Transcript Moisture - Texas A&M University

Moisture
There are several methods of expressing the
moisture content (water in vapor form) of a
volume of air.
Vapor Pressure: The partial pressure exerted by
water molecules in vapor state in a volume. e
Saturation Vapor Pressure: The partial pressure
exerted by water molecules in a vapor state in a
volume when there is a state of equilibrium
between the water vapor and a plane surface of
liquid water. es
The meteorological
definition of saturation
relates to a flat liquid water
surface.
Supersaturation conditions
can exist over a curved
surface; e.g. raindrops or
cloud droplets. Also, if no
condensation nuclei are
present on which the water
can condense.
Evaporation rate
depends on the
temperature of liquid
water.
Condensation rate
depends on moisture
content or air.
Clausius-Clapeyron Equation
Describes the relationship between temperature
and saturation vapor pressure.
The Clausius-Clapeyron equation derives from
using the Carnot cycle (engine) with liquid water
and water vapor as the substance in the engine.
The Carnot cycle considers the expansion and
contraction of an ideal gas both adiabatically (no
loss or gain of energy) and isothermally from an
initial state through processes and back to the
initial state and the resulting temperature changes
that occur.
The relationship is given by :  

L
1
1


es  e0  exp  




Rv T0 T 
where,
es = saturation vapor pressure at T,
e0 = saturation vapor pressure at T0.
e0 = 0.611 kPa when To = 273oK (0oC)
L = latent heat of vaporization or deposition,
Lv 2.5 x 106 J/kg, Ld = 2.83 x 106 J/kg (which is
the value at 0oC) Latent heat actually varies
slightly with temperature.
Rv= gas constant for water vapor = 461 J/oK
kg
Remember, exp means e( )
Teten’s formula
This empirical equation is often used to calculate
saturation vapor pressure since it is relatively
easy to use and gives good results. Teten’s
formula is used by most cloud modelers.
es  e0  e
Where, e0 = 0.611 kPa
b = 17.2694
T1 = 273.16oK,
T2 = 35.86oK
 b T  T1 



 T T 2 

Boiling point - the temperature at which the
equilibrium vapor pressure between a liquid
and its vapor is equal to the external
pressure on the liquid. Usually considered
to be saturation vapor pressure, es, in
meteorology. Note: definition of saturation
vapor pressure refers to flat liquid water
surface. Derivation on page 98 gives:
Lv Rv
a = 14.53
Tboili ng 
Hp= 7.29 km
a  z Hp
Other Humidity Terms
Mixing ratio: Ratio of mass of water vapor
to mass of dry air in the same volume.
mv
r
md
Vapor Pressure: Remember from
chapter 1, vapor pressure can be written as:
 mv



 Mw

e   m
P

mv
d



Md Mw 
If we multiply the numerator and
denominator by Mv and divide both by md


mv
we get:


md
e  M

P
m
v
 v


md 
 Md
Mixing ratio is:
mv
r
md
Mv Rd
The quantity Mv/Md is: M  R    0.622
d
v


Then:
r
e  
P
  r 
e    r  r  P
e  er  r P
e    r  P  e
e 
r
P  e
Where, (P-e) represents the pressure
produced by only the dry air molecules.
Saturation mixing ratio is then the total mass of
water vapor that can exist in a mass of dry air at a
specific temperature. It can be expressed as:
esT 
rs

P
0.62197  rs
rs 
0.62197 esT 
P  es T 
Mixing ratio and saturation mixing ratio values are
usually given as grams of water per kilogram of
dry air.
Mixing ratio and saturation mixing ratio are
pressure dependent. (table 5-1 gives values for sea
level.) Vapor pressure and saturation vapor
pressure are not pressure dependent.
Specific Humidity: The ratio of the mass
of water vapor to the total mass (dry air plus
water vapor) of air.
q
 e
P
Specific Humidity and Saturation
Specific Humidity are pressure dependent.
Absolute Humidity: the ratio of the mass
of water vapor in a volume of air to the total
volume of the mixture. The density of the
water vapor. Units: g/m3
Usually not used in meteorology. It is not
conserved in an adiabatic expansion or
compression.
e
e
v 
Rv  T

P
   d
Relative Humidity: The ratio of the actual
water vapor in the air to the saturation
amount at that temperature.
The ratio is normally multiplied by 100 to
put it in a % value.
e
q

r
RH   100   100   100  100
es
qs
s
rs
Dew Point: The temperature to which a
given parcel of air must be cooled at
constant pressure to become saturated with
respect to a plane surface of water.
 1 R
 e 
v



Td 

 ln 
 

e0 
T0 L
1
e0 = 0.611kPa = es at T0 = 273.15oK
1


 r  P 
or,
1
R


Td    v  ln 



e0  r   
T0 L
For frost point (i.e., when dew point is
below freezing) use Ld (Latent Heat of
deposition) for L instead of Lv.
Dew Point may also be expressed in this
manner:
esT d 
r

