# Solutions to the questions from chapter 1 and 2 in GEF Cloud Physics

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1 Solutions to the questions from chapter 1 and 2 in GEF Cloud Physics Problem 1 (related to figure 1.10) What is the typical size and concentration of a... a) CCN particle? Answer: 0.1 µm and 1000 cm 3 b) Haze drop? Answer: 1 µm and 1000 cm 3 c) Cloud drop? Answer: 10 µm and 1000 cm 3 d) Drizzle drop? Answer: 100 µm and 1 cm 3 e) Raindrop? Answer: 1 mm and 1 L 1 (same as 1000 µm and 0,001 cm 3 ) Problem 2 (related to figure 1.12) a) Why do you think the concentration of cloud droplets in a typical cumulus cloud decrease with height? Answer: Some of the cloud droplets might collide and merge together (coalescence) while being lifted up. We also have entrainment of dry air from the area outside the cloud that evaporates some of the droplets completely the concentration decreases. b) Why do you think the cloud droplets near the top of the cloud are larger than the ones near the cloud base? Answer: Because they have grown while being lifted up. They have grown mainly by condensation (due to higher supersaturation at higher levels because of lower temperatures) but also by some collision and coalescence (since we can see that the number concentration are lower) c) Why do you think we see a broadening of the droplet spectra with height? This is a question many cloud physicists have been trying to answer for a long period of time. It can be that som droplets grows on the expence of other because some droplets evaporates giving rise to more available water vaper for other droplet. It can also be so because some droplets may have collided, even though this mechanism isn t very important before the droplets are bigger. One other reason for this broading can be that we have som giant super-ccns. 1

2 Problem 3 (related to figure 1.13) a) What do we mean by adiabatic LWC? Answer: Adiabatic LWC is the mass of liquid water per unit volume of air in the cloud we would have if air parcels were lifted adiabatically (q=0). This means that there are no interactions with the surroundings other than the work done by the surroundings on the air parcel (or in the other direction) because of expansion/compression b) Why is the LWC in most clouds smaller than the adiabatic value (and never larger)? Answer: The actual liquid water is lower than the adiabatic because of evaporation due to entrainment of dry air. Problem 4 (related to figure 2.2 and 2.27) a) Why do we want to represent the distribution of aerosol particles in a logarithmic form? Answer: to make the wiggles more pronounced. (In other words: so it s easier for us to see how the concentration varies with the sizes of the aerosol particles.). b) Why does the ordinate (the function n(log(d p )))) have other units than the original size distribution, n(d p )? Answer: From the calculations below, we can see that the ordinate function is just the product of the size distribution function, ln10 and the diameter. Since diameter have the units of µm, the units of the ordinate function won t be cm 3 µm 1, but just cm 3 n(log D p ) = dn d(log D p ) dn = d log D p n(log D p ) ( ) ln Dp dn = d n(log D p ) ln 10 dn = 1 ln 10 d(ln D p) n(log D p ) dn = 1 1 dd p n(log D p ) ln 10 D p n(log D p ) = dn ln 10 D p dd p n(log D p ) = n(d p ) ln 10 D p c) What do these two different distributions show, and which units do they have? ds d log D p = n S (log D p ) = A p n(log D p ) = πd 2 p n(log D p ) dv d log D p = n V (log D p ) = V p n(log D p ) = π D3 p n(log D p ) 2

