Transcript No Slide Title

Prepared by
Associate Prof. Dr. Mohamad Wijayanuddin Ali
Chemical Engineering Department
Universiti Teknologi Malaysia
page 0
The applicable conditions are -
Constant mass release rate, Qm = constant,
-
No wind, <uj> = 0,
-
Steady state, <C>/t = 0, and
-
Constant eddy diffusivity, Kj = K* in all directions.
For this case, Equation 9 reduces to the form,
2 C
x
2

2 C
y
2

2 C
z
2
0
(10)
page 1
Equation 10 is more tractable by defining a radius as r² = x² + y² + z².
Transforming Equation 10 in terms of r yields
d  2 d C 
r
0

dr 
dr 
(11)
For a continuous, steady state release, the concentration flux at any point,
r, from the origin must equal the release rate, Qm (with units of
mass/time). This is represented mathematically by the following flux
boundary condition.
d C
2
*
 4r K
 Qm
(12)
dr
The remaining boundary condition is
As r  , C  0
(13)
page 2
Equation 12 is separated and integrated between any point r and r =.

0
C
Qm
d C 
4K *


r
dr
r2
(14)
Solving Equation 14 for <C> yields,
C r  
Qm
4K * r
(15)
It is easy to verify by substitution that Equation 15 is also a solution to
Equation 11 and thus a solution to this case. Equation 15 is transformed
to rectangular coordinates to yield,
C x, y, z  
Qm
4K *
(16)
x2  y2  z 2
page 3
The applicable conditions are -
Puff release, instantaneous release of a fixed mass of material,
Qm* (with units of mass),
-
No wind, <uj> = 0, and
-
Constant eddy diffusivity, Kj = K*, in all directions.
Equation 9 reduces, for this case, to
2 C
2 C
2 C
1  C



*
2
2
t
K
x
y
z 2
(17)
page 4
The initial condition required to solve Equation 17 is
C x, y, z   0 at t  0
(18)
The solution to Equation 17 in spherical coordinates is
C r , t  
Qm*

8 K t
*

3
2

r2 

exp 
* 
 4K t 
(19)
and in rectangular coordinates is
C  x, y , z , t  

Qm*
8 K t
*

3

2

 x2  y2  z2 
exp

*
4
K
t


(20)
page 5
The applicable conditions are
-
Constant mass release rate, Qm = constant,
-
No wind, <uj> = 0, and
-
Constant eddy diffusivity, Kj = K* in all directions.
For this case, Equation 9 reduces to Equation 17 with initial condition,
Equation 18, and boundary condition, Equation 13. The solution is
found by integrating the instantaneous solution, Equation 19 or 20 with
respect to time. The result in spherical coordinates is

Qm
r

C r , t  
erfc

*
4K * r
2 K t




(21)
page 6
and in rectangular coordinates is
C x, y, z, t  

erfc

x2  y2  z 2

Qm
4K *
x 2  y 2  z 2  (22)

2 K *t

As t , Equations 21 and 22 reduce to the corresponding steady
state solutions, Equations 15 and 16.
page 7
This case is shown in Figure 7. The applicable conditions are
-
Continuous release, Qm = constant,
- Wind blowing in x direction only, <uj> = <ux> = u =
constant, and
-
Constant eddy diffusivity, Kj = K* in all directions.
For this case, Equation 9 reduces to
2 C
2 C
2 C
u  C



*
2
2
x
K
x
y
z 2
(23)
page 8
Equation 23 is solved together with boundary conditions, Equation12 and
13. The solution for the average concentration at any point is
Qm
C  x, y , z  
4K * x 2  y 2  z 2
(24)
u


