Water in the Atmosphere
Prof. Fred Remer
University of North Dakota
Reading

Hess
– pp 43 - 44
– pp 58 – 60

Tsonis
– pp 93 – 97

Wallace & Hobbs
– pp 66 – 67
– pp 79 – 84

Bohren & Albrecht
– pp 181-188
Prof. Fred Remer
University of North Dakota
Objectives
Be able to define water vapor
pressure
 Be able to define virtual temperature
 Be able to define specific humidity
 Be able to define mixing ratio

Prof. Fred Remer
University of North Dakota
Objectives
Be able to calculate the water vapor
pressure
 Be able to calculate virtual
temperature
 Be able to calculate specific humidity
 Be able to calculate mixing ratio

Prof. Fred Remer
University of North Dakota
Water In the Atmosphere
Unique Substance
 Occurs in Three Phases Under
Normal Atmospheric Pressures and
Temperatures
 Gaseous State
– Variable 0 – 4%
H
H
O

Prof. Fred Remer
University of North Dakota
Water in the Atmosphere

Remember Dalton’s Law?
– Law of Partial Pressures
p = p1 + p2 + p3 + ….
– Let’s look at the contribution of water
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)

Ideal Gas Law for Dry Air
p d  RdT

p = pressure of dry air
d = specific volume of dry air
Rd = gas constant for dry air
Ideal Gas Law for Water Vapor
ev  Rv T
Prof. Fred Remer
University of North Dakota
e = vapor pressure of water vapor
v = specific volume of water vapor
Rv = gas constant for water vapor
Water Vapor Pressure (e)

Partial pressure that water vapor
exerts
Total Pressure
p = pO2+pN2+pH2Ov
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure
e = pH2Ov
Water Vapor Pressure (e)

Gas Constant of Water Vapor
R
Rv 
MW
H
1
8314 J K kmol
Rv 
1
18 kg kmol
1
Rv  461 J K kg
Prof. Fred Remer
University of North Dakota
1
1
H
O
Molecular Weight (Mw )
Hydrogen = 1kg kmol-1
Oxygen = 16 kg kmol-1
Water = 18 kg kmol-1
Virtual Temperature (Tv)
The temperature dry air must have in
order to have the same density as
moist air at the same pressure
 Fictitious temperature

Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Dry Air
Total Pressure = p
Volume = V
Temperature = T
Mass of Air = md
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Moist Air (Mixture)
Total Pressure = p
Volume = V
Temperature = T
Mass of Air = md + mv
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Density of mixture
md  m v

V
  d   v
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Ideal Gas Law
– For Dry Air
pd  dRdT
or
pd
d 
R dT
– For Water Vapor Alone
e  vRv T
Prof. Fred Remer
University of North Dakota
or
e
v 
RvT
Virtual Temperature (Tv)

Substitute into density expression
e
v 
RvT
pd
d 
R dT
  d   v
pd
e


R dT R v T
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
pd
e


R dT R v T

Dalton’s Law of Partial Pressure
p  pd  e
Prof. Fred Remer
University of North Dakota
or
pd  p  e
Virtual Temperature (Tv)

Substitute
pd  p  e
pe
e


R dT R v T
pd
e


R dT R v T
or
Prof. Fred Remer
University of North Dakota
1 p  e e 
 


T  Rd
Rv 
Virtual Temperature (Tv)
1  (p  e) e 
 


T  Rd
Rv 

Remove Rd
1

TR d
Prof. Fred Remer
University of North Dakota

Rd 
(p  e)  e 
Rv 

Virtual Temperature (Tv)

Define e
R d Mw
e

 .622
R v Md
1
p  e  ee

TR d
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
1
p  e  ee

TR d

Remove p
p

TR d
Prof. Fred Remer
University of North Dakota
 e e 
1  p  p e


Virtual Temperature (Tv)
p

TR d

 e e 
1  p  p e


Rearrange terms
p

TR d
Prof. Fred Remer
University of North Dakota
 e

1   (1  e)
 p

Virtual Temperature (Tv)

By definition, virtual temperature is
the temperature dry air must have in
order to have the same density as
moist air (mixture) at the same
pressure
Instead of
Use
Prof. Fred Remer
University of North Dakota
pd  dRdT
p  RdTv
or
e  vRv T
p = total (mixture) pressure
 = mixture density
Virtual Temperature (Tv)
p
 Substitution of  
TR d


