(Should use spherical coords for Earth, but ideas are similar)

Transcription

1 1 Streamfunction and Vorticity Can decompose 2D vector field U into a potential Φ(φ, θ) and a streamfunction Ψ(φ, θ): Cartesian components: U = ẑ Ψ + Φ. (1) U = Ψ y + Φ x V = Ψ x + Φ y. (2a) (2b) (Should use spherical coords for Earth, but ideas are similar) 1

2 The divergence of U is given by and represents flow out of a given area: U = 2 Φ (3) A UdA = S U ˆnds (4) (line integral around perimeter S, n unit vector normal to perimeter). U = 0 Φ uniform. The relative vorticity, or curl of U, is given by top bot ζdz = ẑ ( U) = 2 Ψ (5) The curl represents the flow around the perimeter of an area: A ẑ ( U)dA = S U ŝds. (6) (ŝ is unit vector parallel to the perimeter). RHS of (6) is circulation. Given ζ and boundary conditions on Ψ, can solve (5) for Ψ. 2

3 In Cartesian coordinates, the divergence and vorticity are given by U = U x + V y = Φ xx + Φ yy ẑ ( U) = V x U y = Ψ xx + Ψ yy. (7a) (7b) In polar coordinates (r, λ), vorticity is given by ẑ ( U) = 1 r (rv ) r 1 r U λ (8) (Batchelor, 1967, Appendix 2). These expressions explain vorticity examples in earlier figure. 3

4 2 Potential Vorticity Budget Start with equations of motion for (u, v, w) in geostrophic layer between surface and bottom Ekman layers, with density ρ uniform (hence no vertical shear in (u, v): u t + uu x + vu y + wu z fv = (1/ρ)p x + A H 2 Hu v t + uv x + vv y + wv z + fu = (1/ρ)p y + A H 2 Hv u x + v y + w z = 0 (9a) (9b) (9c) where p is dynamic pressure, A H is a horizontal eddy viscosity, and 2 H = 2 x + 2 y. (10) 4

5 Uniform ρ & geostrophy u z = 0 and v z = 0. Curl of momentum equations (v x u y ) t + u(v x u y ) x + v(v x u y ) (11) +u x v x + v x v y u y u x v y u y + fu x + fv y + f y v (11) which can be written as = (1/ρ)(p yx p xy ) + A H 2 H(v x u y ). (11) ζ t + uζ x + vζ y + βv + (u x + v y )(f + ζ) = A H 2 Hζ. (12) Continuity (9c) and vector notation gives: ζ t + u (f + ζ) w z (f + ζ) = A H 2 Hζ. (13) 5

6 We added extra variable w, but we know (approx) value at top and bottom of layer. Ekman at top w T = w E = ẑ and Ekman + slope term at bottom (Pedlosky, 1987, eq. (4.5.39)). τ (14) ρf w B = 1 2 δ Eζ u D. (15) 6

7 Integrate (13) from D + δ E to δ E to obtain D[ζ t + u (f + ζ)] = (f + ζ)(w T w B ) + DA H 2 Hζ, (16) (D 2δ E D for δ E D). Substituting for w near the top and bottom, (16) becomes D[ζ t + u (f+ζ)] (f+ζ) u D = (f+ζ)w E 1 2 δ E(f+ζ)ζ+DA H 2 Hζ. or as a potential vorticity budget D f + ζ Dt D = f + ζ D with potential vorticity w E D f + ζ D q = f + ζ D (17) δ Eζ 2D + A H D 2 Hζ. (18) (19) 7

8 3 Scale Analysis of the Vorticity Equation Uniform-ρ simplification still leaves complicated nonlinear equation. Use scale analysis to see if we can expect some terms in the equation to be smaller than others. Not a proof... more of a suggestion. Eliminating small terms might make equation easier to solve. Assume D uniform, steady state. Use u f = βv, so that (18) becomes u ζ + βv = (f + ζ) w E D (f + ζ)δ Eζ 2D + A H 2 Hζ. (20) βv is rate at which the absolute vorticity is changing due to meridional flow. Assume this will be one of the largest terms. Must be at least one other term to balance βv (unless the equation simply implies that v = 0). Wind stress drives motion try w E. 8

