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CHAPTER 4: ATMOSPHERIC MOTIONS and TRANSPORT
WHAT ARE THE FORCES BEHIND ATMOSPHERIC CIRCULATION?
1. Global Circulation as a Giant Sea Breeze.
Concepts: Pressure Gradient Force; visualizing pressure with isobars
2. 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:
CHAPTER 4: ATMOSPHERIC TRANSPORT
Forces in the atmosphere:
• Gravity g
• Pressure-gradient γ p   1/  P
• Coriolis  c  2v sin  to R of direction of motion (NH) or L (SH)
• Friction γ f  kv


Equilibrium of forces:
In vertical: barometric law
p
P
In horizontal: geostrophic flow parallel to isobars
v
c
P + DP
In horizontal, near surface: flow tilted to region of low pressure
p
f
v
c
P
P + DP
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.
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)
R cos 
R

R
 An object on the earth’s
surface at a high latitude has
less angular momentum than
an object on the surface at a
low latitude.
Rotation axis
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
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 parcel’s angular velocity is GREATER THAN the angular
velocity of the earth’s surface at the higher latitude.
It deflects to the right of it’s
original trajectory relative to
the earth’s 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 planet’s 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 !
The air
parcel is
deflected to
the right.
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.
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);  is 42 N (Boston), sin()= 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.
Deflection of an object by the Coriolis force.
Dy = [  (Dx)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 ()=.67; Dx=10
Dy = 9.1  10-4 m
(b) A missile traveling 1000 km at 2000 km/h at 42 N.
v = 555 m/s, Dx=1  106 m; Dy = 9.05  105 m.
At Boston ( = 42N), we find that a snowball traveling 10 m at 20 km/h is
displaced by Dy = 1 mm (negligible), but a missile traveling 1000 km at 2000
km/h is shifted 100 km (important!). Note the importance of (Dx)2
c = 2  v sin () ; t = Dx/v  Dy = ½ c t2
low pressure
N
Pressure
gradient
force
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.
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. )
Vgeostrophic=
1
P
2 sin(  ) X
Vg


Dx
DP
geostrophic wind (m/s)
7.29 10-5 radian/s
latitude
distance (m)
pressure diff. (N/m2)
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.
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 lowpressure region.
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
Near surface circulation around
a low pressure area—March 7,
2006.
Jet Stream
THE HADLEY CIRCULATION (1735): global sea breeze
COLD
HOT
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.
COLD
land
Z
ocean
Z
log (P)
log (P)
low P
high P
low P
high P
ocean
land
hot
cold
Fig 5.2
land
Z
ocean
Fig 5.3
log (P)
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.
THE HADLEY CIRCULATION (1735): global sea breeze
COLD
HOT
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.
COLD
Problems: 1. does not
account for Coriolis force
correctly; 2. circulation does
not extend to the poles.
GLOBAL CLOUD AND PRECIPITATION MAP (intellicast.com)
11 Oct 2005
Today
Images (3) show colder temperatures as brighter colors
Global Circulation and Precipitation as indicated by
satellite images
18 Feb 2007
•
•
•
11 Oct 2005
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
TROPICAL HADLEY CELL
• Easterly “trade winds” in the tropics at low altitudes
• Subtropical anticyclones at about 30o latitude
Global winds and pressures, July
cold
warm
warm
cold
land
Z
Pressure anomaly scale (mb)
ln(P)
sea
CLIMATOLOGICAL SURFACE WINDS AND PRESSURES
(July)
CLIMATOLOGICAL SURFACE WINDS AND PRESSURES
(January)
cold
Pressure anomaly scale (mb)
warm
TIME SCALES FOR HORIZONTAL TRANSPORT
(TROPOSPHERE)
1-2 months
2 weeks
1-2 months
1 year
Buoyancy and Lapse Rate
The 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).
It’s 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.
D2
P1x
D1
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.
The buoyancy force and Archimedes principle.
1. Force on the top of the block: P2  A =  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 =  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 =  water D2 A g +  block (D1 - D2) A g
Unbalanced, Upward force on the block ( [2] – [3] ):
Fb =  water D1 A g – [  water D2 A +  block (D1 - D2) A ] g
=  water g Vblock –  block g Vblock = ( water –  block) V g
weight of block
BUOYANCY FORCE = weight of the water (fluid) displaced by the block
Volume of the block = (D1 – D2) A
VERTICAL TRANSPORT: BUOYANCY
Balance of forces:
γ buoyancy = γ p - g
  

g

z+Dz
Fluid (’)
Object (
z
Note: Barometric law assumed a neutrally buoyant atmosphere with T = T’
γ
T