Into
Produces
Prof. Fred Remer
University of North Dakota
 e

1   (1  e)
 p

p  RdTv
 p
p  TvR d 
 TR d
 e
 
1   (1  e ) 
 p
 
Virtual Temperature (Tv)
 p
p  TvR d 
 TR d

 e
 
1   (1  e ) 
 p
 
Rearrange
Tv 
Prof. Fred Remer
University of North Dakota
p
 p
Rd 
 TR d
 e
 
1   (1  e ) 
 p
 
Virtual Temperature (Tv)
Tv 

p
 p
Rd 
 TR d
 e
 
1   (1  e ) 
 p
 
Start Canceling!
Tv 
Prof. Fred Remer
University of North Dakota
T
 e

1   (1  e)
 p

Virtual Temperature (Tv)
Tv 

T
 e

1   (1  e)
 p

Still looks Ugly! Simplify!
e  .622
Prof. Fred Remer
University of North Dakota
T
Tv 
1  (e / p)(1  .622)
Virtual Temperature (Tv)
T
Tv 
1  (.378 e / p)
p = total (atmospheric) pressure
e = water vapor pressure
T = temperature
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
T
Tv 
1  (.378 e / p)
Moist air (mixture) is less dense than
dry air
 Virtual temperature is greater than
actual temperature
 Small difference

Prof. Fred Remer
University of North Dakota
Specific Humidity (q)

Ratio of the density of water vapor in
the air to the (total) density of the air
v
q

Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)

The mass of water vapor (mv) to the
mass of dry air
Mass of Dry Air = md
Mass of Water Vapor = mv
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)

The mass of water vapor (mv) to the
mass of dry air
mv
w
md
Mass of Dry Air = md
Mass of Water Vapor = mv
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)

Expressed in g/kg
– Dry Air

1 to 2 g/kg
– Tropical Air

Prof. Fred Remer
University of North Dakota
20 g/kg
mv
w
md
Mixing Ratio (w)
mv
w
md

Can mixing ratio be expressed in
terms of water vapor pressure?

Sure as it will rain on a
meteorologist’s picnic!
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)
mv
 By definition w 
md

Divide top and bottom by volume (V)
mv / V
w
md / V
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)
mv / V
w
md / V

But density is   m / V
v
w
d
Prof. Fred Remer
University of North Dakota
so.....
w = mixing ratio
v = density of water vapor in air
d = density of dry air
Mixing Ratio (w)
v
w
d

Ideal Gas Law
pd  dRdT
e  vRv T
Prof. Fred Remer
University of North Dakota
or
pd
d 
R dT
or
e
v 
RvT
Mixing Ratio (w)

Substitute
pd
d 
R dT
v
w
d
pd
e
w
/
R v T R dT
Prof. Fred Remer
University of North Dakota
e
v 
RvT
Mixing Ratio (w)

Simplify
pd
e
w
/
R v T R dT
Rd e
w
R v pd

Remember e  R d  Mw  .622
R v Md
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)
Rd
 Substitute e 
into
Rv
Rd e
w
R v pd
e
we
pd

But pd  p  e
Prof. Fred Remer
University of North Dakota
p = total pressure of air (mixture)
Mixing Ratio (w)
e
 Substitute pd  p  e into w  e
pd

Ta-Da!
Prof. Fred Remer
University of North Dakota
e
we
pe
Mixing Ratio (w)
e
w  .622
pe

Expression for Mixing Ratio (w)
– Water Vapor Pressure (e) in any units
– Atmospheric Pressure (p) in any units
Prof. Fred Remer
University of North Dakota
Mixing Ratio (w)
Can be used to determine
other water variables
 Let’s look at

– Specific Humidity
– Water Vapor Pressure (e)
– Virtual Temperature (Tv)
Prof. Fred Remer
University of North Dakota
Specific Humidity (q)

By definition
v
q


q = specific humidity
v = density of water vapor in air
 = density of air
But     
v
d
Prof. Fred Remer
University of North Dakota
d = density of dry air
Specific Humidity (q)


Substitute   v  d into
Results in
m
 But  
V
Prof. Fred Remer
University of North Dakota
v
q
 v  d
v
q

Specific Humidity (q)
m
 Substitute  
V
into
v
q
 v  d
mv / V
 Results in q 
m v / V  md / V
Prof. Fred Remer
University of North Dakota
Specific Humidity (q)
mv / V
q
m v / V  md / V

