Transcript Sin título de diapositiva

LESSON 3
THERMODINAMICS OF THE ATMOSPHERE
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Equation of state for an ideal gas. Gasses mixture
Work and heat. The first law.
Changes of phase
Air parcel. Adiabatic processes.
Water steam: Moist air. Saturation
Moist air processes. Diagrams
Vertical stability
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Equipo docente:
Alfonso Calera Belmonte
Antonio J. Barbero
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Departamento de Física Aplicada
UCLM
Hurricane Wilma (10/19/2005)
Photo from http://www.nasa.gov/mission_pages/station/multimedia/hurricane_wilma.html
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STATE EQUATION FOR AN IDEAL GAS
pV  nRT

p
n
m RT m R
RT 

T
V
M V
V M
r
R
M
R  8.314 kJ  kmol 1  K 1
m
V
v
KJ  kg
The first law states the conservation of energy
1
V
m
 K 1 
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pv  rT
w  0
FIRST LAW
du  q  w
q  0
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html
q  0
System
w  0
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SYSTEM PROPERTIES
Specific internal energy
u
Specific enthalpy
h  u  pv
Any extensive property has an associated intensive property given by itself
divided by the mass of the system
Trabajo
Specific heat
 u 
cv   
 T v
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 h 
cp   
 T  p
w  p  dv
Relationship for specific heats for an ideal gas
d
d
r  T   r
( p  v) 
dT
dT
Mayer relationship
dh d
u  pv  cv  r

