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Dynamic Meteorology 2015 (lecture 6)

(a.j.vandelden@uu.nl) (http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm)

Topics Stability of geostrophic balance, Inertial oscillations, inertial instability, thermal wind balance, stability of thermal wind balance, baroclinic instability (section 1.18-1.21)

Sections 1.18-1.21 of the lecture notes

Mean zonal flow: polar vortices

http://www.ecmwf.int/research/era/ERA-40_Atlas/

Figure 1.40

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Geostrophic and hydrostatic balance = thermal wind balance

€

θ∂Π∂y

+ fug = 0

€

θ∂Π∂z

+ g = 0and in hydrostatic balance:

€

∂ug∂z

≈ −gfθ∂θ∂y

Thermal wind balance

Check this using scale analysis

Vertical gradient of the geostrophic wind is related to a horizontal gradient in potential temperature

!!!!

If zonal flow is in geostrophic balance:

Manifestation of the thermal wind

FIGURE 1.39: Isentropes (thin solid lines, labelled in Kelvin) and isotachs (isopleths of the velocity) (dashed lines, m s-1) in a vertical section through a cold front. The y-coordinate is positive towards the left. Heavy lines mark the tropopause and frontal boundaries. The section extends approximately 1200 km in the horizontal direction (Palmen, E. and C.W. Newton, 1969: Atmospheric Circulation Systems. Academic Press, 603 pp).

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geostrophic zonal flow

Assume a “geostrophic” zonal flow, ug:

€

fug ≡ −θ∂Π∂y

Assume: no pressure gradient in x-direction Curvature of the flow is also neglected

geostrophic zonal flow

Assume a “geostrophic” zonal flow, ug:

€

fug ≡ −θ∂Π∂y

€

dudt

= fv = f dydt

Equations of motion:

€

dvdt

= −θ∂Π∂y

− fu = f ug −u( )€

du = fdyx-component:

y-component:

Assume: no pressure gradient in x-direction Curvature of the flow is also neglected

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geostrophic zonal flow

Equations of motion:

€

dvdt

= f ug −u( )€

d u − fy( ) ≡ dM = 0x-component:

y-component:

Assume a “geostrophic” zonal flow, ug:

€

fug ≡ −θ∂Π∂y

Assume: no pressure gradient in x-direction Curvature of the flow is also neglected

Absolute momentum per unit mass:

€

M ≡ u − fy

Stability analysis: parcel method Air parcel at

€

y = y0 moves to

€

d u − fy( ) = 0 :Since

€

y = y0 +δy

€

u y0 +δy( )− fδy = u y0( )

€

M ≡ u − fy

Isopleths of M in a meridional/vertical cross section of the atmosphere in the middle latitudes Figure: Markowski and Richardson, 2010

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Stability analysis: parcel method

€

ug y0 +δy( ) ≈ ug y0( ) +∂ug∂y

y0( )δy

€

dv y0 +δy( )dt

= f ug y0 +δy( )−u y0 +δy( )( ) = f ug y0( ) +∂ug∂y

δy−u y0( )− fδy%

& '

(

) *

Geostrophic wind environment:

€

dvdt

= f ug −u( )Because

€

M ≡ u − fy

Isopleths of M in a meridional/vertical cross section of the atmosphere in the middle latitudes Figure: Markowski and Richardson, 2010

Air parcel at

€

y = y0 moves to

€

d u − fy( ) = 0 :Since

€

y = y0 +δy

€

u y0 +δy( )− fδy = u y0( )

Stability of geostrophic balance

€

dv y0 +δy( )dt

= f ug y0 +δy( )−u y0 +δy( )( ) = f ug y0( ) +∂ug∂y

δy−u y0( )− fδy%

& '

(

) *

or

€

d2δydt2

= f∂ug∂y

− f%

& '

(

) * δy

Previous slide:

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Equation governing meridional acceleration: inertial stability

€

d2δydt2

= − f f −∂ug∂y

%

& '

(

) * δy

The solution:

€

δy = Re Aexp −iFt( )[ ] €

F ≡ f f −∂ug∂y

%

& '

(

) *

If F2<0 then inertial instability; If F2>0 then inertial stability (inertial oscillation)

Inertial frequency, F is about 10-4 s-1

€

d2δydt2

= −F2yor with

Previous slide:

Inertial frequency is a function of vorticity!

