Student Presentations: Arctic (Rubber Ducks), Antarctic (who saw the seminar)?
Barotropic and baroclinic pressure gradients
Barotropic—depth independent. Generally associated with tilting sea level
Baroclinic—depth dependent. Associated with horizontal density gradients.
Case 1 Barotropic
Consider a case where the sea-surface tilts down 1 cm over a 1 km distance. This would
produce a barotropic pressure gradient =g dn/dx ~ 10-5 m/s^2.
u(t)=u(t0)+10-4 t
Of course in the ocean other forces would come to play the fluid velocity would not
follow this simple balance. One important point, however, is that the fluid accelerates at
all depths at the same rate. Subsequently a barotropic pressure gradient will not generate
vertical shear in the flow—but rather a depth independent flow.
In contrast, as we will see, a a baroclinic pressure gradient drives vertical shear
Note that the barotropic pressure gradient is driven by the sea-level slope. If is the sea
level above some reference point (z=0) then the pressure a depth z is equal to
=g(z+ assuming that the density is constant. Then the acceleration caused by the
pressure gradient in the x direction is simply the derivative of =g(z+divided by the
density—i.e.
1 P

g
 x
x
So the barotropic pressure gradient is simply gravity times the sea-surface slope. This is
pretty simple. It’s like skiing—the steeper the slope the faster you will accelerate.
Baroclinic Pressure Gradient
Consider a channel filled with fluid on the left side with fluid of density and fluid of
density  on the right side and that  (Figure 2)On the left size the
pressure at depth z is P1(z)= gz, while on the right hand side the pressure at depth z is
P2(z)= gz=g(z. The pressure difference between the two fluid is P2(z)-
1
P2(z)=gz. Therefore the pressure gradient increases with z. The pressure gradient is
zero at the surface and equals gat the bottom. Considering a momentum equation
between the local acceleration and the pressure gradient we write
u g 

z
t  x
Thus the fluid accelerates more rapidly on the bottom than on the at the surface and
as a consequence a vertical shear develops in the water column. Note that the pressure
gradient is directed to the left throughout the water column. This would produce a depth
averaged acceleration of fluid to the left, which in turn would cause sea level on the left
to stand higher than on the right. If the channel is closed on both ends sea level rises
high enough to cancel out the depth averaged pressure gradient so that the depth averaged
flow is zero. This occurs because when depth averaged flow is not zero sea level
continues to rise on the left and the resulting sea level slope eventually produces a
pressure gradient that exactly equals the depth averaged baroclinic pressure gradient.
We’ll calculate this in the next class. But what happens is that a 2 –layer flow develops
that drives fluid to the right in the surface layer and to the left in the lower layer. This 2layer exchange flow is what drives estuarine circulation (Figure 3).

2
x
z


Figure 2 Baroclinic pressure gradient produced by two adjacent fluids of
different density. The arrows represent the baroclinic pressure gradient—that
increases linearly with depth—as well as the acceleration that fluid parcels
undergo in response to the baroclinic pressure gradient.

-.20
Figure 1
shows the
salinity field
in the
Hudson
River
estuary.
Distance is
in km from
the battery.
The white
lines
represent
-.15
-.1
meters
-.05
0
.05
.1
Km from Jersey Shore
isohalines (lines of constant salinity) and
show a salt wedge that extends over 50 km
upstream in the lower layer. The color on
the graphic depicts the concentration of dye,
with red the highest concentration. This dye
was injected into the lower layer 2 days
earlier at ~ km 25—indicating that in 2
days the dye advanced upriver (to the right)
Figure 1
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at a rate of ~ 10 km per day. Note that upstream motion of dye is in the opposite
direction of the river flow (that is of course flows down stream and towards the ocean to
the left). What then drives the bottom layer upstream? It is the baroclinic forcing of the
dense saline ocean waters that slumps underneath the lighter fresh surface water. While a
detailed estimate of the baroclinic pressure gradient can be made with the actual data –
with the schematic shown in Figure 2 we can construct a crude model to estimate the
pressure gradient.
Using the 10 psu isohaline in figure 1 as the interface between the fresh and salty water
we see that the interface between the surface layer and bottom layer slopes downward at
z
approximately 1 meter every 6 km or I  1.6 x10  4 . At the bottom the hydrostatic
x
pressure is equal to the weight of the overlying water which can be written as:
P(x)  g(1 (H  z i )   2 z i ))
Note that the pressure is a function of x. When we take the derivative of this equation
with respect to x the only term that is non zero is the term that has an x in it—all other
terms do not vary in x and thus have a derivative (the rate of change) of zero. Therefore
the along channel pressure gradient in the lower layer due to the sloping isopycnals is:
z
z
z
z
z
P
 g1 i  g 2 i  g1 i  g(1  ) i  g i
x
x
x
x
x
x
So that is the pressure gradient force. In the momentum equation often we express terms
as acceleration—rather than F=ma we write F/m=a—so we want to divide both sides by
the density go yield
z
1 P g z i

