Lesson 5: Shallow water systems and the Ekman Layer
In the late 1800’s biologist Fridtjof Nansen observed that icebergs drifting in
arctic waters did not drift in the same direction as the local winds but rather in a direction
to the right of the wind. Examination of the balance between friction, Coriolis force and
the pressure force show that such a deflection is expected since the Coriolis force will
cause a rightward direction to the pressure induced wind force. Quantitative examination
of the balance between these forces was first proposed by oceanographer Vagn Ekman in
the early 1900s. The following analysis derives the equations that Ekman developed and
discusses their results.
1. Preliminaries - Turbulence and the effects of friction
When we derived the Navier-Stokes equations in Chapter 5, we assumed a
constitutive relationship between the stress tensor the gradient of the velocity field. This
hypothesis is valid provided the flow is laminar; meaning that there are smooth variations
in the velocity field. The upper layer of the ocean, however, exhibits significant
turbulence and mixing and the constitutive relationship derived in Chapter 5 breaks
down. Instead, we must consider mean values of the flow field and assume that
turbulent stresses are proportional to the gradients of this mean flow field.
The most common approach to examining turbulent effects is to split a given
quantity into a mean flow, indicated by an over-bar, and a fluctuating contribution with a
mean value of zero. In other words, for scalar quantity, q, we can represent it in terms of
a mean and fluctuating contribution as:
q  q  q'
where
q'  0
By “mean value” we are referring to an integration over the time scale, T, of the problem
u
of interest. For example, to find the mean flow of
, we average over the integrated
t
flow field as follows:
u 1 T u
u T   u 0
 
dt 
t T 0 t
T
By using this new notation of splitting up the equations of motion into mean and
fluctuating contributions and then taking the time average of these results, we can deduce
the additional contributions that turbulent effects provide to the net frictional forces.
One must be careful in this approach how we represent the product of averaged
v
quantities. For example, for the advective component, u , breaking up the flow into
x
two contributions and taking the average leads to:
1
u

 



v





 u  u'
v  v'  u
v  u'
v  u v'  u ' v'
x
x
x
x
x
x
Mean values are considered constants in the process of integration and can be factored
out of the integration in the above equations. For example, in the second term
v
v
 u'  0
x
x
The term above is equal to zero because the mean of any fluctuation value is zero by
v
definition. The end result is the mean of u
is expressed as two terms:
x
v
v
v'
u
u
 u' .
x
x
x
u'
Applying the above notation and averaging to the Navier-Stokes equation in a
rotating frame with small density fluctuations (Boussinesq approximation), we obtain,
Du
u '
u '
u '
1 p
 u'
 v'
 w'
 fv
  2 u
Dt
x
y
z
 o x
Dv
v'
v'
v'
1 p
 u'
 v'
 w'
 fu 
  2 v
Dt
x
y
z
 o y
Dw
w'
w'
w'
1 p 
 u'
 v'
 w'


g   2 w
Dt
x
y
z
 o z  o
(1)
u  v  w


0
x y z
D




and, due to the Boussinesq
 u v w
Dt t
x
y
z
approximation, the background density is independent of time. Both the conservative
body forces and the Coriolis parameter are also assumed to be independent of time as
compared to the turbulent time scales of our averages.
The continuity equation expressed in equation (1) is in its mean form. But the
original continuity equation also still holds,
In the above equation,









u  u' 
v  v' 
w  w'  0
x
y
z
Consequently we can see that continuity also holds for the fluctuating terms:
u ' v' w'


0
x y z
2
Use of the above continuity equation allows us to rearrange the non-linear terms in
equation (1). For example, the non-linear terms in the east-west equation of motion can
be written as:
u'
 u ' v ' w ' 
u '
u '
u '   u ' u '   u ' v '    u ' w ' 
v'
 w'



 u '



x
y
z
x
y
z
 x y z 
0
 




u 'u ' 
u 'v ' 
u 'w' 
y
z
 x

Similar forms can be written for the non-linear terms in north-south and vertical
directions. The resulting equations of motion then take the form:






