GFD 2007 Boundary Layers:Stratified Fluids

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Lynne Boone
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1 GFD 2007 Boundary Layers:Stratified Fluids The presence of density stratification introduces new and very interesting elements to the boundary layer picture and the nature of the control of the interior flow by the boundary layers. As we discussed previously, the vertical velocity pumped out of (or into) the Ekman layers on horizontal surfaces effected a control on the evolution of the interior, for example, leading to spin-down. One can think of that control as proceeding along the vortex lines of the overall rotation of the fluid as these are the lines along which information is propagated (by inertial waves) for a homogeneous, uniformly rotating fluid. When stratification is added information also propagates laterally, by internal gravity waves (assuming the stratification is gravitationally stable) and frequently this direction is perpendicular to the rotation direction so one can anticipate a kind of competition between rotation and stratification in determining the nature of the boundary layer influence on the bulk of the fluid. This will be the emphasis of our present discussion.

2 The cylinder problem. (tip of the hat to the MJQ) g r o L Ω L Fluid heated to establish a stable stratification when there is no relative motion in the cylinder. T =!T v z / L Assumes that rotation is small and doesn t bend the isotherms, i.e.! 2 L / g << 1

3 Scaling! u * = U! u,!!! x * = L! x,!!!!t * =!T v (z * / L) +!T h T (x, y, z) ( ) " * = " o %& 1 # $!T v z +!T h T ' ( p * = " o g$!t v z * 2 / 2L + " o ful! p(x, y,z) f=2ω choose scaling velocity U =!g"t h f (thermal wind) Motion is incompressible and Boussinesq

4 Equations of motion (steady)!! ui"! u + ˆk #! u = $"p + T ˆk + E 2 "2! u "i! u = 0!! ui"t + ws = E 2% "2 T! = U fl 2",!!E = fl,!!!!# = " 2 $,!!S = %g&t v f 2 L ' N 2 f 2! S = "T h We will generally assume "T this ratio is small. v

5 Boundary conditions No slip, no normal flow One or more boundaries may be moving. Cylinder walls either insulating or at a fixed, given temperature.

6 The linear problem We will consider a problem in which the fluid is driven so gently that the Rossby number is small enough to allow the neglect of all nonlinear terms. When the flow occurs in a cylinder and the forcing and the motion are axially symmetric the neglect of the nonlinearity is generally sensible. It is usually the advent of instabilities of the flow we are going to describe that more seriously limit the validity of the linearization Coordinate system r,θ,z with corresponding velocities u,v,w

7 Linear equations of motion in cylindrical coordinates. Axially symmetric motions!v =! p r + E 2 #$ "2 u! u / r 2 % &, u =!!!!!!!!!!!!! E 2 #$ "2 v! v / r 2 % &, 0 =! p z + T + E 2 "2 w, 1 r ru ( ) r + w z = 0, w's = E 2 "2 T, " 2 = 1 r ( (r r ( (r + (2 (z 2

8 The Ekman layers Again ζ= z/e 1/2, w = E 1/2 W near z=0 the Ekman boundary layer would satisfy!v =! p r u "", u = 1 2 v "", 0 =! p " + E 1/2 T E!W "", if #SE 1/2 W = 1 2 T ""!S " E #1/2 (ru) r + rw " = 0. We can ignore buoyancy forces in the Ekman layer. They are so thin they remain unchanged. True only if boundary is horizontal!!

9 The compatibility condition At lower boundary w I (r,0) = E 1/2 v W (r,!) = E 1/2 v 2r (rv ) I r v T =0 Let the upper boundary, or lid, of the cylinder be rotating with the differential speed v T (r).then the same analysis at the upper boundary yields H L w I (r,1) = E 1/2 v 2 1 r!!r r v T " v I (r,1) ( ) At upper boundary, z =1

10 Interior equations E <<1 v I = p I r,!!!!!t I =! p Iz,! u I = E 2 ("2 v I! v I / r 2 ) Note that u I is O(E) (ru I ) r + w Iz = 0 w I = E 2#S "2 T I Thus w I is equal to its average at z =0 and z =1, i.e. If σs <1 then w I is much larger than u I so that!w I!z = 0 w I = 1 [ 2 w I (r,1) + w I (r,0)] = E1/2 4 1 r!!r r v T " {v I (r,1) " v I (r,0)} [ ]

