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Lesson 11 Frictional Flows

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Lesson 11 Frictional Flows Lesson 11 Frictional Flows Frictional Flows Large scale flow in the atmosphere and Ekman ocean is close to geostrophic and thermal – Layer wind balance, in boundary layers we – Solution observe departures from geostrophy – Spiral due to the frictional terms – Wind Driven Circulation Winds exert a stress on the ocean’s • Ekman Transport surface • Ekman Pumping – Some similarity between the pattern of surface currents and surface winds Frictional Flows: Qualitative Solution Boundary Layer Theory (Nansen, 1898 - sea ice observations) Ekman Layer: Wind accelerates top layers of the water Once in motion, the water (sea ice) moves to the right (NH) due to Coriolis The top layer of water will ‘drag’ the next layer along (due to?) – Momentum flux – Once it is accelerated it will also turn to the right Expect the magnitude to decrease with depth as less momentum is transferred 1 Frictional Flows: Ekman’s Frictional Flows Solution Full turbulent EOM, where: u ⇒ u Scaling balance: – Coriolis, PGF, vertical friction ∂u ∂u ∂u ∂u + u + v + w − fv = Situation ∂ ∂ ∂ ∂ t x y z – Near boundary (H is small) ∂ ∂ ∂ ∂ ∂ ∂ ∂ − 1 P + u + u + u – Steady state AH AH Az ρ ∂x ∂x ∂x ∂y ∂y ∂z ∂z – Advection small (R o< 1) • Scale?? – Horizontal friction small (Ek< 1) Scale as ocean surface boundary layer – Vertical friction large (Ek~ 1) Ekman’s Solution Ekman’s Solution Ekman (1902) solved for the details of Ekman Boundary Layer Equations the flow observed by Nansen – Can be solved to find the currents in the Further Assumptions ocean driven by a surface wind – No significant pressure gradients – Notice: • The pressure gradient term has been removed –Az= constant • No horizontal dependence – uE = uE(z), vE = vE(z) ∂ 2 Result: uE fv + A = 0 2 ∂ 2 E z ∂ 2 ∂ u v z + E = − fu + A E = 0 fv E Az 2 0 E z 2 ∂ 2v ∂z ∂z − fu + A E = 0 E z ∂z2 2 Solution to Ekman BL Equations Solution to Ekman BL Equations Method 1 Define BCs – Solve for uE and vE separately – Align the horizontal axes with the surface Method 2 current – Use complex notation: Q ≡ u + iv u (z = 0) = (u ,0) E E E E0 – And = = = d 2Q if uE vE const at z 0 − = Apply BCs 2 Q 0 – And dz Az f f z i z u → u and v → v as z → −∞ 2A 2A E g E g = z z Q(z) Q0 e e Solution to Ekman BL Equations Ekman Spiral Ekman Depth (D E) Result – The depth of frictional influence – e-π: magnitude decreases by ~ 0.04 – The depth over which these currents occur – e-iπ: direction is -180 ° from the surface • This is the Ekman Depth Define D E such that: Observed 2A = π z – The surface current is 45 ° to the right of DE f the surface wind – With depth the speed will decrease , the Q(−D ) = Q e−π e−iπ direction will continue to deviate from the E 0 wind - Ekman Spiral 3 Ekman Spiral Ekman Spiral Why is this not observed in nature? – It is over long time averages Not really at steady state Waves - cannot make accurate current measurements Az not constant and not accurately represented Horizontal boundaries lead to horizontal variations (neglected) Ekman Solution Ekman Solution We need to include another BC at the surface The solution does not tell us about the – The stress at the air-ocean boundary is continuous relationship between the surface current • The wind stress is equal to the stress in the water – If we calculate the stress in the water, we can get the and the surface wind wind stress and the direction of the wind Use eddy viscosity theory Let’s relate the Ekman solution to the – Result: an equation for the surface Ekman current vs wind... the wind 1( + i) π Q = Τ 0 ρ 0 i 0 fD E 4 Ekman Transport Ekman Transport • Mass Transports (M E) • Explain large scale ocean currents • Not dependant on Az • 90 degress to the right (NH) of the surface stress τ M = 0y Ex f τ M = − 0x Ey f Ekman Transport Ekman Transport 5 Ekman Pumping Lesson 11 Part 2 The vertical response to Ekman transport Sverdrup Circulation Rotates the wind from geostrophic flow – Understand towards low pressure • Theory • Application Physical tendency to push mass out, or • Results/Solutions pull it in, at the bottom of the Ekman layer Streamfunctions Background Background What drives the ocean currents? – Our first choice may be the wind… Sverdrup (1947) helps answer this – He showed that the circulation in the upper kilometer (or so) of the ocean is directly related to the curl of the wind stress – He extended the concept of vertically integrated solutions to total flow (i.e., Ekman and Geostrophic components) 6 Background Ekman vs Sverdrup Assumptions Ekman – Steady state – Ignored the pressure gradient terms –R << 1 o Not