Gaussian plume model and ground concentration

(a) Assumptions

• Emission rate $Q$ is continuous and constant (steady state).
• Wind speed and direction are constant with height and time; the plume travels in the x-direction.
• Concentration distribution in the y (crosswind) and z (vertical) directions is Gaussian (normal).
• No chemical reaction, deposition or decay of the pollutant (conservative).
• The ground reflects the plume completely (an image source is used).
• Terrain is flat and the diffusion in the x direction is small compared with the wind transport.

General equation (point source, effective height $H$):

$$C(x,y,z) = \frac{Q}{2\pi u \sigma_y \sigma_z}\exp\left(-\frac{y^2}{2\sigma_y^2}\right)\left[\exp\left(-\frac{(z-H)^2}{2\sigma_z^2}\right) + \exp\left(-\frac{(z+H)^2}{2\sigma_z^2}\right)\right]$$

At ground level ($z=0$) on the centreline ($y=0$):

$$C(x,0,0) = \frac{Q}{\pi u \sigma_y \sigma_z}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Here $u$ is mean wind speed (m/s), $\sigma_y, \sigma_z$ are dispersion coefficients (m) that depend on stability and distance, $H$ is effective stack height $=$ physical height $+$ plume rise.

(b) Numerical

Data: $Q = 80$ g/s $= 80\times10^6\ \mu$g/s, $u = 5$ m/s, $H = 60$ m, $\sigma_y = 36$ m, $\sigma_z = 18.5$ m.

Pre-exponential term:

$$\frac{Q}{\pi u \sigma_y\sigma_z} = \frac{80\times10^6}{\pi \times 5 \times 36 \times 18.5} = \frac{80\times10^6}{10\,461} = 7647\ \mu\text{g/m}^3$$

Exponential term:

$$\exp\left(-\frac{60^2}{2\times18.5^2}\right) = \exp(-5.259) = 0.005199$$ $$C = 7647 \times 0.005199 = 39.8\ \mu\text{g/m}^3$$

Answer: Ground-level centreline concentration of SO$_2$ at 500 m is about 39.8 $\mu$g/m$^3$ ($3.98\times10^{-5}$ g/m$^3$).