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Module 3 Thermodynamics of the Dry Atmosphere
3.1 Introduction
Thermodynamics essentially deals with energy transformations in
a system and its equilibrium states under such transformations. A
system could be either open (no boundaries) or closed (boundaries).
An open system exchanges energy and matter with its surroundings in
attaining its equilibrium state, whereas a closed system has
boundaries, which are impermeable to matter, but exchange of energy
is possible as the system is not isolated. Thermodynamics plays a
central role across a number of traditional disciplines of physics
and chemistry because pressure and temperature are the key
variables in the understanding the equilibrium states of a given
system. The atmosphere and ocean are also complex systems where the
exchanges of mass and energy are so intricate that their dynamics
cannot be explained satisfactorily without appropriate
consideration of thermodynamics. Thus, simulation of atmospheric
phenomena ranging from cloud microphysics to large-scale
atmospheric motions, involves the basic laws of dynamics and
thermodynamics. The thermodynamic state of the atmosphere is
greatly modified by the water vapour, which on transformation of
phase adds heat to the system. Also, evaporation of liquid water is
accompanied with the cooling of the system. The thermodynamics of
the moist air is thus different from dry air under certain
transformations with phase changes. We first derive expressions for
lapse rate and stability relevant for the dry atmosphere using the
first law of thermodynamics assuming atmosphere in hydrostatic
equilibrium. Finally, the thermodynamics of moist air will be
discussed.
3.2 Equation of State Using the equation of state, first of
thermodynamics and the hydrostatic equilibrium, several expressions
relevant for the dry atmosphere are derived. Air is a mixture of
gases and all gases obey the same equation of state. The ideal gas
law is expressed as
pV = mRT
(3.1) where V is the volume, p the pressure, m the mass, R the gas
constant and T denotes the temperature. The Equation (3.1) can also
be written as
p = ρRT , ρ =
mV
(3.2)
pα = RT
, α
= 1ρ
(Specific volume) (3.3)
A gas is composed of molecules and the number of molecules can
be calculated for its mass m contained in volume V by first
calculating the number of moles n of the gas in the given
volume. Thus, we have n = mM
with M as the molecular weight of the given gas. The
Avogadro’s hypothesis states that gases containing the same
number of molecules occupy the same volume at same temperature.
That is, the number of molecules in one mole of any substance is
constant and it is called the Avogadro Number (NA) NA = 6.022 x
1023 per mole (3.4)
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If we consider one mole of a gas then the gas constant R is same
for one mole of any other gas; that is, R in eqn. (3.3) is same for
one mole of all gases. Hence, it is called the universal gas
constant R* = 8.3145 J K-1 mol-1. Thus, for n moles, the ideal gas
reads as pV = nR*T (3.5) Note that R* is used in calculations with
one mole which contains NA molecules of a gas. Since, NA is also a
constant, the gas constant for one molecule is also a universal gas
constant, known as the Boltzmann’s constant given by,
2310022.63145.8*×
==AN
Rk = 1.3807×10-23
Question 1: Write down the units of k. [Ans. JK-1 molec-1] If
there are no molecules per unit volume, then from equation (3.5),
we obtain p = n0kT (3.6) If pd and α d are respectively the
pressure and specific volume of dry air then with Rd as the gas
constant for 1 kg of dry air, the gas law becomes pdα d = RdT (3.7)
The average molecular weight (Md ) of dry air is equal to 28.97 g
mole
-1; so, one can find the gas constant for one gram of dry air as
R* /Md and therefore the gas constant, Rd as
Rd =R*
Md /1000= 1000 R*
Md= 1000 8.3145
28.97= 287 J kg−1K −1
Question 2: Consider air as the mixture of gases made up almost
entirely by nitrogen (N2), oxygen (O2) and argon (Ar) with
concentration by volume as 78.08%, 20.95% and. 0.93%; calculate the
average molecular weight of air.
