Air Pollution

Modeling
CEV 301 Air Quality Management
Air Pollution Modelling
To build new facilities or to expand existing ones
without harming the environment, it is desirable to
assess the air pollution impact of a facility prior to its
construction, rather than construct and monitor to
determine the impact and whether it is necessary to
retrofit additional controls.

use of air quality simulation models

Source Atmosphere Receptor
Location Transport Location
Wind direction
Local circulations
Configuration Dilution Monitoring
Wind speed
Emission time and Dispersion Sensitivity
duration trubulance Plants, humans
Characteristics Transformation Removal
Height, volume flow, Chemical, physical Dry and wet deposition
temperature
 Transport and Dispersion of Air
Pollutants
• Wind Direction: The initial direction of transport of
pollutants from their source is determined mainly
by the wind direction at the source.

Wind Rose: how wind speed and

direction are typically distributed at a
particular location
• Wind speed: increases with height
• variation of wind speed in the surface boundary
layer:
u(z) = u(za)(z/za)p
• where u(z) is the wind speed at height z,
u(za) the wind speed at height za, and
p an exponent varying from about 0.1 to 0.4
(atmospheric stability, surface roughness, and
depth of the layer).
ESTIMATING CONCENTRATIONS
FROM POINT SOURCES
Gaussian Plume model:
• Steady-state conditions (constant emission source)
• assumes pollutant does not undergo chemical
reactions or other removal processes in traveling
away from the source and
• pollutant material reaching the ground or the top
of the mixing height as the plume grows is eddy-
reflected back toward the plume centerline
• flat surface topography
• no changes in wind direction and speed
 σy St. Dev. of horizontal distribution of plume conc., m
σz St. Dev. of vertical distribution of plume conc., m
z

y x
Mixing
Height, L
 𝑄 𝑦2 𝑧−𝐻 2 𝑧+𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

Lateral dispersion Horizontal dispersion Ground reflection

where
• C is the concentration (g/m3),
• Q is the emission rate of the pollutant from the source (g/s),
• u is the wind speed (m/s) which defines the direction x
• y is the horizontal distance perpendicular to the wind
direction,
• z is the vertical direction,
• H is the effective height of the plume (considering the
additional height ∆h to which the hot gases rise above the
physical height of the source h); i.e., H = h + ∆h, and
• σy & σz are the parameters of the normal distributions in y
and z directions, usually called the dispersion coefficients in
y and z directions respectively.
 Ground level concentration, plume at height H
(z=0)
𝑄 𝑦2 𝑧−𝐻 2 𝑧+𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

Rearrange:
𝑄 𝑦2 𝐻 2 𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

𝑄 𝑦2 𝐻2
𝐶 𝑥, 𝑦, 0 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧
 Ground level center line concentration, plume
at height H (y=0, z=0)
𝑄 𝑦2 𝑧−𝐻 2 𝑧+𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

Rearrange:
𝑄 𝐻 2 𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑧2 2𝜎𝑧2

𝑄 𝐻2
𝐶 𝑥, 0,0 = 𝑒𝑥𝑝 − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑧
 Ground level center line concentration, plume
at height H (y=0, z=0)
𝑄 𝑦2 𝑧−𝐻 2 𝑧+𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

Rearrange:
𝑄 𝐻 2 𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑧2 2𝜎𝑧2

𝑄 𝐻2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑧
 Ground level center line concentration (y=0,
z=0), ground level emission source (H=0)
𝑄 𝑦2 𝑧−𝐻 2 𝑧+𝐻 2
𝐶 𝑥, 𝑦, 𝑧 = 𝑒𝑥𝑝 − 2 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧2 2𝜎𝑧2

Rearrange:
𝑄
𝐶 𝑥, 𝑦, 𝑧 =
𝜋𝑢𝜎𝑦 𝜎𝑧
Max ground level conc. For a given x will occur
directly under the plume centerline (y=0 and
z=0)
𝑄 𝐻2
𝐶 𝑥, 0,0 = exp − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑧

If 𝜎𝑧 𝑎𝑛𝑑 𝜎𝑦 have a constant ratio as x varies,

then max ground level conc will be
2𝑄 𝜎𝑧
𝐶 𝑥, 𝑦, 𝑧 =
𝜋𝐻2 𝑒𝑢 𝜎𝑦
The distance to max concentration is at the
distance where σz = H/√2
• For unstable or neutral conditions, where σz < 1.6
L, where L is mixing height;
𝑄 𝑦2
𝐶 𝑥, 𝑦, 𝑧 = exp − 2
2𝜋𝑢𝜎𝑦 𝐿 2𝜎𝑦

• For large σz values, eddy reflection has occured

repeatedly both at the ground and at the mixing
height, so that vertical expense has been uniformly
mixed through the mixing height. So the conc
profile becomes almost uniform.
 • Max conc is obtained on the vertical centerline
plane (y=0) with the following equation
𝑄
𝐶 𝑥, 0, 𝑧 =
2𝜋𝑢𝜎𝑦 𝐿

• where σz > 1.6 L, then C will be the results of

multiple reflections
𝑄 𝑦2
𝐶 𝑥, 𝑦, 𝑧 = exp − 2 . 𝑔3
2𝜋𝑢𝜎𝑦 𝐿 2𝜎𝑦

where
∞
(𝑧 − 𝐻 + 2𝑁𝐿)2 , (𝑧 + 𝐻 + 2𝑁𝐿)2
𝑔3 = ෍ 𝑒𝑥𝑝 − + 𝑒𝑥𝑝 −
2𝜎𝑧2 2𝜎𝑧2
𝑁=−∞

This infinite series converges rapidly and evaluates with N varying from -4 to +4 is usually sufficient
Another special case is effluent source located at
ground level (H=0);

𝑄 𝑦2 𝑧2
𝐶 𝑥, 𝑦, 𝑧, 𝐻 = 0 = 𝑒𝑥𝑝 − 2 + 𝑒𝑥𝑝 − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦 2𝜎𝑧

For ground level conc. (z=0);

𝑄 𝑦2
𝐶 𝑥, 𝑦, 0, 𝐻 = 0 = 𝑒𝑥𝑝 − 2
𝜋𝑢𝜎𝑦 𝜎𝑧 2𝜎𝑦

Max conc at a given x is found on the centerline and

can be calculated by
𝑄
𝐶 𝑥, 0,0, 𝐻 = 0 =
𝜋𝑢𝜎𝑦 𝜎𝑧
Determination of Dispersion
Parameters, σy,σz
1. By Direct Measurements of Wind Fluctuations
Fluctuation measurements from a fixed wind system can
be used to calcuate plume disperison parameters.
To calculate plume dispersion directly from fluctuations
measurements, Draxler equations can be used:
𝜎𝑦 = 𝑥𝜎𝑎 𝑓𝑦
𝜎𝑧 = 𝑥𝜎𝑒 𝑓𝑧
• Where
σa is the standard deviation of the azimuth angles of a wind vane
over the sampling period τ calculated from average wind
directions averaged over averaging periods of duration s;
σ is the standard deviation of the elevation angle, in radians, over
e

the sampling period τ calculated from averaged elevation angles

over averaging periods s.
• Irwin has used simplified expressions for these
functions where both fy and fz have the form:

1
𝑓= 𝑇
(1+0.9 )
𝑇0

where travel time T is x/u;

T0 is 1000 for fy; T0 is 500 for fz for unstable (including
daytime neutral) conditions;
and
T0 is 50 for fz for stable (including nighttime neutral)
conditions.
2. By Classification of Wind Direction Traces
Where specialized fluctuation data are not available,
estimates of horizontal spreading can be approximated
from conventional wind direction traces. A method
suggested by Smith [2] and Singer and Smith [10] uses
classification of the wind direction trace to determine the
turbulence characteristics of the atmosphere, which are
then used to infer the dispersion. Five turbulence classes
are determined from inspection of the analog record of
wind direction over a period of 1 h. These classes are
defined in below Table.
3. By Classification of Atmospheric Stability
Pasquill categorized the atmospheric stability
according to wind speed, insolation and cloudiness
parameters.

