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.

790 g/s

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.