GFD 2007 Boundary Layers: Sloping bottoms in a stratified, rotating fluid In oceanic coastal regions, e.g. on the shelf regions between the coast and

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1 GFD 2007 Boundary Layers: Sloping bottoms in a stratified, rotating fluid In oceanic coastal regions, e.g. on the shelf regions between the coast and the deep ocean, the bottom generally slopes and the fluid is stratified. We have already seen the way the thermal boundary layers on vertical walls can control the interior flow and how the Ekman layers on horizontal boundaries can do the same for rotating fluids. Sloping boundaries are a type of hybrid of these two. The study of the boundary layer on the sloping bottom boundary introduces some new and fascinating features to the analysis. We can only touch on some of the issues in these lectures.

2 References Chapman, D. and S. Lentz Adjustment of stratified flow over a sloping bottom. J. Phys. Ocean.,27, Garrett, C., P. MacCready, and P.B. Rhines 1993 Boundary mixing and arrested Ekman layers: Rotating stratified flow near a sloping boundary, Annu.Rev. Fluid Mech. 25, MacCready, P. and P.B. Rhines 1993 Slippery bottom boundary layers on a slope. J.Phys. Ocean. 23, Trowbridge, J.H. and S. Lentz Asymmetric behavior of an oceanic boundary layer above a sloping bottom. J. Phys. Ocean.,21,

3 A schematic of the bottom boundary layer: upwelling case N 2 U Ω z y V Flow in x direction U.Ekman flux up the slope from high to low pressure.

4 The along slope density gradient Δz θ Δy Δz = Δysinθ In boundary layer Δρ / Δz = Δρ / Δysinθ g ρ y = ρn2 sinθ, N 2 = g ρ ρ z

5 Thermal wind in boundary layer Fluid moving up the slope a distance will produce a density anomaly Δρ = 1 g N 2 sinθδy f H = U = g Δ ρ z ρ Δ y = N 2 sin θ and over a depth of the bottom boundary layer of the order fu N 2 sinθ f=2ωcosθ 2Ω it would be possible to adjust the speed of the current to zero without Ekman layers and their dissipation. Currents could flow long distances without decay.

6 Steady, laminar boundary layer on a sloping surface Following MacCready and Rhines 1993 Ω g z z y θ y v = v'cosθ + w'sinθ w = v'sinθ + w'cosθ N= ( gαδt v /D) 1/2 Write equations in slant coordinate frame. Far from the lower boundary T = Δ T v z '/ D

7 Equations of motion (1) z ' D + ϑ ( y, z ) where = ΔT v ( z cosθ + y sinθ ) / D + ϑ ( y, z ) ϑ T = ΔT v 0 for large z Search for solutions independent of y and are only functions of z. w identically zero. For large z, u v is independent of y U, a constant Nonlinear terms in momentum equation vanish identically.

8 Equations of motion (2) 2 Ω cosθ u = 1 %p y + Av zz 2 Ω cosθ v = Au zz 2 Ω sinθ u = 1 %p z + b cosθ + b sinθ A is momentum mixing coefficient, κ is the thermal diffusivity. %p is the pressure perturbation. vn 2 sinθ = κ b zz b = α gϑ f =2Ωcos θ α is the coefficient of thermal expansion. u,v,b are independent of y. 2 %p/ y z=0

9 Boundary conditions and boundary layer equation u=0,v=0,b z = N 2 cosθ Insulating condition on z =0 f 2 A v = Av zzzz + b zz sinθ If the fluid were homogeneous or if the bottom were flat we would recover the Ekman layer problem. Using the thermal equation to eliminate b in favor of v yields v zzzz + 4 q 4 v = 0, q 4 = 1 4 f 2 A 2 + N 2 A κ sin 2 θ

10 Solution for laminar boundary layer (1) v=ce qz cosqz+be qz sinqz q 4 = 1 4 If the bottom is flat the bl scale is the Ekman thickness. If the bottom is vertical, i.e. if θ=π/2 the scale is the buoyancy layer thickness v =0 at z =0 means C =0. From the thermal equation b z = B N 2 sinθ 2qκ And from e qz [ cosqz + sinqz] 2 Ω cosθ v = Au zz f u = 2 Aq 2 Be qz cos qz + U f 2 A + N2 2 Aκ sin2 θ

11 Solution for boundary layer (2) Insulating condition b z =-N 2 cosθ B= 2qκ cotθ While the no slip condition on u yields U = f Aq κ cotθ u=u 1 e qz cosqz Flow at infinity is not arbitrary. It is part of the solution!

