AOS145_HW03_KEY - AOS 145 Homework#3 2 QUESTIONS(16 points...

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Unformatted text preview: AOS 145 Homework #3 2 QUESTIONS (16 points total) 1.​ An air mass containing dry particles of pure ammonium sulfate ((NH​4​)​2​SO​4​) is lifted in the atmosphere to a level where the environment saturation ratio is 1.002. The particles have the following size distribution: a) [4 point] Above which dry diameter will the particles be activated and grow? In solution ammonium sulfate dissociates into three ions. The molecular weight and density of ammonium sulfate is 132 g mol​­1​ and 1.77 g cm​­3​, respectively. Assume temperature ​T ​ = 273 K in your calculation. Hints: A dry diameter can be calculated from ​n​s​ or vice versa (see the last slide from Lecture 6). Once the particle takes up water, its diameter will change which follows the Köhler curve. For every dry particle (consisted of solute molecules and will grow to become a solution droplet), there is a corresponding critical saturation ​S​c​. When the environment saturation ​S​env​ > ​S​c​, the particle is said to be activated and starts growing rapidly to become a cloud droplet. The critical saturation ​S​c​ and the dry diameter ​d​s​ of the particle are related: Or, conversely, In this case, any dry particles with corresponding saturation less than the environment saturation of 1.002 will be activated. Let's take ​S​c​ = 1.002 (the upper limit) and calculate the corresponding dry diameter ​d​s​ of the particle. Here ​A ​ ≈ 0.66 / ​T ​ = 0.66 / 273 ​µ​m = 0.0024 ​µ​m (use the approximation given in slide #6 of Lecture 6). Density and molecular weight of ammonium sulfate: ​ρs​​ = 1.77 g cm​­3​ and ​M​s​ = 132 g mol​­1​. Number of ions from the dissociation of an ammonium sulfate molecule in water: ​ν ​ = 3. Density and molecular weight of water: ​ρw​​ = 1 g cm​­3​, ​M​w​ = 18 g mol​­1​. Plug in these values, we get the resulting ​d​s​ = 0.09 ​µ​m. In other words, any dry particles with diameters larger than 0.09 ​µ​m will be activated. b) [2 point] Discuss qualitatively how the size distribution will look like after some time has elapsed. Obviously particles with dry diameters larger than 0.09 ​µ​m will be activated and start growing rapidly. For the rest of the particles, they are on the rising part of their corresponding Köhler curves, and as such, will grow (somewhat slightly) to reach stable equilibrium (at slightly larger wet diameters). c) [2 point] Discuss qualitatively what will happen if, after 5 minutes, the environment saturation drops to 1.001. A drop in the environment saturation to 1.001 means that some of the particles that have not been activated and are in stable equilibrium at ​S​c​ = 1.002 will shrink slightly to reach stable equilibrium at ​S​c​ = 1.001 (at slightly smaller wet diameters). 2.​ Let's calculate how long it would take for a cloud drop with ​Dp​​ = 20 µm to grow to a raindrop with ​Dp​​ = 1 mm. The temperature in the cloud will be constant at ​T ​ = 273 K. We will simply assume an atmospheric pressure of 1013 mbar. a) [4 points] First we consider only condensation. To make a calculation we need to simplify the problem. We will set the environment saturation at ​S​ = 1.002, which is a realistic value in a cloud. We will choose the saturation at the surface of the drop to ​S​ = 1. (Explain why this is a good approximation.). Both saturations remain constant with time. Now we can determine how long it will take for this drop to grow to the size of a rain drop. Is this something that can happen in a cloud? The growth of the drop can be expresses as​ env​ 2​ ­1​ where ​Rp​​ 0 = 10 ​ µ​m, ​Dg​​ = 0.242 cm​ s​ is the diffusion coefficient of water in air, ​ρp​​ = 1 g cm​­3​ ​is ​ the density of the particle and in this case, the water drop.