Weather and the Environment
PX350
The Least You Should Know
These notes are the minimum amount of knowledge required to pass the examination in PX350.
For more than a bare pass more study will be needed!
The notes should be read in conjunction with the PowerPoint slides used in lectures (lecture
numbers below are approx. since some material was spread over several lectures) and the
additional web-based resources linked to from the module pages
http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px350/ .
[References are included to page numbers in the recommended text Introduction to Weather and
Climate, McIlveen, 2nd Ed., Oxford.]
Lecture 1
Weather/meteorology – study of atmosphere and atmospheric processes, daily variation
Climate/climatology –
seasonal average of weather at some location
Climate change –
long term variation in climate
[pp 1-6]
Most weather processes take place in the troposphere, which is approx. 10 km thick. This should
be compared to the Earth’s mean radius of 6370 km.
[Ch3 p54+] Composition of the atmosphere.
Major permanent gases are N2 (78%) and O2 (21%), with about 1% Ar. These ratios stay the same
throughout the well-mixed part of the atmosphere, or turbosphere that includes the troposphere,
stratosphere and mesosphere.
Variable gases include H2O vapour (<4%), CO2 (~0.04%) and other trace gases (see table in Lecture
1 slides). These gases are the ones responsible for absorbing radiation and hence for the
greenhouse effect. Condensation and evaporation of water is a major driver of atmospheric
processes.
[p90-94] Layers of the Atmosphere
Pressure at sea level is 1 bar = 1000 mbar ≡ 105 Pa (calc. from P=hg). Hence, 1 mbar ≡ 100 Pa.
•
•
Troposphere
– pressure drops from 1000 mbar to ~200 mbar,  contains 80% mass
– thickness 10 – 15 km
– temperature drops with height by ~6.5 ºC/km, unstable as hot air rises!
– well mixed: molecules travel up and down in few days
– contains almost all the water vapour of the atmosphere
– most weather in this layer, i.e. most heat transport
Tropopause
– region of stable, isothermal air ~ 200 mbar
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– division between troposphere and stratosphere
– top of “weather”
Stratosphere
– pressure from ~200 mb to ~1 mb, height 10 – 50 km
– temperature increases with height (due to heating in ozone layer)
–  stable air, isothermal strata with little mixing
– stops weather extending upwards, so keeps water in atmosphere
– temperature reaches maximum at Stratopause
Mesosphere
– “middle atmosphere”, pressure 1 mb to 10-2 mb
– temperature drops to lowest values ~ -90ºC
– Mesopause is top of well mixed region, 99.9% of atmosphere below
Thermosphere
– temperature increases rapidly (due to molecular dissociation)
– v.large KE/molecule, but few molecules!  Feels cold
– mean free path so large that no longer well mixed
[p 80+] Origin of the atmosphere.
a) Earth formed 4.6 billion years ago, without atmosphere.
(If there was a proto-atmosphere it would have contained 36Ar, like the rest of the solar
system, but we now only observe 40Ar. So conclude that any proto-atmosphere was
destroyed through collisions.)
b) Early atmosphere created by outgassing from volcanoes (which were very active in the
young hot earth) of N2, CO2, H2O, H2, Cl2, SO2, CH4 , but not oxygen.
c) Water condensed in oceans, H2 escaped, acid rain (from Cl2, SO2 plus water) weathered
rocks, leaving N2-based atmosphere.
CO2 at 100 times present level (i.e. ~4,000 ppm) – larger greenhouse effect kept the
atmosphere warmer and compensated for a 30% weaker solar flux from the young Sun
d) Emergence of life in oceans produced O2 and the process of photosynthesis reduced the
CO2 content of the atmosphere. Death of organisms and sedimentation on ocean bottom
led to sequestering of CO2 in rocks.
e) Once O2 was present O3 could be formed, giving UV protection and allowing life to expand
its range onto the land. Before that the UV level would have been too high for life to
survive on the surface.
[p281-288] Radiation Balance
In equilibrium, the total heat received from the Sun is equal to the total heat radiated by the Earth
2
Flux of solar radiation at Earth’s orbit is S  T
4
Sun
 RSun 
2

  1370 Wm
 d Earth  Sun 
A fraction  of this flux is reflected – known as albedo. This depends on the surface material (see
Lecture 2 slides for values). Average albedo for Earth is ~30%.
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PX350
Area over which solar energy is absorbed is equivalent to a disc of radius REarth This also accounts
for angle that the solar flux strikes the Earth.
4
Radiation emitted from the Earth is given by Stefan’s Law of TEarth
per unit area (assuming black
body radiation and ignoring emissivity variation).
Area over which emission occurs is full sphere of radius REarth
Hence, 4R T
2
4
Earth
 R S 1    and TEarth
2
 S 1    


 4 
1
4
This gives an average Earth temperature of 255 K, which agrees with measurements from high in
the atmosphere (where the radiation eventually leaves the planet, see below), but is much lower
than the mean surface temperature of 288 K. The difference is due to heating in the troposphere
by the Greenhouse Effect, equivalent to an extra 33 K.
[p103-108] The radiation balance can also be performed for a particular latitude . By balancing
the solar radiation received at an angle  along a line of length
the circle of length
2REarth cos  and emitted over
2REarth cos  the zonal temperature T can be calculated.
 S 1    cos  
4
Hence, 2R cos  T  2 R cos  S 1    cos  and T  




