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

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38 CHAPTER 6. GYRE CIRCULATION
6.2 The Ekman Layer
tel-00545911, version 1 - 13 Dec 2010
Strictly speaking, this section belongs to chapter 10 because it deals with how wind forcing
penetrates to the deep ocean, but it just happens that we need to know Ekman theory to
continue our investigation of “Forced Dynamics.”
Ekman’s theory of the adjustment of a fluid in a rotating frame to an equilibrium when
subject to wind forcing, is probably the most cited and most misunderstood theory of ocean
dynamics. To elucidate this Ekman layer dynamics we will advance in small steps, emphasizing
the physical understanding of the process, without neglecting the mathematical derivation.
Suppose we have an infinitely deep layer of a homogeneous fluid subject to wind forcing
τ x , constant in time and space, at its surface that is acting in the x-direction. The flow is
independent of x,y as the forcing has no variations in these variables and as there are no
boundaries. But the flow depends on the vertical coordinate z. In this case the vertical velocity
w vanishes everywhere due to the divergence free condition, eq. (5.4). The Navier-Stokes
equations (5.1) – (5.4), in a rotating frame, then simplify to:
with the boundary conditions:
∂ t u(z,t) − fv(z,t) = ν∂ zz u(z,t) (6.9)
∂ t v(z,t) + fu(z,t) = ν∂ zz v(z,t) (6.10)
ν∂ z u(0) = τ x ; ∂ z v(0) = 0, (6.11)
lim ∂ zu(z) =
z→−∞
lim ∂ zv(z) = 0. (6.12)
z→−∞
The surface boundary condition (6.11) represents the vertical gradient of the horizontal velocity
due to wind stress, while we suppose no frictional forces at the (far away) bottom of the Ekman
layer.
6.2.1 Ekman Transport (one layer)
To further simplify the problem we consider the transport U(t) = ∫ 0
∫ 0
−H
−H
u(z,t)dz and V (t) =
u(z,t)dz of the whole fluid column. Please note, that these variables depend only on time
and we have neglected all vertical structure in the problem. This can be easily done in the
present problem as eqs. (6.9), (6.10) and the boundary conditions (6.11), (6.12) are linear.
Integrating the right hand side of eq. (6.9) we have ∫ 0
ν∂ −H zzu(z,t)dz = ν∂ z u(0) = τ x . When
we further neglect friction at the bottom of the fluid layer eq. (6.9), (6.10) and boundary
conditions (6.12) read:
∂ t U(t) − fV (t) = τ x (6.13)
∂ t V (t) + fU(t) = 0 (6.14)
We now like to consider the spin up of an Ekman transport initially at rest. In the nonrotating
case (f = 0) we have the solution:
U(t) = τ x t (6.15)
V (t) = 0, (6.16)
so the fluid constantly accelerates in the x-direction and no stationary state is reached!