dT dT
c p  cv  r
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THE FIRST LAW APPLIED TO AN IDEAL GAS
q  du  w
q  cv dT  p  dv
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q  cv dT  d ( p  v)  v  dp  (cv  r )dT  v  dp  c p dT  v  dp
d ( p  v)  v  dp  p  dv
q  c p dT  v  dp
q  dh  v  dp
dh  du  p  dv  v  dp
Remark:
q  c p dT  v  dp
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IDEAL GASSES MIXTURE. DALTON’S MODEL
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An ideal gas consists on a set of non interacting particles
whose volume is very little if compared with the total
volume occupied by the gas. Non interacting particles
minds negligible forces from one particle to another.
Every component in the mixture behaves as if it were the
only component occupying the whole volume available at
the same temperature the mixture has.
As a consequence: every component exerts a partial
pressure, being the sum of all partial pressures the total
pressure of the mixture.
pi 
p
ni RT
V
nRT
V
pi ni
ni
  yi 
p n
n1  n2  ...  ni  ...
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Molar fraction
The partial pressure of a component in a
mixture is proportional to its molar fraction.
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PHASE: Aggregation state physically homogeneous having the same properties
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PHASE CHANGES:
Transitions between solid, liquid, and gaseous phases typically involve large
amounts of energy compared to the specific heat
The latent heat is the energy released or absorbed during a change of state
CHANGES OF STATE AT CONSTANT PRESSURE: Entalphy
S L 80 kcal/kg
Water:
L S 540 kcal/kg
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CHANGES OF STATE IN WATER
Let us consider ice at 1 atm
If heat were added at a constant rate to a mass of ice to take it through its phase
changes to liquid water and then to steam, the energies required to accomplish
the phase changes would be as follows:
T (ºC)
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water
+
steam
ice +
water
100
540 kcal/kg
80 kcal/kg
0
1 kcal/kg·ºC
 0.5 kcal/kg·ºC
ice
water
steam
heat
The change líquid  steam involves a great amount of energy!
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MOIST AIR
Moist air: dry air + water steam
(dry air composition: see following slide)
Moist air in contact with liquid water is described according the following model:
1) Dry air and water steam behave as independent ideal gasses (then the
presence of each of them do not affect the behaviour of the other)
2) The equilibrium of the liquid water and steam phases is not affected by the
presence of the air
Steam
Dry air
Moist air
Líquid
Saturated air
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SATURATED AIR
What is the vapour pressure?
Pressure of a vapour given off by
(evaporated from) a liquid or solid,
caused by atoms or molecules
continuously escaping from its
surface. In an enclosed space, a
maximum value is reached when the
number of particles leaving the
surface is in equilibrium with those
returning to it; this is known as the
saturated vapour pressure or
equilibrium vapour pressure
Vapour pressure is increasing up to...
Componentes
mayoritarios
atmósfera
Dry air (majority
components)
(% volumen)
N2 78%
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O2 21%
Ar 0.93%
Otros 0.04%
Saturated vapour pressure: function of T
QUESTION:
WHY IS MOIST AIR LESS DENSE THAN DRY AIR AT SAME TEMPERATURE?
http://www.theweatherprediction.com/habyhints/260/
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Liquid-steam equilibrium (water)
SATURATION:
Presion de vapor del agua (liq) en funcion de la temperatura
0.100
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The phase diagram of water e
n
http://www.lsbu.ac.uk/water/
phase.html
t
a
Properties of Water and
l
Vapour pressure
0.080
P (bar)
0.060
0.040
0.024
0.020
0.000
0
10
Liquid water and
steam phases coexists.
The saturation
vapour pressure is
given by the liquidvapour curve as a
function of
temperature.
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30
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Steam in SI-Units
(Ernst Schmidt)
Springer-Verlag (1982)
T (ºC)
Triple point coordinates: 0.01 ºC, 0.00611 bar
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Linear interpolation
Presion de vapor del agua (liq) en funcion de la temperatura
0.100
T (ºC)
0.01
5.00
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
0.080
P (bar)
0.060
0.040
0.020
1 i
2
0.000
0
10
20
30
40
50
Pi  P1 
P (bar)
0.00611
0.00872
0.01228
0.01705
0.02339
0.03169
0.04246
0.05628
0.07384
0.09593
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Ti  T1
P2  P1  P
T2  T1
h
T (ºC)
P(38º C )  0.06632 bar
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MOISTURE CONTENT OF THE AIR
Specific humidity (or moisture content of air) is the ratio of the mass of water to the mass
of dry air in a given volume of moist air
Specific humidity
Mass of water vapor
or
=
Mass of dry air
Mixture rate
w
mv
ms
kg vapor/kg dry air
Relationship between partial pressure of vapor, total pressure and specific humidity:
The partial pressure from a component of a gasses mixture is proportional to its molar
fraction (Dalton)
mv
Mv
mv
w
ms
pv 
p

p
mv ms
w




mv
1


M v 

Mv Ms
 M v  ms M s 
yv 
mv
Mv
mv ms

Mv Ms
pv 

w
p
w
Mv
 0.622
Ms
Remark:
v indicates vapor
s indicates dry air
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EXEMPLES
.
Determine the vapor pressure in air with specific humidity
6 g kg-1, when the total pressure is 1018 mb.
w
0.006
p 
p
1018  9.7 mb
v w
0.006  0.622
. . .
.
..
.
.
. .. . . .
.
. .
.
. .
Determine the specific humidity of an air mass at a total pressure of 1023 mb if the partial
pressure of vapor is 15 mb.
w
pv
15
 0.622
 0.00926 kg vapor/kg dry air
p  pv
1023  15
Calculator for atmospheric moisture
http://www.natmus.dk/cons/tp/atmcalc/atmocalc.htm
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RELATIVE HUMIDITY
Relative humidity: quotient between the molar fraction of water vapor in a given
sample of damp air and the molar fraction of water vapor in saturated air at the
same temperature and pressure.
 y 
   v 
 yv , sat T , p
As the molar fraction and vapor partial pressure are proportional,
the relative humidity can be also expressed as
pv  yv p
pv , sat  yv , sat p
 pv 

