Question

A uniform dispersion of spheres with diameter of 0.8 µm attenuates a light source by 92 percent at a distance of 1000 m. The particle material density is 1.15 g/cm3 and the particulate concentration in the air is 745 µg/m3 .

Determine

the scattering coefficient in m-1 if absorption is neglected

the value of the scattering ratio, K, for the substance and

the limit of visibility in kilometers

Answer 1

Scattering Coefficient :

C : concentration :

: Scattering Coefficient

distance = 1000 m

Area =

Substituting values

For limit of visibility :

consider the following equation :

F : fraction of diminished particles

= attenuation coefficient = 1.64 (1/m) for air (available on internet)

x : distance over attentuation happens

Limits :

at 1000 m , the fraction of diminished particles is 0.08

at x m , the fraction will be 1 i.e. at the end of this visibility all particles are diminished.

Integrating the equation:

2.5257 = -1.64 (x-1000)

- 1.54 = x - 1000

x = 998.46 m

This means at 1000+998.46 = 1998.46 m the visibility will be zero.

= 0.199846 KM