Dynamic Meteorology: lecture 3 - staff.science.uu.nl

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Dynamic Meteorology: lecture 3 Working on project 1 Any other questions (rooms 201 & 205 BBG) Next lecture: Wednesday 16 September 2020, 13:15-15:00 (online) Next: first tutorial on Friday 18 Sep. 11:00-12:45 : Buoyancy-oscillations and hydrostatic instability Brunt-Vaisala frequency CAPE Introduction to project 1 (problem 1.6) Sections 1.3-1.5 and Box 1.5 (see Blackboard / Assignments) 64

Transcript of Dynamic Meteorology: lecture 3 - staff.science.uu.nl

Page 1: Dynamic Meteorology: lecture 3 - staff.science.uu.nl

Dynamic Meteorology: lecture 3

Working on project 1 Any other questions

(rooms 201 & 205 BBG)

Next lecture: Wednesday 16 September 2020, 13:15-15:00 (online)

Next: first tutorial on Friday 18 Sep. 11:00-12:45:

Buoyancy-oscillations and hydrostatic instability Brunt-Vaisala frequency CAPE Introduction to project 1 (problem 1.6)

Sections 1.3-1.5 and Box 1.5

(see Blackboard / Assignments)

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Page 2: Dynamic Meteorology: lecture 3 - staff.science.uu.nl

Analysis of the stability of hydrostatic balance employing the “parcel method”

Last week (lecture 2) (Section 1.5)

An air parcel at z=z* has a potential temperature,

θ =θ *Potential temperature of this air parcel is conserved!

Assume that in the environment:

θ0 = θ *+ dθ0dz

δz

: “lapse rate” of the potential temperature

dθ0dz

δz ≡ z − z*

What happens when the vertical position of the air parcel is perturbed?

d2zdt2

= g θ 'θ0

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Page 3: Dynamic Meteorology: lecture 3 - staff.science.uu.nl

Analysis of the stability of hydrostatic balance employing the “parcel method”

d2zdt2

= g θ 'θ0

An air parcel at z=z* has a potential temperature,

θ =θ *Potential temperature of this air parcel is conserved!

Previous slide:

θ 'θ0

=θ *−θ *− dθ0

dzδz

θ0=− dθ0dz

δz

θ0Buoyant force is proportional to

d2δzdt2

= − gθ0

dθ0dz

δzTherefore

θ0 = θ *+ dθ0dz

δzEnvironment:

Section 1.5

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Previous slide:

Acceleration due to buoyancy Section 1.5

d2δzdt2

= − gθ0

dθ0dz

δz ≡ −N2δz

N2 ≡ gθ0

dθ0dz

(1.29)

(1.31)

N: Brunt-Väisälä-frequency

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Acceleration due to buoyancy

The solution to eq. 1.29:

δz = exp ±iNt( )

N2 = gθ0

dθ0dz

< 0If Exponential growth instability

N2 = gθ0

dθ0dz

> 0If oscillation stability

Section 1.5

d2δzdt2

= − gθ0

dθ0dz

δz ≡ −N2δz

N2 ≡ gθ0

dθ0dz N: Brunt-Väisälä-frequency

(1.29)

(1.31)

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Stability of hydrostatic balance

dθ0dz

< 0If exponential growth instability

dθ0dz

> 0If oscillation stability

Section 1.5

depends on sign of vertical gradient of potential temperature

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Brunt-Väisälä frequency

N 2 ≡ gθ0

dθ0dz

Extra problem:

Demonstrate that the Brunt-Väisälä frequency is constant in an isothermal atmosphere.

What is the typical time-period of a buoyancy oscillation in the isothermal lower stratosphere?

N is about 0.01-0.02 s-1

Associated period is

2πN

≈ 300 − 600 s

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Page 8: Dynamic Meteorology: lecture 3 - staff.science.uu.nl

Mountain induced vertical oscillations

manifestation of static stability

manifestation of static instability

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Cumulus clouds

Figure 1.26: (Espy, 1841) (Leslie Bonnema, 2011)

manifestation of static instability

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Convective Available Potential Energy (CAPE)

dwdt

≈ w dwdz

= Bg.