3.3 Hydrostatic equation Consider a column of air in the
vertical direction (z) with positive direction pointing upward. The
weight of a slab at a height z of thickness δ z is ρgδ z , g is the
acceleration due to gravity taken as constant. Now calculate the
vertical force that act on this slab of air between z and z +δ z
with atmospheric pressure p(z) and p(z +δ z) = p +δ p . Since
pressure decreases with height, so the vertical force on unit
cross-sectional area is −δ p . The hydrostatic balance requires
that vertical force be balanced by the weight of the
cross-section,
−δ p = gρδ z or ∂p∂z
= −gρ (3.8) Eqn. (3.8) is known as the hydrostatic equation and
negative sign in this equation ensures that pressure decreases with
height. Eq. (3.8) can be integrated from surface (z = 0) to a
height z to obtain the following result
⎭⎬⎫
⎩⎨⎧−= ∫ H
dzpzpZ
Oo exp)( (3.9)
Fig. 3.1. Hydrostatic balance
z + δ z
z
Ground level
g
z= 0
−δ p
p(z)
p(z)+ δ p
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H is known as the scale height, and it is that altitude where
pressure at surface reduces by a factor e−1 (e = 2.71828183) . H
varies between 6 km at T = 210 K to 8.5 km at T = 290 K for the
lower atmosphere.
Sea Level Pressure: If the mass of the earth were distributed
uniformly over the globe with actual topography, the sea level
would be 1.013 x 105 Pa or 1013.25 hPa, which is referred to as 1
atmosphere (or 1 atm).
Question 3: Show that in an atmosphere of uniform temperature
(To ), p(z) = po e− z/H ; what
is the expression for H?
3.4 Geopotential The geopotential Φ at any point in the earth’s
atmosphere is defined as the work done
to raise a mass of 1 kg from sea level to that point against the
force of gravity. The geopotential at sea level Φ(0) = 0. In our
calculations, we can therefore take, Φ = gz with z as the
geometrical height in meters (units of geopotential J kg-1 or m2
s-2).
Φ Z( ) = gdzO
Z
∫ (3.10) The geopotential height is defined as,
Z = Φ(z)go
= 1go
gdzo
Z
∫ (3.11)
z(km) Z(km) g0 0 9.811 1 9.8010 9.99 9.77
g0 = 9.81ms−2 is the globally averaged value of g. In the lower
atmosphere go ≈ g.
Geopotential thickness: From eqns. (3.10) and (3.8) with α =
RdTp
, we have
dΦΦ1
Φ2
∫ = gdzZ1
Z2
∫ = − RdTdppp1
p2
∫ ⇒ Φ2 −Φ1 = −Rd Tdppp1
p2
∫ , which gives
Z2 − Z1 = Rdg0
T dppp2
p1
∫ or Z2 − Z1 =RdTg0
ln p1p2
⎛⎝⎜
⎞⎠⎟
(T = mean temperature) (3.12)
The difference, Z2 − Z1 is referred to as the geopotential
thickness or simply the thickness of an atmospheric layer bounded
by pressure levels p1 and p2. If [T ] is the average
temperature
of this atmospheric layer, then with H = Rd[T ]go
≡ RdTgo
Z2 − Z1 = H lnp1p2
⎛⎝⎜
⎞⎠⎟
(3.13)
p2 = p1 exp −Z2 − Z1H
⎧⎨⎩
⎫⎬⎭
(3.14)
Eqn. (3.13) is known as the hypsometric equation in meteorology
parlance.
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Exercise 1: Calculate the height (geopotential) of 1000 hPa and
900 hPa when atmospheric pressures at sea level is: (a) 1014 hPa
and (b) 1020 hPa. Take H=8 km. (Ans.: 112m & 160m).
Question 4: Define T in (3.13). Ans. T =T dppp1
p2
∫dppp1
p2
∫ (weighted mean layer temperature).
Exercise 2: Calculate ΔZ of a layer between 500 hPa and 1000 hPa
at a point in (i) tropics with T =15°C (Ans.: 5849 m); (ii) polar
latitudes with T = −40°C (Ans.: 4732 m). An important conclusion
from eq. (3.12) is that geopotential thickness is proportional to
mean temperature of the layer between pressures p1 and p2 as seen
in Fig. 3.2; when moisture is present in the layer T is replaced by
T v (the mean virtual temperature. which will be defined
later).
Fig. 3.2 The solid lines indicate isobars. Thickness of layer
between isobars is larger in the warm area than over cold area.
Panel (a): Winds in a warm core lows are strongest at the surface
and decrease with height. Panel (b): Upper level lows sometimes do
not extend down; so at lower levels, there is cold core (i.e. the
high is topped by a low)
3.5 System of Units There are two system of units, viz., (i)
MKS: Metre (m), Kilogramme (kg) and
Second (s); and (ii) cgs: centimetre (cm), gramme (g) and second
(s).
