Space Science: AtmospheresPart- 8

Read CH Chap. 2dePL 4.5

GENERAL CIRCULATION HADLEY CELL ROTATING FRAME CORIOLIS FORCE TOTAL TIME DERIVATIVE GEOSTROPHIC FLOW GRADIENT WIND CYCLOSTROPHIC FLOW INERTIAL FLOW CYCLONIC

General Circulation

DIFFERENTIAL HEATING => FLOW Both longitudinal and latitudinal

GlobalPercentage Heat Input Per Day E V* M JdT/Te 0.68% 0.36% 38% 1.5x10-3 %*If one use the planet rotation rate then the ratio becomes >100%

Horizontal Pressure Differences

In troposphere T ~ [Ts - z](p/po) = (T/To); x = /(-1); =cp/cv

 

pH

Low Lapse

pL

High Lapse

pH pL

=>Z

Often called a thermal wind General picture: Circulation CellsLocal Scale : Land / Sea BreezesPlanetary Scale : Hadley Cell

remember: d2z/dt2 ~ - (gz/T)[ w - ]

po po

Sea Breeze

Land Breeze

clouds

T

z

LAND SEA

cloud

Early am

Noon

Evening

Night

Sea Breeze

Land Breeze

] - [ ) T / z g ( z w −=&&

Buoyancy Equation

Mars Troposphereheating and cooling

are directly from surfaceand rates are high

20

30

200

800

200

1200

1600 2400

T T200Rapid

convection200

Rapid surface heating

Like a land/ sea breeze but on a planetary scale          Te is essentially at the surface

VENUS TROPOSPHEREGIANT HADLEY CELL

Clouds

IR

Hot air rises in equatorial regionscooled air falls in polar regions

But Venus also has horizontalday-night circulation!

Therefore, even in absence of rotationthe circulation can be complex

Equator

Pole

DAY – NIGHT

VENUS ROTATION PERIOD ~ 243 days (retrograde)

Night Day

Tropospheric Winds ~ 4 day rotation Also thermospheric Winds (Cyclostrophic )

Hadley Cells on Earth

Effect of Rotation

H = High Pressure; L= low Pressure ITCZ =Intertropical Convergence Zone

Dominate by circulation cells and zonal winds

marked by different colored cloud layers

Black small circle is Io

Jupiter’s Atmosphere

Planet Rotation

Write Newton’s Law in the rotating frame in which we measure positions, speeds

and accelerations

Acceleration of a parcel of air= - Pressure Gradient + Gravity + Coriolis + Centrifugal + Viscosity

Usually obtain approximate solutionsHydrostatic Law = Gravity and Pressure

Compare forces (accelerations)Ekman Number = Viscous / Inertial

Ekman = 1/ReynoldsBoundary Layer --> viscous

Rossby Number = Convection/Coriolis

Rotating Reference Frame: Simple Case

€

Inertial (ˆ i ,ˆ j , ˆ k ); Rotating(ˆ i ',ˆ j ', ˆ k ')

ˆ i ' = ˆ i cosθ + j sinθ

j ' = − ˆ i sinθ + j cosθ

€

vΩ =Ω ˆ k ; θ = Ω t ; Ω = dθ/dt = angular speed

dˆ i '

dt= [−ˆ i sinθ + ˆ j cosθ] Ω

dˆ j '

dt= [−ˆ i cosθ − ˆ j sinθ] Ω

Note : ˆ k × ˆ i '= ˆ j cosθ − ˆ i sinθ

ˆ k × j '= −ˆ j sinθ − ˆ i cosθ,

Therefore

dˆ i '

dt= Ω ˆ k × ˆ i ' =

r Ω × ˆ i '

similarly - -the change in the j ' direction

dˆ j '

dt=

r Ω × j '

Ωv

k

i j

'j

'i

Rotating Reference Frame (Cont.)Force or Momentum Equation Applies in an Inertial Frame

€

m r a = m

dv v

dt =

v F ∑

Inertial frame

v v = ˆ i vx + ˆ j vy + ˆ k vz

Rotating Frame (primes); v Ω = angular speed

v v = ˆ i ' vx ' + ˆ j ' vy ' + ˆ k ' vz '

dv v

dt ⇒ ˆ i '

dvx '

dt +

dˆ i '

dt vx ' , etc.

