Transcript Pressure

Atmospheric
Thermodynamics
TMD July-Aug 2008 L03
Atmospheric Motion
Dynamics
Newton’s second law
m
F
x
Mass × Acceleration = Force
d2x
m 2 F
dt
Thermodynamics
Concerned with changes in the internal
energy and state of moist air.
The Atmosphere as an ideal gas
 Atmosphere as an ideal gas.
 Volume V, pressure p,
p TV
 Temperature T, mass m
 Specific volume = volume of 1 kg a = V/m.
 Density r = mass per unit volume = 1/a
 Equation of state for dry air
pa = RdT
Pressure – Partial pressure
The molecules of a gas are in constant random motion
 In an ideal gas each molecule has kinetic energy
 Occasionally molecules collide with each other or with the
walls of the containing vessel.
The nature of pressure
v
Momentum change due to a
perfectly elastic collision
= m(-v) - m(v) = -2mv
v
Unit area
The momentum change by many collisions averaged over
unit time and unit area represents a force.
The force per unit area is the pressure.
The unit is the Pascal. 1 Pa = 1 Nm-2.
Partial pressure
O2
N2
Unit area
The O2 molecules exert a partial pressure p1 and the N2
molecules a partial pressure p2 on the unit area.
The pressure is thee sum of the partial pressures.
Air motion
V
 The molecules of a fluid (gas or liquid) are in a constant
state of motion. The mean motion of all molecules in a
fluid parcel is the macroscopic velocity of the parcel, V.
 The (internal) energy residing in the random motion is
characterized by the absolute temperature of the fluid, T.
The nature of temperature
According to the kinetic theory of gasses, the absolute
temperature of a gas is proportional to the mean kinetic
energy of the molecules (including rotational energy). This
energy is called the internal energy.
The state of unsaturated moist air
 The state of moist air is characterised by:
• pressure, p
• absolute temperature, T
• density, r (or specific volume a = 1/r), and
• Some measure of the moisture, for example.
• The water vapour mixing ratio, r, defined as the mass
of water vapour per unit mass of dry air.
Equation of state for moist, unsaturated air
V
Water vapour
mv
Dry air
md
Now
p d V  md R d T
and
eV  m v R v T
pV  (pd  e)V  (md R d  mv R v )T
Divide by m
 mv R v 
1  m R 
(md R d  m v R v )
d
d 
pa 
T  Rd 
T
mv
md  m v
1
md
Equation of state for moist, unsaturated air
 mv R v
1 
md R d
(m d R d  m v R v )

pa 
T  Rd
mv
md  m v
1
md


 T  R 1  r /   T
d
1 r
Let  = Rd/Rv = 0.622
r = mv/md is the water vapour mixing ratio (typically << 1, max
0.04)
pa  R d Tv
Tv  1  0.61rT is the virtual temperature
The density of a sample of moist air is characterized by its
pressure and its virtual temperature, i.e.
p
r
RTv
Moist air (r > 0) has a larger virtual temperature than dry air
(r = 0) => the presence of moisture decreases the density of
air --- important when considering the buoyancy of an air
parcel!
For cloudy air
Tr  T
1 r / 
1  rT
p
r
RTr
is the density temperature
is the total water mixing ratio
The virtual temperature
a  R dTv / p
Tv  T(1  0.61r)
 Dry air with the virtual temperature Tv has the same
specific volume as moist air with temperature T at the
same pressure.
 To determine Tv one must convert r into kg/kg and convert
T into Kelvin.
The hydrostatic equation
 Except for motion on small scales, e.g. thunderstorms,
the atmosphere is to a very good approximation in
hydrostatic balance.
The hydrostatic equation 2
Hydrostatic balance
Mass = rAdz
p(z)A - p(z + dz)A = grAdz
As dz  0
Cross section A
dp

g (z)
dz
The minus sign is because the
The hydrostatic equation
pressure decreases with height
The hydrostatic equation 3
dp

g (z)
dz

 When r(z) is known, we can integrate with
respect to z :
p(z) 


z
g r(z)dz
z
 Note that p(z)  0 as z  .
 The pressure at height z is just the weight of a column
of air with unit cross section.
The hydrostatic equation 4
 Mean sea level pressure:
p(z) 



g r(z)dz
0
0
 The product of the mean sea level pressure (= 105 Pa)
times the area of the Earth‘s surface (= 5  1014 m2) gives
approximately the mass of the atmosphere (5  1019 kg).
The hydrostatic equation 5
 The vertical density profile r(z) is difficult to measure: p
and T are easier to measure.
 r(z) can be obtained from the ideal gas equation
p = rRdTv :
dp
 -g r(z)
dz
1 dp
g
p dz
R d Tv (z)