Temperature: °C and °F; TF =95T + 32 , TF (°F) and T (°C)
Pressure: 1 bar = 105 Pa = 103 hPa; 1 millibar (mb) = 102 Pa
Torricelli or mm Hg (torr); 1 torr = 133.322 Pa 1 atmosphere (atm)
= 1.01325 bar = 1.01325 x 105 Pa = 760 torr Pound-per-square-inch:
1 psi = 6894.76 Pa
Derived units
Acceleration m s-2 cm s-2 Density kg m-3 g cm-3 Force Newton (N)
dyne (dyn) N=kg m s-2 dyn = g cm s-2
Pressure Pascal (Pa) microbar (µbar) Pa = N m-2 µbar=dyn cm-2
Energy Joule (J) erg J=N m erg=dyn cm Specific energy J kg-1 = m2
s-2 erg g-1 = cm2 s-2 Power Watt (J s-1) erg s-1
760 mm of Hg at 0°C = 1 atm 1 cal = 4.1840 J
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3.6 Reduction of Pressure to Sea Level (PSL) In mountainous
regions, the difference in surface pressure (ps) from one station
to
another is largely due to difference in their elevation above
the sea level. In order to determine the pressure change due to the
passage of weather systems, the pressure are reduced to sea level
(reference level). The sea level pressure is calculated using the
hypsometric equation (3.14),
p2 = p1 exp −Z2 − Z1H
⎧⎨⎩
⎫⎬⎭
; ps = psl exp −Zs − Z1H
⎧⎨⎩
⎫⎬⎭
with p1 = psl and p2 = ps , Z1 = 0
psl = ps expZsH
⎧⎨⎩
⎫⎬⎭= ps exp
gZsRdT
⎧⎨⎩
⎫⎬⎭
(3.15)
3.7 Concept of a Parcel of Air We have assumed atmosphere in
hydrostatic equilibrium. However, atmosphere is
constantly heated by solar radiation at the surface, so hot
surface air will rise upwards, which may disturb this kind of
equilibrium. When will atmosphere turn unstable? Here, the concept
of an air parcel helps in determining the necessary conditions. We
define an air parcel as a part of the atmosphere which is not
different from other parts at the same level, but becomes a
distinct portion once it is displaced from its original position.
Its dimensions are sufficiently large in comparison to the
mean-free path of air molecules but small enough to represent the
average properties of the atmosphere of its location. Most
importantly, a parcel is insulated from its surrounding during its
displacements but instantaneously adjusts its pressure with the
environmental pressure. Large expanses of air having consistent
characteristics of temperature and moisture of regions where they
originate are called “air masses”. The air mass retains its
characteristics and does not mix even during large movements with
the wind from places of its origin to other distant parts on the
globe. The parcel movement is generally considered in the vertical
direction, but air masses move horizontally over long distances.
Parcels drawn from an air mass can give the characteristics and
origin of air masses with the help of retro trajectories.
3.8 Adiabatic Lapse Rate Consider the vertical movement of a
parcel at pressure p, temperature T and of specific
volume α (=1/ρ) in the atmosphere. Assume that atmosphere is in
hydrostatic equilibrium; that is, the upward force due to pressure
gradient in vertical is balanced by the weight of the parcel. The
gravitational force and the buoyancy force are balanced; hence
there is no need to consider them in the first law of
thermodynamics. For a unit mass, the first law of thermodynamics
states that a quantity dQ of heat added to the system is utilized
in increasing the internal energy of the system and as the work
done on the system. Stated mathematically, the First Law of
Thermodynamics is given as
dQ = dU + dW (3.16) Internal energy increment: dU = CvdT
; Work done by pressure: dW = p dα dQ = Cv dT
+ pdα (3.17) Cv is the specific heat of air at constant
volume. Differentiation of the equation of state gives d(pα ) = d
RdT( ) ⇒ p dα +αdp = RddT
(3.18) Using α = 1
ρ and Rd = Cp – Cv in (3.18), we can write it in the
following form
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pdα + dpρ
= RddT = Cp − Cv( )dT . Substituting the value of pdα in
(3.17), we get
dQ = CpdT − dpρ ⇒ dQ = Cp dT + gdz
(3.19)
If the parcel undergoes adiabatic expansion or compression
during its movement in the vertical, there is no exchange of heat
(i.e. heat neither enters or leaves the parcel during motion);
hence dQ = 0 and (3.19) gives
dQ = CpdT + gdz = 0 ⇒ −dTdz
= gCp
= Γd (3.20)
Equation (3.20) gives the temperature decrease with height that
can be calculated with g = 9.81ms−2 and Cp = 1005 J kg
−1 as
Γd =gCp
= 9.76 K km−1 (3.21)
This is the first important result in atmospheric
thermodynamics. The adiabatic lapse rate Γd is the rate of change
of temperature with height of a parcel of dry air when it
adiabatically rises up or sinks down in a dry atmosphere. However,
the profile of temperature from an
ascent of a radiosonde gives the actual lapse rate Γ = −
dTdz
of the atmosphere. It varies
widely due to the presence of water vapour and the average value
of Γ is 6-7 K km-1 in the troposphere; that is Γ < Γd .
