Transcript Document

The Potential Temperature 
In order to able to compare water at different depths, it is necessary
to remove the effect of pressure (e.g. compression) on the
temperature of water.
We define the potential temperature, , as the temperature that a
parcel of water would have if transported to the surface
adiabatically (no heat transfer).
Potential temperature is a
theorectical temperature
and is calculated based on
the properties of the fluid.
(Knauss - figure 2.3)
Magnitude of the Compressibility Effect on
Temperature
is
Oceanographers use 4 Different Representations for Density
in situ density - 103 kg m-3
calculated using in situ salinity, temperature and pressure
density calculated taking in account the effect of compression on T
density with contribution from compression removed - calculated by
assuming surface pressure
calculated using the in situ temperature; hence is in error as a result of the
change in temperature due to compression, the adiabatic effect
density with contribution from compression removed
density calculated taking in account the effect of compression on T
provides the easiest density to use for stability calculations
density with contribution from compression removed
density calculated taking in account the effect of compression on T
density calculated taking in account the effect of pressure on compressibility
provides the most accurate density to use for stability calculations of deep
water that is very nearly neutrally stable
Stratification
Stability is often expressed in terms of N, the Brünt-Väisälä
frequency, also known as the Buoyancy Frequency
N has units of s-1 (i.e., radians per second), it is the (angular)
frequency at which a parcel of fluid would oscillate up and
down if displaced vertically
Typical values of 2/N, i.e., the period of oscillation:
Seasonal thermocline: 5 minutes
Main thermocline: 20 minutes
Deep water: 20 hours
The Hydrostatic Equation
A motionless fluid has w = 0 always, hence Dw/Dt = 0 and
This is the Hydrostatic Equation
It is an equation with which you should all become familar
But it’s an approximation; the correct equation is:
i.e., the error we make using the hydrostatic equation is:
But, even when w ≠ 0 this equation will be an excellent
approximation if:
What does Hydrostatic mean?
Pressure variation with depth is approximately hydrostatic
At a depth z, pressure is equal to the weight of the overlying
water plus the atmospheric pressure
Simplifications
gravitational acceleration g is essentially constant
 varies by only a few percent with depth
Thus,
where avg is the average density of the water column between
the surface and z (Recall z<0)
Conservation Laws
Conservation laws play a critical role in physics
Loosely speaking, conservation laws state that what goes into a
system must come out of it, if there are no sinks or sources of
the property within the system
Fundamental conservation equations that we will deal with are:
Conservation of mass, volume, salt, and other substances
Conservation of energy (e.g. heat)
Conservation of linear momentum (mass x velocity)
Conservation of angular momentum (vorticity, like angular
momentum)
The Substantial Derivative
This is the acceleration of a parcel of fluid, Lagrangian acceleration
Lagrangian accel. = Eulerian accel. + Advective accel.
More, generally
The Lagrangian D/Dt term is the rate of change experienced by
a given tagged water parcel
The Eulerian term
term is the local rate of change at a
fixed point.
is what you get from a current meter at a fixed
point in space
The advective term is
which contributes to the change
in the property due to the displacement of the parcel. It
converts between Eulerian and Lagrangian rates of change
Steady Flow in a Pipe
Example
Consider the steady (
everywhere) flow of an
incompressible fluid in a narrowing pipe
A water parcel enters the pipe with velocity u1
And leaves it with velocity u2
u2 > u1 since the pipe narrows
The parcel clearly accelerates as it moves into the narrower region,
but the local acceleration is zero, so:
Temperature in a Channel
After some small time t, the float has moved x. How much has the
float temperature of the float changed?
If nothing heats or cools the water as it is advected to the right, the drifter
will not experience any temperature change.
t=0
t=t
That is,
Temperature in a Channel
If nothing heats or cools the water as it is advected to the right, the drifter
will not experience any temperature change. That is,
The local rate of change in T will equal the negative of the advective
flux. An observer at a fixed location would see an apparent cooling.
Temperature in a Channel
If the water parcel is heated, then the Lagrangian change is
where H is the heating rate. Imagine the spatial structure doesn’t
change with time:
Moving Parcels Veer to the Right (Left) in the
Northern (Southern) Hemisphere
varies with latitude 
Parcels veer to the right in the Northern hemisphere, veering
increases with latitude
In the Southern hemisphere, things are reverse, f < 0, parcels
veer to the left
f = 0 at the equator and
at the poles
To simplify things
if the meridional displacement is very small, make the fplane approximation: f ≈ constant or
if the meridional displacement is moderately small, make
the beta-plane approximation: f = f0 + y. At 45oN,
Momentum / Continuity Equations
Scaling the Momentum Equation
Consider the x-momentum equation for a small-scale flow (i.e., neglecting
Coriolis and turbulent flow terms):
We want to know the relative importance of the advective terms to the
friction terms.