applicable near western – Assumed homogeneity and level isobars –EH << 1 boundary currents • Non-geostrophic –E ~ 1 Z Sverdrup – Lateral friction and molecular viscosity are small – Ignores horizontal friction terms – Baroclinic flow – Surface turbulence described by eddy viscosity – Adds wind stress away from a coastal boundary – Level of no motion – the level at which the wind – Retains the pressure gradient terms driven circulation vanishes – Isn’t concerned with the velocity as a function of – Mean flow of deep ocean must be weak depth • Very small frictional stress due to bottom Sverdrup wants to solve for depth integrated • Very small vertical motion mass transport by the wind Sverdrup Circulation Sverdrup Circulation ∂ ∂τ 1 P 1 x ∂ ∂ Equations: − fv = − + M x My ρ ∂ ρ ∂ + = 0 0 x 0 z ∂x ∂y 1 ∂P 1 ∂τ fu = − + y – Cross differentiate EOM ρ ∂ ρ ∂ 0 y 0 z ∂u ∂v ∂w + + = 0 Result is Sverdrup Relation (Equation) ∂x ∂y ∂z β = ⋅∇×τ – No closure model required M y zˆ 0 Process Important and fundamental result! – Integrate the equations vertically – The northward mass transport of wind driven – Assume no motion at z = -H (H>>D E) currents is equal to the curl of the wind stress 7 Sverdrup Relation Sverdrup Relation Magnitude of Mass Transport Solve for Mx: (x<0 due to BCs) φ ∂ = − x tan ⋅∇×τ + ⋅∇×τ M x (x, y) zˆ 0 (zˆ 0 ) τ β R ∂y x 0 0 1 ≈ )( ≡ − )( dx M y x ∫ β ∆ x y = Average )( between x and 0 ∂ 2τ x x ≈ 0 curvature – Recall: M x (x, y) 2 β ∂y ∂f 2Ωcos φ β ≡ = ∂τ 1 x ∂y R M (x, y) = − 0 curl y β ∂y Sverdrup Relation Streamfunction ( ψ) Streamline – Lines that are everywhere parallel to the instantaneous wind velocity (steady flow) Streamfunction – Defines 2D flow – Satisfies the 2D continuity equation u = ∇ ×ψ where ψ = ,0,0( ψ ) if u = (u, v )0, ∂ψ ∂ψ or u = , v = − ∂y ∂x 8 Streamfunction ( ψ) Sverdrup Results ∂ψ M = x ∂y ∂ψ M = − y ∂x Can now say Sverdrup equation is: ∂ψ − β = zˆ ⋅∇×τ ∂x 0 Streamlines of mass transport in the eastern Pacific calculated from Sverdrup’s theory using mean annual wind stress. From Reid (1948). Sverdrup Results Comments on Sverdrup Mass transport in the Recall the assumptions: EPAC calculated from Sverdrup’s theory using – Geostrophic internal flow mean wind stress (solid lines) and pressure calculated from – Uniform level of no motion hydrographic data from ships (dots). – Ekman’s transport is correct Transport is in tons per second through a section one meter wide extending Solutions are limited to the east side of basins from the sea surface to a depth of one kilometer. – This stems from neglecting friction Note the difference in scale between My and Mx . No information on the vertical distribution of the From Reid (1948). current Only one BC can be satisfied – no flow through the eastern boundary – Need more complete equations for more complete description of the flow 9 Lesson 11 Part 3 Review Ekman Stommel – Shallow layer (Ekman layer) – Understand his solution – Flow was to the right of the wind – Western intensification – Divergence (±) and Pumping Monk Sverdrup – Understand his solution – Deep layer (not to bottom, only to z= -H) – Western intensification and – General circulation - counter currents countercurrents – Fails to represent western boundary currents (WBC) Stommel’s Solution Stommel’s Solution Stommel (1948) adds a simple bottom ∂ψ stress , which is proportional to velocity β = zˆ ⋅ ∇ × τ − k∇ 2ψ – Incorporate frictional terms - so we get the ∂x 0 WBC – The result - basic Sverdup with ad hoc friction ‘Interior’ balance Basic Sverdrup ∂ψ Boundary current β = zˆ ⋅ ∇ × τ − k∇2ψ Balance ∂x 0 10 Stommel’s Solution Stommel: Summary It’s second order – i.e. can solve two BC • No normal flow on east and west boundaries Uses ad hoc friction β causes western intensification – Since β does not change sign in the southern β=0 β=constant hemisphere we get western intensification there too Streamlines for simplified wind-driven circulation Stommel’s Quote Monk’s Solution Extension of Stommel and similar in logic “The artificial nature of this theoretical to Sverdrup model should be emphasized, Allows for lateral friction particularly the form of the dissipative Assumes term (the ad hoc friction). The writer –Ro << 1 thinks this work as suggestive, certainly –EH ~ 1 not conclusive. The many features of –EZ ~ 1 actual ocean structure omitted in this – Closure (for lateral friction) model should be evident.” – No motion at z = -H 11 Monk’s Solution Monk: Summary Result: 4th order - can satisfy 4 BCs – No normal flow east and west ∂ψ −β = ⋅ ∇×τ − ∇4ψ – No slip east and west zˆ 0 AH ∂x Good qualitative agreement Friction Gulf Stream transport too small Basic Sverdrup 12.
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