Question

SO$_2$ is emitted at a rate of 20 kg/s from a 10 km $\times$ 10 km airshed in an industrial area. Wind blows at a speed of 4 m/s from one direction in that area. Radiation inversion restricts the mixing height to 1200 m. Neglect SO$_2$ concentration in the incoming air. Assuming emitted SO$_2$ to be conservative, the steady state SO$_2$ concentration in the airshed, in $\mu g/m^3$, is _______ (rounded off to 2 decimal places).

Hint:

When calculating steady-state pollutant concentration in an airshed, remember to convert units of emission rate to $\mu$g/s and apply the correct area and mixing height.

Answer 1

Step 1: Use the formula for steady-state concentration in an airshed. \[ C = \frac{Q}{A \cdot H \cdot V} \] Where: \(C\) = steady-state concentration ($\mu g/m^3$) 
\(Q\) = emission rate (20 kg/s) 
\(A\) = area of the airshed (10 km × 10 km = \(100 \, {km}^2 = 10^8 \, {m}^2\)) 
\(H\) = mixing height (1200 m) 
\(V\) = wind velocity (4 m/s) 
Step 2: Convert emission rate to $\mu$ g/s. \[ Q = 20 \, {kg/s} = 20 \times 10^6 \, {g/s} = 20 \times 10^9 \, \mu{g/s} \] Step 3: Calculate the concentration. \[ C = \frac{20 \times 10^9}{10^8 \times 1200 \times 4} = \frac{20 \times 10^9}{4.8 \times 10^{11}} = 410 \, \mu{g/m}^3 \] Answer: The steady-state SO$_2$ concentration is 410 $\mu$g/m$^3$.