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Moisture  There are several methods of expressing the moisture content (water in vapor form) of a volume of air.  Vapor Pressure: The partial pressure.

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Presentation on theme: "Moisture  There are several methods of expressing the moisture content (water in vapor form) of a volume of air.  Vapor Pressure: The partial pressure."— Presentation transcript:

1 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. e s

2  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.

3 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.

4  The relationship is given by : where, e s = saturation vapor pressure at T, e 0 = saturation vapor pressure at T 0. e 0 = 0.611 kPa when T o = 273 o K (0 o C) L = latent heat of vaporization or deposition, L v 2.5 x 10 6 J/kg, L d = 2.83 x 10 6 J/kg (which is the value at 0 o C) Latent heat actually varies slightly with temperature. R v = gas constant for water vapor = 461 J/ o K kg  Remember, exp means e ( )

5 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. Where, e 0 = 0.611 kPa b = 17.2694 T 1 = 273.16 o K, T 2 = 35.86 o K

6  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, e s, in meteorology. Note: definition of saturation vapor pressure refers to flat liquid water surface. Derivation on page 98 gives: a = 14.53 H p = 7.29 km

7 Other Humidity Terms  Mixing ratio: Ratio of mass of water vapor to mass of dry air in the same volume. Vapor Pressure: Remember from chapter 1, vapor pressure can be written as:

8  If we multiply the numerator and denominator by M v and divide both by m d we get:

9  Mixing ratio is:  The quantity M v /M d is:  Then:  Where, (P-e) represents the pressure produced by only the dry air molecules.

10  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:  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.

11  Specific Humidity: The ratio of the mass of water vapor to the total mass (dry air plus water vapor) of air. Specific Humidity and Saturation Specific Humidity are pressure dependent.

12  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/m 3  Usually not used in meteorology. It is not conserved in an adiabatic expansion or compression.

13  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.

14  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. e 0 = 0.611kPa = e s at T 0 = 273.15 o K or,

15  For frost point (i.e., when dew point is below freezing) use L d (Latent Heat of deposition) for L instead of L v.  Dew Point may also be expressed in this manner:

16  Lifting Condensation Level  As air is forced upward, it cools at the dry adiabatic lapse rate,  d = 9.8 o C/km, and:  The dew point also changes, decreasing at  dew = 1.8 o C/km, and.  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 = T d.

17  So, replacing T d with T in the second equation and subtracting the second from the first gives: and,  (  -  dew = 8 o C/km. If the LCL above ground is desired, then z 0 = 0 and,

18 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.

19  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.

20  Then we can write: 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.

21  Dividing both sides by m d and remembering that mixing ratio is: Integrating gives: C pd, C pv and L v are all functions of temperature. The left integral is very difficult to solve as written.

22  If we ignore the temperature change of the water molecules in the volume of air and consider C pd and L v as constants (changes so small they can be ignored) then: which results in: where, r w is the saturation mixing ratio at T w.

23  Another, more complicated expression for the Wet Bulb Temperature is given as: where, L v(T) = heat of vaporization of water at T, L v(Tw) = heat of vaporization of water at Tw, C w = specific heat of liquid water, r w(Tw) = saturation mixing ratio at Tw, C p = specific heat of dry air at constant pressure, C pv = specific heat of water vapor at constant pressure.

24  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.

25 Total Water Mixing Ratio  The grams of water of all phases per gram of dry air. Loss by chemical action is considered negligible. Change in total occurs only by precipitation. If no precipitation, then the total must remain constant.

26 Thermodynamic Diagrams

27 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.

28  The moist lapse rate is:  Remember,  If we let:and, Then,

29  However, If we want to know the change in temperature as pressure changes, we have to change to  T/  P. Using: the equation can be converted to:

30 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 atmosphere.

31 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.

32 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


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