Fig. 3.3 Atmospheric lapse rates: Panel (a):- The line marked OA
is dry adiabat. The environmental temperature profile is the dotted
line CXPYD. The part CX represents steep fall in temperature with
height in the environment. Negative lapse rate corresponds to an
adiabat sloping on the right; Panel (b):- Profile OA corresponds to
dry adiabatic lapse rate; the dotted line PQ shows rise in
temperatures with height (inversion or negative lapse rate); the
dashed line CQ corresponds to “superadiabatic lapse rate” in the
atmosphere.
The meaning of constant lapse rate: If the atmosphere is heated
by contact with the earth’s surface and vertical motions set in,
then heat will be distributed up in such a manner that the vertical
temperature profile will have a constant gradient (Γd ) of ~10 K
km
-1 with altitude.
Y
X
z
T
Γ = Γd Γ <Γd
Γ>Γd
O
A B
C
DΓ = 0
PQ
(a)
z
T
Γ = Γd
O
A
CP
Q
(b)
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Now, consider the stability of the atmosphere with respect to
vertical motions with the
help of Fig. 3.3 that shows different vertical profiles of
temperature (T vs z ) in the atmosphere. In Fig. 3.3a, different
temperature profiles are given: profile marked OA is the dry
adiabat (i.e. the temperature profile with constant lapse rate Γd
); profile CXYD is the actual temperature profile of the
environment (actual lapse rate); and profile OB shows the lapse
rate Γ greater than Γd , and dashed line profile shows that
temperature of the atmosphere is uniform with height (isothermal
layer) and Γ= 0. If a parcel initially at point X rises, then it
will follow the dry adiabat line OA and would arrive at a location
Q, where it is surrounded by ambient atmosphere with conditions at
P on the profile CXYD. That is, parcel is warmer than the ambient
atmosphere and shall continue to rise; hence the point X on profile
CXYD is unstable. In other words, the parcels that are displaced
from point X on the profile CXYD would never return to X. By
similar argument point Y on profile CXYD is stable if it is
displaced vertically; because parcels that are pushed up (down),
being heavier (lighter) than the surrounding, shall return back to
point Y. That is, atmosphere is stable above the point Y. Thus we
have
(i) Atmosphere stable if − ∂T∂Z
< Γd (3.22)
(ii) Atmosphere unstable if − ∂T∂Z
> Γd (3.23)
3.9 Potential Temperature: The potential temperature θ of an air
parcel is defined as the temperature that an air
parcel would have if it were expanded or compressed
adiabatically from its existing pressure (p) and temperature (T) to
a standard pressure p0 (generally taken as 1000 hPa). From the
first law of Thermodynamics, we have the relation
dQ = CpdT − dpρ (3.24)
Eqn. (3.24) can be written for an adiabatic transformation ( dQ
= 0 ) as
dQ = CpdT − α dp = 0 ⇒ CpRd
dTT
− dpp
= 0 (3.25)
On integrating equation (3.25), we obtain
CpRd
dTTθ
T
∫ =dppp0
p
∫ ( θ = T0 at p = po ) ⇒CpRd
lnTθ= ln p
po
θ = T pop
⎛⎝⎜
⎞⎠⎟
Rd /Cp
(3.26)
The equation (3.26) is called the Poisson equation.
Define κ = RdCp
and γ =CpCv
then κ = (1− 1γ
) = 0.2856
In calculating the value of κ , we have used Rd = 287 J K−1kg−1
and Cp = 1005J K
−1kg−1 .
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Exercise 3: Starting with eq. (3.26), show that: Tγ
pγ −1= T0
γ
poγ −1 =
θγ
poγ −1 = const.
