Question

An industry with an effective stack height of 80 m emits 1200 g/h of CO. The windrose indicates wind speed $u=3$ m/s. At a downwind distance of 2 km (plume centerline), the dispersion coefficients $(\sigma_y,\sigma_z)$ depend on wind direction as per the table below.  
 

Air Pollution Concentration Calculation

During the maximum duration of the year (i.e., the most frequent wind direction from the windrose), the ground-level PM_2.5 concentration is to be computed at the plume centerline at 2 km. Find the concentration (in μg/m³, rounded off to two decimal places).

Dispersion Coefficients (in m)

Wind Direction	Dispersion Coefficients (in m)
Northeast	Crosswind: 50, Vertical: 20
North	Crosswind: 45, Vertical: 30
Northwest	Crosswind: 40, Vertical: 35
East	Crosswind: 45, Vertical: 30
Southeast	Crosswind: 55, Vertical: 35
South	Crosswind: 60, Vertical: 40
Southwest	Crosswind: 65, Vertical: 45
West	Crosswind: 70, Vertical: 50

Table 1: Dispersion coefficients based on wind direction

 

Hint:

For ground-level, centerline concentration from a tall stack with reflection, use $C=\dfrac{Q}{\pi u \sigma_y \sigma_z}\exp\!\left(-\dfrac{H^2}{2\sigma_z^2}\right)$. Pick $(\sigma_y,\sigma_z)$ at the given downwind distance and the \emph{most frequent} wind direction when the question asks for “during the maximum duration of the year.”

Answer 1

Correct Answer: 1.75

Step 1: Identify the most frequent wind direction and corresponding $(\sigma_y,\sigma_z)$.
From the windrose, the Northwest direction has the maximum frequency.
From the table (for 2 km): $\sigma_y=40\ \text{m}$, $\sigma_z=35\ \text{m}$.
Step 2: Convert source strength to consistent units.
Emission rate $Q=1200\ \text{g/h}=\dfrac{1200}{3600}\ \text{g/s}=0.3333\ \text{g/s}=3.333\times 10^5\ \mu\text{g/s}$.
Step 3: Use the centerline ground-level Gaussian plume formula (with image reflection).
For a completely mixed atmosphere and ground reflection, centerline $(y=0)$, ground level $(z=0)$: \[ C(x,0,0)=\frac{Q}{\pi\,u\,\sigma_y\,\sigma_z}\; \exp\!\left(-\frac{H^2}{2\sigma_z^2}\right), \] where $H=80\ \text{m}$ is the effective stack height and $u=3\ \text{m/s}$.
Step 4: Evaluate the exponential term.
\[ \frac{H^2}{2\sigma_z^2}=\frac{80^2}{2\times 35^2}=\frac{6400}{2450}=2.6122 \Rightarrow \exp(-2.6122)\approx 0.0733. \]
Step 5: Compute the denominator $\pi u \sigma_y \sigma_z$.
\[ \pi u \sigma_y \sigma_z=\pi\times 3 \times 40 \times 35 =\pi\times 4200 \approx 13194.69. \]
Step 6: Calculate concentration.
\[ C=\frac{3.333\times 10^5}{13194.69}\times 0.0733 =25.27\times 0.0733 =1.852\ \mu\text{g/m}^3. \] Final Answer: \[ \boxed{1.85\ \mu\text{g/m}^3} \]