Etudes et évaluation de processus océaniques par des hiérarchies ...

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6.2. THE EKMAN LAYER 39
In the rotating case (f ≠ 0) the solution is given by:
U(t) = τx f
sin(ft) (6.17)
V (t) = τx f
(cos(ft) − 1) (6.18)
Initially the solution behaves as in the non rotating case, that is, it accelerates in the x-direction
with an acceleration given by τ x . But in the rotating case eqs. (6.13) and (6.14) also have the
stationary (time-independent) solution:
tel-00545911, version 1 - 13 Dec 2010
U = 0;
V = − τ x
f , (6.19)
which has no counter part in the non-rotating case. This solution is a force balance between
the Coriolis force and the wind stress. The depth averaged Ekman transport is at 90 o to the
right of the wind force as this is the only possibility for the Coriolis force to balance the wind
stress.
✻
wind stress
Coriolis force
❄
Ekman
✲
transport
Figure 6.3: Depth averaged Ekman transport
The solutions for the rotating case given in eqs. 6.17 and 6.18 are in fact a sum of the
stationary solution plus inertial oscillation. When friction is included the oscillations will be
damped and the transport will converge towards a (modified) stationary solution.
Exercise 38: What is the energetics of the Ekman transport?
Exercise 39: Does the Ekman transport depend on the viscosity?
It is the Ekman transport, and only the Ekman transport, that determines the influence of
the wind forcing on the deep ocean. For completeness we will discuss in the next subsection
the vertical structure of the Ekman dynamics.
6.2.2 The Ekman Spiral
We start this section with two instructive exercises.
Exercise 40: What happens when we include bottom (Rayleigh) friction in eqs. 6.13 and
6.14? The stationary solution is governed by:
−fV = −rU + τ x (6.20)
fU = −rV (6.21)
and we obtain the solution:
U =
V =
r
r 2 + f 2τ x (6.22)
−f
r 2 + f 2τ x. (6.23)