Have shown dˆ i '

dt =

v Ω × ˆ i '

Therefore,

dv v

dt=

dv v

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′+

r Ω × (ˆ i ' vx '+ˆ j ' vy '+ ˆ k 'vz ') =

dv v

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′+

r Ω ×

r v

€

Therefore, measuring r and v in the rotating frame

r v =

dr r

dt =

dr r

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′ +

r Ω ×

r r

r v =

r v ' +

r Ω x

r r

Where r v ' is the rate of change of

r r

measured in the rotating frame

Similarly

d

v v

dt =

d2v r

dt 2 =

d

dt

dv r

dt

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟′ +

v Ω ×

dv r

dt

= d

r v

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′ +

v Ω ×

r v

Rotating Frame (cont.)

Substitute the Velocity

€

dv v

dt=

d

dt(v v '+

r Ω ×

v r )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

′+

r Ω × (

v v '+

r Ω ×

v r ) ;

r Ω = constant

=d

v v '

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′+

r Ω ×

dv r

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′+

r Ω ×

v v '+

r Ω × (

r Ω ×

r r )

v a =

v a ' + 2

r Ω ×

v v ' +

r Ω × (

r Ω ×

r r )

r Ω × (

r Ω ×

v v ) =

r Ω (

r Ω ⋅

v r ) − (

r Ω ⋅

r Ω )

v r

=r Ω Ω z - Ω2r r

= - Ω2 r R centripetal accel.

v R = (

r r -

r z ) ⊥ component

v a =

v a ' + 2

r Ω ×

v v ' − Ω2

v R

Accel = Accel. Obs. + Coriolis - Centripetal

Ωv

Rv

zv

rv

Acceleration in Inertial Frame has 3 components in a Rotating Frame

€

Momentum (Force) Equation

d

v v

dt=

dv v '

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′ + 2

v Ω ×

v v ' − Ω2

r R

= − 1

ρ∇p +

v g + ν ∇ 2v

v Navier Stokes

{

ν = viscosity; ν ∇ 2v v = viscous force per unit mass

(will examine this form for the viscous force later)

Momentum Equation in the Rotating Frame

d

v v '

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′ = −

1

ρ∇p − 2

v Ω ×

v v ' +

v g e + ν ∇ 2v

v '

r g e =

r g + Ω2(

r R ) ;

r R =

r r -

r z

Combining Centrifugal with gravity -

- - -means gravity is not simply radial,

but then the earth is not quite spherical in any case

Therefore, get one new "force" in the rotating reference

frame - 'Coriolis force'

Continuity Equationvariations in air density

€

∂ρ∂t

= −∇ ⋅(ρ v v )

Rate of change = - divergence of of density the flowIf we follow a parcel of air, then it is useful

to rewrite density changes on left

€

∂ρ∂t

= −ρ ∇ ⋅v v −

v v ⋅∇ρ

∂ρ

∂t+

v v ⋅∇ρ = −ρ ∇ ⋅

v v or

dρ

dt= −ρ ∇ ⋅

v v ; d/dt =

∂

∂t +

r v ⋅∇

For an incompressible fluid

dρ

dt= 0 ; ∇ ⋅

v v = 0

Total time derivative for a parcel of air

d/dt = ∂/∂t + r v ⋅∇ (dePL use

D

Dt)

local rate of change +change due to flow

€

Aside : ADIABATIC LAPSE RATE (again)

Using the total time derivative

Heat Equation

d[q] / dt = Work + Conduction + Sources − Losses

Follow a parcel of air :

Rate of change = local rate of change + flow

d

dt =

∂

∂t + v ⋅ ∇

ρcv

dT

dt = ρcv[∂T/∂t + v ⋅ ∇T]

Repeat what we did before to use cp

Keep only work against gravity

ρ cp

dT

dt = ρ v ⋅ g

OR

ρ cp ∂T

∂t+ v ⋅∇T

⎡ ⎣ ⎢

⎤ ⎦ ⎥ = ρ v ⋅ g

∂T

∂t = 0 ; v ⋅ [ cp ∇T − g ] = 0

Solution cp ∇T = g !

