Transcript Part-8 (Monday, April 10, 2006)

Space Science: Atmospheres
Part- 8
Read CH Chap. 2
dePL 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
Global
Percentage Heat Input Per Day
E
V*
M
J
dT/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
High Lapse
Low Lapse
pH
Z
po
pH
=>
pL
pL
po
Often called a thermal wind
General picture: Circulation Cells
Local Scale : Land / Sea Breezes
Planetary Scale : Hadley Cell
remember: d2z/dt2 ~ - (gz/T)[ w - ]
Sea Breeze
clouds
Land Breeze
LAND
Early
am
SEA
z
T
cloud
Sea Breeze
Noon
Evening
Land Breeze
Night
z   ( g z / T ) [ w -  ]
Buoyancy Equation
Mars Troposphere
heating and cooling
are directly from surface
and rates are high
1200
800
20
30
200
200
Rapid surface heating
2400
1600
200
Rapid
convection T
200
T
Like a land/ sea breeze but on a planetary
scale
Te is essentially at the surface
VENUS TROPOSPHERE
GIANT HADLEY CELL
Clouds
Pole
IR
Equator
Hot air rises in equatorial regions
cooled air falls in polar regions
But Venus also has horizontal
day-night circulation!
Therefore, even in absence of rotation
the circulation can be complex
DAY – NIGHT
VENUS ROTATION
PERIOD ~ 243 days (retrograde)
Day
Night
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
Jupiter’s Atmosphere
Dominate by circulation
cells and zonal winds
marked by different colored
cloud layers
Black small circle is Io
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 solutions
Hydrostatic Law = Gravity and Pressure
Compare forces (accelerations)
Ekman Number = Viscous / Inertial
Ekman = 1/Reynolds
Boundary Layer --> viscous
Rossby Number = Convection/Coriolis
Rotating Reference Frame: Simple Case
k̂
Inertial ( ˆi , ˆj, kˆ ); Rotating( ˆi ', ˆj ', kˆ ')
ˆi '  ˆi cos  ˆj sin 


ĵ '
iˆ

ˆj'   ˆi sin   ˆj cos
ĵ
iˆ' ˆ
  k ;    t ;   d/dt = angular speed
d ˆi ' ˆ
 [ i sin   ˆj cos ] 
dt
dˆj '
 [ˆi cos  ˆj sin  ] 
dt
Note :
kˆ  ˆi ' ˆj cos  ˆi sin 
kˆ  ˆj '  ˆj sin   ˆi cos,
Therefore
d ˆi '
  kˆ  ˆi '    ˆi '
dt
similarly - -the change in the ˆj ' direction
dˆj'
   ˆj'
dt
Rotating Reference Frame (Cont.)
Force or Momentum Equation
Applies in an Inertial Frame
dv
m a =m

dt
F
Inertial frame
v  iˆ v x  ˆj v y  kˆ v z
Rotating Frame (primes) ;  angular speed
v  iˆ ' v x '  ˆj ' v y '  kˆ ' v z '
dv
dt

Have shown
ˆ
ˆi ' dv x ' + di ' v ' ,
x
dt
dt
etc.
d ˆi '
  iˆ '
dt
Therefore,






dv dv
dv
ˆ
ˆ
ˆ
     ( i ' v x ' j ' v y 'k' v z ') =     v
dt 
dt dt 

Rotating Frame (cont.)
Therefore, measuring r and v in the rotating frame
dr 
dr
v=
     r
dt 
dt
v = v' + x r
Where v' is the rate of change of r
measured in the rotating frame
Similarly
 d dr 
dv
d2r
dr




  
dt
dt 2
dt
dt dt 
dv 
     v
dt 
Substitute the Velocity


dv d
  (v' r )   (v' r ) ;  constant

dt dt
dv' 
dr 
         v' ( r )
 dt 
dt 
a  a'  2  v'   ( r)


