Advanced Dynamical Meteorology

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a stratified shear flow U = (U(z),0,0), including the special case of unifor flow

1 Advanced Dynaical Meteorology Roger K. Sith CH03 Waves on oving stratified flows Sall-aplitude internal gravity waves in a stratified shear flow U = (U(z),0,0), including the special case of unifor flow U(z) = constant. Linearized anelastic equations: ut + Uux + wuz = Px wt + Uwx = Pz + b b + Ub + N w = 0 t x ux + wz w /Hs = 0 1

2 Travelling wave solutions Substitute (u,w,p,b) = (u(z),w(z),p(z),b(z))exp[i(kx ˆ ˆ ωt)] a set of ODEs or algebraic relationships between i( ω Uk) u + wu = ikp z uz ( ) etc. i( ω Uk)wˆ = Pˆ + bˆ ω ˆ + ˆ = i( Uk)b N w 0 w iku + w z = 0 H s z û,p,b ˆ ˆ Relate quantities to : ˆ ˆ ˆ (u,p,b) ˆ = w ˆ, ω wˆ + ku w ˆ, w * i w i * ω N w z z z * k Hs k Hs iω ˆ ω* = ω Uk is the intrinsic frequency of the wave. The frequency easured by an observer oving with the local flow speed U(z). Then the second equation gives an ODE for w

3 L NM F 1 w * * H w k k U U z N zz z + zz + k w H H * s H G 1I K J + 1 = 0 ω ω ω s s z F HG I KJ O QP Put wz ( ) = exp( z/ H ) wz ~ ( ) c = ω / k s Then ω* = kc ( U) ~ w ( l ( z) k ) w ~ zz + = 0 Scorer s equation Scorer paraeter l ( z) = a N U c f Uzz + Uz / Hs 1 U c 4H s Free waves When w satisfies hoogeneous boundary conditions (e.g. w = 0) at two particular levels, we have an eigenvalue proble. w = 0 at z = 0, H w~ = 0 at z = 0, H For a given horizontal wavenuber k, free waves are possible only for certain phase speeds c. Alternatively, when c is fixed, there is a constraint on the possible wavenubers. The possible values of c(k), or given c, the possible values of k, and the corresponding vertical wave structure are obtained as solutions of the eigenvalue proble. Often the eigenvalue proble is analytically intractable. 3

4 Exaple 1: Stratified shear flow between rigid horizontal boundaries w = 0 z = H z N(z) x U(z) w = 0 z = 0 Exaple : Wave-guide for waves on a surface-based stable layer underlying a deep neutrally-stratified layer no vertical wave propagation! neutrally-stable layer N = 0 z U(z) stable layer N(z) x A odel for the Morning-Glory wave-guide 4

5 The Morning-Glory 5

6 The Gulf of Carpentaria Region Morning-Glory cloud lines A Morning Glory over Bavaria 6

7 Forced waves If the waves are being generated at a particular source level, the boundary conditions on Scorer s equation are inhoogeneous and a different type of atheatical proble arises: - Exaple 1: Waves produced by the otion of a (sinusoidal) corrugated boundary underlying a stably-stratified fluid at rest. c g k z x c corrugated boundary 7

8 Assue: 1. no basic flow (U = 0). the Boussinesq approxiation is valid (1/H s = 0) 3. N and c are given constants the vertical wavenuber is deterined by the dispersion relation = N / c k c is the speed of the boundary k is the horizontal wavenuber the wavenuber of the corrugations The group velocity of the waves is 3 c c g = N (, k ) The vertical flux of ean wave energy is F = pw = Ew g the ean wave energy density the vertical coponent of the group velocity Here c < 0 sgn(w g ) = sgn(k) these waves propagate energy vertically phase lines k c 8

9 The solution of ~ w ( l ( z) k ) w ~ zz + = 0 is deterined by 1. A condition at z = 0 relating w ~ to the aplitude of the corrugations, and the speed of boundary otion, c.. A radiation condition as z the condition at z = 0 is an inhoogeneous condition the condition as z ensures that the direction of energy flux at large heights is away fro the wave source, i.e. vertically upwards. This condition is applied by ensuring that only wave coponents (in this case there is only one) with a positive group velocity are contained in the solution. Conditions 1 and deterine the forced wave solution uniquely Exaple : Waves produced by the flow of a stably-stratified fluid over a corrugated boundary. U c g z x This flow configuration is a Galilean transforation of the first: the boundary is at rest and a unifor flow U (> 0) oves over it. 9

Mountain waves When a stably-stratified airstrea crosses an isolated ridge or, ore generally, a range of ountains, stationary wave oscillations are frequently

10 Mountain waves When a stably-stratified airstrea crosses an isolated ridge or, ore generally, a range of ountains, stationary wave oscillations are frequently observed over and to the lee of the ridge or range. These so-called ountain waves, or lee waves, ay be arked by sooth lens-shaped clouds, known as lenticular clouds. 10

11 Linear Theory Assue: 1. Unifor flow U. Boussinesq fluid (1/H s = 0) 3. Sall aplitude sinusoidal topography: z = h( x) = h coskx Surface boundary condition: strealine slope equals the terrain slope (U + u, w) = (dx, dh) linearize w dh U+ u dx on z h x w = U dh dx at z = 0 11

