ATOC/CHEM 5151 Problem 30 Converting a Particle Size Distribution into Something Useful (read answers before Final Exam)

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1 ATOC/CHEM 5151 Problem 30 Converting a Particle Size Distribution into Something Useful (read answers before Final Exam) The Ultra-High Sensitivity Aerosol Spectrometer (UHSAS) measures submicron particles over the diameter range nm ( m) in 99 bins. This measurement is accomplished by using Mie scattering of a 1.05 m infrared laser and classifying the particles on the basis of the magnitudes of the pulses of light scattered off the particles. The following sub-micron (i.e., PM1) size distributions were observed in several plumes of burning biomass in Mexico from a Twin Otter craft in March Download the data file for a typical plume from the following location: NOTE The data in the table are not the same values as the figure above. The units for the Y axis are number of particles per cubic centimeter PER dlogd. In other words, you will need to multiply the dn/dlogd term by dlogd (which is about for each bin) to get particles per cm 3 for each bin. 1. Calculate the following quantities for this size distribution. Total number of particles per cubic meter (N): Total particulate cross sectional area (A) in square centimeters per cubic meter of : Total particulate surface area (SA) in square centimeters per cubic meter of : Total particulate volume (V) in cubic centimeters per cubic meter of : Total particulate mass (M, assume a density of 1.4 g cm -3 for condensed matter): See entries in the bottom row of the table below. V = 29 x 10-6 x 1.4 g m -3 = 41 g m -3

2 2. What information do you need to convert these quantities into mass mixing ratios (i.e., X per kg of )? We need to know the pressure and temperature of the i.e., altitude! We don t know where these measurements were made, and a cubic meter of at sea level doesn t have the same mass as a cubic meter of at a different altitude. 3. Speculate on the amounts of total particulate number, surface area, and mass that are missing in your calculations. One issue we need to consider is how we define the diameter. For calculating number, we don t worry we just count up all the particles in all the bins. For area and volume, we need to pick a diameter for R 2 and R 3. Do we take the low end of the range, the high end, the middle? Let s estimate the error from this choice. I picked the low end in my calculations. Picking the high end would increase area by 6% (1.03) 2 and increase volume by 9%. We probably should have picked an area-weighted and volume weighted diameter. So I have missed 4% of the area and 6% of the volume using my choice of diameter. You may have chosen the midpoint, in which case your errors would be smaller. Another quantity to consider is whether we are capturing the full size distribution or not, or if we are missing some important sizes at the < 55 nm or > 1000 nm ends. The following figure shows the distributions of dn, da, and dv. I ve added reasonable guesses (as dotted lines) for the quantities outside the range of the instrument. Using the same approach as for part 1, we can estimate the missing values. For N, I get around 8% or so is missing (mostly at the smallest sizes). For A, I get about 0.7% (split equally from the small and large sizes), and for V, I get about 3% (dominated by the larger sizes). So for this particular size distribution, which peaks around 100 nm, the small diameter cutoff has some impact on the total number density, but doesn t have a large impact on the area or volume (mass). See table on the following pages for detailed calculations.

3 D nm dn/dlogd per cm3 dlogd dn (per da (cm2 per dsa (cm2 per dv (cm3 per E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-13

4 E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-14

5 E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E N = per cm3 If I had used a midpoint diameter (i.e., instead of 55, etc.) N = 1.5(10) N = 1.5(10) A = 2.1(-6) cm2 per cm3 A = 2.1 cm2 A = 2.2 cm2 SA = 8.6(-6) cm2 per cm3 SA = 8.6 cm2 per m3 SA = 8.8 cm2 per m3 V = 2.7(-11) V = 2.7(-5) V = 2.9(-5)