 ( v)  ( r )  ( ) r

R

z

r
  z - 2 r
 - 2 R centripetal accel.
R  (r - z)  component
a 
a'
 2  v'
 2R
Accel = Accel. Obs. + Coriolis - Centripetal
Acceleration in Inertial Frame has
3 components in a Rotating Frame
Momentum (Force) Equation

dv dv' 
    2  v'  2R
dt  dt 
1
  p  g    2 v

Navier Stokes
 = viscosity;   2 v = viscous force per unit mass
(will examine this form for the viscous force later)
Momentum Equation in the Rotating Frame
dv' 
1


p  2  v'  ge    2 v'
 
 dt 

g e  g  2 (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 Equation
variations in air density

   ( v)
t
Rate of change = - divergence of
of density
the flow
If we follow a parcel of air, then it is useful
 to rewrite density changes on left

    v  v  
t

 v        v
or
t
d

    v ; d/dt =
+ v 
dt
t
For an incompressible fluid
d
0 ;
v0
dt
Total time derivative for a parcel of air
D
d/dt = /t + v  
(dePL use
)
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


 v  
dt
t
dT
c v
 c v [T/ t  v  T]
dt
Repeat what we did before to use c
p
Keep only work against gravity
 cp
dT
 v g
dt
OR
T

 c p   v  T   v  g
t

T
 0;
t
v  [ c p T  g ]  0
c p T  g !
Solution
1- D
cp
T
 g
z
In Rotating Coordinate
follow an air mass
Therefore,
dv v

 (v  )v
dt t
Simple Euler' s Equation
dv
  p   g
dt
dv
0
hydrostatic
dt
Navier Stokes : With Viscosity
v
1
 (v  )v     g   2 v
t

dynamic viscosity
  
kinematic viscosity
 has dimensions

L v (like D)
Write Eqs: in Scaled Variables
typically non - dimensional
r
   kˆ
 Lr
t  -1 t ; v  Uv ; p    U L (p)
New r, v, t, p etc. do not have dimensions
First: Meteorologists use reduced pressure
(hydrostatic piece of Eq.)

1
1
 p  g e   p r

pr  p 

V
gravitational
potential
-

2
(  r )  ( r )
centrifugal as
a potential energy
“Surface” no longer // to planet’s surfaee
Momentum Eq:new variables(incompressible)
dv' v
1
2
    v  v  2   v  pr   v
dt  t

v
1
   v  v  2 kˆ  v  pr  E 2 v
t

only two terms have scaling coeficients
Rossby Number
Ekman Number
(dimensionless)
convection
U 2 /L
U
 


coriolis
U L
1
viscous force
E 

Reynolds
rotational inertia force
 (U/L 2 )


=
U
 L2
Size Scales
Horizontal vel.
Vertical vel.
U
W
~ 10m/s
~ 10cm/s
Horizontal L
~ 1000km  10 8 cm
Vertical
~ 10km  10 6 cm
H
(p) H
~
(p) V
~
10mb  10 4 dynes/cm 2
1b  10 6 dynes/cm 2

~ 10 -4 s
g
~ 10 3 cm/s

~ 1mg/cm 3

~ 0.15 cm2 /s
Scales for vertical terms (accelerations : cm/s 2 )
dW/dt
W2
H
2U
10 -3
10 -4
10 -1
g
10 3
(p) V
H
 2H
10 3
10
Scales for horizontal terms (accelerations : cm/s 2 )
dU/dt
U2
L
2U
2 W
(p) H
L
10 -1
10 -2
10 -1
10 -3
10 -1
Rossby Number  Finertia / Fcoriolis
(U 2 /L) 102

~ 1  101
U
10
Ekman Number  Fviscous / Fcoriolis
 (U /H 2 )

0.1
 107
2
8
U
H
10
Reynolds number ~ Re = inertial/viscous


> ~ 6000 turbulent
Here, inertial ~ coriolis : Re ~ 10 7 mostly turbulent
Geostrophic Flow
Typically :
E  1
except near surface  boundary layer
U

 1
L
Steady Flow (vÝ~ 0)
  1, E  1
we obtain the remakably simple result
- p  2 kˆ x v; kˆ = planet rotation axis
r
Geostrophic Approximation
- p r  force
Force  to flow (like a magnetic field)
OR
v  p r
Flow // to isobars
Typically drop the primes,
measure on the rotating planet,
and just use p in examples.
Geostrophic Flow
Simplify – Go back to Regular Units
Vertical Equation (largest terms)
1 p
g
hydrostatic
 z
(actually pr  0)
Horizontal (largest terms : flow nearly along isobars)
0  
v'
1
 2  v' (p) // (used primes again temporarily)
t

