---
title: "Gaussian plume model and ground concentration"
subject: "Environment and Pollution Control"
chapter: 2
key: "semester_7/environment_and_pollution_control:gaussian-plume-ground-concentration"
source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/gaussian-plume-ground-concentration
site: BCT Engineering
---

# Gaussian plume model and ground concentration

Environment and Pollution Control · Chapter 2: Metrological aspects of air pollution dispersion

- Asked in: year not given, 3+5 marks
- Video: [Gaussian plume reflection and final equation](https://www.youtube.com/watch?v=WmTy4_MVTZc) — Linsey Marr
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/gaussian-plume-ground-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-2

## Question

(a) State the assumptions of the Gaussian plume model and write the equation for the concentration at ground level along the plume centreline. (b) A chimney emits SO2 at 80 g/s. The wind speed at the stack top is 5 m/s and the effective stack height is 60 m. At a downwind distance of 500 m, for the given stability class, sigma_y = 36 m and sigma_z = 18.5 m. Calculate the ground-level centreline concentration of SO2.

## Short answer

The Gaussian plume assumes steady continuous emission, constant wind, Gaussian spread, no reaction, and ground reflection. Ground-level centreline concentration is C = Q/(πuσyσz) exp(-H²/(2σz²)). For the data it gives 39.8 μg/m³ of SO2 at 500 m.

## Answer

#### (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$^3$).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is an assumption of the Gaussian plume model?
   A. Pollutant decays quickly in air
   B. Ground absorbs all pollutant that reaches it
   C. Steady continuous emission with Gaussian distribution in y and z, and total ground reflection
   D. Wind speed changes strongly with height
   Answer: C. The model assumes steady emission, constant wind, Gaussian spread, no reaction or deposition, and full ground reflection. (Key point: Model assumptions)

2. At z = 0 and y = 0 the concentration is:
   A. Q/(2 pi u sy sz) x exp(-H^2/2sz^2)
   B. Q/(pi u sy sz) x exp(-H^2/2sz^2)
   C. Q/(pi u sy sz) x exp(H^2/2sz^2)
   D. Q/(u sy sz)
   Answer: B. The reflection term doubles the value, so the factor becomes Q/(pi u sy sz) times exp(-H^2/(2 sz^2)). (Key point: Ground-level centreline equation)

3. For Q = 80 g/s, u = 5 m/s, sy = 36 m, sz = 18.5 m, Q/(pi u sy sz) is about:
   A. 7647 ug/m3
   B. 764.7 ug/m3
   C. 76 470 ug/m3
   D. 39.8 ug/m3
   Answer: A. 80 x 10^6 / (pi x 5 x 36 x 18.5) = 80 x 10^6 / 10 461, about 7647 ug/m3. (Key point: Pre-exponential term)

4. With exp(-60^2/(2 x 18.5^2)) = 0.0052, the ground-level concentration is about:
   A. 7647 ug/m3
   B. 3.98 ug/m3
   C. 39.8 ug/m3
   D. 398 ug/m3
   Answer: C. 7647 x 0.0052 = 39.8 ug/m3. (Key point: Final concentration)

## Related in Environment and Pollution Control

- [Maximum ground-level concentration from a stack](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Amaximum-ground-level-concentration) — Environment and Pollution Control, chapter 2
- [Types of plume behaviour from a stack](https://www.bctengineer.com/question.md?key=semester_7%2Fenvironment_and_pollution_control%3Aplume-behaviour-types) — Environment and Pollution Control, chapter 2