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The atmospheric
compartment
How much does it weigh?
Temperature and pressure
Circulation and mixing
Where did Oxygen come from
Particle emissions
Emissions of other pollutants
The total atmosphere weights
5.13x1018 kg
Assume that an average car weighs 1000 kg
or one metric tonne
how many cars are equal to the weight of the
atmosphere?
if you could cover the earth in cars how high
would the stack be? Assume the average car
has a “foot print” of 4x2 meters and is 2
meters high?
atmosphere weighs
5.13E+18 kilograms
car weighs
1000 kilograms
# cars = to atmosphere=
radius of the earth =
5.13E+15 cars
4000 miles
radius of earth =
6437.376 kilometers
surface area of earth=
area= 4 pi xr^2
area of a car =
5.20E+08 square kilometers
# cars/surface
0.000008 square kilokometer
6.50605E+13
weight of one surface 6.50605E+16
of cars
# of layers of cars=
height of cars =
78.8
158 meters
Two important features
the atmospheric
Compartment are
temperature and pressure
The atmosphere is usually
divided into the following:
• Troposphere
• tropopause
• Stratosphere
• stratopause
• mesosphere
• Thermosphere
0-10 km
~10km
10-50km
~50 km
50-80kn
80 + km
These divisions come about because of
temperature differences as one increases in
altitude:
The troposphere contains about 80% of the
atmospheric mass.
Air cools with altitude in the troposphere.
The top; 10-15 km is at ~-60oC; which
means very little water vapor.
In the stratosphere, temp. increases with
height because O3 absorbs uv radiation.
Thermal mixing of air (heat) is responsible
for global circulation in the lower
atmosphere.
The atmosphere is held to
earth’s surface
by the gravitational attraction of the earth
At a given altitude the downward force (F)
is related to the mass (M) of the atmosphere
above that point.
F= M (g); where g is the
gravitational acceleration constant
The pressure or force per unit area
decreases with increasing altitude
The decline in pressure (P) with altitude is
approximately = to
log P= - 0.06 (z); where z is the
altitude in km and P is bars
How thin is the air at the top of
Mt. Everest?
Mt. Everest is 8882 meters high or 8.88 km
high
log P = -0.06 x 8.88
P = 10-0.06x 8.88 = 0. 293 bars
Assume there are 1.01bars/atm.
This means there is < 1/3 of the air
Where does log p = - 0.06 (z) come from;
Force = mass x acceleration
acceleration = g
The mass of air over a surface, A, equals height x
A x mass/volume; the mass / volume = density, r;
 So
Force = -z x A x r x g
The change in force at any altitude
dF = -dz x A x r x g;
F/A = pressure, p
So the change in pressure with height is
dp= -dz x r x g
dp = -dz x r x g
What is the ideal gas law
pV = nRT
Show that r = Mw x p/RT
Substituting for r in
dp = -dz x r x g
dp = -dz x Mw x p/RT x g
So dp/p = -dz x Mw /RT x g
 Integrating
 So p
= po exp {-Mw g z /RT}

p = po exp {-Mw g z /RT}
If we set H = (Mw g /RT)-1 it has the units of
length
 and we get a simple expression
p= po exp {-z /H}
Solving for H at 290K; R = 8.3 joules/(K mole)
one joule = 1kg meter2/sec2
average Mw of air = 28.9 g/mole
g = 9.8 meter/sec2
H = 8.5 km; people actually find that 7km
works best; when 7 km is used we end up with
 log p = - 0.06 (z);
Is it possible in the troposphere to
calculate the rate that temperature
of the air decreases with altitude?