We do this by scaling the equation
The velocity scale is U
The spatial scale is L
The velocity varies by U over the spatial scale L
With these scales, this means the time scale is L/U
Defining dimensionless velocity, scale and time variables by dividing
the numerical scales by these scales:
Scaling the Rest of the Momentum Equation
Scaling the other terms the same way yields
So the ratio of the advective terms to the viscous term is
is called the Reynolds number, Re
Without solving for the complete flow field, we can use the
Reynolds number to get a qualitative picture:
small Re  friction-dominated flow (if flow is steady, so
)
large Re (say Re > 104)  turbulent flow (for a non-rotating
fluid)
Characteristics of the Rossby Number
and
small Ro  Coriolis-dominated flow (e.g., Coriolis and
pressure gradient forces balance) rotation is important
large Ro  rotation is not important
For typical values of large scale circulation
U ~ 0.1 m s-1, f = 10-4 s-1, L = 105 m
Geostrophic Flow
Next, we assumed that
Flow is steady 
Friction terms are small Rossby number is small
Then
The balance between the Coriolis terms and the pressure gradient
terms is called the Geostrophic Balance
In the northern hemisphere (f > 0)
Geostrophic flow is parallel to
isobars with high pressure to the
right
Winds (and currents) circle clockwise around a high-pressure area
Counter-clockwise around a low-pressure area
Geostrophic Flow in a Two-Layer Ocean: Pressure
in the Upper Layer
Now let’s look at geostrophic flow in a two-layer ocean
First, look at the upper layer (subscript 1)
At location A, the pressure at depth H (= -z), where (H < H1 - h2),
in the upper layer is
At location B, the pressure at this depth is
Geostrophic Flow in a Two-Layer Ocean: Velocity
in the Upper Layer
The velocity in the upper layer is:
Geostrophic Flow in a Two-Layer Ocean: Pressure
in the Lower Layer
Pressure at two points in the lower layer:
In the lower layer at z = -(H1 + H2), the pressure at A where is
At the same level, at location B, the pressure is
Geostrophic Flow in a Two-Layer Ocean: Velocity
in the Lower Layer
So with
and
Thus, the velocity in the lower layer is
Thermal Wind: Differentiating the Horizontal Momentum
Equations in the Vertical
Now, we will show that for geostrophic flow, vertical gradients of
velocity are related to horizontal gradients of density
Differentiating the geostrophic balance equations with respect to z:
But, the flow is hydrostatic
and f is not a function of depth, so
Simplifications to The Thermal Wind Equations
These eqns can be simplified using the Boussinesq Approximation
Expand the left-hand side of the x-equation:
Now consider the ratio of the two terms on the left-hand side:
Thus
is generally negligible compared with
So, with this Boussinesq approximation (that density variations are
unimportant except in pressure and buoyancy terms), we have
The Boussinesq Approximation
Taking  over to the right-hand side, we get:
and
These equations are called the Thermal Wind Equations
To neglect the
term is to make the Boussinesq
approximations - density variations are not important except in
the pressure and buoyancy terms
Note that if
to balance
is small compared with
, then
cannot also be small, otherwise there is nothing
The Geopotential Height
A geopotential (horizontal or level) surface is everywhere 
We define the change in geopotential
thus,
We can relate the change in geopotential (geopotential height) to
pressure via the hydrostatic equation:
Hence changes in geopotential are related to changes in pressure
So the geopotential at pb relative to pa is
Determining the Geostrophic Velocity from the
Geopotential Height
Consider the simple case of a sloping isobar. Imagine at some
depth, the velocity is zero. At that depth, the isobar is parallel to
the geopotential surface.
Determining the Geostrophic Velocity (continued)
At point A, let the pressure at a
depth located zA above this
surface be pA=p0
At the same depth for location B,
the pressure is pB=p0+gz
The geostrophic velocity at this depth
is
At point A, the distance between the two isobars is zA, so A=gzA. At B, the distance
between the same isobars is zB. Thus, B=gzB. The change in geopotential height
between B and A is B- A=g(zB- zA)=gz. Thus, the geostrophic velocity at this
pressure is
We can determine the geostrophic velocity from the along-pressure gradients of
geopotential height.