// means in the plane parallel to isobars
k
ˆj'
v' uiˆ 'vˆj '
kˆ ' u  fv  1 p
t


 x
v
1 p
  fu 
t
 y
f  2sin 
f 0
at equator
f  1 at north pole
f  1 at south pole
Geostrophic Flow
Prob.H 7.1, 7.4, 7.5
(Steady flow)
1 p
 x
1 p
0  fu 
 y
1 p
1 p
v =
u 
f x
f y
Note : From boats on ocean : measure p
0  fv 
use equation and get 
v below surface!
H
Non-rotating
L
j’
Rotating
H
i’
L
j’
i’ H
L
Simple Picture
Coriolis ‘Force’
Start

Observer
v
v

v
Geostrophic Flow (cont.)
(approach to steady state)
 1 p 
u  t
  x 
t=0
H
L
1 p p
Calling c 
;
0
 x y
v
u
f uc ;
 f v ; c  const
t
t
Solve
2
Ý  fuÝ substitute
Ý
Ý
Ý
vÝ
u
;
v


f
v

like a harmonic oscillator equation !
Becomes circular flow around a high or a low
(we' ll change to radial coordinates soon)
Geostrophic (Northern hemisphere)
L
L
H
L
L
Southern Hemisphere
Flow is Opposite
FORCES
Steady flow
Pressure grad.
Coriolis
H
OPPOSITE ABOUT A LOW


-Pressure Gradient = 2  v x 
-counter clockwise: cyclonic
-clockwise: anti cyclonic
Near surface pressures
suggest flow pattern
2 –D Flow: ‘Flat’ Surface
Gradient Wind Equation
dv
1
 f kˆ x v  (p) // : pressure grad. in plane
dt

dv
 0, Pure Geostrophic
dt
no acceleration along isobars
Curved Motion
r
n̂
ˆ
v

v
r  radius of curvature
kˆ  vˆ  nˆ

v
dv
dv dvˆ
v2

 0 ; Write
 vˆ
 v  vˆ vÝ nˆ
t
dt
dt dt
r
measure along pathlength, s :
1)
vˆ
dv
1 p

; along the flow direction
dt
 s
v2
1 p
ˆ
2) n
fv 
( curvative)
r
 n
Flow nearly // to isobars : ignore vˆ terms
Gradient Wind Equation
Ignore Eq. 1:
discuss direction normal to the flow
v2
r
Case1:
1 p
 fv  
 n
Pure inertial flow
v2
 fv  0
r
If there is a velocity, the motion is curved
Northern
Southern
centrifugal
coriolis
coriolis
v̂
centrifugal
Clockwise
anticyclonic
n̂
Counterclock
cyclonic
n̂
Period = ½ period of the focault
pendulum
CYCLOSTOPHIC FLOW
Case 2
centrifugal = - press. grad.
cent.
p.
L
v2
1 p
Gradient Wind Eq.
 fv  
r
 n
Case 2: no coriolis force f ~ 0
(near equator or r small or v very large)
v2
r
1 p
 
 n
Venus : Polar Vorticity
GEOSTROPHIC FLOW
Centripetal term is zero
(e.g., large scale flow)
L
L
H
L
L
( anticyclonic )
Nothern Hemisphere
Southern Hemisphere
(cyclonic)
L
H
L
L
v2
1 p
Gradient Wind Eq.
 fv  
r
 n
Case 3: no centrifugal force, r very large
1 p
 fv  
 n
GRADIENT WIND
North ( f > 0 )
Regular
cor.
L
p.


p. c.
cor.
H 
c.
Anamolous
p.

L 
c.
cor.
Problems
• Geostrophic Flow
p = 5 mb between
Washington and Charlottesville
Find v.
• Cyclostrophic Flow
p = 5 mb over 10 km
Find v.
Inertial Flow at Charlottesville
v = 20 m/s
r=?
Pressure Force =0