To do this we need to start with
simple thermodynamics
 The first law of thermo says that the change internal energy of a
system is the sum of its changes in heat content and work that is
done.
 dU = dq - dw
 A change in work can only occur if a force moves through a
distance; dw = d (fxz) = d (pV) for work there must be movement
 Hence dw = Vdp and dU = dq - Vdp
 Another form of energy is call enthalpy (H) which is the sum of the
internal energy and pV from pV=nRT
 so H = U + PV or dH = dq –Vdp + Vdp +pdV = Vdp (dq is
assumed to be zero for a process that does not have a heat loss)
 The change in the heat of a mass, per change in a degree
centigrade, is called its specific heat Cp and Cp = dH/dT
 we said the enthalpy dH = Vdp
 specific heat capacity; Cp = dH/dT
 So Cp dT= dH = Vdp
 Before we said that the change in pressure with height, z was
dp= -dz x r x g
 So substituting for dp we get
 Cp dT= - V x dz x r x g
 So the change in temp with height is dT/dz = - V x r x g/ Cp
 Density r is mass/V; so for an air mass of one gram, r = 1/V
 This puts Cp in units of energy gram-1 deg-1;
for dry air this is 0.24 cal gram-1 o K-1 and we will call it cp
 So - dT/dz = g/ cp where g = 9.8 m sec-2
 1 cal = 4.1 joules so cp = ~1 joule gram-1 o K-1
 One joule = 1 kg m2sec-2
 So - dT/dz = g/ cp = 0.0098 oK/meter or 9.8 oK/kilometer
 The fact that - dT/dz = g/ cp = 9.8 oK/kilometer is constant is
consistent with observations
 And this is called the dry adiabatic lapse rate so that - dT/dz = d
 When - dT/dz > d the atmosphere will be unstable and air will
move (convection) to re-establish a stability
The quantity d is called the dry
adiabatic lapse rate
Air that contains water is not as heavy and has
a smaller lapse rate  and this will vary with the
amount of water
If the air is saturated with water the lapse rate
is often called s
 Near the surface sis ~ -4 oK/km and at 6 km
and –5oK/km it is ~-6K/km at 7km high
At midday, there is generally a reasonably wellmixed layer lying above the surface layer into
which the direct emissions are injected.
As the sun goes down, radiative cooling results in
the formation of a stable nocturnal boundary layer,
corresponding to a radiation inversion.
Sun-down
earth cools
midday

more cooling at
surface at night

}
temp
temp
temp
Inversion layer
What happens to the material above the inversion layer??
more cooling at
surface at night
}
} Inversion layer
residual layer
temp
These materials are in a residual layer that contains the
species that were well-mixed in the boundary layer during
the daytime. These species are trapped above and do not
mix rapidly during the night with either the inversion
boundary layer below or the free troposphere above.
When the sun comes up the next day it heats the earth an
the air close to the earth.
more cooling at
surface at night
}
temp
Heating at surface
during the next day
} Inversion layer
Inversion layer
temp
During the next day heating of the earth's surface results
in mixing of the contents of the nocturnal boundary layer
and the residual layer above it
Mixing height in the morning
• We will start with the balloon temperature
curve that is taken at the airport each
morning.
• In the morning the temperature usually
increases with height for a few hundred
meters and then starts to decrease with
height (see the green curve) according to the
temperature sensor on the balloon
• The the break in the curve usually defines the
inversion height in the early morning
Mixing height in the morning
Balloon temperature
}Inversion height
Temp in oC
Mixing height in the morning
• There are another set of lines called the dry adiabatic lines, which
are thermodynamically calculated, and represent the ideal
decrease in temperature with height for dry air starting from the
ground.
• In the morning, the mixing height is estimated by taking the
lowest temperature just before sunrise and adding 5oC to it, and
then moving up the dry adiabatic line at that temperature until it
intersects the balloon temperature line or the green curve.
• Let’s say the lowest temperature just before sunrise was 20oC. We
would add 5oC to it and get 25oC. We then move up the 25oC dry
adiabatic line. We then go straight across to the right, to the
height in kilometers and get a morning mixing height of ~350
meters (0.35 km). This is illustrated in the next slide. It is
animated so you can see it more easily
Mixing height in the morning
Balloon temperature
1.5
1.1
Dry adiabatic
lines
20
25
30
Temp in oC
35
0.4
0.3
0.2
0.1
0.0
Mixing height in the afternoon
• To get the mixing height in the afternoon, you just
take the highest temperature between 12:00pm and
15:00 pm
• Do not add anything to it, but as before run up the dry
adiabatic curve and intersect the morning balloon
temperature curve.