Du
1 p



 fv
  2 u  u ' u '  u ' v'  u ' w'
Dt
 o x
x
y
z
Dv
1 p



 fu 
  2 v  v' u '  v' v'  v' w'
Dt
 o y
x
y
z
(2)
Dw
1 p 





g   2 w  w' u '  w' v'  w' w'
Dt
 o z  o
x
y
z
We now apply a turbulent form of the constitutive hypothesis to the turbulent
non-linear components of the above equations of motion. The following analysis is from
the work of Kirwan, A. D. 19691. We can see from equations (2) that the eddy terms are
second order tensors that may be related to the dissipation or frictional effects of the fluid
medium. We relate the above turbulent velocity components by introducing the
Reynold’s stress tensor
 ij  u 'i u ' j
(3)
The individual terms, uu, uv, uw, vv, vw, and ww are called Reynolds stresses or
eddy stresses.
Recall from chapter 5, we derived the frictional term from the deviatoric term of the
stress tensor,  ij   p ij  d ij . The Reynold’s stress tensor is analogous to the deviatoric
1
Kirwan, A. D., Formulation of Constitutive Equations for Large-Scale Turbulent Mixing, Journal of
Geophysical Research, 74, 6953-6959, 1969
3
tensor in laminar flows. As such, we can establish the same constiuitive hypothesis as
before and assume that the Reynold’s stress tensor is linearly proportional to the velocity
gradients of the fluid medium:
 ij  K ijkl ekl
Where
ekl 
(4)
1  u k u l


2  xl xk




Notice there is an important additional point in the above hypothesis. In relating the
stresses to the velocity gradients or strain rates, we assume that they are related to the
mean velocity field elements as opposed to the turbulent contributions. The fourth order
tensor, K ijkl , can be simplified by utilizing
1. The symmetry of the Reynold’s stress tensor
2. Assuming incompressibility
3. Making a physical assumption about the distribution of the molecular medium.
By making these assumptions, we see that
 u i u j

 x
xi
j

 ij   p ij   

   0 uiu j .

(5)
The Reynolds stress tensor 0 uiu j has 9 Cartesian components:
  0 u 2  0 u v  0 u w


 0 vu  0 v2  0 vw  .


 0 wu  0 wv 0 w2 


(6)
The diagonal components are the normal stresses, and the off-diagonal components are
the shear stresses.
Using this expression for the Reynolds stress tensor, the equation of motion for the mean
flow of an incompressible fluid in the inertial reference frame is:
ui
u
1  ij g
uj i 
 i3 ,
t
x j 0 x j 0
(7)
4
1  u i u j 

Where  ij   p ij  2 eij  0 uiuj and the shear stress eij  
.
2  x j xi 
By multiplying equation (7) by ui , averaging, and rearranging terms, we arrive at an
expression for mean kinetic energy:
 1 2
 1 2 1 
1
u i g
ui  ij   ij

u  .
 ui   u j
 ui  
t  2 
x j  2  0 x j
0 x j 0 i i 3


If we multiply equation (7) by ui , and averaging, it can be shown (without proof) that the
result is the turbulence kinetic energy equation:
D 1 2 
 ui  
Dt  2  x j
 1

1
u i
puj  ui2uj  2 uieij   uiuj
 g wT   2 eij eij .

2
x j
 0

The right-hand-side terms are:
1. Spatial transport of turbulence kinetic energy (TKE; note no production or
dissipation, only transport)
2. Shear production of TKE
3. Buoyancy production of TKE
4. Viscous dissipation of TKE
The Navier-Stokes equations in modified isotropy flow
In the laminar case (Chapter 5), we assumed that the entire molecular medium was
isotropic. For the turbulent case, we observe that the mixing scales (molecular motion)
between the horizontal and vertical are different. Thus we assume horizontal planar
isotropy instead of uniformity throughout the whole fluid medium. The resulting
mathematics of how this modifies K ijkl (equation 5) is beyond the scope of this lesson.
Understanding the impact of the underlying physical assumptions is all that is important
for now. After assuming incompressibility and planar isotropy, the resulting equations of
motion become:
  2u  2u 
  2u 
Du
1 p
 fv  
  h  2  2    v  2 
Dt
 o x
y 
 x
 z 
  2v  2v 
  2v 
Dv
1 p
 fu  
  h  2  2    v  2 
Dt
 o y
y 
 x
 z 
5
(8)
 2w 2w 
 2w 
Dw
1 p 


