the state of the atmosphere 1.atmospheric mass and pressure 2.temperature structure 3.geopotential...
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The State of the Atmosphere
1. Atmospheric mass and pressure
2. Temperature structure
3. Geopotential
4. Circulation
5. Water in the atmosphere
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Atmospheric mass and pressure
• From the equation of continuity, after integration for a polar cap limited by
WallSVV
vdsdsncdvcdivdvt
.)(
In the long-term mean: 0Wall
dsv
Assuming that the atmosphere is a ideal gas (p=RT) in hydrostatic equilibrium
z
dzgzp )(
we haveRT
dzg
p
dp and by integrating: )exp()(
0
0 z
z
dzRT
gpzp
We can then rewrite the first equation as:
Wallsurfh
vdsdsptg
_
0
1(1)
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The rapid decrease of density with height distinguishes the atmosphere from the ocean
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Distribution of mass in terms of pressure
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• If H_surf=2.56x1014 m2
• PNorth_h=983.6mb, PSouth_h=988.0mb, PGlobe=983.6mb• g=9.8 m/s2
Then
m_North_h=2.57x1018 kg
m_Souh_h=2.58x1018 kg
m_Globe=5.15x1018 kg
...Distribution of mass in terms of pressure
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The interhemispherictransport
From Eq. (1) we can calculate the transport trough the equator:
0 0
/][][z
dzdzvv
If p=1mb/month
smRmbmonth
mbR
v
/002.021012
1
1
12
][
2
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Global distribution of temperature
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Vertical and meridional change
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Temporalvariabilityof temperature
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Geopotential heights
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Mean circulation
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• Correlation coefficient between northern hemisphere stratospheric geopotential at 50hPa and an index representing the tropospheric 500hPA NAO.
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Hadley model
Hadley (1735) wanted to explain trade wind circulationComplements of E. Kant and J. DaltonFerel (1856): Coriolis force and Geostrophic windHelmholtz: the role of friction-deviation of wind includes turbulent viscosity for the first time
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Actual vertical circulation
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• Kinetic energy of the atmosphere
• K=KTE +KSE +KM
• K=0.5[u2m+v2
m]
• KTE=0.5[(u‘2+v‘2) m]
• KSE=0.5[u m *2+v m *2]
• KM=0.5([u m] 2 +[v m] 2)
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Precipitation
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