  

 pv , sat T , p
Another form
In the troposphere p >> pv,sat

w
wsat
pv
p
 v
p  pv
p
pv , sat
pv , sat


p  pv , sat
p
w
wsat
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EXEMPLE
Consider an air mass at 1010 mb and 20 ºC in which the partial vapor
pressure is 10 mb. Calculate its relative humidity, actual specific humidity
and saturation specific humidity.
 pv 
  10  0.428 (43%)

 pv , sat T , p 23.39
  
w
wsat  
pv
10
 0.622
 0.00622 kgkg-1
p  pv
1010  10
pv , sat
23.39
 0.622
 0.0147 kgkg-1
p  pv , sat
1010  23.39
T (ºC)
0.01
5.00
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
P (bar)
0.00611
0.00872
0.01228
0.01705
0.02339
0.03169
0.04246
0.05628
0.07384
0.09593
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P
wsat
w
pv,sat
pv
T
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the temperature at which air must be cooled at constant pressure in
order for it to become saturated with respect to a plane surface of
water. Atmospheric Science: An Introductory Survey by Wallace & Hobbs
Dew point:
Presion
de vapor
delofagua
funcion
la temperatura
Pressure
vapor
water(liq)
as aen
function
of de
temperature
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pv
w40 ºC  

o
p  pv
n
m
0.020
 0.622
 0.0126 kg  kg 1 e
1.010  0.020
n
t
a
pv
w10 ºC  

l
0.100
The air keeps its
specific humidity
but increases its
relative humidity
0.080
P (bar)
0.060
Exemple. Damp air mass
cooling down from 40 ºC up
to 10 ºC (pv = 20 mb, total
pressure 1010 mb)
0.040
0.020
0.012
p  pv
0.000
P
0.012
h
T (ºC)
 0.622
 0.0748 kg  kg 1
y
1.010  0.012
Dew point  17.5 ºC
s
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What is the initial relative humidity? And the final one?
c
16
s
http://weathersavvy.com/Q-dew_point1.html
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10
20
30
40
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TEMPERATURE AND HUMIDITY DAILY CYCLE
If the vapor of water in the air
remains constant...

100
35
24 mb
MAXIMUM OF
TEMPERATURE
25
80
60
MINIMUM
OF
MOISTURE
MINIMUM OF
TEMPERATURE
Relative humidity %
TEMPERATURE
DAILY CYCLE
MAXIMUM
OF
MOISTURE
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Temperature ºC
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Constant vapor pressure
MOISTURE
DAILY CYCLE
40
20
0
3
6
9
12
Hora
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ENTHALPY
Specific internal energy
u
Specific enthalpy
h  u  pv
The heat content, usually called the enthalpy, of air rises with increasing water
content.
This hidden heat, called latent heat by meteorologists and air conditioning engineers, has to be
supplied or removed in order to change the relative humidity of air, even at a constant
temperature. This is relevant to conservators. The transfer of heat from an air stream to a wet
surface, which releases water vapour to the air stream at the same time as it cools it, is the
basis for psychrometry and many other microclimatic phenomena. Control of heat transfer can
be used to control the drying and wetting of materials during conservation treatment.
The enthalpy of dry air is not known. Air at zero degrees celsius is defined to have
zero enthalpy. The enthalpy, in kJ/kg, at any temperature, t, between 0 and 60C is
approximately:
below zero: h = 1.005t
h = 1.007t - 0.026
http://www.natmus.dk/cons/tp/atmcalc/ATMOCLC1.HTM#enthalpy
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Enthalpy of air-water vapor mixture
Remark:
v indicates vapor
s indicates dry air
H  H s  H v  ms hs  mv hv
H Hs Hv
m


