1) Show ppt file figures.
2) Derive vorticity equation (Box 5.6 in Knauss)
Begin with discussion of curl
3) resume discussion of pressure gradient (Below)
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 and by decomposing the pressure gradient into a barotropic component and a baroclinic component we can come up with a baroclinic pressure gradient that drives vertical shears—but no depth averaged flow. Later we’ll learn that friction can also drive vertical shear in the flow.
Note that the barotropic pressure gradient is driven by the sea-level slope. If h is the sea level above some reference point (z=0) then the pressure a depth z is equal to =rg(z+h) -- assuming that the density is constant. Then the acceleration caused by the pressure gradient in the x direction is simply the derivative of =rg(z+h) divided by the density—i.e.
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 accelera
Baroclinic Pressure Gradient
Consider a channel filled with fluid on the left side with fluid of density r1 and fluid of density r2 on the right side and that r2 = r1+dr> r1 (Figure 2). On the left size the pressure at depth z is
P1(z)= gr1z,
while on the right hand side the pressure at depth z is
P2(z)= gr2z=g(r1+dr)z
. The pressure difference between the two fluid is
P2(z)- P2(z)=g drz.
Therefore the pressure gradient increases with z. The pressure gradient is zero at the surface and equals gdrH at the bottom. Considering a momentum equation between the local acceleration and the pressure gradient we write
This can also be shown in a case where the density only varies in the x direction.
P(z)=rgz
Which is the same as above.
Would this pressure gradient generate vertical shear?
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, the 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. What developes 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 2-layer exchange flow is what drives estuarine circulation (Figure 3).
Figure 1 shows the salinity field in the Hudson River estuary. Distance is in km from the battery. The white lines represent
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) 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 the above analysis derived an expression for the baroclinic pressure gradient in terms of a horizontal density gradient—it is often useful (and simpler) to simplify the density field as two layers. For example CTD section above we can construct a crude model to estimate the pressure gradient.
Using the 10 psu isohaline as the interface between the fresh and salty water we see that the interface between the surface layer and bottom layer slopes downward at approximately 1 meter every 6 km or . At the bottom the hydrostatic pressure is equal to the weight of the overlying water which can be written as:
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:
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
where g’ is reduced gravity. Reduced gravity is the effective gravity that a body feels in a fluid. Note that if the two densities (r1 & r2) 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 Dr= 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.
where () is the slope of the sea surface slope. In rivers the sea-surface slopes 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
gradient. This would occur when
Which means that the ratio of the slope of the interface to the slope of the sea surface is r/Dr. In the example shown in figure 1 Dr~ 10 kg/m3 and since r~ 1000 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)
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=rg(H+Dh)
P2==(r+Dr)gH
where Dh is the sea-surface difference between the two stations r is the average density for that profile 1, Dr 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
rg(H+Dh)=(r+Dr)gH
pgH + rgDh=pgH +DrgH
rDh= DrH
Dh= H Dr/ r
So if H is 4000m and Dr is .1 kg/m3 Dh would be 40 cm.
If the horizontal density gradient were constant—then the pressure gradient would increase linearly from zero at the bottom to g dn/dx at the surface.
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).
If Dr is the density difference between the two fluids and r is the mean density of the two fluids then r1=r-Dr/2 and r2=r+Dr/2. 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(r1V1H/2 +r2V2H/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(r+Dr/2 + r-Dr/2)=rgVH/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(r1V13H/4 +r2V2H/4 )
=gVH/8(3r1 +r2 )
=gVH/8(3(r-Dr/2)+(r+Dr/2))
=gVH/8(4r-Dr)
= r gVH/2 - Dr gVH /8
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
Dr gVH /8 (3)
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 rV thus if we assume that all the potential energy has turned into kinetic energy we can equate the kinetic energy to 3
Dr gVH /8=½ mv2 = 1/2 rVv2
which can be rearranged to read
(4)
where g’ is the reduced gravity and equal to gDr/r.
Equation 4 is the speed that the front of the gravity current, such as the one released in the lab, propagates.
Equation 4 is the speed that the front of the gravity current, such as the one released in the lab, propagates.
So the velocity difference across the interface is .
Stratification in the ocean is measured by the buoyancy frequency N2 (also called the Brunt-Vaisala frequency)
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
where Dz is the thickness of the interface between the two fluids. The velocity difference (or vertical shear) between the two fluids is
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.
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:
Thus the Richardson number is independent of the density difference. In fact interface thickness (Dz) 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.
The following show laboratory, natural and artistic renditions of the Kelvin-Helmoltz instability.
IF you make it here begin with inertial motion…..
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