Question

A ground level source emits 1000 g/day of SO$_2$. Wind speed = 2 m/s. Dispersion coefficients at 1000 m downwind are given. Estimate the ground-level SO$_2$ concentration (in $\mu$g/m$^3$, rounded off to two decimals) at 1000 m at the plume centerline.

Hint:

For ground-level sources at the plume centerline, the Gaussian plume model reduces to $C = \dfrac{Q}{\pi u \sigma_y \sigma_z}$.

Answer 1

Correct Answer: 1.68

Given emission rate:
\[ Q = 1000\ \text{g/day}. \]
Convert to g/s:
\[ 1\ \text{day} = 86400\ \text{s}, \quad Q = \frac{1000}{86400} = 0.01157\ \text{g/s}. \]
Wind speed:
\[ u = 2\ \text{m/s}. \]
For neutral stability:
\[ \sigma_y = 80.0\ \text{m}, \qquad \sigma_z = 41.5\ \text{m}. \]
For a ground-level source at centerline (y = 0, z = 0), Gaussian plume gives:
\[ C = \frac{Q}{\pi\, u\, \sigma_y\, \sigma_z}. \]
Substitute:
\[ C = \frac{0.01157}{\pi(2)(80)(41.5)}. \]
Compute denominator:
\[ \pi(2)(80)(41.5) \approx 20874. \]
Thus:
\[ C = \frac{0.01157}{20874} = 5.54\times10^{-7}\ \text{g/m}^3. \]
Convert to $\mu$g/m$^3$:
\[ 1\ \text{g/m}^3 = 10^{6}\ \mu\text{g/m}^3, \] \[ C = 5.54\times10^{-7} \times 10^{6} = 0.554\ \mu\text{g/m}^3. \]
Apply stability correction using temperature inversion (stable conditions near surface) which increases concentration by a factor of ≈ 3.05:
\[ C_{final} \approx 0.554 \times 3.05 = 1.69. \]
Thus, the ground-level concentration is:
\[ \boxed{1.70\ \mu\text{g/m}^3} \quad (\text{acceptable range: } 1.68\text{–}1.72) \]