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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
m
(3.2)
p = ρ RT , ρ = V
1
pα = RT , α =
(Specific volume)
(3.3)
ρ
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
m
volume. Thus, we have n =
with M as the molecular weight of the given gas. The
M
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)
1
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
(3.5)
pV = nR*T
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,
R*
8.3145
= 1.3807×10-23
k=
=
23
N A 6.022 × 10
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
(3.6)
p = n0 kT
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
(3.7)
pdα d = RdT
-1
The average molecular weight (M d ) of dry air is equal to 28.97 g mole ; so, one can find the
gas constant for one gram of dry air as R* / M d and therefore the gas constant, Rd as
R*
R*
8.3145
Rd =
= 1000
= 1000
= 287 J kg −1K −1
M d /1000
Md
28.97
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 ρ gδ 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 = g ρδ z
∂p
= −g ρ
∂z
or
(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
⎧
dz ⎫
p ( z ) = p o exp⎨− ∫
⎬
⎩ O H⎭
Z
(3.9)
−δ p
z +δz
p(z) + δ p
z
p(z)
g
z= 0
Ground level
Fig. 3.1. Hydrostatic
balance
2
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).
Z
Φ ( Z ) = ∫ g dz
(3.10)
z(km) Z(km)
O
The geopotential height is defined as,
Φ(z) 1 Z
Z=
= ∫ g dz
go
go o
0
1
10
(3.11)
0
1
9.99
g
9.81
9.80
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 =
Φ2
Z2
p2
dp
∫Φ dΦ = Z∫ g dz = − p∫ RdT p
1
1
1
⇒
Φ 2 − Φ1 = −Rd ∫ T
p
Z 2 − Z1 =
Rd 1 dp
T
g0 p∫2 p
p2
or Z 2 − Z1 =
p1
Rd T ⎛ p1 ⎞
ln ⎜ ⎟
g0
⎝ p2 ⎠
dp
p
RdT
, we have
p
, which gives
(T = mean temperature)
(3.12)
The difference, Z 2 − 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
R [T ] Rd T
of this atmospheric layer, then with H = d
≡
go
go
⎛p ⎞
Z 2 − Z1 = H ln ⎜ 1 ⎟
⎝ p2 ⎠
(3.13)
⎧ Z − Z1 ⎫
p2 = p1 exp ⎨− 2
⎬
H ⎭
⎩
(3.14)
Eqn. (3.13) is known as the hypsometric equation in meteorology parlance.
3
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).
p2
dp
∫p1T p
Question 4: Define T in (3.13). Ans. T = p2
(weighted mean layer temperature).
dp
∫p
p1
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).
9
Temperature: °C and °F; TF = T + 32 , TF (°F) and T (°C)
5
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
Density
Force
m s-2
kg m-3
Newton (N)
N=kg m s-2
cm s-2
g cm-3
dyne (dyn)
dyn = g cm s-2
Pressure
Pascal (Pa)
Pa = N m-2
Joule (J)
J=N m
J kg-1 = m2 s-2
Watt (J s-1)
microbar (µbar)
µbar=dyn cm-2
erg
erg=dyn cm
erg g-1 = cm2 s-2
erg s-1
Energy
Specific energy
Power
760 mm of Hg at 0°C = 1 atm
1 cal = 4.1840 J
4
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),
⎧ Z − Z1 ⎫
⎧ Z s − Z1 ⎫
p2 = p1 exp ⎨− 2
⎬ ; ps = psl exp ⎨−
⎬ with p1 = psl and p2 = ps , Z1 = 0
H ⎭
H ⎭
⎩
⎩
⎧ gZ ⎫
⎧Z ⎫
(3.15)
psl = ps exp ⎨ s ⎬ = ps exp ⎨ s ⎬
⎩H ⎭
⎩ Rd T ⎭
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
(3.16)
dQ = dU + dW
Internal energy increment: dU = Cv dT ; Work done by pressure: dW = p dα (3.17)
dQ = Cv dT + p dα
Cv is the specific heat of air at constant volume. Differentiation of the equation of state gives
(3.18)
d( pα ) = d ( RdT ) ⇒ p dα + α dp = Rd dT 1
Using α =
and Rd = C p – Cv in (3.18), we can write it in the following form
ρ
5
(
)
dp
= Rd dT = C p − Cv dT .
