Atmospheric & Oceanic Processes
Lecture 2: Characteristics of the atmosphere
Fresh snow… Feb/06/2016
Some parameters of Earth.
Earth’s rotation rate
Surface gravity
Earth’s mean radius
Surface area of Earth
Area of Earth’s disc
Mass of Earth
Ω
g
a
4πa2
πa2
ME
7.27 × 10−5s−1
9.81 ms−2
6.371 × 106 m
5.09 × 1014 m2
1.27 × 1014 m2
5.972 × 1024 kg
By Newton’s law of universal gravitation, force exerted by mass ME on mass m is:
Fg = -(GMEm/r2)(r/r)
where G (= 6.67408 × 10-11 m3 kg-1 s-2) is the universal gravitational constant, r is the distance
vector measured from the center of mass of ME, the earth, to an air mass parcel m of the
atmosphere, i.e. r = |r| = a + z, where z is the height of the atmospheric air parcel measured
from the mean sea level (where z=0). Then, earth’s gravity g (i.e. acceleration due to gravity) is:
g = (Fg/m) = -(GME/r2)(r/r)  -(GME/a2)(r/r)  9.81 m s-2 (r/r)
where we have made use of the fact that the atmosphere is “thin”…
Most of the atmosphere is contained in a “shell” of about 10 km high (see HW1, & below),
which is very thin compared to the radius of the earth
Figure: The thinness (to scale) of a shell of 10km thickness on the Earth of radius 6370 km
Atmospheric wind is
never completely
blocked by
topography
Ocean current is
confined to basins
A north-south section of topography relative to sea level (in meters) along the Greenwich meridian (0◦ longitude) cutting
through Figure below (red line). Antarctica is over 2km high, whereas the Arctic Ocean and the south Atlantic basin are
about 5km deep. Note how smooth the relief of the land is compared to that of the ocean floor.
World relief showing elevations over land and ocean depth. White areas over the continents mark
the presence of ice at altitudes that exceed 2 km. The mean depth of the ocean is 3.7 km, but depths sometimes
exceed 6 km. The thin white line meandering around the ocean basins marks a depth of 4 km.
TABLE 1.2. Some of the most abundant atmospheric constituents.
Chemical
species
Molecular
weight (g mol−1)
Proportion
by volume
N2
O2
Ar
H2O (vapor)
CO2
28.01
32.00
39.95
18.02
44.01
78%
21%
0.93%
∼0.5%
380 ppm
Water vapor (H2O) is mainly from ocean evaporation.
CO2 are due to photosynthesis and respiration, ocean-atmosphere exchange & anthropogenic
activities.
H2O & CO2 are important in radiative transfer (the passage of radiation through the
atmosphere), because they strongly absorb and emit in the infrared, the region of the spectrum
(wavelengths about 10 μm) at which Earth radiates energy back out to space.
H2O & CO2 are present in very small concentrations, and so are sensitive to anthropogenic
activities.
Atmospheric CO2 concentrations observed at Mauna Loa, Hawaii (19.5◦ N, 155.6◦ W). Note the seasonal cycle superimposed on the long-term
trend. The trend is due to anthropogenic emissions. The seasonal cycle is thought to be driven by the terrestrial biosphere: net consumption
of CO2 by biomass in the summertime (due to abundance of light and heat) and net respiration in wintertime.
Over the course of Earth’s history, CO2 levels have greatly fluctuated; they are different in warm
as opposed to cold periods of Earth’s climate. For example:
20,000 years ago,
30 million years ago:
220 million years ago:
CO2 ~ 180 ppm; (cold period: last glacial maximum)
CO2 ~ 600 ppm; (very warm period)
CO2 ~ 2,000 ppm; (very very warm)
If the CO2 curve shown above continues to rise, then by year 2100, CO2 ~ 600ppm—not seen
since 30 million years ago, a period of great warmth in Earth history.
