Today We Re Going to Complete the Momentum Equation by Adding the Frictional Term

Today we’re going to complete the momentum equation by adding the frictional term. The frictional term is probably the most complex term in the momentum equation and certainly the one that we understand the least.

We are all familiar with friction. It occurs when one thing is rubbed against another—like two sticks to make fire. So in the ocean we can get friction when the water is rubbed by the atmosphere (wind friction) where the water rubs the bottom (bottom friction) and when water rubs against itself (internal friction). Friction is characterized as a tangential stress.

The wind inputs energy into the ocean via a surface wind stress. This energy is exchanged between the layers via friction (although internal wave motion also does this) and is drained from the ocean at the bottom via bottom friction.

In fluid mechanics we often use the Greek Letter for friction

Reynold’s stress parameterization

More Rigor (even more on page 103)

We call these stresses Reynolds Stress after the famous fluids guy Osborne Reynolds (1842-1912). A more formal derivation of the Reynolds stress comes from a decomposition of the flow from a mean and turbulent part.

Following Knauss page 102-104 u=+u’.

1 T u   udt T 0

1 T u'  udt  0 T 0

Consider the averaging of separate components

(do this with uw in class)

1 uv  (u  u')(v  v')  uv  uv'vu'u'v =

If we average this over a time period we get uv  uv  u'v'

When is the second term zero? When is it not zero?

On page 103 – if you can follow the derivation – you can see that this term arrises from the advective term in the momentum equation.

Start here October 12th

1) Homework Posted—may not be able to do this now. By Monday you will.

On Monday we talked about the Reynolds Stress

(where brackets indicate the average over time)

Note that if there is as mean vertical shear that when w is upwards this will advect slower velocity upward and thus reduce u.

Thus when w’ is positive, u’ will tend to be negative.

Likewise when the w’ is downward this will tend to bring faster fluid down and thus increase u’

Thus when w’ is negative u’ tends to be positive.

Thus u’ and w’ are correlated and thus the average of their product over time is is not zero:

Draw time series that are correlated:

Multiply them together and show average is not zero:

2 Similar analogy can be made with a tracer (such as salinity or temperature)

DO THIS AND WRITE (w’s’) represents a flux of salt associate with turbulent fluctuations.

Recall that the diffusion of material is k * dC/dx: does anybody remember this?

A similar analogy is used for the transform of momentum (stress) using an eddy viscosity

u   A xz v z

If the transfer of momentum is actually done by the friction of water molecules rubbing against each other then

Internal Friction—is the friction between the fluid elements and in principle it can either laminar or turbulent though for all practical purposes friction in the ocean is due the interaction of turbulent eddies. In laminar flow (movie in class) friction arises due to the actual rubbing of molecules— and the frictional force is equal to the density times the kinematic viscosity times the vertical

v    z where is the molecular kinematic viscosity of sea water where has a value of 1*10-6 m2/s (note those units—we’ll talk about them later). We’ll see later that turbulent friction, while considerably more complicated can be expressed in an identical form expect is replaced with the turbulent eddy viscosity (Av) also with unites of m2/s--what makes turbulent friction more complicated is that Av varies by orders of magnitude in the ocean depending on the state of the flow. For numerical mores of ocean circulation this requires a “turbulent closure” since we know have another variable in the equation. Discussion of turbulent closures is beyond the scope of this class.

So we write:

v   A v z

3 The transition between Laminar flow and Turbulent flow is determined by a Reynolds number—which is essentially the ratio of the advective term to the viscous term.

Monday I’ll try to show some movies of turbulent flows and laminar flows and the transition.

Now that you know the units of viscosity you can calculate the units of friciton—and they are kg/(m*s2) which is a force per unit area

F=ma/AREA= (kg m/s^2)/ m^3=kg (m s^2)

Thus friction has the same units pressure (force per unit area like your tires – pounds per square inch) and can be expressed in Pascals, but we call frictional forces stress.

1 Pascal = 1 Nm-2 = 10 dyne cm-2

Question: How does Pressure enter the momentum equation? (i.e. how is pressure related to acceleration?). Since Stress has the same units as pressure how then must the stress (frictional) term enter the momentum equation.