11 The vertical velocity and the thermal wind relation!v I!z =!T I!r, Thermal wind Leading to: v I (r,1)! v(r,0) = " "r 1 # 0 T I (r,z')dz' So the vertical velocity becomes given directly in terms of the temperature and external forcing w I =! E1/2 4 1 r " $ "r r " & % "r 1 # 0 ' T I (r, z') ) ( + E1/2 4 1 r " ( ) "r rv T!S!<< 1

12 Interior thermal equation E 2!S "2 T I + E1/2 4 "2 # T I (r, z')dz' = E1/ $ r $r rv T ( ) we define the pseudo temperature θ! = T I + " T I dz' 1 #.! = "S 0 2E 1/2! 2 " = # 1 r $ ( ) $r rv T

13 The temperature in terms of the pseudo temperature T I =! " # 1 $ 1+ #!dz 0! 2 " = # 1 r $ ( ) $r rv T We need boundary conditions on all the side walls of the cylinder to determine the solution.

14 For σs <1 the streamfunction the temperature deviation in the Ekman layer is O(E 1/2 σs), and, as we shall see, there is no other boundary layer on the horizontal boundaries for σs <<1 so the interior temperature must satisfy the boundary conditions on the horizontal boundaries. However, we need to consider the boundary layers on the side wall to find the appropriate boundary conditions for the pseudo temperature equation on r=r 0. Useful to introduce streamfunction for u and w ru =!" z, For the interior,from rw = " r w I =! E1/2 4 1 r " $ "r r " & % "r 1 # 0 ' T I (r, z') ) ( + E1/2 4 1 r " ( "r rv T ) r! I (r) = E1/2 4 rv T " E 1/2 4(1+ #) r $ $r 1 & 0 %dz

15 The side wall boundary layers σ S<<1 Write all variables as their interior values plus a boundary layer correction that by definition vanishes for large values of the boundary layer coordinate e.g. u = u I (r) + u bl (!),!!!!!!!!!!!!!u bl " 0,!!!! " # v = p r u = E 2 v rr 0 =! p z + T + E 2 w rr u r + w z = 0 Boundary layer balances for the correction functions. The azimuthal velocity v remains in geostrophic balance. "Sw = E 2 T rr

16 References Barcilon,V. and J. Pedlosky 1967 a, Linear theory of rotating stratified fluid motions. J. Fluid. Mech.,29, b, Unified linear theory of homogeneous and stratified rotating fluids. J. Fluid Mech., 29, Pedlosky,J., J.A. Whitehead, and G. Veitch, Thermally driven motions in a rotating stratified fluid. Theory and experiment. J. Fluid Mech., 339, Veronis, G. Analogous behavior of rotating and stratified fluids. Tellus, 19, Whitehead, J.A. and J. Pedlosky 2000, Circulation and boundary layers in differentially heated rotating stratified fluid. Dyn. Atmospheres and Oceans, 31, 1-21

17 The Master Equation Eliminate all variables with respect to the pressure to obtain:!# " a $# E 2 p 6r / 4 +!Sp rr b% c% + pzz = 0 We assume that all z derivatives are O(1) and r derivatives give us a term! "1 So term (a) is O( E 2 /! 6 ) To satisfy all the boundary conditions the full order of the equations must be maintained.

18 The Stewartson Layer Now we can ask what balances are possible and we note that the full order of the equation (6 th order in r) must be preserved in any parameter setting. Let s suppose that terms (a) and (c) are of the same order. This clearly implies that E 2 /! 6 = O(1),!!!!!!"! = E 1/3 This layer exists for a homogeneous fluid when S is zero. To find the parameter limit of validity of that balance, we can evaluate the order of term (b) and compare it to terms (a) and (c). Term (b) is!s / E 2/3 So for the balance a-c to hold we need!s < E 2/3!# " a $# E 2 p 6r / 4 +!Sp rr b% c% + pzz = 0

19 The buoyancy layer(1) If the inequality is reversed the a-b balance is no longer valid.!# " a $# b% c% E 2 p 6r / 4 +!Sp rr + pzz = 0! = E1/2 From the a-b balance ("S) 1/4 This is the buoyancy layer thickness! * = L! = ("# )1/4 N 1/2 And so the buoyancy layer balance is valid when!s > E 2/3 Dimensional: independent of rotation The ratio of term c to term b is 1!S / " = E 2!S ( ) 3/2