 hs  v hv
ms ms ms
ms
Sensible heat:
Specific
(kJ/kg dry air)
h  hs  w  hv
Sensible heat is defined as the heat energy stored in a substance as a result of
an increase in its temperature
Really the sensible heat is the same as enthalpy; the heat absorbed or transmitted by a
substance during a change of temperature which is not accompanied by a change of state
Units: kJ/kg dry air o en kcal/kg dry air (specific magnitude).
Specific heat of dry air is 0.24 kcal/kg
Latent heat:
The heat released or absorbed per unit mass by a system in a reversible isobaricisothermal change of phase. In meteorology, both the latent heats of evaporation
(or condensation), fusion (melting), and sublimation of water are important
http://www.shinyei.com/allabout-e.htm#a19
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ADIABATIC SATURATION PROCESS
Adiabatic isolation
T2
T1
2
1
Air flows across a pipe or duct adiabatically insulated where there is an open water
reservoir. As air moves around, its specific humidity increases. It is assumed that air
and water are in contact time enough until saturation is reached.
The enthalpy of the wet air remains constant because the adiabatic isolation. As a
consequence, the air temperature decreases when the saturated air leaves the pipe.
T2 = Tsa
About adiabatic saturation and humidity
http://www.taftan.com/xl/adiabat.htm
http://www.shinyei.com/allabout-e.htm
Adiabatic saturation temperature
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PSYCHROMETER
Determination of w specific humidity of the air from three properties:
pressure p, temperature T and adiabatic saturation temperature Tsa
dry
wet
T
Tsa
It consists of two
thermometers, one of
which (the dry bulb)
is an ordinary glass
thermometer, while
the other (wet bulb)
has its bulb covered
with a jacket of clean
muslin which is
saturated with
distilled water prior
to an observation.
When the bulbs are
suitably ventilated,
they indicate the
thermodynamic wetand dry-bulb
temperatures of the
atmosphere.
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hs (Tsa )  hs (T )  w' hv (Tsa )  hliq (Tsa )
w
r
hv (T )  hliq (Tsa )
o
n
m
pv (Tsa )
w'  
e
p  pg (Tsa )
n
t
Wet bulb temperature  Adiabatic saturation temperature
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

Psichrometric chart
M J Moran, H N Shapiro. Fundamentos de Termodinámica Técnica. Reverté (1994)
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Psychrometric chart
VALID FOR A
GIVEN PRESSURE
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
h
v
w, pv
T (wet bulb)
T (dry)
Density of the moist air
Specific volume
(kg/m3)
(m3/kg)
http://www.taftan.com/thermodynamics/SPHUMID.HTM

v
1

ms  mv
V

V
ms  mv
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EXEMPLE
An air mass at 30 ºC having a 30% relative
humidity undergoes an adiabatic saturation
process. Then it is cooled down to 13.5 ºC, and
after it is warmed up to 19 ºC. What is its final
relative humidity? How much has been changed
its specific humidity ?
 = 0.0095-0.0080 =
= 0.0015 kg·kg-1
70%
18 ºC
30%
13.5 ºC
0.0095
0.0080
30 ºC
19 ºC
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AIR PARCEL
An air parcel is a glob of atmospheric air containing the usual mixture of
non-condensing gases plus a certain amount of water vapor.
Its composition remains roughly constant whereas this glob moves across
the atmosphere from one site to another.
Water in the gaseous form exists in any parcel of air with relative humidity
greater than 0%. If the amount of water vapor (and other environmental
conditions) is such that the relative humidity of the air parcel reaches 100%, it
is said that the parcel is saturated.
Under conditions of saturation, water in the vapor form can change its phase
from vapor to liquid. Then that water vapor condenses into liquid water
droplets
When phase changes occur, there is a readjustment of the forms of energy
associated with the water molecules. In the vapor phase, water has a relatively
large amount of internal energy represented by its looser internal molecular
structure. When vapor condenses to liquid, the internal energy of the water
molecule in the liquid phase is smaller, owing to its tighter molecular
structure.
If liquid freezes into solid water (ice), the internal energy is reduced further
owing to the much tighter packing of water molecules in the solid phase.
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AIR PARCEL: MODEL
We’ll consider that air parcels obey the following model:
E
1 Any air parcel is adiabatically isolated and its temperature changes adiabatially
n
when it rises or descends.
v
i
Are we able to calculate temperatures from that or some related formula?
PV   constant
r
o
2 The movement of air parcels is slow enough to make possible to assume that their n
m
kinetic energy is a very little fraction of their total energy.
e
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3 It is assumed that any air parcel is in hidrostatic equilibrium with its environment: t
it has the same pressure that its environment has.
a
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What does hidrostatic equilibrium mean?
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HIDROSTATIC EQUATION
Air column, density 
Air mass contained in dz:
S  dz
Weight of air contained in dz:
gS  dz
S
p+dp
Pressure forces:
-Sdp
Ascending:
dz
z
Descending:
gSdz
p
Net pressure force:
pS
S  ( p  dp)
S  p  S  ( p  dp)   S  dp
The net pressure force goes upwards, because dp is a
negative quantity
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HIDROSTATIC EQUATION (Continued)
We assume that every air layer is near equilibrium
S
Weight equilibrates the pressure forces
p+dp
-Sdp
dp
  g
dz
 S  dp  gS  dz
dz
z
gSdz
p
As a function of specific volume:

