1Cloud Condensation NucleiActivationActivationCloud ProcessingEvaporationEvaporation0.110Diameter (µm)0.110Diameter (µm)dNdlogDDiameter (µm)0.110Hoppel Minimum• Particle evolution inremote marineconditions• cloud processing –growth of particlesdue to coalescenceand solutecondensation incloud7.1Seinfeld and Pandis, Fig. 15.23 (Hoppel et al., 1990)Number Distributions vs.Population Distributions1.2Age0 60PopulationDp (µm)0.01 10dNdlogDpAerosol ParticleSize Distribution(Manhattan)Human PopulationAge Distribution(Manhattan)20 400.1 1Particle Size Distributions1.2• Number concentration– Total number N– Differential number n• Mean size– Geometric– Arithmetic– Number-based– Mass-based• Size variability– Standard deviation σ– Geometric standarddeviation σgDpg0.01 10Dp (µm)dNdlogDp∼ σgn(Dp)0.1 1Particle Characteristics• Concentration and size• Chemical composition• Light scatteringDp (µm)0.01 0.10dndlogDpDpg∼ σgn(Dp)Size Characterization of Particles• clusters of molecules• starting at 100molecules/cluster• growth bycondensation ofmolecules is nearlycontinuous• multiple ways to graphsame distribution2468103246810424681052Concentration (cm-3)2 3 4 5 612 3 4 5 6102Diameter (µm) ni Ni Lognormal ni dN/dlogDp dN/dDp ScaledNi2Particle SizesDp (µm)0.01 0.10fine modecoarse modeAitken nuclei1.0 10.0 100.0• range of particle sizes isapproximately from 1nm to 1 mm in diameter• range of approximately6 orders of magnitude• concentrations at each ofthese sizes also varySize Distribution ModesDp (µm)0.010.10dndlogDpfine modeaccumulation modenucleation modecoarse modeAitken nucleiprecipitation-sized dropletscloud droplets1.0 10.0 100.0recently nucleated particles• modes of aerosol aredistinguished by– size– sources– behaviorLog-Normal NumberDistributions• DifferentialDp (µm)0.010.10Cumulative Distribution (% less than Dp)DpgσgCumulative Number10005084.1%15.9%}Cumulative SurfaceCumulative VolumeDp (µm)0.010.10dndlogDpDpg∼ σgn(Dp)• CumulativeMicrophysics• Aerosol includes both particles and vapor• Number, area, volume, mass vary nonlinearly• Deposition velocity depends on size (nano, micro, milli)• Scavenging, coalescence, activation and condensation change the sizedistributionGlobal Aerosol DistributionCapaldo et al., Nature, 1999• Regionalvariations inaerosolmass andcomposition[NARSTO,2002]Toronto (1997-99)Egbert (1994-99)Abbotsford (1994-95)Quaker City OH (1999)Arendstville PA (1999)Atlanta (1999)Yorkville (1999)Mexico City - Pedregal (1997)Los Angeles (1995-96)Fresno (1988-89)Kern Wildlife Refuge (1988-89)SulfateNitrateAmmoniumBlack carbonOrganic carbonSoilOther12.3 ug m-38.9 ug m-37.8 ug m-312.4 ug m-310.4 ug m-319.2 ug m-314.7 ug m-355.4 ug m-330.3 ug m-323.3 ug m-339.2 ug m-3Washington DC (1996-99)14.5 ug m-3Colorado Plateau (1996-99)3.0 ug m-3 Mexico City - Netzahualcoyotl (1997)24.6 ug m-3Esther (1995-99)St. Andrews (1994-97)5.3 ug m-34.6 ug m-33ROAST Reviews• Review comments due to editor: Nov. 6– Reviews are completed by individuals not groups.– Reviewer’s name should not appear in review.– Reviewer’s name should be in filename.• "GroupAp1v1reviewLASTNAME.pdf"• http://www.elsevier.com/wps/find/editorsinfo.editors/ebu/issue5aHomework Ch. 5 Prob. 3€ r* =3baS* =1+4a327brho_l 1000 kg/m3sigma_lv 0.075015 N/m Eqn. 5.7Rv 461.478686 J/K/KGT 280 KI 2M_v 18.016 g/molM_sol 132.1 g/molm_sol 1.00E-18 KGa 1.1611E-09 m a=2sigma/(rho*Rv*T)b 6.5117E-23 m3 b=3i*Mv*msol/(4pi*Msol*rho)r* 4.1018E-07 m r*=sqrt(3b/a)S* 1.00188713 S*=1+sqrt(4a^3/27b)Example Ch. 5 Prob. 7Curry and Webster, p. 158, Problem 7a-c€ xmeaxdx =∫xmeaxa−maxm −1eaxdx∫= eax−1( )rm!xm − rm − r( )!ar +1r= 0m∑Integral Tables (521), CRC 1986 p. 330Example Ch. 5 Prob. 7€ xmeaxdx =∫eax−1( )rm!xm− rm − r( )!ar +1r= 0m∑x2eaxdx =∫eaxax2−2a2ax −1( ) Integral Tables (521)CRC 1986 p. 330€ N = Ar2e−Brdr0∞∫= Ae−Br−Br2−2B2−Br −1( ) 0∞= 0 − −AB2B2 =2AB3Example Ch. 5 Prob. 7€ xmeaxdx =∫eax−1( )rm!xm− rm − r( )!ar +1r= 0m∑x3eaxdx =∫eaxax3−3x2a+6a3ax −1( ) Integral Tables (521)CRC 1986 p. 330€ r =1NAr3e−Brdr0∞∫=ANe−Br−B r3+3r2B−6B3−Br −1( ) 0∞= 0 − −ABN6B3 =3BExample Ch. 5 Prob. 7€ r =3B= 10 ×10−6B = 3 ×105N =2AB3= 200 ×106A = 2.7 ×10244Example Ch. 5 Prob. 7€ xmeaxdx =∫eax−1( )rm!xm− rm − r( )!ar +1r= 0m∑= 120eaxx5120a−x424a2+x36a3−x22a4+xa5−1a6 Integral Tables (521)CRC 1986 p. 330€ wl=4πAρl3ρa r5e−Brdr∫=4πAρl3ρa 120e−Brr5−120B−r424B2+r3−6B3−r22 B4+r−B5−1B6 0∞=4πAρl3ρa 0 −−120B6 =160πAρlB6ρa Cloud in a Jar Demonstrationhttp://groups.physics.umn.edu/demo/old_page/demo_gifs/4B70_20.GIFLecture Ch. 6a• Saturation of moist air• Relationship between humidity and dewpoint– Clausius-Clapeyron equation• Dewpoint– Temperature– Depression• Isobaric coolingCurry and Webster, Ch. 6For Tuesday: Read Ch. 7How does saturation occur?• By increasing water vapor– Evaporation of water at surface– Evaporation of falling rain• By cooling– Isobaric– Radiative cooling of rising air• By mixing of two unsaturated air parcelsCurry and Webster, Ch. 6Saturation of Moist Air• Dew point temperatureCurry and Webster, Ch.
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