρV
m
1 

so, a =
ΣF
ρV
Oceanographers willconsider force per unit volume.

 du
Acceleration a =
dt
du 1
dv 1
dw 1
So,
= ΣF
= ΣF
= ΣF
dt ρ x
dt ρ y
dt ρ z
u,x: west to east v,y: south to north w,z: up to down
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Ocean Dynamics
What are the forces that drive the Ocean?
• 
• 
• 
• 
• 
• 
• 
Gravity (earth, sun and moon) and centrifugal force
Pressure forces
Friction – predominantly at boundaries*
Wind – only at surface boundary
Seismic Forces – occasional impulse only
Coriolis* - due to the earth’s rotation
Buoyancy – due to vertical T,S differences (related to
heat and water fluxes at the ocean surface)
* These forces are only experienced by fluid in motion
Forces on a Parcel of Water



du 1  
= ( Fg + FC + FP + F f + ...)
dt ρ
• 
• 
• 
• 
Gravity
€
du 1
= ∑ Fx
dt ρ
dv 1
= ∑ Fy
dt ρ
dw 1
= ∑ Fz
dt ρ
Coriolis
Pressure
Friction
€
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Newton’s Laws of Motion
Vertical 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
ρ
Δx
ΔV
What are the forces acting in the vertical direction?
Δz
P=F/A
Δy
P+ΔP
maz = ΣFz
ρΔVaz = mg + PΔA − (P + ΔP)ΔA
ρ(ΔxΔyΔz)az = ρ (ΔxΔyΔz)g + P(ΔxΔy) − (P + ΔP)(ΔxΔy)
1 ΔP
az = g −
Surface: ΔA=ΔxΔy
ρ Δz
Volume: ΔV=ΔxΔyΔz
€
Mass:€m=ρΔV=ρΔxΔyΔz
Weight (mg)
But acceleration
€ in the z direction is just: dw/dt, so
Newton’s Laws of Motion
Vertical direction
Vertical acceleration
Buoyancy force
Weight
We can normally make this even simpler:
We can us SCALE ANALYSIS to evaluate the size of each term,
compare their magnitudes, and neglect the small terms.
e.g. Suppose A = B + C
If we know that B=0.00001 and C=10, then to a good approximation
we could simplify this equation to A ≈ C
e.g. Suppose D = E + F
If D=0.00001 and E=10, then then to a good approximation
we could simplify this equation to E ≈ −F
€
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Exercise
•  Note that the Equatorial Pacific is ~ 10.000 km wide and the upwelling
is concentrated in a 50 km wide band at the equator. Once the winds
start blowing it takes about a day for the upwelling to spin up
•  What is dw/dt?
•  How does it compare in size to gravity?
Exercise
•  Note that the Equatorial Pacific is ~ 10.000 km wide and the upwelling
is concentrated in a 50 km wide band at the equator. Once the winds
start blowing it takes about a day for the upwelling to spin up
•  What is dw/dt?
•  How does it compare in size to gravity?
Upwelling =100,000,000m3/s. This is coming up through an area of:
A=10,000,000x50,000=5x1011m2. So (remember volume transport):
w= (100,000,000m3/s)/(5x1011m2) = 0.0002m/s (very small!)
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Exercise
•  What is dw/dt?
•  How does it compare in size to gravity?
w= (100,000,000m3/s)/(5x1011m2) = 0.0002m/s (very small!)
If the wind were to stop and then start again it would take about a day to
start upwelling at that rate again i.e. It takes 1 day to reach a velocity of
w=0.0002m/s if we start from w0=0m/s.
So we can approximate our acceleration to
dw w − w 0 (0.0002 − 0.0)ms−1
≈
=
= 2 × 10 −9 ms−2 << g = 10ms-2
dt
T
1 × 24 × 60 × 60s
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
€
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
This is called the
hydrostatic equation
We can integrate this equation since density and g are essentially constant.
h
∫
0
dp
dz =
dz
h
∫ ρgdz
0
Or simply
Which you are hopefully
familiar with already!
