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Why are there upwellings on the northern shelf of Taiwan under northeasterly winds? *

Why are there upwellings on the northern shelf of Taiwan under northeasterly winds? *. L.-Y. Oey lyo@princeton.edu Outline : Introduction – upwelling, effects of strong currents A simple model of upwelling at the western edge of Kuroshio in East China Sea Observational evidences

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Why are there upwellings on the northern shelf of Taiwan under northeasterly winds? *

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  1. Why are there upwellings on the northern shelf of Taiwan under northeasterly winds?* L.-Y. Oey lyo@princeton.edu Outline: Introduction – upwelling, effects of strong currents A simple model of upwelling at the western edge of Kuroshio in East China Sea Observational evidences Generalizations Conclusions *Chang, Oey, Wu & Lu, 2009 – J. Phys. Oceanogr. Submitted.

  2. Wind & Ocean Currents Wind mixing surface layer: 10~100m Y:v/t + fu = y/z fu = y/z (..)dz  fU = oy

  3. U = oy /f oy U Ekman Layer 10~100m

  4. Non-uniform Wind, constant f oy < 0 oy > 0 X Wind from north Wind from south U < 0 U= oy /f > 0 Ekman Layer Upwelling

  5. Constant wind, non-uniform f Trade Wind oy < 0 f > 0 f < 0 x U= oy /f < 0 U= oy /f > 0

  6. Constant wind & f, but with a coast:Upwelling oy Coast Wind U Warm Cool

  7. Coastal Downwelling oy Coast Wind U Warm Cool

  8. Why are there upwellings on the northern shelf of Taiwan under northeasterly winds?

  9. In the Presence of a Spatially Non-uniform Ocean Current vo(x), the Ekman transport U = oy /f ~ Period, where f = fo+ o; o = vo/x NE Monsoon oy<0 Jupiter = 10 hrs/rotation Wind China Mars = 24.6 hrs/rotation z y o>0 o<0 o=0 vo x Warm Cool Kuroshio

  10. A Simple Model +QsNT For oy < 0 Warming Cooling A oy T/t = A eit Consider Oscillatory Wind: T/t is in phase with wind if A > 0; T/t is 1800 out of phase with wind if A < 0.

  11. Wind T/t ~ oy(t) [s/xT/f ] T/t T/t ~ oy(t) [To/x] Wind T/t

  12. Observational evidences East China Sea Kuroshio LongTung Study Region

  13. Ro Rossby number (Ro) Ro↓, Ue↑ Ro↑, Ue↓ Southward wind: oy<0

  14. Effects of Kuroshio: Long-Tung SST & wind stress Southward wind: cooling T/t ~ oy(t)/(fE)[To/x +s/xT/f ]

  15. Annual (~30yrs) Mean Wind 2001 wind

  16. Summary Current shears near strong ocean jets play a significant role in controlling the vertical motions in the ocean.

  17. Generalizations 1. 2. (fh1/d)/t + voxxu = (f+vox)wE/d (1) In SS, using wE = oy/(f +vox)2.voxx, (2) we have, u = - oy/[d(f +vox)]. (3)

  18. Idealized Calculations

  19. A “bulge” is defined as a near-surface buoyant fluid that moves across shelf as a result of Ekman transport by downwelling wind and its interaction with ocean’s vorticity across, say, a front. It is “2d-like” when |/y| << |/x| where y = alongshore and x = cross-shore. Idealized Calc.: oy < 0 Day 3 Day 1 Day 2 Day 4 Caption: V-contours (black:0.2, 0.4, ..; grey: -0.05; white:-0.1,-0.15,-0.2,..) m/s, on color T (oC) from day 1 through 4after an up-front wind is applied.

  20. A Nonlinear Model Assume /H << 1, |/ y| << |/ x|; Within the bulge, temperature T = Tb(x,z,t) is weakly stratified: Tb = T- + (z + ), for 0z(x,t); g/N2 << 1 (A.1) T- is related to the temperature Ti(z)beneath the bulge: T- = Ti(), and Ti(z) = Tdeep + [N2/(g)](z + zdeep), for (x,t) z zdeep (A.2a,b)

  21. /t = n[-1/2/x+ 21/23/x3]; where  = 2 • = F(),  = -1(x + ct), c = nconstant > 0 Figure A2. The bulge solution according to equation (A.21) for C1 = 1 and various indicated values of c. Both the ordinate and abscissa are non-dimensionalized: ordinate is the bulge thickness (“”) below the free surface while the abscissa is  = -1(xcnt); see text. For each c, the dotted line indicates where the solution terminates at the head of the bulge where a front is formed.

  22. PV at day 80

  23. Conclusions • Down-front wind leads to slantwise instability with intense mixing ~300m • Up-front wind leads to propagating “bulge” solution that produces deep recirculation cells

  24. Thank you.

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