9 u U and v U horizontal length scale over which u changes by a significant amount in either the x or y direction is L. f f 0 = f(θ 0 ) and β β 0 = β(θ 0 ), Compare relative magnitude of terms in f + ζ: ζ/f U fl Ro, (21) Ro 1 for large scale flow away from the equator, so relative vorticity planetary vorticity. 9

10 Compare advection to advection of planetary potential vorticity: u ζ βv with inertial length scale, U 2 /L 2 = U β 0 U β 0 L = δ I 2 L 2 (22) δ I U /β 0 (23) If δ I L, neglect the advection terms. May not be weak even if Ro 1. Rewrite (22) as (R = Earth s radius). U β 0 L = U 2 f 0 L f 0 β 0 L = Ro tan θ 0 (24) L/R 10

11 The relative strength of the bottom friction term is given by rζ βv ru /L β 0 U = r β 0 L = where we have defined a frictional parameter δ S (25) L r f 0 2 and a bottom friction length scale δ E D (26) δ S r/β 0 = 1 2 δ E D R tan θ 0 (27) (Stommel length scale after Stommel (1948)) Relative strength of horizontal friction term is given by A H 2 Hζ βv A HU/L 3 β 0 U 0 where the horizontal friction length scale is = A H β 0 L = δ M 3 L 3 (28) (Munk length scale after Munk (1950)). δ M (A H /β 0 ) 1/3, (29) 11

12 To summarize scaling results: advection terms: (δ I /L) 2, δ I U /β 0. bottom friction terms: δ S /L, δ S r/β 0. horizontal friction terms: (δ M /L) 3, δ M (A H /β 0 ) 1/3. When are these small? δ I, need U and L. βv fw E /D U = f 0 β 0 w E D. From definition of Ekman: U = τ /ρ β 0 DL. ρ 1000 kg/m 3, β = m 1 s 1, τ =.1 N/m 2, L = m, D 1000 m δ I m. WBC: speeds are (M/L W ) faster (gyre width / WBC width). For L W = 100 km and M km, δ I 100 km, so inertial terms are important. 12

13 Friction comparisons independent of U. Mixing coefficients unclear. Stommel: δ E = 30 m, D = 4000 m, R = 6400 km δ S = 24 km. Munk: β 0 = m 1 s 1. A mixing-length argument based on mesoscale eddies A H 10 6 m 2 /s δ M 400 km. Thus our scale analysis implies inertial and friction terms negligible for gyre-scale motion, but since we know from observations that the velocity scale is larger and the length scale is smaller near the western boundary than they are for the gyre as a whole, these terms are potentially important for the WBC dynamics. 13

14 4 Sverdrup Relation and the Need for a Western Boundary Current u ζ+βv = f w E D + ζ w E D (f + ζ)δ Eζ 2D + A H 2 Hζ. (30) Neglecting all small terms, get Sverdrup Relation. Can quickly derive two versions of Sverdrup from simplified equations of motion. Geostrophic flow only: ρfv G = p x ρfu G = p y. (31a) (31b) take the curl to get f(u Gx + v Gy ) + βv G = 0 (32) and combine with the continuity equation (9c) to form the linear, inviscid form of the vorticity equation, βv G = fw z. (33) Vertical integral, assume w = 0 at bottom and w = w E at top: βv G = fw E = ẑ τ, (34) ρf Did not assume ρ uniform. Did assume w = 0 at bottom. 14

