CEE 262A H YDRODYNAMICS Lecture 18 Surface Ekman layer.
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Transcript of CEE 262A H YDRODYNAMICS Lecture 18 Surface Ekman layer.
![Page 1: CEE 262A H YDRODYNAMICS Lecture 18 Surface Ekman layer.](https://reader036.fdocuments.in/reader036/viewer/2022081519/56649e7b5503460f94b7bf18/html5/thumbnails/1.jpg)
CEE 262A
HYDRODYNAMICS
Lecture 18
Surface Ekman layer
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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
x1x2
x3
u1u2
0x3 = 0
Assumptions: Steady flow, uniform density, constant viscosity, no pressure gradients
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Steady Linear
Constant density, only hydrostatic pressure with no motion
Coriolis – Friction Balance
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The governing equations reduce to (f-plane)2
12 2
3
22
1 23
ufu
x
ufu
x
With the BCs at x3 = 0:201
*3 0
2
3
0
uu
x
u
x
Again, we wish to make the momentum equations dimensionless, and again, we look at the surface BC
unknown velocity scale
unknown length scale
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These imply that
Let’s define * * *1 1 2 2 3 3u Uu u Uu x Zx
2 * 2 *2* *1 1
2 22 *2 *23 3
2 * 2 *2* *2 2
1 12 *2 *23 3
u uU ZfUu f u
Z x x
u uU ZfUu f u
Z x x
Thus if we choose
2 ** 1
2 *23
2 ** 2
1 *23
uu
x
uu
x
the momentum equations are parameter free
The Ekman layer thickness
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These must be solved subject to the dimensionless BCs:
*1*
3
*2
3
1
0
u
x
u
x
Both at x3*=0
We also require that the flow decay to zero as *
3x
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
* *1 2u iu
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In terms of , the two real o.d.e.s become one complex equation:
which has the solution
where A,B, 1, and 2 are all complex
Using the ode itself, we see that for either i
2i i
The two roots are given by
1 2
1 11 1
2 2i i
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With
1 2
1 11 1
2 2i i
the condition that the flow disappear at great depth implies that B=0 since
* * *2 3 3 3exp exp 2 as B x x x
The condition on the water surface
1
1 11 1 1
2 2A A i A i
Thus, 3* *
1 2
3 3 3
11exp
2 2
1exp cos sin
2 2 2 2
i xiu i u
x x xii
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Taking the real part to find u1*
* 3 3 31
3 3
1exp cos sin
2 2 2 2
exp sin42 2
x x xu
x x
(after using some trigonometric identities)
* 3 32 exp sin
42 2
x xu
Likewise, u2* is found from the imaginary part of and a
little more trigonometry to be
Thus, the velocity vector decays and rotates with depth – aka the “Ekman Spiral”
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The vertical structure of the Ekman spiral, as originally plotted by Ekman
At the surface (x3=0):
E
x2
x1
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Friction + wind:
Wind stress
motion
Friction (stress on bottom of parcel)
Friction + wind + Coriolis:
Wind stress
motion
Friction (stress on bottom of parcel)
Coriolis
In order to balance Coriolis and wind stress, motion must be at some angle to the wind
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Finally, imagine we go deep enough into then ocean to have the stress be zero on the bottom of our slab
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: 00 0
21 12 3 3 *
3 3 3
00 02 2
1 3 33 3 3
0
du dudf u dx dx u
dx dx dx
du dudf u dx dx
dx dx dx
Wind stress
motion
Coriolis
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0 2*
2 2 3
0
1 1 3 0
uq u dx
f
q u dx
Thus the net transports are found to be
i.e., to the right of the wind in the Northern hemisphere!
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California
Wind from north
Upwelling (downwelling): The effects of variations in Ekman transport
Ekman drift
Upwelling from depth
California
Wind from south
Ekman drift
Downwelling from surface
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Why are California coastal waters so COLD? Upwelling!
Upwelling-favorable winds duringspring/summer.