1- D cp ∂T

∂z = − g

In Rotating Coordinate follow an air mass

€

Therefore,

d

v v

dt=

∂v v

∂t+ (

v v ⋅∇)

v v

Simple Euler's Equation

ρd

v v

dt= −∇p + ρ

r g

d

v v

dt= 0 hydrostatic

Navier Stokes : With Viscosity

∂

v v

∂t+ (

v v ⋅∇)

v v = −

1

ρ∇ρ +

r g + ν∇ 2v

v

dynamic viscosity η = ρν

kinematic viscosity ν

ν has dimensions L v (like D)

Write Eqs: in Scaled Variables typically non - dimensional

€

r r ⇒ L

r r

v Ω ⇒ Ω ˆ k

t ⇒ Ω-1 t ; v v ⇒ U

r v ; p ⇒ ρ Ω U L (p)

New r r ,

r v , t, p etc. do not have dimensions

First: Meteorologists use reduced pressure (hydrostatic piece of Eq.)

€

−1

ρ∇p +

v g e = −

1

ρ∇pr

pr = p + ρ Vgravitational potential

{ - ρ

2 (

v Ω ×

v r ) ⋅(

v Ω ×

r r )

centrifugal asa potential energy

1 2 4 4 3 4 4

Momentum Eq:new variables(incompressible)

€

dr v '

dt

⎛

⎝ ⎜

⎞

⎠ ⎟′→

∂v v

∂t+

v v ⋅ ∇

v v = −2

r Ω ×

v v −

1

ρ∇pr + ν∇ 2v

v

→∂

v v

∂t+ ε

v v ⋅ ∇

v v = −2 ˆ k ×

v v −

1

ρ∇pr + E∇ 2v

v (dimensionless)

only two terms have scaling coeficients

Rossby Number ε ≡ convection

coriolis→

U2/L

Ω U=

U

Ω L

Ekman Number E ≡1

Reynolds =

viscous force

rotational inertia force

→ν (U/L2)

Ω U =

ν

Ω L2

“Surface” no longer // to planet’s surfaee

Size Scales

€

Horizontal vel. U ~ 10m/s

Vertical vel. W ~ 10cm/s

Horizontal L ~ 1000km =108cm

Vertical H ~ 10km =106cm

(Δp)H ~ 10mb = 104 dynes/cm2

(Δp)V ~ 1b = 106 dynes/cm2

Ω ~ 10-4 s

g ~ 103cm/s

ρ ~ 1mg/cm3

ν ~ 0.15 cm2 /s

Scales for vertical terms (accelerations : cm/s2)

dW/dt W 2

H 2 Ω U g

(Δp)V

ρH Ω2H

10-3 10-4 10-1 103 103 10

Scales for horizontal terms (accelerations : cm/s2)

dU/dt U2

L 2 Ω U 2 ΩW

(Δp)H

ρL

10-1 10-2 10-1 10-3 10-1

Rossby Number = Finertia / Fcoriolis

ε ≡(U2/L)

ΩU~

10−2

10−1=10−1

Ekman Number = Fviscous / Fcoriolis

ν (U /H 2)

ΩU≈

ν

ΩH2≈

0.1

108≈10−7

Reynolds number ~ Re = inertial/viscous > ~ 6000 turbulent

Here, inertial ~ coriolis : ∴ Re ~ 107 mostly turbulent

€

Typically : E << 1

except near surface ⇒ boundary layer

ε =U

ΩL << 1

Steady Flow (r v ~ 0) ε << 1, E << 1

we obtain the remakably simple result

-∇pr ∝ 2 ˆ k x v v ; ˆ k = planet rotation axis

Geostrophic Approximation

-∇pr ⇒ force

Force ⊥ to flow (like a magnetic field)

OR

r v ⊥ ∇pr

Flow // to isobars

Typically drop the primes,

measure on the rotating planet,

and just use p in examples.