1
v
g  dz  v  dp
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VIRTUAL TEMPERATURE
V
ms
Virtual temperature is an adjustment applied to the
real air temperature to account for a reduction in air
density due to the presence of water vapor
mv
Moist air =
= dru air +
+ water vapor
Density of
moist air:

ms  mv
  s  v
V
s: density that the same dry air mass ms would have if
the dry air were the only component in the volume V
“Partial” densities
v: density that the same water vapor mass mv would
have if the water vapor were the only component
ps  rs sT
Ideal gas
pv  rv vT
Dalton’s law
p  ps  pv
p  pv pv


rsT
rvT
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
Tvirtual
p  pv pv

rsT
rvT
T
T


p
w
1   
1  v 1    1 
p
we

p
rsT
 pv  rs  p
1  1   
p  rv  rsT


Tvirtual 
 pv



1

1




p


rs M v

 0.622
rv M s
T
1
pv
1   
p
The ideal gas equation can be written as:
p  rs Tvirtual
Moist air
pressure
Virtual temperature definition Tvirtual
Constant of
dry air
Density of
moist air
Virtual temperature is the temperature that dry air should have so that its
density be the same than that of the moist air at the same pressure.
Moist air is less dense than dry air  virtual temperature is larger than
absolute temperature
Virtual temperature calculator: http://www.csgnetwork.com/virtualtempcalc.html
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POTENTIAL TEMPERATURE
Potential temperature  is the temperature that a parcel of dry air would
have if it were brought dry adiabatically from its original position to a
standard pressure p0 (generally the value 1000 mb is taken as the reference
p0).
ADIABATIC PROCESS
q  c p dT  v  dp  0
c p dT dp