1
4
This shows that the temperature near the Equator ( = 0) will be greater than at higher latitude.
The point where T  TEarth is at 38ºN (or S). At this latitude the heat received is balanced by the
heat emitted. Excess heat received at lower latitudes is transported to higher latitudes (with a
maximum flux through 38º) by the weather systems and by ocean heat transport.
Lecture 2
[p288-299] The Greenhouse Effect.
Greenhouse gases (esp. H2O and CO2) absorb strongly in the infrared (IR), much less so in UV. The
absorption windows can be added for each gas to see the total absorption by the atmosphere.
Some windows are completely opaque, but at other wavelengths there is only partial absorption,
hence changing the amount of greenhouse gases will have a non-linear effect on the temperature.
Most of the Greenhouse Effect arises from infrared absorption by water vapour (30K), with the
remainder mainly from carbon dioxide absorption.
High energy UVc and UVb are absorbed in the stratosphere by ozone, causing heating there.
Heat is said to be “trapped” by the atmosphere. There is still an equilibrium so the amount of
radiation arriving and leaving is still balanced; however, the surface of the Earth is warmer than
the surface from which most radiation leaves the planet.
An energy balance argument, involving a slab model for the atmosphere, was used to explain the
Greenhouse warming. All values of energy fluxes are normalised to the incoming solar radiation as
100 units. Further averaging over day/night and angle makes this S/4 = 342 Wm-2.
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30% of incoming solar radiation is reflected by clouds (20), atmospheric dust (6) and
directly from the surface (4).
19% of incoming solar radiation is absorbed in the atmosphere (mostly in clouds)
51% (only) is absorbed directly by the surface.
Next consider the IR radiation.
• Surface is at Ts=288 K and radiates σTs4 = 400 Wm-2 (117 units).
• Atmosphere is at 250 K so radiates 220 Wm-2 (64), both up and down.
• Establish equilibrium to space: Input 19+51=70; Output 64+x=70 i.e. x=6 units of radiation
directly lost from the surface.
• Hence, 117-6=111 units emitted from the surface to the atmosphere, but 29% of this is
reflected by the atmosphere (32 units), giving a total downward flux of 64+32=96 units that
is absorbed by the surface.
• Hence, the surface receives 51 (UV) + 96 (IR) = 147 units.
• For equilibrium the surface must also emit 147 units, made up of 117 (IR) + a non-radiative
flux of 30 units. The non-radiative bit is made up of convection currents (kinetic energy)
and water evaporation (latent heat).
• Check the atmosphere also is in equilibrium: Input 19 (UV) + 79 (IR) + 30 (non-rad.) = 128;
Output 64 (IR up) + 64 (IR down) = 128.
• The non-radiative 30 units, i.e. 100 Wm-2, drives the atmospheric circulation and weather.
This is a significant fraction of the input solar energy and dwarfs any man-made energy
generation process. [cf. typical nuclear power station 1GW occupies a site of area 10km2,
so power output is 100 Wm-2, but these are not built over entire surface of Earth!]
With increased greenhouse gas concentration in the atmosphere the downward IR flux to the
surface will increase and hence the surface equilibrium will be at a higher surface temperature. If
there is an additional flux F the temperature will increase by T. Known as “radiation forcing”.
4
4
Tnew
 Toriginal
F
4
new
T
T
4
original
 4T T  F 
3
or T 
F
Temp. increases by 1K for extra 5Wm-2.
4T 3
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[Ch10 p391+] Surface Warming and Cooling
Essential fact: Oceans have very large heat capacity, so temperature remains constant throughout
day and seasons. Land has much smaller heat capacity, so has large daily and seasonal T variation.
Solar radiation received during the course of a day varies with time of day, latitude and season,
due to the angle of incidence of the radiation, tilt of the earth and the length of daylight hours N,
as well as due to local albedo effects.
It can be show that Daily Insolation = Imax2N/.
For the equator at an equinox (i.e. sun directly overhead at noon). Imax = S(1-) and N is a 12 hour
day, giving D.I. of about 26 MJm-2.
This is reduced by ~25% due to cloud cover i.e. 20 MJm-2
For mid-latitude (e.g. UK at 52ºN) mid-summer days are longer, but sun is lower in sky. These two
effects approx. cancel (c.f. 18/12 cos(52º)~1) so daily insolation is again ~20 MJm-2. However in
winter the sun is much lower and days are shorter leading to daily insolation of only ~2 MJm-2.
Terrestrial radiation is lost due to ~blackbody emission from the surface of T4~390Wm-2, but
offset by infrared radiation received from the atmosphere (see last section) of varying amount
dependant on cloud cover. Overall there is an average net emission of ~50Wm-2, i.e. 4.3 MJm-2
over a 24 hour period.
For the purpose of the following calcs. we can assume that heat input and output of a particular
location are balanced at ±5 MJm-2 per day. Similarly, seasonal changes of ±500 MJm-2 can be
estimated as about this amount of warming (cooling) for 100 days of summer (winter).
[p396] To find the change in temperature we need to know the heat capacity per square metre, H:
For a depth d affected by temperature change H[JK-1m-2] = [kgm-3] d[m] C[JK-1kg-1]
Ocean: H =  d 4200 = 4.2 d MJK-1m-2 , with depth of layer responding to temperature change
depending on exponential drop of absorption of sunlight through water
Top ~10m is affected by daily warming.  Daily change T = D.I./H = 5/(10x4.2)= ±0.12ºC
Layer of ~50m affected by seasonal change.  T = 500/(50x4.2)= ±2.4ºC
Land: depth of layer responding depends on heat conduction through layer of soil/rock/sand; air
pockets in dry soil reduce heat transport, water in damp soil will increase it.
1500 kgm-3, C~1000 JK-1kg-1  H = 1.5 d MJK-1m-2
On average only ~10cm layer sees the daily change T = 5/(0.1.x1.5)= ±33ºC
and ~1m for annual changes T = 500/(1x1.5)= ±333ºC (which is clearly ridiculous!)
However, also need to consider the atmospheric boundary layer over the surface which also
gains/losses heat energy.
For daily changes a ~300m thick layer is involved so H = 1.2x300x1000 = 0.36 MJK-1m-2
Combining this with the oceanic heat capacity makes virtually no difference, but over land there
will be a combined H = 0.5 MJK-1m-2 which with perfect coupling would moderate T = ±10ºC.
For annual changes the whole atmosphere thickness is involved so H = 12 MJK-1m-2 which
dominates the land’s heat capacity and means T = 500/(13.5)= ±37ºC ; however, in tropical
regions this change is much less because the sun remains relatively high all year round and so the
seasonal change is more like ±100 MJm-2 leading to T ~ ±7ºC i.e. less than the daily change.
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PX350
Lecture 3
Essential Thermodynamics for those who missed Thermal Physics 1 & 2!
State variables, e.g. p, V, T, describe the “state” of a system. To change state usually requires an
energy change, which could be heat input or work done on the system.
Pressure of a gas is due to rate of collision of the particles within the gas, so is proportional to the
number density and their kinetic energy, which in turn is fixed by the temperature.
The specific heat capacity is the amount of heat (energy) required per degree of temperature
change per unit mass, but the conditions of the change need to be specified e.g. cv at constant
volume or cp at constant pressure (the usual case in the atmosphere).
The specific latent heat, l, is the energy required to change unit mass of liquid to gas (latent heat
of vaporisation), from solid to liquid (latent heat of fusion), or from solid to gas (latent heat of
sublimation). The change happens at a fixed temperature, but that temperature must be specified
because it is easier to break the chemical bonds concerned at higher temperature when the atoms
have more energy.
[Ch4 p86+] State of the Atmosphere
[p88] Treat atmosphere as containing an ideal gas. p  RT
Use pressure and temperature (in kelvin) as the independent state variables, and  is the density.
R*
8.31

 287 Jkg 1K 1 is the gas constant per unit mass NOT per mol and is
3
M 28.97 10
different for each gas depending on their molecular mass M. The value 28.97g/mol is a weighted
average of the molecular mass of N2 (28) and O2 (32).
For air R 
Total pressure is the sum of the partial pressures i of each constituent gas.
p   i pi  i i R iT  mix
 x R T
i
i
i
and the specific mass xi 
i
air
[p96-102] Hydrostatic equilibrium – change of pressure with height.
Consider pressure dp change due to weight of a slab of atmosphere of thickness dz in an
isothermal atmosphere.
Show that dp  gdz , which is the equation for hydrostatic equilibrium
Then divide by the ideal gas law to get
dp
 g
dz
dp
g

dz
p
RT
Integrate this differential equation to show p  z   p0e

gz
RT
, where p0 is the pressure at z=0 i.e.
ground level.
Define H = RT/g as the (exponential) scale height, i.e. height for pressure to drop by factor of e.
Also decadal scale height H10= ln10 RT/g is height for pressure to drop by factor of 10.
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Scale height is approx.. measure of the thickness of the troposphere (which has pressure drop
from 1000mbar to ~200mbar so Htropopause= ln5 RT/g).
Scale height is a function of temperature, so warmer layers of atmosphere containing the same
mass (i.e. same pressure difference) will be thicker. This means the tropopause is higher above
the equator than the poles, which creates a potential energy imbalance that leads to flow of air
from the top of the warmer equatorial columns towards the poles (see global circulation later).
Height of tropopause has been observed to increase by ~10m p.a. indicating global warming of
~0.2 K p.a.
[Ch 5 p120] Thermodynamics of atmosphere.
Consider an “air parcel” as a certain bit of air, of a particular gas mixture and fixed number of
particles (which we don’t need to know), that travels around the atmosphere exchanging energy,
but not particles, and changing its temperature and pressure.
1
The change in heat energy of the air parcel is dq  c p dT  dp

[p145] When no heat energy is exchange the process is termed adiabatic.
Rapid process in the atmosphere can be approximated as adiabatic.
First, consider the case for dry air (i.e. no water content to condense/evaporate)
1
dp which can be changed using the eqn. for hydrostatic equilibrium

dT
g
dp

 g (above) to get the rate of temperature change with height
dz
dz
cp
In this case: c p dT 
The rate of DECREASE of temperature with height is called the Lapse Rate.
In this case the Dry Adiabatic Lapse Rate is  D  
dT
g 9.81
 
 9.8 K/km
dz
c p 1005
i.e. the temperature of the dry air parcel will drop by ~10 K for every 1000 m it is lifted up.
If there is some water vapour in the air parcel, but it remains as a vapour without condensing, the
air is termed “damp”.
The lapse rate will change from the DALR, as the heat capacity is different, e.g. for 1% water
c p  1  x  c pair  xc pwater
vapour
, but not by much as 1%  9.67 K/km
 0.99*1005  0.01*1911  1014 Jkg 1K 1
Therefore, for all cases we use the Dry Adiabatic Lapse Rate, except …
… when the air is saturated with water vapour. Then the lapse rate is less, because cooling the air
parcel leads to condensation into liquid water which releases latent heat and so offsets some of
the cooling. Depending on the water content the Saturated Adiabatic Lapse Rate is 5-9 K/km.
Note this is still adiabatic because the energy change from latent heat to sensible heat is all within
the parcel and does not involve heat exchange with the surroundings.
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Saturated Adiabatic Lapse Rate (SALR)
If water condenses the mixing ratio changes by -dx.
Latent heat dq  l dx is released, so
1
dq  l dx  c p dT  dp