p
= -g 
T’ would produce bouyant acceleration
Vertical
transport:
Pressure,
work, and
Temperature
Question: Where does the energy come from for
an air parcel to do this work on the atmosphere?
Change of atmospheric temperature with altitude (  pressure )
Atmospheric pressure vs altitude follows the barometric law, DP=-gDz .
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.
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
Boyles law
P1V1
V
(P1 + DP1)( V1 + DV1)
= P2V2
P1DV = — V1DP, or DP/DV = — P1/V1
P
P1+DP1 = 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.
DV/V1 = ─ DP/P1
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 ]
P
= m cp DT
Dh
[ 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
Piston
with top
area A,
volume
Ah
h
- mcp DT = P DV
(basic energy balance)
VDP = - P DV
(Boyle’s law) =>> - mcp DT = - VDP
DP = - g  DZ
(Barometric law) =>>- mcp DT = (V g DZ
V = 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 =>> - g / cp = — 9.8 x 10-3 K/m or — 9.8 K/km.
The change in temperature with
altitude in the atmosphere. The
example is from 30 degrees north
latitude in summer.
ATMOSPHERIC LAPSE RATE AND STABILITY
“Lapse rate” = -dT/dz
Consider an air parcel at z lifted to z+dz and released.
It cools upon lifting (expansion). Assuming lifting to be
adiabatic, the cooling follows the adiabatic lapse rate G :
z
stable
G = 9.8 K km-1
g
G  dT / dz 
 9.8 K km-1
Cp
z
unstable
inversion
unstable
What happens following release depends on the
local lapse rate –dTATM/dz:
ATM
• -dTATM/dz > G e upward buoyancy amplifies
(observed) initial perturbation: atmosphere is unstable
• -dTATM/dz = G e zero buoyancy does not alter
perturbation: atmosphere is neutral
• -dTATM/dz < G e downward buoyancy relaxes
T
initial perturbation: atmosphere is stable
• dTATM/dz > 0 (“inversion”): very stable
The stability of the atmosphere against vertical mixing is solely determined
by its lapse rate.
EFFECT OF STABILITY ON VERTICAL STRUCTURE
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
ATM
G
z
final
G
ATM
T
Initial equilibrium
state: - dT/dz = G
G
initial
T
Solar heating of
surface/radiative
cooling of air:
unstable atmosphere
T
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.
The change in temperature with
altitude in the atmosphere. The
example is from 30 degrees north
latitude in summer.
IN CLOUDY AIR PARCEL, HEAT RELEASE FROM
H2O CONDENSATION MODIFIES G
Wet adiabatic lapse rate GW = 2-7 K km-1
z
T
RH
“Latent” heat release
as H2O condenses
RH > 100%:
Cloud forms
GW  2-7 K km-1
G  9.8 K
km-1
100%
GW
G
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.
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).
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
Ambient T
15 (->35)
-13
-58
Dry Adb.
-9.8
-9.8
-9.8
G= -g/(cp +  Dw/DT )
= latent heat of vaporization (J/kg); Dw/DT=change in spec humidity/K
Convective cloud over Amazonia
3
Z
km
latent heat
release
2
1
Tdew
0
283 293 303
Temperature
K
[Photo: S. Wofsy,
Manaus, Brazil, 1987.]
cloud base
Tdew = Tair
VERTICAL PROFILE OF TEMPERATURE
Mean values for 30oN, March
Altitude, km
Radiative
cooling (ch.7)
- 3 K km-1
2 K km-1
Radiative heating:
O3 + hn e O2 + O
O + O2 + M e O3+M
heat
Radiative
cooling (ch.7)
- 6.5 K km-1
Latent heat release
Surface heating
DIURNAL CYCLE OF SURFACE HEATING/COOLING:
ventilation of urban pollution
z
Subsidence
inversion
MIDDAY
1 km
G
Mixing
depth
0
NIGHT
MORNING
T
NIGHT
MORNING AFTERNOON
SUBSIDENCE INVERSION
typically
2 km altitude
FRONTS
WARM FRONT:
WIND
WARM AIR
Front boundary;
inversion
COLD AIR
COLD FRONT:
WIND
WARM AIR
COLD AIR
inversion
TYPICAL TIME SCALES FOR VERTICAL MIXING
•
Estimate time Dt to travel Dz by turbulent diffusion:
Dz 

Dt 
2
2K z
with K z
5
10 cm s
tropopause
(10 km)
10 years
5 km
“planetary 2 km
boundary layer”
0 km
1 month
1 week
1 day
2 -1