A uniform dispersion of spheres with a diameter of 0.8µm attenuates a light sour

Question

A uniform dispersion of spheres with a diameter of 0.8µm attenuates a light source by 92% at a test distance of 1000m. The particle material density is 1.15 g/cm3and the particulate concentration in the air is 745µg/m3.   (a) Determine the scattering ratio (K) for these test particles defined above. (b) Assume the presence of NO2 in the atmosphere with a concentration of 0.08ppm. The wavelength of visible light can be taken as 0.45µm.    Determine the extinction coefficient in m-1 due to the presence of air molecules, NO2 and test particles defined in this problem above. You may use the K value calculated in step (a). (c) What is limit of visibility due to the presence of air molecules, NO2 and the particles in meters?

Explanation / Answer

(a) For scattering ratio (K)

K = 2pir/l

where,

r = radius = 4 x 10^-7 m

l = wavelenght = 0.45 micrimeter = 4.5 x 10^-7 m

Feed values,

K = 2 x 3.1416 x 4 x 10^-7 / 4.5 x 10^-7 = 5.59

(b) For extinction coefficient e,

I = ecl

for air,

I = 8

c = 745 microgram/m^3 = 7.45 x 10^-4 g/m^3

l = 1000 m

Feed values,

e = 8/7.45 x 10^-4 x 1000 = 10.74

For NO2

c = 0.08 ppm = 8 x 10^-2 g/m^3

Feed value,

e = 8/8 x 10^-2 x 1000 = 0.1

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