prof. fred remer university of north dakota water in the atmosphere

Prof. Fred RemerUniversity of North Dakota

Water in the AtmosphereWater in the Atmosphere


ReadingReading

Hess– pp 43 - 44– pp 58 – 60

Tsonis– pp 93 – 97

Wallace & Hobbs– pp 66 – 67– pp 79 – 84

Bohren & Albrecht– pp 181-188


ObjectivesObjectives

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


ObjectivesObjectives

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


Water In the AtmosphereWater In the Atmosphere

Unique Substance Occurs in Three Phases Under

Normal Atmospheric Pressures and Temperatures

Gaseous State– Variable 0 – 4%

HH HHOO



Remember Dalton’s Law?– Law of Partial Pressures

– Let’s look at the contribution of water

p = p1 + p2 + p3 + ….p = p1 + p2 + p3 + ….


Water Vapor Pressure (e)Water Vapor Pressure (e)

Ideal Gas Law for Dry Air

Ideal Gas Law for Water Vapor

TRp dd

TRe vv

p = pressure of dry airp = pressure of dry airdd = specific volume of dry air = specific volume of dry air

RRdd = gas constant for dry air = gas constant for dry air

e = vapor pressure of water vapore = vapor pressure of water vaporvv = specific volume of water vapor = specific volume of water vapor

RRvv = gas constant for water vapor = gas constant for water vapor



Partial pressure that water vapor exerts

Total PressureTotal Pressurep = pp = pOO22

+p+pNN22+p+pHH22OOvv

Water Vapor PressureWater Vapor Pressuree = pe = pHH22OOvv



Gas Constant of Water Vapor

HH HHOO

Wv M

RR

Molecular Weight Molecular Weight (M(Mw w ))

Hydrogen = 1kg kmolHydrogen = 1kg kmol-1 -1

Oxygen = 16 kg kmolOxygen = 16 kg kmol-1-1

Water = 18 kg kmolWater = 18 kg kmol-1-1

1

11

v kmolkg18

kmolKJ8314R

11v kgKJ461R


Virtual Temperature (TVirtual Temperature (Tvv))

The temperature dry air must have in order to have the same density as moist air at the same pressure

Fictitious temperature



Dry Air

Total Pressure = pTotal Pressure = p

Volume = VVolume = V

Temperature = TTemperature = T

Mass of Air = mMass of Air = mdd



Moist Air (Mixture)

Total Pressure = pTotal Pressure = p

Volume = VVolume = V

Temperature = TTemperature = T

Mass of Air = mMass of Air = md d ++ mmvv



Density of mixture

V

mm vd

vd



Ideal Gas Law

– For Dry Air

– For Water Vapor Alone

TRp ddd

TRe vv

ororTR

p

d

dd

TR

e

vv oror



Substitute into density expression

vd

TR

p

d

dd

TR

e

vv

TR

e

TR

p

vd

d



Dalton’s Law of Partial Pressure

TR

e

TR

p

vd

d

epp d oror eppd



Substitute eppd

TR

e

TR

p

vd

d TR

e

TR

ep

vd

oror

vd R

e

R

ep

T

1



Remove Rd

vd R

e

R

)ep(

T

1

v

d

d R

Re)ep(

TR

1



Define 622.

M

M

R

R

d

w

v

d

eepTR

1

d



Remove p

eepTR

1

d

p

e

p

e1

TR

p

d



Rearrange terms

p

e

p

e1

TR

p

d

)1(

p

e1

TR

p

d



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

vdTRp

TRp ddd TRe vvInstead ofInstead of oror

UseUsep = total (mixture) pressurep = total (mixture) pressure

= mixture density= mixture density



Substitution of

)1(

p

e1

TR

p

d

vdTRp Into

)1(

p

e1

TR

pRTp

ddv

Produces



Rearrange

)1(

p

e1

TR

pRTp

ddv

)1(pe

1TRp

R

pT

dd

v



Start Canceling!

)1(pe

1TRp

R

pT

dd

v

)1(

pe

1

TTv



Still looks Ugly! Simplify!

)1(

pe

1

TTv

)622.1)(p/e(1

TTv

622.