• Let say the highest afternoon temperature is 35oC, we
would estimate an afternoon mixing height of ~1.67
km
Afternoon Mixing height
Balloon temperature
1.5
1.1
Dry adiabatic
lines
20
25
30
Temp in oC
35
0.4
0.3
0.2
0.1
0.0
How does air circulate
At the equator air is heated and rises and
water is evaporated.
As the air rises it cools producing large
amounts of precipitation in equatorial
regions.
Having lost its moisture the air mass moves
north and south.
It then sinks and compresses (~30oN and S
latitude) causing deserts
Circulation currents
30oN
Hadley
cell
equator
A similar system at the poles
occurs where cold air sinks and
flows south (in the northern
hemisphere.
 These sinking columns of air
create a circulation system called
Hadley cells
Circulation currents
indirect
cells
Hadley
cell
30oN
Hadley
cell
equator
In between the Hadley cells, north
and south of the equator, and at the
poles, counter current or indirect
cells are set up.
These drive circulation between 4060o latitude, producing westerly
winds and storms
Circulation currents
indirect
cell
Hadley
cell
30oN
Hadley
cell
equator
The air in each hemisphere mixes with a
time constant , t, of a few months.
The air between the north and south
hemispheres completely mix on the order of
one year.
Air mixes into the stratosphere from rising
Hadley cells in the tropics, storms and eddy
diffusion.
exchange between the troposphere and the
stratosphere can be thought of in terms of
mean residence times (MRT)
The mean residence time (MRT) can be
expressed as:
MRT = mass / flux
where flux is mass/time
If 75% of the mass/year in the stratosphere
comes from the troposphere

1
MRT = ----------------- = 1.3 years
–
0.75/year
Mt. Pinatubo in the Philippines erupted in
June 1991, and added a huge amount of
SO2 and particulate matter the
stratosphere. After one year how much
SO2 was left?
For a 1st order process
C= Coe -1 year/ MRT
C/Co= e -1 year/ MRT = e -1/1.3= 0.47 or ~ 50%
in 4 years, C/Co= e -4 years/1.3 years =
~5%
What happened to global
temperatures after the Pinatubo
eruption
A lot of SO2 was injected into the
atmosphere
What we will learn later that SO2
forms fine sulfate particles that
reflect light back into the
atmosphere and this cools the
upper troposphere
Atmospheric Composition
Three gases, O2, N2, and argon make up
99% of the atmosphere mass of 5.14x1021 g
 These gases are relatively un-reactive and
their mean residence times are much longer
than the rate of atmospheric mixing. Hence
the conc. of N2, O2, and the Nobel gases
(He, Ne, Ar, Kr, and Xe) are relatively
uniform.
Atmospheric Composition
N2
O2
78.084%
20.946
3.87x1021 grams
1.19x1021
Ar
CO2
Ne
0.934
0.036
18.2 ppm
6.59x1021
2.80x1018
6.49x1016
H2
CFC 11
MeBr
510 ppb
280 ppt
11 ppt
1.82x1014
6.79x1012
1.84x1011
Where does oxygen come
from in our atmosphere?
 ~3.8 billion years ago the earliest
bacteria were able to take acetic acid
and metabolize it to CO2 and water.
CH3COOH ->CO2+ H2O
A later form of bacteria could obtain
energy from the reduction of H2S to S
CO2 +2H2S->CH2 O+ 2S + H2O
As supplies of H2S were consumed in the
oceans other energy generating metabolic
processes became more competitive
one was photosynthesis
H2O +CO2 -> CH2O + O2
A good summary of the “Rise
of Life on Earth”
is given in National Geographic, vol 193, p 54,
March 1998
Experiments in the early 1950’s (Dr. Stanley
Miller, U. Cal. San Diego) showed that is is
possible to generate amino acids in an
atmosphere of CH4, NH3 and H2 over a pool of
water with an electrical discharge
This gave rise to the theory that
“life” could have originated in
warm tidal pools of the oceans,
since amino acids are the building blocks of
“life”.
