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chapter 4 Copyright © 2011, 2015 by Roland Stull. Meteorology for Scientists and Engineers, 3rd Ed. Moisture contents Water vapor is one of the gases in air. Unlike nitrogen and oxygen, water-vapor concen- Saturation Vapor Pressure 87 tration can vary widely in time and space. Vapor Pressure 87 4 Most people are familiar with relative humidity as Saturation 87 a measure of water-vapor concentration because it Humidity Variables 91 affects our body’s moisture regulation. But other Mixing Ratio 91 humidity variables are much more useful in other Specific Humidity 91 contexts. Absolute Humidity 91 Storms get much of their energy from water va- Relative Humidity 92 por, because when water vapor condenses or freezes Dew-Point Temperature 92 Saturation Level or Lifting Condensation Level it releases latent heat. For this reason we carefully (LCL) 93 track water vapor as it rises in buoyant thermals or Wet-Bulb Temperature 94 is carried by horizontal winds. The amount of mois- More Relationships Between Moisture Variables 96 ture available to a storm also regulates the amount Total Water 97 of rain or snow precipitating out. Liquid and Solid Water 97 What allows air to hold water as vapor in one Total-Water Mixing Ratio 97 case, but forces the vapor to condense in another? Precipitable Water 98 This depends on a concept called “saturation”. Lagrangian Budgets 99 Water Conservation 99 Lagrangian Water Budget 99 Thermo Diagrams – Part 2: Isohumes 100 Heat Conservation for Saturated Air 101 saturation Vapor pressure Saturated Adiabatic Lapse Rate 101 Thermo Diagrams – Part 3: Saturated Vapor pressure Adiabats 102 All the gases in air contribute to the total pres- Liquid-Water, Equivalent, and Wet-Bulb Potential sure. The pressure associated with any one gas in a Temperatures 104 mixture is called the partial pressure. The partial Eulerian Water Budget 107 pressure of water vapor in air is called vapor pres- Horizontal Advection 107 sure. Symbol e is used for water-vapor pressure. Precipitation 107 Surface Moisture Flux 108 Units are pressure units, such as kPa. Turbulent Transport 109 Humidity Instruments 111 saturation Summary 112 Air can hold any proportion of water vapor. Threads 112 However, for humidities greater than a threshold Exercises 113 called the saturation humidity, water vapor tends Numerical Problems 113 to condense into liquid faster than it re-evaporates. Understanding & Critical Evaluation 115 This condensation process lowers the humidity to- Web-Enhanced Questions 117 ward the equilibrium (saturation) value. The pro- Synthesis Questions 118 cess is so fast that humidities rarely exceed the equi- librium value. Thus, while air can hold any portion of water vapor, the threshold is rarely exceeded by more than 1% in the real atmosphere. Air that contains this threshold amount of wa- “Meteorology for Scientists and Engineers, 3rd Edi- tion” by Roland Stull is licensed under a Creative ter vapor is saturated. Air that holds less than that Commons Attribution-NonCommercial-ShareAlike amount is unsaturated. The equilibrium (satura- 4.0 International License. To view a copy of the license, visit http://creativecommons.org/licenses/by-nc-sa/4.0/ . This work is tion) value of vapor pressure over a flat surface of available at http://www.eos.ubc.ca/books/Practical_Meteorology/ . 87 88 chapter 4 MoISTURE pure liquid water is given the symbol es. For un- saturated air, e < e . s Air can be slightly supersaturated ( e > es ) when there are no surfaces upon which water vapor can condense (i.e., unusually clean air, with no cloud condensation nuclei, and no liquid or ice particles). 4VQFSTBUVSBUFE L1B Temporary supersaturation might occur when the T F threshold value drops so quickly that condensation E F does not remove water vapor fast enough. However, BU VS photographs of air flow over aircraft wings for both BU 4 subsonic and supersonic flight through humid air 6OTBUVSBUFE indicate that condensation to form cloud droplets occurs almost instantly. m m m Even at saturation, there is a continual exchange 5 $ of water molecules between the liquid water and Figure 4.1 the air. The evaporation rate depends mostly on the Saturation vapor pressure over a flat surface of pure water. A temperature of the liquid water. The condensation blowup of the portion of this curve colder than 0°C is given in rate depends mostly on the humidity in the air. At the next figure. equilibrium these two rates balance. If the liquid water temperature increases, then evaporation will temporarily exceed condensation, FT<XBUFS>mFT<JDF> L1B FT L1B and the number of