Divide through by dz
l
dx
dT 1 dp
 cp

dz
dz  dz
g l  dx 
 dT 
Final term is g (hydrostatic eqn). Rearrange to get  
    
 dz  sat c p c p  dz  sat
S  D 
LHS is the SALR
l  dx 
 
c p  dz  sat
Since xsat decreases with height (as it gets colder with height) final term is negative and  S   D
Lecture 4
[p147-154] Atmospheric Stability
Consider an air parcel at equilibrium at an original height z0 and temperature T0. The gravitational
force on the parcel, mpg, is balanced by an up-thrust due to weight of fluid (air) displaced, mag.
Now raise the parcel to a new height and calculate the acceleration in +ive z (i.e. upwards)
d 2z
m p 2  ma g  m p g
dt
Volume of parcel and air displaced is same, so replace mass by density; then note density inversely
proportional to temperature from ideal gas law.
 ma

 a

 Tp 
d 2z

g

1

g

1

g




  1
m



dt 2
 Ta

 p 
 p 
The air parcel will change temperature according to the DALR: Tp  T0   D dz
 dT 
The atmosphere will have its own temperature profile: Ta  T0   
 dz
 dz a
d 2z
g
dT 
Hence





 z
D
dt 2
T0 
dz a 
If the quantity in brackets is negative the atmosphere is unstable, since any displacement is
amplified.
Atmosphere is unstable if
dT
dz
  D , i.e. drop in temperature with height is more than 10 K/km
a
If the quantity in brackets is positive this represents S.H.M., which means the atmosphere is stable
since any displacement experiences a restoring force. However, need to consider whether air is
dry or saturated since  D   S
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Weather and the Environment
Atmosphere is stable if
If  S 
PX350
dT
  S , i.e. the drop in temperature with height is less than ~6 K/km
dz a
dT
  D the atmosphere is conditionally stable, i.e. stable if dry, turbulent if saturated.
dz a
An isothermal atmosphere is always stable, as is one where temperature increases with height.
This is known as a temperature inversion, as seen in the stratosphere and often near ground level
after a cold night. The temperature inversion at the tropopause stops moist air ascending into the
stratosphere and thereby retains the water content on Earth. Some vigorously convecting clouds
can extend a few hundred metres above the tropopause, due to overshoot in SHM.
Take care in evaluating these inequalities from graphs that show height on the y-axis and
1
dz  dT 
temperature on the x-axis, as (i) the slope of that graph is

 and (ii) temperature is
dT  dz 
decreasing with height, but it is the magnitude of decrease that has to be compared to  .
[p 475-480] For a stable atmosphere there will be an SHM frequency, N (Brunt-Vaisala frequency)
2
g
N 2   D Leading to oscillations of period  
500s .
N
T
For typical wind speeds (~20 m/s) this represents a wavelength of 3-20 km and shows up as lines
of lenticular clouds parallel to and in the lee of mountain ridges, known as “Lee Waves”.
Lectures 5/6/7
[p 126] Water in the atmosphere
Fraction of water vapour in air is given by the mass mixing ratio, x 
mass water vapour v

mass air parcel

Typically x ~ 10gkg 1   1% 
[The book defines both the specific humidity q and humidity mixing ratio r , depending on whether
the denominator in x is the total air parcel mass or just the dry air mass, but since x<<1 these are
essentially interchangeable. Book also treats a lot of this material in terms of potential
temperature and tephigrams, which we will not use in this module, so ignore these sections.]
The partial pressure exerted by water vapour is given by the ideal gas law pv  v RvT ,
where Rv = 462 Jkg-1K-1 is the gas constant for water.
Or in terms of the total air pressure. The ratio of gas constants is the inverse of the molecular
mass ratio. Often see the ratio   R  M v  18  0.622
Rv
M air
29
[p126 and 158] Saturated vapour pressure (svp) is the vapour pressure over a flat surface where
the liquid and vapour are in contact. In this equilibrium the rate of evaporation of molecules from
the liquid equals the rate of condensation of molecules from the vapour. The former is a function
of temperature only, so the svp is a function of temperature only.
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The svp line divides the liquid and vapour phases on a P-T phase diagram and has an equation of

L
psvp  p0 exp 

 Rv
 1   1 

      , where p0 is the svp at T0 , usually chosen as 6.1 mbar at 0ºC.
 T0   T  

(This is derivable from the Clausius Clapyron equation, but you don’t need to.)
[p158-161] Ice and Water.
Maximum water vapour content of air is given by svp(T). From graph this is ~1% (12 mbar) at 10ºC
and ~4% at 30ºC. At very low temperatures air is essentially dry since svp(-30ºC) < 1 mbar.
Since the latent heat of sublimation is greater than the latent heat of vaporisation the svp over ice
is lower than the svp over supercooled water. (This can be seen as due to bonds in ice being
stronger so harder for molecules to escape.) Hence, vapour that is just saturated with respect to
water is supersaturated with respect to ice, leads to preferential condensation on ice crystals over
water droplets. Air over ice is thus even drier.
Relative humidity defined as R.H. 
vapour pressure
svp at same T
If the vapour pressure equals the svp, RH=1, and condensation can occur.
This could be achieved by either increasing the water vapour pressure by adding more water
(increasing x), or by reducing the temperature so that svp drops to reach the existing water vapour
pressure (well actually the svp drops exponentially which is faster than the linear drop in vp from
ideal gas law so they meet).
Supersaturation SS=RH-1. In atmosphere SS is rarely above 1%.
Condensation in atmosphere occurs whenever region of damp air is cooled below saturation
temperature (aka dew point). Examples include:
• Radiation loss
Radiation fog: land cools during cloud-free night by radiation to
space; cools overlying layer of air below dew point; fog forms.
• Ascent (of air mass)
Usual case is that temperature drops with height, so lifting up
air mass will cool it. Hill fog/clouds: air moving towards hills
has to ascend, so forms cloud above certain height. Frontal
systems: warm moist air is forced above cooler more dense air,
cools and forms cloud.
• Descent, through inversion
Sometimes temperature at ground level is lower than above,
known as a temperature inversion. Air descending, typically
into a valley at night, will cool and condense. Steam fog: forms
over lakes in valleys.
• Advection, over cool region
Sideways motion of air mass to cooler region. Advection fog
(Haar): moist maritime air blown over cool land in winter.
Newfoundland fog banks: warm moist air over Gulf Stream
moves over cold Labrador Current.
• Influx of cold dry air
Blast of cold air (eg. Katabatic flow off glacier) cools warmer
moist air over water. Arctic Sea Smoke.
• Convection
Ascent of air parcels in convection circulation, forms individual
or regular array of cumulus clouds.
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Lifting Condensation Level (LCL), is the level at which air would become saturated if lifted dry
adiabatically from the surface. It is the height at which clouds form.
Example to calculate LCL when moisture content at ground level is known.
At ground level, temperature =T1 , water vapour mixing ratio = x , so water vapour partial pressure
is related to total air pressure p0 by p(T1) = x p0 and RH = r  p T1  psvp T1 
At LCL, RH=1, but water vapour partial pressure has also reduced as temperature is less at T2 ,
p T2  T2
cooling of an ideal gas follows
At LCL have 1  p T2  psvp T2 

p T1  T1
T1 psvp T2  T1 exp   L RwT2 

T2 psvp T1  T2 exp   L RwT1 
where gas constant for water vapour is Rw = 461 Jkg-1K-1.
Combining, r 
If change in T is small then
T1
T2
 L
1 and r  exp  
 R
w