Answer. Gradients in stress dt/dz dt/dy dt/dx

In general the greatest stress gradients are in the vertical.

Q Why?

Recall the aspect ration of the Ocean. Lots of structure in the vertical—scales are much smaller and there fore gradients tend to be much larger!

Internal friction transfers energy between layers of fluid. Consider the case of wind blowing on the surface of the ocean (later we’ll quantify how much friction the wind imparts on the ocean’s surface’)

4 By doing what is called dimensional analysis we can estimate how long it would take for for wind friction to penetrate down 100 meters via molecular viscosity alone.

Note that the depth squared divided by the viscosity has units of time.

H2 L2   T  L2 T

This is something I would like you to be able to do—to use DIMENSIONAL ANALYSIS to provide an order of magnitude estimate of time, length, velocity ect. It’s often a good idea to do the before attacking a problem in detail:

What time scale would this be? It would be the time scale for the wind forced motion to be transferred vertically by friction . In the molecular viscosity case it would be the top layer of molecules accerating the layer beneath it and then that layer accelerating the layer beneath it—and so on. Given the low value of molecular viscosity

100/1.e-6=1.e8 seconds ~ 3 years

It take three years for the wind to effect the water 10 meters down.

From observations we know that this is not the case. In the 1800’s a researcher named Zoppritz argued that the winds could not be a driving mechanism for ocean circulation because laboratory measurements of water’s viscosity were so low that it would take 1000’s of years for the wind to drive ocean circulation. Obviously Zoppritz spent too much time in the lab. In contrast Nansen, who spent years stuck on a boat in trapped in the ice in the Arctic, who was contemporary Zoppritz ,noted that Icebergs—which can extend 100 meters below the ocean’s surface—are driven by the wind and suggested that the ocean is in fact driven by the wind. It was Nansen’s observations that gave birth to the modern field of physical oceanography by the development of a theory for wind-driven flow’s by Ekman who was inspired by Nansen’s observations.

5 S

Rather than using the molecular viscosity Ekman proposed that the vertical transfer of momentum occurs through turbulent eddies and proposed a vertical eddy viscosity that was many u times the magnitude of molecular z viscosity. Here the value of the  viscosity can be thought of as the size of the turbulent eddy times the current velocity in the turbulent eddy i.e.

Av=L*q  where L and q are characteristic length and velocity scales of turbulent eddies. (Explain what characteristics scales are), and Av is the vertical eddy viscosity. In the ocean this can vary from near molecular levels to over 1 m2/s in unstratified turbulent flows.

Incorporating friction into the momentum equation

How do we incorporate frictional stresses into the momentum equation? The stress represents the force applied to an area—but to have it enter the momentum equation we need to consider the force on a mass of fluid. To do this consider the box

Consider a case where the flow is vertically sheared and that the flow is stronger near the surface. Intuition would tell you that the frictional forces on the top of the middle box would drag the box to the right thus increasing its speed , while the slower moving fluid beneath the middle box would tend to drag the fluid to the left and decelerate the flow. The total force on the box then would be the stress on top of the box trying to drag the box to the right minus the stress on the bottom of the box which is trying to drag the box to the left

6 (2  1 )s (1)

where s is the surface area of the box. Recall the Taylor expansion whereby

     z (2) 2 1 z

Putting 2 into 1 yields

  (sz)  (V) z z where V is the volume of the box. The force per unit volume becomes

  v   w'u'   (A )   = z z v z z

Which is the vertical stress divergence.

An identical procedure can be used to characterize the stresses on the vertical sides of the box—and they would take the form   v   (A ) z z H y

Note that Av and Ah are fundamentally different—because in the ocean the vertical turbulent eddies are suppressed by stratification, while horizontal eddies are not.

Now we’re ready – for the first time—to write the full blown momentum equation that includes the effects of the local and field accelerations, pressure gradients, the earth’s rotation and the vertical and horizontal stress divergence.