water molecules in the air will in- crease until a new equilibrium is reached. Thus, the equilibrium (saturation) humidity increases with temperature. The net result is that warmer air can hold more water vapor at equilibrium than cooler air. The Clausius-Clapeyron equation describes the relationship between temperature and satura- tion vapor pressure, which is approximately: m m m m m 5 $ L 1 1 MJRVJE es≈ e o ·exp · − •(4.1a) ℜv TT o JDF where eo = 0.611 kPa and To = 273.15 K are con- m m m m m stant parameters, and ℜ = 461 J·K–1·kg–1 is the gas 5 $ v constant for water vapor. Absolute temperature in Figure 4.2 Kelvin must be used for T in eq. (4.1). Saturation vapor pressure over flat surfaces of pure liquid water Because clouds can consist of liquid droplets and ice, at temperatures below 0°C. The insert shows difference and ice crystals suspended in air, you must consider between saturation vapor pressures over water and ice. saturations with respect to water and ice. Over a flat water surface, use the latent heat of vaporiza- 6 –1 Solved Example tion L = Lv = 2.5x10 J·kg in eq. (4.1), which gives Find the saturation vapor pressure for T = 21°C? Lv/ℜv = 5423 K. Over a flat ice surface, use the 6 –1 latent heat of deposition L = Ld = 2.83x10 J·kg , Solution which gives Ld/ℜv = 6139 K. Given: T = 21°C = 294 K Saturation vapor pressure is listed in Table 4-1 Find: es = ? kPa. Use eq. (4.1) for liquid water. and plotted in Figs. 4.1 and 4.2. These were produced 1 1 by solving eq. (4.1) on a computer spreadsheet. The es = (.0 611kPa)·exp (5423K)· − 273KK303 exponential-like increase of water-vapor holding ca- pacity with temperature has tremendous impact on = (0.611 kPa)·exp(1.419) = 2.525 kPa clouds and storms. Supercooled (unfrozen) liquid water can exist between temperatures of 0 to –40°C, Check: Units OK. Physics OK. Agrees with Table 4-1. hence, the curve for saturation over water includes Discussion: Compared to average sea-level pressure temperatures below 0°C. P = 101.3 kPa, this is roughly 2.5% of the total. R. STULL • METEoRoLogy FoR SCIENTISTS AND ENgINEERS 89 The Clausius-Clapeyron equation also describes the relationship between water-vapor pressure e and BeYOND aLGeBRA • clausius-clapeyron dew-point temperature (T , to be defined later): d Historical Underpinning To build better steam engines during the indus- L 1 1 (4.1b) e= e ·exp · − trial revolution, engineers were working to discover o ℜv TT o d the thermodynamics of water vapor. Steam engi- neers B.-P.-E. Clapeyron in 1834 and R. Clausius in 1879 applied the principles of S. Carnot (early 1800s), to study isothermal compression of pure water vapor Table 4-1. (1) Saturation values of humidity vs. actual in a cylinder. They measured the saturation vapor air temperature (T). -or- pressure at the point where condensation occurred. (2) Actual humidities vs. dew-point temperature (Td). Values are for over a flat surface of liquid water. Variation of Vapor Pressure with Temperature Mixing ratio r and specific humidity q values are for By repeating the experiment for various tempera- sea-level pressure. Vapor-pressure e and absolute hu- tures, they empirically found that: midity ρv values are for any pressure. Subscript s de- notes a saturation value. −1 de L 1 1 s= v − dT T ρv ρ L For P = 101.325 kPa where ρ is the density of water vapor, and ρ is the T es rs qs ρvs v L density of liquid water. But ρ >> ρ (see Appendix or L v B and Table 4-1), thus 1/ ρL << 1/ρv , allowing us to Td e r q ρv neglect the ρL term: 3 (°C) (kPa) (g/kg) (g/kg) (g/m ) de L s≅ v ρ –20 0.127 0.78 0.78 1.09 dT T v –18 0.150 0.92 0.92 1.28 The ideal gas law can be used to relate the wa- –16 0.177 1.09 1.09 1.50 ter vapor pressure to the vapor density: es = ρv·ℜv·T, –14 0.209 1.28 1.28 1.75 where ℜ is the gas constant for water vapor. Solv- –12 0.245 1.51 1.51 2.04 v ing for ρv and plugging the result into the eq. above –10 0.287 1.77 1.76 2.37 gives: –8 0.335 2.07 2.06 2.75 de Lv· e s –6 0.391 2.41 2.40 3.18 s ≅ dT 2 –4 0.455 2.80 2.80 3.67 ℜv ·T –2 0.528 3.26 3.25 4.22 Use the Separation of Variables method: move all 0 0.611 3.77 3.76 4.85 e terms to the left, all T terms to the right: 2 0.706 4.37 4.35 5.57 s 4 0.814 5.04 5.01 6.37 des Lv dT ≅ 2 6 0.937 5.80 5.77 7.28 es ℜv T 8 1.076 6.68 6.63 8.30 Then integrate, taking care with the limits: 10 1.233 7.66 7.60 9.45 12 1.410 8.78 8.70 10.73 es T des Lv dT 14 1.610 10.05 9.95 12.17 ≅ 2 ∫es ℜv ∫ T 16 1.835 11.48 11.35 13.77 eo To 18 2.088 13.09 12.92 15.56 where eo is a known vapor pressure at reference tem- 20 2.371 14.91 14.69 17.55 perature T .
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