​ ​M​w​ = 18 g mol​­1​ is the molecular weight of water. Now we need to figure out ​c∞​​ and ​cs​​ , which are the ​molar concentrations​ of water molecules in the environment and near the surface of the drop, respectively. At ​T ​ = 273 K, ​p​ = 1013 mbar = 101.3 kPa, using the ideal gas law ( ​pV​ = ​nRT​ → molar concentration ​n/V​ = ​p/R/T​, number of moles of a molecule per unit volume of air) we can calculate the molar concentration of air ​n​air​ = 4.47×10​­5​ moles cm​­3​. The water saturation vapor pressure (over a flat surface) at ​T ​ = 273 K is ​psat​ ​ (∞) = 0.61129 kPa, according to the Clausius­Clapeyron equation (or see HW#2, question #1). The environment vapor pressure is p​env​ = ​S​env​ × ​p​sat​(∞). Therefore ​c∞​​ = (​p​env​ / ​p​) ​n​air​ = ​S​env​ (​p​sat​(∞) / ​p​) ​n​air​. By the same token, by definition the vapor pressure at the surface of the drop is ​S​ × ​p​sat​(∞). Thus, ​cs​​ = ​S​ (​psat​ ​ (∞) / ​p​)​ ​nair​ ​ . Therefore ​c∞​​ ­ ​cs​​ = (​Senv​ ​ ­ ​S​) (​psat​ ​ (∞) / ​p​) ​nair​ ​ = (1.002 ­ 1) (0.61129 kPa / 101.3 kPa) (4.47×10​­5​ moles cm​­3​) = 5.39×10​­10​ moles cm​­3​. Here the assumption of ​S​ = 1 is a good approximation because at ​Dp​​ = 20 ​µ​m, the curvature effect is already fairly small. Plug these values in the above equation, we have t ≈ 6 days. Clearly this is way too slow and not something that can happen in a cloud – the cloud is long gone before this ever has a chance. b) [4 points] We can also look at growth through gravitational settling. Here we need to assume that we already have an existing large particle. We will use a particle of 0.2 mm diameter, which falls through a cloud with 100 drops cm​­3​ of ​Dp​​ = 20 µm at a speed of 30 cm s​­1​. Calculate the growth rate (​dD​p​/d​t) for this drop. How much will the droplet grow at this rate when falling through a cloud that is 1000 m thick? The initial diameter of the large particle, ​Dp​​ 1 = 0.2 mm. The diameter of the small cloud drops ​ D​p​2 = 20 ​ µ m ​ . The number concentration of the cloud drops ​ N​2​ = 100 cm​­3​. The gravitational ​ ­1​ settling speed of the large particle ​ut​​ = 30 cm s​ . The gravitational settling speed of the small cloud drops are negligible. The coagulation coefficient between the large particle and the small cloud drops due to gravitational settling is ​K​12​ = π/4 × ​Dp​ 12​​ ×​ u​t​. ​Therefore the collision rate​ is J​12​ = ​K​12​ × ​N​2​.​ ​As the large particle takes up the small cloud drops, its volume ​V1​​ ​= π/6×​Dp​​ 13​​ increases. On the one hand, ​dV​1​/​d​t = d(π/6×​Dp​​ 13​​ )/dt = π/2×​Dp​​ 12​​ × ​dD​p​1/​​ d​t. (recall that ​dx​3​/​d​t = 3​x2​​ dx​/d​ ​t) On the other hand, the volume increase of the large particle is equal to the combined volume of all the small cloud drops taken up by the large particle which is ​J12​ ​ × ​V2​​ = ​J12​ ​ × π/6 ×​Dp​​ 23​​ . Therefore, we have (this is very much analogous to the thought process for question 3 in HW#1) dV​1​/​d​t = d(π/6×​Dp​​ 13​​ )/dt = π/2×​Dp​​ 12​​ × ​dD​p​1/​​ d​t. = ​J12​ ​ × π/6 ×​Dp​​ 23 ​​ = ​K​12​ × ​N​2​ ×π/6×​Dp​​ 23 ​ Rearrange the above equation, we have ​dD​p​1/​​ d​t = π/12 × ​u​t​ × ​N​2​ × ​Dp​​ 23 ​ Simple integration gives ​Dp​​ 1(t) = ​ Dp​​ 1(0) + (π/12 × ​ ut​​ × ​N2​​ × ​Dp​​ 23​​ ) × t ​ ​ where ​Dp​​ 1(0) = 0.2 mm is the initial diameter of the large particle. For the large particle to fall ​ through a 1000 m thick cloud, it takes 1000 m / 30 cm s​­1​ ≈ 3333 seconds, after which the large particle has grown in diameter (π/12 × 30 cm s​­1​ × 100 cm​­3​ × 20​3​ ​µ​m3​​ ) × 3333 s ≈ 0.21 mm, i.e., doubling its diameter. ...
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