reynolds averaging - ucd · reynolds averaging we separate the dynamical fields into slowly...
TRANSCRIPT
Reynolds Averaging
Reynolds AveragingWe separate the dynamical fields into slowly varying meanfields and rapidly varying turbulent components.
Reynolds AveragingWe separate the dynamical fields into slowly varying meanfields and rapidly varying turbulent components.
For exampleθ = θ + θ′
Reynolds AveragingWe separate the dynamical fields into slowly varying meanfields and rapidly varying turbulent components.
For exampleθ = θ + θ′
We assume the mean variable is constant over the periodof averaging. By definition, the mean of the perturbationvanishes:
θ′ = 0 θ = θ θθ′ = θ
Reynolds AveragingWe separate the dynamical fields into slowly varying meanfields and rapidly varying turbulent components.
For exampleθ = θ + θ′
We assume the mean variable is constant over the periodof averaging. By definition, the mean of the perturbationvanishes:
θ′ = 0 θ = θ θθ′ = θ
Thus, the mean of a product has two components:
wθ = (w + w′)(θ + θ′)
= (wθ + wθ′ + w′θ + w′θ′)
= wθ + wθ′ + w′θ + w′θ′
= wθ + w′θ′
Thus, we have:
wθ = wθ + w′θ′
2
Thus, we have:
wθ = wθ + w′θ′
We assume that the flow is nondivergent (see Holton §5.1.1):
∂u
∂x+
∂v
∂y+
∂w
∂z= 0
2
Thus, we have:
wθ = wθ + w′θ′
We assume that the flow is nondivergent (see Holton §5.1.1):
∂u
∂x+
∂v
∂y+
∂w
∂z= 0
The total time derivative of u is
du
dt=
∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z+ u
(∂u
∂x+
∂v
∂y+
∂w
∂z
)=
∂u
∂t+
∂uu
∂x+
∂uv
∂y+
∂uw
∂z
2
Thus, we have:
wθ = wθ + w′θ′
We assume that the flow is nondivergent (see Holton §5.1.1):
∂u
∂x+
∂v
∂y+
∂w
∂z= 0
The total time derivative of u is
du
dt=
∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z+ u
(∂u
∂x+
∂v
∂y+
∂w
∂z
)=
∂u
∂t+
∂uu
∂x+
∂uv
∂y+
∂uw
∂z
Writing sums of mean and eddy parts and averaging:
du
dt=
∂u
∂t+
∂
∂x
(u u + u′u′
)+
∂
∂y
(u v + u′v′
)+
∂
∂z
(u w + u′w′
)
2
Repeat:
du
dt=
∂u
∂t+
∂
∂x
(u u + u′u′
)+
∂
∂y
(u v + u′v′
)+
∂
∂z
(u w + u′w′
)
3
Repeat:
du
dt=
∂u
∂t+
∂
∂x
(u u + u′u′
)+
∂
∂y
(u v + u′v′
)+
∂
∂z
(u w + u′w′
)But the mean flow is nondivergent, so
∂
∂x(u u) +
∂
∂y(u v) +
∂
∂z(u w) = u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
3
Repeat:
du
dt=
∂u
∂t+
∂
∂x
(u u + u′u′
)+
∂
∂y
(u v + u′v′
)+
∂
∂z
(u w + u′w′
)But the mean flow is nondivergent, so
∂
∂x(u u) +
∂
∂y(u v) +
∂
∂z(u w) = u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
We define the mean operator
d
dt=
∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z
3
Repeat:
du
dt=
∂u
∂t+
∂
∂x
(u u + u′u′
)+
∂
∂y
(u v + u′v′
)+
∂
∂z
(u w + u′w′
)But the mean flow is nondivergent, so
∂
∂x(u u) +
∂
∂y(u v) +
∂
∂z(u w) = u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
We define the mean operator
d
dt=
∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z
Then we have
du
dt=
∂u
∂t+
∂
∂x
(u′u′
)+
∂
∂y
(u′v′
)+
∂
∂z
(u′w′
)
3
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
4
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
The thermodynamic equation may be expressed in a similarway:
d θ
dt+ w
dθ0
dz+
[∂
∂x
(u′θ′
)+
∂
∂y
(v′θ′
)+
∂
∂z
(w′θ′
)]= 0
4
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
The thermodynamic equation may be expressed in a similarway:
d θ
dt+ w
dθ0
dz+
[∂
∂x
(u′θ′
)+
∂
∂y
(v′θ′
)+
∂
∂z
(w′θ′
)]= 0
The terms in square brackets are the turbulent fluxes.
