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Transcript of The Potential Temperature In order to able to compare water at different depths, it is necessary to...
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).
(Knauss - figure 2.3)
Potential temperature is a theorectical temperature and is calculated based on the properties of the fluid.
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 with contribution from compression removed - calculated by assuming surface pressure
density with contribution from compression removeddensity calculated taking in account the effect of compression on Tprovides the easiest density to use for stability calculations
density with contribution from compression removeddensity calculated taking in account the effect of compression on Tdensity calculated taking in account the effect of pressure on compressibilityprovides the most accurate density to use for stability calculations of deep water that is very nearly neutrally stable
density calculated taking in account the effect of compression on T
calculated using the in situ temperature; hence is in error as a result of the change in temperature due to compression, the adiabatic effect
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 EquationIt 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
Simplificationsgravitational 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 accelerationLagrangian 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 iswhere 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 polesTo simplify things
if the meridional displacement is very small, make the f-plane 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 UThe spatial scale is LThe velocity varies by U over the spatial scale LWith these scales, this means the time scale is L/UDefining 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 areaCounter-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 is small compared with , then cannot also be small, otherwise there is nothing to balance
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
The geostrophic velocity at this depth is
At the same depth for location B, the pressure is pB=p0+gz
At point A, the distance between the two isobars is zA, so A=gzA. At B, the distance between the same isobars is zB. Thus, B=gzB. The change in geopotential height between B and A is B- A=g(zB- zA)=gz. 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 haveto change This means that pressure surfaces must parallel density surfaces. Isobars parallel isopycnals
This means that pressure surfaces must parallel density surfaces. Isobars parallel isopycnalsSeveral 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 bywhich we call the relative vorticityThe 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 VorticityAssuming that no torques are applied, the conservation of angular momentum states that:
We define as the potential vorticity, , of the fluid parcelThe 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 eastwardsThe 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 stateFlow 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 layerThe experimental conditions were:Table rotation rate Glycerine in water solution
What is the Ekman depth?
Surface stress is from the right to the left
QuickTime™ and aSorenson Video decompressorare needed to see this picture.
QuickTime™ and aSorenson Video decompressorare needed to see this picture.
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
Sverdrup transport is largest where is a maximum: 30oN (southward)
Ekman transport is largest where has maxima: at 40oN (southward) and at 20oN (northward)
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 equatorwardThe 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 sameIn 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 changeIntegrating 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 PVThis 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, , orBy 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 ParcelBUT 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