Unstable
Most stable
σy horizontal dispersion parameter
σz vertical dispersion parameter estimation
x is distance in km
𝜎𝑦 = 𝑎𝑥 0.894

𝜎𝑧 = 𝑐𝑥 𝑑 + 𝑓
Example 1
Assume that a source releases 0.37 g/s of a pollutant
at an effective height of 40 m into the atmosphere
with the wind blowing at 2 m/s. What is the
approximate distance of the maximum concentration,
and what is the concentration at this point if the
atmosphere is appropriately represented by Pasquill
stability class B?
Example 2
An industrial site emitting H2S which has a pungent smell
like rutton eggs, at a rate of 10g/min at ground level.
Suppose that between 9:00-10:00 hours in the morning
there is a wind from the north at about 4 m/s. A slight
sensible heat flux has been measured. There is a
temperature inversion at 700 m and the surface
roughness of the area is 40 cm. What would the average
conc in air be during this hour at the boundary fence 200
metres to the south? Would the odor be detectable?
What difference would it make if the release was from a
30 m stack, assuming negligible heat and momentum for
the release? Note that the smell of H2S is detectable at
about 7 µg/m3.
Example 3
The odor threshold of H2S is about 7 µg/m3. If an
industry emits 0.8 g/s of H2S out of a 40 m stack
during an overcast night with a wind speed of 3 m/s,
estimate the area (interms of x and y coordinates)
where H2S would be detected.
Example 4
The concentration of H2S at a location 200 m
downward from an abondoned oil well is 3.2 mg/m3.
What is the emission rate, on a particularly overcast
afternoon if the wind speed is 2.5 m/s?
Example 5
A power plant burns 1000 tons of coal/day, 2% of
which is sulphur. This is emitted from the 100 m
stack. For a wind speed of 10 m/s, calculate
a)the max. GLC of SO2 10 km downwind from the
plant,
b)the max GLC and the point at which this occurs for
stability categories A, C and F.
Plume Rise
Gases leaving the tops of stacks rise higher than the stack top
• when they are either of lower density than the surrounding
air (buoyancy rise or thermal rise ,as T↑, ρ↓) or
• ejected at a velocity high enough to give the exit gases
upward kinetic energy (momentum rise).
 Stack Downwash
Stack tip downwash is affected with the stack gas exit
velocity
• When the stack gas exit velocity vs /u (the wind
speed u(m/s) at the stack top) ≥1.5,
h’=h (no stack downwash effect)
• If vs /u <1.5 ,
Vs <1.5 u, plume will sink
h’ = h+2d[(vs/u)-1.5],
where d is stack-top dimeter, m.
Vs/u >1.5 Vs/u<1.5
Buoyancy and Momentum Fluxes
Buoyancy Effect: Rise due to the temperature difference
between stack plume and ambient air.
𝑔𝑣𝑠 𝑑2 𝑇𝑠 − 𝑇 𝑔𝑄ℎ
𝐹= =
4𝑇𝑠 𝜋𝐶𝑝 𝜌𝑇

Momentum Rise: Rise due to exit velocity of the effluents

(emissions).
Buoyancy Flux Parameter
For most plume rise estimates, the value of the buoyancy flux parameter F
in m4s3 is needed:
𝑔𝑣𝑠 𝑑 2 𝑇𝑠 − 𝑇 𝑔𝑄ℎ
𝐹= =
4𝑇𝑠 𝜋𝐶𝑝 𝜌𝑇

𝐹 = 2.45𝑣𝑠 𝑑 2 𝑇𝑠 − 𝑇 /𝑇𝑠
where g is the acceleration due to gravity, about 9.806ms2,
Ts is the stack gas temperature in K,
T is ambient air temperature in K,
d is inside stack top diameter, m
𝑄ℎ is heat emission in watts,
𝐶𝑝 is specific heat at constant pressure, J/kg K
ρ is mass density of ambient air, kg/m3
Unstable or Neutral - Crossover
Between Momentum and Buoyancy
For cases with stack gas temperature greater than or
equal to ambient temperature, it must be determined
whether the plume rise is dominated by momentum or
buoyancy.
1/3
𝑉𝑠
For F <55, ∆𝑇𝑐 = 0.0297 𝑇𝑠 2/3
𝑑𝑠