Geostrophic Flow

Geostrophic FlowSimplify – Go back to Regular Units

€

Vertical Equation (largest terms)

0 = −1

ρ

∂p

∂z− g hydrostatic

(actually ∇pr = 0)

Horizontal (largest terms: flow nearly along isobars)

∂

v v '

∂t= −2

v Ω ×

v v '−

1

ρ(∇p)// (used primes again temporarily)

// means in the plane parallel to isobars

€

vv '= uˆ i '+vˆ j '

∂u

∂t= fv −

1

ρ

∂p

∂x

∂v

∂t= − fu −

1

ρ

∂p

∂y

f = 2Ωsinφ

f = 0 at equator

f =1 at north pole

f = −1 at south pole

k

€

ˆ k '

€

ˆ j '

Geostrophic Flow Prob.H 7.1, 7.4, 7.5 (Steady flow)

€

0 = f v − 1

ρ

∂p

∂x

0 = − f u − 1

ρ

∂p

∂y

v = 1

ρf

∂p

∂x u = −

1

ρf

∂p

∂y

Note : From boats on ocean : measure p

use equation and get ⇒ r v below surface!

Non-rotating

Rotating

H

L

H

L

i’

j ’

H L

i’

j’

Simple Picture

Coriolis ‘Force’

Startv

Ω

Ωv

v

Observer

Geostrophic Flow (cont.) (approach to steady state)

€

Δu = -1

ρ

∂p

∂x

⎡

⎣ ⎢

⎤

⎦ ⎥Δt

€

Calling c = 1

ρ

∂p

∂x ;

∂p

∂y= 0

∂v

∂t= − f u − c ;

∂u

∂t= f v ; c = const

Solve

˙ v = − f ˙ u substitute ˙ u ; ˙ v = − f 2v

like a harmonic oscillator equation!

Becomes circular flow around a high or a low

(we'll change to radial coordinates soon)

Geostrophic (Northern hemisphere)

HL

t=0

HL

L

L

L

Southern HemisphereFlow is Opposite

Coriolis H

-Pressure Gradient = 2 ρ v x Ω

-counter clockwise: cyclonic-clockwise: anti cyclonic

FORCESSteady flow

Pressure grad.

OPPOSITE ABOUT A LOW

Near surface pressuressuggest flow pattern

2 –D Flow: ‘Flat’ Surface Gradient Wind Equation

€

d

v v

dt= −f ˆ k x

v v −

1

ρ(∇p) // : pressure grad. in plane

d

v v

dt= 0, Pure Geostrophic

no acceleration along isobars

Curved Motion

€

r = radius of curvature

ˆ k × ˆ v = ˆ n

€

∂ r

v

∂t= 0 ; Write

dv v

dt= ˆ v

dv

dt+

dˆ v

dtv = ˆ v ˙ v + ˆ n

v2

r

measure along pathlength, s :

1) ˆ v dv

dt= −

1

ρ

∂p

∂s ; along the flow direction

2) ˆ n v2

r± f v = −

1

ρ

∂p

∂n (± curvative)

Flow nearly // to isobars : ignore ˆ v terms

n

€

ˆ v

vv

r

Gradient Wind Equation Ignore Eq. 1:

discuss direction normal to the flow

€

v2

r ± f v = −

1

ρ

∂p

∂n

Case1 : Pure inertial flow

v2

r ± f v = 0

If there is a velocity, the motion is curved

Northern

Clockwiseanticyclonic

Southern

Counterclockcyclonic

Period = ½ period of the focault pendulum

coriolis

n

vn

centrifugal centrifugal

coriolis

CYCLOSTOPHIC FLOWCase 2

centrifugal = - press. grad.

cent.

L

Venus : Polar Vorticity

p.

€

Gradient Wind Eq. v2

r ± f v = −

1

ρ

∂p

∂n

Case 2 : no coriolis force f ~ 0

(near equator or r small or v very large)

v2

r = −

1

ρ

∂p

∂n

GEOSTROPHIC FLOWCentripetal term is zero (e.g., large scale flow)

HL

L

L

L

( anticyclonic ) Nothern Hemisphere Southern Hemisphere (cyclonic)

HL

L

L

€

Gradient Wind Eq. v2

r ± f v = −

1

ρ

∂p

∂n

Case 3 : no centrifugal force, r very large

± f v = −1

ρ

∂p

∂n

GRADIENT WIND North ( f > 0 )

Regular

p.

p.

p.

c. c.

c.

cor.

cor.

cor.

Anamolous

L

L

H

Problems

• Geostrophic Flow Δp = 5 mb between Washington and Charlottesville Find v.

• Cyclostrophic Flow Δp = 5 mb over 10 km Find v.

Inertial Flow at Charlottesville v = 20 m/s r = ? Pressure Force =0