20 The Hydrostatic layer (1) The balance b-c yields! = ("S) 1/2!# " a $# E 2 p 6r / 4 +!Sp rr b% c% + pzz = 0 And this too requires!s > E 2/3 and, of course, σs <1 When σs =E 2/3 the hydrostatic layer and the buoyancy layer merge to become the E 1/3 layer! buoyancy! hydrostatic = or, as the stratification increases the Stewartson layer splits into two sublayers, the hydrostatic layer and the buoyancy layer. Note that the ratio of their two thicknesses is: E 1/2 ("S) 1/4 ("S) = # E 2/3 & 1/2 $ % "S ' ( 3/4 << 1 δ b δ h

21 The overall balances Buoyancy layer!# " a $# E 2 p 6r / 4!S > E 2/3 +!Sp rr Hydrostatic layer b% c% + pzz = 0 Stewartson layer!s < E 2/3 The overall sixth order system is preserved even when the boundary layer splits in two

22 The hydrostatic layer balances (1) $! h L = " ' 1/2 N % & # ( ) f L In dimensional units. Aside from Prandtl number, this is the Rossby deformation radius.! = r " r Boundary layer coordinate o (#S) 1/2 Correction functions in hydrostatic layer u h = E!u(", z) Momentum equation.!s v h =!v(!,z) implies p h = (!S) 1/2!p From geostrophy It is easy to show that the frictional term in the vertical equation is negligible, hence the layer is in hydrostatic balance. T h =!S ( )1/2! T

23 Hydrostatic layer balances (2) From continuity equation w h = E (!S)!w! 3/2 h = E! (#, z) "S Ignoring terms of order E 2/3!S!v =!!p ",!!!u = 1 2!v ""!! or smaller the governing equations in this layer are:!t =!p z,!!!!!!w = 1 2!u " =!w z so!t ""! = " r o 2!r o!w =!" # =! r o 2!v!! +!v zz = 0 bdy layer eqn. is a pde.!t #!T ##

24 The hydrostatic layer eigenvalue problem (1) The width of the hydrostatic layer is much greater than the Ekman layer thickness as long as σs >>E (!S) 1/2 E 1/2 So the Ekman layer will act as if the hydrostatic layer is part of the interior and the ordinary compatibility condition will apply, w h = E (!S)!w = E 3/2 2!S ( ) 3/2! T "" = # E ( ) 3/2!v z" 2!S heat eqn. + thermal wind = E1/2 2 v h r = # E 1/2 ( ) 1/2!v " 2!S Ekman pumping related to vorticity!!v z = ±2!v At z = 0 # 1 " $ % &

25 Hydrostatic layer eigenvalue problem (2) letting!v = V(z)e!a" leads to an eigenvalue problem for {a,v(z)} V zz + a 2 V = 0 " V z =!2!V!!!at!!z = # 1 $ 0 If λ is very small it is easy to show that # the lowest eigenvalue is a o! 2"S 1/2 & $ % E 1/2 % & ' ' (! = "S 2E 1/2 Note that if the stratification is large v=0 at the boundary and if the stratification is weak it is the shear that vanishes. and this gives rise to a solution nearly independent of z with a characteristic scale in the radial direction of E 1/4. This calculation is left to the student. This is the second boundary layer scale of Stewartson for homogeneous fluids

26 The buoyancy layer (1)! b = E1/2 Given the scale we can deduce the relative ("S) 1/4 balances in this thin sub layer T b =! b ˆ T,!!wb = v b = E ŵ,!u b = E "S û,! "S! b E ("S) ˆv,!!!!! E p 3/2 b =! b "S ( ) 3/2 ˆp Boundary layer coordinate:! = (r o " r) #b ˆv =! ˆp ",!!û = 1 2 ˆv "",!!ŵ = 1 2 ˆ T "",!!û " = ŵ z, 0 = ˆ T ŵ"" The vertical equation is a balance between buoyancy and friction: The layer is non-hydrostatic

27 The buoyancy layer (2)! b = E! "S ˆ,!!!!!!!! ˆ = # r o T ˆ $ 2 T4! ˆ + 4T ˆ = 0 ode and like Ekman layer ˆ T = Ae!" cos" + Be!" sin", ŵ = Ae!" sin"! Be!" cos" yields velocity normal to boundary and depends on variation vertically along the side wall û =!A z e!" ( cos" + sin" ) / 2 + B z e!" cos"! sin" ( ) / 2 equivalently! ˆ = "r o 2 { } + Be "# { cos# " sin# } $% "Ae "# cos# + sin# & '