0
r T
p
rT
c p dT  dp  0
p
p  v  rT
T
c p dT

r T


Dry air
p

p0
dp
p
cp T
p
ln  ln
r 
p0
r
287 J  K 1  kg 1

 0.286
cp
1004 J  K 1  kg 1
cp
p
T  r
  
p0
 
r
cp
 p0

 p
  T
T  constante   p0.286
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PSEUDOADIABATIC CHART
T  constante   p0.286
0
All points in this line have the same
potential temperature
Exemple. An air parcel at 230
K lies in the 400 mb level and
goes down adiabatically up to
the 600 mb level. Calculate its
final temperature.
10
P (mb)
100
230 K
Adiabatic dropping
200
=100K
=200K
=300K
=400K
=500K
300
 constant
400
600
259 K
800
1000
100
200
300
400
T (K)
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Continuous lines in K: Dry adiabatics
Along these lines the potential temperature is constant ( cte)
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Dashed lines in K: Pseudoadiabatics
(for moist air,  wet bulb cte)
Continuous lines in g/kg:
Saturation mixing ratio lines
(specific humidity for saturation ws)
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USE OF PSEUDOADIABATIC CHART
Exemple
An air mass at 1000 mb and 18 ºC has a mixed ratio of 6 gkg-1. Find out its
relative humidity and its dew point temperature.
* Plot the T, p coordinates on the thermodynamics diagram (red point)
* Read the saturation mixing rate. See that ws = 13 gkg-1
* Relative humidity  
w
6
  0.46 (46%)
wsat 13
* Dew point temperature: draw a horizontal line across the 1000 mb ordinate up
to reaching the saturation mixing ratio line corresponding to the actual mixing
ratio (6 gkg-1). Its temperature is 6 ºC, that is to say, when reaching this
temperature a water vapor content of 6 gkg-1 become saturating, then liquid
water condenses.
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Exemple
An air parcel at 1000
mb and 18 ºC has a
mixing ratio of 6 gkg-1.
Find out its humidity
and dew point
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w 6
    0.46 (46%) n
ws 13
t
a
ws = 13
l
gkg-1
1000 mb
Dew point
6 ºC
18 ºC
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LIFTING CONDENSATION LEVEL
The level where a wet air parcel ascending adiabatically becomes saturated
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During the lifting process the mixing ratio w and the potential temperature 
remain constant, however the saturation mixing ratio ws drops because the
temperature is decreasing. Saturation is reached when the the saturation
mixing ratio equals the actual mixing ratio w.
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NORMAND’S RULE
• The ascending condensation level of an air parcel can be
found in the pseudoadiabatic chart in the intersection
point of the following lines:
• the potential temperature line (dry adiabatic line) in the point
determined by the temperature and pressure of the air parcel;
• the equivalent potential temperature line (pseudoadiabatic line) passing
through the point indicating the wet bulb temperature and the pressure
of the air parcel;
• the saturation mixing ratio line (constant humidity line) passing
through the point indicating the dew point and the pressure of the air
parcel;
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Air parcel at pressure p, temperature T, dew point TR and wet bilb
temperature Tbh.
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Condensation level
dry adiabatic line
p
 constant
wsat constant
Sat. mixing ratio line
p
Tbh
T
pseudoadiabatic line sat constant
1000 mb
TR
bh
T
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EXEMPLE 1. Lifting condensation level
A) An air parcel at 15 ºC has a dew point temperature of 2 ºC. It lifts adiabatically
from the 1000 mb level. Find out its lifting condensation level and the temperature
in this level.
B) If this air parcel goes on ascending over the condensation level and reaches a
level 200 mb above, find out its final temperature and the amount of water
condensed during the ascending process.
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B) If this air parcel goes
on ascending over the
condensation level and
reaches a level 200 mb
above, find out its final
temperature and the
amount of water
condensed during the
ascending process.
2.0 g/kg
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630 mb
Condensado:
4.5-2.0=2.5 g/kg
4.5 g/kg
830 mb
A) An air parcel at 15 ºC
has a dew point
temperature of 2 ºC. It
lifts adiabatically from
the 1000 mb level. Find
out its lifting
condensation level and
the temperature in this
level.
1000 mb
-15 ºC
TR=2 ºC
-1 ºC
15 ºC
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
6
 0.5
12
(50%)
770 mb
12 g·kg-1
6 g·kg-1
TR=4.5 ºC
T=15 ºC
EXEMPLE 2
An air parcel at
900 mb and 15 ºC
has a dew point
temperature of
4.5 ºC. Find out
the lifting
condensation
level, the mixing
ratio, relative
humidity, its wet
bulb temperature,
the potential
temperature and
the wet bulb
potential
temperature.
8.5 ºC
13 ºC
23.5 ºC
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THE FIRST LAW APPLIED TO AN IDEAL GAS
q  du  w
q  cv dT  p  dv
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q  cv dT  d ( p  v)  v  dp  (cv  r )dT  v  dp  c p dT  v  dp
d ( p  v)  v  dp  p  dv
q  c p dT  v  dp
q  dh  v  dp
dh  du  p  dv  v  dp
Remark:
q  c p dT  v  dp
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LAPSE RATE
Rate at which temperature decreases with height   
dT
dz
K/km ºC/km
A positive value indicates decrease of T with height
Troposphere: general decrease in T with height
Environmental lapse rate (ELR): it is the actual temperature of air we can measure
(observed air temperature at any height)
Dry adiabatic lapse rate (DALR): rate which a non-saturated air parcel cools at it rises
Saturated (wet) adiabatic lapse rate (SALR): rate which a saturated air parcel cools at it
rises
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DRY ADIABATIC LAPSE RATE (DALR)
Rate which a non-saturated air parcel cools at it rises
Consider a raising air parcel
First law
q  c p dT  v  dp
q  c p dT  g  dz  0
Pressure and density
decrease with height, then
as an air parcel rises it
expands and cools
 dT 
 