2/3
𝑉𝑠
For F > 55, ∆𝑇𝑐 = 0.00575 𝑇𝑠 1/3
𝑑𝑠

If 𝑇𝑠 − 𝑇𝑎 ≥ ∆𝑇𝑐 ; plume rise is bouyancy dominated

𝑇𝑠 −𝑇𝑎 < ∆𝑇𝑐 ; plume rise is momentum dominated
Unstable-Neutral Buoyancy Plume Rise
The final effective plume height H, in m, is stack height plus plume
rise. Where buoyancy dominates, the horizontal distance xf from the
stack to where the final plume rise occurs is assumed to be at 3.5x*,
where x* is the horizontal distance, in km, at which atmospheric
turbulence begins to dominate entrainment.
3
21.425𝐹 4
For F < 55 , 𝐻 = ℎ′ + 𝑎𝑛𝑑 𝑥𝑓 = 0.049𝐹 5/8
𝑢
3
38.71𝐹 5
F ≥ 55 , 𝐻 = ℎ′ + 𝑎𝑛𝑑 𝑥𝑓 = 0.119𝐹 2/5
𝑢

Distance from the

stack where the final
plume rise occur
xf=3.5 x*
Unstable or Neutral - Momentum
Rise
For situations where the Ts ≤ Ta and 𝑇𝑠 − 𝑇𝑎 < ∆𝑇𝑐 ,
the plume rise is dominated by momentum.
The plume height is calculated from

′
𝑣𝑠
𝐻 = ℎ + 3𝑑𝑠
𝑢𝑠

𝑣𝑠
This equation is most applicable when >4.
𝑢𝑠
Stable Conditions
• Stability parameter s is calculated by

𝜕𝜃
𝑔
𝜕𝑧
𝑠=
𝑇𝑎
𝜕𝜃
Where, is the change in potential temperature with height.
𝜕𝑧
𝜕𝜃
For Stab. Class E, 𝜕𝑧 =0.020 K/m, and
𝜕𝜃
Stab, Class F, 𝜕𝑧 =0.035 K/m.
Stable - Crossover Between
Momentum and Buoyancy
For cases with Ts> Ta, it must be determined whether
the plume rise is dominated by momentum or
buoyancy.
If Ts-Ta ≥ ∆𝑇𝑐 plume is buoyancy dominated,
otherwise momentum dominated.

∆𝑇𝑐 = 0.019582𝑇𝑠 𝑣𝑠 𝑠
Stable – Buoyancy Rise
• Buoyancy plume rise at stable conditions:
𝐻 = ℎ′ + 2.6 𝐹/(𝑢𝑠) 1/3
𝑥𝑓 = 0.00207𝑢𝑠 −1/2

• Buoyancy plume rise at stable and calm (u<1 m/s)

conditions:
𝐻 = ℎ′ + 4𝐹1/4 𝑠 −3/8
Stable Momentum Plume Rise
For low-buoyancy plumes in stable conditions, plume
height due to momentum is given by

′ 1/3
𝐻 = ℎ + 1.5 𝐹𝑚 /(𝑢𝑠 𝑠)

Where
Gradual Rise: Buoyancy Conditions
• Plume rise for distances closer to the source than
the distance to the final rise can be estimated from

𝐻 = ℎ′ + 160𝐹1/3 𝑥 2/3 /𝑢
Where, 𝑥 is the source-to-receptor distance, km. If
this height exceeds the final effective plume heght,
that height should be substituted.
Example
Calculate the final plume rise from a power plant for
the following conditions:
Solution:
1) Deacon power law for calculating wind speed at stack height
u = u1 * (z/z1)p
Where,
u = desired but unknown wind speed, (us)
u1 = wind speed at known height, (u10)
z = height where wind speed is unknown, hs
z1 = height where wind speed is known, 10m
p = exponent from table 3-3 in the text = 0.15
Therefore, u = u1 * (z/z1)p = 4 * (67/10)0.15 = 5.3 m/sec
2) Check for downwash:
Vs / u >= 1.5 (downwash conditions need not be considered) = 19.0/5.3 = 3.571 >1.5 (therefore downwash need
not be considered)
Where, Vs = stack velocity in m/sec and u = wind speed at plume elevation
3) Calculate buoyancy flux parameter

Fb = g * vs * d2 * ΔT / (4 * Ts) = 9.81 * 19* 32 * (400 - 283) / (4 *400) = 123 m4/s3 (Fb > 55m4/s3)

