CHAPTER 4: ATMOSPHERIC MOTIONS and TRANSPORT

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CHAPTER 4 ATMOSPHERIC MOTIONS and TRANSPORT

• WHAT ARE THE FORCES BEHIND ATMOSPHERIC
  CIRCULATION?
• Global Circulation as a Giant Sea
  Breeze.Concepts Pressure Gradient Force
  visualizing pressure with isobars
• Introduction to the Coriolis Force (with a
  supporting role played by angular momentum).We
  want to explain circulation patterns like these,
  which take place over large enough scales that
  the rotation of the earth has an effect on moving
  air parcels

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CHAPTER 4 ATMOSPHERIC TRANSPORT

• Forces in the atmosphere
• Gravity
• Pressure-gradient
• Coriolis
• Friction

to R of direction of motion (NH) or L (SH)
Equilibrium of forces
In vertical barometric law In horizontal
geostrophic flow parallel to isobars
gp
P
v
P DP
gc
In horizontal, near surface flow tilted to
region of low pressure
gp
P
v
gf
P DP
gc
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Illustration of the Coriolis force. (Left
panel). An observer sitting on the axis of
rotation (North Pole) launches a projectile at
the target. The curved arrow indicates the
direction of rotation of the earth. (Right
panel) The projectile follows a straight-line
trajectory, when viewed by an observer in space,
directed towards the original position of the
target. However, observers and target are
rotating together with the earth, and the target
moves to a new position as the projectile travels
from launch to target. Since observers on earth
are not conscious of the fact that they and the
target are rotating with the planet they see the
projectile initially heading for the target, then
veering to the right. The Coriolis force is a
fictitious force introduced to the equations of
motion for objects on a rotating planet,
sufficient to account for the apparent pull to
the right in the Northern hemisphere or to the
left in the southern hemisphere.
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The geometry of the earth, showing the distance
from the axis of rotation as a function of the
latitude ? .
r R cos ?(the distance from the axis of
rotation) ? An object on the earths surface at a
high latitude has less angular momentum than an
object on the surface at a low latitude.
v 2?r cos( ? ) / t where t 1 day (86400
seconds). The latitude of Boston is 42? plugging
in numbers, you will find that you are traveling
at a constant speed v 1250 km/h (800 mph!).
1667 km/hr at the equator. Note sound speed
1440 km/hr
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• Coriolis Force (Northern Hemisphere)
• An air parcel (mass) begins to move from the
  Equator toward North Pole along the surface
  of the earth.
• The parcel moves closer to the axis of
  rotation r decreases
• The parcels angular velocity is GREATER THAN
  the angular velocity of the earths surface at
  the higher latitude.

It deflects to the right of its original
trajectory relative to the earths surface. In
the Southern Hemisphere, the parcel would appear
to deflect to the left.
The angular momentum of an object on the earth
due to the planets rotation L mr2 ?. The
requirement that L be conserved implies that, if
r changes, ? must change so as to counteract the
change in r, i.e. ? L/(mr2). For example, if r
were to decrease by factor 2, ? would increase by
factor 4 so that L would stay unchanged. Example
skater "spinning up" note that the skater
really does spin up, by doing work (adding energy
to the spinning motion !
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The air parcel is deflected to the right.
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We thus find in all cases that the Coriolis force
is exerted perpendicular to the direction of
motion, to the RIGHT in the Northern Hemisphere
and to the LEFT in the Southern Hemisphere.
Coriolis acceleration F/m 2?v sin( ? ).
Coriolis acceleration increases as ?
(latitude) increases, is zero at the equator.
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Coriolis acceleration F/m 2?v sin( ? ).
A sample calculation ? 7.5 ? 10-5 s-1 v
10 m/s (36 km/hr, 21.6 mph) l is 42 N
(Boston), sin(l) 0.67 Coriolis acceleration 1
? 10-3 ms-2 The change in velocity is
3.6 m s-1 in 1 hour (3600 s), during which the
parcel travels 36 km in its original
direction. The change in velocity would be 86 m
s-1 in 24 hours if the Coriolis acceleration
stayed the same over the whole period. Obviously
this will not be the case.
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Deflection of an object by the Coriolis
force. ?y ? (?x)2 / v sin(?) (a) A
snowball traveling 10 m at 20 km/h in Boston
(42N) 20 km/hr 5.5 m/s ? 7.5 ? 10-5 s-1
sin (l).67 Dx10 ?y 9.1 ? 10-4 m (b) A
missile traveling 1000 km at 2000 km/h at 42 N. v
555 m/s, Dx1 ? 106 m ?y 9.05 ? 105 m. At
Boston (? 42?N), we find that a snowball
traveling 10 m at 20 km/h is displaced by ?y 1
mm (negligible), but a missile traveling 1000 km
at 2000 km/h is shifted 100 km (important!). Note
the importance of (?x)2
gc 2 W v sin (l) t Dx/v ? Dy ½ gc t2
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low pressure
Pressure gradient force
N
high pressure
S
Motion of an air subjected to a north/south
pressure gradient. Pt. A1, initially at rest
Pt. A3, geostrophic flow. The oscillatory motion
depicted in the previous slide is usually not
observed in the real atmosphere, because
atmospheric mass will be redistributed to
establish a pressure force balanced by the
Coriolis force, and motion parallel to the
isobars.
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• Geostrophy
• For air in motion, not on the equator,
• Coriolis Force ? Pressure gradient force
• Air motion is parallel to isobars

