Exam 2: Cloud Physics April 16, 2008 Physical Meteorology Questions 1-10 are worth 5 points each. Questions are worth 10 points each.

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1 Exam : Cloud Physics April, 8 Physical Meteorology 344 Name Questions - are worth 5 points each. Questions -5 are worth points each.. Rank the concentrations of the following from lowest () to highest (3): cloud condensation nuclei (3) cloud droplets () raindrops (). Match the following particles with their typical size cloud condensation nuclei cloud droplets raindrops μm. μm μm 3. Why do ice crystals grow at the expense of supercooled water droplets? The saturation vapor pressure over liquid is higher than the saturation vapor pressure over ice. Therefore, the environment will be more supersaturated with respect to ice than with respect to liquid, and the ice crystals will grow more quickly. At some point, the ice crystals may bring S (with respect to ice) down to, in which case S l (with respect to liquid) is less than, causing the liquid drops to evaporate. 4. Match the following particles with their most likely means of growth 5 μm cloud drop 5 μm ice crystal μm dendritic ice crystals in ice cloud 5 μm ice crystal in mixed-phase cloud μm cloud drop accretion aggregation depositional growth condensational growth collision-coalescence 5. Fill in the blanks: Not all clouds with temperatures below the freezing point of water contain ice, because of the scarcity of ice nuclei in the atmosphere, and the temperature at which they nucleate ice. As the cloud temperature decreases, the probability of ice increases to the

2 temperature of -4 C, at which point homogeneous freezing occurs. A recent study, featured in one of the blog articles, claims that bacteria can serve as efficient ice nuclei, nucleating ice at temperatures of - C.. Aerosols can have an effect on Earth s climate both directly, through their interaction with radiation, and indirectly, through their effects on cloud properties. The so-called first aerosol indirect effect is a process by which increased aerosol loading (i.e., CCN concentration) decreases the drop sizes in clouds and increases the cloud reflectivity, causing a negative radiative forcing (cooling) on climate. The second aerosol indirect effect is a process by which increased aerosol loading affects the lifetime of a cloud. All other things being equal, do you expect more aerosols in a cloud would increase or decrease its lifetime? Why? One would expect an increase in cloud lifetime. With more aerosols, there are more particles to compete for the available water vapor. This inhibits the growth of cloud drops to precipitation-sized droplets, which would tend to increase cloud lifetime. 7. The equation for the equilibrium saturation ratio over a water droplet containing dissolved NaCl is given by 7 3.x cm.5x cm S = + 3 r r Calculate the maximum size for haze particles in this distribution From the equation above, a =. x -7 cm, and b =.5 x - cm 3. The maximum size for haze particles is given by the critical radius, r*. 3 * 3b 3(.5x cm ) 5 r = = =.x cm=.μm 7 a.x cm

3 8. Explain qualitatively, in a few sentences, the terms collision efficiency, collection efficiency, and coalescence efficiency. Collision Efficiency The probability that drops in the path of a collector drop will intercept the drop Coalescence Efficiency The probability that drops will stick together if they hit Collection Efficiency The probability that drops in the path of a collector drop will both intercept the drop and stick to it. In other words, collection efficiency is the product of collision efficiency and coalescence efficiency. 9. Explain the three physical processes involved in droplet growth for particles of size ~ μm. Droplets grow when water vapor diffuses to the drop. Latent heat is released, and this heat is diffused away from the drop.. For droplets of pure water, droplets with a radius smaller than the critical radius will evaporate, whereas larger droplets will grow.. Calculate the cutoff point in the problem above for a saturation ratio of.5 and temperature of C. The critical radius is the cutoff point that is referred to above. For pure water, the critical radius is given by r c = σ/( R v ρt ln S) (7.5x Nm ) 9 3 rc = =.3x m=.3x μm 3 (4 Jkg K )( kgm )(73 K)(ln.5) 3

4 . The Marshall-Palmer raindrop size distribution is given by the equation N( D) = Ne D, with N =.8 cm -4 and (R) = 4 R -., where R is the rain rate in mm hr -, and (R) is in units cm -. Calculate the LWC of a cloud with a rain rate of mm hr -. LWC is defined as π 3 LWC = ρ N( D) D dd l So one needs to calculate the 3 rd moment of the size distribution. For the Marshall- Palmer distribution, the n th moment is given by n n! N( D) D dd= N n+ Thus, the 3 rd moment is given by 3 3! 4 3! 4 7 N( D) D dd= N = (.8 cm ) = (.8 cm ) = 3.4x (4* cm) π LWC = (. gcm )(3.4x ) =.x gcm =.gm 3. A small rain drop with a radius of μm completely evaporates in minutes as it falls from cloud base towards the ground, having fallen about 5 meters. Assume that the saturation ratio beneath cloud base is constant. If the cloud base is at 7 mb and - C, what is the relative humidity? Here, we start with the growth equation r = (r + ξt) /. We want to solve for relative humidity, but first well solve for the saturation ratio, S. r = r + ξ( S ) t r S = ξ t We get ξ from the chart, using 7 kpa and T=- C. Plugging this in, we get ( μm) S = =.. ( μms )( s) RH = % 4

5 4. A 5 μm diameter collector drop is located at a position km above cloud base, where the liquid water content is.7 g m -3, the average droplet size is 5 μm, and a steady updraft of m s - exists. Calculated collection efficiency is 9%. The terminal velocity of the collector drop is 4. ms -. Estimate the change in radius with altitude (in μm per meter). The change in radius with altitude is given by dr EM u( R) = dz 4 ρ U u( R) L Plugging in the values given in the problem, we get 3 dr (.9)(.7 gm ) 4ms 7 = = 7.5x m/ m=.75 μm/ m 3 dz 4(x gm ) ms 4ms 5. Assuming mm diameter precipitation drops land on one square meter of the surface every second, estimate the equivalent rain rate in mm hr -. Based on this, do you think the assumption is reasonable? In terms of the number of drops hitting the square meter per hour, we have drops 3 s 5 = 3.x drops ms hr mhr There are many ways to do this, but let s calculate the volume of a single drop, assuming mm diameter (.5 ) Vdrop = πr = π cm = x cm drops 4 cm m 4 3 Vtotal = 3.x 5.4x.9x m 3 = mhr drop cm The height of water is the volume divided by the area ( m ), so the height is.9 x -4 m =.9 mm. So the rainrate would be.9 mm hr -. Contrary to what you re instructor thought, drops per square meter per second is a perfectly reasonable assumption!!! 5

6 Equations: r c = σ/( R v ρt ln S) dr/dt = (S )/[F k + F d ] = (S ) ξ =ξ r(t) = (r + ξt) / S = + a b 3 r r r* = (3b/a) / S* = + (4a 3 /7b) ½ dr EM u( R) = dz 4 ρ U u( R) L dr EM = ur ( ) dt 4ρ L rgρl u = 9 ( C R / 4) μ D e n n! N( D) D dd= N (Marshalln+ Palmer dist.) 3 LWC π = ρ N( D) D dd l ext ( D) d π β = Q D D 4 g = 9.8 m s - σ = 7.5 x - N/m R = 8.34 J mol - K - R v = 4 J kg - K - ρ l = g cm -3 ρ ice =.9 g cm -3

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