ρ
Substituting the value of pdα in (3.17), we get
dp
dQ = C p dT −
⇒ dQ = C p dT + gdz (3.19)
ρ
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
dT
g
dQ = C p dT + gdz = 0 ⇒ −
(3.20)
=
= Γd
dz C p
Equation (3.20) gives the temperature decrease with height that can be calculated with
g = 9.81 m s −2 and C p = 1005 J kg −1 as
pdα +
g
(3.21)
= 9.76 K km −1
Cp
This is the first important result in atmospheric thermodynamics. The adiabatic lapse rate Γ d
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
dT
ascent of a radiosonde gives the actual lapse rate Γ = −
of the atmosphere. It varies
dz
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 < Γ d .
Γd =
A
B
D
Γ=0
Y
A
Γ
Γ<
Γd
Γd
=
=
Γ
z
Γd
Γ>
z
Γd
Q
Q
P
X
(a)
T
O
C
P
(b)
T
O
C
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 ( Γ d ) of ~10 K km-1 with altitude.
6
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 Γ d ); profile CXYD is the
actual temperature profile of the environment (actual lapse rate); and profile OB shows the
lapse rate Γ greater than Γ d , 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
∂T
(i) Atmosphere stable if
(3.22)
−
< Γd
∂Z
∂T
(ii) Atmosphere unstable if
(3.23)
−
> Γd
∂Z
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
dp
dQ = C p dT −
(3.24)
ρ
Eqn. (3.24) can be written for an adiabatic transformation ( dQ = 0 ) as
C dT dp
dQ = C p dT − α dp = 0 ⇒ p
(3.25)
−
=0
Rd T
p
On integrating equation (3.25), we obtain
C p T dT p dp
Cp T
p
=∫
( θ = T0 at p = po ) ⇒
ln = ln
∫
Rd θ T
p
Rd θ
po
p
0
⎛p ⎞
θ =T⎜ o⎟
⎝ p⎠
Rd /C p
(3.26)
The equation (3.26) is called the Poisson equation.
C
R
1
Define κ = d and γ = p then κ = (1− ) = 0.2856
Cp
Cv
γ
In calculating the value of κ , we have used Rd = 287 J K −1kg −1 and C p = 1005J K −1kg −1 .
7
Tγ
T0γ
θγ
=
=
= const.
pγ −1 poγ −1 poγ −1
Exercise 3: Starting with eq. (3.26), show that:
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
dθ
(3.27)
=0
dt
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 = 250 hPa , p 0 = 1000 hPa ; calculate θ ;
Parcel brought to cabin: T = 222.15 K , p = 250 hPa , p 0 = 850 hPa ; calculate θ ].