Assuming a stationary state and horizontal homogeneity we can write:

B ≡ θ 'θ0

B = buoyancy

or

wdw = Bgdz.

Box 1.5, page 39-43

dwdt

= g θ 'θ0

≡ BgGoverned by:

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Integrate this equation from a level z1 to a level z2. An air parcel, starting its ascent at a level z1 with vertical velocity w1, will have a velocity w2 at a height z2 given by

w22 = w1

2 + 2 ×CAPE,

CAPE ≡ g Bdzz1

z2∫ .

wdw = Bgdz.

Convective Available Potential Energy (CAPE)

What is CAPE in the model-problem in project 1 (problem 1.6)? What is the associated maximum value of w? Does your model reproduce this value?

B ≡ θ 'θ0

Box 1.5, page 39-43

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Updraughts in cumulus clouds

The vertical motion in a fair weather cumulus cloud 1.5 km deep, over a track about 250 m below cloud top

The relatively sharp downdraughts at the edge of the cumulus cloud are a typical feature of cumulus clouds. What effect is responsible for these downdraughts?

Box 1.5, page 39-43

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Updraughts in cumulus clouds

The vertical motion in (b) a cumulo-nimbus cloud more than 10 km deep, over a track 6 km above the ground

Upward velocities of 50 m/s ! What value of CAPE is required to get

this upward velocity?

Box 1.5, page 39-43

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What theoretical concepts do you need to know?

Introduction to Project 1

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Parcel model of a buoyancy oscillation in a stratified environment

Construct a numerical model (in e.g. Python; see next slide), which calculates the vertical position and vertical velocity of a dry air-parcel, which is initially at rest just above the earth’s surface. Initially this air parcel is warmer than its environment. The governing equations are

The potential temperature of the environment of the air parcel is

Project 1: problem 1.6, page 53-54, lecture notes

dzdt

= w; dwdt

= g θ 'θ0

; dθdt

= 0 .

θ0 =θ0 0( ) + Γz.

You start with a case in which the lapse rate is constant. In the second case the lapse rate in the environment of the air parcel is varies with height (see the lecture notes)

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Reference for Python language: https://www.python.org

Book about use of Python in Atmospheric Science: http://www.johnny-lin.com/pyintro/

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Dynamic Meteorology: lecture 4 Next lecture: Wednesday 23 September 2020, 13:15-15:00 (online)

Relative humidity, mixing ratio Clausius Clapeyron equation Distribution of water vapour Precipitable water Scale height Dew point temperature Lifted condensation level

Moisture in the atmosphere and its effect (parts B and C)

Working on project 1 Any other questions

Fri. 18 Sep. 2020 (rooms 201 & 205 BBG) Tutorial 1

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Radia%veequilibriumtemperature

Seeboxes1.1-1.4(youdonotneedtostudytheseboxesfortheexam)

Inan“idealatmosphere”(wellmixed,onegreenhousegas,Solarradia=onabsorbedonlybyearth’ssurface):

Radia=veequilibriumtemperatureis(accordingtothesolu=onofSchwarzschild’sequa=on:

T = Q2σ

σ aqa pg

+1⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎧ ⎨ ⎩

⎫ ⎬ ⎭

1/4

(eq.1.27a)

AbsorbedSolarradia=on

Stefan-Boltzmannconstant

absorp=oncross-sec=onspecificconcentra=onoftheabsorber(CO2)

Appendix

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δ s = 1.81�

T = Q2σ

δ +1( )⎧ ⎨ ⎩ ⎫ ⎬ ⎭

1/4

δ = σ aqa p /gOp=calpath:

δ s = σ aqa ps /gOp=calpathatsurface:

Surfacepressure

σ a = 0.3 m2kg-1; ra = 390 ppmvαp = 0.3;

S0= 1366 W m-2;ps = 1000 hPa

Radia%veequilibriumtemperatureAppendix

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δ s = 1.81�

T = Q2σ

δ +1( )⎧ ⎨ ⎩ ⎫ ⎬ ⎭

1/4

Radia%veequilibriumtemperatureItiseasytoshowthatthissolu=oncorrespondstoasta=callystablestra=fica=oninwhichpoten=alincreaseswithheighteverywhere.

Appendix

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