€

F ≡ f f −∂ug∂y

%

& '

(

) *

Inertial frequency, F is about 10-4 s-1 (in middle latitudes)

€

ζ = f −∂ug∂y

vorticity:

relative vorticity: planetary vorticity:

Inertial Instability if vorticity<0 (anticyclonic flow)

(section 1.22)

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Inertial stability

Parcel method:

Air parcel is in geostrophic balance at A. It is pushed toward position B. Pressure distribution remains fixed. Because momentum is conserved this leads to a change in zonal velocity. Therefore, also a change in Coriolis force in the meridional direction, which either drives the parcel away from equilibrium or not, depending on the pressure gradient force at the new position, B.

Figure: Markowski and Richardson, 2010

Inertial stability

Figure: Markowski and Richardson, 2010

How far must an air parcel, which is initially at rest at the equator, be displaced poleward in the meridional direction in order to acquire a zonal velocity of 10 m s-1? Pressure in the environment of the air parcel is constant (Repeat this exercise for an air parcel that is initially at rest at 50°N).

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Stability of thermal wind

Figures: Markowski and Richardson, 2010

Air parcel at

€

y, z( ) = y0, z0( )

moves to

€

y, z( ) = y0 +δy, z0 +δz( )

isentrope

M=constant

Section 1.20

Geostrophic wind in the environment:

€

ug y0 +δy, z0 +δz( ) ≈ ug y0, z0( ) +∂ug∂y

δy+∂ug∂z

δz

Zonal velocity of the air parcel:

€

u y0 +δy, z0 +δz( ) ≈ ug y0, z0( ) + fδy

Meridional acceleration is governed by:

€

dvdt≈ f ug −u( )

Air parcel at

€

y, z( ) = y0, z0( ) moves to

€

y, z( ) = y0 +δy, z0 +δz( )

Stability of thermal wind Section 1.20

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Geostrophic wind in the environment:

€

ug y0 +δy, z0 +δz( ) ≈ ug y0, z0( ) +∂ug∂y

δy+∂ug∂z

δz

Zonal velocity of the air parcel:

€

u y0 +δy, z0 +δz( ) ≈ ug y0, z0( ) + fδy

Acceleration:

€

dvdt≈ f ug −u( )

Air parcel at

€

y, z( ) = y0, z0( ) moves to

€

y, z( ) = y0 +δy, z0 +δz( )

€

dvdt≈ f

∂ug∂y

δy− fδy+∂ug∂z

δz&

' (

)

* +

Stability of thermal wind Section 1.20

Stability of thermal wind acceleration:

€

dvdt≈ f

∂ug∂y

δy− fδy+∂ug∂z

δz&

' (

)

* + = f

∂ug∂y

− f +∂ug∂z

δzδy&

' (

)

* +

&

' (

)

* + δy

Assume that parcel moves along an isentrope:

€

dθ =∂θ∂ydy+

∂θ∂zdz = 0

€

dzdy⎛

⎝ ⎜

⎞

⎠ ⎟ θ

= −∂θ /∂y∂θ /∂z

=fθg∂ug /∂z∂θ /∂z

€

∂ug∂z

≈ −gfθ∂θ∂y

with thermal wind balance:

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Stability of thermal wind Acceleration along isentrope:

€

dvdt≈ f

∂ug∂y

δy− fδy+∂ug∂z

δz&

' (

)

* + = f

∂ug∂y

− f +∂ug∂z

δzδy&

' (

)

* +

&

' (

)

* + δy

€

dvdt

=d2δydt2

≈ f∂ug∂y

− f +fθg

∂ug /∂z( )2

∂θ /∂z

'

