CHAPTER 5
Concentration Models:
Diffusion Model
Diffusion model
• Using the Gaussian plume idea.
• Consideration:
– The point source is the chimney or smoke
stack.
– One need to measure
concentration downwind
form the point source
• The Gaussian Plume.
Physical stack height = h
The plume rise = h
Effective stack height, H = h + h
Figure A
• The Gaussian Plume.
Assumptions:
• Wind blows in the x direction, with
velocity, u and emission rate, Q, and
it is independent
of time, location
or elevation.
Figure A
• Through material balance around a cube of space
near the center of the plume, and considering the
dispersion due to turbulent mixing:
z
x
y
Diffusion Model – Gaussian Plume
• Gaussian puff, 3D spreading
• Applicable to an instantaneous shot-term
release of pollutants from the chimney shown
in previous figure, i.e. at x = y = 0 and z = H
  1  x 2 y 2 z  H2
Qt
c
exp   


3/2
1/2

Kz
8t  K xK yKz 
  4 t  K x K y
where
• K = turbulent dispersion coefficient
• x = the distance upwind or downwind from the
center of the moving puff
• t = time since release
•  t = time duration of release




Diffusion Model – Gaussian Plume
• Gaussian plume, 2D spreading
• Applicable to steady-state release of plume.
• Assume negligible net transfer of material in the
x direction
2 
2


z  H 
Q /u
 1  y
c
exp   


3/2
1/2


Kz 
4t  K yKz 
  4 t  K y
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• The above equation is generally used by making
the following substitutions:
u
K y  0.5
x
2
K z  0.5 z
2
y
x
t
u
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Where:
y = horizontal dispersion coefficient
z = vertical dispersion coefficient
Diffusion Model – Gaussian Plume
• Making the substitutions, we find:
2 
2


z  H 
Q
y

c
exp


2
2


2π uσ y σ z
2σ z 
  2σ y
 y2 
 z  H2 
Q

c
exp  2  exp
2
 2σ y 
2π uσ y σ z
 2σ z 


• Basic 2D Gaussian Plume equation
Example 5:
• A factory emits 20 g/s of SO2 at height H. The
wind speed is 3 m/s. At a distance of 1 km
downwind, the values of σy and σz are 30 and 20
m, respectively.
What are the SO2 concentrations at the centerline
of the plume, and at a point 60 meters to the side
and 20 meters below the centerline?
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Solution
• At centreline, y = 0 and z = H (refer Fig. A). Thus,
at centreline:
20g / s
g
g
c
 0.00177 3  1770 3
23m / s 30m20m
m
m
• At the point away from the centreline,
 1  60m 2 1   20m 2 
20g / s
g
c
exp  
  
   145 3
23m / s 30m20m
2  20m  
m
 2  30m 
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Diffusion Model – Gaussian Plume
• The basic Gaussian plume equation predicts a
plume that is symmetrical with respect to y
and with respect to z.
• Different values of σy and σz mean that spreading
in the vertical and horizontal directions is not
equal.
• To find the approximated values for σ y and σ z ,
Diffusion Model – Gaussian Plume
Surface
Wind Speed
(at 10 m), m/s
Day
Night
Incoming Solar radiation
Thinly
Clear or
overcast or
 3/8 cloud
 4/8 cloud
Moderat
Strong
e
Slight
0–2
A
A–B
B
–
–
2–3
A–B
B
C
E
F
3–5
B
B–C
D
D
E
5–6
C
C–D
D
D
D
6
C
D
D
D
D
Note: The neutral class D should be assumed for overcast conditions
during day or night
• Horizontal dispersion coefficient
Horizontal
dispersion
coefficient, y, as
a function of
downwind
distance from the
source for various
stability
categories
• Vertical dispersion coefficient
Vertical dispersion
coefficient, z, as
a function of
downwind
distance from the
source for various
stability
categories
Diffusion Model – Gaussian Plume
Some modifications
The effect of the ground
• The ground damps out vertical dispersion and
vertical spreading terminates at ground level.
• Commonly assumed that any pollutants that
would have carried below z = 0 if the ground
were not there; are ‘reflected’ upward as if the
ground is a mirror
Diffusion Model – Gaussian Plume
Some modifications
• Therefore:

 y
Q

c
exp 0.5

2u y z

 y





2





2
2





 z H 
 z H 



   exp 0.5
 
 exp 0.5
z  
 z  










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Example 6:
• If z = 10 m, repeat the calculation in Example 5 for
the cases where H = 20 m and where H = 30 m.
Solution:
• H = 20 m
2