 1 1 
1 1 Rw
 
ln r
    and
T2 T1 L
 T2 T1  
1
461
 1

Putting in values for T1=280K, r=70% T2  

ln 0.7   275K
6
280
2.5

10


Hence, temperature at LCL is 5 K lower than at ground level and, following DALR, LCL is at ~500m.
[If full change in T included psvp T2  should be lower (colder) and LCL slightly higher at 530m. ]
Lee winds
If damp air mass is forced over a ridge of mountains it will cool and condense during the ascent
and form clouds on the windward side of the ridge. Temperature will then drop by SALR for
further ascent to the top of the ridge. On descending in the lee of the mountains the air will by
dry and warm as DALR. This can lead to warmer temperature at sea level in the lee of mountain
range and a warm “lee wind” blowing down the hill, known as Föhn (Alps), Chinook (Rockies),
Zonda (Andes) etc., which can be up to 20ºC warmer and persist for several days.
During a hot day, air in a valley will warm and expand, giving rise to a warm breeze blowing up the
slopes of the valley, known as an Anabatic wind.
During a clear night, air in a valley will cool and contract, giving rise to a cool breeze blowing down
the slopes of the valley, known as an Katabatic wind. This effect is exaggerated if there are
glaciers present.
By collecting cool air in the base of the valley there will be a temperature inversion over-night and
cloud (mist) may form in the valley floor. This cloud gradually rises as the morning sunshine heats
the air.
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[Ch 6] Warm Rain Process
•
•
•
•
•
Saturation
Nucleation
Growth by condensation
Growth by accretion
Rainfall
[p170-176] Nucleation. svp over curved surface is greater than over flat surface, because
molecules less tightly bound by neighbours so can escape the liquid state more easily.
1
2 
 1.2 
psvp  r   psvp    exp 
  psvp    1 
 for water at 273K, with r in nm.
r 

 r water RT 
A water molecule of 0.13nm radius would need RH~1020% to get spontaneous condensation. But
observe SS of only 1%, so condensation must be happening on much larger (>100 nm) particles.
Nucleation on particles from soot, sea salt, pollen, dust, existing droplets etc. More small particles
over polluted urban areas. Typical atmospheric concentration of particles:
Radius
Name
Conc. (cm-3)
SS
< 0.1 µm
Aitken nuclei
10,000
1%
0.1 - 1 µm
large nuclei
100
0.1%
> 1 µm
giant nuclei
1
0.01%
Hydroscopic droplets, containing sea salt, have lower svp because there are fewer water
molecules in their surface and these are held more tightly than in pure water droplets. So


combining with above expression for curved surface, psvp  r   psvp    1 
1.2 Bm 
 3 .
r
r 
The result is that small droplets can form on hygroscopic nuclei at RH of only 80-90%. The Köhler
curves [p172 & Lecture 5 slide #10] plot this eqn. and show a maximum SS as a function of radius.
This is the critical radius rc above which droplets are said to be ‘activated’ and will grow rapidly by
condensation, without limit, forming clouds. Below rc droplets are stable, but don’t change size
unless the RH changes. They appear as haze, where the 0.1-1um droplets scatter sunlight and
reduce visibility, but without clouds developing.
[p181] Growth by Condensation.
Once a droplet has nucleated it grows by diffusion of water vapour from the saturated air (SS>0)
to the liquid surface (SS=0). The rate of diffusion is given by Fick’s Law.
In the diffusion process the rate of change of surface area of the drop is constant, which means
d  4 r 2 
dt
 const. i.e. r
dr
dr 1
 const . or

dt
dt r
A full derivation is given on p181 and on Lecture 5 slide #12.
dr D  svp  SS 1

dt
w
r
Hence, initial growth is very rapid, reaching ~20um in a few minutes, but subsequent growth is
slow and would take days to grow to the size of full rain drops (r~1mm).
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How long does it take a droplet to grow from r1 to r2?

r2
w
r1
D  svp  SS

   r22  r12 

Putting in svp at 11ºC, SS=0.2% and r in um,   1.1 r22  r12 seconds.
dr D  svp  SS 1

dt
w
r
Time to reach 20um ~ 7.5 minutes – typical cloud drop size.

Time to reach 100um ~ 3 hours – small rain drop size.
0
Time to reach 1mm ~ 13 days – large rain drop size.
rdr   dt
w
2 D  svp  SS
Therefore, must have additional process operating to produce rain
within timeframe of hours.
Evaporation of the drops takes place outside the saturated region of cloud, using the same
formulae with SS<0. So time to evaporate in a region with RH=90% (SS=-10%) is   0.02 r12  r22 ,


or less than 10 seconds for a 20um drop, but many hours for a rain drop.
Hence, clouds have well defined edges because the small cloud drops essentially evaporate as
soon as they leave the saturated region.
[p184] For precipitation to reach the ground its fall time must be less than the time to evaporate.
The terminal velocity of drops in turbulent flow from about 30um to 1mm is vt[m/s] ~ 8 r[mm].
To calculate distance that drop falls before evaporation:
1
1
 dr 
vt~r and    r , so h~r4 With realistic values, a 100um radius drop
 dt 
(drizzle) can fall only about 250m before disappearing, so drizzle is only
1
0
 dr 
experienced from low clouds; grey streaks seen below higher clouds where
h   vt   dr
r
 dt 
small rain drops are falling, but fail to reach the surface, is known as virga;
by contrast a rain drop of 1mm could fall hundreds of km.
 dr 
dz  vt dt    dr
 dt 
[p185] Growth by collision
From the initial nucleation phase, and through the growth by condensation phase, there will be a
distribution of drop sizes within a cloud. Larger cloud drops (R) fall through the cloud (at speed V)
and grow as a result of colliding with other smaller drops (r) (which are falling more slowly v).
Rate of mass increase is related to rate of size increase by
dm
dR
  w 4 R 2
dt
dt
Rate of mass increase is also given by mass of smaller droplets swept up per unit time, which is the
mass of all those within a volume   R  r  V  v  . If efficiency of collection is E and mass of
dR
2
water per unit volume given by xs air , then  w 4 R 2
   R  r  V  v  xs  air E
dt
dR xs  air E
dR
if R  r and V  v

V . For medium size drops V ~ R so
R
dt
dt
4w
2
R 
Following similar approach to above, time to grow from R1 to R2 is   B ln  2 
 R1 
With realistic values it can take ~30 minutes to grow from cloud drop to rain drop.
Rain drops larger than ~2mm will break up due to turbulence in the cloud.
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[Ch 6 p186-189] Cold Rain (Bergeron-Findeison) Process
Clouds generally contain supercooled water drops, rather than ice, because a temperature of
-40ºC is need for spontaneous ice crystallisation and because there are ~10 6 fewer nuclei in clouds
that ice will form on than water (due to need for correct crystal shape).
However, once some ice crystals do nucleate, at temperature between -10ºC and -20ºC, they will
grow rapidly at the expense of the water droplets. This is because the svp over ice is less than
over water at the same temperature (due to greater latent heat of sublimation than of
evaporation), hence there is a vapour pressure gradient that encourages mass transport from
water drops to ice crystals.
Ice crystals can grow to size of mm’s within a few minutes, especially as they are very efficient
collectors for the collision process.
Large ice crystals will break up in turbulence of cloud and so provide more ice nucleation sites.
It was observed that rain falls from cumulonimbus clouds soon after the top of the cloud shows
glaciation (ice formation).
Precipitation types
In tropics, most precipitation is by warm rain process and falls as rain drops, except in very intense
storms where the clouds extend the full height of the atmosphere into the very cold regions.
In mid-latitudes, most precipitation starts as ice crystals in the cloud. These can coagulate to form
snow-flakes, or melt and refreeze to produce hail. In active clouds with vertical up/down-draughts
hail stones can experience multiple freezing cycles and so emerge with a layered structure. Rapid
freezing can trap air in the hail, which gives it a white appearance.
If the sub-cloud layer is warm the snow/hail will melt to fall as rain. (see (i) below) If it stays < 0ºC
then snow falls (ii). When snow partially melts the result is sleet (requires thinner warm region
and thicker sub-zero region). Hail that partially melts in the sub-cloud layer is called graupel.
Occasionally the precipitation will have completely melted to rain drops and then pass through a
very cold layer of air that leaves it supercooled, but not frozen as there are no freezing nuclei (iii).
When this supercooled water lands on the surface it freezes instantly. Known as freezing rain.
(i)
(ii)
(iii)
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Cloud Types
[p437] Cumulus clouds generally form in convective system i.e. in unstable air. Damp air rises in
the sub-cloud layer until reaches LCL where condensation starts. Convection continues through
the cloud, as release of latent heat makes air parcel more buoyant. Once all water content has
condensed convection will stop and dry air descends around the edge of the cloud. This
descending dry air then mixes in the sub-cloud layer to retain a sub-saturated vapour, and hence
cloud free layer despite drawing in additional moisture from the ground.
Base of cloud is well defined (appears to be flat) by LCL and fact that any droplets falling from the
cloud evaporate very quickly in the sub-cloud layer. Similarly edges are well defined by warm
descending air, but tops continue to grow in turbulent updraughts.
Often these cumulus clouds are in a pattern determined by the convection cells. Individual clouds
are initially ~100m across. As the cloud drifts through the convection patter it may grow to a few
hundred meters across or start to dissolve over a time period of ~ten minutes.
Individual cumulus humilis can grow and merge to form cumulus congestus. After 20-30 minutes
cumulus can have grown to form towering cumulonimbus, (Cb) with potential for rain.
With strong convection in the tropics, very tall individual clouds can form that reach the
tropopause. In the presence of vertical wind shear these tall clouds will appear slanted.
As cumulus dissolve the remnants are known as fractocumulus or scud.
If there is a stable layer of atmosphere above where the cumulus is developing, convection will
stop and cumulus appear flattened. These can become stratocumulus.
Stratus are layered clouds that indicate a stable layer of atmosphere. Often form at warm fronts,
or other places where an air is lifted en-mass.
Cirrus clouds consist entirely of ice crystals, high in troposphere. Ice crystals evaporate more
slowly than water droplets so often see thin, wispy cirrus.
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A tropical rain calculation:
Tropical climate typically has Cb clouds form each day and release torrential rain in afternoon.
With ~3m annual rainfall that means ~10mm per day in a single half hour downpour.
Water falling per square metre is thus 1cm/m2 with a volume of 104cm3/m2 or mass of 10kg/m2.
If rain comes from a cloud that is 8000m high, water content in whole cloud is 10/(8000*1.2)=0.1%
How many cloud droplets?
Typical radius of 20um. Vol. of droplet = 43   20 106   33 1015 m3  33pl
3
3 1011
10 103
11
 3.8 107 m 3

