air properties

19
Atmospheric moisture Gas laws Air & vapor pressure Moisture content in air Evaporation/condensation

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Page 1: Air Properties

Atmospheric moisture

Gas lawsAir & vapor pressure

Moisture content in air

Evaporation/condensation

Page 2: Air Properties

The ideal gas law

The ideal gas law:

P = pressure [mb]ρ = density of gas [g/cm3]T = absolute temperature [°K]R = gas constant (depends on the molecular weight of the

gas and the other units)

Decrease in pressure ⇒ Decrease in temperature and density (and vice versa).Water vapor can be considered an ideal gas

RTP ρ=

Page 3: Air Properties
Page 4: Air Properties

Dalton's law

Each gas in a mixture creates pressure as if the other gases were not present. The total pressure is the sum of the pressures created by the gases in the mixture

Total air pressure (P) is the sum of dry air pressure (pd) and water vapor pressure (e).Water vapor pressure is typically 1-2% of total air pressure.

Dalton's Law of Partial Pressures:

Page 5: Air Properties

Example

Determine the mass of 1 m3 of dry air at 20°C and a pressure of 1013 mb (= 1 atmosphere).

Solution:The dry air gas constant is given at page 21 as

R = 2.87x103 mb cm3/g °KFrom the ideal gas law:

1 m3 of dry air weighs 1.2 kg.

3

d

cm/g0012.0)20273(2870

1013RTP

=

+×==ρ

Page 6: Air Properties

Definition:Mass of water vapor contained in a unit of moist air [g/g]. It can be calculated as ρw/ ρm.

Calculation:Ideal gas law:

Dry air:

Vapor:

Specific humidity

RTP ρ=

TReP

TRp

dd

dd

−==ρ

]622.0/RR[TR

e622.0TR

edw

dww ===ρ

Page 7: Air Properties

Specific humidity

Density of moist air:

Specific humiditye378.0P

e622.0qm

w

−=

ρρ

==

[ ]e378.0PTR

1d

wdm −=ρ+ρ=ρ

Page 8: Air Properties

Saturation vapor pressure & dew point temperature

Saturation vapor pressure (= maximum vapor pressure)

(Eq. 1.6)

Td = dew point temperature [°C]es = saturation vapor pressure [mb]

Saturation vapor pressure depends only on temperature. It increases with temperature.

Dew point:Temperature to which a parcel of moist air would have to be cooled (under constant pressure and water content) before condensation starts.

+

−×=79.242T

6.4278exp107489.2ed

8s

Page 9: Air Properties

Relative humidity

Relative humidity is a measure of the degree of saturation of the air. Precipitation is associated with a relative humidity of 100%.

see

pressurevaporsaturationpressurevaporactualRH ==

Page 10: Air Properties

Example

At a weather station, the air pressure is measured to be 101.1 kPa, the air temperature is 22 °C and the dew point temperature is 18 °C . Calculate the corresponding

a) Vapor pressure

b) Relative humidity

c) Specific humidity

d) Air density

Page 11: Air Properties

Conversion: 101.1 kPa = 1011 mb (Appendix B)

First calculate vapor pressure e and saturation vapor pressure es using:

+

−×=79.242T

6.4278exp107489.2ed

8s

Solution

Relative humidity = RH = 20.60/26.40 = 78%

mb40.2679.24225

6.4278exp107489.2e

mb60.2079.24218

6.4278exp107489.2e

8s

8

=

+−×=

=

+−×=

Page 12: Air Properties

Specific humidity:

Air density:

airkg/waterkg0128.06.20378.01011

6.20622.0e378.0P

e622.0q

=

×−×

=−

=

[ ]

[ ]

3

33

3

ddwm

m/kg2.1cm/g102.1

60.20378.01011)22273(1087.2

1

e378.0PTR

1

=

×=

×−+×

=

−=ρ+ρ=ρ

Page 13: Air Properties

Phase changes

Phase change requires/produces heat (energy)

Latent heat of evaporation = Latent heat of condensation

Latent heat of melting / freezing:

CinT],g/cal[T57.03.597LL oce −=−=

]g/cal[7.79LL fm =−=

SolidIce

FluidWater

GasVapor

Page 14: Air Properties

Example

Calculate the energy required to heat 1 liter of water from 0ºC to 100ºC. Then calculate the energy required to evaporate the water (at 100ºC).

Solution

Specific heat: energy required to raise the temperature of 1g of a substance by 1ºC.

Specific energy of water: C)-cal/(g1C Ow =

Page 15: Air Properties

1 liter of water = 1000g water. Energy required to heat 1 liter of water from 0ºC to 100ºC:

Evaporation energyLatent heat of evaporation at 100ºC:

Required evaporation energy:

cal000,100C100C)cal/(g1g1000TMC oow =××=∆

cal/g3.540)0100(57.03.597Le =−−=

cal300,540cal/g3.540g1000MLe =×=

Page 16: Air Properties

Measurement of vapor pressure: The psychrometer

Psychrometer:– Two identical glass thermometers– One has a wet fabric applied to the liquid bulb– A fan blows air over the thermometers

Dry-bulb thermometer measures air temperature.Wet-bulb temperature is reduced due to evaporation.

t: dry-bulb temperature, [°C]tw : wet-bulb temperature, [°C]ew: saturation vapor pressure corresponding to the

wet-bulb temperature, [mb]: psychrometer constant

)tt(ee ww −γ−=

C/mb66.0 o=γ

Page 17: Air Properties

Sling psychrometer

Page 18: Air Properties

Example

A psychrometer indicates a dry-bulb temperature of 40°C and a wet-bulb temperature of 30°C. What are the vapor pressure and the relative humidity?

Solution:Saturation vapor pressure corresponding to the wet-bulb temperature:

The psychrometer equation yields:

mb4.4279.242)30T(

6.4278exp107489.2e 8w =

+=

−×=

mb8.35)3040(66.04.42)tt(ee ww

=−−=

−γ−=

Page 19: Air Properties

The saturation vapor pressure at air temperature:

Relative humidity = RH = 35.8/73.9 = 48.5%.

mb9.7379.242)40T(

6.4278exp107489.2e 8s =

+=

−×=