4
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
The thermodynamic equation may be expressed in a similarway:
d θ
dt+ w
dθ0
dz+
[∂
∂x
(u′θ′
)+
∂
∂y
(v′θ′
)+
∂
∂z
(w′θ′
)]= 0
The terms in square brackets are the turbulent fluxes.
For example, u′w′ is the vertical turbulent flux of zonal mo-mentum, and w′θ′ is the vertical turbulent heat flux.
4
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
The thermodynamic equation may be expressed in a similarway:
d θ
dt+ w
dθ0
dz+
[∂
∂x
(u′θ′
)+
∂
∂y
(v′θ′
)+
∂
∂z
(w′θ′
)]= 0
The terms in square brackets are the turbulent fluxes.
For example, u′w′ is the vertical turbulent flux of zonal mo-mentum, and w′θ′ is the vertical turbulent heat flux.
In the free atmosphere, the terms in square brackets aresufficiently small that they can be neglected.
4
Now we can write the momentum equations as
du
dt− fv +
1
ρ0
∂p
∂x+
[∂
∂x
(u′u′
)+
∂
∂y
(v′u′
)+
∂
∂z
(w′u′
)]= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
[∂
∂x
(v′u′
)+
∂
∂y
(v′v′
)+
∂
∂z
(v′w′
)]= 0
The thermodynamic equation may be expressed in a similarway:
d θ
dt+ w
dθ0
dz+
[∂
∂x
(u′θ′
)+
∂
∂y
(v′θ′
)+
∂
∂z
(w′θ′
)]= 0
The terms in square brackets are the turbulent fluxes.
For example, u′w′ is the vertical turbulent flux of zonal mo-mentum, and w′θ′ is the vertical turbulent heat flux.
In the free atmosphere, the terms in square brackets aresufficiently small that they can be neglected.
Within the boundary layer, the turbulent flux terms arecomparable in magnitude to the remaining terms, and mustbe included in the analysis.
4
In the boundary layer, vertical gradients are generally or-ders of magnitude larger than variations in the horizontal.Thus, it is possible to omit the x and y derivative terms inthe square brackets.
5
In the boundary layer, vertical gradients are generally or-ders of magnitude larger than variations in the horizontal.Thus, it is possible to omit the x and y derivative terms inthe square brackets.
Then the complete system of equations becomes
du
dt− fv +
1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
d θ
dt+ w
dθ0
dz+
∂
∂z
(w′θ′
)= 0
∂p
∂z+ gρ0 = 0
∂u
∂x+
∂v
∂y+
∂w
∂z= 0
5
We note that this system of five equations, for the variables(u, v, w, p, θ), also contains the variables (u′, v′, w′, θ′) in theturbulent fluxes.
6
We note that this system of five equations, for the variables(u, v, w, p, θ), also contains the variables (u′, v′, w′, θ′) in theturbulent fluxes.
Thus, the system is not self contained. To make it soluble,we must make some closure assumption, in which the fluxesare parameterized in terms of the mean fields.
6
We note that this system of five equations, for the variables(u, v, w, p, θ), also contains the variables (u′, v′, w′, θ′) in theturbulent fluxes.
Thus, the system is not self contained. To make it soluble,we must make some closure assumption, in which the fluxesare parameterized in terms of the mean fields.
Atmospheric modellers make various closure assumptions indesigning parameterization schemes for numerical models.
6
We note that this system of five equations, for the variables(u, v, w, p, θ), also contains the variables (u′, v′, w′, θ′) in theturbulent fluxes.
Thus, the system is not self contained. To make it soluble,we must make some closure assumption, in which the fluxesare parameterized in terms of the mean fields.
Atmospheric modellers make various closure assumptions indesigning parameterization schemes for numerical models.
Next, we will consider a simple parameterization of the ver-tical momentum flux, ∂(w′u′)/∂z.
6
Parameterization of Eddy Fluxes
7
Parameterization of Eddy FluxesThe momentum equations will be taken in the form
du
dt− fv +
1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
7
Parameterization of Eddy FluxesThe momentum equations will be taken in the form
du
dt− fv +
1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
We have dropped the overbars on the mean quantities.
