space science: atmospheres part- 8 read ch chap. 2 depl 4.5 general circulation hadley cell rotating...
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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
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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%
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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
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Sea Breeze
Land Breeze
clouds
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T
z
LAND SEA
cloud
Early am
Noon
Evening
Night
Sea Breeze
Land Breeze
] - [ ) T / z g ( z w −=&&
Buoyancy Equation
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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
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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
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DAY – NIGHT
VENUS ROTATION PERIOD ~ 243 days (retrograde)
Night Day
Tropospheric Winds ~ 4 day rotation Also thermospheric Winds (Cyclostrophic )
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Hadley Cells on Earth
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Effect of Rotation
H = High Pressure; L= low Pressure ITCZ =Intertropical Convergence Zone
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Dominate by circulation cells and zonal winds
marked by different colored cloud layers
Black small circle is Io
Jupiter’s Atmosphere
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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
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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
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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
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€
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.)
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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
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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'
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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
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€
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
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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)
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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
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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
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€
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
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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 '
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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’
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Simple Picture
Coriolis ‘Force’
Startv
Ω
Ωv
v
Observer
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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
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Coriolis H
-Pressure Gradient = 2 ρ v x Ω
-counter clockwise: cyclonic-clockwise: anti cyclonic
FORCESSteady flow
Pressure grad.
OPPOSITE ABOUT A LOW
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Near surface pressuressuggest flow pattern
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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
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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
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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
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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
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GRADIENT WIND North ( f > 0 )
Regular
p.
p.
p.
c. c.
c.
cor.
cor.
cor.
Anamolous
L
L
H
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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