The geostrophic approximation is a simplification
of very complicated atmospheric motions. This
approximation is applied to synoptic scale
systems and circulations, roughly 1000 km. (It is
easiest to think about measuring the pressure
gradient at a constant altitude, although other
definitions are more rigorous. )
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Circulation of air around regions of high and low
pressures in the Northern Hemisphere. Upper
panel A region of high pressure produces a
pressure force directed away from the high. Air
starting to move in response to this force is
deflected to the right (in the Northern
Hemisphere), giving a clockwise circulation
pattern. Lower panel A region of low pressure
produces a pressure force directed from the
outside towards the low. Air starting to move in
response to this force is also deflected to the
right, rotating counter-clockwise. Directions of
rotation of the wind about high or low centers
are reversed in the Southern Hemisphere, as
explained earlier in this chapter.
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The effect of friction around a high pressure
region is to slow the wind relative to its
geostrophic velocity. This causes the pressure
force to slightly exceed the Coriolis force. The
three forces add together as shown in the figure.
Air parcels gradually drift from higher to lower
pressure, in the case shown here, from the center
of a high pressure region outward. An analogous
flow (inward) occurs in a low-pressure region.
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Air converges near the surface in low pressure
centers, due to the modification of geostrophic
flow under the influence of friction. Air
diverges from high pressure centers. At altitude,
the flows are reversed divergence and
convergence are associated with lows and highs
respectively, closing the circulation through
analogous processes noted in the sea breeze
example
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Near surface circulation around a low pressure
areaMarch 7, 2006.
Jet Stream
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THE HADLEY CIRCULATION (1735) global sea breeze

• Explains
• Intertropical Convergence Zone (ITCZ)
• Wet tropics, dry poles
• General direction of winds, easterly in the
  tropics and westerly at higher latitudes
• Hadley thought that air parcels would tend to
  keep a constant angular velocity.
• Meridional transport of air between Equator and
  poles results in strong winds in the longitudinal
  direction.

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Reminder of the sea breeze Distribution of
pressure with altitude. The atmosphere expands
as it is heated over the land, generating
buoyancy and increasing the scale height H. The
rate of pressure decline with altitude is
reduced, therefore at altitude, the pressure is
higher over land than over the adjacent sea,
which causes mass to be transferred to the air
column over the sea. Surface pressure over the
ocean is therefore increased, giving rise to the
distribution of pressure shown in the figure.
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THE HADLEY CIRCULATION (1735) global sea breeze

• Explains
• Intertropical Convergence Zone (ITCZ)
• Wet tropics, dry poles
• General direction of winds, easterly in the
  tropics and westerly at higher latitudes
• Hadley thought that air parcels would tend to
  keep a constant angular velocity.
• Meridional transport of air between Equator and
  poles results in strong winds in the longitudinal
  direction.
• Problems 1. does not account for Coriolis force
  correctly 2. circulation does not extend to the
  poles.

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GLOBAL CLOUD AND PRECIPITATION MAP
(intellicast.com)
11 Oct 2005
Today
Images (3) show colder temperatures as brighter
colors
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Global Circulation and Precipitation as indicated
by satellite images
11 Oct 2005
18 Feb 2007

• ITCZ location, strength
• Wet and Dry season in Amazônia
• Strength and location of polar and subtropical
  jet streams
• Images show colder temperatures as brighter colors

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TROPICAL HADLEY CELL

• Easterly trade winds in the tropics at low
  altitudes
• Subtropical anticyclones at about 30o latitude