Eliminate V
mv
q
m v  md
Prof. Fred Remer
University of North Dakota
Specific Humidity (q)
mv
q
m v  md

Divide top and bottom by md
m v / md
q
m v / md  md / md
Prof. Fred Remer
University of North Dakota
Specific Humidity (q)
m v / md
q
m v / md  md / md

But
Prof. Fred Remer
University of North Dakota
mv
w
md
so
w
q
w 1
Specific Humidity (q)
w
q
w 1

Expression for specific humidity (q)
– Mixing Ratio (w) in kg kg-1
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)

Pressure exerted by water vapor is a
fraction of total pressure of air
e  f p

e = water vapor pressure
f = fractional amount of water vapor
p = total pressure of air
Fraction is proportional to # of moles
in mixture
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)

How many moles of water are in a
sample of air?

Number of moles of water
mv
nv 
Mw
Prof. Fred Remer
University of North Dakota
nv = # of moles
mv = mass of water molecules
Mw = molecular weight of water
Water Vapor Pressure (e)

How many moles of dry air are in a
sample of air?

Number of moles of dry air
md
nd 
Md
Prof. Fred Remer
University of North Dakota
nd = # of moles
md = mass of dry air
Md = mean molecular weight of dry air
Water Vapor Pressure (e)

How many moles of air are in a
sample of air?

Number of moles of air
m v md
n

Mw Md
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)

What is the molar fraction of water
vapor in the air?
m v / Mw
f
mv / Mw  md / Md

Substitute into
e  f p
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)
m v / Mw
f
mv / Mw  md / Md
e  f p


mv / Mw
p
e  
 mv / Mw  md / Md 

Yikes! Let’s make this more
manageable!
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)


mv / Mw
p
e  
 mv / Mw  md / Md 

Multiply top and bottowm by Mw/md

 Mw / md 
mv / Mw

p
e  
 mv / Mw  md / Md  Mw / md 
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)

 Mw / md 
mv / Mw

p
e  
 mv / Mw  md / Md  Mw / md 

Canceling out


mv / md
p
e  
 mv / md  Mw / Md 
Prof. Fred Remer
University of North Dakota
Water Vapor Pressure (e)


mv / md
p
e  
 mv / md  Mw / Md 

But
mv
w
md
Mixing Ratio
Prof. Fred Remer
University of North Dakota
and
R d Mw
e

 .622
R v Md
Water Vapor Pressure (e)
mv
w
md
Mw
e
Md


mv / md
p
e  
 mv / md  Mw / Md 
Prof. Fred Remer
University of North Dakota
 w 
e
p
w e
Water Vapor Pressure (e)
 w 
e
p
w e

Expression for water vapor pressure (e)
– Mixing Ratio (w) in kg kg-1
– Atmospheric Pressure (p)
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Derive an expression for virtual
temperature (Tv) using mixing ratio (w)
Tv 
Prof. Fred Remer
University of North Dakota
T
 e

1   (1  e)
 p

Virtual Temperature (Tv)
Tv 

T
 e

1   (1  e)
 p

Expression for water vapor pressure
 w 
e
p
w e
Prof. Fred Remer
University of North Dakota
or
e  w 


p w e
Virtual Temperature (Tv)

Substituting
Tv 
e  w 


p w e
T
 e

1   (1  e)
 p

Prof. Fred Remer
University of North Dakota
T
Tv 
  w 

1   w  e (1  e)

 

Virtual Temperature (Tv)
T
Tv 
  w 

1   w  e (1  e)

 


Expand
Tv 
Prof. Fred Remer
University of North Dakota
T
  w   w 
1   w  e   e w  e 
 

 
Virtual Temperature (Tv)
Tv 

T
  w   w 
1   w  e   e w  e 
 

 
Common denominator w+e
Tv 
Prof. Fred Remer
University of North Dakota
T
 w  e  w   w 
 w  e   w  e   e w  e 

 


Virtual Temperature (Tv)
Tv 

T
 w  e  w   w 
 w  e   w  e   e w  e 

 


Group
T
Tv 
 w  e  w  ew 


we
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
T
Tv 
 w  e  w  ew 


we

Simplify
T
Tv 
 e  ew 
 w  e 
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
we
Tv  T
e(1  w )

Divide numerator by denominator
(polynomial division) and eliminate w2 terms
 1 e 
Tv  T1 
w
e


Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)
 1 e 
Tv  T1 
w
e



Substitute e = .622
Tv  T1 .61w 
Prof. Fred Remer
University of North Dakota
Virtual Temperature (Tv)

Expression for virtual temperature
– Mixing Ratio (w) in kg kg-1
Tv  T1 .61w 
Prof. Fred Remer
University of North Dakota
Review of Water Variables

Water Vapor Pressure
e  vRv T
 w 
e
p
w e
Prof. Fred Remer
University of North Dakota
Review of Water Variables

Virtual Temperature
Tv  T1 .61w 
T
Tv 
1  (.378 e / p)
Prof. Fred Remer
University of North Dakota
Review of Water Variables

Mixing Ratio
mv
w
md
v
w
d
e
w  .622
pe
Prof. Fred Remer
University of North Dakota
Review of Water Variables

Specific Humidity
v
q

Prof. Fred Remer
University of North Dakota
w
q
w 1
Water in the Atmosphere

Moisture Variables
– Water Vapor Pressure
– Virtual Temperature
– Mixing Ratio
– Specific Humidity

Amount of Moisture in the
Atmosphere
Prof. Fred Remer
University of North Dakota
Water in the Atmosphere

Unanswered Questions
– How much water vapor can the air hold?
– When will condensation form?
– Is the air saturated?

The Beer Analogy
Prof. Fred Remer
University of North Dakota
The Beer Analogy
You are thirsty!
 You would like a
beer.
 Obey your thirst!

Prof. Fred Remer
University of North Dakota
The Beer Analogy

Prof. Fred Remer
University of North Dakota
Pour a glass
but watch the
foam
The Beer Analogy
Wait!
 Some joker put
a hole in the
bottom of your
Styrofoam cup!
 It is leaking!

Prof. Fred Remer
University of North Dakota
The Beer Analogy

Prof. Fred Remer
University of North Dakota
Having had
many beers
already, you are
intrigued by the
phenomena!
The Beer Analogy
Rate at beer flows
from keg is constant
Prof. Fred Remer
University of North Dakota
The Beer Analogy
Rate at beer flows
from keg is constant
Rate at beer flows from
cup depends on height
Prof. Fred Remer
University of North Dakota
The Beer Analogy
The higher the level of
beer in the cup, the faster
it leaks!
Prof. Fred Remer
University of North Dakota
The Beer Analogy
The cup fills up
 Height
becomes
constant
 Equilibrium
Reached
Leakage

Inflow
(Constant)
(Varies with
Height)
Prof. Fred Remer
University of North Dakota
The Beer Analogy

What do you
do?
Inflow
(Constant)
Leakage
(Varies with
Height)
Prof. Fred Remer
University of North Dakota
The Beer Analogy

Prof. Fred Remer
University of North Dakota
Get a new cup!
Overview

Similar to what
happens to water in
the atmosphere
Prof. Fred Remer
University of North Dakota
Overview
Molecules in
liquid water
attract each other
 In motion

Prof. Fred Remer
University of North Dakota
Overview
Collisions
 Molecules near
surface gain
velocity by
collisions

Prof. Fred Remer
University of North Dakota
Overview
Fast moving
molecules leave
the surface
 Evaporation

Prof. Fred Remer
University of North Dakota
Overview

Soon, there are
many water
molecules in the
air
Prof. Fred Remer
University of North Dakota
Overview
Slower molecules
return to water
surface
 Condensation

Prof. Fred Remer
University of North Dakota
Overview

Net Evaporation
– Number leaving
water surface is
greater than the
number returning
– Evaporation
greater than
condensation
Prof. Fred Remer
University of North Dakota
Overview
Molecules leave
the water surface
at a constant rate
 Depends on
temperature of
liquid

Prof. Fred Remer
University of North Dakota
Overview
Molecules return
to the surface at
a variable rate
 Depends on
mass of water
molecules in air

Prof. Fred Remer
University of North Dakota
Overview

Rate at which
molecule return
increases with
time
– Evaporation
continues to pump
moisture into air
– Water vapor
increases with time
Prof. Fred Remer
University of North Dakota
Overview
Eventually, equal
rates of
condensation and
evaporation
 “Air is saturated”
 Equilibrium

Prof. Fred Remer
University of North Dakota
Overview

Derive a
relationship that
describes this
equilibrium
Prof. Fred Remer
University of North Dakota
Clausius-Clapeyron
Equation
Prof. Fred Remer
University of North Dakota