7
Parameterization of Eddy FluxesThe momentum equations will be taken in the form
du
dt− fv +
1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
We have dropped the overbars on the mean quantities.
In the free atmosphere, scale analysis shows that the inertialterms and turbulent flux terms are small compared to theCoriolis term and pressure gradient terms.
7
Parameterization of Eddy FluxesThe momentum equations will be taken in the form
du
dt− fv +
1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
dv
dt+ fu +
1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
We have dropped the overbars on the mean quantities.
In the free atmosphere, scale analysis shows that the inertialterms and turbulent flux terms are small compared to theCoriolis term and pressure gradient terms.
This leads us to the relation for geostrophic balance:
u ≈ uG ≡ − 1
fρ0
∂p
∂y, v ≈ vG ≡ +
1
fρ0
∂p
∂x
7
In the boundary layer, it is no longer appropriate to neglectthe turbulent flux terms. However, the inertial terms maystill be assumed to be relatively small.
8
In the boundary layer, it is no longer appropriate to neglectthe turbulent flux terms. However, the inertial terms maystill be assumed to be relatively small.
Thus we get a three-way balance between the forces:
−fv +1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
+fu +1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
8
In the boundary layer, it is no longer appropriate to neglectthe turbulent flux terms. However, the inertial terms maystill be assumed to be relatively small.
Thus we get a three-way balance between the forces:
−fv +1
ρ0
∂p
∂x+
∂
∂z
(w′u′
)= 0
+fu +1
ρ0
∂p
∂y+
∂
∂z
(w′v′
)= 0
8
Using the definition of the geostrophic winds, we write themomentum equations as
−f (v − vG) +∂
∂z
(w′u′
)= 0
+f (u− uG) +∂
∂z
(w′v′
)= 0
9
Using the definition of the geostrophic winds, we write themomentum equations as
−f (v − vG) +∂
∂z
(w′u′
)= 0
+f (u− uG) +∂
∂z
(w′v′
)= 0
To progress, we need some means of representing the tur-bulent fluxes in terms of the mean variables.
9
Using the definition of the geostrophic winds, we write themomentum equations as
−f (v − vG) +∂
∂z
(w′u′
)= 0
+f (u− uG) +∂
∂z
(w′v′
)= 0
To progress, we need some means of representing the tur-bulent fluxes in terms of the mean variables.
The traditional approach to this closure problem is to as-sume that the turbulent eddies act in a manner analogousto molecular diffusion.
9
Using the definition of the geostrophic winds, we write themomentum equations as
−f (v − vG) +∂
∂z
(w′u′
)= 0
+f (u− uG) +∂
∂z
(w′v′
)= 0
To progress, we need some means of representing the tur-bulent fluxes in terms of the mean variables.
The traditional approach to this closure problem is to as-sume that the turbulent eddies act in a manner analogousto molecular diffusion.
Thus, the flux of momentum is assumed to be proportionalto the vertical gradient of the mean momentum. Then wecan write
u′w′ = −K
(∂u
∂z
)where K is called the eddy viscosity coefficient.
9
The negative sign ensures that a positive vertical shearyields downward momentum flux.
9
The negative sign ensures that a positive vertical shearyields downward momentum flux.
This closure scheme is often referred to as K-theory. Alter-natively, we may call it the Flux-gradient theory.
9
The negative sign ensures that a positive vertical shearyields downward momentum flux.
This closure scheme is often referred to as K-theory. Alter-natively, we may call it the Flux-gradient theory.
We will assume that K is constant. Thus, for example,
∂
∂z
(w′u′
)= − ∂
∂z
(K
∂u
∂z
)= −K
∂2u
∂z2
9
The negative sign ensures that a positive vertical shearyields downward momentum flux.
This closure scheme is often referred to as K-theory. Alter-natively, we may call it the Flux-gradient theory.
We will assume that K is constant. Thus, for example,
∂
∂z
(w′u′
)= − ∂
∂z
(K
∂u
∂z
)= −K
∂2u
∂z2
The momentum equations then become
−f (v − vG)−K∂2u
∂z2= 0
+f (u− uG)−K∂2v
∂z2= 0
In the following lecture, we will use these equations to modelthe Ekman Layer.
9
End of §5.3
10