g   h  2  2    v  2 
Dt
 o z  o
y 
 x
 z 
u v w


 0.
x y z
Further, Reynold’s stress tensors are defined as



 ij   i
u j
xi
 j
u i
x j




Before moving on to Ekman’s quantitative analysis, let us examine how the
Reynold’s stress tensor is applied to real world analysis. First looking at the Reynold’s
stress index notation itself we can see it that the first index indicates the direction of the
forcing term. But note that the Reynold’s stress tensor is symmetric. In the lower
m2
m2
atmosphere, eddy coefficients have typical values of  h  10 5
and  v  10
. In
s
s
m2
m2
the upper ocean, typical coefficient values are  h  100
and  v  0.01
.
s
s
^
Example: Let us now consider a mean flow field of the form, u  tanh( 2y ) i 2. A
graph of this field is shown in figure 1 below.
U(y)=tanh(2*pi*y)
1
y
0
-1
0
x
^
Figure 1 – Flow profile of velocity field u  tanh( 2y ) i
2
The following analysis follows from similar discussion in McWilliams, Fundamental of Fluid Dynamics,
Cambridge University Press, Cambridge UK, 2006, 83-85.
6
The associated Reynold’s stress for this mean flow is
u
 yx   y
 2 y 1  tanh 2 2y  and is shown in figure 2. Part of the forcing
y
associated with this flow is in the direction of the arrow and is observed to have a
spreading or diffusive effect on the flow by its lateral direction to the initial direction of
the flow field.


Associated Reynold's Stress Tensor to the flow u(y)=tanh(2*pi*y)
0.5
0.4
0.3
0.2
y
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
-7
-6
-5
-4
-3
-2
-1
0
x
2. Ekman’s solution
Given the equation of Geostrophic balance derived in the chapter 7, we know that
the Coriolis force contribution of the Geostrophic flow will perfectly balance with the
pressure gradient provided the Coriolis vector is assumed constant.
f ug  
1
o
p
This leaves the friction flow contribution of the Coriolis force in balance with the viscous
terms of the equation of motion. We will assume that the mixed layer is horizontally
homogeneous so that we neglect all terms for coefficients proportional to  h and  v is
spatially constant. This leads us to the most basic form of the Ekman flow equation
which assumes a steady state balance of the Coriolis force with the vertical viscous forces
in the open ocean.
  2u F 

 fvF   v 
2 
 z 
(6)
  2vF 
fu F   v  2 
 z 
^
Now consider a constant wind stress,  , acting in the east-west or i direction on the free
surface of the ocean, z=0.
7
  constant   v
u F
at z=0
z
(7)
We assume another boundary condition that the flow is negligible as we go to large
depths in the water to avoid any bottom friction contributions and to avoid any issues
with energy sources at large distances (The only energy source is due to the wind stress at
the surface given by equation (7)). A negligible flow field at large ocean depths is
quantitatively expressed as
u F , vF  0 as z  
(8)
Equations (6) – (8) provide us all the information we need to solve for the Ekman or
frictional contribution to the flow field.
To solve for the v component of the frictional flow field, take the second
derivative of the north-south component of equation (6) with respect to z:
 2u F  v   4 vF 
  4 
f  z 
z 2
Now substitute the east-west component of equation (6) to obtain:
2
    4 v 
v F   v   4F 
 f   z 
A similar equation can be obtained for u F :

u F   v
 f



2
  4u F

4
 z



Both equations have a solution of the form3:
u F  Ae 1i  E z  Be 1i  E z
v F  Ce1i  E z  De 1i  E z
where
E 
f
2 v
We can see right away that the deep water boundary condition, equation (8),
requires B =D=0. Coefficients A and C are then related by the original coupled
equations by substituting in the general solutions:
3
There is actually a more general solution with 2 additional terms in each velocity component. For
presentation purposes, we have only given the above reduced form to stick with main physical feature of
the problem. The alternative is to present the full general solution and impose another boundary condition;
likely an imposition to the velocity field at the surface which would seem too restrictive.
8
  2 vF 
2
2
2
1i  E z


fuF   v  2   fAe
  v 2 Ce 1i  E z   v Ce 1i  E z 1  i   E
z
 z 
Therefore A  iC
Finally, using the free surface boundary condition, equation (7), we see that
v
A
du F
dz
    v A1  i  E    constant and
z 0

 v 1  i  E

 1  i 
2 v  E
With some algebraic manipulation and application of common trigonometric identities,
the general solutions of the equations of motion are:
uF 
 1  i  1i 
e
2 v  E

