ENVR 116 Introduction to Aerosol Science December 17, 2001 FINAL EXAMINATION

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1 ENVR 116 Introduction to Aerosol Science December 17, 2001 FINAL EXAMINATION Please answer all questions on the attached sheets Answer Question No. 1 and 4 of the remaining 6 questions. No extra credit will be given for working extra questions. Each question has a weight of 20 points. If you work more than four of the optional questions, only the?rst four will be graded. USEFUL CONSTANTS viscosity 1.83 x 10 4 poises mean free path 6.80 x 10 2 µm k 1.38 x erg/k T 293 o C Es 33,000 statvolts/cm M (molecular weight of water)18 gm R 8.31 x 10 7 erg/(k mole) density of water 1.0 g/cm 3 1. Answer each question by circling the appropriate letter: A for always true S for sometimes true N for never true. A S (N) i. At least as much light scatters in the forward direction as at ninety degrees. A (S) N ii. Sixteen times as much light with wavelength of 0.4 µm (blue light) scatters from a particle as light with wavelength of 0.8 µm (red light). A (S) N iii. A particle for which size parameter has a value of 20 scatters more light than a particle for which size parameter has a value of 1. A (S) N iv. A particle will grow if the saturation ratio is greater than 1.0. (A) S N v. Increasing the atmospheric pressure will decrease the slip correction factor, if all else remains the same. (A) S N vi. The concentration of aerosol particles in a room is 10 3 /cm 3. The particles are monodisperse, and 0.5 µm in aerodynamic diameter. This concentration will probably remain fairly constant over a half hour. A S (N) vii. Consider two aerosols: one is monodisperse with aerodynamic diameter of 10 µm and the other is polydisperse with count median diameter of 10 µm and geometric standard deviation of 3. The number concentration of both aerosols is the same. Light from the same source and with the same intensity shines through each aerosol. More light will pass through the monodisperse aerosol than through the polydisperse aerosol. (A) S N viii. Smaller particles in outdoor air are less likely to carry an electrical charge than larger particles in outdoor air. (A) S N ix. Liquid sprays out the end of a small-diameter tube that has a high negative voltage applied to it. The liquid forms droplets that go into a vacuum. These droplets might explode before they fully evaporate. 1

2 (A) S N x. A particle 0.01 µm in diameter that has a density of 1 g/cm 3 is more likely to collect on a horizontal surface by diffusion than by gravitational settling. A S (N) xi. Two particles of different size each hold seven electronic charges. In the same electric?eld, the terminal electrical velocity for the larger particle will be greater than the terminal electrical velocity for the smaller particle. A (S) N xii. Two particles of different size are placed in a high concentration of unipolar ions within an electric?eld. The terminal electrical velocity for the larger particle will be greater than the terminal electrical velocity for the smaller particle. A (S) N xiii. Particles diffuse upward and downward at the same rate. A (S) N xiv. Given an ignition source and suf?cient oxygen, any material that will burn in open air can explode if its particle size is small enough. (A) S N xv. A cascade impactor is an excellent device to collect a sample of viable aerosol. (A) S N xvi. If aerosol concentration doubles, visual range is halved if all else remains the same. (A) S N xvii. A basketball 25 cm in diameter thrown horizontally into still air with a velocity of 500 cm/s and at a height of 1 meter will travel farther in air than a particle 1 µm in diameter thrown horizontally into the same gas from the same place and with the same velocity. (A) S N xviii. A?lter downstream of a probe that samples at a?ow twice isokinetic will collect more mass than it would if?ow were isokinetic. (A) S N xix. Water droplets that are 1 µm in diameter will in dry air at 20 C will take less than one second to evaporate. A (S) N xx. Electrostatic attraction is more effective than inertial impaction for collecting particles less than about 1 µm in diameter. 2. For air at 20 o C and 760 mm pressure, at what particle diameter is the apparent mean free path a minimum? Assume the particle density is 1 g/cm 3. Compute Cc using Cc = din µ m ( [ ]) d exp for several values of d (say, 1 µm, 0.5 µm, 0.1 µm, and 0.05 µm). For d = 1µm, Cc = Then calculate τ for the same values of d using For d = 1µm, τ = s. τ = 1 d 2 ρ p Cc Next, compute the mass of an aerosol particle from m = π 6 d3 ρ p for each of the d s chosen. ( For d = 1µm, m = gm) Finally, the apparent mean free path can be computed from 8kT l b = τ πm 2