 dz aire

seco
g
 s
cp
g  dz  v  dp
Adiabatic process
Hydrostatic equation
g = 9.81 ms-2
cp = 1004 Jkg -1K-1
A physical change of the state of the air parcel
that does not involve exchange of energy with
the air surrounding the air parcel.
s = 0.0098 Km-1 = 9.8 Kkm-1
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SATURATED ADIABATIC LAPSE RATE (SALR)
When the air is saturated, condensation occurs
Latent heat of vaporization is released
That keeps the air parcel warmer than it would otherwise
Decrease in temperature with height is not as great as it would be for dry air
SALR is dependent on the amount of moisture on the atmophere
 dT 
sat   
 dz aire
More moisture in
the atmosphere…
… the greater will be the
release of condensation heat
4 K km-1
Range for sat
sat
MORE MOISTURE
… warmer the air
will remain
9 K km-1
LESS MOISTURE
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STATIC STABILITY FOR NON-SATURATED AIR
Environmental lapse rate  (ELR)
Height
Case  <s
A
B
s - >0
dT
dz
Initial conditions

TA
TB
 STABLE ATMOSPHERE
When raising, the air parcel pressure
evens up to that of its environment
dT
 0  0
dz

What will happen if we consider an ascending
movement of the non-saturated air parcel ?
Density of raising air (A is colder) is
bigger than density of environmental air B
s
Temperature
A restoring force inhibiting the
vertical movement appears
Positive static stability
When the ELR  is SMALLER than dry
adiabatic lapse rate s
The air parcel tends to return to its original
level instead of remaining on A
Could you figure out the same problem for descending non-saturated air?
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STATIC STABILITY FOR NON-SATURATED AIR (2)
Environmental lapse rate  (ELR)
Height
Case  <s ...and  < 0
A
B
s - >0  STABLE ATMOSPHERE
dT
 0 0
dz

When raising, the air parcel pressure
evens up to that of its environment
dT
dz
Initial conditions
Density of raising air (A is colder) is
bigger than density of environmental air B
s

TA
What will happen if we consider an ascending
movement of the non-saturated air parcel ?
TB
Temperature
Negative static stability
The ELR is negative (of course, it is
SMALLER than that of the dry air)
A restoring force inhibiting the
vertical movement appears
The air parcel tends to return to its original
level instead of remaining on A
Could you figure out the same problem for descending non-saturated air?
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THERMAL INVERSION
Very cold air
Warm air layer
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Cold air
Thermal inversions play a significant
role on the contaminants gathering
http://www.sma.df.gob.mx/sma/gaa/
meteorologia/inver_termica.htm
About thermal inversions
http://www.aviacionulm.com/meteotemperatura.html
http://www.sagan-gea.org/hojared/hoja20.htm
http://www.rolac.unep.mx/redes_ambientales_cd/capacitacion/Capitulo1/1_1_2.htm
http://en.wikipedia.org/wiki/Thermal_inversion
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STATIC UNSTABILITY FOR NON-SATURATED AIR
Environmental lapse rate  (ELR)
Height
Case  >s
B
A
s - < 0
dT
 0  0
dz