7 u u u u 1 P   u    u    u   u  v  w    fv  A v   AH   AH  t x y z  x z  z  x  x  y  y  similar equation for v

Boundary Friction

One way to think of surface and bottom friction would be to depth average the momentum equation. This week’s homework problem (due next week) involves the depth averaged momentum equation (sometimes called the BAROTROPIC MODE) Most of the terms have a pretty simple form in the depth averaged sense (in detail both the pressure gradient term and the field acceleration term require more care than I’m going to give them here) and read. the stress term becomes.

0 1  1 z      H  z H s b H

Where s and b are the surface wind stress and bottom stress respectively.

Surface Wind Stress

Observations suggest are surface currents are 3% of the wind—a 20 knot wind would drive a .6 knot current and that the stress applied to the ocean’s surface increases with the 2 speed of the wind squared i.e. =aC w .

8 3  is the surface wind stress where a is the density of air (1.2 kg/m ) , C is a “constant” and w is wind speed. The reason that constant is in quotes is that it’s not really a constant but varies with wave height and shape, but typically is somewhere between 1-2 10-3.

Assume a 10 m column of fluid starts at rest and a 10 m/s wind begins blowing and Cd is .001. If we neglect all other terms the water will begin accelerating at a rate of

u  .002*102  s  =1.2*10-5m/s2 t H 103 *10

So after 10 hours (36000 seconds) the current speed would be .43 m/s

But after 100 hours the current speed would be 4.3 m/s

But observations tell us that after 4 days of a 20 knot wind the current does not flow at 4.3 m/s (8 knots!!).

Why not?

Because the assumption of other terms not being important is not valid. In this case both the earth’s rotation, bottom friction and in the case of enclosed seas—the pressure gradient will become important.

Bottom Stress

The term Av du/dz has units m2/s2 and its square route is called a friction velocity u*.

Thus stress is:

2 u*

(we can also use a u* at the surface for the surface friction velocity)

The breakthrough by Von Karman was that the eddy viscosity would vary proportionally to u*z

9 i.e.

Av  u*z

Because larger the friction velocity—the larger the turbulent velocity flucutations—and the larger the distance from the bottom (z) the larger the size of the turbulent eddies.

Thus the stress can written as

U   u z  u2 * z *

Which yields

U u  * z z which can be integrated to yield the famous “Law of the Wall”

u z U(z)  * ln( )  z0

where zo the bottom roughess is a constant of integration obtained by assuming that the velocity goes to zero at height zo above the bottom.

If you square the above equation and rearrange the terms you find

2 2 2 2 u*  ln(z0 / z U (z)  Cd U

10 So like the wind the bottom friction is proportional to the current speed squared/.

Frictional forces on the bottom will tend to slow down the fluid motion.

b  Cdu | u | note that we use u times the absolute value of u so that the bottom frictional force always opposes the flow. While numerous things can impact the value or k it is also tends to fall in the range of .002- .003.

To simply the equations—sometimes we will linearize the friction term so that

 b  ru

No—that’s not Rutgers University. Its often surprising how well this linearized version works in simple physical models.

So in the case of the 10 m/s wind blowing if we included bottom friction the flow would become steady when bottom stress equals surface stress—i.e.

 Cw 2 C u | u | a  d   yields

1/ 2 aC  u    w ~ .03w Cd   which works out to the current speed u is approximately 3 percent of the winds speed.

11 Talk about the the barotropic pressure gradient.

Balance this with friction

Estimate the depth averaged flow.

This is a River!

Could be the shelf.

What would the flow be like if the balance were between the Coriolis Force and the wind stress?

Briefly discuss where we’re going with the equation on the board

Inertial

Geostrophic

I think you can do this now!! But we’ll spend the next few lecture going over them in gory detail.

This sort of balance would occur in the ocean away from boundaries and it is exactly the momentum balance that Nansen observed and Ekman developed 100 year’s ago.

Draw Free Body Diagrams and Show movies

TRACERS

12 Same formulation. Usually use Kv.

EXAM RESULTS

90-100 4 70-80 5 50-69 2