Thus, a new temperature θ can be defined with (3.26) which a
parcel will have if brought adiabatically to a reference pressure
po , taken as 1000 hPa. It must be noted that for an incompressible
medium, temperature is the appropriate variable; but for the
compressible atmosphere, it is the potential temperature θ. To
study the dynamic evolution of the atmospheric flows θ is an
appropriate variable, because it is a measure of the sum of
potential and internal energy and it is conserved in adiabatic
motions of the atmosphere. Quantities that remain invariant during
any transformation are said to be conserved variables. The
potential temperature is therefore an extremely important parameter
in atmospheric thermodynamics. The conservation of θ simplifies the
treatment of the dynamics of compressible fluids in the absence of
heat sources and sinks, much like the dynamics of incompressible
fluids that is rendered simpler (divergence free motions) because
density remains constant. The conservation of θ is mathematically
expressed as
dθdt
= 0 (3.27)
Exercise 3: A parcel of air has a temperature -51°C at 250 hPa
level. What is its potential temperature? What temperature will the
parcel have if it were brought into the cabin of a jet aircraft and
compressed adiabatically to a cabin pressure of 850 hPa? [Hint:
Start with T = (−51+ 273.15) K , p = 250hPa , p0 = 1000hPa ;
calculate θ ; Parcel brought to cabin: T = 222.15 K , p = 250hPa ,
p0 = 850hPa ; calculate θ ]. Note: Taking the logarithm of (3.26),
it becomes easier to differentiate (3.26); we obtain,
lnθ = lnT + RCpln po −
RCpln p . Now differentiate on both sides to get
1θdθdz
= 1TdTdz
− RCp1pdpdz
(3.28)
3.10 Static Stability In equation (3.28), if hydrostatic
equation is used then we have,
1θdθdz
= 1TdTdz
− RCp1p(−gρ) ; ρ = p
RT
Tθdθdz
= dTdz
+ gCp
(3.29)
For an atmosphere with constant θ (i.e. an adiabatic
atmosphere), the atmospheric lapse rate
dzdT− is obtained from (3.29) by setting dθ
dz= 0 , and we obtained the earlier result,
− dTdz
= gCp
= Γd (3.30)
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Hence, if potential temperature θ is a function of height, the
atmospheric lapse rate
⎟⎠⎞⎜
⎝⎛ −=Γ
dzdT will differ from adiabatic lapse rate Γd and we have from
(3.29),
Tθdθdz
= Γd − Γ (3.31)
If Γ < Γd , it implies that dzdθ is positive i.e. θ increase
with height. Thus, an air parcel that
undergoes an adiabatic displacement from its equilibrium
position will be positively buoyant when displaced vertically
downward. The air parcel will be negatively buoyant (sinks) when
displaced vertically upward from its equilibrium position. Such as
atmosphere is said to be statically stable or stably stratified
atmosphere 3.11 Buoyancy Oscillations
Adiabatic oscillations of a fluid parcel about its equilibrium
position in a stably stratified atmosphere are called buoyancy
oscillations. The frequency of such oscillations could be derived
by vertically displacing a parcel upward or downward a small
distance δz from its equilibrium position at a height z. The
atmosphere is always considered in hydrostatic equilibrium. So any
imbalance between the upward pressure gradient and its weight will
result in vertical acceleration; but when there is a balance, we
have
dp0dz
= −gρo => −1ρo
∂po∂z
− g = 0 (3.32 )
Here po and ρo are pressure and density of the environment. If p
and ρ are the pressure and density of the parcel, any imbalance
between the vertically acting pressure gradient force
and the weight of the parcel will result in vertical
accelerations ( dwdt
) of the parcel. The
Newton’s second law gives the equation of motion of the parcel
as,
dwdt
= − 1ρdpdz
− g or ρ dwdt
= − dpdz
− gρ (parcel) and w = ddt(δ z) ; hence
ρ d2 (δ z)dt 2
= − dpdz
− gρ . (3.33)
Since environment is in hydrostatic equilibrium, dpdz
in (3.33) is replaced by dp0dz
and we get
ρ d2 (δ z)dt 2
= − dp0dz
− gρ
Now replace dp0dz
by −gρo in the above equation using (3.32), and (3.33)
becomes,
ρ d2 (δ z)dt 2
= g(ρ0− ρ) (3.34)
Note that the term g(ρ0− ρ) on the right hand side of (3.34) is
the buoyancy force acting on
the parcel of unit volume. We can now write (3.34) as
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d2
dt 2(δ z) = g ρo − ρ
ρ (3.35)
As pointed out earlier that a parcel adjusts its pressure
instantaneously with the surrounding pressure. Thus, for a parcel
undergoing small displacements δz from its equilibrium position
without disturbing its environment, we set p (parcel) = p0
(environ) . Hence, the pressure gradient
term in (3.33) is replaced by dpodz
in parcel method.