15 Can include Ekman (using vertical viscosity): ρfv = p x + (ρν V u z ) z ρfu = p y + (ρν V v z ) z, (35a) (35b) (ν V is vertical eddy viscosity). Since τ = ρν V u z, (36) vertical integral of the momentum equation and continuity is ρfv = P x + τ ρfu = P y + σ, U x + V y = 0 (37a) (37b) (37c) where assume w T = w B = 0, τ = ˆxτ + ŷσ, and P = 0 p dz. (38) D Taking the curl of (37) and combining with (37c), get another Sverdrup Relation: ρβv = ẑ τ (39) 15

16 5 Gyres with Western Boundary Currents Sverdrup s Dilemma: Sverdrup formula may imply all fluid going north or all going south. Other dynamics needed for return flow. Drop nonlinear terms, assume D uniform, and try to solve : fv = 1 ρ P x + τ/ρ + A H 2 HU 1 2 fδ E(u B v B ) fu = 1 ρ P y + σ/ρ + A H 2 HV 1 2 fδ E(u B + v B ) (40a) (40b) (see Pedlosky, 1987, eq (4.3.30) for bottom friction). Take curl of momentum equations and use continuity (set u B = U/D): βv = 1 ρ (σ y τ x ) + A H 2 (V x U y ) 1 2 f(δ E/D)(V x U y ). (41) H U = 0 use streamfunction Ψ: βψ x f(δ E/D) 2 Ψ A H 4 Ψ = 1 ρ (σ y τ x ). (42) 16

17 Linear PDE in Ψ. Stommel (1948) finds exact solution for idealized geometry and A H = 0. More generally, use boundary layer approximation. 1) Find interior solution Ψ S (x, y) (Sverdrup) outside boundary layer. 2) Assume WBC width δl (δ 1 but value unknown). Friction important in this region only. 3) Set Ψ(0, y) = Ψ(L, 0) (no net transport). 4) In boundary layer, scale analysis gives Ψ y Ψ x δ and Ψ Sx Ψ x δ. (43) 5) Then in boundary layer, our vorticity equation becomes βψ x f(δ E/D)Ψ xx A H Ψ xxxx = 0. (44) We turned PDE in (x, y) into ODE in x. 17

18 Solve A H = 0 case. In boundary layer: which is solved by βψ x f(δ E/D)Ψ xx = 0. (45) Ψ = A(y)(1 e x/δ S ). (46) This satisfies B.C. Ψ(0, y) = 0. For x δ S, Ψ A(y), so take A(y) = Ψ S (0, y). (47) Solution also shows boundary layer must occur on west side. The solution is illustrated in earlier lecture. In WBC, get V = (Ψ S /δ S )e x/δ S, (48) which is max at x = 0, has opposite sign to V S, and is 1/δ S larger than V S. 18

19 Solve r = 0 case. In boundary layer: βψ x A H Ψ xxxx = 0. (49) Extra derivatives extra constant B(y) extra BC. General solution is Ψ = Ψ S (x, y) 1 e x/2δ M 3x cos + B(y)e x/2δ M 3x sin, 2δ M 2δ M (50) What is extra BC? No one knows, but here use no slip. Can rewrite solution as which is solved by Ψ = Ψ S (x, y)θ(x) and V = v S Θ + Ψ S Θ x. (51) Θ = 1 e x/2δ M cos 3x 2δ M sin 3x, (52) 2δ M 19

20 0.5 Munk Streamfunction (δ M /L =.02) y/m x/l 1 v(x) at y=0, western region 1 ψ(x) at y=0 v/v max ψ/ψ max x/l x/l Figure 1: (a) Streamfunction, (b) V (x, y) at y = 0, and (c) Ψ(x, y) at y = 0 for solution to (49) for ẑ τ/(βρ) = V 0 cos(πy/m) and δ M =.02L. The WBC similar to bottom-friction case. In both cases, WBC width given by δ (δ S for bottom friction and δ M for horizontal friction). Often in numerical experiments, A H is adjusted to give WBC width consistent with resolution. 20