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Global winds and pressures, July
warm
warm
cold
cold
land
sea
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CLIMATOLOGICAL SURFACE WINDS AND PRESSURES(July)
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CLIMATOLOGICAL SURFACE WINDS AND
PRESSURES(January)
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warm
cold
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TIME SCALES FOR HORIZONTAL TRANSPORT(TROPOSPHERE)
1-2 months
2 weeks
1-2 months
1 year
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Buoyancy and Lapse RateThe concept of an air
parcel
1) It's a distinct 'block' of air in an
environment of air we often assume it has
volume of 1 m3. It has to be small enough so
that it has uniform properties (T, P, etc).
Its a fictional entity that helps us to think
through a physical process. 2) We can follow it
(as if it were colored with dye) and it stays
together (the same molecules are inside at the
end of a process as there originally). 3) At the
beginning of any of thought exercise, it has the
same characteristics as its surrounding
environment. 4) The parcel can change with time,
by moving, emitting or absorbing heat radiation,
etc --usually in a way we can
describe with equations. 5) The environment of
the parcel can change too. The parcel changes as
a parcel NOT necessarily with the environment.
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Buoyancy force Forces on a solid body immersed
in a tank of water. The solid is assumed less
dense than water and to area A (m2 ) on all
sides. P1 is the fluid pressure at level 1, and
P1x is the downward pressure exerted by the
weight of overlying atmosphere, plus fluid
between the top of the tank and level 2, plus the
object. The buoyancy force is P1 P1x (up ?) per
unit area of the submerged block.
D2
P1x
D1
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The buoyancy force and Archimedes principle.
1. Force on the top of the block P2 ? A r
water D2 A g (A area of top) weight of the
water in the volume above the block 2. Upward
force on the bottom of the block P1 ? A r
water D1 A g 3. Downward force on the bottom of
the block weight of the water in the volume
above block weight of block r water D2 A g
r block (D1 - D2) A g Unbalanced, Upward force
on the block ( 2 3 ) Fb r water D1 A g
r water D2 A r block (D1 - D2) A g r
water g Vblock r block g Vblock (r water r
block) V g
weight of block BUOYANCY
FORCE weight of the water (fluid) displaced by
the block
Volume of the block (D1 D2) A
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VERTICAL TRANSPORT BUOYANCY
Balance of forces
zDz
Object (r)
Fluid (r)
z
Note Barometric law assumed a neutrally buoyant
atmosphere with T T
T
T would produce bouyant acceleration
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Vertical transport Pressure, work, and
Temperature
Question Where does the energy come from for an
air parcel to do this work on the atmosphere?
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Change of atmospheric temperature with altitude (
? pressure ) Atmospheric pressure vs altitude
follows the barometric law, DP-rgDz . Let's
think of an ideal case where the buoyancy forces
and the weight of an air parcel are perfectly
balanced at every altitude, and we neither add or
remove heat as the parcel moves. Because an air
parcel expands as pressure is lowered, it must do
work on the atmosphere as it moves up. The only
source of energy is the motion of the molecules,
and therefore the air parcel must get colder as
it moves up. Two steps are needed to understand
how an air parcel that moves up or down changes
it temperature. Step 1. Figure out the
exchanges of energy between the air parcel and
the environment as the parcel changes its
pressure, using the definition of heat capacity
and Boyle's law. Step 2. Relate this energy
balance to the change in altitude, using the
barometric law.
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Boyle's Law P1V1 P2V2 How can we use Boyle's
Law to determine the change in V when P changes,
for a parcel of air (at constant temperature)?
Boyles Law P2V2 P1V1 P1DV V1DP DPDV
P1V1 P1DV V1DP, or DP/DV P1/V1
P1V1
V
(P1 DP1)( V1 DV1) P2V2
Boyles law
P
P1DP1 P2 V1 DV1 V2
This is an example of how we can understand the
relationship between two properties of air (or
any gas), when both change together, by dividing
the process into very small steps where one
changes while the other is held constant, then
hold the first constant and change the one
initially held fixed.
?V/V1 - ?P/P1
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How do we get energy out of molecular
motion Heat capacity or Specific heat of a
substance The specific heat (Cp) of a substance
is defined as the energy needed to raise the
temperature of 1 kg by 1o K (the "p" denotes that
the pressure is held constant). This energy goes
into the thermal motions of the atoms and
molecules (think of a "golf-ball atmosphere").
The specific heat is a quantity we can measure
for any gas. It tells us how much energy we
extract from the motion of the molecules to lower
the temperature of 1 kg by 1o K. The energy
obtained by lowering T is the negative of this
amount Energy that must be added to a parcel
to change T by DT

m cp DT Energy obtained (total) by lowering
T by DT m cp DT. Work done against (or by)
atmospheric pressure to change the pressure of
an air parcel by DP is given by P DV. ( e.g.,
for the cylinder at the right, Work Dh F P A
Dh P DV ) - m cp DT P DV (basic energy
balance)
P
P
Dh
Piston with top area A, volume Ah
h
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- mcp DT P DV (basic energy balance)
VDP - P DV (Boyles law) gtgt - mcp DT
- VDP DP - g r DZ (Barometric
law) gtgt- mcp DT (Vr) g DZ rV m mass
of parcel We see that for an air parcel moving
vertically in a hydrostatic atmosphere
(barometric law applies), - cp DT
g DZ DT / Dz
-g/cp - 9.8 oK/km
This change in temperature with altitude is
called the "adiabatic lapse rate". cp 1005
J/kg/K g 9.8 m s-2 gtgt - g / cp 9.8 x
10-3 K/m or 9.8 K/km.
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The change in temperature with altitude in the
atmosphere. The example is from 30 degrees north
latitude in summer.
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ATMOSPHERIC LAPSE RATE AND STABILITY
Lapse rate -dT/dz
Consider an air parcel at z lifted to zdz and
released. It cools upon lifting (expansion).
Assuming lifting to be adiabatic, the cooling
follows the adiabatic lapse rate G
z
G 9.8 K km-1
stable
z
unstable