3

10
Or
8000
33 1015
This is determined by the density of nucleation sites.
Number of droplets =
How much energy? Total energy required = evaporation + lifting
mL  mgh  10* 2.4 106  10*10*5000
 2.4 107  5 105  24.5 MJm 2
Compare this with the daily insolation in tropics found earlier of 20 MJm-2. And note that only
about ¼ of area is covered by cloud so energy can be “collected” from 4x cloud area.
Non-tropospheric Clouds
Occasionally clouds form above the tropopause. These are very thin ice clouds and are only visible
after sunset in reflected sunlight.
Nacrecous clouds are 15 - 25 km high, in the stratosphere well above tropospheric clouds, and
occur mostly in polar regions and in winter at high latitudes. Seen after dusk. Also called motherof-pearl or polar stratospheric clouds (PSC). Responsible for the catalysed destruction of ozone.
Noctilucent clouds are thought to be composed of small ice-coated particles; their precise nature
remains a mystery. They form at very high altitudes ~ 82 km - and are a quite separate
phenomenon from normal tropospheric clouds.
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Lectures 8/9
[Ch 7] Atmospheric Dynamics
Consider motion of air parcel with respect to a frame of reference fixed on the rotating Earth.
Assume northern hemisphere unless told otherwise.
Position r = (x, y, z) – x points East; y points North; z is “up”, perpendicular to the surface.
Velocity v = (u, v, w)
Lines of constant pressure (e.g. on a weather map) are called isobars.
Pressure gradient force [p208]
Consider forces acting on a box of side dx, dy, dz in a region where there a pressure gradient.
In the x-direction, there will be a net force dFx  m
du
p
  px  dx  px  dy dz   dx dy dz
dt
x
The parcel has a mass m   dx dy dz , so acceleration in the x-direction is
du
1 p

dt
 x
1
This is called the Pressure Gradient Force (strictly a force per unit mass). In 3-D PGF   p

Air flow will initially be from high pressure to low pressure, but ...
Coriolis Force [p216]
The Earth is a non-inertial frame as it rotates once per day (anticlockwise looking at the N-pole)
i.e. the angular velocity  
2
 7.27x10 5 rads 1
24 * 60 * 60
Time derivatives in a rotating frame are related to those in an inertial frame by:
 dr 
 dr 
       r and, with all terms on RHS in rotating frame, the acceleration is
 dt  In  dt  Rot
 d 2r 
d
  dr

 2        r 
  dt

 dt  In  dt
d 2r
dr
 2       r
2
dt
dt
dv

 2  v      r
dt

The actual forces that cause acceleration in an inertial frame are gravity g, friction F and PGF.
 d 2r 
1
  p  g  F
 2

 dt  Inertial
So, in the rotating frame of reference the acceleration is:
dv
1
  p  2 v  g    r  F
dt

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The second term on RHS is the Coriolis Force (CoF). Note the “force” is proportional and at right
angles to the velocity. In the N-hemisphere this causes a moving air parcel to deviate to the right.
The third and fourth terms can be combined as an apparent gravity g’ [p220]. At the equator the
fourth term is 2 R , which is the centripetal acceleration needed to stay on the surface of the
rotating Earth. However, the Earth itself needs to stay together and has distorted from a sphere
with a larger equatorial radius, and should be considered as a gravitational equipotential, i.e. the
surface is always perpendicular to the local apparent gravity which varies with latitude. At the
poles g’ = 9.83 ms-2 while at the equator g’ = 9.78 ms-2.
The acceleration can be written in components for latitude ϕ:
du
1 p

 2 sin  v  2 cos  w  Fx
dt
 x
dv
1 p

 2 sin  u  Fy
dt
 y
dw
1 p

 g   2 cos  u  Fz
dt
 z
Scale Analysis [p224]
The equations of motion can be simplified by estimating the size of each term and only keeping
those of similar size. Two length scales will be considered: large scale (called synoptic, as it refers
to features that appear on synoptic charts aka weather maps) and small scale. Typical values of
relevant quantities appear in the table below. [Friction ignored for now]
Quantity
Synoptic scale
Small scale
Length x,y
1000 km
Height z
10 km
Time
1 day ~ 105s
Horizontal pressure change
10 mbar
Vertical pressure change
1000 mbar
10-4 s-1
2sin
1 kgm-3
Density 
From these values other quantities can be estimated:
10 km
10 km
1000 s
1 mbar
1000 mbar
10-4 s-1
1 kgm-3
Horizontal velocity = x/t
Vertical velocity = z/t
Synoptic Scale
du
1 p

fv
dt
 x
dv
1 p

fu
dt
 y
1 p
0
 g
 z
10 ms-1
0.1 ms-1
10 ms-1
Small Scale
du
1 p

dt
 x
dv
1 p

dt
 y
1 p
0
 g
 z
Note: in each case, the horizontal and vertical components separate;
Coriolis is only important for large scale motion and f  2 sin  is the Coriolis parameter.
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Geostrophic Balance [p229]
In the synoptic scale analysis, above, the acceleration terms are an order of magnitude smaller
than the Coriolis and PGF terms, therefore a steady state can be found where acceleration is
ignored. In that case the horizontal equations become:
u
1 p
f  y
v
1 p
f  x
Note that the velocity is perpendicular to the pressure gradient: a geostrophic wind blows parallel
to the isobars (in the absence of friction). The closer the isobars, the stronger the wind.
Calculate typical geostrophic wind speed
At 50ºN the isobars are separated by 4 mbar and appear 200 km apart running east-west on a
weather chart, with low pressure to the south, what is the wind speed and direction?
1
4 102
1
0
u   4
 15ms 1 v  4
0
3
10 *1.3 200 10
10 *1.3 x
The wind is a 15m/s Easterly (it blows from East to West).
Cyclostrophic Balance
For small scale systems the Coriolis terms are negligible. E.g. in a tornado the centripetal
acceleration at a radius r is balance by the PGF (modulus sign used as there is low pressure at
centre).
v 2 1 p