4) Calculate temperature difference

ΔT = Ts - Ta = 400 - 283 = 117 K

5) Calculate cross over temperature difference (ΔT)c

for Fb > 55m4/s3

(ΔT)c = 0.00575 * Ts * vs 2/3 / ds 1/3 = 0.00575 * 400 * 19 2/3 / 3 1/3 = 11.4 K

6) Evaluate temperature differences

if ΔT > (ΔT)c plume rise is buoyancy dominated or else momentum dominated

Here, ΔT > (ΔT)c hence the plume rise is buoyancy dominated

7) Calculate final plume rise Δh

for Fb > 55m4/s3 Δh = 38.71 * (Fb3/5 / u ) = 38.71 * (1233/5 / 5.3) = 130m

8) Calculate final effective plume height H

H = 130 + 67 = 197m

This is less than the typical 300m night time inversion height; so plume rise may be
reasonably accurate.
 Example:

The following source data and meteorological data are given for a flare stack:

Meteorological Data are available from a nearby tower at a height 75 m. above the ground. Two cases are given:

Compute the plume rise for

both stabilities using above
information. State your
assumptions.
Solution:
1) Deacon power law for calculating wind speed at stack height (25m)
u = u1 * (z/z1)p
Where,
u = desired but unknown wind speed, (us)
u1 = wind speed at known height, (u10)
z = height where wind speed is unknown, hs
z1 = height where wind speed is known, 10m
p = exponent from table 3-3 in the text = 0.15
Therefore, stable u25 = u1 * (z/z1)p = 7.6 * (25/75)0.55 = 4.2 m/sec
unstable u25 = 8.9* (25/75)0.07 = 8.2m/sec
2) Check for downwash:

Vs / u >= 1.5 (downwash conditions need not be considered)

stable = 10.0/4.2 = 2.408 >1.5 (therefore downwash need not be considered)

unstable = 10.0/8.2 = 1.213 >1.5 (therefore downwash is to be considered)

Where,

Vs = stack velocity in m/sec

u = wind speed at plume elevation

3) Calculate reduced stack height hs'

hs' = hs + 2 * ds [(Vs / us - 1.5] = 25 + 2 * 0.2[(10 /8.2 - 1.5] = 24.9 m

Where,

hs = stack height

us = wind speed at top of the stack

4) Calculate buoyancy flux parameter

Fb = g * vs * d2 * ΔT / (4 * Ts) = 9.81 * 10* 0.152 * (1200 - 300) / (4 *1200) = 0.4134 m4/s3 (Fb < 55m4/s3)

5) Calculate temperature difference

ΔT = Ts - Ta = 1200 - 300 = 900 K

6) Calculate stability parameter, S

S = g * Δθ/Δz / Ta = 9.8 * 0.035 K/m / 300 K = 0.0011

7) Calculate cross over temperature difference (ΔT)c

for Fb < 55m4/s3

(ΔT)c = 0.019582 * Ts * vs * √S = 0.019582 * 1200 * 10 * √0.0011 = 7.95 K (stable case)

(ΔT)c = 0.0297 * Ts * vs 1/3 / ds 2/3 = 0.0297 * 1200 * 101/3 / 0.15 2/3 = 271.98 K (unstable case)
8) Evaluate temperature differences

if ΔT > (ΔT)c plume rise is buoyancy dominated or else momentum dominated

Here, ΔT > (ΔT)c for both the cases hence the plume rise is buoyancy dominated

9) Calculate stable plume height Δh

for Fb < 55m4/s3 Δh = 2.6 * (Fb / u S)1/3 = 2.6* ( 0.41 / 8.2 * 0.0011)1/3 = 9.2m

final effective plume height H

H = 9.2 + 25 = 34m

10) Calculate unstable plume height Δh

for Fb < 55m4/s3 Δh = 21.425 * (Fb3/4 / u ) = 21.425 * (0.413/4 / 4.2 ) = 2.7 m

final effective plume height H

H = 2.7 + 25 = 28m
Example

A paper mill has a plant in a valley, and it wants to build a stack to push the
plume centerline over the mountain so it can reduce the sulfur dioxide
concentration in its own valley. The mountain is 4 km away, and its elevation is
3400 ft. The valley floor, where the plant is, is at 1400 ft. If the emission
temperature is 200◦C, the prevailing temperature in the valley is 20◦C, the wind
velocity can be assumed at 2 m/s, and a prevailing lapse rate of 0.006◦C/m can
be assumed, what stack height must they build to achieve their objective?
 18.1
= 2.6 = 106 𝑚
2𝑥1.33𝐸 − 4