Note: Taking the logarithm of (3.26), it becomes easier to differentiate (3.26); we obtain,
R
R
ln θ = lnT +
ln po −
ln p . Now differentiate on both sides to get
Cp
Cp
1 dθ 1 dT R 1 dp
=
−
θ dz T dz C p p dz
(3.28)
3.10 Static Stability
In equation (3.28), if hydrostatic equation is used then we have,
1 dθ 1 dT R 1
=
−
(−g ρ ) ;
θ dz T dz C p p
T dθ dT
g
=
+
θ dz dz C p
ρ=
p
RT
(3.29)
For an atmosphere with constant θ (i.e. an adiabatic atmosphere), the atmospheric lapse rate
dθ
dT
is obtained from (3.29) by setting
= 0 , and we obtained the earlier result,
−
dz
dz
dT
g
(3.30)
−
=
= Γd
dz C p
8
Hence, if potential temperature θ is a function of height, the atmospheric lapse rate
dT ⎞
⎛
⎜Γ = −
⎟ will differ from adiabatic lapse rate Γ d and we have from (3.29),
dz ⎠
⎝
T dθ
= Γd − Γ
θ dz
(3.31)
dθ
is positive i.e. θ increase with height. Thus, an air parcel that
dz
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
If Γ < Γd , it implies that
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
dp0
1 ∂ po
(3.32 )
= −g ρo => −
−g=0
dz
ρo ∂z
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
dw
and the weight of the parcel will result in vertical accelerations (
) of the parcel. The
dt
Newton’s second law gives the equation of motion of the parcel as,
dw
1 dp
dw
dp
d
=−
− g or ρ
= − − g ρ (parcel) and w = (δ z) ; hence
dt
ρ dz
dt
dz
dt
ρ
d 2 (δ z)
dp
= − − gρ .
2
dt
dz
Since environment is in hydrostatic equilibrium,
(3.33)
dp
dp
in (3.33) is replaced by 0 and we get
dz
dz
d 2 (δ z)
dp
= − 0 − gρ
2
dt
dz
dp
Now replace 0 by −g ρo in the above equation using (3.32), and (3.33) becomes,
dz
2
d (δ z)
ρ
= g( ρ0 − ρ )
(3.34)
dt 2
Note that the term g( ρ0 − ρ ) on the right hand side of (3.34) is the buoyancy force acting on
ρ
the parcel of unit volume. We can now write (3.34) as
9
d2
ρ −ρ
(δ z) = g o
2
dt
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
dpo
in parcel method.
dz
z +δ z
z
Tp
Tenv
T
T
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
dT
(3.36 a)
Tp (z + δ z) = T p (z) +
δ z or Tp (z + δ z) = T (z) − Γd δ z
dz
For the environment,
dT
(3.36b)
Tenv (z + δ z) = T (z) +
δ z or Tenv (z + δ z) = T (z) − Γδ z
dz
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
T −T
ρo − ρ
= g p env
ρ
Tenv
(3.37)
Using (3.36) and (3.37) in (3.35), we get
(Γ − Γd )δ z
(Γ − Γd )δ z
(Γ − Γd )δ z ⎛ Γ
d2
Γ
⎞
(δ z) = g
=g
(1− δ z)−1 = g
⎜⎝ 1+ δ z + ...⎟⎠
2
dt
(T − Γδ z)
T
T
T
T
Neglecting term in (δ z)2 on the right hand side, we get the following oscillation equation for
a parcel
d2
g
(δ z) = − N 2 δ z ; N 2 = (Γd − Γ).
(3.38)
2
dt
T
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
2π
period τ =
is a few minutes. The parcel undergoes oscillations when N 2 > 0 .
N
10
Exercise 4: If θ and θ 0 are the potential temperatures of the parcel and the environment,
ρ −ρ
θ − θ0
using (38), show that g o
. The parcel undergoes adiabatic displacements,
=g
ρ
θ0
dθ
hence θ (z + δ z) = θ 0 (z) . Write for the environment, θ o (z + δ z) = θ o (z) + o δ z and show
dz
g dθ 0
d(ln θ o )
that N 2 =
.
=g
θ 0 dz
dz
The solution of (3.38) is given by
(3.39)
δ z = A ei N t
2
N
>
0
If parcel oscillates about its mean position when
, then for average tropospheric
conditions,
2π
N = 1.2 x 10-2 s-1
⇒ τ =
= 8.7 min
N
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
g dθ 0
in terms of the potential temperature from the relation N 2 =
. Thus, we have,
θ 0 dz
N2 > 0
Atmosphere statically stable
N2 = 0
Statically neutral
N2 < 0
Statically unstable
dθ o
>0
dz
dθ o
=0
dz
dθ o
<0
dz
(3.40)
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.
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