)p/e378(.1

TTv

p = total (atmospheric) pressurep = total (atmospheric) pressure

e = water vapor pressuree = water vapor pressure

T = temperatureT = temperature



Moist air (mixture) is less dense than dry air

Virtual temperature is greater than actual temperature

Small difference

)p/e378(.1

TTv


Specific Humidity (q)Specific Humidity (q)

Ratio of the density of water vapor in the air to the (total) density of the air

vq


Mixing Ratio (w)Mixing Ratio (w)

The mass of water vapor (mv) to the mass of dry air

Mass of Dry Air = mMass of Dry Air = mdd

Mass of Water Vapor = mMass of Water Vapor = mvv



The mass of water vapor (mv) to the mass of dry air

Mass of Dry Air = mMass of Dry Air = mdd

Mass of Water Vapor = mMass of Water Vapor = mvv

d

v

m

mw



Expressed in g/kg– Dry Air

1 to 2 g/kg

– Tropical Air 20 g/kg

d

v

m

mw



Can mixing ratio be expressed in terms of water vapor pressure?

Sure as it will rain on a meteorologist’s picnic!

d

v

m

mw



By definition

Divide top and bottom by volume (V)

d

v

m

mw

V/m

V/mw

d

v



But density is so.....

V/m

V/mw

d

v

V/m

d

vw

w = mixing ratiow = mixing ratio

vv = density of water vapor in air = density of water vapor in air

dd = density of dry air = density of dry air



Ideal Gas Law

d

vw

TRp ddd

TRe vv

TR

p

d

dd

TR

e

vv

oror

oror



Substitute

d

vw

TR

p

d

dd

TR

e

vv

TR

p/

TR

ew

d

d

v



Simplify

Remember

TR

p/

TR

ew

d

d

v

dv

d

p

e

R

Rw

622.M

M

R

R

d

w

v

d



Substitute into

But

v

d

R

R

dv

d

p

e

R

Rw

dp

ew

eppd p = total pressure of air (mixture)p = total pressure of air (mixture)



Substitute into

Ta-Da!

dp

ew eppd

ep

ew



Expression for Mixing Ratio (w)– Water Vapor Pressure (e) in any units– Atmospheric Pressure (p) in any units

ep

e622.w



Can be used to determine other water variables

Let’s look at– Specific Humidity – Water Vapor Pressure (e)

– Virtual Temperature (Tv)



By definition

But

vqq = specific humidityq = specific humidity

vv = density of water vapor in air = density of water vapor in air

= density of air = density of air

dv dd = density of dry air = density of dry air



Substitute into

Results in

But

vqdv

dv

vq

V

m



Substitute into

Results in

dv

vq

V

m

V/mV/m

V/mq

dv

v



Eliminate V

dv

v

mm

mq

V/mV/m

V/mq

dv

v



Divide top and bottom by md

dv

v

mm

mq

dddv

dv

m/mm/m

m/mq



But so

dddv

dv

m/mm/m

m/mq

d

v

m

mw

1w

wq



Expression for specific humidity (q)– Mixing Ratio (w) in kg kg-1

1w

wq



Pressure exerted by water vapor is a fraction of total pressure of air

Fraction is proportional to # of moles in mixture

pfe e = water vapor pressuree = water vapor pressure

f = fractional amount of water vapor f = fractional amount of water vapor

p = total pressure of airp = total pressure of air



How many moles of water are in a sample of air?

Number of moles of water

w

vv M

mn

nnvv = # of moles = # of moles

mmvv = mass of water molecules = mass of water molecules

MMww = molecular weight of water = molecular weight of water



How many moles of dry air are in a sample of air?

Number of moles of dry air

d

dd M

mn

nndd = # of moles = # of moles

mmdd = mass of dry air = mass of dry air

MMdd = mean molecular weight of dry air = mean molecular weight of dry air



How many moles of air are in a sample of air?

Number of moles of air

d

d

w

v

M

m

M

mn



What is the molar fraction of water vapor in the air?

