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MSE3 Ch07 Precipitation

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MSE3 Ch07 Precipitation Chapter 7 Copyright © 2011, 2015 by Roland Stull. Meteorology for Scientists and Engineers, 3rd Ed. preCipitation Contents Hydrometeors are liquid and ice parti- cles that form in the atmosphere. Hydrom- Supersaturation and Water Availability 186 eter sizes range from small cloud droplets Supersaturation 186 7 and ice crystals to large hailstones. Precipitation Water Availability 186 occurs when hydrometeors are large and heavy Number and Size of Hydrometeors 187 enough to fall to the Earth’s surface. Virga occurs Nucleation of Liquid Droplets 188 when hydrometers fall from a cloud, but evaporate Cloud Condensation Nuclei (CCN) 188 before reaching the Earth’s surface. Curvature and Solute Effects 189 Precipitation particles are much larger than cloud Critical Radius 192 particles, as illustrated in Fig. 7.1. One “typical” Haze 192 Activated Nuclei 193 raindrop holds as much water as a million “typical” cloud droplets. How do such large precipitation Nucleation of Ice Crystals 194 Processes 194 particles form? Ice Nuclei 195 The microphysics of cloud- and precipitation- particle formation is affected by super-saturation, Liquid Droplet Growth by Diffusion 196 nucleation, diffusion, and collision. Ice Growth by Diffusion 198 Supersaturation indicates the amount of excess Ice Crystal Habits 198 water vapor available to form rain and snow. Growth Rates 200 The Wegener-Bergeron-Findeisen (WBF) Nucleation is the formation of new liquid or Process 201 solid hydrometeors as water vapor attaches to tiny dust particles carried in the air. These particles are Collision and Collection 202 Terminal Velocity of Hydrometeors 202 called cloud condensation nuclei or ice nuclei. Cloud Droplets 202 Diffusion is the somewhat random migration Rain Drops 203 of water-vapor molecules through the air toward ex- Hailstones 204 isting hydrometeors. Conduction of heat away from Collection & Aggregation Processes 204 the droplet is also necessary to compensate the la- Warm-cloud Process 204 tent heating of condensation or deposition. Cold-cloud Processes 205 Collision between two hydrometeors allows Precipitation Formation 205 them to combine into larger particles. These pro- Warm Clouds 205 cesses affect liquid water and ice differently. Cold Clouds 206 Precipitation Characteristics 207 Rain-Drop Size Distribution 207 Rainfall Rates 208 Snowfall Rates & Snow Accumulation 208 7PMVNF NN Precipitation Distribution 209 m m m m Precipitation Measurement 210 Summary 211 TNBMMDMPVE 3N Threads 211 UZQJDBMDMPVE 3N Exercises 212 MBSHFDMPVE 3N Numerical Problems 212 Understanding & Critical Evaluation 214 ESJ[[MF 3N Web-Enhanced Questions 217 TNBMMSBJO 3N Synthesis Questions 217 )ZESPNFUFPS UZQJDBMSBJO 3N “Meteorology for Scientists and Engineers, 3rd Edi- MBSHFSBJO tion” by Roland Stull is licensed under a Creative 3N Commons Attribution-NonCommercial-ShareAlike 4.0 International License. To view a copy of the license, visit http://creativecommons.org/licenses/by-nc-sa/4.0/ . This work is Figure 7.1 available at http://www.eos.ubc.ca/books/Practical_Meteorology/ . Drop volumes and radii, R. 185 186 chapter 7 PRECIPITATIoN S HLH 1IZTJDBM S HLH 4JUVBUJPO supersaturation and Water BEJBCBUNPJTU availability 1 S& S L1B ST 9 5 supersaturation When there is more water vapor in the air than it can hold at equilibrium, the air is said to be super- saturated. Supersaturated air has relative humid- -$- JTPIVNF ity RH greater than 100%. Define this excess relative ESZBEJBCBU BJS QBSDFM humidity as a supersaturation fraction S: SJTF S = RH – 1 •(7.1) 9 m m or as a supersaturation percentage S%: 5 $ S% = 100% · S = RH% – 100% •(7.2) Figure 7.2 Thermo diagram (emagram) showing how excess-water mixing Using the definition of relative humidity from ratio rE (thick line with arrows at each end) increases with in- the Moisture chapter, rewrite supersaturation in creasing height above cloud base (LCL) for an air parcel rising terms of total vapor pressure e and saturation vapor adiabatically. r = total water mixing ratio; r = saturation T s pressure e : mixing ratio for water vapor. r is the amount of water that can s e E S 1 form hydrometeors. = − •(7.3) es for e > es. It can also be approximated using mixing ratios: r S ≈ − 1 •(7.4) r Solved Example s An air parcel starts at 100 kPa with temperature for r > rs. 20°C and mixing ratio 4 g/kg. Use a thermo diagram to find the adiabatic value of excess-water mixing ratio when the parcel reaches a pressure of 60 kPa, and find Water availability the max available supersaturation. Suppose that an initially-unsaturated air parcel has total water