Løsningsforslag: oppgavesett kap. 6 (2 av..) GEF2200

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1 Løsningsforslag: oppgavesett kap. 6 (2 av..) GEF2200 s.m.blichner@geo.uio.no Oppgave 1 Vi har to like store dråper; en som bare består av rent vann og en som består av en løsning. Hvor vil metningstrykket for vanndamp være størst, og hva innebærer dette? Metningstrykket til vanndamp over en løsning er lavere enn metningstrykket til vanndamp over rent vann. Følgelig vil man lettere oppnå overmetning over løsningen, og dermed lettere kunne danne dråper. Reduksjonen i vanndampstrykket er gitt ved Raoults lov, som sier at e = f e, hvor e er mettet vanndampstrykket over løsningen, men e er vanndampstrykket over det rene vannet. f angir hvor stor andel vannet utgjør av det totale antallet molekyler løsningen (antall mol vannmolekyler delt på antall mol totalt i løsningen). Oppgave 2 (a) Hva kjennetegner en varm sky, og hvilke parametre er vi interesserte i når vi skal beskrive mikrostrukturen til disse skyene? En varm sky har temperatur over null grader i hele skyen. For å beskrive mikrostrukturen til en varm sky, vil vi blant annet være interessert i liquid water content (LWC), altså hvor mye vann i væskefase vi har per volum (ofte målt i g/m 3, antallet dråper (cloud droplet concentration) og størrelsesfordelingen til disse (droplet size spectrum). Disse egenskapene vil bl.a avgjøre skyens radiative egenskaper og potensialet for regn. 1

2 (b) Hvilken sammenheng har vi mellom skyers optiske tykkelse, LWP (liquid water path) og dråperadius i en varm sky? Vi har at τ c LW P r c hvor LW P (Liquid Water Path) er massen til det ytende vannet i en vertikal søyle mellom to høydenivåer i atmosfæren, med en grunnate på 1 m 2, mens τ c er skyens optiske tykkelse og r e er den eektive radiusen til skydråpene (som er nesten det samme som den gjennomsnittlige radiusen. Mer om dette i senere kurs). Jo mindre dråper, jo høyre optisk tykkelse. Jo mer vann, jo høyere optisk tykkelse. (c) Hva illustrerer Figurene 6.6 a) og b)? Figur 6.6 a) viser de vertikale hastighetene i en sky. Positive verdier betyr at vi har oppstigning, negative nedsynkning. Oppstigning kreves for å danne en sky (lufta stiger, utvides, temperaturen synker og overmetningen stiger). Vi ser at det er en del ugjevnheter, men at vi i hovedsak har oppgående bevegelser i midten av skyen, og nedgående bevegelser ved sidene. Figur 6.6 b) viser det ytende vanninnholdet i skyen. Vi ser at vanninnholdet et størst der oppstigningen er størt, og motsatt. Dette er naturlig fordi vi i disse områdene har fuktig luft fra bakken som løftes og dermed kondenserer vanndampen og danner dråper. Dermed stiger også LWC her. I områder med nedsynkning vil det motsatte skje (evt dråper vil evaporere med stigende temperaturer). (c) Hva illusterer Figur 6.7? Hvorfor får vi disse forskjellene mellom marine og kontinentale skyer? Figurene 6.7 a) og c) illustrerer hvor stor andel av hhv de marine og de kontinentale skyene som har de ulike dråpekonsentrasjonene. Vi ser at alle de marine skyene har dråpekonsentrasjoner under 200 cm 3, og at de este ligger under 100 cm 3, mens for de kontinentale skyene er resultatet mer spredt (helt opp til 900 cm 3 ), men at de este har dråpekonsentrasjoner under 400 cm 3. Figurene 6.7 b) og d) illustrerer størrelsesfordelingen til skydråpene i hhv en marin og en kontinental sky. Vi ser at skydråpene i den marine skyen er større enn i 2