We focus on the lowest 50km of the atmosphere, wherein
the mean free path of atmospheric molecules is so short
and molecular collisions so frequent that the atmosphere
can be regarded as a continuum fluid; it satisfies gas law:
p = ρRT
where p is pressure, T absolute temperature in K, and the
gas constant R = 287 J kg−1K−1.
Note:
For dry air, any 2 of p, ρ, and T is sufficient to define its
state;
If p, then ρ, for T = constant  air is compressible;
If T, then ρ-1, for p = constant  air expands.
Some atmospheric numbers
“Standard”
Pressure,
temperature &
density
Properties of dry air at standard temperature and pressure
Gas constant for water vapor
Rv
461.39 J kg−1 K−1
For a mixture of gases, the gas law is applied individually to the
different components; e.g. dry air & water vapor:
Atmospheric air = dry air + H2O = moist air
Let V = some volume of moist air, then partial densities are:
ρv = mass of H2O per unit V and
ρd = mass of dry air per unit V,
Then:
Partial pressure of H2O = e, is the pressure that H2O would
exert, at the same T, if it alone occupies V;
Similarly:
Partial pressure of dry air = pd, is the pressure that the dry air
would exert, at the same T, if it alone occupies V.
Apply the gas law individually to dry air & H2O:
e = ρvRvT
&
pd = ρdRdT
Total pressure is then:
p = pd + e
(a)
(b)
e = ρvRvT (partial pressure of vapor)
pd= ρdRdT (partial pressure of dry air)
p = pd + e (total pressure of vapor & dry air)
Air over water in a closed box at temperature T. At equilibrium the rate of
evaporation equals the rate of condensation. The air is saturated with water vapor,
and the pressure exerted by the vapor is e = es, the saturated vapor pressure. On
the right we show the mixture comprising dry ‘d’ and vapor ‘v’ components.
es is higher for warmer T
So if the air in (b) above is cooled, then water vapor must
condense out to return to the water below, and a new
lower es in the “atmosphere” above is established at the
lower T
Warm water is poured into a carboy to a depth of 10 cm or so, as shown on the left. We leave it
for a few minutes and throw in a lighted match to provide condensation nuclei: small particles
on which the vapor can condense. We rapidly reduce the pressure in the bottle by sucking at the
top. The adiabatic expansion of the air reduces its temperature and hence the saturated vapor
pressure, causing the vapor to condense and form water droplets, as shown on the right.
Condensation nuclei in atmosphere: sulfate aerosols, dust, smoke from fires, ocean salt…
A photograph of the sound barrier being broken by a US Navy jet as it crosses the Pacific Ocean at the speed of sound just
23 m above the ocean. Condensation of water is caused by the rapid expansion and subsequent cooling of air parcels
induced by the shock (expansion/compression) waves caused by the plane outrunning the sound waves in front of it.
When we see “mist” the partial water-vapor pressure e
has reached the saturated vapor pressure es
To reach higher es, we need to raise the T…
(b) T profile for the ‘‘US standard atmosphere’’ at
40◦ N in December.
(a) es (in mbar) as a function of T in ◦C (solid
curve)
Clausius-Clapeyron
Relation:
es  A exp(T)
T in oC,
A = 6.11 hPa,
 = 0.067 oC-1
-10oC
(a)
3hPa
(b)
Temperature (K)
-30
-10oC
• Vapor is in the lowest few km’s
• Tropics are more moist than poles
• Precipitation when moist air cools by convection, forcing H2O concentrations back
to saturation at the lower T
• Warm climate is more moist
The Clausius-Clapeyron Relation:
es  A exp(T)
T in oC, A = 6.11 hPa,  = 0.067 oC-1
has important implications for global warming.
It predicts a strong increase of near-surface water vapor content with temperature
(about 6~7% per K of warming)
So, if T increases by 1K, does the CC-relation mean that precipitation also increases
by 6~7%?
We will see that precipitation indeed increases as the planet warms, but the
increases is less 1~3%...
Why?