12 Thermal balance Frictionally driven flow up the slope z 0 ( ) ψ(z)= v dz =κcotθ 1 e qz cosqz +sinqz The total is ψ ( ) = κ cotθ = Aq f U And this follows directly from the integral of the thermal equation vn 2 sinθ=κb zz b z = N 2 cosθ, z=0

13 The control of the interior This result, while at first glance non intuitive, is really just a manifestation of the control mechanisms we have already met in our discussion of the linear flow in the cylinder, although here in a more extreme form. Part of our unease, is related to the sense that we ought to be able to drive the system as we would like and establish some equilibrium velocity along the isobaths that will differ from (3.2.14) or that we should, at least initially specify a different far field along shore flow. We continue our presentation, following the work of MacCready, Rhines and Garrett by first considering the latter possibility.

14 The slow diffusion equation (1). Same equations, w =0, but with time dependence. v t + 2Ω cosθ u = 1 %p y + Av zz + b sinθ u t 2Ω cosθ v = Au zz 2Ω sinθu = 1 %p z + b cosθ b t + vn 2 sinθ = κ b zz Scaling: ( u,v)=u(u',v'), y = Ly', z = Dz', t = D2 κ t' %p = ful%p', b = ful D b'

15 Non dimensional equations δ =D/L E v 2 σ t + u = %p y + E 2 v zz + b sin θ δ, E 2 σ u t v = E 2 u zz tan θ δ u = %p z + b cosθ E 2 σ b t + sin θ δ Sv = E 2 σ b zz f = 2Ω cosθ E = 2A fd 2, S = N2 δ 2 f 2

16 Interior equations u I = p I y + b I sinθ δ b I cosθ = p I z δ u I tanθ Eliminating v between E 2 σ u t v = E 2 u zz yields E 2 σ b t + sin θ δ Sv = E 2 σ b zz Eliminating the pressure in the first two equations and noting that b is independent of y 1 σ t u I δ Ssinθ b I =1 2 2 z u 2 I δ σssinθ b I u I z = b I z sinθ δ

17 The slow diffusion equation (2) t u I z = σ S sin2 θ = N2 δ 2 μ diff = σ δ σ S sin 2 θ 2 δ 2 z S sin 2 θ f 2 sin2 θ S * 1 σ + S * 1 + S * ( μ diff ) dimensional = A 1 σ + S * 1 + S * u I z Effective diffusion coefficient dimensional if σ >1, the diffusion coefficient would be smaller than in the absence of stratification

18 Final form of slow diffusion eqn. u I t δ σ S sin 2 θ = σ δ S sin 2 θ 2 u I z 2 If u is independent of z and t at z--> inf. To obtain boundary conditions for s.d.e need to consider boundary layer at sloping bottom. Boundary layer coordinate ζ =z E 1/2 Label correction variables with e

19 Boundary layer equations for correction functions u e = p e y v sinθ e ζζ + b e, δ v e = 1 2 u e ζζ, These are the same steady equations we dealt with before and they yield δ tan θ e sinθ v e = δ = 1 E 1/2 1 2 σ S b e ζζ p e ζ + b e cosθ, 4 v e θ z σSsin2 δ 2 v e =0.