 g' i
 x
 x
x
where g’ is reduced gravity. Reduced gravity is the effective gravity that a body feels in a
fluid. Note that if the two densities ( are the same that reduced gravity is zero.
This is the weightlessness one feels in water. For small differences between the two
densities the effective gravity is significantly smaller than g. For example if = 10
kg/m3 as in the case in the Hudson, then reduced gravity is about 100 times smaller than
gravity. So while a particle exposed to the full force of gravity would accelerate at 9.8
m/s2, accleration in this reduced gravity field would be .098 m/s2. Also note that this
baroclinic pressure gradient has an identical form as the barotropic pressure gradient
associated with a sloping sea-surface i.e.
P

g
x
x
4

) is the slope of the sea surface slope. In rivers the sea-surface slopes
x
downward towards the ocean and accelerates the river water towards the ocean.
Therefore the barotropic pressure gradient is in the opposite sign of the baroclinic
pressure gradient. In the surface layer the total pressure gradient is only due to the
barotropic pressure gradient, while in the lower layer the total pressure gradient is the
sum of the barotropic and baroclinic pressure gradients. A special case occurs when the
barotropic pressure gradient is equal to but of the opposite sign of the baroclinic pressure
where (
Surface

H
Interface
zI

z
Bottom
x
Figure2. Simplfied model of a two layer system to estimate baroclinic pressure
gradient of salt field shown in figure 1
gradient. This would occur when
g'
z i

 g
x
x
Which means that the ratio of the slope of the interface to the slope of the sea surface is
In the example shown in figure 1 kg/m3 and since  kg/m3 if the
slope of the sea-surface were 1/100th of that of the slope of the interface the total pressure
gradient in the lower layer would be zero and the barotropic pressure gradient would
arrest the salt wedge because the flow in it would be zero. IN the case above this
corresponds to a 1 cm rise in sea level every 1-6 km—not an unreasonable number in
rivers and estuaires. On the otherhand observations of currents in the lower layer and of
the movement of dye shows that there is a upstream velocity in the lower layer.
Subsequently the total pressure gradient in the lower layer is directed up stream and
suggests that the sea-level slope is something less than 1 cm every 6 km—but still tilted
seaward for the surface flows are infact seaward (see Figure 9 from Sept 16th notes)
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In the ocean both the slope of the surface and the slope isopycnals tends to be smaller
than what one sees in the estuary. However there is a tendency for the density field to
adjust in such a way that at depth the total pressure gradient becomes quite small. If we
assume that the pressure gradient does infact go to zero at some depth and we have CTD
profiles we can estimate the actual slope of the sea surface.
The pressure at the two stations can be written
P1=g(H+

P2==(gH
where  is the sea-surface difference between the two stations  is the average density
for that profile 1, is the difference in the mean density between profile 1 and profile 2
and H is the presumed level of no motion. Since at this depth P1=P2 we find
g(H+(gH
pgH + gpgH +gH
 H



So if H is 4000m and is .1 kg/m3 would be 40 cm.
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START WITH UPCOMING CLASS SCHEDULE
THEN MENTION ROSSBY NUMBER—RATION OF ROTATION
The Gravity Current Adjustment Problem
One way to think about the tank experiment is that when we pulled the gate we released
potential energy that turned into kinetic energy in the form of a gravity current. This was
seen to slosh back and forth as the kinetic energy was changed back into potential energy
as the dense fluid rose up—due to inertia—on the other side of the tank only to be
released as kinetic energy again as the baroclinic field sloshed back. This process is
similar to a pendulum whose energy changes from all kinetic, when the wire is straight up
and down, to all potential energy, when the pendulum has stalled at the highest point and,
for the moment, has zero kinetic energy.
Figure 3 shows the initial set up of the tank experiment (upper panel) and the situation
just before the fronts hit then ends of the tank (lower panel).
Initial Condition (High PE state)


L
Final Condition (Lowered PE state & High KE state)

H

L
Figure 3 Schematic showing the start and end of the baroclinic adjustment
problem conducted in lab October, 13 2003.
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If is the density difference between the two fluids and is the mean density of the two
fluids then and In the initial set-up (upper panel) the center of
mass of each fluid is at Zcm=H/2 and the volume of each fluid is W*L/2*H (where W is
the width of the tank). Potential energy is equal to gravity * Mass * height of the center
of mass. Mass is equal to the density * the volume. Therefore the potential energy just
prior to the release of the baroclinic field is:
PE=g(V1H/2 +V2H/2 )
(1)
Since V1= V2=V/2 where V is the total volume of the tank equation 1 can be written as
PE=gVH/4(gVH/2
(2)
After the fluid has adjusted the dense water is in the bottom layer and the center of mass
of the lower layer fluid resides at H/4, while the center of mass of the lighter uperlayer
fluid is at 3H/4. The volume of each layer remains the same and the potential energy is
PE=g(V13H/4 +V2H/4 )
=gVH/8(3 + )
=gVH/8(3(


