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