Before this can question can be addressed, we need to digress and first study some
basic properties of the atmosphere…
Definitions and nomenclatures
“Top” of atmosphere, e.g. pressure ~ 1 hPa
Wind or ocean current velocities:
u: zonal (west to east)
v: meridional (south to north)
w: vertical (>0 upward)
Land
orography
Coordinate axes:
x: zonal (west to east)
y: meridional (south to north)
z: vertical (>0 upward, =0 at mean sea level)
Sea surface
Mean ocean depth
Ocean’s bottom
ZT(x,y)
Hydrostatic balance
If the atmosphere were at rest, or static, then pressure at any level would depend on the
weight of the fluid above that level. This balance is called hydrostatic balance.
FT = − (p + δp)δA
z
Fg = −gM
FB = pδA
Mass of the cylinder is
Gravitational force
Pressure force at top face
Pressure force at bottom face
M = ρδAδz
Fg = −gM = −gρδAδz
FT = −(p + δp)δA
FB = + pδA
Net force
Fg + FT + FB = 0  δp + gρδz = 0:
∂p/∂z + gρ = 0
Hydrostatic Equation
An Application of the Hydrostatic Equation:
∂p/∂z = −gp/RT
(using hydrostatic equation ∂p/∂z = -gρ and the Gas Law: p = ρRT
R = 287.05 J kg-1 K-1 = 287.05 m2 s-2 K-1)
Assume temperature is constant = T0, then we have an “isothermal” atmosphere
∂p/∂z = −gp/RT0 = −p/Hs;
Hs = RT0/g = the Scale Height
So,
p(z) = ps exp(−z/Hs).
or,
z = Hs ln(ps/p).
If we choose a representative value T0 = 250 K, then Hs = 7.31 km. Therefore, for example,
in such an atmosphere p is 100 hPa, or one tenth of surface pressure, at a height of z = Hs
× (ln 10) = 16.83 km.
The isothermal assumption is actually not too
bad, if we choose an appropriate Hs, as shown
in the figure at right, which shows observed
profile of pressure (solid) plotted against a
theoretical profile (dashed) based on the
above equation for p(z) with Hs = 6.8 km.
For isothermal atmosphere (from previous slide),
p(z) = ps exp(−z/Hs).
The corresponding density is then, from the Gas Law:
ρ(z) = [ps/(RTo)].exp(−z/Hs).
Mass of the atmosphere (per unit m2) can be estimated:
𝑧
𝜌𝑑𝑧′ = (ps/g).[1 - exp(−z/Hs)].
0
Set z = ∞:
Total mass/m2 = (ps/g).
Because of the exponential decay with height, most of the mass of the atmosphere is very
near the surface. Thus 80% of it is below z80% where:
Therefore,
1 - exp(−z80%/Hs) = 0.8
z80% = -ln(0.2)Hs  1.6 × 7.3km  12 km;
i.e. about 80% of the mass of the atmosphere is below an altitude of about 12 km, i.e.
within the troposphere.
Pressure gradients
Homework #2
All HWs are due in 2 weeks from the time they are distributed.
1. Derive or write down the formula for the Rossby number Ro .
2. Explain how the magnitude of Ro can describe if a given atmospheric or oceanic flow can
“feel” the earth’s rotation.
3. Explain how the behaviors of atmospheric or oceanic flow can be very different on equator
and over the poles?
4. Do these different behaviors explain why typhoons or hurricanes do not develop near the
equator? Explain as best as you can.
5. Derive or write down the formula for the hydrostatic equation?
6. Explain why for large-scale atmospheric or oceanic flow the hydrostatic equation is valid?
7. Then also explain why over the hill near the front gate of National Central University, the
hydrostatic equation is not likely to apply to describe the wind that blows over the hill?
8. What are the approximate densities of air and water?
9. On earth’s surface, we experience the weight of air some 10 km high. Estimate how deep
into the ocean do we need to dive in order to experience the same pressure.
10. What is the saturation vapor pressure es at 15oC?
11. Then show that the corresponding ρv = mass of water vapor per unit volume at 15oC is
equal to 0.0126 kgm−3. This is the maximum amount of water vapor per unit volume that
the atmosphere can hold at this temperature.