ln p(z) - ln p(0) 
z
0

p(z)  p(0) exp 


z
0
g
dz
R d Tv (z)

g
dz
R dTv (z) 
Other moisture variables
 The partial pressure of water vapour, e = rp/( + r)
 The relative humidity, RH = 100  e/e*(T).
• e* = e*(T) is the saturation vapour pressure, which is
the maximum amount of water vapour, that an air
parcel can hold, without condensation occurring.
 The specific humidity, q = r/(1 + r), is the mass of water
vapour per unit mass of moist air.
Saturation vapour pressure e*(T).
A more accurate empirical formula is (see E94, p117):
ln e* 53.67957 
6743.769 / T 
4.8451 ln T
e* in mb and T in K
A corresponding expression for ice-vapour equilibrium is:
ln e* 23.33086 
6111.72784 / T 015215
.
ln T
These formulae are used to calculate the water vapour content
of a sample of air. If the air sample is unsaturated, the dew point
temperature (or ice point temperature) must be used.
Water vapour the relationship between e, r and p
Water vapour
mv
Dry air
md
Now
p d V  md R d T
and
eV  m v R v T
Divide the equations of state =>
pd V m d R d T

eV m v R v T
p-e 

e
r
e
r
p-e
More moisture variables
 The dew point temperature, Td, is the temperature at
which an air parcel first becomes saturated as it is cooled
isobarically.
 The wet-bulb temperature, Tw, is the temperature at which
an air parcel becomes saturated when it is cooled
isobarically by evaporating water into it. The latent heat
of evaporation is extracted from the air parcel.
Vertical distribution of r and RH from a radiosonde sounding on a
humid summer day in central Europe.
Aerological (or thermodynamic) diagrams
T = constant
p
ln p
.
.
(p, a)
(p, T)
T = constant
a
pa  R dT
T
Aerological diagram with plotted sounding
T = constant
ln p
. .
(p, Td)
(p, T)
The first law of thermodynamics
The increase in the internal energy of a system is equal to the
amount of energy added by heating the system, minus the
amount lost as a result of the work done by the system on its
surroundings.
 The internal energy of a system includes the kinetic and
potential energy of the molecules or atoms.
 When the kinetic part of the internal energy increases (i.e.
the molecules move faster on average), the temperature of
the gas increases Materials.
 The potential energy of the molecules is determined by
their position relative to neighbouring molecules.
 James Joule (1818-1889). The mechanical
equivalent of heat:
 Joule was a very enthusiastic experimenter.
 During his honeymoon in Switzerland he
tried to determine the temperature change
of water at a waterfall.
Waterfall
Water gains kinetic Energy from potential Energy.
The kinetic Energy of the Water is in
converted first to turbulence and
later to heat.
Der erste Hauptsatz der Thermodynamik 9
 Joule showed that for a thermally isolated system,
dU = dQ + dW
• dU = the increase of the internal energy
• dQ = the heat input
• dW = the work done on the gas.
U
dQ, dW
U + dU
dU = dQ + dW
dQ = dU - dW
Not all the heat is available to
increase the internal energy
When the gas expands (i.e.
dV > 0), it does work on its
surroundings.
dQ
U
=
U + dU - dW
A thought experiment
cylinder
gas
Pressure p
Temperature
T
Piston
Volume V
Graphical representation of state changes
 The thermodynamic state of dry air can be represented by
a point in a pV- or pa-diagram.
p
p
A
(a,p)
B
a
 The change in state can be represented by a
curve in such a diagram.
a
Ein Gedankenexperiment 2
Area A
dx
Pressure force = pA
Pressure p
Work done pA  dx = pdV
Volume change dV = Adx
Work done pdV/(unit mass) = pda per unit mass
p
A
Work done dW = pdV
p
V1 dV
B
V2
Total work =
V
-W

V2
V1
pdV
The first law for 1 kg of an ideal gas
dq = du - dw = du + pda
Heat input
The work done by
the gas
Change in internal energy
 Temperature increase
For a sample of moist air du  cv dT
Then
or
dq  cvdT  pda
dq  cp dT - adp
d(pa)  pda  adp  d(R T)
cv  cvd (1  0.94r) c vd
cp  cpd (1  0.85r) cpd
Adiabatic processes
An adiabatic process is one in which there is zero heat input
dq = 0


d ln T  R / cp d ln p
RT
0  c p dT dp
p
dq  c p dT - adp
d ln T  d ln p
where
ln T   ln p  ln A
  R d / c p  0.2865
a constant
Define A such that, when p equals some standard pressure, po,
usually taken to be 1000 mb, T = .
The quantity  is called the potential temperature
The potential temperature
The potential temperature and is given by:
 po 
  T 
 p 

