Reynolds Averaging. We separate the dynamical fields into slowly varying mean fields and rapidly varying turbulent components.

Transcription

1 Reynolds Averaging

2 Reynolds Averaging We separate the dynamical fields into sloly varying mean fields and rapidly varying turbulent components.

3 Reynolds Averaging We separate the dynamical fields into sloly varying mean fields and rapidly varying turbulent components. For example θ = θ + θ

4 Reynolds Averaging We separate the dynamical fields into sloly varying mean fields and rapidly varying turbulent components. For example θ = θ + θ We assume the mean variable is constant over the period of averaging. By definition, the mean of the perturbation vanishes: θ = 0 θ = θ θθ = θ

5 Reynolds Averaging We separate the dynamical fields into sloly varying mean fields and rapidly varying turbulent components. For example θ = θ + θ We assume the mean variable is constant over the period of averaging. By definition, the mean of the perturbation vanishes: θ = 0 θ = θ θθ = θ Thus, the mean of a product has to components: θ = + )θ + θ ) = θ + θ + θ + θ ) = θ + θ + θ + θ = θ + θ

6 Thus, e have: θ = θ + θ 2

7 Thus, e have: θ = θ + θ We assume that the flo is nondivergent see Holton 5.1.1): u + v y + = 0 2

8 Thus, e have: θ = θ + θ We assume that the flo is nondivergent see Holton 5.1.1): u + v y + = 0 The total time derivative of u is du dt = u t + u u + v u y + u + u = u t + uu + uv y + u u + v y + ) 2

9 Thus, e have: θ = θ + θ We assume that the flo is nondivergent see Holton 5.1.1): u + v y + = 0 The total time derivative of u is du dt = u t + u u + v u y + u + u = u t + uu + uv y + u u + v y + ) Writing sums of mean and eddy parts and averaging: du dt = u t + u u + u u ) + u v + u y v ) + u + u ) 2

10 Repeat: du dt = u t + u u + u u ) + y u v + u v ) + u + u ) 3

11 Repeat: du dt = u t + u u + u u ) + y u v + u v ) + u + u ) But the mean flo is nondivergent, so u u) + y u v) + u ) = u u + v u y + u 3

12 Repeat: du dt = u t + u u + u u ) + y u v + u v ) + u + u ) But the mean flo is nondivergent, so u u) + y u v) + We define the mean operator u ) = u u + v u y + u d dt = t + u + v y + 3

13 Repeat: du dt = u t + u u + u u ) + y u v + u v ) + u + u ) But the mean flo is nondivergent, so u u) + y u v) + We define the mean operator u ) = u u + v u y + u d dt = t + u + v y + Then e have du dt = u t + u u ) + y u v ) + u ) 3

14 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 4

15 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 The thermodynamic equation may be expressed in a similar ay: [ d θ dt + dθ 0 dz + u θ ) + v y θ ) + θ )] = 0 4

16 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 The thermodynamic equation may be expressed in a similar ay: [ d θ dt + dθ 0 dz + u θ ) + v y θ ) + θ )] = 0 The terms in square brackets are the turbulent fluxes. 4

17 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 The thermodynamic equation may be expressed in a similar ay: [ d θ dt + dθ 0 dz + u θ ) + v y θ ) + θ )] = 0 The terms in square brackets are the turbulent fluxes. For example, u is the vertical turbulent flux of zonal momentum, and θ is the vertical turbulent heat flux. 4

18 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 The thermodynamic equation may be expressed in a similar ay: [ d θ dt + dθ 0 dz + u θ ) + v y θ ) + θ )] = 0 The terms in square brackets are the turbulent fluxes. For example, u is the vertical turbulent flux of zonal momentum, and θ is the vertical turbulent heat flux. In the free atmosphere, the terms in square brackets are sufficiently small that they can be neglected. 4

19 No e can rite the momentum equations as du dt fv + 1 [ p ρ 0 + u u ) + v y u ) + dv dt + fu + 1 [ p ρ 0 y + v u ) + v y v ) + u )] = 0 v )] = 0 The thermodynamic equation may be expressed in a similar ay: [ d θ dt + dθ 0 dz + u θ ) + v y θ ) + θ )] = 0 The terms in square brackets are the turbulent fluxes. For example, u is the vertical turbulent flux of zonal momentum, and θ is the vertical turbulent heat flux. In the free atmosphere, the terms in square brackets are sufficiently small that they can be neglected. Within the boundary layer, the turbulent flux terms are comparable in magnitude to the remaining terms, and must be included in the analysis. 4

20 In the boundary layer, vertical gradients are generally orders of magnitude larger than variations in the horizontal. Thus, it is possible to omit the x and y derivative terms in the square brackets. 5

21 In the boundary layer, vertical gradients are generally orders of magnitude larger than variations in the horizontal. Thus, it is possible to omit the x and y derivative terms in the square brackets. Then the complete system of equations becomes du dt fv + 1 p ρ 0 + u ) = 0 dv dt + fu + 1 p ρ 0 y + v ) = 0 d θ dt + dθ 0 dz + θ ) = 0 p + gρ 0 = 0 u + v y + = 0 5

22 We note that this system of five equations, for the variables u, v,, p, θ), also contains the variables u, v,, θ ) in the turbulent fluxes. 6