12 Let ζ( xzt,, ) be the vertical displaceent of a fluid parcel. ζ( xzt,, ) w D ζ = Dt ζ linearize w = U x w = ikuζ U = constant F HG d ζ N + k dz U ζ satisfies the sae equation as w I ζ = 0 KJ = N / U k The general solution is ζ ( z) Ae iz = + Be iz constants ζ ( z) Ae iz = + Be iz The boundary condition at the ground is ζ( x, 0 ) = h e ikx ( ) ζ 0 = h The upper boundary condition and hence the solution depends on the sign of = N / U k There are two cases: call l 0 < k < l l N = > 0 U real l < k iaginary 1

13 Two cases: I. 0 < k < l real if, k > 0, we ust choose B = 0 to satisfy the radiation condition. The phase lines slope upstrea with height sgn(k) > 0 Then the coplete wave solution is (, ) i( kx+ z) ζ xz = h e The real part is iplied II. l < k iaginary (= i 0 say) where =+ k l 0 Then ζ ( z) Ae z = + Be 0 0 z The upper boundary condition requires that reains bounded as z B = 0 Then the coplete wave solution is (, ) ( ) ζ xz = h e k l z+ ikx Here, the phase lines are vertical ( ) ζ z 13

14 Strealine patterns for unifor flow over sinusoidal topography vertical propagation: U k < N 0 < k < l U vertical decay: N < U k (, ) i( kx+ z) ζ xz = h e l < k U (, ) ( ) ζ xz = h e k l z+ ikx Exercise Calculate the ean rate of working of pressure forces at the boundary z = 0 for the two solutions (, ) i( kx+ z) Show that for ζ xz = he drag on the airstrea the boundary exerts a whereas for (, ) ( ) ζ xz = h e k l z+ ikx it does not. Show also that, in the forer case, there exists a ean downward flux of horizontal oentu and that this is independent of height and equal to the drag exerted at the boundary. 14

15 Flow over isolated topography U h b Consider flow over an isolated ridge with h(x) = h /[1 + (x/b) ] axiu height of the ridge = h characteristic half width = b The ridge is syetrical about x = 0 where it ay be expressed as a Fourier cosine integral z 0 h( x) = h( k)cos( kx) dk z 0 = Re hke ( ) ikx dk z h e dx h( k) = z 0 h( x) cos kx dx = Re π π x/ b ikx ikbu h b z e du = Re π u a f 15

16 The solution for flow over the ridge is just the Fourier synthesis of the two solutions for 0 < k < l and l < k. L NM zl 0 z 1 / 1 / ikx [ + ( l k ) z] ikx ( k l ) z l ζ( xz, ) = Re hke ( ) dk+ hk ( ) e dk Put kb = u ζ( xz, ) = h Re exp u1 L NM zlb 0 z lb L NM L NM F H F H + exp u 1 = I 1 + I, say ix b ix b I ilb K + I K d d u 1 / u 1 / l b i i O QP O QP z b du z b du O QP O QP Two liiting cases: narrow ridge, lb << 1 I 1 << I and ix z ζ( xz, ) h Re exp u1 + z0 H b b b h = F I H b+ zk [ 1+ x /( b+ z) ]. L NM F IO KQP du each strealine has the sae general shape as the ridge the width of the disturbed portion of a strealine increases linearly with z its axiu displaceent decreases in proportion to 1/(1 + z/b), becoing relatively sall for heights a few ties greater than the barrier width. 16

17 narrow ridge, lb << 1 Nb/U << 1 Steady flow of a hoogeneous fluid over an isolated two-diensional ridge, given by ζ( xz, ) = F H I b h b+ zk [ 1 + x /( b+ z) ]. The highest wind speed and the lowest pressure occurs at the top of the ridge. broad ridge, lb >> 1 Nb/U >> 1 I 1 >> I and ζ( xz, ) h Re{ e exp u1 ix/ b du} = h Re ilzz0 = h L NM a f cos lz ( x / b)sin lz 1+ ( x/ b) O QP. L NM ilz be b ix O QP In this liit, essentially all Fourier coponents propagate vertically. 17

18 Buoyancy-doinated hydrostatic flow over an isolated twodiensional ridge, given by ζ (x,z) = h cos lz (x / b)sin lz 1 + (x/b) The pressure difference across the ountain results in a net drag on it. The drag can be coputed either as the horizontal pressure force on the ountain z dh D = p(x, 0) dx dx or as the vertical flux of horizontal oentu in the wave otion D= ρ 0 ( z) uwdx z Boussinesq approxiation ρ0 ( z ) = ρ * 18

19 oderate ridge, lb 1 Nb / U 1 The integrals I 1 and I are too difficult to evaluate analytically, but their asyptotic expansions at large distances fro the ountain (copared with 1/l) are revealing. Let Then u = lbcosα where 0 α π / x = rcos θ, z = rsin θ where 0 θ π π lb cos α irl cos( θ α) z I1 = hlbre (sin αe ) e dα 0 zπ irl g( α) I1 = Re h( α) e dα 0 Asyptotic expansion for large r by stationary phase ethod z I = h lbre (sin αe ) e dα 1 π 0 lb cos α irl cos( θ α) Far field behaviour of I 1 19