It is now thought that the atmosphere 4-3.5
billion years ago consisted of CO2 and N2
sparking CO2 and N2 does not generate
large amounts of organics
A more recent theory is that “life”
could have originated in deep hot
pools of water heated by volcanic
rock.
Some of the most primitive live forms,
called thermophil bacteria, can live at
190oC and consume iron and sulfur (3.8
billion years ago).
 Another theory involves ice. Trapped
H2C=O, NH3, HCN and water--->glycine
(amino acid) followed by meteorite impact
Photosynthetic cyanobacteria (blue green
algae) may go back 3.46 billion years. They
were recently found in a rock from NW
Australia.
It is felt these types of bacteria invaded other
cells and their chloroplasts“stayed” there.
Some bacteria were also invaded by
mitochondria which burn sugar
These photosynthetic cells gave off huge
amounts of oxygen.
Oxygen did not really start to build up in the
earth's atmosphere until ~2 billion years ago.
Why??
The oceans contained much dissolved iron. It
had to be oxidized before O2 could persist in the
atmosphere.
Aerosols
The atmosphere contains particles which
are known as aerosols. These come from a
number of different sources
Soil minerals are dispersed by wind erosion
from arid (dry) and semi-arid regions
Particles with a diameter of 1.0 mm (the
smallest we can see is 20 mm) are held aloft
by turbulent motion and can be transported
long distances
It is estimated that 1x1015grams/year of soil
particles enter the atmosphere and 20% are
involved in long range transport.
Dust from the deserts of central Asia falls
on the Pacific ocean where it contributes
much of the iron (Fe) needed by oceanic
phytoplankton
Dust from the Sahara desert supplies
nutrients to phytoplankton in the Atlantic
ocean and phosphorous to the Amazon rain
forest.
The oceans create bubbles which burst and
provide a huge quantity of large and small
aerosols.
About 200x1012 grams of sea salt are
carried to land each year
Forest fires in the Amazon are thought to
produce 1x1013 grams of fine aerosols.
These aerosols affect regional rain fall
patterns and influence global warming
Volcanoes emit particles into the atmosphere
which contribute to soil development down
wind of major eruptions.
Small particles are also produced by reactions
of gases in the atmosphere
SO2 ----> H2SO4
H2SO4 + NH3 --> [NH4]2SO4 aerosols
These aerosols and others from the sea
(dimethylsulfoxide) influence global warming
Natural sources
Primary aerosols
Soil dust
sea salt
volcanic dust
organic particles
Secondary aerosols
sulfates from organics
sulfates from SO2
organic condensates
nitrates from NOx
sum of natural sources
1500x 1012 g/year
1300
33
50
90
12
55
22
3070 x1012g/year
Anthropogenic sources
Primary aerosol
Industrial particles
soot
forest fires
100x 1012 g/year
20
80
Secondary aerosols
sulfates from SO2
organic condensates
nitrates from NOx
140
10
36
sum of Anthropogenic
390 x1012g/year
sum of natural sources
3070 x1012g/year
If more particles are emitted from
natural sources, why are we worried
about man made or anthropogenic
sources; they are 10 times less????
sum of Anthropogenic
390 x1012g/year
sum of natural sources
3070 x1012g/year
Primary pollutants emitted
into the atmosphere- sources
Oxides of nitrogen (NOx = NO + NO2)
Sulfur dioxide (SO2)
Carbon monoxide (CO)
Volatile hydrocarbonds (VOCs)
Approximately 72 Tg (1Tg = 1x1012grams ) of
NOx emitted into the atmosphere
The US emits ~30% of this
On the next slide we will look at global trends
in emissions of NOx
NOx Emissions Trends
 Much of the NOx comes from
 Cars
 Power plans
 Off the road vehicles
 The same emitted mass of NOx from different
sources can have different atmospheric chemistry
effects.
NOx
dilution aloft
NOx emitted near
the ground
NOx Emissions in the US
SO2 Emissions trends
SO2 Emissions in the US
VOC Emissions in the US
CO Emissions in the US
Lead Emissions in the US
Lead Emissions in the US