(

) )

*

+

, , δy

Air parcel follows isentrope, therefore

€

δzδy

=dzdy⎛

⎝ ⎜

⎞

⎠ ⎟ θ

= −∂θ /∂y∂θ /∂z

=fθg∂ug /∂z∂θ /∂z

Next slide

Stability of thermal wind

€

d2δydt2

= − f −∂ug∂y

+ f − fθg

∂ug /∂z( )2

∂θ /∂z

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ δy

Acceleration along isentrope:

Stability is governed by sign of factor in front of δy on right hand side. Parcel will return to its equilibrium position if

(1.145a)

€

F2 ≡ f f −∂ug∂y

%

& '

(

) *

Remember:

or: €

f −∂ug∂y

+ f − fθg

∂ug /∂z( )2

∂θ /∂z

%

&

' '

(

)

* *

> 0

€

F2 − f 2∂ug /∂z( )2

N 2

$

%

& &

'

(

) )

> 0

€

N 2 ≡gθ∂θ∂z

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Thermal wind is stable if:

Problem: Show that this stability-criterion can be expressed in terms of the slope of potential temperature surfaces (isentropes) relative to the slope of momentum surfaces as,

€

F2 − f 2∂ug /∂z( )2

N 2

$

%

& &

'

(

) )

> 0

€

∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ θ

<∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ M

Thermal wind is unstable if:

€

F2 − f 2∂ug /∂z( )2

N 2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

< 0

θ Μ60 m s-1

50 m s-1

40 m s-1

303 K 300 K

297 K ¢

Are parcel will always try to match its potential temperature and its momentum with the potential temperature and momentum in the environment.

“push from A to B” Equilibrium position

€

∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ θ

>∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ M

or

see the figure below

B

A

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“Baroclinic instability”

Figure: Markowski and Richardson, 2010

Instability-criterion:

€

F2 < f 2∂ug /∂z( )2

N 2

€

∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ θ

>∂z∂y⎛

⎝ ⎜

⎞

⎠ ⎟ M

or

Zonal mean isentropes

FIGURE 1.26: zonal-mean, annual-mean potential temperature distribution (ERA-40 reanalysis) and possible baroclinically unstable motions (double arrow)

330 K

300 K

350 K

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Energy conversions in baroclinic motion

Exchange of air parcels

Section 1.21

Energy conversions in baroclinic motion

ΔPE = M1gz2 +M2gz1( )− M1gz1 +M2gz2( ) =M1g z1 − z2( ) M2

M1

−1#

$%

&

'(

Change in potential energy:

Exchange of air parcels

Section 1.21

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Energy conversions in baroclinic motion

Change in potential energy is:

€

ΔPE = M1g z1 − z2( )θ1 −θ2θ2

⎛

⎝ ⎜

⎞

⎠ ⎟

Exchange of air parcels

Section 1.21

M2

M1

=ρ2V2ρ1V1

=ρ2p1

1/γ

ρ1p21/γ =

θ1θ2

Ratio of masses is:

C=constant€

p1V1γ = p2V2

γ (Problem 1.2)

With:

€

θ =Cρp1/γ Show this

Energy conversions in baroclinic motion

Change in potential energy:

Section 1.21

Exchange of air parcels

€

ΔPE < 0

€

ΔPE > 0

€

ΔPE = M1g z1 − z2( )θ1 −θ2θ2

⎛

⎝ ⎜

⎞

⎠ ⎟

Increase of KE Decrease of KE (kinetic energy)

€

θ1 >θ2 z1>z2

z1<z2

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Typical sky in a region of slantwise upward motion due to baroclinic instability

Jan Sluijters 1881-1957 “Landschap in tegenlicht”

Next week

Wednesday: Problems on the seabreeze (exercise session 4) Problems in slides of today’s lecture Evaluation of the weather prediction by team “Sunshine”

Friday: Continue with section 1.21 Introduction to circulation, vorticity and potential vorticity (section 1.22)

Story is continued next week