Transcript Lecture #2 - Atmospheric and Oceanic Science

LECTURE 2
AOSC 637
Atmospheric Chemistry
R. Dickerson
Copyright R. R. Dickerson 2011
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Solubility of CO2
0.065
0.06
H (M/atm)
0.055
0.05
0.045
0.04
0.035
0.03
275
280
285
290
295
300
305
Temp (K)
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Solubility of CO2 near mean surface ocean
temperature
Solubility of CO2
0.048
H (M/atm)
0.0475
0.047
0.0465
0.046
0.0455
0.045
287
287.5
288
288.5
289
Temp (K)
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ATMOSPHERIC PHYSICS
Seinfeld & Pandis: Chapter 1
Finlayson-Pitts: Chapter 2
1.Pressure: exponential decay
2.Composition
3.Temperature
The motion of the atmosphere is caused by differential
heating, that is some parts of the atmosphere receive more
radiation than others and become unstable.
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VERTICAL PROFILES OF PRESSURE AND TEMPERATURE
Mean values for 30oN, March
Stratopause
Tropopause
Composition of the Earth’s Troposphere
H2
O2
CH4
N2
N2O
PM
CO
O3
←SO2, NO2,
CFC’s, etc
Ar
CO2
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Inert gases
6
Banded iron formation.
The great Oxygen Catastrophe
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Units of pressure:
1.00 atm. = 14.7 psi = 1,013 mb = 760 mm Hg (torr) = 33.9 ft H₂O
= 29.9” Hg = 101325 Pa
(a Pascal is a Nm⁻² thus hPa = mb)
Units of Volume: liter, cc, ml, m³
Units of temp: K, °C
Units of R: 0.08206 1 atm mole-1 K-1
8.31 J mole-1 K-1
R’ = R/Mwt = 0.287 J g-1K-1
For a mole of dry air which has the mass 29 g.
Problem for the student: calc Mwt. Wet (2% H₂O vapor by volume) air.
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Derive the Hypsometric Equation
Start with The Ideal Gas Law
PV = nRT or P = ρR’T or P = R’T/α
Where R’ = R/Mwt
Mwt = MOLE WT. AIR
ρ = DENSITY AIR (g/l)
α = SPECIFIC VOL AIR = 1/ρ
We assume that the pressure at any given altitude is due to the weight of
the atmosphere above that altitude. The weight is mass times acceleration.
P = W = mg
But
m = Vρ
For a unit area V = Z
P = Zρg
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For a second, higher layer the difference in pressure can be related to the
difference in height.
dP = − g ρ dZ
But
ρ = P/R’T
dP = − Pg/R’T * dZ
For an isothermal atmosphere g/R’T is a constant. By integrating both sides of
the equation from the ground (Z = 0.0) to altitude Z we obtain:
PZ
Z
Pg
P dP  0  R' T dz
0
PZ
Z
1
g
dp