3 and the results plotted on the graph (see attached. For d = 1µm, l B = cm). From the graph, estimate the minimum mean free path value.this is about 0.2 µm. 3a. Considering the maximum negative charge (in electrons) which a water droplet can carry, and also the Rayleigh limit, at what diameter will the Raleigh limit exceed the electron limit, or in other words, what is the minimum water droplet diameter at which a charged droplet can disintegrate on evaporation? Use 76.1 dyne/cm for the surface tension of water. The Raleigh limit is given by and the electron limit by Equating the two and solving for d gives n R = (1/e) 2πγ d 3 n e = d 2 4e d H2 O = 32πγ E 2 s = µm 3b. Suppose instead of water we are interested in mercury droplets. What is this minimum diameter for mercury droplets? Use dyne/cm for the surface tension of mercury. One can determine this minimum diameter for mercury by following the steps given in 3a above. Or an easier way is to divide the surface tension of water out of the 3a solution and multiply the result by the surface tension of mercury, i.e., d Hg = = 0.402µm 4a. Using Kelvin s equation, determine the particle diameter at which an ion will start to have water condense on it. This appears to occur when S = 4. With Kelvin s equation d = 4γ M RT lns = µm 4b. Will a sodium chloride particle (density = 1.77 g/cm 3 ) having a mass of 1.30 x grams and a single electrical charge grow when the saturation ratio is the same as in 4a above? Explain. A sodium chloride particle with a mass of 1.30 x grams will have a diameter of about 1.12 x 10 3 µm and it would not grow on its own. Even with the single electrical charge, the saturation ratio is not suf?cient to cause the particle to grow. 5. A raindrop (150 µm diameter) falls and as it falls it evaporates. According to Pruppacher and Klett (1978) one can estimate the increase in evaporation rate from the following two equations: For Re < 2.5 3

4 f v = (Sc 1/3 Re 1/2) 2 and for Re> 2.5 f v = (Sc 1/3 Re 1/2) where Re is the Reynolds number, and Sc the Schmitt number, de?ned as the ratio of the kinematic viscosity to the gas diffusion coef?cient The term f is the ratio of the evaporation rate for the droplet moving in air to the evaporation rate for the droplet in still air. Estimate how much faster the falling raindrop will evaporate compared to a non-moving droplet of the same diameter. Since a 150 µm drop will have a terminal settling velocity outside of the Stokes region, it is necessary to use the relationship for the drag coef?cient, C D and Reynolds number, And then from Reist s equation estimating Re, C D Re 2 = ( 4 3 )( ) 3 ρ m ρ p 980 µ 2 = 158 Sc = Re = 10 ( log(c D Re 2 ) ) = 4.61 kinematic viscosity gas diffusion coef?cient = = f v = ( / /2 = 1.37 This indicates that a falling drop will evaporate 37% faster than a stationary drop. 6a. Using the plot given on the attached sheet, determine the ratio of Q ext for 0.75 µm particles when exposed to blue light (λ = 0.4 µm) as compared to red light (λ = 0.7 µm). Use m = Recalling that α = πd/λ, α for a wavelength of 0.7 µm is and for a wavelength of 0.4 µm is From the attached?gure, for m=1.33, Q ext = 2.5 when the wavelength is 0.7 µm and 4 when the wavelength is 4 µm. This gives a ratio of b. Now repeat the calculation for the same size particle but use m = i. What conclusions can you draw (if any) regarding the role of clean clouds and polluted clouds in global warming? Again, from the attached?gure, for m= i, Q ext = 2.5 when the wavelength is 0.7 µm and 3 when the wavelength is 4 µm. This gives a ratio of The presence of an absorbing component implies anthropogenic pollution. Light passing through clean clouds (no absorbing component) will have more of the short wavelength light scattered than the longer wavelength light. Since most of the incoming light is short wavelength, this means that the sun s energy is distributed more or less uniformly around the globe and retained since there is less long-wavelength light transmitted out of the earth s atmosphere through the clean clouds and energy (heat) accumulates. With dirty clouds incoming short wavelength energy is almost equally transmitted out of the atmosphere by the longer wavelengths so that there is little heat accumulation. Thus dirty air would be expected to enhance cooling of the earth s atmosphere, clean air would enhance its warming. 4

5 7. Given the following data collected in an experiment of coagulating particles: Seconds Concentration, p/cm Determine a value for the average coagulation coef?cient. What can you say about particle size from these data? The inverse of the concentration is plotted as a function of time. This plot should result in a straight line with the slope being equal to the coagulation coef?cient. The attached?gure shows such a plot for these data. The slope is determined to be cm 3 /s which should be compared to a value of cm 3 /s which is constant for all large aerosol particles (> 10µm). The particle diameter estimated to give a coagulation coef?cient of cm 3 /s is about 0.3 µm. 5

THE UNITED REPUBLIC OF TANZANIA MINISTRY OF EDUCATION AND VOCATIONAL TRAINING FORM TWO SECONDARY EDUCATION EXAMINATION, 2005.

THE UNITED REPUBLIC OF TANZANIA MINISTRY OF EDUCATION AND VOCATIONAL TRAINING FORM TWO SECONDARY EDUCATION EXAMINATION, 2005 0031 PHYSICS Time: 2 Hours 1. This paper consists of sections A, B and C. 2.