• What happens following release depends on the
  local lapse rate dTATM/dz
• -dTATM/dz gt G e upward buoyancy amplifies
  initial perturbation atmosphere is unstable
• -dTATM/dz G e zero buoyancy does not alter
  perturbation atmosphere is neutral
• -dTATM/dz lt G e downward buoyancy relaxes
  initial perturbation atmosphere is stable
• dTATM/dz gt 0 (inversion) very stable

ATM (observed)
inversion
unstable
T
The stability of the atmosphere against vertical
mixing is solely determined by its lapse rate.
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EFFECT OF STABILITY ON VERTICAL STRUCTURE
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WHAT DETERMINES THE LAPSE RATE OF THE ATMOSPHERE?

• An atmosphere left to evolve adiabatically from
  an initial state would eventually tend to neutral
  conditions (-dT/dz G ) at equilibrium
• Solar heating of surface and radiative cooling
  from the atmosphere disrupts that equilibrium and
  produces an unstable atmosphere

z
z
z
final G
ATM G
ATM
initial
G
T
T
T
Initial equilibrium state - dT/dz G
Solar heating of surface/radiative cooling of
air unstable atmosphere
buoyant motions relax unstable atmosphere back
towards dT/dz G

• Fast vertical mixing in an unstable atmosphere
  maintains the lapse rate to G.
• Observation of -dT/dz G is sure indicator of
  an unstable atmosphere.

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The change in temperature with altitude in the
atmosphere. The example is from 30 degrees north
latitude in summer.
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IN CLOUDY AIR PARCEL, HEAT RELEASE FROM H2O
CONDENSATION MODIFIES G
Wet adiabatic lapse rate GW 2-7 K km-1
z
T
RH 100
GW
Latent heat release as H2O condenses
GW 2-7 K km-1
RH gt 100 Cloud forms
G
G 9.8 K km-1
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Atmospheric temperature and dewpoint for a
typical summer day shows the "planetary boundary
layer" or "atmospheric mixed layer", that
develops as the sun heats the ground in the
daytime. This graph is drawn from actual data
obtained by Harvard's Forest and Atmosphere
Studies group during an experiment (code name
"COBRA") over North Dakota in August, 2000.
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What you see Puffy little clouds, called fair
weather cumulus, occurring over land on a typical
afternoon. The lapse rate in the mixed layer is
approximately adiabatic, and air parcels heated
near the ground are buoyant. Each little cloud
represents the top of a buoyant plume.
(Photograph courtesy University of Illinois Cloud
Catalog).
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Moist pseudo-adiabatic lapse rate Air is heated
by release of latent heat when water condenses T
will decline less rapidly than the dry adiabat
Pressure (Mb)
Temperature (C)
-40 0 40
1000 -9.5 -6.4 -3.0
600 4.2km -9.3 -5.4
200 11.8km -8.6
Dry Adb. -9.8 -9.8 -9.8
Ambient T 15 (-gt35) -13 -58

• -g/(cp l Dw/DT )
• l latent heat of vaporization (J/kg)
  Dw/DTchange in spec humidity/K

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Photo S. Wofsy, Manaus, Brazil, 1987.
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VERTICAL PROFILE OF TEMPERATUREMean values for
30oN, March
Radiative cooling (ch.7)
- 3 K km-1
Altitude, km
2 K km-1
Radiative heating O3 hn e O2 O O O2 M e
O3M
heat
Radiative cooling (ch.7)
Latent heat release
- 6.5 K km-1
Surface heating
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DIURNAL CYCLE OF SURFACE HEATING/COOLINGventilat
ion of urban pollution
z
Subsidence inversion
MIDDAY
1 km
G
Mixing depth
NIGHT
0
MORNING
T
NIGHT
MORNING
AFTERNOON
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SUBSIDENCE INVERSION
typically 2 km altitude
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FRONTS
WARM FRONT
WIND
Front boundary inversion
WARM AIR
COLD AIR
COLD FRONT
WIND
WARM AIR
COLD AIR
inversion
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TYPICAL TIME SCALES FOR VERTICAL MIXING

• Estimate time Dt to travel Dz by turbulent
  diffusion

tropopause
(10 km)
10 years
5 km
1 month
1 week
2 km
planetary boundary layer
1 day
0 km