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Exam 2: Physics April 16, 2008 Physical 3440

Questions 1-10 are worth 5 points each. Questions 11-15 are worth 10 points each.

1. Rank the concentrations of the following from lowest (1) to highest (3):

cloud nuclei (3) cloud droplets (2) raindrops (1)

2. Match the following particles with their typical size

cloud condensation nuclei 10 μm cloud droplets 0.1 μm raindrops 1000 μm

3. Why do grow at the expense of supercooled droplets?

The saturation 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 1, in which case Sl (with respect to liquid) is less than 1, causing the liquid drops to evaporate.

4. Match the following particles with their most likely means of growth

5 μm cloud drop accretion 5 μm ice crystal aggregation 100 μm dendritic ice crystals in ice cloud depositional growth 500 μm ice crystal in mixed-phase cloud condensational growth 100 μm cloud drop collision-

5. Fill in the blanks:

Not all with below the freezing point of water contain ice, because of the scarcity of ice nuclei in the , and the at which they nucleate ice. As the cloud temperature decreases, the probability of ice increases to the

1 temperature of -40 °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 -2 °C.

6. Aerosols can have an effect on Earth’s 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 . This inhibits the growth of cloud drops to -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

1.21x 10−−7163cm 1.5 x 10 cm S =+1 − rr3 Calculate the maximum size for haze particles in this distribution

From the equation above, a = 1.21 x 10-7 cm, and b = 1.5 x 10-16 cm3. The maximum size for haze particles is given by the critical radius, r*. 3bxcm 3(1.5 10−16 3 ) rxcmm*5== =6.1 10− = 0.61μ axcm1.21 10−7

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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 ~ 10 μm.

Droplets grow when water vapor diffuses to the drop. Latent heat is released, and this heat is diffused away from the drop.

10. For droplets of pure water, droplets with a radius smaller than the critical radius will evaporate, whereas larger droplets will grow.

11. Calculate the cutoff point in the problem above for a saturation ratio of 2.5 and temperature of 0 °C.

The critical radius is the “cutoff point” that is referred to above. For pure water, the critical radius is given by rc = 2σ/( Rv ρT ln S)

2(7.5xNm 10−−21 ) rxmxm===1.3 10−−93 1.3 10 μ c (462Jkg−−11 K )(1000 kgm − 3 )(273 K )(ln 2.5)

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12. The Marshall-Palmer raindrop size distribution is given by the equation −ΛD -4 -0.21 ND()= Ne0 , with N0 = 0.08 cm and Λ(R) = 41 R , where R is the rate in mm hr-1, and Λ(R) is in units cm-1. Calculate the LWC of a cloud with a rain rate of 2 mm hr-1.

LWC is defined as π ∞ LWC= ρ N() D D3 dD l ∫ 6 0 So one needs to calculate the 3rd moment of the size distribution. For the Marshall- Palmer distribution, the nth moment is given by ∞ n! NDDdD()n = N ∫ 0 n+1 0 Λ Thus, the 3rd moment is given by ∞ 3! 3! 6 NDDdD( )3447== N (0.008 cm− ) = (0.008 cm−− ) = 3.04 x 10 ∫ 0 ΛΛ31+− 4(41*2 0.21cm ) 4 0 π LWC===(1.0 gcm−−37 )(3.04 x 10 ) 1.6 x 10 −−− 733 gcm 0.16 gm 6

13. A small rain drop with a radius of 200 μm completely evaporates in 10 minutes as it falls from cloud base towards the ground, having fallen about 500 meters. Assume that the saturation ratio beneath cloud base is constant. If the cloud base is at 700 mb and -10°C, what is the relative ?

2 2 1/2 Here, we start with the growth equation r = (r0 + 2ξt) . We want to solve for relative humidity, but first well solve for the saturation ratio, S. 22 rr=+012(ξ St − 1) r 2 S =−1 0 2ξ1t

We get ξ1 from the chart, using 70 kPa and T=-10 C. Plugging this in, we get (200μm )2 S =−10.16 = 2(101.6μms 2− 1 )(600 s ) ∴RH =16%

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14. A 500 μm diameter collector drop is located at a position 1 km above cloud base, where the liquid water content is 1.7 g m-3, the average droplet size is 15 μm, and a steady updraft of 2 m s-1 exists. Calculated collection efficiency is 90%. The terminal velocity of the collector drop is 4.0 ms-1. Estimate the change in radius with altitude (in μm per meter).

The change in radius with altitude is given by dR EM u() R = dz4()ρL U− u R

Plugging in the values given in the problem, we get

dR(0.9)(1.7 gm−−31 ) 4 ms ==−=−7.65x 10−7 mm / 0.765μ mm / dz4(1 x 1063 gm−−− ) 2 ms 1− 4 ms 1

15. Assuming 100 1 mm diameter precipitation drops land on one square meter of the surface every second, estimate the equivalent rain rate in mm hr-1. 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 100drops 3600 s drops = 3.6x 105 ms22 hr mhr There are many ways to do this, but let’s calculate the volume of a single drop, assuming 1 mm diameter 44 Vr==ππ3343(0.05 cmxcm ) = 5.24 10− drop 33 33 ⎛⎞54drops⎛⎞⎛⎞−− cm1 m 43 Vxtotal ==⎜⎟3.6 10263⎜⎟⎜⎟ 5.24 x 10 1.9 xm 10 ⎝⎠mhr⎝⎠⎝⎠ drop10 cm The height of water is the volume divided by the area (1 m2), so the height is 1.9 x 10-4 m = 0.19 mm. So the rainrate would be 0.19 mm hr-1. Contrary to what you’re instructor thought, 100 drops per square meter per second is a perfectly reasonable assumption!!!

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Equations:

∞ rc = 2σ/( Rv ρT ln S) n! NDDdD()n = N (Marshall- ∫ 0 n+1 0 Λ dr/dt = (S – 1)/[Fk + Fd] = (S – 1) ξ1=ξ 2 1/2 Palmer dist.) r(t) = (r0 + 2ξt) π ∞ ab LdWDC = ρ NDD()3 S =+1 − l 3 6 ∫ rr 0 1/2 ∞ r* = (3b/a) π 2 3 ½ β = QDD(Dd) S* = 1 + (4a /27b) ext ∫ 4 0 dR EM u() R -2 = g = 9.8 m s -2 dz4()ρL U− u R σ = 7.5 x 10 N/m -1 -1 dR EM R = 8.314 J mol K = uR() -1 -1 Rv = 462 J kg K dt 4ρL -3 ρl = 1 g cm 2 rg2 ρ -3 u = l ρice = 0.9 g cm 9 (CRDe / 24)μ

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