Ez


e  E z cos   E z  
f v
4

1
 1  i  1i  z
1  z 

vF  
e
 
e sin    E z  
2 v  E
f v
4

For z  0
A graph of horizontal velocity with increasing depth is shown in Figure 1.
E
(9)
E
Scaled units of horizontal velocity with depth
0

2 E
v
z
z=0
0
u
Figure 1 – Graph of solution of equation (9) with depth for a constant wind field in the
east-west direction. Values have been normalized such that uz  0  1 . This type of
graph is also called a Hodograph.
9
Notice the following properties of the solution and from figure 1:
1. The velocity field at the surface makes a 45 degree angle to the right of the direction
of the wind stress at the surface in this steady state condition.
2. The exponential dependence shows that the flow decreases to 0 as we go deeper in the
ocean.

3. At a distance z  
, the flow has reversed direction from southeast to northwest.
E
This depth is arbitrarily taken as the effective depth of the Mixed layer and is called the
Ekman layer.
Unfortunately, it is very difficult to experimentally observe the above solutions in
the ocean. The primary causes of the inaccuracy of the above solution are due to the
assumption of a spatially independent vertical viscosity v which was somewhat arbitrary
for simplification purposes and the assumption of a constant stress field at the surface (we
know that winds are continuously changing with time). Further, no horizontal boundaries
were assumed so the results above are only applicable in the open ocean away from the
continents. Needless to say it is still quite on open area of research to relax some of the
above assumptions and still obtain a nice compact analytic solution. We can, however,
still appreciate the above results. Equation (9) and figure (1) show us a simple but
physically plausible coupling of the atmospheric conditions to the ocean. Further,
observations of the flow field in the mixed layer over a span of multiple days, confirm
approximation representations of the above solutions.
3: Transport and Upwelling
Often we are interested in examining the transport of a vertical fluid column
throughout the ocean basin. The appropriate quantity to consider is the horizontal mass
transport and is defined as
0
M H    u H dz
h
The mass transport is a measure of the quantity of a vertical fluid column that passes a
point per unit time. The depth value, h, is determined by the scaling of the mixed layer, or
alternatively where horizontal motions are assumed to vanish.
For the Ekamn solution discussed above, we can assume the ocean is
incompressible and thus examine a volumetric flow rate. Integration over the vertical
Ekman layer of equation (9) provides us with the components of the volume of fluid
transported within the mixed later:
0
Volume transport in the east-west direction 

10
 u
E
0
F
dz   u F dz  0

0
 v F dz 
Volume transport in the north-south direction 


E
0
v

F
dz  

f
Recall that the applied stress was assumed in the east-west direction while the volume
transport is solely in the north-south direction. These results indicate that the transport of
fluid volume is at right angles to the direction of the applied wind stress at the surface.
The net volume transport or mass transport to the right of the wind stress derived above
has significant implications on the ocean current patterns in the Atlantic and Pacific
basin.
The average effects of the semi-permanent Bermuda high in the Atlantic Ocean
leads to a long duration southerly wind flow along the east coast of the United States.
The water volume in the mixed layer is therefore transported to the east as calculated
above. Conservation of mass requires the above transported volume to be replaced by the
ocean water below the mixed layer. This leads to an upwelling or net vertical current
along the coastline. These upwelling effects have major implications on the ocean
climate as well as the biological conditions of the coastline environment. One can
imagine that nutrient rich conditions below the mixed layer provide the surrounding
marine life at the surface with an abundant source of food.
The arguments leading to upwelling along the coast are qualitatively sound but
are not quantitatively relevant yet since we assumed no continental boundaries in section
2. In the next chapter, we will include these horizontal boundary conditions as well as
relax some of the above assumptions to obtain a more accurate representation of the
horizontal current along the coastline.
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