P
0.62197 r
Lifting Condensation Level
As air is forced upward, it cools at the dry
adiabatic lapse rate, Gd = 9.8oC/km, and:
T  T0  Gz  z0 
The dew point also changes, decreasing at Gdew
= 1.8oC/km, and. Td  Td0  Gdewz  z0 
As the air rises, the temp. and dew point move
closer to each other. At the LCL, the
temperature and dew point are equal, T = Td.
So, replacing Td with T in the second
equation and subtracting the second from
the first gives: 0  T0  Tdo  Z LCL  Zo G  Gdew 
and,T  Tdo  z  z
LCL
0
G  G de w

(G-Gdew = 8oC/km. If the LCL above
ground is desired, then z0 = 0 and,
zLCL  0.125km C T  Tdew 
o
Wet-bulb Temperature
The temperature a parcel of air would have
if cooled adiabatically (no transfer of heat
into or out of the parcel) to saturation at
constant pressure by evaporating water into
the parcel, with all latent heat being
supplied by the parcel.
A simplified expression for wet-bulb
temperature can be found from a heat
budget expression as follows.
Consider that the wet-bulb thermometer is
measuring the temperature of the air. As the
air temperature changes the thermometer
changes. As water is evaporated into the air
the air temperature changes. The
temperature change occurring in the volume
of air (about the thermometer and measured
by the thermometer) results from latent heat
being extracted from the air to evaporate
water into the air about the wet bulb
thermometer.
Then we can write:
md  Cpd  mvCpv  dT  Lv  dmv
Heat loss resulting
Latent heat gain
in a change in temp
by water
of the air
Considering the units on both sides shows
energy on both sides.
o
 kg J

 J 
kg J


 o 
 K  
 
k g
o
kg K kg K 
kg 
Dividing both sides by md and
mv
remembering that mixing ratio is: r 
md
Cpd  rCpv dT  Lv  dr
Integrating gives:
Tw

T
C
pd
 rCpv 
dT   dr
Lv
r
rw
Cpd, Cpv and Lv are all functions of
temperature. The left integral is very
difficult to solve as written.
If we ignore the temperature change of the
water molecules in the volume of air and
consider Cpd and Lv as constants (changes
so small they can be ignored) then:
Tw
rw
T
r
Cpd  dT  L  dr
which results in:
Cp  T  Tw   Lv  r  rw 
where, rw is the saturation mixing ratio
at Tw.
Another, more complicated expression for the Wet
Bulb Temperature is given as:


Lv Tw  Cpv  Cw
Cp  Cpv r 

log

log

Lv T 
C pv
C

C
r
pv s Tw  
 p
where,
Lv(T) = heat of vaporization of water at T,
Lv(Tw) = heat of vaporization of water at Tw,
Cw = specific heat of liquid water,
rw(Tw) = saturation mixing ratio at Tw,
Cp = specific heat of dry air at constant pressure,
Cpv = specific heat of water vapor at constant
pressure.
Most modelers use the equations derived
from Teten’s empirical formula for
saturation vapor pressure to calculate
mixing ratio and from that other moisture

expressions.
rw 
 c T 
w 

 T   
 w

bPe
1
r  rw    T  Tw 
Total Water Mixing Ratio
The grams of water of all phases per gram
of dry air. r  r  r  r
T
L
i
Loss by chemical action is considered
negligible. Change in total occurs only by
precipitation. If no precipitation, then the
total must remain constant.
Thermodynamic Diagrams
Saturated Adiabatic Lapse Rate
A saturated parcel of air rising in the
atmosphere will cool at the saturated
(moist) adiabatic lapse rate. It is less than
the dry adiabatic lapse rate because latent
heat released by condensing water adds heat
to the volume of air.
 r  L 
s
v 

1

 R  T 


d
The moist lapse rate is:
T
g

 Gs 
 

2

z
Cp
L
r


1  v s 2 
 C  R  T 
g


p
d
Gd 
Remember,
Cp
Lv
Lv2  
If we let: a 
and, b 
Cp  Rd
Rd
Then, G s  G d
1  a  r T 


1  b  r
s
s
T
2

However, If we want to know the change in
temperature as pressure changes, we have to
P g
change to T/P. Using: P

z
Rd  T
the equation can be converted to:
Rd T  Lv rs  C p
T


2

P
L

r


v
s

P  
1
 C  R  T 2 


p
d


Liquid Water Potential Temp.
Temperature a saturated parcel of air would have
if brought to 100 kPa (1000 hPa). It has the
advantage that it is conserved over all phase
changes as a parcel ascends or descends in the
L  
atmosphere.
v
 rL
 L    
C  T 
 p 
L  
g
v
 rL
L  T 
 z  


Cp
Cp  T 
Equivalent Potential Temperature
Temperature a saturated parcel of air would
have if raised to a level where all water was
condensed out (P = 0) and then brought
down to 100 kPa.
Lv
Qe  QL 
 rT
Cp
Questions & Problems
N1, N2 (d), N3, N5 (d, e, h), N6 (c, e), N8
(b, e, f), N9 (a, c), N10 (b, e, h), N11(c, f,
h), N18 (c)
SHOW ALL EQUATIONS USED AND
CALCULATIONS