TB TA
When raising, the air parcel pressure
evens up to that of its environment
dT
dz
Initial conditions
s
 UNSTABLE ATMOSPHERE
Density of raising air (A is warmer) is
smaller than density of environmental air B

Temperature
A force in favor of further
vertical movement appears
Static unstability
ELR is BIGGER than the dry air
lapse rate
The air parcel tends to move away
from its original level
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STATIC STABILITY NON-SATURATED AIR (SUMMARY)
Stable
Positive static stability
 <s
Negative static stability
 <s
(inversion)
Unstable
Neutral stability:
Convective mixing
 <0


s
 >s
 =s
s

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EXEMPLE ON STABIBLITY AND UNSTABILITY
Environmental lapse rate  (ELR)
Dry Adiabatic Lapse Rate (DALR)
What will happen to an unsaturated air
parcel initially lying on the ground?
The air parcel temperature
here is lower than
sourrounding air temperature
Any air parcel which overtakes
this level sinks to equilibrium level
Inversion level
ELR  > DALR s
As the air parcel is
unsaturated, it lifts
along the dry adiabatics
Ground
temperature
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BIBLIOGRAPHY
Basic books:
John M Wallace, Peter W Hobbs, Atmospheric Science. An introductory survey. Academic Press (1997)
M J Moran, H N Shapiro. Fundamentos de Termodinámica Técnica. Reverté (1994)
On humidity and its measure http://www.usatoday.com/weather/whumdef.htm
Enthalpy of change of state data
http://www.adi.uam.es/docencia/elementos/spv21/sinmarcos/graficos/entalpiadevaporizacion/evapor.html
http://www.adi.uam.es/docencia/elementos/spv21/sinmarcos/graficos/entalpiadefusion/efusion.html
http://www.gordonengland.co.uk/conversion/specific_energy.htm
On specific heat
http://www.engineeringtoolbox.com/36_339qframed.html
Stability and unstability discussions
http://www.geocities.com/silvia_larocca/Temas/emagrama2.htm
http://www.cesga.es/telecursos/MedAmb/medamb/mca2/frame_MCA02_3.html
http://www.qc.ec.gc.ca/meteo/Documentation/Stabilite_e.html
http://www.usatoday.com/weather/wstabil1.htm (usa unidades inglesas)
http://www.indiana.edu/~geog109/topics/08_stability/LapseStability_CB_L.pdf
Clouds
http://seaborg.nmu.edu/Clouds/types.html
Other related sites
http://www.usatoday.com/weather/whumdef.htm
http://www.usatoday.com/weather/wwater0.htm
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PSEUDOADIABATIC CHART
700
320
pressure mb
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750
300
800
290
850
280
900
950
1000
2.0
3.0
4.0
6.0
8.0
10.0 12.0
15.0
20.0
25.0
30.0
1050
-10
-5
0
5
10
15
Continuous red lines labelled in K: dry adiabatics
20
25
30
35
40
temperature ºC
Continuous black lines labelled in g·kg -1: saturation mixing ratio
Discontinuous grey unlabelled lines: pseudoadiabatics
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PSEUDOADIABATIC CHART
700
320
pressure mb
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750
300
800
290
850
280
900
950
1000
2.0
3.0
4.0
6.0
8.0
10.0 12.0
15.0
20.0
25.0
30.0
1050
-10
-5
0
5
10
15
Continuous red lines labelled in K: dry adiabatics
20
25
30
35
40
temperature ºC
Continuous black lines labelled in g·kg -1: saturation mixing ratio
Discontinuous grey unlabelled lines: pseudoadiabatics
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