chapter 5 concentration models: diffusion model. diffusion model using the gaussian plume idea....
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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
Plume of contaminated air
• The Gaussian Plume.Physical stack height = hThe plume rise = hEffective stack height, H = h + h
Figure A
Plume of contaminated air
• The Gaussian Plume.
Assumptions:• Wind blows in the x direction, with
velocity, u and emission rate, Q, andit 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
z
2
y
2
x
2
2/1zyx
2/3 KHz
Ky
Kx
t41
expKKKt8
tQc
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
z
2
y
2
2/1zy
2/3 KHz
Ky
t41
expKKt4
u/Qc
• The above equation is generally used by making the following substitutions:
ux
t
5.0Kxu
5.0K
2zz
2yy
Where:y = horizontal dispersion coefficient z = vertical dispersion coefficient
Diffusion Model – Gaussian Plume
• Making the substitutions, we find:
• Basic 2D Gaussian Plume equation
2
z
2
2y
2
zy 2σHz
2σy
expσuσ 2π
Qc
2
z
2
2y
2
zy 2σHz
exp 2σy
expσuσ 2π
Qc
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?
Solution
• At centreline, y = 0 and z = H (refer Fig. A). Thus, at centreline:
• At the point away from the centreline,
33 mg
1770mg
00177.0m20m30s/m32
s/g20c
3
22
mg
145m20m20
21
m30m60
21
expm20m30s/m32
s/g20c
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 overcast or 4/8 cloud
Clear or 3/8 cloud
StrongModerat
eSlight
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:
2
z
2
z
2
yzy
Hz5.0exp
Hz5.0exp
y5.0exp
u2Q
c
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
3
22
2
mg
289
202010
5.0exp20
20105.0exp
3060
5.0exp203032
20c
Solution:
• H = 30 m
3
22
2
mg
178
203010
5.0exp20
30105.0exp
3060
5.0exp203032
20c
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
• 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
2
z
2
2y
2
zy σ2Hz
expσ2y
exp σ σ u2
Qc
Diffusion Model – Gaussian Plume Ground level concentration,
simplified• At ground level, z = 0.• Substituting into the previous equation:
2
z
2
yzy
H5.0exp
y5.0exp
uQ
c
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
• Rearrange:
2
zzy
H5.0exp
1Qcu
2
zzy
H5.0exp
uQ
c
Diffusion Model – Gaussian Plume Ground level concentration,
simplified• We can plot a graph of cu/Q vs. distance x.
2
zzy
H5.0exp
1Qcu
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
• To estimate h, Holland’s formula is:
where h = plume rise, mVs = stack exit velocity, m/sD = stack diameter, mu = wind speed, m/sP = pressure, mbarTs = stack gas temperature, KTa = atmospheric temperature, K
s
as3s
TTT
PD10x68.25.1uDV
h
Plume Rise
• 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:
m
x x x x
101373
288373310131068.25.1
2320
h 3
Example
End of Lecture