Etudes et évaluation de processus océaniques par des hiérarchies ...

Etudes et évaluation de processus océaniques par des hiérarchies ... Etudes et évaluation de processus océaniques par des hiérarchies ...

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204 30 CHAPTER 5. DYNAMICS OF THE OCEAN 5.9 Energetics of flow in Geostrophic Equilibrium For the shallow water dynamics the total energy is composed of kinetic and available potential energy (the part of the potential energy which is available in the layered model by reducing the surface anomaly η, if η = 0 everywhere the available potential energy vanishes): E total = E kin + E pot = ρ ∫ H(u 2 + v 2 )dxdy + gρ ∫ η 2 dxdy (5.49) 2 A 2 A = g2 ρ H 2f ∫A ( (∂ 2 x η) 2 + (∂ y η) 2) dxdy + gρ ∫ η 2 dxdy (5.50) 2 A tel-00545911, version 1 - 13 Dec 2010 where we used (eqs. 5.44 and 5.45). If the surface perturbation has the simple form η = η 0 sin(x/L) then the energy is given by: E total = E kin + E pot = gη2 0 4 ∫ A ( Hg f 2 L 2 + 1 ) dxdy (5.51) Were the first term is the kinetic and the second term the available potential energy. We see that in a geostrophic flow the kinetic energy is larger than the available potential energy when the structure is smaller than the Rossby radius R = √ gH/f 2 . So for large geostrophic structures most of the energy is in the potential part and for small structures in the kinetic part. The Rossby radius is of the order of a few thousands of kilometers for the shallow water dynamics of the ocean (the barotropic Rossby radius) but only several tenths of kilometers when the reduced gravity dynamics of the layer above the thermocline are considered (the baroclinic Rossby radius). 5.10 Linear Potential Vorticity and the Rossby Adjustment Problem If we take ∂ x (eq. (5.42)) - ∂ y (eq. (5.41)) we see that: ∂ t ζ + f(∂ x u + ∂ y v) = 0. (5.52) relating vorticity ζ = ∂ x v − ∂ y u to divergence ∂ x u + ∂ y v. Using eq. (5.43) we get: ( ζ ∂ t f − η ) = 0. (5.53) H One usually calls Q lin sw = ζ H − fη H 2 the linear shallow water potential vorticity. The above equations show, that at every location linear shallow water potential vorticity (PV) is conserved, when the dynamics is governed by the linearised shallow water equations. The Rossby adjustment problem considers the adjustment of an initially step-like perturbation (see fig. 5.10), and we would like to know the final, geostrophically balanced state of this perturbation. To this end we use the conservation of potential vorticity and we further require the final state to be in geostrophic equilibrium. The initial potential vorticity is given by sgn(x)(fη 0 )/H 2 the PV of the adjusted state is the same, we thus have, g/(Hf)∂ xx η a − fη a /H 2 = sgn(x)(fη 0 )/H 2 , (5.54) R 2 ∂ xx η a − η a = η 0 sgn(x), (5.55)

205 5.10. LINEAR POTENTIAL VORTICITY AND THE ROSSBY ADJUSTMENT PROBLEM31 ✻ η Z ✲ X Figure 5.5: Initial condition tel-00545911, version 1 - 13 Dec 2010 which has the solution: η a = sgn(x)η 0 (exp(−|x|/R) − 1) (5.56) with R = √ gH/f 2 is called the Rossby radius of deformation. It is the distance, a gravity wave travels in the time f −1 . η ✻ Z Figure 5.6: Adjusted state We have calculated the final geostrophically adjusted state from an initial perturbation using geostrophy and conservation of linear PV, but we have not shown how this adjustment happens. For this a numerical integration of the linear shallow water equations are necessary, eqs. (5.41) – (5.43). Exercise 25: calculate the final velocity field (u,v). Exercise 26: what happens when rotation vanishes? Exercise 27: in section 5.2 we saw that if rotation is vanishing, an initial perturbation of the free surface moves away in both directions leaving an unperturbed free surface and a zero velocity behind. Does this contradict the conservation of linear potential vorticity? Exercise 28: calculate the loss of available potential energy and compare it to the gain in kinetic energy during the adjustment process. Exercise 29: calculate the (barotropic) Rossby radius of deformation (H = 5km), calculate the reduced gravity (baroclinic) Rossby radius of deformation (H = .5km, g ′ = 3. · 10 −2 m/s −2 ) ✲ X

205<br />

5.10. LINEAR POTENTIAL VORTICITY AND THE ROSSBY ADJUSTMENT PROBLEM31<br />

✻<br />

η<br />

Z<br />

✲ X<br />

Figure 5.5: Initial condition<br />

tel-00545911, version 1 - 13 Dec 2010<br />

which has the solution:<br />

η a = sgn(x)η 0 (exp(−|x|/R) − 1) (5.56)<br />

with R = √ gH/f 2 is called the Rossby radius of <strong>de</strong>formation. It is the distance, a gravity wave<br />

travels in the time f −1 .<br />

η<br />

✻<br />

Z<br />

Figure 5.6: Adjusted state<br />

We have calculated the final geostrophically adjusted state from an initial perturbation<br />

using geostrophy and conservation of linear PV, but we have not shown how this adjustment<br />

happens. For this a numerical integration of the linear shallow water equations are necessary,<br />

eqs. (5.41) – (5.43).<br />

Exercise 25: calculate the final velocity field (u,v).<br />

Exercise 26: what happens when rotation vanishes?<br />

Exercise 27: in section 5.2 we saw that if rotation is vanishing, an initial perturbation of<br />

the free surface moves away in both directions leaving an unperturbed free surface and a zero<br />

velocity behind. Does this contradict the conservation of linear potential vorticity?<br />

Exercise 28: calculate the loss of available potential energy and com<strong>par</strong>e it to the gain in<br />

kin<strong>et</strong>ic energy during the adjustment process.<br />

Exercise 29: calculate the (barotropic) Rossby radius of <strong>de</strong>formation (H = 5km), calculate<br />

the reduced gravity (baroclinic) Rossby radius of <strong>de</strong>formation (H = .5km, g ′ = 3. · 10 −2 m/s −2 )<br />

✲ X

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