Fig. 3.4 Displacement of a parcel
Consider a parcel in Fig. 3.4 at a height z and it is displaced
vertically by δ z to a new position z +δ z . As the parcel moves
from its equilibrium position z to new position z +δ z , we can
estimate the change in the thermodynamic variables T, θ and ρ.
Initially, the parcel and environment are at the same temperature
T, density and pressure, but when displaced its temperature is
given by
Tp (z +δ z) = Tp (z)+dTdz
δ z or Tp (z +δ z) = T (z)− Γd δ z (3.36 a)
For the environment,
Tenv(z +δ z) = T (z)+dTdz
δ z or Tenv(z +δ z) = T (z)− Γδ z (3.36b)
The density ρ of parcel and ρo of environment can also be
written at new position z1 . Now, one may derive the following
expression that holds also at the new position of the parcel,
g ρo − ρρ
= gTp −TenvTenv
(3.37)
Using (3.36) and (3.37) in (3.35), we get
d 2
dt 2(δ z) = g
(Γ − Γd )δ z(T − Γδ z)
= g(Γ − Γd )δ z
T(1− Γ
Tδ z)−1 = g
(Γ − Γd )δ zT
1+ ΓTδ z + ...⎛⎝⎜
⎞⎠⎟
Neglecting term in (δ z)2 on the right hand side, we get the
following oscillation equation for a parcel
d 2
dt 2(δ z) =−N 2 δ z ; N 2 = g
T(Γd − Γ). (3.38)
In eqn. (3.38), N 2 is the buoyancy frequency also called the
Brunt-Väisälä frequency; N 2 is a measure of static stability of
the atmosphere; for the lower atmosphere, the corresponding
period τ = 2πN
is a few minutes. The parcel undergoes oscillations when N 2
> 0 .
T T
Tp Tenv
z
z+δ z
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Exercise 4: If θ and θ0 are the potential temperatures of the
parcel and the environment,
using (38), show that g ρo − ρρ
= gθ −θ0θ0
. The parcel undergoes adiabatic displacements,
hence θ(z +δ z) = θ0 (z) . Write for the environment, θo(z +δ z)
= θo(z)+dθodz
δ z and show
that N 2 = gθ0
dθ0dz
= g d(lnθo )dz
.
The solution of (3.38) is given by δ z = A eiN t (3.39) If
parcel oscillates about its mean position when N 2 > 0 , then
for average tropospheric conditions,
N = 1.2 x 10-2 s-1 ⇒ τ = min7.82 =Nπ
If N = 0 , There will be no oscillations of the displaced parcel
about its equilibrium position and parcel will be in neutral
equilibrium (i.e. no accelerating force on parcel).
If N 2 < 0 , the potential temperature will decrease with
height; which means the displacement of the parcel will increase
exponential with time; in other words, parcels continue to move
through the atmospheric column and the column is unstable. The
above condition also lead us to write the conditions of static
stability of the atmosphere
in terms of the potential temperature from the relation N 2 =
gθ0
dθ0dz
. Thus, we have,
N 2 > 0 Atmosphere statically stable 0>dzd oθ
N 2 = 0 Statically neutral dθodz
= 0 (3.40)
N 2 < 0 Statically unstable 0<dzd oθ
An important note: On the synoptic scale, atmosphere is always
stably stratified and when the stratification is unstable,
convective overturning removes quickly the unstable regions that
develop. The convective overturning essentially mixes the air in
overlaying layers and it redefines the distribution of isentropes
(surfaces of equal θ ) in the stratification. However, for moist
atmosphere, situation is more complicated. If the temperature
actually increases with height (temperature inversion) the
atmosphere is gravitationally stable as warm air overlays the cold
air. Temperature inversions inhibit mixing; and water vapour,
aerosols and pollutants are trapped in the layer. During winter,
temperature inversions occur during nights, therefore in the big
cities, vehicular emissions will be trapped in this layer and there
will be episodes of high pollution and smog after sunrise. However,
it is worth mentioning that over global deserts, in day time hours,
negative stability can exist in the air next to the ground with
super-adiabatic lapse rates.