20 Boundary layer solution v e = Ae αζ cosαζ + Be αζ sinαζ, b e = σ S sinθ e αζ {( A B )sinαζ ( A + B )cosαζ} δ, u e = 1 α 2 Ae αζ sinαζ Be αζ cosαζ, α = 1 + σ S sin 2 θ δ 2 1/4

21 Matching conditions u I + u e = 0, v I + v e = 0, v I is O(E) B =0 b I z + E 1/2 b e ζ = S ε cosθ, A=0 ε = U / fl The frictional boundary layer vanishes to lowest order. u I must satisfy the no-slip bc at z =0

22 Diffusion solution u I = U 2 π ζ /(μ dif t ) 1/2 0 e ϕ2 dϕ Next order boundary layer solution still has v e =0--> A=0 Boundary layer contribution to buoyancy flux yields δα S B = 2σ S sinθ ε cosθ + δu sinθ πμ diff t

23 Long time solution in boundary layer As t goes to infinity b e = E 1/2 S εα cosθe αζ cosαζ which is the steady state solution (in n.d. form) already attained. Hence it is possible to consider arbitrary interior flows but, at least with the simple physics here, the boundary layer control eventually expunges the along isobath flow and yields an asymptotically weak frictional boundary layer. This, in one sense resolves the conundrum posed by the steady boundary layer solution in which the interior flow and the cross shelf flow depended only on the stratification and the vertical thermal diffusion coefficient. Nevertheless, the solution presented here eventually approaches that very constrained solution.

24 Richer Physics Stress driven flow τ δ τ/ρf N 2 z y θ Off shore Ekman flux in upper Ekman layer

25 Equations of motion variations in the wind stress with y will force an Ekman pumping at the base of the Ekman layer w e (y). Assume θ is small. fu = 1 p y + bsinθ + A v v zz + A H v yy, fv + fw tanθ = A v u zz + A H u yy, In slant frame At base of Ekman layer f tanθu = 1 p z + b cosθ v y + w z = 0, w = w e = 1 f τ y v N 2 sinθ + b y + w N2 cosθ + b z = κ V b zz +κ H b yy

26 Interior solution (1) Ignore friction in momentum equations. v is O(E)=A v /fd 2. Implies w is independent of z and hence equal w e everywhere in the interior. The thermal equation becomes.(we have assumed an interior solution for b independent of z) wn 2 = κ H b yy 1 f τ y N 2 = κ H b yy b I y = 1 f N 2 κ H τ (y) Assuming b I vanishes at large negative y where the stress vanishes

27 Interior solution (2) u I = N 2 κ H τ f 2 z + u o (y) From thermal wind eqn. Unknown barotropic contribution The total interior buoyancy diffusive flux perpendicular to the lower boundary is: I z = N 2 κ v cosθ + τ sinθ f o

28 Bottom boundary layer (again) v b zzzz + 4 v 4 b = 0, l l 4 = f N 2 sin 2 θ 2 4 A v v b =e z/l [ Acosz/l +Bsinz /l] f 2 κ v / A v u b = f A v l 2 2 e z/l [ Asinz/l Bcosz/l ] b b z = N2 2κ v δsinθe z/l A( cosz/l sinz/l )+B(cosz/l +sinz/l ), w b = l 2 e z/l A y ( cosz/l sinz/l )+B y (cosz/l +sinz/l )

29 Matching u o f A v l 2 2 B = 0 For u 1 τ sinθ + A= 0, For f y 1 τ f y +l 2 (A+B)=0, v For w redundant κ v N 2 cosθ+ N2 τsinθ N 2 o ρf l2 sinθ(a+b) For b z

30 Solution l 2 (A+ B) = τ f +κ v cotθ Boundary layer mass flux Off shore flux balancing on shore Ekman flux Transport in bottom boundary layer induced by stratification and diffusion as in earlier solution finally A = 1 f τ y sinθ, B = 1 f τ y sinθ + 2 l τ f + 2 l κ v cotθ u o = f A v l 2 2 B

31 Definitions vertical Ekman # horizontal Ekman# E v = 2 A v fd 2, σ v = A v κ v E H = 2 A H fl 2, σ H = A H κ H vertical and horizontal Prandlt numbers