20 ix b = h bexp( lb)/z. (u lb)z/b I h ax exp u 1 e du lb u lb I is at ost O(b/r) Since lb Ο(1), I 1 + I is always the sae order as I 1 not surprising since I 1 contains all the vertically propagating wave coponents. Then for x > 0 and θ not to close to 0 or π 1 / lbcos θ π ζ( xz, ) hlbsinθe F I H cos lr lr K F H π 4 I K Lee waves in the case where lb 1 0

21 Scorer s Equation ~ w ( l ( z) k ) w ~ zz + = 0 l ( z) = a N U c f Uzz + Uz / Hs 1 U c 4H s Trapped lee waves It was pointed out by Scorer (1949) that, on occasion, a long train of lee waves ay exist downstrea of a ountain range. Scorer showed that a long wave train is theoretically possible if the paraeter l (z) decreases sufficiently rapidly with height. Eigensolutions of ~ w ( l ( z) k ) w ~ zz + = 0 are difficult to obtain when l varies with z. 1

22 A two-layer odel perturbation streafunction U l1 = N1 / U, ψ 1 layer 1 U l = N / U, ψ layer In the Boussinesq approxiation d ψ ~ dz n n ~ ψ satisfies the sae ODE as + ( l k ) ψ ~ = 0, ( n = 1,) u + w = 0 w = ψ w~ = ikψ ~ x z x n ~ w d ψ ~ dz n n + ( l k ) ψ ~ = 0, ( n = 1,) Assue stationary wave solutions so that c = 0 then l = N /U as before n Solutions: ψ= ~ exp( iz) where = l k Assue that > 0, so that vertical wave propagation is possible in the lower layer.

23 Case (a): upper layer ore stable (l 1 > l ) k 1 layer 1 k 1 transitted wave z = 0 layer k k k - k layer interface reflected wave incident wave Streafunctions: transitted wave ψ 1 = α + 1 e i( kx z) incident plus reflected wave 1, > 0 ψ i( kx + z) i( kx z) = ae + βe a given Wave solutions are coupled by conditions expressing the continuity of interface displaceent and pressure at z = 0. 3

24 Continuity of interface displaceent at z = 0 ζ = ζ at z = 1 0 w = w at z = 1 0 ψ = ψ at z = 1 0 When the density is continuous across the interface, continuity of pressure iplies that ψ provided, as here, that U 1 = U ψ z = z at z = see Ex a α = 1 β= + and + a 1 F HG 1 I KJ always a transitted wave no trapped "resonant" solutions in the lower layer Case (a): upper layer less stable (l 1 < l ) Assue that 1 < 0 exponential decay layer 1 k z = 0 layer k k layer interface reflected wave incident wave 4

25 The appropriate solution in the upper layer is the one which decays exponentially with height: ψ 1 = α 1 e ikx z ψ i( kx + z) i( kx z) = ae + βe β = a ψ = Ae sin z ikx w = ikψ = ikae sin z ikx w = 0 at z = h if sinh = 0 tan h h = put x = h h y 1 π < h < π 3π < h < π π π π 3π x x y = h 1 y = tan x 5

26 π < h < Recall that 1 < 0 π π 4h < = l k l 1 < k l1 < k < l π 4h To illustrate the principles involved in obtaining solutions for trapped lee waves, we assue that the upper layer behaves as a rigid lid for soe appropriate range of wavelengths w = 0 ψ = 0 z = H U l = N / U, ψ z = 0 Solution ikx ψ ( xz, ) = Ae sin H ( z) constant = l k 6

27 The general solution ay be expressed as a Fourier integral of ikx ψ ( xz, ) = Ae sin H ( z) z 1 ikx ψ( xz, ) = Ake ( ) sin H ( zdk ) π Uh = + 1 ( x/ b) 1 π z ikx Ake ( ) sinhdk 7

28 A solution showing trapped lee waves. Suary Lee waves for lb 1 Nb/U 1 8

29 narrow ridge, lb << 1 Nb/U << 1 Steady flow of a hoogeneous fluid over an isolated two-diensional ridge, given by ζ( xz, ) = F H I b h b+ zk [ 1 + x /( b+ z) ]. The highest wind speed and the lowest pressure occurs at the top of the ridge. Broad ridge 1 << lb Nb/U << 1 ζ (x,z) = h cos lz (x / b)sin lz 1 + (x/b) 9

30 Cross-section of potential teperature field (K) during a severe downslope wind situation in the Colorado Rockies on 11 January 197. Fro Klep and Lilly, The End 30

Theory of linear gravity waves April 1987

Theory of linear gravity waves April 1987 April 1987 By Tim Palmer European Centre for Medium-Range Weather Forecasts Table of contents 1. Simple properties of internal gravity waves 2. Gravity-wave drag REFERENCES 1. SIMPLE PROPERTIES OF INTERNAL