3 Answer: The first one, n S (log D p ), is the surface area distribution, and the second one, n V (log D p ), is the volume distribution. n S (log D p ) has the unis µm 2 cm 3 (The units in Figure 2.27 in my book are wrong), and shows the total surface area of the areosols with diameter between log D p and log D p + d(log D p ). n V (log D p ) has the unis µm 3 cm 3 and shows the total volume of aerosol particles in the same range. d) What do we mean when we say that the aerosol distribution are multi-modal, how many modes do the different distributions usually have, what do we call them and in which size range do we find them? The multi-modal nature means that we have more than one maxima in the distributions. This means that it isn t just one size range that contributes the most to both the total number, total surface and total volume. This shows that many different sizes of particles can play an inportant role in different physical processes. The number distribution usually have one mode, the Aitken mode between 0.01 and 0.1 µm. The surface and volume distributions usually have two or more modes. The maximum in the Aitken range (remember that the Aitken range was small aerosols up to 0.1 µm) is called accumulation mode and the maximum above 2 µm is called coarse mode. e) What does the equation below represent? Why do we choose this equation to represent this? 3 3 [ ] N i n(log D p ) = n i (log D p ) = exp log2 (D p /D pg,i ) log σ g,i 2π 2 log 2 σ g,i i=1 i=1 Answer: This is the three mode size distribution. It is log-normal (since we use the logarithm of D p instead of D p ) to easier see how the concentration varies with the sizes of the particles. It also uses the geometric mean diameter instead of the arithmetic to avoid negative values of D p. It uses three different log-normal distributions, one for each mode. f) What do the equations below represent? M 2.5 = πρ p T SP = πρ p D 3 p n(d p )dd p D 3 p n(d p )dd p. Answer: The first one shows how to calculate the mass concentration of particles in the PM 2.5 cathegory, which means that their diametres are below 2.5 µm. The units becomes µgcm 3. The second one shows the same, but for all aerosols. We call the total mass concentration of all particles for total suspended particulate matter (TSP). g) What does it means that the equations above are based on the third moment of the aerosol distribution? (And what is the zeroth, the first and the second one?) It means that we have multiplied n(d p ) by D 3 p. If we muliply by D 2 p, we call it the second moment, and so on. 3

4 Problem 5 - About aerosols and Figure 2.28 a) What do we call hygroscopic aerosols that absorb water from the atmosphere and help initiate cloud formation? CCN: cloud condensation nuclei b) What is the difference between primary and secondary sources of aerosols? Answer: The primary sources (like mechanical lifting of particles from the groud or the ocean) moves the particles directly into the air. A secondary source means that the aerosols comes from chemical reactions of gases (like condensation of volatile orcanic compounds or oxidation of DMS or SO 2 to sulfate). c) Describe what you see in Figure Why do we see this difference? We can see that the aerosol concentrations over land are higher than over ocean, mainly becuase the sources are stronger here. We can also see that the aerosols are larger over ocean than over land. This is because the main source over ocean are sea spray of sea salt, which results in large aerosols. d) Explain briefly the vertical distribution of large particles in Figure The concentration have a maximum next to the groud because this is where we find most of the sources. We also see a maximum in the stratosphere. (but note the scale on the x-axis. These concentrations are much lower than the ones next to the ground). This maximum is called The Junge layer and concists of enhanched concentration of sulfuric-acid particles. These comes from oxidation of SO 2 from volcanic emissions. Problem 1 from the midterm exam in 2015 a) Answer: Total cloud droplet number concentration can be determined by this formula: n tot = N bins j=1 n j = = 55 (The book says 54, but it s not that easy to see how many droplets there are in each bin...) 4

5 b) Answer: the average cloud droplet size can be determined by this formula: D = 1 N bins D j n j n tot j=1 = 1 ( (2.5 9) + (7.5 2) + (12.5 3) + (17.5 7) + ( ) 55 + (24 14) + (2.5 1) + (30 1) ) µm = µm = 17.7 µm c) Answer: the liquid water content (LWC) can be determined by the formula below. LWC is usually given with the units gm 3 This means that we want the unis of the calculations after the summation to be unitless. This means that we need to do these convertations: (µm 3 = m 3 ) and (cm 3 = 10 m 3 ), so we need to multiply by ω L = πρ L N bins j=1 D 3 j n j = π 10 gm 3 (( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + (30 3 1) ) = gm 3 Problem in the book a) Answer: We can find the approximate distance between the centers of the droplet by first calculating how many droplets we on average have per cm, and then take the inverse of this: cm 3 = 4.4 cm 1 1 = cm = 2.15 mm 4.4 cm 1 We find out how many droplet radii this spacing represent: 2.15 mm 15 µm = = 143 b) Answer: To find out how large fraction of the air is occupied by liquid water, we 5

6 need to calculate the volume of one droplet and multiply by the concentration V d = 4 3 πr3 r = 4 3 π ( m ) 3 = m 3 = cm 3 We have 100 droplets in 1 cm 3. The total volume of these droplets is V liquid = cm 3 V liquid = cm 3 This mean that if we have 1 cm 3 of air, cm 3 of this is liquid. This is the same as %. (So the dominant phase of air is gas, even if we are in the midle of a cloud). c) Answer: The liquid water concentration (LWC), ω L is ω L = 0 = 4 3 πr3 ρ L n d 4 3 πr3 ρ L n(r)dr = cm 3 10 gm cm 3 = 1.41 gm 3

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