r  r
v 
r p
 r
The relative sizes of acceleration and Coriolis terms determines the Rossby Number. (Same
U
v2 r
person, but unconnected with Rossby waves.) Ro 
or for an estimate Ro ~
fL
fv
When Ro >> 1 cyclostrophic balance prevails and when Ro << 1 get geostrophic balance.
Calculate typical tornado wind speed
What is the wind speed around a tornado of radius 500 m if the central pressure is 100 mbar lower
than in the surrounding atmosphere?
Small scale system, so use cyclostrophic balance.
You have been given Δp and r but not ∂p/∂r, so there is insufficient information unless you assume
the pressure changes uniformly over the radial distance, then v 
500 100*100
 88 ms-1
1.3
500
More likely, pressure drops over a shorter distance so the velocity would be higher.
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Lecture 10/11
Gradient Wind [p241]
Large systems such as mid-latitude cyclones and anticyclones (low and high pressure systems,
respectively) can have intermediate values of Ro , so all terms are needed.
For a N-hemisphere Low pressure system, the centripetal acceleration is provided by the PGF
minus the CoF and the wind blows anticlockwise along the isobars.
[Draw a diagram to convince yourself!]
v 2 1 p

fv
r  x
v 2  fr v 
rf
 rf  r p
vgr      
2
 2   r
r p
 0 , then solve quadratic
 x
2
CYCLONE
There is no inherent limitation on the size or pressure gradient in a low pressure system.
For a given PGF the wind speed is lower than for straight isobars, or sub-geostrophic.
For a N-hemisphere high pressure system the wind blows clockwise and the centripetal
acceleration is given by CF minus PGF. Similar maths gives:
rf
 rf  r p
vgr      
2
 2   r
2
ANTICYCLONE
For a given PGF the wind speed is higher than for straight isobars, or super-geostrophic.
Note that the contents of the square root must be positive, with first term larger than second.
So high pressure systems
(i) do not form at the equator where f=0;
(ii) cannot have a very large pressure gradient or consequently very high wind speeds;
(iii) tend to be large systems
Effects of Friction [p241]
Friction occurs in lower atmosphere due to proximity of the surface and its boundary layer.
Exactly at the surface there is no wind and its velocity will increase with height through the
boundary layer until the free atmosphere is reached at about 500 m. For laminar flow in stable air
this will be approx. linear increase with height, whereas in turbulent, unstable air, the wind speed
increase more rapidly in a log fashion.
Friction always acts in opposite direction to velocity
vector, and therefore perpendicular to CoF. As velocity is
reduced by friction, the CoF becomes smaller for a given
PGF, leading to the force balance shown below, which
means the wind direction has backed by an angle θ
compared to the geostrophic wind vg. Compared to the
high level wind in the free atmosphere, the surface level
wind will blow from further to the left (in N-hemisphere
for both high and low pressure systems). This is known as
wind shear.
As a function of height the wind direction will change
continually in an Ekman spiral. Consequently there will
v
CoF=fv
vg
F 

