1. CEE 262A H YDRODYNAMICS Lecture 17 Geostrophy

2. Often in analyzing large-scale flows we find that the momentum equations simplify greatly, i.e. we can neglect lots of terms.. Consider the first of these e.g. Ocean basin: Coriolis force & large-scale flows: Geostrophy

3. Or, for the unsteady part: Next we turn to the second Coriolis term: Finally, we can deal with the viscous term: Inertial period: f -1 =1/(2  sin  Equator:  f -1 =∞, Poles: f -1 =12 hours ~ ~

4. The x 2 momentum equation behaves the same. What about the vertical momentum balance? Here we only need look at hydrostatic pressures, but we must be careful! First we note that In general, this could be associated with free surfaces or with density variations or both. Let us consider the latter case, such that in the interior of the fluid we have Already small compared to Coriolis ~ ~

5. We have geostrophic flow Thus, if: Time scale is longer than f -1 Rossby number is small We are not within an Ekman layer (i.e. near a boundary) Velocity (v) Coriolis force (  f v) High PLow P Pressure force -∂p/∂x

6. These can be used to explain weather charts & to prove a useful (weird) theorem: The Theorem Take This gives If at some level in fluid it is 0 everywhere Taylor Proudman Theorem

7. Strongly rotating (Geostrophic) flow Side view Bump on bottom Plan view (any level) 2D streamlines Taylor column: Bottom boundary Top of Ekman layer

8. Because the flow is 2D, we can use a stream function: A second nice use of geostrophy: The weather map

9. Imagine we look at a pressure field that looks like this ( a “high”) P2P2 P4P4 What does flow look like? P3P3 P1P1 P 1 >P 2 >P 3 >P 4 H

10. P2P2 P3P3 x y Since ∂p/∂x 1 < 0, u 2 <0 Consider situation on east side of high u2u2 P2P2 P3P3 u1u1 Whereas on north side of high Since ∂p/∂x 2 0 High Result (northern hemisphere): Flow is clockwise around a high (counterclockwise around low) High = anticyclone Low = cyclone

12. Suppose now we consider geostrophic flows with a free surface We can use sea-surface topography to infer flows

14. Consider a homogeneous ocean, e.g. Florida Strait between Florida and the Bahama. Since flow is only to north (+x 2 ), we only need zonal (x 1 ) momentum equation. We include now our hydrostatic pressure gradient due to a sloping free surface u 2 = 1 m/s (to North) (x1)(x1) 80 km FloridaBahamas Section at 25º North (f =6.1 x 10 -5 s -1 ) Rearrange to solve for slope Thus, the change in  across the Florida strait = 6 x 10 -6 x 8 x 10 4 m = 0.5 m x1x1

15. Now, let’s suppose that the Atlantic in the Florida strait consists of two layers (a better model): (x1)(x1) 80 km Florida Bahamas x1x1 h 1 (x) Our geostrophic momentum equations in the 2 layers read: u 2 =1.5 m/s  1 =1024 u 2 =0.5 m/s  2 =1025.5 h1h1 How do we compute the hydrostatic pressure gradient in the lower layer?

16. We return to hydrostatics and evaluate pressure in lower layer: We now compute ∂p 2 /∂x 1 So, the momentum equations for the two layers read: Thus the pressure gradient comes from the free surface and from the interface

17. With the Boussinesq approximation, we find that So, if we take the difference of these two expressions, we can compute the interface slope - a formula known as the Margules relation - in terms of the shear: Which translates into  h 1 = -0.004 x 80,000 = -320m. Thus interface is 300m lower on Bahamas side than on the Florida side. Note that this is -  /(  ) -

18. The Gulf Stream between Florida and the Bahamas From: “Ocean Circulation” - Open University 20 40 60 (km)