We define the virtual potential temperature, v by
F
p I
 T G J
Hp K

o
v
v
take the value of  for dry air.
Enthalpy
The first law of thermodynamics can be expressed as
dq  d(u  pa) - adp = dk - adp
k = u + pa is called the specific enthalpy.
The enthalpy is a measure of heat content at constant
pressure.
For an ideal gas, k = cpT.
Entropy
An excellent reference is Chapter 4 of the book:
C. F. Bohren & B. A. Albrecht
Atmospheric Thermodynamics
Oxford University Press
Specific
entropy
dq
ds 
 c pd ln 
T
s  c p ln   cons tan t
The equivalent potential temperature
dq  c p dT - adp
dq
dT
dp
ds 
 cp
- Rp
 cp d ln 
T
T
p
Suppose dq results from latent heat release, i.e. dq = -Lvdr
Lv
c p d ln  
dr  0
T

Lv r
Lv r 
ln  
 ln e
d  ln  
0



cp T
c
T
p 

 Lv r 
e   exp 
 c T 
 p 
a constant
e is called the equivalent
potential temperature
Some notes
 e, L, and Lv are conserved in reversible adiabatic
processes involving changes in state of unsaturated or
cloudy air.
 e, L, and Lv are not functions of state - they depend on
p, T, r and rL
  Curves representing reversible, adiabatic processes
cannot be plotted in an aerological diagram
 In a saturated process, r = r*(p, T)
The pseudo-equivalent potential temperature
 The formula e*  exp (Lvr*/cpdT) is an approximation for
the pseudo-equivalent potential temperature ep .
 A more accurate formula is:
p I
F
 TGJ
Hp K
0.2854 / (1 - 0.28 r )
 ep
o
L
O
F
I
3376
exp M
r(1  0.81 r ) G - 2.54J
P
T
H
K
N
Q
LCL
Temperature at the LCL
 TLCL is given (within 0.1°C) by the empirical formula:
L
ln(T / T ) O
1
M

 56
P
T - 56
800 Q
N
-1
TLCL
K
d
d
TK and Td in Kelvin
The reversible equivalent potential temperature
The reversible equivalent potential temperature is defined by
 po 
er  T  
 pd 
R d /(cpd  rT cL )
(RH)
- rR v /(cpd  rT cL )


Lv r
exp 

 (cpd  rT cL )T 
It is based on the assumption that all water vapour is carried
with an air parcel.
Lines of constant er cannot be plotted on an aerological
diagram.
 If an air parcel is lifted pseudo-adiabatically to the high
atmosphere until all the water vapour has condensed out,
ep = .
 The pseudo-equivalent potential temperature is the
potential temperature that an air parcel would attain if
raised pseudo-adiabatically to a level at which all the
water vapour were condensed out.
 The isopleths of ep can be plotted on an aerological
diagram. These are sometimes labelled by their
temperature at 1000 mb which is called the wet-bulb
potential temperature, w.
The adiabatic lapse rate
 Lift a parcel of unsaturated air adiabatically
 it expands and cools, conserving its v
 its T decreases with height at the dry adiabatic lapse rate, d :
g
1 r
 dT 
d  - 


 dz dq 0 c pd 1  r(c pv / c pd )
 Note that r is conserved, but r* decreases because e*(T)
decreases more rapidly than p.
 Saturation occurs at the lifting condensation level (LCL)
when T = TLCL and r = r*(TLCL, p).
Above the LCL, the rate of which its temperature falls, m,
is less than d because condensation releases latent heat.
For reversible ascent:
g 1  rT
 dT 
m  -   
 dz s c pd 1  r c pv
cpd



Lv r
1


R dT



L2v (1  r / )r 
cL

1  rL

2
c

rc
R
T
(c

rc
)

pd
pv
v
pd
pv 

When rT is small, the ratio m/d is only slightly less than unity,
but when the atmosphere is very moist, it may be appreciably
less than unity.
The moist static energy and related quantities
The first law gives
dq = dk - addp
where dq is expressed per unit mass of dry air.
Adiabatic process (dq = 0)  dk - addp = 0
ad = a(1 + rT)
For a hydrostatic pressure change, adp = -gdz.
Under these conditions:
dh  (c pd  rT c L )dT  d(L v r)  (1  rT )gdz  0
Some notes
If rT is conserved, we can integrate
h  (c pd  rTc L )T  L v r  (1  rT )gz  cons tan t.
 The quantity h is called the moist static energy.
 h is conserved for adiabatic, saturated or unsaturated
transformations in which mass is conserved and in which
the pressure change is strictly hydrostatic.
 h is a measure of the total energy:
(internal + latent + potential)
The dry static energy
 Define the dry static energy, hd.
 Put rT = r
=>
h d  (c pd  rc L )T  (1  r)gz.
 This is conserved in hydrostatic unsaturated
transformations.
 h and hd are very closely related to e and .
Vertical profiles of dry conserved variables
Dry static energy
z (km)
z (km)
 , v