Substitute into

ddwv

wv

M/mM/m

M/mf

pfe



Yikes! Let’s make this more manageable!

ddwv

wv

M/mM/m

M/mf

pfe

pM/mM/m

M/me

ddwv

wv


pM/mM/m

M/me

ddwv

wv


Multiply top and bottowm by Mw/md

pm/M

m/M

M/mM/m

M/me

dw

dw

ddwv

wv



Canceling out

pm/M

m/M

M/mM/m

M/me

dw

dw

ddwv

wv

pM/Mm/m

m/me

dwdv

dv



But

pM/Mm/m

m/me

dwdv

dv

d

v

m

mw

Mixing RatioMixing Ratio

andand 622.M

M

R

R

d

w

v

d



pM/Mm/m

m/me

dwdv

dv

d

v

m

mw

d

w

M

M

pw

we


pw

we


Expression for water vapor pressure (e)– Mixing Ratio (w) in kg kg-1

– Atmospheric Pressure (p)



Derive an expression for virtual temperature (Tv) using mixing ratio (w)

)1(

pe

1

TTv



Expression for water vapor pressure

pw

we

)1(

pe

1

TTv

oror

w

w

p

e


)1(

pe

1

TTv

w

w

p

e


Substituting

)1(

ww

1

TTv



Expand

)1(

ww

1

TTv

ww

ww

1

TTv



Common denominator w+

ww

ww

1

TTv

ww

ww

ww

TTv



Group

ww

ww

ww

TTv

wwww

TTv



Simplify

wwww

TTv

ww

TTv



Divide numerator by denominator (polynomial division) and eliminate w2 terms

)w1(

wTTv

w1

1TTv



Substitute = .622

w1

1TTv

w61.1TTv


Virtual Temperature (Tv)Virtual Temperature (Tv)

Expression for virtual temperature– Mixing Ratio (w) in kg kg-1

w61.1TTv


Review of Water VariablesReview of Water Variables

Water Vapor Pressure

TRe vv

pw

we



Virtual Temperature

w61.1TTv

)p/e378(.1

TTv



Mixing Ratio

d

v

m

mw

ep

e622.w

d

vw



Specific Humidity

vq1w

wq



Moisture Variables– Water Vapor Pressure– Virtual Temperature– Mixing Ratio– Specific Humidity

Amount of Moisture in the Atmosphere



Unanswered Questions– How much water vapor can the air hold?– When will condensation form?– Is the air saturated?

The Beer Analogy


The Beer AnalogyThe Beer Analogy

You are thirsty! You would like a

beer. Obey your thirst!



Pour a glass but watch the foam



Wait! Some joker put

a hole in the bottom of your Styrofoam cup!

It is leaking!



Having had many beers already, you are intrigued by the phenomena!



Rate at beer flows from keg is constant



Rate at beer flows from keg is constant

Rate at beer flows from cup depends on height



The higher the level of beer in the cup, the faster it leaks!



The cup fills up Height

becomes constant

Equilibrium Reached

Inflow(Constant)

Leakage(Varies with

Height)



What do you do?

Inflow(Constant)

Leakage(Varies with

Height)



Get a new cup!


OverviewOverview

Similar to what happens to water in the atmosphere


OverviewOverview

Molecules in liquid water attract each other

In motion


OverviewOverview

Collisions Molecules near

surface gain velocity by collisions


OverviewOverview

Fast moving molecules leave the surface

Evaporation


OverviewOverview

Soon, there are many water molecules in the air


OverviewOverview

Slower molecules return to water surface

Condensation


OverviewOverview

Net Evaporation– Number leaving

water surface is greater than the number returning

– Evaporation greater than condensation


OverviewOverview

Molecules leave the water surface at a constant rate

Depends on temperature of liquid


OverviewOverview

Molecules return to the surface at a variable rate

Depends on mass of water molecules in air


OverviewOverview

Rate at which molecule return increases with time– Evaporation

continues to pump moisture into air

– Water vapor increases with time


OverviewOverview

Eventually, equal rates of condensation and evaporation

“Air is saturated” Equilibrium


OverviewOverview

Derive a relationship that describes this equilibrium


Clausius-Clapeyron Clausius-Clapeyron EquationEquation

prof. fred remer university of north dakota water in the atmosphere

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