Unformatted text preview: AOS 145 Homework #1 3 QUESTIONS (10 points total) 1. The following drop size distribution was measured in fog: a) [1 point] Draw a graph of the drop size distribution. b) [1 point] Draw a graph of dN/d (logDp ) of the drop size distribution. Keep in mind that dN/d (logDp ) = ln(10) Dp dN/d Dp =
2.303 D
p d
N/d
Dp For Dp in the above expression, you can use the average diameter for each size interval. c) [1 point] Calculate the total liquid water (in g m3) contained in the fog. We know that the volume size distribution is related to the number size distribution as follows dV/d Dp = (π/6)(Dp ) 3 dN/d Dp Therefore the total volume of the fog droplets (per unit volume of air) is ∑(dV/d Dp) ×∆D
p =∑[(π/6)(Dp ) 3 dN/d Dp] ×∆D
p which has the unit of (µm)3 (cm3 µm1) µm = µm3 cm3. Here ∆Dp i s t he width of each size interval. For Dp in the term (Dp ) 3 in the above expression, you can use the average diameter for each size interval. The resulting volume of fog droplets (per unit volume of air) is 3.25×106 µm3 cm3, which upon multiplication with the density of water, 1 g cm3, gives 3.25 g m3. 2. [2 points] Use the graph below to calculate how many particles per cubic centimeter can be found between 0.1–0.2 µm. Just eyeball that dN/d (logDp ) is about 13000 cm3 on average between 0.1 and 0.2 µm. Therefore the number of particles per cm3 between 0.1 and 0.2 µm is dN/d (logDp ) × ∆(logDp ) = 13000 cm3 × [ log(0.2) log(0.1) ] = 3913 cm3 3. Let's look at the fate of particles in a room: A cubic room with a wall length of L (room volume = L3 ) contains particles of diameter Dp with a concentration of N(t = 0, Dp ) at time t = 0. The air in the room is well mixed, i.e., particles are distributed uniformly in the room. A laminar layer of thickness h = 1 mm exists on all wall surfaces. The walls are perfectly absorbing. Particles hitting the wall will therefore stick permanently on them. a) [2 points] Derive an expression for N(t) if particles are solely lost by diffusion onto the walls. Hint: Because the walls are perfectly absorbing the particle concentration on the wall is zero. On the one hand, we can express the number of particles lost per unit time as dN(t)/dt × L3 where dN(t)/dt is the rate of particle loss and has the unit of cm3 s1. I multiply that with the volume of the room, L3 (unit: cm3), and the resulting dN(t)/dt × L3 has the unit of s1, i.e., the number of particles lost per second. On the other hand, the loss of particles is due to diffusion and we know how to express that according to Fick's first law. Keep in mind that Fick's first law only applies to laminar flow (smooth, orderly flow, as opposed to turbulent flow), hence I mentioned in the question the h = 1 mm thick laminar layer on the wall. The concentration of particle at the surface of the laminar flow is N(t) – it is the same throughout the room (air in the room is well mixed) except in the laminar layer. The concentration at the surface of the wall is zero the wall is perfectly absorbing so any particle that collides with the wall is gone. Therefore, the flux of particles lost through diffusion to the wall is J = – D dC/dx = – D [ 0 – N(t) ] / h = D N(t)/h where D is the particle diffusion coefficient. The dx in Fick's first law is the distance traveled by the particle in the laminar flow in our case, it is h = 1 mm. In the above expression, J is the flux of particles loss to the wall through diffusion and has the unit of cm2 s1, i.e., number of particles per unit area per second. Now if I multiple J with the total surface area of the walls (six of them, including the floor), 6L2 , we have D × N(t)/h × 6L2 which has the unit of s1, i.e., the number of particles lost to the wall per second (due to diffusion). Now the above two expressions of the loss of particles should be equal – they are both describing the number of particles lost per second. Therefore, dN(t)/dt × L3 = – J × 6L2 = – D × N(t)/h × 6L2 Note that I put a minus sign on the right hand side. That is because dN(t)/dt is negative – we are losing particles, while as the flux term is positive therefore we put a minus sign so that it expresses the actual loss of particles. You can then rearrange the above expression as dN(t)/N(t) = – (6D/hL)dt Simple integration of the above expression gives you lnN(t) – lnN(0) = – (6D/hL) t here N(0) is the particle concentration at t = 0. or N(t) = N(0) exp[ – (6D/hL) t ] It is an exponential decay for N(t). b) [2 points] Can you do the same calculation for gravitational settling? The particles reach a terminal settling velocity in no time. Let's use ut to represent the terminal velocity. Before we get started, let me throw this out there: the flux of particles (or molecules) in a flow with velocity u and particle (or molecule) concentration N is just N × u. In fact, chemical flux or transport flux are defined as such. With that being said, then the loss of particles to the floor due to gravitational settling can be expressed as N(t) × ut × L2 here L2 is the surface area of the floor. This expression has the unit of s1, i.e., number of particles lost due to gravitational settling. Similar to what we did in 3a), we have dN(t)/dt × L3 = – N(t) × ut × L2 (note again the minus sign). Which can be easily rearranged as dN(t)/N(t) = – (ut/L)dt Simple integration gives N(t) = N(0)exp[ – (ut/L)t ] It is an exponential decay (loss) for N(t). c) [1 point] Discuss qualitatively (without calculations) how the two loss processes depend on Dp . The particle diffusion coefficient D ∝ Dp 1 – the larger the particle, the smaller the diffusion coefficient hence the slower the diffusional loss, reflected here as a slower decrease of N(t) with time. This makes intuitive sense. The particle gravitational terminal settling velocity ut ∝ Dp 2 the larger the particle, the larger the terminal velocity hence the faster the gravitational settling loss, reflected here as a faster decrease of N(t) with time. Again, this makes intuitive sense. ...
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 Winter '17
 Quinbin LI
 Fick, drop size distribution

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