28 Matching at r = r o : an example Our goal is to find how the boundary layers translate the physical conditions on the boundary to conditions on the interior problem, i.e. how do the boundary layers control the interior (and vice-versa) At the rim,using the sum of the interior and boundary layer corrections: last term negligible E v I (r o,z) +!v(0,z) + (!S) ˆv(0, z) = 0 no slip on v 3/2 T Ir!!T "! ˆ T # = 0, side wall insulated! I + E "S w I + ( )! + E! ("S) ˆ = 0 no normal flow E E!w 2/3 +!S ( )!S" b ŵ = 0 no slip on w

29 Matching (2)! I " E r o (#S) 2!T $ " E r o (#S) 2 ˆ T % = 0 from! = " r o 2!T #! b = E! "S ˆ,!!!!!!!! ˆ = # r o T ˆ $ 2 But from thermal wind eqn. and v I (r o,z) +!v(0,z) = 0 for!s >> E 2/3 T Ir!!T " = 0 Taking z derivative thus also ˆ T! = 0 on r = r o ŵ = 0 Buoyancy layer absent to lowest order (analogous to slippery Ekman layer)

30 Matching (3) thus! I (r o ) " r o 2 E #S!T $ = 0, on r = r o! I (r o ) = E #S r o 2 T I r using r! I (r) = E1/2 4 rv T " E 1/2 4(1+ #) r $ $r E 1/2 4 r ov T! E1/2 4 r o" T Ir dz! r o Boundary condition expressed now entirely in terms of interior variables.note the streamfnc. is independent of z so T ir must be also independent of z on the rim 1 & 0 %dz 1 E #S " T I r dz = 0 0

31 Side wall boundary condition With!T I!r (r Independent of z we obtain the final boundary o ) condition for the interior thermal equation. T Ir (r o ) =! 1 +! v T (r o ),! = "S 2E 1/2 For the pseudo temperature! r (r o ) = "v T (r o ) that satisfies! 2 " = # 1 r $ ( ) $r rv T

32 Example 1:The purely mechanical driven circulation Suppose (although it is a bit artificial) or! z = 0,!!z = 0,1 then a solution to! 2 " = # 1 r $ $r rv T T Iz = 0,!!!!!z = 0,1 ( ) satisfying! r (r o ) = "v T (r o )! r (r) = "v T (r) # T Ir = " 1+ " v T (r) From the thermal wind equation v z = T r v I =! 1 +! zv T + V o (r)

33 The mechanically driven velocity To determine V O (r) use the boundary condition at z =0 w(z = 0) =! E1/2 2 1 r!rv(r, 0)!r = E 2"S T I rr = E 2"S # # r!rv T!r! = "S 2E 1/2 $ V o = 1 2(1+ #) v T v I =! 1 +! v T (z " 1 / 2) + v T / 2

34 v I as a function of λ v T λ=0 λ >>1 w I = E 1/2 1 4(1+!) r "rv T "r 0! = "S 2E 1/2 For large stratification the interior flow satisfies the no slip boundary conditions on z=0, 1 expunging the Ekman layers. The Ekman layers are choked off as the stratification increases and the vertical velocity diminishes.w I falls from O(E 1/2 ) to order E/σS as the stratification increases.

35 Experiment: Paul Linden, JFM , Linden used a sugar solution to stratify the fluid. The flow was in an annulus,not a cylinder, but otherwise the theory should apply. The Prandtl number is large σ is about ! S / E 1/ 2 ~ But σs was large S=0 Large Stratification, no Ekman layers

36 Example 2 The heated cylinder, theory and experiment Consider the same cylindrical geometry but now the top is not moving differentially. Instead the fluid is driven by an applied temperature at the upper boundary. recall that T refers to the temperature anomaly around the basic stratification. T = T u (r),!!!!!!!z = 1, T = 0,!!!!!!!!!!!!!z = 0

37 The (σs) -1/2 layer This system was examined experimentally and spanned a rather wide range of values of S. Indeed, it was rather natural in the laboratory situation to consider values of σs that were large compared to unity as well as small values. When σs is large, the hydrostatic layer has already filled the interior. Indeed, as we shall show below, a metamorphosis of that layer occurs and a boundary layer of scale (!S) "1/2 now exists in the vicinity of the upper, heated boundary and it plays a role similar to the (!S) 1/2 layer on the side wall for small values of S. (σs) -1/2 interior

38 Equations of motion v = p r,!!!!!!!u = E # 2!2 v " $ v r 2 % &, Outside the Ekman layers T = p z,!!!!!w = E 2'S!2 T #$ % & 1 r ( (r (ru) + (w (z = 0 $! 2 &"S! 2 h + #2 % #z 2 ' ) ( p = 0,! 2 = 1 r # #r r # #r + #2 #z 2,!!!! h2 = 1 r # #r r # #r

