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
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. This then requires a closure
  
  A v
v
z
Av(flow)
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.
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Tomorrow I’ll try to show some movies of turbulent flows and laminar flows and the
transition.
Re=UL/Av (Dimensionless)
For example so far we’ve seen equations that define
du/dt = f*v + dP/dx + Friction
dv/dt=-fu + dP/dx + friction
Pressure = rgh
So we’ve got three equations but 4 unknowns! (we could continue adding equations for
salt and for temperature to get density but we’ll still have more unkowns than equations)
So we need to close the problem
friction =F(density, velocity) . This is the “Turbulent Closure problem” that has not been
completely solved.
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. Thus friction has the same units
pressure and can be expressed in Pascals, but we call frictional forces stress.
Units of stress are kg/m3 * m2/s * 1/s= kg (s-2m-1)=
N=kg ms-2
(F=ma –Newton’s Famous equation – this is why we call it a Newton)
Divide N by area (m2)
So units of Stress are Nm-2
Or Weight per unit area
What’s that? -- A pressure!!!
1 Pascal = 1 Nm-2 = 10 dyne cm-2
More Rigor (even more on page 103)
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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.
U=<u>+u’.
T
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u   udt
T0
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’)
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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
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.
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 times the magnitude of molecular 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
Many turbulent closures have equations for L and q ( .5*q^2 is called the turbulent
kinetic energy)
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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.
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
( 2  1 )s
(1)
S

u
z

where s is the surface area of the box.
Recall the Taylor expansion whereby
 2  1 

z
z
(2)
Putting 2 into 1 yields


(  s z ) 
( V )
z
z
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where V is the volume of the box. The force per unit volume is then


v
  (A v )
z
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 H )
z
z
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.
u
u
u
u
1 P
 
u   
u   
u 

u
v
w

 fv   A v
   AH
   A H
t
x
y
z
 x
z 
z  x 
x  y 
y 
similar equation for v
Non-Dimensionalize equation
U/T + U^2/L.. = … fU
AvU/H^2
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Ratio of “viscous” term to “inertial term” is
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.
Tomorrow I’ll try to show some movies of turbulent flows and laminar flows and the
transition.
Re=UL/Av (Dimensionless)
UH^2/AvL
And if H=aL
=Ua^L/Av
Re=UL/Av
A more formal derivation of the Reynolds stress comes from a decomposition of the flow
from a mean and turbulent part.
U=<u>+u’.
A more complete discussion on page 103 but to simply
In momentum equation we have term
wdu/dz ( discuss this term and how it accelerates the flow)
Continuity equation
 


 ( u )  ( v)  ( w)  0
t x
y
z
Multiply continuity by u and add to momentum equation we find term


(uw)
z
.. there are a bunch of other terms—but in the end this ends up as a momentum equation.
And this term becomes two terms
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d/dz(wu) and d/dz (w’u’)
By definition <u’w>=0
But <u’w’> n.e. 0 if they are coorelated
This is a stress—and it can accelerate or decelerate the flow
<w’u’>=tau=Av du/dz
Boundary Friction
One way to think of surface and bottom friction would be to depth average the
momentum equation. 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.
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H
0


1
 s   b 
z 
z
H
H
Where s and b are the surface wind stress and bottom stress respectively.
Surface Wind Stress
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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 speed of the wind squared i.e. =aC w2.
 is the surface wind stress where a is the density of air (1.2 kg/m3) , 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 s .002 *10 2
=1.2*10-5m/s2


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t H
10 *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:
u*2
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The breakthrough by Von Karman was that the eddy viscosity would vary proportionally
to u*z
i.e.
A v  u *z
(note that the surface wind stress is also often characterized by a “friction velocity”
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
  Av
U
U
 u* z
 u*2
z
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
(note that
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 x  log( x)  c  log( x * c)
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
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u *2   ln( z 0 / z 2 U 2 (z)  Cd U 2
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  C d u | 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.
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.
 a Cw 2 C d u | u |



yields
 C
u a 
 C d 

1/ 2
w ~ .03w
which works out to the current speed u is approximately 3 percent of the winds speed.
How do you think the depth of the water would come into play here?
What about stratification? What if eddy viscosity becomes real small?
What would the flow be like if the balance were between the Coriolis Force and the wind
stress?
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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
Same formulation. Usually use Kv.
EXAM RESULTS
90-100 4
70-80 5
50-69 2
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