3 den kontinentale, hvilket skyldes at man har ere CCN over land, og dermed større konkurranse om vanndampen. (d) What does gure 6.11 show? This showes the LWC measured at dierent levels in 802 cumulus clouds and the theoretical LWC (red) LWC based on adiabatic lifting of an air parcel (adiabatic LWC). Squares show the largest neasyred LWC. This shows us that the theoretically computed LWC is rarely correct, but more so at cloud base. As we move upward, entrainment of dry air from the surroundings becomes more and more important, thus lowering the LWC. (f) Figure 6.16 in Wallace and Hobbs shows theoretical computations of the growth of dierent droplets with time. Why doesn't the lower two droplets grow? The plot shows the developement in radius of dierent haze droplet (unactivated droplets) with dierent solutions. It also shows the developement of supersaturation with time. We can observe that while supersaturation increases, all the hazedroplets grow (though some more than others). Some of these are (all but the lowest two) are activated and grow spontaniously, while others (the bottom two) are never activated. Thus, when the supersaturation decreases again (due to water condensating on the droplets), these evaporate again. Oppgave * (a) Hva betyr adiabatisk vanninnhold (adiabatic LWC)? This is the theoretical LWC assuming adiabatic conditions (b) Fortell hva Figurene 6.10 og 6.11 illustrerer, og forklar hvorfor det målte vanninnholdet i en sky ikke når opp til det beregnede, adiabatiske vanninnholdet. See 2d. 3

4 (c) Kan du, med bakgrunn i det du skrev som svar på oppgave b), forklare hvorfor Figur 6.6 a) og 6.6 b) er såpass hakkete? Entrainment of dry air from the surroundings will be responsible for this. This air will mix with the air in the cloud. This can lead to evaporation of droplets which will lead to evaporative cooling of the air. It can then sink and drop for several kilometers. Oppgave 3 (a) What types of growth do we have for warm clouds? ˆ Growth mechanisms for warm clouds: Condensation dm = 4πrD [ρ v ( ) ρ v (r)] (1) dt For a spherical droplet, this equation may be written as growth by radius where r dr dt = G ls (2) G l = Dρ v( ) ρ l (3) Since the rate of growth by condensation is inversely proportional to the droplets radius, smaller droplets grow faster than larger droplets. Growth by condensation alone in warm clouds are too slow to produce raindrops with radii of several millimeters. See gure Collision and coalescence Two droplets collide and form a larger drop, need droplets more than about 20 micrometers in radius. Can produce droplets sizes of raindrops. (b) Dene collision eciency. larger than other droplets? Why is it low for a collector drop being much 4

5 ˆ Collision eciency is dened as the ratio between the collision cross section (πy 2 ) and the geometrical cross section π(r 1 + r 2 ) 2 : E = y 2 (r 1 + r 2 ) 2 (4) y is the largest distance the center of a collection droplet can have to the center of the collector droplet and still collide. Small droplets have low inertia (because they have how mass) and will thus follow the streamlines of the air, easily escaping collision. Thus y (and E) is small for small collision droplets. As the collected droplets increase in size, E increases. When r 2 /r 1 increases to about 0.6 to 0.9, E decreases again (particularly for small collector droplets) because the droplets terminal fall speed become similar. Finally, when r 2 /r 2 approaches unity, E increases again because the two droplets interact and wake eects can increase the the collision eciency. (c) Dene coalescence eciency. Explain how this eciency changes when the collector drop and droplets approach the same size. ˆ Coalescence eciency is dened as the fraction of collisions which result in a coalescence (E ). E = #successfullcollisions #collisionsintotal For a particular collector droplet, the coalescence eciency rst decreases with the size of the collected droplet, and then increases again when they approach the same size. When r 1 and r 2 approach the same size, the coalescence eciency will increase sharply,see gure The impact energy is small and less able to prevent contact for intermediate values of the size radio of the droplet to the drop. The impact energy is a measure of the deformation of the collector drop due to the impact. (5) (d) Dene the collection eciency. ˆ Collection eciency is dened as E c = EE. 5