32 Interior long shore velocity u I = N2 τ κ H f f z + f l 2 A v 2 term 1 2 τ l f 123 term 2 1 τ f y sinθ+2κ vcotθ Ratio of terms f A v l term 1 term 2 = N 2 f Dl f 2 κ H = N 2 D 2 f 2 L 2 fl 2 A H (κ H / A H ) = σ H S E H E v 1/2 = σ S H 1/2 E v l D E v E H If E V /E H = O(1) then if σ H S >> E v 1/2 Then the interior velocity very nearly satisfies the no slip condition on z=0 without the b.l. As before

33 The turbulent bottom boundary layer τ δ τ/ρf Coast N 2 In this previous example we arrange to have cold water riding up the shelf under warmer water (upwelling). z y θ This enhances the static stability and a laminar model is at least plausible.the situation is different in downwelling

34 Chapman Lentz Model Light water driven under denser water. Expect convective mixing and a thick turbulent boundary layer. See Trowbridge and Lentz (1991) and Chapman and Lentz (1997). The Chapman Lentz (CS ) model:boundary layer is well mixed and isopycnals are vertical z z=-h(y) δ Off shore frictional flow u x ρ y

35 CL model: equations of motion fv = p x / + τ x z / fu = p y / 0 = p z ρ g 0 = u x + v y + w z u ρ x + vρ y + w ρ z = B z At the bottom, z=-h b (y) B z =0 and Linear momentum eqn. Non linear thermal eqn. τ x is stress in fluid in x direction. B is the vertical turbulent density flux. τ x ( h)= ru b

36 The CL adjustment problem CL examine the evolution of a coastal current. It starts as a narrow current upstream and spreads laterally due to friction. The boundary layer thickens as the current flows downstream. The current outside the boundary layer is not sheared vertically. Using integral budgets for mass, momentum and buoyancy, eqns for the boundary layer thickness δ are found as well as the interior u.

37 Results of CL problem (1) The boundary layer evolves until finally the thermal wind brings u to zero and eliminates the bottom stress. The scaling for the bl thickness is, as before, H = fu N 2 sinθ

38 Results of CL problem (2) Dashed curves for N=0. Note the stratified current flows without further decay after u b goes to zero. Want a model to discuss this equilibrium state that does not require a 3-d numerical calculation

39 Stress driven turbulent bottom boundary layer: The CL model τ w (y) z y z=0 ρ b (y) ρ Ι (y,z) z=-h(y) z=-h+δ Note: interior density is fnc. of y and z

40 Equations of motion Motion is assumed two dimensional--- independent of x Boundary layer must carry off shore flow. fu = 1 p y, fv = 1 τ z, τ is turbulent stress in fluid ρ g = 1 v y + w z = 0. p z, ρ density anomaly v ρ y +w ρ z = y κ ρ H y + z κ ρ V z

41 Interior(1) Onshore flux in upper Ekman layer M E = τ w / f Ekman pumping velocity w = w e = 1 f τ w y Buoyancy flux normal to bottom vanishes and all perturbations go to zero for large y. The density is continuous at z= αy+δ Below the Ekman layer and above the bbl the turbulent stress in the fluid interior is zero. Thus v I = 0 w I = w e (y) = 1 ρf τ w y

42 Interior (2) ρ w I e z = y κ H I ρ I y + z κ V I ρ I z κ VI ρ I z = H z=0 let ρ I = + ρ I (z)+ %ρ I (y) 1 ρ I z = H 1 κ VI g N 2

43 Interior density equation N 2 w e g = y κ H I %ρ I y ρ I = z N 2 g τ w N 2 dy' +1 f gκ y H I u I z = g f %ρ I y = τw N 2 f 2 κ HI u I = z κ HI N 2 f 2 τ w 1 f p s y unknown

44 In the bottom boundary layer Mixing is so intense density is only a function of y but is continuous with interior. Thus ρ b (y)=ρ I (y,z= h(y)+δ) ρ b = (h δ ) N 2 g τ w N 2 dy' + 1 f gκ y H I Pressure is hydrostatic and continuous with interior: p b = (h δ) z+(h δ)/2 u b = N2 f y [ ]N 2 +z τw f y ( z+[h δ]/2)(h δ) +z τw N 2 gκ HI dy' gz+ p s N 2 1 f 2 κ HI f p s y