20


 60  
c
exp  0.5
 
233020

 30  


2
2







 10  20  
 10  20  
 exp  0.5
   exp  0.5
 


 20  
 20  





g
 289 3
m
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Solution:
• H = 30 m
2

20
60



 
c
exp  0.5
 
233020

 30  


2
2







 10  30  
 10  30  
 exp  0.5
   exp  0.5
 


 20  
 20  





g
 178 3
m
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Example 7
A large, poorly controlled copper smelter has a
stack 150 m high and a plume rise of 75 m. It is
currently emitting 1000 g/s SO2. Estimate the
ground level concentration of SO2 from this source
at a distance 5 km directly downwind when the
wind speed is 3 m/s and the stability class is C.
Solution
 y2
Q
c
exp 
2

2u σ y σ z
 2σ y
•
•
•
•
•
Q = 1000 g/s
u = 3 m/s
y = 438 m – from Figure 1
z = 264 m – from Figure 2
y = h + h = 225 m
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
  z  H2 
 exp  

2
 2σ 

 
z


Diffusion Model – Gaussian Plume
Ground level concentration, simplified
• At ground level, z = 0.
• Substituting into the previous equation:
2
 y  


Q
H
c
exp  0.5  exp  0.5 
 y  
u y  z
 z 
  
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2



Diffusion Model – Gaussian Plume
Ground level concentration, simplified
• At y = 0 and z = 0
 correspond to the line on the ground directly
under the centerline of the plume
2

H 
Q
c
exp  0.5  
u y z 
z  



H
cu
1
• Rearrange:

exp  0.5 
Q y  z
 z 
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2
Diffusion Model – Gaussian Plume
Ground level concentration, simplified
• We can plot a graph of cu/Q vs. distance x.
H
cu
1

exp  0.5 
Q y  z
 z 
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2
Ground-level
,
directly under the plume
centreline, as a function
of downwind distance
from the source an
effective stack height, H,
in meters, for stability
Class C only
Example 8
A plant is emitting 750 g/s of particulates. The stack
height is 100 m and the plume rise is 50 m. The
wind speed is 7 m/s and the stability category is C.
a) What is the maximum estimated ground-level
concentration ?
b) How far downwind it does occur?
Plume Rise
• Figure below shows the plume rising a
distance h, called the plume rise,
above the top of the stack before
leveling out.
Plume Rise
• Plumes rise buoyantly because they
are hotter than the surrounding air
and also because they exit the stack
with a vertical velocity that carries
them upward.
Plume Rise
• They stop rising because:
(i) they mix with surrounding air
(ii) they lose velocity
(iii) they cool by mixing
Plume Rise
•
To estimate h, Holland’s formula is:

VsD 
Ts  Ta  
3
 1.5  2.68x10 PD

h 
u 
Ts

where h
Vs
D
u
P
Ts
Ta
= plume rise, m
= stack exit velocity, m/s
= stack diameter, m
= wind speed, m/s
= pressure, mbar
= stack gas temperature, K
= atmospheric temperature, K
Example
• Estimate the plume rise for a 3 m diameter stack whose
exit gas has a velocity of 20 m/s when the wind velocity is
2 m/s, the pressure is 1 atm, and the stack and
surrounding temperatures are 100oC and 15oC,
respectively.
• Solution:
373  288 
20 x 3 
3
h 
 1.5  2.68 x 10 x 1013 x 3

2 
373

 101 m
End of Lecture