Transcript Lecture18

CEE 262A
HYDRODYNAMICS
Lecture 18
Surface Ekman layer
Another exact solution:
Coriolis   Viscous Stress: The Ekman Layer
A rotational boundary layer in an infinite ocean: flow driven by
a wind stress at the surface (x3 = 0), acting in the x1 direction
0
x3
x1
x2
u2
x3 = 0
u1
Assumptions: Steady flow, uniform density, constant viscosity, no
pressure gradients
Steady
Linear
Constant density,
only hydrostatic pressure with no motion
Coriolis – Friction Balance
The governing equations reduce to (f-plane)
 2u1
 fu2   2
x3
 2 u2
fu1  
x32
With the BCs at x3 = 0:
u1 0


 u*2
x3 0
u2

0
x3
Again, we wish to make the momentum equations dimensionless,
and again, we look at the surface BC
unknown velocity scale
unknown length scale
These imply that
Let’s define
u1  Uu1*
u2  Uu2*
x3  Zx3*
2 *
2 *
2

u

u1
U
Z
*
*
1
 fUu2   2
f
u2 
*2
Z x3

x3*2
U  2 u 2*
Z 2 *  2 u 2*
fUu   2
 f
u1 
*2
Z x3

x3*2
*
1
Thus if we choose
The Ekman layer thickness
the momentum equations are parameter free
2 *

u1
*
u2 
x3*2
2 *

u2
*
u1 
x3*2
These must be solved subject to the dimensionless BCs:
u1*
1
*
x3
Both at x3*=0
u2*
0
x3
*
We also require that the flow decay to zero as x3  
This completes the physics part of the problem; what remains is
the mathematical problem.
To solve this system of two equations, we follow Ekman and
define
  u1*  i u2*
In terms of , the two real o.d.e.s become one complex equation:
which has the solution
where A,B, l1, and l2 are all complex
Using the ode itself, we see that for either li l i 2  i
The two roots are given by
 1 
 1 
l1  
1

i
l

 2    1  i 

 2
 2
With
 1 
 1 
l1  
1

i
l

 2    1  i 

 2
 2
the condition that the flow disappear at great depth implies that
B=0 since
B exp  l 2 x3*   exp  x3* 2   as x3*  


The condition on the water surface
 1 
 1 
A l1  A 
 1  i   1  A  
 1  i 
 2
 2
 1  i  x3 
 1 i 
*
*
Thus, u1  i u2  

 exp 
2 
 2

 1 i 
 x3  
 x3 
 x3  

 exp 
  cos 
  i sin 

 2
 2 
 2
 2 
Taking the real part to find u1*
 1 
 x3  
 x3 
 x3  
u 
 exp 
  cos 
  sin 

2
2
2
2








 x3   x3  
 exp 
 
 sin 
 2   2 4
*
1
(after using some trigonometric identities)
Likewise, u2* is found from the imaginary part of  and a
little more trigonometry to be
 x3   x3  
u2  exp 
 
 sin 
 2   2 4
*
Thus, the velocity vector decays and rotates with depth – aka
the “Ekman Spiral”
x1
The vertical structure of
the Ekman spiral, as
originally plotted by
Ekman
x2
At the surface (x3=0):
dE
Friction + wind:
Friction
(stress on
bottom of
parcel)
Wind stress
motion
Friction + wind + Coriolis:
Friction
(stress on
bottom of
parcel)
Wind stress
motion
Coriolis
In order to balance Coriolis and wind stress, motion must be at some
angle to the wind
Finally, imagine we go deep enough into then ocean to have the stress
be zero on the bottom of our slab
Wind stress
Coriolis
motion
Thus, we see that in an integral sense the motion we produce must
be at right angles to the wind - this motion is known as Ekman drift.
To see this more formally, we integrate the dimensional momentum
equations:
0
0
0
du1
d  du1 
2
 f  u2 dx3  

dx



u

 3
*
dx
dx
dx
3 
3 
3 


du2
d  du2 
f  u1dx3  

dx



 3
dx
dx
dx3
3 
3 


0
0
0
0

Thus the net transports are found to be
0
u*2
q2   u2 dx3  
f

0
q1 
 u dx
1
3
0

i.e., to the right of the wind in the Northern hemisphere!
Upwelling (downwelling):
The effects of variations in Ekman transport
Wind from north
Ekman drift
Upwelling from depth
California
Wind from south
Ekman drift
Downwelling from surface
California
Why are California coastal waters so COLD? Upwelling!
Upwelling-favorable winds during
spring/summer.