mixing ratio of rT. If the air parcel Solution rises adiabatically (i.e., no mixing with the environ- Given: Initially r = rT = 4 g/kg, T = 20°C, P = 100 kPa ment, and no precipitation in or out), then it cools Finally: P = 60 kPa. and its saturation mixing ratio decreases. You can Find: rE = ? g/kg, SA = ? (dimensionless) use a thermo diagram (from the Stability chapter) to find these changing mixing ratios, as sketched here Use a full size thermo diagram from the Stability chap- in Fig. 7.2. ter. Starting with the initial conditions, and following Above the parcel’s lifting condensation level the air parcel up dry-adiabatically from the surface, we (LCL) the saturation mixing ratio r is less than r . find the LCL is atP = 75 kPa and T = –3°C, which marks s T This supersaturated air has an excess water mix- cloud-base. Above the LCL, follow the moist adiabat, ing ratio rE of: which gives T ≈ –15°C at rs = 2 g/kg at P = 60 kPa. As- suming no entrainment and mixing, rT = 4 g/kg. rE = rT – rs •(7.5) Use eq. (7.5): rE = (4 – 2) g/kg = 2 g/kg. Then use eq. (7.7): SA = rE/rs = 1 = 100%. The excess water is available to condense onto liquid drops and to deposit onto ice crystals: Check: Units OK. Physics OK. Agrees with Figs. 7.2 & 7.3) rL + ri = rE (7.6) Discussion: A supersaturation of 100% corresponds to a relative humidity of RH = 200% from eq. (7.2). Thus, where rL is liquid-water mixing ratio, and ri is ice there is twice as much water in the air as can be held at mixing ratio. At temperatures T in the range –40 < T equilibrium. This excess water vapor drives conden- < 0°C, supercooled liquid water and ice crystals can sation. For real clouds, we expect r in the range of 0.5 E co-exist in the air. to 1.5 g/kg, based on the curves in Fig. 7.3. R. STULL • METEoRoLoGy FoR SCIENTISTS AND ENGINEERS 187 Available supersaturation fraction is defined 1IZTJDBM as: 4JUVBUJPO rE rT S = = − 1 & A r r (7.7) s s 1 As hydrometeors grow and remove water vapor, r L1B 5ZQJDBMS becomes less than rT, and S becomes less than SA. & Droplets and ice crystals stop growing by condensa- tion and deposition when they have consumed all "EJBCBUJDS the available supersaturation (i.e., when r → rs and S → 0). BJS The adiabatic estimate of r increases with QBSDFM E SJTF height above the LCL, as sketched in Figs. 7.2 and 7.3. However, in many real clouds, diabatic (i.e., non-adiabatic) processes such as entrainment and mixing of clear drier air into the top and sides of the cloud cause the total-water mixing ratio to not be S& HLH conserved with height. As a result, typical amounts of excess-water mixing ratio are less than adiabatic Figure 7.3 (dashed lines in Fig. 7.3). Example of the increase of excess water mixing ratio rE vs. Beware that the adiabatic rE curve varies from height above cloud base for the scenario in the previous figure. situation to situation, depending on the thermody- The adiabatic value (solid line) represents the max likely value in namic initial conditions of the air parcel. So Fig. 7.3 non-precipitating clouds, while the dashed curves show ranges is just one example (based on the Fig. 7.2 scenario). of actual max values. P is pressure. number and size of hydrometeors Suppose we partition the available excess wa- ter equally between all hydrometeors (for example, for all liquid water droplets). In this way, we can Solved Example estimate the average radius R for each droplet due Within a cloud, suppose air density is 1 kg/m3 and only to condensation (i.e., before collisions between the excess water mixing ratio is 4 g/kg. Find the final droplets allow some to merge and grow into larger drop radius for hydrometeor counts of (a) 108/m3, and drops): (b) 109/m3 . 1/ 3 3 ρ r R = ··air E •(7.8) Solution 3 4π ρwater n Given: rE = 0.004 kgwater/kgair, ρair = 1 kg/m n = (a) 108/m3, and (b) 109/m3 . Find: R = ? µm where excess-water mixing ratio rE is in kgwater/ 3 Assume: ρwater = 1000 kg/m , as listed in Appendix A. kgair, ρ is density, and n is the number density of hydrometeors (the count of hydrometeors per cu- Use eq. (7.8). Part (a),: bic meter of air). Typical values are R = 2 to 50 µm, which is small compared to the 1000 µm separation 3 1/ 3 3·( 1kgair/m )·(.0 004kgwater/kg air ) between droplets, and is too small to be precipita- R = 3 38 -3 4π·( 10 kkg /m )·()10 m tion. water This is an important consideration. Namely, even if we ignore the slowness of the diffusion process = 2.12x10–5 m = 21.2 µm. (described later), the hydrometeors stop growing by condensation or deposition before they become (b) Similarly, R = 9.8 µm. precipitation. The reason is that there are too many hydrometeors, all competing for water molecules, Check: Units OK.
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