P p 0 R' T dZ
0
ln(P/P 0 )   Z/H 0
H  T R' / g
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Where H₀ = R’T/g we can rewrite this as:
PZ  P0 exp( Z / H 0 )
*HYPSOMETRIC EQUATION*
Note: Scale Height: H₀ ~ 8 km for T = 273K
For each 8 km of altitude the pressure is down by e⁻¹ or one “e-fold.”
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Problems left to the student.
1. Show that the altitudes at which the pressure drops by a factor of ten
and two are 18 and 5.5km. Rewrite the hypsometric Eq. for base 2
and 10.
2. Calculate the scale-height for the atmospheres of Venus and Mars.
3. Derive an expression for pressure as a function of altitude for an
atmosphere with a temperature that varies linearly with altitude.
4. Calculate the depth of a constant density atmosphere.
5. Calculate, for an isothermal atmosphere, the fraction of the mass of
the atmosphere between 200 and 700 hPa.
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Temperature Lapse Rate
Going to the mountains in Shenandoah National Park the summer is a nice
way to escape Washington’s heat. Why? Consider a parcel of air. If it
rises it will expand and cool. If we assume it exchanges no heat with the
surroundings (a pretty good assumption, because air is a very poor
conductor of heat) it will cool “adiabatically.”
CALCULATE: ADIABATIC LAPSE RATE
First Law Thermodynamics:
dU = DQ + DW
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WHERE
U = Energy of system (also written E)
Q = Heat across boundaries
W = Work done by the system on the surroundings
H = Internal heat or Enthalpy
ASSUME:
a)
Adiabatic (dH = 0.0)
b)
All work PdV work
(remember α = 1/ρ)
dH = Cp dT – α dP
CpdT = α dP
dT = (α/Cp) dP
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Remember the Hydrostatic Equation
OR
Ideal Gas Law
dP  ( g )dZ
dP  (gP/R'T)dZ
  R'T / P
 R ' T ( gP )
dT 
dZ
PC p R ' T
Result:
dT / dZ   g / C p
This quantity, -g/Cp, is a constant in a dry atmosphere. It is called the dry
adiabatic lapse rate and is given the symbol γ₀, defined as −dT/dZ.
 0  9.8K / km
For a parcel of air moving adiabatically in the atmosphere:
T2  T1   0 ( Z2  Z1 )
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Where Z₂ is higher than Z₁, but this presupposes that no heat is added to or
lost from the parcel, and condensation, evaporation, and radiative heating can
all produce a non-adiabatic system.
The dry adiabatic lapse rate is a general, thermodynamic property of the
atmosphere and expresses the way a parcel of air will cool on rising or warm
on falling, provided there is no exchange of heat with the surroundings and no
water condensing or evaporating. The environmental lapse rate is seldom
equal to exactly the dry adiabatic lapse rate because radiative processes and
phase changes constantly redistribute heat. The mean lapse rate is about 6.5
K/km.
Problem left to the students: Derive a new expression for the change in
pressure with height for an atmosphere with a constant lapse rate,
TZ  T0  Z
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Stability and Thermodynamic Diagrams
(Handout)
Solid lines – thermodynamic property
Dashed (colored) lines – measurements or soundings
day
γa > γ₀ unstable
γb = γ₀ neutrally stable
γc < γ₀ stable
On day a a parcel will cool more slowly than surroundings – air will be
warmer and rise.
On day b a parcel will always have same temperature as surroundings –
no force of buoyancy.
On day c a parcel will cool more quickly than surroundings – air will be
cooler and return to original altitude.
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This equation is easily corrected for a wet air parcel. The heat capacity must be
modified: Cp' = (1-a)Cp + aCp(water), where "a" is the mass of water per mass of dry
air. If the parcel becomes saturated, then the process is no longer adiabatic.
Condensation adds heat.
DQ = -Lda
Where L is the latent heat of condensation
-dT/dZ = g/Cp + (L/Cp) da/dZ
Because the amount of water decreases with altitude the last term is negative, and the
rate of cooling with altitude is slower in a wet parcel. The lapse rate becomes the wet
adiabat or the pseudoadiabat.
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(HANDOUT)
Very important for air pollution and mixing of emissions with free
troposphere. Formation of thermal inversions.
(VIEWGRAPH)
In the stratosphere the temperature increases with altitude, thus stable or
stratified.
DFN: Potential temperature, θ : The temperature that a parcel of air would
have if it were brought to the 1000 hPa level (near the surface) in a dry
adiabatic process. You can approximate it quickly,   T  Z
z
Z
0
or a proper derivation yields:
(
)
θz = T 1000 hPa/Pz ** R’/Cp
= T (1000 hPa/Pz) ** 0.286
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DIURNAL CYCLE OF SURFACE
HEATING/COOLING:
z
Subsidence
inversion
MIDDAY
1 km
Mixing
depth
NIGHT
0
MORNING
T
NIGHT
MORNING AFTERNOON
Plume looping, Baltimore ~2pm.
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Plume Lofting, Beijing in Winter ~7am.
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Atmospheric Circulation and Winds
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ITCZ
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Mid latitude cyclone with fronts
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Warm Front
Fig. 9.13
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Cold front
Fig. 9.15
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Fig. 30
1-17, p. 21
Detailed weather symbols
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Fig. 32
1-16, p. 19
Electromagnetic Radiation
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Electromagnetic spectrum
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Planck’s Law for blackbody radiation:
E ( , T ) 
2hc
5
2
1
e hc / kT  1
( Wm -2 m -1 )
Take first derivative wrt T and set to zero: Wien’s Law:
max
3
a 2.9 10 m K
 
(m)
T
T
-1
Integrate over all wavelength: Stephan-Boltzmann Law:
ET  T ( W/m )
4
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UV-VISIBLE SOLAR SPECTRUM
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Planck’s Law says that the energy of a photon is proportional
to its frequency.
E  h (J)
Where Planck’s Const., h = 6.6x10-34 Js
The energy associated with a mole of photons at a given
wavelength in nm is:
E  1.2 x10 /  (kJ/mole )
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