PGF=fvg
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Weather and the Environment
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be a component of velocity across the isobars towards lower pressure.
Friction will be greater over land than oceans, so θ will also be greater over land than sea.
Clockwise rotation of wind direction is known as veering, anticlockwise rotation is backing.
Frictional Convergence [p265]
In a mid-latitude cyclone (anticyclone) the result of friction causing air flow across the isobars is an
influx (out flow) of air at surface level. Air can be treated as incompressible so this causes air to
rise (fall) within the system and spiral out (in) at higher levels.
Rising (and cooling) air in a low pressure system will lead to clouds forming and hence overcast
weather and rain. The inflow of fresh air at the surface improves air quality.
For a high pressure system the descending air leads to clear skies, making the associated weather
cold in winter and warm/sunny in summer. Air quality can be poor if the high stays static.
Basic principles determining climate []
Warmer air is less dense than cooler air. A vertical air column containing a fixed mass of air (i.e.
with the same surface pressure difference from top to bottom) will be taller if it is warmer. This
explains why the tropopause is higher at the equator than at the poles. Alternatively, a warmer air
column of fixed height will contain less mass of air and produce a lower surface pressure.
Oceans have a very large heat capacity, so stay at same temperature all year round.
Regions close to an ocean will therefore have small annual temperature variation.
Land masses, away from the tropics, change temperature dramatically between summer and
winter. In winter (summer) the land is cold (hot), so the overlying air is also cold (hot), more (less)
dense and has a higher (lower) surface pressure. [Remember “winter” is June in S-Hemisphere!]
Prevailing wind directions are, on average, set by the global circulation (see below). Particular
surface wind direction depends on pressure distribution, blowing approx. parallel to isobars.
Moist air forced to rise, for instance by a mountain range, will produce clouds and hence rain
(orthographic effect). In the lee of mountains the air will be descending and dry, so less rainfall.
Air Masses are bodies of air that retain their properties, such as temperature, stability and
moisture content, whilst moving over quite large distances.
These properties are characteristic of the origin of the air mass: polar - cold, tropical – warm,
continental – dry, maritime – wet. Remember that warmer air can hold more water vapour.
Where dissimilar air masses meet there will be a front. Warmer (less dense) air mass will rise over
colder air mass, often leading to clouds associated with the front. If the frontal slope is gentle and
the air relatively stable these will be stratus family, if steep/turbulent then cumulus result.
The surface pressure distribution, and hence direction air masses move, is intimately connected
with processes in the upper atmosphere and the exact positions of Jet Streams. This is very
complicated, if not currently impossible, to predict!
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Lecture 12/13
Global Circulation [p108-117, 314-323]
Maximum solar heat input where the Earth’s surface is normal to the solar flux, i.e. on the equator
at the equinoxes and otherwise within the tropics. A major result of the global atmospheric
circulation is to redistribute heat towards the poles; heat also flows polewards via the oceans.
Meridional flow = N-S motion, along lines of constant longitude e.g. the Greenwich Meridian.
Zonal flow = E-W motion, along lines of constant latitude.
Half of the surface area of the Earth is within 30ºN/S, known as the tropics. In this region 75% is
ocean.
Hadley Cell [p110]
Heating at the Equator leads to moist air rising, forming towering cumulus clouds.
Ascending air reaches the tropopause, where there is a temperature inversion, it cannot ascend
further and must therefore flow N or S.
At ~30ºN (or S) the air descends to the surface and then returns towards the equator.
Descending air leads to a surface high pressure region (subtropical high STH) and will be dry so the
surface regions below this descent will be arid. Most of the world’s deserts are in these latitudes
~30ºN/S. Air flow around the STH produces a W to E flow of surface winds in mid-latitudes.
Surface level return flows (~1 km thick layer) from each hemisphere meet at Inter-Tropical
Convergence Zone (ITCZ) and help drive the circulation by forcing upward motion in the ITCZ. The
ITCZ is approx. 5º wide, although this varies in real observations.
Effect of rotation [p112]
An air parcel on the Equator, stationary with respect to the Earth, has a velocity of ΩR = 464 ms-1
relative to the stars. At 30ºN the tangential velocity is ΩRcosθ = 402 ms-1. Therefore, if the air
flows N without any friction (which should be a good approx. in the upper atmosphere) it will
acquire a west to east velocity relative to the Earth’s surface.
The quantity to conserve in an inertial frame is angular momentum about the Earth’s axis i.e. mvr .
So the relative velocity v acquired is given by ΩR.R = (v+ Ω Rcosθ).Rcosθ , or v = ΩR(1/cosθ – cosθ)
At 30ºN, v = 134 ms-1 . This high speed means the Hadley Cell could not extend all the way to the
pole and limits it to about 30º. It also sets up East flowing Subtropical Jet Streams at the poleward
edge in the upper atmosphere.
The surface return flow of the Hadley Cell is slowed by friction, but still acquires a West flowing
component due to the Earth’s rotation. This sets up the Trade Winds that blow from the NE in the
N-hemisphere and SE in the S-hemisphere towards the ITCZ. Exactly at the ITCZ the air is moving
upwards and there is no N-S component, although there is a West flowing component.
Two other cells are suggested, Ferrel and Polar cells, but not strongly evident due to influence of
extra-tropical cyclones in mid-latitudes. However, the Polarfront Jet Stream that appears above
the Polar Front between the warm air of the Ferrel Cell and the cold air of the Polar Cell is a real
feature with large influence on mid-latitude climate (see below).
Tropical Climate
Daily weather within the ITCZ (e.g. Singapore) tends to be consistent and repeatable. Intense solar
heating, and high moisture content of air (because it is warm), leads to cumulonimbus formation
over the course of 4-8 hours. These clouds can reach tropopause, or in case of strong convection,
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overshoot into the stratosphere. Usually, a thunderstorm and heavy rainfall follow, >1mm falling
in one go.
The 23.5º tilt of the Earth means that the ITCZ moves seasonally north and south of the equator,
which creates a tropical wet and dry climate for locations that have a pronounced wet season
when under the ITCZ and dry season for the other half year. Movement of the ITCZ lags the time
when the Sun is directly overhead by about two months and only reaches about 14ºN. It is also
highly variable, due to interaction with other weather systems, and means that annual rainfall can
change dramatically towards the limit of ITCZ excursion e.g. Sub-Saharan Africa (Sahel).
Asian Monsoon [p535-8]
“Monsoon” means “season”. In the Asian Monsoon there are two pronounced seasons: one very
wet and the other dry, the timing and intensity of which varies from place to place depending on
local land/sea alignment. In N-hemisphere winter, cold air over the Asian land mass leads to a
surface high pressure there that forces air to flow S over India. The air is quite dry as it originates
over land and dries further through descent over the Himalayas. This is the dry season in IndoChina. Extension of the southerly airflow over the Indian Ocean leads to a wet season over
Indonesia and NW Australia.
In the summer, the land heats up forming a low pressure region over the Tibetan plateau, which
draws in air from the Indian Ocean. The air is very moist due to evaporation over the ocean under
intense sunlight and from being warm. Ascent over land then triggers cloud formation and intense
rainfall in this wet season. The system also pulls in moist air from the Pacific over Japan and China.
Hurricanes [p541-554]
Hurricanes (Tropical Cyclones or Typhoons) are tropical revolving storms that form over large
areas of warm oceans (water >27ºC to encourage sufficient evaporation and release sufficient
latent heat on condensation to drive the system). They occur during the autumn (whichever
hemisphere) after the oceans have received maximal heating. Rotation is central to formation and
requires a contribution form the Coriolis force, so hurricanes form at 5º-15º, not the Equator.
Hurricanes develop from tropical depressions, with lines of tall Cb “hot towers”, through tropical
storms, having some rotation and wind speeds >17ms-1. Category 1 hurricanes have surface wind
speeds of > 33 ms-1 in a ring surrounding a relatively calm eye, with a surface pressure drop of
> 20 mbar. The strongest Category 5 hurricanes have winds > 70 ms-1 and pressures <920 mbar.
Although the central pressure of a hurricane (~950 mbar) is similar to some mid-latitude cyclones,
the pressure drop is over a small distance (mostly the eye wall thickness) meaning that the
pressure gradient is much larger (up to 1 mbar.km-1).
Spin-up Process: moist air over a warm ocean converges towards the core of the hurricane; the air
rises, cools and water condenses into tall clouds, which releases latent heat; heating of air in the
core lowers the central pressure, which encourages further convergence …
Air converges from a radius of ~1000 km. Air drawn from N travels more slowly (in an inertial
frame) and that from S faster than the centre of the hurricane, leading to rotation. Conserving
angular momentum about the hurricane core, mvr is the same at the eye as at a long distance r0:
mveye reye  m r0 sin  r0
veye  f
r02
2reye
Typical values: r0~500 km, reye~20 km
Once over land, hurricanes decay in intensity as the source of latent heat is removed.
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Hurricanes cause extreme damage due to: intense rainfall, of up to 100 mm in a few hours; the
strong winds, to which the westward motion of the system (~10ms-1) should be added; storm
surges of several metres, from waves whipped up by the winds and the low central pressure
causing a local rise in sea level.
The eye of the hurricane is warmer than the surroundings by ~8ºC. Air in this region is therefore
less dense and there is a drop in pressure compared to the surroundings. Air in the core is slowly
descending (as a balance to strong up-draughts within the clouds of the eye-wall) which means it
is dry and a cloud free region.
Lecture 14
Mid-Latitude Climate [p559, Ch 12]
Much of the weather experienced in mid-latitudes (~30-60ºN/S) can be explained by motion of air
masses (see above), bringing characteristic wet, dry, warm and cool weather from their source
region. The direction of travel of air masses is related to the surface pressure distribution. Their
effect over land will be enhanced by orthographic effects, i.e. if a moist air mass is forced to rise
over hills then clouds will form, resulting in much rain, whereas in the lee of a ridge the air will be
drier and consequently lower rainfall.
In addition extra-tropical cyclones, and their associated fronts, play an important role and their
direction of travel is related to the location of the polar front, with its contrast between warm and
cool air masses, which itself is ‘steered’ by jet streams in the upper atmosphere.
For most of mid-latitudes the prevailing surface winds are from the West. Western coastal
margins therefore have mild, damp climates due to predominance of maritime air masses that
don’t vary in temperature during the year. By contrast, continental regions have large seasonal
swings in temperature due to heating and cooling of the land surface and tend to be drier.
UK Weather associated with different wind directions
Westerly
Unsettled weather with variable wind directions as depressions cross the
country. Mild and stormy in winter, generally cool and cloudy in summer (mP, mPw, mT).
NW Cool, changeable conditions. Strong winds and showers affect windward coasts
especially, but the southern part of Britain may have dry, bright weather (mP, mA).
Northerly Cold weather at all seasons, often associated with polar lows. Snow and sleet
showers in winter, especially in the north and east (mA).
Easterly
Cold in the winter half-year, sometimes very severe weather in the south and
east with snow or sleet. Warm in summer with dry weather in the west.' Occasionally
thundery (cA, cP).
Southerly
Warm and thundery in summer. In winter it may be associated with a low in
the Atlantic giving mild, damp weather especially in the south-west or with a high over central
Europe, in which case it is cold and dry (mT or cT, summer; mT or cP, winter).
Cyclonic
Rainy, unsettled conditions often accompanied by gales and thunderstorms.
This type may refer either to the rapid passage of depressions across the country or to the
persistence of a deep depres¬sion (mP, mPw, mT).
Anticyclonic Warm and dry in summer, occasional thunderstorms (mT, cT). Cold and frosty in
winter with fog, especially in autumn (cP).
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Extra-tropical cyclones [p485-500]
The formation process of an extra-tropical (mid-latitude) cyclone (low pressure system) is known
as cyclogenesis. First, consider air near the surface (see diagrams in lecture notes). The polar
front provides a marked contrast between cold, relatively dry, polar air and warm moist tropical
air. A kink that appears in the polar front will be amplified as the cold air mass sinks and the warm
air mass is forced to rise. Two distinct fronts that divide the different air masses are associated
with the cyclone. Lifting of the warm air at the fronts leads to regions of cloud. The gradual slope
(1:100) of the warm front produces layered stratus clouds and an extended region of light rain.
The steep (1:20) cold front forms tall Cb clouds and a more intense band of rain. At the fronts
there are sharp changes in wind direction, associated with bends in the isobars. Between the
fronts is the warm sector with parallel isobars and mostly cloud free.
A fixed observer might make the following observations of a cyclone approaching from the west,
with the centre of the low pressure due to pass to the north of the observer:
Initial view of high level cirrus clouds associated with the warm front. {If this is seen around
sunrise, the sunlight from the east will be reflected from the clouds of the warm front giving the
classic warning of an approaching storm: ‘red sky in the morning’.} Wind is cyclonic from the SW
and slowly veering. Pressure is dropping slowly. Temperature is cool. As the warm front
approaches more lower level stratus cloud appears and light rain follows. The warm front may be
~300 km wide and so take several hours to pass if the cyclone is moving at ~10ms-1.
When the surface projection of the warm front arrives the observer will enter the warm sector;
there will be a sharp increase in temperature, wind shift to be from the W, rain stops, sky mostly
clears, pressure stays constant.
When the cold front arrives there will be heavy rain, maybe thunderstorms. Then a marked drop
in temperature, clear skies, pressure starts to rise again, wind shift to be from NW (and continues
to veer). {After the cold front has passed the observer could look E at sunset and see a ‘red sky in
the evening’ showing that the storm had passed.}
Frontal Slope
Consider pressure change along a front sloped at angle α through the warm and cold air masses,
which are above and below the front. The pressure change must be the same by both routes:
 p 
 p 
 p 
 p 
p    dx    dz   
dx   
dz
 x cold
 z cold
 x  warm
 z  warm
Then rearrange and use equations for geostrophic
p
p
  fv and hydrostatic
   g balance.
x
z
fvc c dx  c gdz  fvw  w dx   w gdz
tan  
dz f  vc c  vw  w  f  vcTw  vwTc 
 