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Newton’s Laws of Motion
P
Horizontal direction
Just like vertical pressure gradients can create a
force on the water, so can horizontal gradients
ρ
Δz
ΔV
Δy
Δx
Surface
P+ΔP
We can now forget the
vertical – we’ve done
that already. What
about forces in the
horizontal
Bottom
The pressure on the right of the box will be larger than the pressure on the left of the box
There will be a net force from right to left. Let’s zoom in …
Newton’s Laws of Motion
max = ΣFx
ρΔVax = P1ΔA − P2 ΔA
Horizontal direction
ρ (ΔxΔyΔz)ax = (P)ΔA − (P + ΔP)ΔA
€
ρ (ΔxΔyΔz)a
x = (P)ΔyΔz − (P + ΔP)ΔyΔz
ρ ax Δx = P − P − ΔP
du
1 ΔP
ax =
=−
dt
ρ Δx
€
ρ
€
P
P+ ΔP
Δz
Δx
€
du
1 dp
=−
dt
ρ dx
Δy
€
And, doing the same in the y-direction:
Surface: ΔA=ΔyΔz
Volume: ΔV=ΔxΔyΔz
Mass: m=ρΔV=ρΔxΔyΔz
€
€
dv
1 dp
=−
dt
ρ dy
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Barotropic and Baroclinic Motion
Remember, p = ρgz ,
and
5cm
5cm
1cm
2cm
2cm
2cm
2cm
2cm
2cm
ρ1=1026
ρ0=1027
<
ρ2=1028 kgm-3
Barotropic and Baroclinic Motion
Remember, p = ρgz
p1=1027*10*0.01=102.7kg/m/s2 or Pa
p2=1027*10*0.03=308.1 Pa
p3=513.5 Pa
p1=205.4 Pa
p2=410.8 Pa
p3=616.2 Pa
5cm
1
1cm
du
1 ∂p
1 Δp
=−
=−
dt
ρ ∂x
ρ Δx
2cm
2cm
2
2cm
3
€
ρ0=1027
du1
1 205.4 −102.7
=−
= −2m /s2
dt
1027
0.05
du2
1 410.8 − 308.1
=−
= −2m /s2
dt
1027
0.05
du3
1 616.2 − 513.5
=−
= −2m /s2
dt
1027
0.05
€
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Barotropic and Baroclinic Motion
Remember, p = ρgz ,
p1=1026*10*0.02=205.2Pa
p2=1026*10*0.04=410.4Pa
p3=615.6Pa
and
p1=205.6Pa
p2=411.2Pa
€
p3=616.8Pa
du
1 ∂p
1 Δp
=−
=−
dt
ρ ∂x
ρ Δx
du1/dt=-(1/1027)*(205.6-205.2)/0.05 = -0.0078ms-2
du2/dt=-(1/1027)*(411.2-410.4)/0.05 = -0.0156ms-2
du3/dt= -0.0234ms-2
5cm
2cm
2cm
2cm
ρ1=1026
<
ρ2=1028 kgm-3
Barotropic and Baroclinic Motion
Remember, p = ρgz ,
Barotropic
velocity is constant with depth
and
du
1 ∂p
=−
dt
ρ ∂x
Baroclinic
velocity changes with depth
€
ρ0
Motion due to surface slopes
ρ1
<
ρ2
Motion due to density differences
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Barotropic and Baroclinic motion
Barotropic Currents move the whole water column. Can be
induced by a horizontal tilt of the water surface, which
produce a pressure gradient throughout the whole water
column.
Baroclinic Currents vary with depth. Can be induced by
horizontal density gradients. This effect is known as the
thermal wind balance (more later).
BAROTROPIC
BAROCLINIC
Barotropic and Baroclinic Motion
Remember, p = ρgz ,
and
Mixed situation
ρ1
<
ρ2
Motion due to density differences
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Barotropic:
isobaric (equal pressure) surfaces are parallel to
isopycnic (equal density) surfaces
Baroclinic:
isobaric and isopycnic surfaces are NOT parallel
Barotropic:
Isobaric surfaces are parallel to isopycnic surfaces
Baroclinic:
isobaric and isopycnic surfaces are NOT parallel
Where are we likely to find barotropic conditions in the ocean?
•  Well-mixed surface layers
•  Shallow shelf seas (particularly where shelf waters are well mixed by tidal
currents)
•  Deep ocean (below the permanent thermocline)
Where are we likely to find baroclinic conditions in the ocean?
•  Regions of fast surface currents
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Barotropic Ocean
For a constant density ocean, we can write the pressure
gradient in an easier way.
Remember, p = ρgz , and
η2
η1
d
h1
h2
P1=h1ρg
h1=d+η1
P2=h2ρg
Δx
h2=d+η2
So we are left with
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Barotropic Motion
For a constant density ocean, the acceleration of
water just depends on what is happening at the
surface i.e. the slope of the sea surface
Δη (0.02 − 0.01) 1
=
=
Δx
0.05
5
du
1
= −10( ) = −2ms−2
dt
5
5cm
1cm
2cm
2cm
2cm
€
Same answer as we got before (using pressure)
It’s magic!