In the Course Notes, Section 4.4.3 provides another example where the lower layer is not
at rest.
Summary of Geostrophic Balance
Barotropic Flow and Property Surfaces
Barotropic flow is flow that does not depend on depth
This means that the horizontal pressure gradient does not change with depth
Which in turn means that the horizontal gradient of the weight of the
overlying fluid must be independent of depth; if it changed, then the pressure
gradient would have to change
This means that pressure surfaces must parallel density surfaces.
Isobars parallel isopycnals
This means that pressure surfaces must parallel density surfaces.
Isobars parallel isopycnals
Several surfaces defined: isobaric - surface of constant pressure
isopycnal - surface of constant density
geopotential - `level’ surface, pendicular to gravity
Baroclinic Flow and the Reference Velocity
Baroclinic flow occurs when isobars and isopycnals cross each
other;
The geopotential method allows us to estimate the velocity shear,
the baroclinic part of the flow
It does not allow us to determine the barotropic component
We must obtain the barotropic component in some other fashion:
Historically, oceanographers assumed that there would be a
`level of no motion’ somewhere deep in the ocean where the
flow is quiescent
This assumption could be wrong, especially in
shallow coastal regions or in areas where deep ocean
currents are present
Use actual currents at a particular depth with an ADCP or
current meter
The Vertical Component of Vorticity - the Relative Vorticity
In most large-scale flows, we only consider the vertical component
of vorticity.
The vertical component of vorticity is given by
which we call the relative vorticity
The vorticity is a measure of the tendency for a fluid parcel to rotate
Planetary and Total Vorticity
The vorticity due to the rotating earth, (the planetary vorticity) is f
(the Coriolis parameter)
The total or absolute vorticity (vorticity viewed from an inertial
coordinate system) is f + 
For large-scale oceanic flows, the relative vorticity is typically
much smaller than the planetary vorticity
The ratio of the relative vorticity to planetary vorticity,
, is the Rossby Number, Ro, which is typically <<1
Conservation of Potential Vorticity
Assuming that no torques are applied, the conservation of angular
momentum states that:
We define
as the potential vorticity, , of the fluid
parcel
The potential vorticity  of a water parcel is conserved unless
external torques are applied to it (wind, friction, etc)
If you stretch a water column of water, its radius decreases, its
moment of inertia decreases and its angular momentum increases
Rossby Adjustment Process
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The Basic Assumptions for the Adjustment Problem
How does the ocean adjust to an applied wind stress?
We want to determine
for the mixed layer as a whole, while
the fluid is accelerating from
to its final steady state
value.
Assume: A surface mixed layer of thickness H
The wind blows steadily exerting a uniform stress, , independent of
x, y and t.
Non-linear terms are small compared to the other terms:
Level surface, constant density
Friction is linearly related to the mixed-layer velocity:
The Steady State Solution
To determine
as the fluid accelerates we need to solve:
We begin by finding the steady solution, then the transient solution
The steady state solution,
, must satisfy the above with
Solving for u0 and v0 yields, for steady state values
and
The wind is blowing eastward but the flow is not eastwards
The flow is to the right of the wind for
If
(no friction),
Flow is southward, perpendicular to
the wind
Deviation from the Steady State Solution
Any imbalance in these forces will lead to an acceleration, a time
dependence in the velocity
Note, the Coriolis force and frictional force depends on the velocity
of the water, which is initially zero
The Time-Dependent Solution - The Adjustment to
Steady State
Assume the complete solution is the sum of the steady solution
already found and a transient part
where
is the time-dependent velocity
Substituting for u, v in the original equations
we obtain
since
The Time-Dependent Solution - The Adjustment to
Steady State
With the initial velocity being zero at t=0, then
and
at this time
The other boundary conditions are that u1=0 and v1=0 as t∞ in
order to get the steady state solution for the total velocity
The solution to these differential equations with these boundary
conditions is
The Physics of the Complete Solution
The solution contains two separate physical behavioral patterns:
Inertial oscillations given by
This is harmonic motion at angular frequency
If one arbitrarily disturbs the ocean (gives it a kick), it will tend to
oscillate at the local inertial period
With friction present, this oscillation decays away
Tendency toward steady state
Flow almost at right angles to the wind if the friction is small
The Solution
With these boundary conditions, the solution for u and v to the
equations of motion
is
where
+ for f positive, - for f negative
The surface velocity U0
and the Ekman depth
A Schematic of the Solution
Surface flow is 45o to the
right of the wind
Deflection increases to the
right in successively deeper
layers
The depth at which is
important is on the order of
the Ekman depth - our
definition is where the
velocity is in the opposite
direction of the surface
velocity and 4% of it.