v
deg K
hda
hd
hd = (cpd+ rcL)T
+ (1 + r)gz
105 J/kg
hda = cpdT +gz
Vertical profiles of moist conserved variables
Moist static energy
z (km)
z (km)
, pseudo e
e ep
esp
ha
h
hs
epa
epa   exp (Lvr*/cpdT)
deg K
105 J/kg
ha = cpT + Lvr + gz
The stability of the atmosphere
 Consider the vertical displacement of an air parcel from its
equilibrium position
 Calculate the buoyancy force at its new position
 Consider first an infinitesimal displacement ; later we
consider finite-amplitude displacements
 Parcel motion is governed by the vertical momentum
equation
d 2
b
2
dt
 r p - ra
b( )  - g 
 rp


 a p - aa 
  g 

a
a



is the buoyancy force
per unit mass
Newton's law for an air parcel:
d 2 buoyancy force

2
dt
unit mass
buoyancy force
b
: b() 

unit mass
z z 0
d 2 b
0
2
dt z z 0
The motion equation for small displacements is:
d2
2

N
0
2
dt
where
b
N z
2
The motion equation for small displacements is:
d2
2

N
0
2
dt
b
N z
2
where
For an unsaturated displacement, vp is conserved and we
can write
 Tvp - Tva 
 vp - va 
b()  g 
  g

T

va
va




Since vp = constant  va,
vp va
b
g va
N - g 2

z
va z
va z
2
Stability criteria
 Parcel displacement is:
• stable if
va/z > 0
• unstable if
va/z < 0
• neutrally-stable if
va/z = 0
 A layer of air is stable, unstable, or neutrally-stable if these
criteria are satisfied in the layer.
 In a saturated (cloudy) layer of air, the appropriate
conserved quantity is the moist entropy s (or the
equivalent potential temperature, e, or L)
 Must use the density temperature to calculate b.
 Replace ap in b by the moist entropy, s.
 In this case
1
N 
1  rT
2
 a p - aa 
b()  g 

a
a


rT 
 s
m z -  cL m ln T  g  z 
 A layer of cloudy air is stable to infinitesimal parcel
displacements if s (or e) increases upwards and the total
water (rT) decreases upwards. It is unstable if e decreases
upwards and rT increases upwards.
Some notes
 The stability criterion does not tell us anything about the
finite-amplitude instability of an unsaturated layer of air that
leads to clouds.
 Parcel method okay, but must consider finite displacements
of parcels originating from the unsaturated layer.
Potential Instability
 A layer of air may be stable if it remains dry, but unstable
if lifted sufficiently to become saturated.
 Such a layer is referred to as potentially unstable.
 The criterion for instability is that de/dz < 0.
unstable
lift
stable
unsaturated
saturated/cloudy
Conditional Instability
 The typical situation is that in which a displacement is
stable provided the parcel remains unsaturated, but which
ultimately becomes unstable if saturation occurs.
 This situation is referred to as conditional instability.
 To check for conditional instability, we examine the
buoyancy of an initially-unsaturated parcel as a function of
height as the parcel is lifted through the troposphere,
assuming some thermodynamic process (e.g. reversible
moist adiabatic ascent, or pseudo-adiabatic ascent).
 If there is some height at which the buoyancy is positive,
we say that the displacement is conditionally-unstable.
 If some parcels in an unsaturated atmosphere are
conditionally-unstable, we say that the atmosphere is
conditionally-unstable.
 Conditional instability is the mechanism responsible
for the formation of deep cumulus clouds.
 Whether or not the instability is released depends on
whether or not the parcel is lifted high enough.
 Put another way, the release of conditional instability
requires a finite-amplitude trigger.
 The conventional way to investigate the presence of
conditional instability is through the use of an
aerological diagram.
200
LNB
300
dry
adiabat
pseudoadiabat
500
10 g/kg
700
850
LFC
LCL
1000
20 oC 30 oC
Positive and Negative Area
Convective Inhibition (CIN)
The positive area (PA)
PA 
1 u2
2 LNB
-
1 u2
2 LFC
p
z cT

