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Coriolis

CURRENTS WITH FRICTION. y. y. z. y. Pressure Gradient. Coriolis. x. x. x. x. Pressure Gradient. Pressure Gradient. Pressure Gradient. Coriolis. Coriolis. Coriolis. Nansen’s qualitative argument on effects of friction. ?. EKMAN SOLUTION (1905). Assumptions: homogeneous fluid

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Coriolis

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  1. CURRENTS WITH FRICTION y y z y Pressure Gradient Coriolis x x x x Pressure Gradient Pressure Gradient Pressure Gradient Coriolis Coriolis Coriolis Nansen’s qualitative argument on effects of friction ?

  2. EKMAN SOLUTION (1905) Assumptions: homogeneous fluid no horizontal pressure gradients infinitely deep and wide ocean no horizontal friction constant eddy viscosity steady wind northward

  3. EKMAN DEPTH or depth of frictional influence Wind-induced current speed Boundary conditions:

  4. DE V0 is 45° to the right of the wind (in the northern hemisphere) V0 decreases exponentially with depth as it turns clockwise (NH) At depth z = -DE the flow speed falls to e-π = 0.04 that at the surface and in opposite direction

  5. Alternatively, Von Kármán constant = 0.4 Charnock constant = 0.0185 sea surface roughness

  6. From observations outside 10º by Ekman:

  7. and Combining Indicates that surface currents are ~ 1% of the wind speed at the poles ~2.5% of the wind speed at 450 ~11% of the wind speed at 200 Empirically, it is seen that V0 / W oscillates between 1 and 5%

  8. Ekman Transport

  9. Rearranging: Ekman Equations Integrating over two Ekman layers, from depth z = 2DE to z = 0: Ekman Transport

  10. y y x x Ekman Transport m2/s Ekman transport is inversely proportional to f Water is replaced from the side – but what happens at the coast?

  11. Equatorward winds on ocean eastern boundaries Poleward wind on ocean western boundaries Poleward winds on ocean eastern boundaries Equatorward wind on ocean western boundaries

  12. Consequence of Upwelling

  13. Bottom Friction and Shallow Water Effects Assumptions: Bottom at z = 0 u, v = 0 at z = 0 (no flow at the bottom) u = ug, v= 0 at the top of the bottom Ekman layer (z = DB) ug is geostrophic flow

  14. PERSPECTIVE VIEW ug East Distance from bottom (m) DB = 15 m North 45° PLAN VIEW N BOTTOM EKMAN LAYER ug

  15. DE DB Flow at interior? Flow at bottom? z -x Overlap of bottom and surface Ekman layers Importance of shelf break depth Problems with Ekman theory – constant Az, constant wind, linear flow, steady state, infinite ocean, no pressure gradients

  16. t x y x SVERDRUP SOLUTION Assumed gradients in the wind field -- in contrast to Ekman’s spatially uniform wind Ekman Transport Convergence Divergence Differentiating Ekman equations with respect to y and x, to look at gradients in wind field:

  17. t x 0 Sverdrup’s Equation y Convergence Divergence x Adding the two yields:

  18. DE Qx DB Sverdrup’s Equation z -x

  19. 0 using Meridional transport of water given by the curl of the wind How about zonal (in ‘x’) transport?

  20. Trades 20° 10° Doldrums EQUATOR y Trades x Integrating from Eastern Boundary (x=0) to the west ‘-x’

  21. Trades NORTH EQUATORIAL CURRENT 20° Divergence NORTH EQUATORIAL COUNTERCURRENT 10° Doldrums Convergence EQUATOR y Divergence SOUTH EQUATORIAL CURRENT Trades x + - +

  22. NORTH EQUATORIAL CURRENT NORTH EQUATORIAL COUNTERCURRENT SOUTH EQUATORIAL CURRENT Streamlines of mass transport from mean wind stress (Reid, 1948)

  23. Conservation of mass is forced by including north-south currents confined to a thin, horizontal boundary layer. From Tomczak and Godfrey (1994).

  24. SVERDRUP SOLUTION Limited to the east side of the oceans because Qx grows with x. Neglects friction which would eventually balance the wind-driven flow Solutions may be used for describing the global system of surface currents. Conservation of mass is forced at the western boundaries by including north-south currents confined to a thin, horizontal boundary layer Only one boundary condition can be satisfied, no flow through the eastern boundary. More complete descriptions of the flow require more complete equations. Solutions give no information on the vertical distribution of the current. What happens on the western part of the oceans?

  25. STOMMEL SOLUTION (1948) using a streamfunction definition: Solution with: Currents are fast and narrow on the western part of ocean basins slow and broad on the eastern part of basins

  26. STOMMEL SOLUTION (1948)

  27. Addition of linearized bottom friction allows a closed circulation in the basin 45°N f = const f = 0 Westerlies Easterlies streamlines 15°N sea surface height Change of f with latitude is the main responsible for western intensification of ocean currents

  28. u u v v - + v v N u u E Western Intensification can also be understood with vorticity arguments Vorticity – tendency for portions of fluid to rotate We will consider: RELATIVE, PLANETARY, ABSOLUTE, AND POTENTIAL

  29. u u v v - + v v N, y u u E, x Relative vorticity: Planetary vorticity: Equals f. A stationary object on the earth has planetary vorticity that varies with latitude

  30. Absolute vorticity: Planetary plus relative vorticities Changes in absolute vorticity are useful to help understand tendencies for fluids to rotate To describe changes in absolute vorticity, take: Changes of absolute vorticity in time are related to divergences

  31. Plan View Side View Convergence = Gain of Absolute Vorticity = Column Stretching

  32. _ Plan View Side View Divergence = Loss of Absolute Vorticity = Column Squashing

  33. Combining with: CONSERVATION OF POTENTIAL VORTICITY Now consider a layer of thickness D, whose equation of continuity is: Changes of layer thickness are given by divergences/convergences Convergences = increase in D Divergences = decrease in D

  34. 45°N Westerlies - Easterlies 15°N D constant and f changing Consider: D changing and f constant Use concept of Potential Vorticity Conservation to explain Western Intensification

  35. MUNK’S SOLUTION (1950) Integrating, using a streamfunction definition: Sverdrup’s solution and cross-differentiating: biharmonic operator: friction Extended domain from Stommel’s Added horizontal friction More realistic wind

  36. MUNK’S SOLUTION (1950) subtropical gyre NEC NECC SEC (from Munk, 1950) 4th order d.e. with b.c.:

  37. L = 6000 km J = 3000 km Beta = 2x10-11 m-1 s-1 r = 2x10-6 s-1 (from George Veronis) /J /J

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