The Ekman Depth - A Laboratory Experiment
An experiment was undertaken to visualize the Ekman layer
The experimental conditions were:
Table rotation rate
Glycerine in water solution
Surface stress is from the right to the left
What is the Ekman depth?
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Total Transport in the Ekman Layer
We call
the Ekman transport
Note that this transport is entirely independent of the eddy
viscosity - it doesn’t depend on how the stress is distributed in
the mixed layer
Mass Conservation Eliminates a Term
But
from continuity
So, the vertically integrated momentum equations become
where
This is the Sverdrup Relation
which is the fundamental equation of large-scale wind-forced
circulation - north-south integrated mass transport is proportional
to the curl of the wind stress
Meridional Dependences of Ekman and Sverdrup Transports
for a Sinusoidal Wind Stress
x has a maximum at 40oN, zero at 30oN and a minimum at 20oN
Ekman transport is largest where
has maxima: at 40oN
(southward) and at 20oN (northward)
Sverdrup transport is largest where
(southward)
is a maximum: 30oN
Sverdrup transport is zero at 20oN and 40oN
Ekman and Sverdrup Transports for a Sinusoidal Wind Stress
At 40oN in our example, the Sverdrup transport is zero and the
Ekman transport is to the south, so the deep geostrophic must be to
the north and the same magnitude of the Ekman transport
Two ways of Looking at This
There are really two ways of looking at this:
The Ekman convergence squashes the lower layer and
conservation of PV forces it equatorward
The negative wind-stress curl applies a negative (clockwise)
surface torque and the angular momentum of a parcel
consequently dccreases (entire water column)
These are effectively the same
In PO we tend to view the balance as describe in the first case
The Ekman layer is considered to be VERY thin, hence, does
not contribute significantly to the angular momentum balance
The Ekman layer applies no horizontal stress to the lower
layer
at the base of the Ekman layer
Instead, the Ekman layer communicates the surface torque to
the water below by vertical pumping - “Ekman pumping”
Wind Stress in the Subtropical North Atlantic
How does potential vorticity relate to Sverdrup flow?
Suppose that the wind stress has negative curl, as in the subtropical
North Atlantic
Changes in Vorticity Due to Convergences in the Subtropical
The Ekman transport is
greater at C than at A
u(C) > u(A)
Surface water accumulates
between A and C (e.g., at
B)
The water columns in the
lower layer are squashed
Since
is conserved in the lower layer, the reduction in
H must be matched either by a reduction in  or in f
Parcels Must Move South
(small Rossby
number)
So, changes in  cannot
alter +f significantly
f must decrease  the fluid
moves South
Negative wind stress curl  Ekman convergence  squashing
water columns beneath the Ekman layer  Southward flow
The Physics
The wind stress curl.
, applies a
torque to cylinders of water to
This torque changes the angular
momentum, L, of the cylinders
For L to change, I (1/H) and/or  must change
Sverdrup says that  must change
Integrating to a fixed depth, H, keeps I from changing
But assuming
means that the water column will not
spin up relative to a coordinate system fixed to Earth
So the only way to change  is to change  sin; i.e. change 
Need Either Another Torque or a Change in PV
This violates the fundamental conservation law for angular
momentum:
To resolve this,
Either the PV must change as the parcel goes around the gyre
By decreasing the relative vorticity, , or
By decreasing its moment of inertia, 1/H
Or there must be a positive torque applied to the parcel
somewhere along its trajectory that compensates for the wind
stress curl
Need an Additional Torque Applied to the Fluid Parcel
BUT in the long term, a continual decrease in PV as the parcel goes
around and around the gyre is unreasonable:
Either a parcel’s vorticity would have to decrease continually,
Or its thickness would have to increase continually,
So we need an additional torque . Where does it come from?
The flow must be zero at a coastal boundary,
So the boundary applies a stress to the fluid
If this provides the needed positive torque, on which side of the
basin will we find the return flow?
Additional Torque Cannot Come from the Eastern Boundary
Let’s assume that the return flow is on the eastern boundary
The return flow on the eastern boundary is inconsistent with
PV = 0
Additional Torque Must Come from the Western Boundary
Now assume that the return flow is on the western boundary
The return flow on the western boundary is consistent with
PV = 0