23 We note that this system of five equations, for the variables u, v,, p, θ), also contains the variables u, v,, θ ) in the turbulent fluxes. Thus, the system is not self contained. To make it soluble, e must make some closure assumption, in hich the fluxes are parameterized in terms of the mean fields. 6

24 We note that this system of five equations, for the variables u, v,, p, θ), also contains the variables u, v,, θ ) in the turbulent fluxes. Thus, the system is not self contained. To make it soluble, e must make some closure assumption, in hich the fluxes are parameterized in terms of the mean fields. Atmospheric modellers make various closure assumptions in designing parameterization schemes for numerical models. 6

25 We note that this system of five equations, for the variables u, v,, p, θ), also contains the variables u, v,, θ ) in the turbulent fluxes. Thus, the system is not self contained. To make it soluble, e must make some closure assumption, in hich the fluxes are parameterized in terms of the mean fields. Atmospheric modellers make various closure assumptions in designing parameterization schemes for numerical models. Next, e ill consider a simple parameterization of the vertical momentum flux, u )/. 6

26 Parameterization of Eddy Fluxes 7

27 Parameterization of Eddy Fluxes The momentum equations ill be taken in the form du dt fv + 1 p ρ 0 + u ) = 0 dv dt + fu + 1 p ρ 0 y + v ) = 0 7

28 Parameterization of Eddy Fluxes The momentum equations ill be taken in the form du dt fv + 1 p ρ 0 + u ) = 0 dv dt + fu + 1 p ρ 0 y + v ) = 0 We have dropped the overbars on the mean quantities. 7

29 Parameterization of Eddy Fluxes The momentum equations ill be taken in the form du dt fv + 1 p ρ 0 + u ) = 0 dv dt + fu + 1 p ρ 0 y + v ) = 0 We have dropped the overbars on the mean quantities. In the free atmosphere, scale analysis shos that the inertial terms and turbulent flux terms are small compared to the Coriolis term and pressure gradient terms. 7

30 Parameterization of Eddy Fluxes The momentum equations ill be taken in the form du dt fv + 1 p ρ 0 + u ) = 0 dv dt + fu + 1 p ρ 0 y + v ) = 0 We have dropped the overbars on the mean quantities. In the free atmosphere, scale analysis shos that the inertial terms and turbulent flux terms are small compared to the Coriolis term and pressure gradient terms. This leads us to the relation for geostrophic balance: u u G 1 p fρ 0 y, v v G + 1 p fρ 0 7

31 In the boundary layer, it is no longer appropriate to neglect the turbulent flux terms. Hoever, the inertial terms may still be assumed to be relatively small. 8

32 In the boundary layer, it is no longer appropriate to neglect the turbulent flux terms. Hoever, the inertial terms may still be assumed to be relatively small. Thus e get a three-ay balance beteen the forces: fv + 1 p ρ 0 + u ) = 0 +fu + 1 ρ 0 p y + v ) = 0 8

33 In the boundary layer, it is no longer appropriate to neglect the turbulent flux terms. Hoever, the inertial terms may still be assumed to be relatively small. Thus e get a three-ay balance beteen the forces: fv + 1 p ρ 0 + u ) = 0 +fu + 1 ρ 0 p y + v ) = 0 8

34 Using the definition of the geostrophic inds, e rite the momentum equations as fv v G ) + u ) = 0 +fu u G ) + v ) = 0 9

35 Using the definition of the geostrophic inds, e rite the momentum equations as fv v G ) + u ) = 0 +fu u G ) + v ) = 0 To progress, e need some means of representing the turbulent fluxes in terms of the mean variables. 9

36 Using the definition of the geostrophic inds, e rite the momentum equations as fv v G ) + u ) = 0 +fu u G ) + v ) = 0 To progress, e need some means of representing the turbulent fluxes in terms of the mean variables. The traditional approach to this closure problem is to assume that the turbulent eddies act in a manner analogous to molecular diffusion. 9

37 Using the definition of the geostrophic inds, e rite the momentum equations as fv v G ) + u ) = 0 +fu u G ) + v ) = 0 To progress, e need some means of representing the turbulent fluxes in terms of the mean variables. The traditional approach to this closure problem is to assume that the turbulent eddies act in a manner analogous to molecular diffusion. Thus, the flux of momentum is assumed to be proportional to the vertical gradient of the mean momentum. Then e can rite ) u u = K here K is called the eddy viscosity coefficient. 9

38 The negative sign ensures that a positive vertical shear yields donard momentum flux. 9

39 The negative sign ensures that a positive vertical shear yields donard momentum flux. This closure scheme is often referred to as K-theory. Alternatively, e may call it the Flux-gradient theory. 9

40 The negative sign ensures that a positive vertical shear yields donard momentum flux. This closure scheme is often referred to as K-theory. Alternatively, e may call it the Flux-gradient theory. We ill assume that K is constant. Thus, for example, u ) = K u ) = K 2 u 2 9

41 The negative sign ensures that a positive vertical shear yields donard momentum flux. This closure scheme is often referred to as K-theory. Alternatively, e may call it the Flux-gradient theory. We ill assume that K is constant. Thus, for example, u ) = K u ) = K 2 u 2 The momentum equations then become fv v G ) K 2 u 2 = 0 +fu u G ) K 2 v 2 = 0 In the folloing lecture, e ill use these equations to model the Ekman Layer. 9

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