Question

Calculate the rate of emission of SO2 in g/s that results in a centerline (y = 0) concentration at ground level of 1.786 x 10-3 g/m3 one kilometer from the stack. The time of measurements was 1 P.M. on a clear summer afternoon. The wind speed was 2.0 m/s measured at a height of 10 m. The effective stack height is 96 m. No inversion is present.

Answer 1

790 g/s

Step-by-step explanation:

The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.

The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;

Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].

Where g = wind direction, y and h are the distance in metres.

Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;

Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.

Emission rate of pollutant,E = 790 g/s.