ρ0=1027kgm-3
Pressure
Gradients
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Summary:
• Start with Newton’s Law: the acceleration of a parcel of water is related to the
sum of forces acting on that water parcel.
• We can break down the acceleration and forces into those acting in the x,y and z
directions
• In the z-direction, vertical velocities/accelerations are small, which leaves us with
the hydrostatic equation
This just means that the weight of water is balanced by the
vertical pressure gradient (in other words the buoyancy force)
The water acceleration is driven by horizontal
differences in pressure (these differences may
be related to horizontal differences in surface
height (barotropic) or in density (baroclinic))
• For a barotropic (constant density) situation
we can simplify the equation to:
• In the x-direction
• Similarly in the y-direction
• But there are other forces out there …
The Coriolis Force
 1 
a = ΣF
m
du
1 ∂p
=−
+ Coriolis force + ...
dt
ρ ∂x
€
dv
1 ∂p
=−
+ Coriolis force + ...
dt
ρ ∂y
∂p
= ρg
∂z
€
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18
Earth’s rotation
Observations of moving parcels are judged relative to coordinates that
are fixed to the Earth
If we were somewhere on a spaceship we would see
the atmosphere’s independent motion and
the Earth turning out from under the moving parcels of air
Deflection to the right
in the Northern
Hemisphere
Northern
hemisphere
Deflection to the left in
the Southern
Hemisphere
Southern
hemisphere
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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
Ω= 2π / 24 hours = 7.272x10-5 rad/s
The Coriolis Force
Earth spins counter clockwise looking at it from the north pole
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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
f=2Ωsin(Φ)
The acceleration due to the Coriolis force is:
fv - in the x-direction (east-west), and
-fu - in the y-direction (north-south)
What is the value of Ω (the earth rotates 360º every day)?
What is f at the north pole, the equator, 30N and 30S?
What is the acceleration of water flowing at 1m/s at 30S? What pathway will the
water trace out?
The Coriolis Force Summary
• Acceleration due to the Coriolis force is
fv (x direction) and -fu (y direction)
• acts only if water/air is moving
• 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
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The Equations of Motion

du 1 
= ΣF
dt ρ
€
du 1
= ΣF
dt ρ x
dv 1
= ΣF
dt ρ y
dw 1
= ΣF
dt ρ z
Horizontal Equations:
€
Acceleration = Pressure Gradient Force + Coriolis
du
1 dp
=−
+ fv
dt
ρ dx
dv
1 dp
=−
− fu
dt
ρ dy
du
dη
= −g
+ fv
dt
dx
dv
dη
= −g
− fu
dt
dy
Or, for a Barotropic Ocean:
Vertical Equation:
€
€
Pressure Gradient Force = Gravitational Force
du
1 dp
=−
+ fv
dt
ρ dx
dv
1 dp
=−
− fu
dt
ρ dy
If the only force acting on a body is Coriolis,
the Navier Stokes equations can be simplified to:
€
What does this mean?
du
= fv
dt
dv
= − fu
dt
u<0
u>0
€
y
v>0
v<0
x
22
If the only force acting on a body is Coriolis, the Navier Stokes equations
can be simplified to:
du
= fv
dt
dv
= − fu
dt
Northern Hemisphere:
f>0
What does this mean?
u<0
u>0
€
v>0
y
v<0
x
If the only force acting on a body is Coriolis, the Navier Stokes equations
can be simplified to:
du
= fv
dt
dv
= − fu
dt
Northern Hemisphere:
f>0
What does this mean?
u<0
u>0
€
y
v>0
v<0
x
23
If the only force acting on a body is Coriolis, the Navier Stokes equations
can be simplified to:
du
= fv
dt
dv
= − fu
dt
Southern Hemisphere:
f<0
What does this mean?
u<0
u>0
€
y
v<0
v>0
x
Horizontal Equations:
Acceleration = Pressure Gradient force + Coriolis
du
1 dp
=−
+ fv
dt
ρ dx
dv
1 dp
=−
− fu
dt
ρ dy
If pressure gradients are small:
Inertia currents
€
du
= fv
dt
dv
= − fu
dt
the water flows
around in a circle
with frequency |f|.
T=2π/f
T(Sydney) = 21 hours 27 minutes
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