=gVH/8(4

 gVH/2  gVH





Which represents the potential energy at the end of the experiment. Comparison with
equation 2 we find that the potential energy of the system has decreased by
 gVH 




which mostly turned into kinetic energy. Kinetic energy = ½ mv2 where v is the velocity
of the fluid and m mass of the fluid. The mass of the fluid equals V thus if we assume
that all the potential energy has turned into kinetic energy we can equate the kinetic
energy to 3
 gVH½ mv2Vv2
which can be rearranged to read
v
1
g' H  
2




8

where g’ is the reduced gravity and equal to g


Equation 4 is the speed that the front of the gravity current, such as the one released in
the lab, propagates.
In the tank experiment we did.
L=1.4 m H=W=.0.25 m
V=0.075 m^3
S=0.9 kg
Thus kg
g’=g m/s^2
Reduced gravity is about 80 times slower than g
.5*sqrt(g’h)=8.6 1.cm/s
L=1.4m
T=L/c= 16.3 s
I think we measured 18 seconds. A little slower than we observed.. why
1)Had to accelerate
2) friction
3) experimental error
1 & 2 would tend to make is slower—3 less accurate.
So the velocity difference across the interface is vdiff  g' H 
Stratification in the ocean is measured by the buoyancy frequency N2 (also called the
Brunt-Vaisala frequency)
N2 
g 
 z
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Note that this has the units of (1/Time)2 so that the square of it has the units of 1/Time
which is frequency. This time represents the highest frequency internal motion that
occurs in a stratified fluid. In the case of the gravity current we can write this in finite
from as
g 
N2 
 z
where z is the thickness of the interface between the two fluids. The velocity difference
(or vertical shear) between the two fluids is
g' H
v


z
z

Note that the vertical shear also has the units of frequency (1/T) so that shear-squared
has the same units as N2. Therefore if we divide N2 by shear squared we come up with a
non-dimensional number know as the Richardson Number.
g 
 z

Ri 
(v ) 2
z


The Richardson number is a measure of the stability of the flow and when the Richardson
drops below ¼ the mixing occurs via the Kelvin Helmholtz instability. This is a common
phenomena, it occurred in our experiment in the atmosphere and it’s beauty has been
noted my many artists (see figures on next panel). Note how the instability rolls up—
much like the animation I showed earlier this year of a random walk in an eddy. In this
case that the rolling up of the layers allows the molecular viscosity to eventually take
over and homogenize the rolled up fluid.
In the case of our gravity current tank test we can write the Richardson Number as:
Ri 
g 
 z
 v 
 
 z 
2

g' z z


g' H
H
In this case the Richarson number is independent of the density difference. This occurs
because the current speed is set by the density difference and the interface thickness
10
(z) tends adjust itself so that the Richardson number is around ¼. Consequently the
interface thickness tends to be around ¼ of the total water column depth.
Note that the term sqrt(g’h) is typically the phase speed of internal waves. In the case of
surface gravity waves (which we’ll discuss later) the phases speed is sqrt(gh). In the case
of the gravity current there is a ½ in front of the “phase speed” this occurs because two
layere are active. A more general solution would be
c=
g ' h1 h2
h1  h2
Which in the case here where (h1=h2=1/2 H) gives
c
1
g' H 2
1
4

g' H
H
2
In the case of, say the equatorial Pacific, which we’ll discuss when we talk about El –
Nino h2>> h1
c
g ' h1h2
~
h1  h2
g ' h1h2
 g ' h1 
h2
So the phase speed of the internal wave is set by the thickness of the upper layer.
An important non-dimensonal number that arises here is the Froude number--- which is
the ration of F=u/c
When flow is super-critical F>1, subcritical F<= and Critical F=1
There is an internal (using the internal wave) and external (barotropic wave) Froude
number
Hydrulic jump occurs in transition from sub-critical to critical—this is the leading edge
of the gravity current. But these details are for a much advanced class.
In the ocean if h1=200 m and kg/m^3 yields
c~2 m/s
11
Which takes ~ 60 days to cross the entire equatorial pacificocean.
Away from the equator where rotation is important this sets a length scale
What do we need to combine with c to get length scale
If c= 2 m/s and f =1.e-4 give a horizontal length scale of 20 km—or approximately the
scale of the eddies we saw in the ocean yesterday. This is the Rossby Number
R=c/f
There is a baroclininc (internal) and barotropoic (external) ROssby Number
Ekman Pumping (page 128)
Upwelling case
The following show laboratory, natural and artistic renditions of the Kelvin-Helmoltz
instability.
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