LFC
p LNB
vp
h
- Tva R d d ln p
The negative area (NA) or convective inhibition (CIN)
NA  CIN  
p parcel
p LFC
T
vp
- Tva  R d d ln p
Convective Available Potential Energy - CAPE
The convective available potential energy or CAPE is the net
amount of energy that can be released by lifting the parcel from
its original level to its LNB.
CAPE = PA - NA
We can define also the downdraught convective available
potential energy (DCAPE)
DCAPE i 
z R (T
po
pi
d
ra
- Trp )d ln p
The integrated CAPE (ICAPE) is the vertical mass-weighted
integral of CAPE for all parcels with CAPE in a column.
Reversible e
z (km)
z (km)
Pseudo e
b m s-2
b m s-2
Liquid water
z (km)
z (km)
Buoyancy
zL km
zL km
Height (km)
reversible
with ice
reversible
pseudo-adiabatic
Buoyancy (oC)
Downdraught convective available potential energy (DCAPE)
DCAPE i 
Td
z R (T
po
pi
d
ra
- Trp )d ln p
qw = 20oC
T
700
LCL
Tw
800 mb, T = 12.3oC
Td
850
r* = 6 g/kg
1000
Trp
DCAPE
Tra
The End
Summary: Various forms
of the equation of state

Für m kg

pV = mRT

Für 1 kg

pa = RT

Für eine beliebige Menge

Für ein Kmole

pV = MRT oder pV = R*T

Für n Kmole

pV = nR*T

Für 1 kg feuchte Luft 

p = rRT
pa = RdTv
Example

p = 990 mb 

T = 26 C

w = 8 g/kg 

99000 Pa
299 K
0,008 kg/kg
pa  R dTv
 p  rR dTv

Tv = 299  (1 + 0,61  0,008) = 300,46 K

r = p/RdTv = 99000/(287  300,46) = 1,15 kg/m3

a = 1/r = 1/1.15 = 0,87 m3/kg
Der erste Hauptsatz der Thermodynamik 6
Wasser
Rührwerk
Gewicht
 Joule schloß daraus
Mechanische Energie  Wärme
The equation of state for cloudy air
Consider cloudy air as a single, heterogeneous system
specific volume = (total volume)/(total mass)
a   Va  Vl  Vi  /  M d  M v  M l  M i 
Divide by Md
total mixing ratio of
water substance
a  a d 1  rl (a l / a d )  ri (a i / a d ) / 1  rT 
RdT 1
R d T pd  e 1
RdT 1 r / 
a


p d 1  rT
p
p d 1  rT
p 1  rT
 = Rd/Rv = 0.622
defines the density temperature for cloudy air:
Tr = T(1 + r/)/(1 + rT)
specific heat of
water vapour
The reversible equivalent potential temperature
 po 
e  T  
 pd 
R d /(cpd  rT cL )
(RH)
=p
- rR v /(cpd  rT cL )
=1


Lv r
exp 

 (cpd  rT cL )T 
=1
 For dry air (r = 0, rL = 0, rT = 0), e reduces to .
 Note that e is not a state variable  isopleths of constant
e cannot be plotted on an aerological diagram.
The (virtual) liquid water static energy
 Define two forms of static energy related to L and Lv.
 These are the liquid water static energy:
h w (c pd rT c pv )T L v rL (1 rT )gz
and the virtual liquid water static energy
h Lv
   rT 
Lv rL
 cpd 
 gz
 Tr 1  rT
   rT - rL 
 hLv is almost precisely conserved following slow adiabatic
displacements.
 If rL = 0, hLv = cpdTv + gz (just as L reduces to ).
Buoyancy and e
z
Lifted parcel
b  -g
Tvp rp
Environment
(rp - ro )
ro
To(z), ro(z), p(z), ro(z)
or
bg
(Tvp - Tvo )
To
Lifted parcel
bg
Trp rp
Environment
(Trp - Tvo )
To
To(z), ro(z), p(z), ro(z)
Below the LCL (Trp= Tvp)
sgn (b) = sgn { Tp(1 + rp) - To(1 + ro) = Tp - To + [Tprp - Toro]}
At the LCL (Trp= Tvp)
 = 0.61
sgn (b) = sgn [Tp(1 + r*(p,Tp)) - To(1 + ro)]
= sgn [Tp(1 + r*(p,Tp) - To(1 + r*(p,To)) + To(r*(p,To) - ro)]
sgn (b) = sgn [Tp(1 + r*(p,Tp) - To(1 + r*(p,To)) + To(r*(p,To) - ro)]
Since Tv is a monotonic function of e, b  (*ep - *eo  )
small
z
parcel saturated
eo
*eo
LFC
LCL