6 Oppgave 4 Midterm About droplet growth in warm clouds. (a) For heterogeneous nucleation, which physical processes are described by the equations r dr dt = G ls (6) and dr 1 dt = v w l E 1 (7) Explain the eciencies of these two processes as a function of droplet size, where v 1 = kr 1, k = const. ˆ For heterogeneous nucleation, droplets grow by two processes, condensation from vapor phase and collision with other droplets. Condensation is described by Equation (6), while collision is described by Equation (7). Growth by condensation is most eective for smaller droplets, with decreasing eectivity as the droplet radius increase. For radii larger than 10µm, this process is almost insignicant. This can be seen from Equation (6), which is inversely proportional to the radius. Growth by collision increase with droplet radius, as can be seen from Equation (7), where the growth is proportional with v 1 = kr 1. (b) Given equation (7), let w l = kgm 3, v 1 = s 1 r 1, E = 0.8 and ϱ l = kgm 3. Starting with a large cloud droplet of radius 100µm, how long will it take to form a typical rain drop (radius 1mm) by this process? ˆ Integrate Equation (7) from t 0 = 0, r(t 0 ) = r 0 til t,r 2, after inserting v 1 = 6

7 kr 1 : r2 dr 1 = r 0 r 1 ( ) r2 ln r 0 t 0 = k w le t k w le dt t = kw l E ln Inserting the values given, we get t = s. ( r2 r 0 ) (8) Oppgave 5 Eksamen To prosesser for dråpevekst i varme skyer er beskrevet ved og r 1 dr 1 dt = G ls (9) dr 1 dt = (v 1 v 2 )w l E c (10) (a) Hvilke to prosesser beskriver disse ligningene? Angi parametrene i ligningene. ˆ Prosessen i ligning (9) beskriver dråpevekst ved kondensasjon/avsetning. Ligning 10 beskriver dråpevekst ved kollisjon. r 1 : Droplet radius G l : Parameter discribing the diusion of water vapor S: Supersaturation. v 1 : The fallspeed of the collection droplet. v 2 : The fallspeed of the collected droplets. w l : LWC. E c : Collection eciency. ϱ l : Density of liquid water (the droplet). (b) Hvilken av disse prosessene er viktigst for henholdsvis små og store skydråper? Forklar kort hvorfor. 7

8 ˆ Consider the equations: for growth by condensation dr 1 decreases as r dt 1 grows, but the dierence in fallspeed (v 1 v 2 ) increases as the collector droplet grows. (c) En skydråpe antas på et tidspunkt å vokse like raskt ved de to prosessene (gitt ved likning 9 og 10). Sett opp et uttrykk for dråperadien i dette tilfellet. Forklar hvordan dråperadien i dette tilfellet påvirkes av overmetningen og vanninnholdet. ˆ Seeing as the droplets grow equally fast by both processes, we can set the growth rate in eq. 9 equal to the growth rate of eq. 10. Solving for r eq we get the following expression: G l S r eq = (v 1 v 2 )w l E c (11) If the supersaturation S increases, r eq also increases. This is due to the fact that supersaturation increases the growth rate by condensation, but increasing droplet size increases the relative eciency of collection. On the other hand, if liquid water content increases this will make the growth rate by collection greater, and thus r eq must decrease to keep the equality of the two processes. Oppgaver fra 6.8 i boka: (k) Because we usually have entrainment of dry air from the surroundings. This dry air then evaporates cloud droplets and makes holes in the cloud, as shown in gure 6.6. Conditions producing LWC greater than adiabatic values could be expected e.g. if we have precipitation introducing liquid water formed above the sample volume. (o) In marine clouds there are fewer CCN than in continental airmasses. Thus the vapor condenses onto fewer CCN and these grow to sizes greater than in continental airmasses (where the liquid is spread out between many CCN). Since collision coalescence is more ecient for larger droplets, this will enhance the growth of droplets to precipitation relevant sizes. (p) This has to do with the denitions of either: collision eciency is dened as the ratio of the crossection (πy 2 ) within a smaller droplet center can be and still collide with the collector to π(r 1 + r 2 ) 2. The crossection πy 2 may be greater than π(r 1 + r 2 ) 2 due to wake eects (wake turbulence). The coalescence eciency on the other hand hand is dened as the likelyhood of a collision ending in coalescence, and this can be no greater than unity. 8