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

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40 CHAPTER 6. GYRE CIRCULATION
We see that when including bottom friction the fluid motion is still deviated to the right (on
the northern hemisphere) with respect to the wind stress but the angle is smaller than the 90
of the frictionless case. So the friction induces fluid motion in the direction of the wind stress.
Exercise 41: Two layers: We now suppose that the Ekman layer can be decomposed into
two layers of thickness H 1 and H 2 . This “poor man’s vertical structure” does not correspond
to any real situation but helps us to understand the physics of the Ekman spiral treated in the
next subsection. The governing equations for the stationary solution are:
−fV 1 = r(U 2 /H 2 − U 1 /H 1 ) + τ x (6.24)
fU 1 = r(V 2 /H 2 − V 1 /H 1 ) (6.25)
−fV 2 = r(U 1 /H 1 − U 2 /H 2 ) (6.26)
fU 2 = r(V 1 /H 1 − V 2 /H 2 ) (6.27)
tel-00545911, version 1 - 13 Dec 2010
Where r times the velocity difference represents the linear friction between the two layers. You
can write the linear system (6.24) – (6.27) in the form,
AU = B, (6.28)
where U = (U 1 ,V 1 ,U 2 ,V 2 ). Verify that all solutions have: U 1 + U 2 = 0, and f(V 1 + V 2 ) = τ x ,
which is the Ekman transport already calculated above. You can use this to eliminate U 2 and
V 2 from the problem and simplify eq. (6.28) to:
with Ũ = (U 1 ,V 1 ). Find the solution and give an interpretation.
ÃŨ = ˜B (6.29)
What is the vertical structure of the Ekman transport? We can approximate the vertical
structure by including more and more layers in the vertical. The first layer being subject to
wind forcing, the Coriolis force and the friction induced by the second layer. Every other layer
is driven by the frictional force transmitted by is upper neighbour and feels the friction of its
lower neighbour. All layers are subject to the Coriolis force. Using the results from subsection
6.2.1 we estimate that every layer will move to the right of the movement of its upper neighbour,
at a smaller pace. Such motion will lead to a spiral motion in the vertical decaying with depth.
To render this qualitative arguments into a quantitative theory we go back to eqs. (6.9), (6.9)
and boundary conditions (6.11), (6.12). To simplify the problem we will only consider the time
independent solution of these equations neglecting the inertial oscillations. Equations (6.9) and
(6.10) can be combined to form a single equation of fourth order:
−f 2 u(z) = ν 2 ∂ zzzz u. (6.30)
If we suppose that the viscosity is independent z-component this equation is easily solved and
the solution satisfying the boundary conditions 6.11 6.12 is:
u(z) = V 0 exp(z/δ) cos(z/δ + π/4) (6.31)
v(z) = V 0 exp(z/δ) sin(z/δ + π/4) (6.32)
where δ = √ 2ν/|f| is the Ekman layer thickness and V 0 = τ x δ/(ν √ 2). The solution shows that
the current at the surface is deviated 45 o to the right with respect to the wind velocity (on the
northern hemisphere, to the left on the southern hemisphere).
Exercise 42: What is the energetic balance of the Ekman spiral?