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A Ekman transport

  1. Jan 19, 2017 #1

    ATY

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    Hey guys, I got this problem:
    We had the derivation of the ekman transport today in class. And what I wondered about is this:
    Usually the equation for the ekman transport looks similiar to this (depends on the author)
    [tex] u = V_0 e^{az} cos(\frac{\pi}{4}+az) [/tex]
    [tex] v = V_0 e^{az} sin(\frac{\pi}{4}+az) [/tex]
    [tex] a= \sqrt{\frac{f}{2 A_z}} [/tex]

    This is fine for the northern part of the earth, but what happens when I go to the southern hemisphere ? the coriolis parameter f should become negative (since f is [tex] f = 2 \Omega sin(\phi) [/tex])
    So I can not use the equations above. I am really confused because none of the derivations that I found talked about this. Or am I missing a really obvious point ?
    best wishes
    ATY
     
  2. jcsd
  3. Jan 19, 2017 #2

    Andy Resnick

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    The derivation I have is in Pozrikidis ("Introduction to Theoretical and Computational Fluid Mechanics"), and there is no ambiguity- I wonder if you have a choice-of-coordinates sign change hidden somewhere in your derivation. The Coriolus force can written as -2Ω×u, where Ω is the rotation rate (Ωez) of the fluid and u=(ux(z),uy(z),0) is the "horizontal" velocity. In the end, the velocity components of u are found to be:

    ux+iuy=(Ux+iUy)exp(-(1+i) |z|/δ)

    where Ux and Uy are the horizontal velocity components on the fluid surface (taken to be z = 0) and δ is the Ekman layer thickness.

    Does this help?
     
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