39 Balances in the σs -1/2 layer The temperature in this layer is O(1). From the thermal wind equation this means the vertical shear of v is order 1 v is order σs -1/2.This in turn implies that u =O(E(σS) 1/2 ),w=o(e) and the lowest order circulation in the vertical plane is limited to the upper thermal layer for large stratification. The Ekman layers are also, again, absent for large σs. Below this layer u and w are so small, in the interior, that the azimuthal equation satisfies,! 2 v I " v I / r 2 = 0 And so is a Couette flow driven by the condition that the total v vanish on both z =0 and 1. Details left for the student.

40 Full system outside boundary layers: σs>> E 1/2 $! 2 "S! 2 h + #2 ' & % #z 2 ) ( p = 0,! 2 = 1 r # #r r # #r + #2 #z 2,!!!! h2 = 1 r # #r r # #r p r = 0,!!!!!!!!z = 0,1 (no Ekman layers) p z = 0,!!!!!!!!z = 0, p z = T u (r)!!!z = 1, p r = 0,!!!!!!!!!r = r o! P n = A n e!q n (z!1) + B n e!q n z + C n e!q nd(z!1) + D n e!q ndz p = P n (z)j o (k n r / r o ) n=1 q n = k n / r o,!!!!d = (!S) 1/2

41 An idealized problem suppose T u = e r /r o! 1 λ= (a) 0.01, (b) 1.0, and (c) 100 For small and moderate λ the pseudo temperature equation is used.

42 An experiment Whitehead and Vetch carried out an experiment in which the upper surface was forced by a temperature anomaly whose measured form was

43 The velocity as seen in dye streaks

44 Theoretical predictions

45 Comparison with experiment λ =42 Good agreement

46 Example 3: Side wall heating:driven by buoyancy layer We consider now a case in which the flow is driven by the side wall buoyancy layer. (Whitehead, J.A. and J. Pedlosky 2000, Circulation and boundary layers in differentially heated rotating stratified fluid. Dyn. Atmospheres and Oceans, 31, 1-21) Ω heating coil H(z) is shape of heating function on side wall

47 The buoyancy layer suction! ˆ = E "S r o 2 H [ (z)e#$ sin$ + cos$ ]! I = " E #S r o 2 H (z) u I H(z) $ u I (r o, z) = " 1 r o %! I %z = E 2#S dh dz this circulation will drive, via the Coriolis torque a strong azimuthal velocity

48 Equations and boundary conditions $! 2 &"S! 2 h + #2 % #z 2 ' ) ( p = 0,! 2 = 1 r from!!r " 2 h p = 1 #S # #r r # #r + #2 #z 2,!!!! h2 = 1 r v = p r,!!!!!!!u = E # 2!2 v " $ v r 2!H!z,!!!!!!r = r o % &, # #r r # #r and v = p r = 0 on r o on z =0,1! 2 p z =! "S E 1/2! h2 p,!!!!!!!! h2 = 1 r # #r r # #r Again, solution obtained by a Bessel expansion in r.

49 Results of side heating Contours of v Contours of ψ S= 13.44, E=

50 Velocity profiles v(z) at various radii Theory Experiment S= 13.44, E=

51 Velocity profiles, larger stratification Theory Experiment S = and E =

52 Further comparison A comparison of the maximum azimuthal velocity from 4 parameter setting between theory and experiment. See Whitehead &Pedlosky (2000) for details

53 Further thoughts on the role of stratification. The Ekman layers and the side wall boundary layers combine to set up a vertical circulation in the interior that is closed in the side wall layers for small S. In our cylinder example it is a two stage process. First the flux in the Ekman layer forces vertical motion in the interior and also feeds the side wall layers. Then the side wall layers flux fluid vertically and establish a heating and cooling of the interior altering the strong, O(1) circulation through the thermal wind balance. For large S this can choke off the bottom Ekman layers and the frictional dissipation there. If the side walls and the bottom should combine, i.e. should the bottom be sloping with respect to the stratification, this 2 step program could coalesce.

Boundary Layers: Stratified Fluids

Boundary Layers: Stratified Fluids Lecture 3 by Jeroen Hazewinkel continued from lecture 2 Using w I = /(2σS) 2 T I, the interior of the cylinder is be described by 2σS 2 T I + 1/2 4 2 This result can