 

dx g  c   w  g  Tw  Tc 
tan  
fT v 104  300 10

 0.01 rad
g T
10  3
where the inverse proportion of temperature and pressure and ΔT<<T are invoked. The change in
v arises from the shift in wind direction at the front.
This front has a slope of 1:100, so if it extends through the lower 2 km of atmosphere it will be
200 km long.
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Rossby waves [p513-517]
Long wavelength oscillations in the middle/upper atmosphere (above 700 mbar) are known as
planetary or Rossby waves. These can lead to a wave structure being superposed on the polar
front and the direction of the polar jet stream. When the wave amplitude is small, flow is
predominantly easterly termed a ‘high zonal index’, whereas large wave amplitude leads to
substantial meridional flow in the ‘low index’ state. Air mass transport will follow the direction of
the wave, since this determines the warm/cold air division and the direction of isobars, so in the
low index state cold polar air will be driven south in one part of the globe, while warm tropical air
is driven north elsewhere. The positions of antinodes in the Rossby waves tend to be fixed by
topographic features of the landforms e.g. the Rocky mountains, which can mean the waves stay
stationary for a period of weeks during which time opposite extremes of unseasonal weather can
be experienced in places half a wavelength (several thousand km) apart.
This mixing of air acts to transport heat from Equator to poles more effectively than in the illdefined Ferrel cell.
Consider an air parcel travelling from west to east (+ive x-direction) in the N-hemisphere under
geostrophic balance with a PGR towards the pole (+ive y-direction) balanced by a CoF towards the
equator. Now displace the air parcel towards the pole. This makes the CoF (which varies with
sinθ) greater than the PGF and acts as a restoring force resulting in SHM.
1 p
d 2 y dv
1 p
but at the original latitude there is no y-motion so 0   f 0u 

  fu 
2
 y
dt
dt
 y
Therefore,
d2y
df
  fu  f 0u  u  y
2
dt
dy
and since f  2 sin  and y  R ,
df
d
1
 2 cos 
 2 cos 
dy
dy
R
d2y
2 cos 
2
R
 y , which is SHM with a period  
Hence, 2  u
 2
dt
R

2 cos  u
Putting in numbers, this gives a wavelength of ~7000 km, which is a substantial fraction of the
circumference of a line of latitude.
Cyclogenesis Again
Rossby waves in the jet stream provide another way to describe cyclogenesis. For air flowing
eastwards along such a wave, the anticlockwise rotation around the high-latitude antinode implies
motion around a high pressure region (in the upper atmosphere) which is super-geostrophic. By
contrast the clockwise rotation around the low-latitude antinode will be sub-geostrophic.
In going from super- to sub-geostrophic the air must decelerate which leads to convergence and
downward motion towards the surface, since the air is considered incompressible and has to go
somewhere. This creates a high pressure region at the surface (anticyclone).
Similarly, going from sub- to super-geostrophic the air must accelerate which leads to divergence
and upward motion from the surface. This creates a low pressure region at the surface (cyclone).
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North Atlantic Oscillation [p562]
As well as the large/small amplitude of Rossby waves, mid-latitude climate is influenced by the
strength of windspeed flowing along the Polar Front. In winter this can be driven by a Polar Vortex
in the stratosphere surrounding cold air that doesn’t see the sun. In summer, the polar vortex
disappears because all the air is again heated by the sun and more mixing occurs.
There are two stable states, corresponding to high or low speed, that can persist for several
months and account for marked different seasonal variations from year to year. These states
were originally identified in terms of the degree of pressure difference between the Icelandic Low
(in N Atlantic) and Azores High (Atlantic subtropical high) by the index of the North Atlantic
Oscillation (NAO).
In Low Index years of the NAO there is small pressure difference and a weak polar vortex. The
Trade Winds at the equator and mid-latitude Westerlies are also weak. Winter depressions follow
a track further to the south and lead to cool, wet conditions in the Mediterranean as well as drier
conditions in northern Europe.
In High Index years there is a large pressure difference and strong polar vortex. Trade Winds and
Westerlies are strong. The storm track is further north, leading to wet conditions in northern
Europe with more intense winter storms, meanwhile the Mediterranean is dry.
The NAO is now thought to be connected to the ENSO and other global phenomena.
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Lecture 15
Oceans
Oceans have a huge heat capacity compare to the atmosphere, but only respond on a much
slower timescale. Annual temperature changes in the oceans are small and are only seen in the
surface layer of less than 100 m deep. Oceans account for about ¼ of the heat flow from tropics
to high latitudes.
Surface currents generally follow the directions of prevailing winds, with an E-W flow at the
equator and W-E flow in mid latitudes, especially around the Southern Ocean where there is no
land to hinder the flow. Also circulation around the relatively permanent low and high pressure
regions, which leads to currents towards the Equator on E side of ocean basin, which are cold
currents, and away from the Equator on W sides in warm currents such as the Gulf Stream.
There is also a huge system of deep water currents called the Thermohaline Conveyor , which is
driven by gradients in temperature and salinity. In polar regions (North Atlantic and Antartica, but
not Pacific due to land of Alaska and Russia), sea ice is formed from fresh water which leaves cold
salty water to sink to the ocean floor. This cold saline bottom current flows around the globe:
south through the Atlantic, east around the Southern Ocean and north into the Indian and Pacific
Oceans. The return flow is via warm surface currents.
There is a concern that if sea ice stops forming in the N Atlantic, due to climate change, then this
global circulation could slow or stop which would reduce the warming experienced in northern
Europe from the Gulf Stream.
El-Nino Southern Oscillation (ENSO)
A correlation has been observed between high sea surface temperature in the eastern Pacific
(known as the El Nino effect in Peru), high pressure in Eastern Australia and high rainfall in the
central Pacific, with an erratic period of about 4 years. (See diagrams in lecture notes.)
During Normal conditions, in the December to February period, E-W surface (trade) winds at the
Equator drive the surface waters of the Pacific away from the coast of S America. This causes
upwelling of cool, nutrient rich water (which is good for fishing). The Pacific sea surface
temperature increases towards the western side of the basin, which leads to most evaporation
and cloud formation being at the western side, so little rain in the central Pacific and low pressure
under the clouds.
During El Nino conditions, the Trade Winds are much weaker (which may also be connected with
the NAO). The cold upwelling off S America stops (bad for fishing). The sea surface temperature is
then uniform across the Pacific, not cold on the eastern part, leading to more uniform evaporation
and rainfall across the whole area, but less rain and higher surface pressure in Australia, which
may experience drought.
There is an opposite extreme with stronger than normal Trade Winds called La Nina, where the
rainfall is pushed even further west than under normal conditions.
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Characteristic values for atmospheric quantities
Mean Earth radius
Approx. distance from pole to equator
6370 km
10,000 km
Mean planetary temperature
Mean surface temperature
Solar Flux
Stefan’s constant, 
255 K
288 K
1370 Wm-2
5.67x108 Wm-2K-4
Mean molecular mass of dry air
Gas constant for dry air, Rd
Gas constant for water vapour, Rv
  Rd Rv  M H2O M air  0.622
28.97 gmol-1
287 Jkg-1K-1
461 Jkg-1K-1
Density of dry air at stp
Mean surface pressure
Scale height at 250 K
1.29 kgm-3
1013 mbar
16.8 km
Heat capacity at constant pressure of dry air, cp
1005 Jkg-1K-1
Heat capacity at constant pressure of water vapour 1911 Jkg-1K-1
Heat capacity at constant pressure of 1% moist air 1013 Jkg-1K-1
Latent heat of vaporisation at 273 K Lw,v
Latent heat of vaporisation at 373 K Lw,v
Latent heat of sublimation at 273 K Li,v
2.50x106 Jkg-1
2.25x106 Jkg-1
2.83x106 Jkg-1
Dry adiabatic lapse rate, d = g/cp
Saturated adiabatic lapse rate, s
9.8 Kkm-1
5~9 Kkm-1
Angular velocity, 
Coriolis parameter, f = 2sin at 50N
2/day = 7.27x10-5 rads-1
 h-1= 1.1x10-4 s-1
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