HW4soln

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SIO 172 Physics of the Atmosphere Homework #4 1. A stratocumulus cloud layer is 300 m thick, has liquid water content equal to 0.7 g m −3 , and has droplets with radius equal to 14 μm. The liquid water content and droplet radius are the same everywhere in the cloud. The scattering efficiency of the droplets at visible wavelengths is equal to 2. a) What is the optical thickness of the cloud layer at visible wavelengths? τ = (3/2) w l ∆z / ( r e ρ l ) = ( 1.5) (0.0007 kg m −3 ) (300 m) / ((14×10 −6 m) (1000 kg m −3 )) = 22.5 b) What is the number concentration of droplets in units of cm −3 ? N (4/3) π r e 3 ρ l = w l N = w l / ((4/3) π r e 3 ρ l ) = (0.0007 kg m −3 ) / ((4/3) π (14×10 −6 m) 3 (1000 kg m −3 )) N = 60.9×10 6 m −3 = 60.9 cm −3 c) Drizzle then forms in the stratocumulus cloud layer and precipitates out, which reduces the liquid water content to a value equal to 0.6 g m −3 . If the cloud droplet number concentration stays the same (new droplets are activated to replace those collected by the drizzle drops), what is the new cloud droplet radius in units of 14 μm? N = w l_new / ((4/3) π r new 3 ρ l ) = w l_old / ((4/3) π r old 3 ρ l ) w l_new / r new 3 = w l_old / r old 3 r new = r old ( w l_new / w l_old ) 1/3 = 14 μm (0.6 g m −3 / 0.7 g m −3 ) 1/3 = 13.3 μm d) What is the new optical thickness of the cloud layer at visible wavelengths? τ = (1.5) (0.0006 kg m −3 ) (300 m) / ((13.3×10 −6 m) (1000 kg m −3 )) = 20.3

2. A larger drop falls through a cloud while collecting smaller droplets along the way. The fall speed of the drop v is linearly proportional to its radius r according to the relationship below , where C = 6000 s −1 and r is in units of meters. v = C r Answer the following questions using the following assumptions: • The temperature of the cloud layer is above 0 °C everywhere • The liquid water content w l of the cloud is 1 g m −3 everywhere and at all times • The collection efficiency E c of the drop is 0.8 and does not change over time • Condensation or evaporation of water vapor on the drop is negligible • The cloud droplets are much smaller than the larger drop • The smaller cloud droplets have negligible fall speed a) Provide an equation that expresses the rate of growth of the radius of the drop in terms of the fall speed of the drop, the liquid water content of the smaller droplets, the collection efficiency, and any other necessary parameters. Use the above relationship for velocity in the equation. dr / dt = C r w l E c / (4 ρ l ) b ) Derive an equation that expresses the radius of the drop r in terms of the length of time t it has been growing through collection from an initial time when it had radius r 0 , along with any other necessary parameters. dr / dt = C r w l E c / (4 ρ l ) dr / r = C w l E c dt / (4 ρ l ) The above equation can be integrated from an initial time = 0 and radius = r 0 to a later time = t and radius = r . log( r ) − log( r 0 ) = C w l E c t / (4 ρ l ) r = r 0 exp[ C w l E c t / (4 ρ l )] c) If the initial radius of a drop is 20 μm, how much time does it take to grow to a radius of 400 μm? t = 4 ρ l log( r / r 0 ) / ( C w l E c ) t = (4) (1000 kg m −3 ) log(400 μ m / 20 μm) / ((6000 s −1 ) (0.001 kg m −3 ) (0.8)) = 2496 s t = 41.6 min

3. Below is a simple relationship for converting cloud optical thickness τ to cloud albedo α (fraction of radiation that is reflected by the cloud back to space) . α = τ / (7 + τ) This relationship is most applicable at visible wavelengths, and all cloud optical properties described below are for visible wavelengths. The scattering efficiency of cloud droplets is equal to 2. Assume that liquid water content and cloud droplet radius is the same everywhere in the cloud. a) If the optical thickness of a cloud layer is 7 , what is the albedo of the cloud layer? α = τ / (7 + τ) = 7 / (7 + 7 ) = 0. 5 b) In order to reflect more sunlight back to space and partially offset greenhouse warming caused by increased atmospheric CO 2 , a proposed geoengineering experiment has the goal of increasing the albedo of the cloud layer by 10% (e.g., α new = 1.1α old ). What percent change in the cloud optical thickness is needed to produce a 10% increase in cloud albedo if the starting cloud optical thickness is 7? Does cloud optical thickness need to increase or decrease? α = τ / (7 + τ) τ = 7α / (1 − α) new α = 1.1α = (1.1) (0.5) = 0.55 new τ = (7) (0.55) / (1 – 0.55) = 8.56 percent change = (8.56 / 7) – 1 = 0.22 = 22% , an increase c) Assuming that the geometric thickness of the cloud layer and the cloud droplet radius do not change, what percent change in liquid water content is needed to produce the percent change in cloud optical thickness calculated in (b)? Does liquid water content need to increase or decrease? τ = (3/2) w l ∆z / ( r e ρ l ) τ / w l = (3/2) ∆z / ( r e ρ l ) If ∆z and r e do not change, then τ new / w l_new = τ old / w l_old w l_new / w l_old = τ new / τ old = 1.22 percent change = 1.22 – 1 = 0.22 = 22%, an increase d) Assuming that the geometric thickness of the cloud layer and the liquid water content do not change, what percent change in cloud droplet radius is needed to produce the percent change in cloud optical thickness calculated in (b)? Does cloud droplet rad ius need to increase or decrease? τ = (3/2) w l ∆z / ( r e ρ l ) τ r e = (3/2) w l ∆z / ρ l If ∆z and w l do not change, then

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τ new r e_new = τ old r e_old r e _new / r e _old = τ old / τ new = 0.82 percent change = 0.82 – 1 = −0.18 = −18%, a decrease 4. An ice particle starts at the top of a cold cloud layer and then falls through cloud while accreting (riming) supercooled droplets along the way. Answer the following questions using the following assumptions: • The temperature of the cloud layer is below 0 °C everywhere • The (supercooled) liquid water content w l of the cloud is 2 g m −3 everywhere • The cloud layer is 2 km thick • The ice particle falls at a constant speed v equal to 200 cm s −1 • The collection efficiency E c of the ice particle is 0.8 • Deposition of water vapor on the ice particle is negligible • M elting or sublimation of the ice particle is negligible • The cloud droplets are much smaller than the ice particle • The cloud droplets have negligible fall speed • The ice particle can be approximated as a sphere • Ice has the same density as liquid water • No air bubbles are present in the ice particle • The ice particle starts at the top of the cloud with a radius equal to 200 μm a) How long does it take for the ice particle to fall from the top of the cloud layer to the bottom of the cloud layer? time = distance / speed = 2000 m / 2 m s −1 = 1000 s = 16.7 minutes b) What is the radius of the ice particle when it reaches the bottom of the cloud layer? dr / dt = v w l E c / (4 ρ l ) Since the variables on the right side of the equation are all constant, it can be integrated to yield ∆ r = ∆t v w l E c / (4 ρ l ) = (1000 s) (2 m s −1 ) (0.002 kg m −3 ) (0.8) / ((4) (1000 kg m −3 ) ∆ r = 0.0008 m = 800 μm Add this to the size at the top of the cloud to get the size at the bottom of the cloud 800 μm + 200 μm = 1000 μm = 1 mm