# air properties

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

The ideal gas lawThe ideal gas law:P = RT

P = pressure [mb] = 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

Dalton's lawDalton's Law of Partial Pressures: 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 mixtureTotal 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.

ExampleDetermine the mass of 1 m3 of dry air at 20C 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 K From the ideal gas law:P 1013 d = = RT 2870 (273 + 20) = 0.0012 g / cm3

1 m3 of dry air weighs 1.2 kg.

Specific humidityDefinition: Mass of water vapor contained in a unit of moist air [g/g]. It can be calculated as w/ m. Calculation:Ideal gas law: P = RTpd Pe d = = RdT RdT e 0.622e w = = RwT RdT [Rw = Rd / 0.622]

Dry air:

Vapor:

Specific humidityDensity of moist air: m = d + w = 1 [P 0.378e ] RdT

0.622 e w = Specific humidity = q = m P 0.378 e

Saturation vapor pressure & dew point temperatureSaturation vapor pressure (= maximum vapor pressure) 4278.6 es = 2.7489 10 exp T + 242.79 d8

(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.

Relative humidityactual vapor pressure e RH = = saturation vapor pressure es

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

ExampleAt 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

SolutionConversion: 101.1 kPa = 1011 mb (Appendix B) First calculate vapor pressure e and saturation vapor pressure es using: 4278.6 es = 2.7489 10 exp T + 242.79 d 8

4278.6 e = 2.7489 108 exp = 20.60 mb 18 + 242.79 4278.6 es = 2.7489 108 exp = 26.40 mb 25 + 242.79 Relative humidity = RH = 20.60/26.40 = 78%

Specific humidity:q= 0.622 e 0.622 20.6 = P 0.378 e 1011 0.378 20.6

= 0.0128 kg water / kg air

Air density:1 [P 0.378e ] m = w + d = RdT =

1 [1011 0.378 20.60] 3 2.87 10 (273 + 22)

= 1.2 10 3 g / cm3 = 1.2 kg / m3

Phase changes Solid Ice Fluid Water Gas Vapor

Phase change requires/produces heat (energy)Latent heat of evaporation = Latent heat of condensation Le = Lc = 597.3 0.57 T [cal / g], T in oC Latent heat of melting / freezing: Lm = Lf = 79.7 [cal / g]

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

SolutionSpecific heat: energy required to raise the temperature of 1g of a substance by 1C. Specific energy of water: Cw = 1 cal/(g -O C)

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

MCw T = 1000g 1 cal/(goC) 100oC = 100,000 calEvaporation energy Latent heat of evaporation at 100C:Le = 597.3 0.57(100 0) = 540.3 cal/g

Required evaporation energy:MLe = 1000g 540.3 cal/g = 540,300 cal

Measurement of vapor pressure: The psychrometerPsychrometer: 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.

e = ew (t tw )t: dry-bulb temperature, [C] tw : wet-bulb temperature, [C] ew: saturation vapor pressure corresponding to the wet-bulb temperature, [mb] = 0.66 mb /oC : psychrometer constant

Sling psychrometer

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

Solution: Saturation vapor pressure corresponding to the wet-bulb temperature: 4278.6 8 ew = 2.7489 10 exp = 42.4 mb (T = 30) + 242.79 The psychrometer equation yields: e = ew (t tw )= 42.4 0.66( 40 30) = 35.8 mb

The saturation vapor pressure at air temperature: 4278.6 es = 2.7489 10 exp = 73.9 mb (T = 40) + 242.79 8

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