45 Integral of x momentum eqn in bl V b δ = τ(z = h) f V b = 1 δ h+δ h v b dz As in CL V b δ = τ( h) f = ru b ( h)/ f If we know V b we then know the surface pressure gradient for a given boundary layer thickness (still unknown)

46 Mass budget for boundary layer It is obvious that the off shore mass flux must balance the on shore Ekman flux. But it is illuminating to examine the detailed budget in preparation for the buoyancy budget. ds A n z t =-h+δ y y+δy z= -h

47 Mass balance z b y v bdz+w b (z t ) v b (z t ) z t z t y w b(z b )+v b (z b ) z b y =0 cancel ˆn = ˆk ĵ z t y Flux across upper dy 2 2 ( + dz t ) dy 1/2 or bdy of bbl V b δ y + w * = 0 w * = w b (z t ) v b (z t ) z t y

48 Mass continuity with interior Since v I =0 w * must match w e V b δ y + w e = 0, or V b δ y τ w / f y = 0. V b δ = τ w f

49 Buoyancy budget dyv b δ ρ b y =I bgˆn ds +dy y κ H b δ ρ b y Diffusive mass flux at upper bdy of bbl V b δ ρ b y = κ ρ b V b z κ ρ b H b y z t y + z=z t y δκ ρ b H b y Continuity of diffusive flux at top of bbl yields first term on rhs in terms of interior variables.

50 Final budget for buoyancy in bbl V b δ ρ b y =κ ρ I V I z κ ρ I H I y After some algebra z t y + y δ κ ρ b H b y y κ H b δ ( y h δ)+κ H b δ τw fκ HI =κ V I + τw f 2 /κ HI Eqn. for δ in terms of which 1 f ( ) p s y = N2 f δ h δ + h τ w N 2 y f 2 κ HI + τ w r

51 velocity in boundary layer, interior velocity u b = N2 f ( ) ( z +h) h δ y + τw fκ HI + τw r u I = N2 f τ w (z +h) fκ HI + N 2 f δ ( y h δ τw )+ r Continuous at z=-h+δ Note that in the absence of stress the bottom velocity in the bbl is zero as in MacCready and Rhines

52 ODE for δ (1) Take bottom of form h=-αy. Scale lengths with L and thickness of bbl with αl. Consider a stress of form. τ w = τ o e a ( y / L ) d dy δ dδ dy (1+ Fe ay )δ = Σ V F 2 Σ H e 2ay F = τ o α fκ H I, Σ V = κ V I κ H b α 2, Σ H = κ H I κ H b

53 ODE for δ (2) δ dδ dy (1+ Fe ay )δ = Σ V (y y o ) F2 2a e 2ay e 2ay o Σ H +C, C = δ dδ dy (1+ Fe ay o ) y=yo y o is starting point for integration (model not valid in apex of wedge) g ρ C = δ b αn 2 y y=y o < 0

54 Results The boundary layer thickness with respect to the sloping bottom for a =1, Σ v = 0.1, Σ H = A starting value of δ of half the depth at y = y o = 0.01 is chosen and C is F =1 has been used.

55 Results (2) For larger diffusion coefficients in the interior Σ V =0.2 Σ H = 0.1 δ goes to zero well beyond the region of wind stress. The interior velocity then satisfies the zero velocity condition

56 When δ goes to zero u I = N2 f τ w (z +h) fκ HI + N 2 f δ ( y h δ τw )+ r Ratio of last term to first term then is: N 2 H 2 f 2 L 2 fl 2 κ HI So again, when r Hf σs E 1/2 v = σs 1/2 E H E v E v E H σ S E v 1/2 >>1 interior velocity satisfies bc.

Stress Driven Flow with a Sloping Bottom

Stress Driven Flow with a Sloping Bottom Lecture 6 by Frédéric Laliberté In the preceding chapter, we investigated an isolated Ekman boundary layer for a sloping bottom in a stratified, rotating fluid.