9.8 𝑥 3 𝑥 22 (200−20)
= 4(20+273)
= 18.1 𝑚4 /𝑠 3

9.81
= −0.006 + 0.01 = 1.33 𝑥 10−4
293
Example
A 10 MW power plant with a stack of 70 m continiously
releases effluents into the ambient atmosphere of
temperature 298 K. Stack velocity is 15 m/s.
a)Consider a neutral environment and an ambient wind of 20
km/hr. How high above the stack is the plume 200 m
downwind?
b)Consider a ground-based inversion 150m thick through
which the temperature changes 0.3°C. If the wind is blowing
at 10 km/hr will the effluent plume penetrate the ground-
based inversion? What if no wind is blowing?
c)Consider an elevated inversion between 100-1000m in
height through which the temperature changes 2°C. Will the
plume penerate the elevated inversion?
Example
3 gr/s of an air pollutants are emitted from a 12 meter stack
that has a diameter of 0.5 m. The exit velocity is 15 m/s and
the exit temperature is 315 K. The local environment is
considered rural.
a)On an afternoon when the Pasquill stability class is judged
to be B, what is the plume rise if the wind speed at stack top is
4 m/s? Assume that the ambient air temperature is 20 °C.
b)For these atmospheric conditions and the plume rise found
in part a, use Pasquill-Gifford, σy diagram to approximate the
distance to the maximum groundlevel conc and the max conc.
c) Using the distance to the max approximated in part b, find
the Pasquill-Gifford dispersion parameters from Table.
Determine max GLC along the plume centerline.
Questions
(1) Sulphur dioxide is emitted at a rate of 2kg/s from the top
of a chimney that is 120m high. The plume initially rises
vertically a further 10m above the chimney exit, before
being convected horizontally by a wind speed of 15m/s
under conditions of neutral stability. The surrounding
terrain is flat with a roughness length zo of 0.01m.
Calculate;
(a) The concentration (kg/m3 ) on the plume centre-line at a
distance of 800m downwind of the chimney.[Answer:
5311 µg/m3 ] (b)
(b) The ground level concentration at a distance of 800m
downwind of the chimney (that is, along the x-axis).
[Answer: 257 µg/m3 ]
(c) The location (x) where the maximum ground level
concentration occurs downwind of the chimney on the
xaxis. [Answer: 1542.3 m]
(d) The concentration at this location. [Answer: 1051
µg/m3)
 (2) A 100m tall chimney stack emits hydrogen chloride
(density = 1.64kg/m3 ) at a rate of 1m3 /s. The plume
initially rises a further 5m directly above the exit before
being convected horizontally by a wind blowing at a
speed of 10m/s under neutral atmospheric conditions.
The terrain has a roughness length of zo = 0.03m. A small
housing development commences at a location which is a
distance of x = 1500m downwind of the stack and y =
500m from the centre-line of the plume.
(a) What is the pollution concentration (kg/m3 ) at
ground level at the start of the housing development?
[Answer: 43 µg/m3 ]
(b) If the wind direction changed so that the plume axis
pointed directly towards the housing development,
what would be the new ground level concentration at
the same location as before ? (Assume that the wind
speed and the rate of emission remain unchanged).
[Answer: 1537 µg/m3 ]
 (3) A 100m tall chimney emits sulphur dioxide at a
rate of 2.5kg/s in a highly buoyant plume. The plume
rise (∆h) after exiting the chimney follows a
trajectory which is given by the equation: ∆h = 0.13
x2/3 (all dimensions in metres). The wind speed is
10m/s and the local ground roughness length is zo =
0.01m. What is the ground level concentration along
the x-axis at a distance of 2.5km down wind of the
chimney? [Answer: 1400 µg/m3 ]
 (4) A ground level accidental release of 0.5 g/s
occurs at a 4 m height in terrain with a roughness
length of zo = 0.01m. Assuming a wind speed of 1
m/s what is the maximum ground level
concentration 4 km downwind estimated from the
gross screening method and from the Gaussian
plume equation? [Answers: 25 µg/m3 , 0.56 µg/m3 ]