2
2
2
x

z

y

x
exp 

*

 2K
If a slender plume is assumed (the plume is long and slender and is not far
removed from the x-axis),
y2  z 2  x2
(25)


and, using
1  a  1  a 2 , Equation 24 is simplified to
Qm
u

2
2 


C x, y, z  
exp

y

z
*
*

4K
 4K x

(26)
Along the centreline of this plume, y = z = 0 and
C x  
Qm
4K * x
(27)
page 9
This is the same as Case 2, but with eddy diffusivity a function of
direction. The applicable conditions are -
Puff release, Qm* = constant,
-
No wind, <uj> = 0, and
- Each coordinate direction has a different, but constant eddy
diffusivity, Kx, Ky and Kz.
Equation 9 reduces to the following equation for this case.
 C
t
 Kx
2 C
The solution is
C x, y, z, t  
8t 
3 2
x
2
 Ky
2 C
y
2
 Kz
2 C
(28)
z 2
2
2 
 1  x2
y
z

(29)
exp





Ky
K z 
KxKyKz
 4t  K x
page 10
Qm
This is the same as Case 4, but with eddy diffusivity a function of
direction. The applicable conditions are -
Puff release, Qm* = constant,
-
Steady state, <C>/t = o,
-
Wind blowing in x direction only, <uj> = <ux> = u = constant,
- Each coordinate direction has a different, but constant eddy
diffusivity, Kx, Ky and Kz, and
-
Slender plume approximation, Equation 25.
page 11
Equation 9 reduces to the following equation for this case.
u
 C
x
 Kx
2 C
x
2
 Ky
2 C
y
2
 Kz
(30)
2 C
z 2
The solution is
2
 u  y2
z

C x, y, z  
exp


Kz
4x K y K z
 4 x  K y
Along the centreline of this plume, y = z = 0 and the average
concentration is given by
Qm
C x  
Qm




(31)
(32)
4x K y K z
page 12
This is the same as Case 5, but with wind. Figure 8 shows the geometry.
The applicable conditions are -
Puff release, Qm* = constant,
-
Wind blowing in x direction only, <uj> = <ux> = u = constant, and
- Each coordinate direction has a different, but constant eddy
diffusivity, Kx, Ky and Kz,.
The solution to this problem is found by a simple transformation of
coordinates. The solution to Case 5 represents a puff fixed around the
release point.
page 13
If the puff moves with the wind along the x-axis, the solution to this case is
found by replacing the existing coordinate x by a new coordinate system, x
- ut, that moves with the wind velocity. The variable t is the time since the
release of the puff, and u is the wind velocity. The solution is simply
Equation 29, transformed into this new coordinate system.
C  x, y , z , t  
8t 
3 2
Qm*
KxKyKz
2

y2
z 2 
 1   x  ut 

exp




4t  K x
Ky
K z  



(33)
page 14
This is the same as Case 5, but with the source on the ground. The ground
represents an impervious boundary. As a result, the concentration is twice
the concentration as for Case 5. The solution is 2 times Equation 29.
C  x, y , z , t  
4t 
3 2
Qm*
KxKyKz
2
2 
 u  x2
y
z


exp


Ky
K z 
 4t  K x
(34)
page 15
This is the same as Case 6, but with the release source on the ground, as
shown in Figure 9. The ground represents an impervious boundary. As a
result, the concentration is twice the concentration as for Case 6. The
solution is 2 times Equation 31.
C x, y, z  
Qm
2x K x K y
2 
 u  y2
z


exp

K z 
 4 x  K y
(35)
page 16
Figure 9 Steady-state plume with source at ground level. The
concentration is twice the concentration of a plume without the ground.
page 17
For this case the ground acts as an impervious boundary at a distance H
from the source. The solution is

uy 2 

C  x, y , z  
exp 
4x K y K z
 4K z x 
Qm



u
u

2
2 
z  H r    exp
z  H r   
 exp

 4K z x

 4K z x


(36)
page 18
Cases 1 through 10 above all depend on the specification of a value for
the eddy diffusivity, Kj. In general, Kj changes with position, time, wind
velocity, and prevailing weather conditions. While the eddy diffusivity
approach is useful theoretically, it is not convenient experimentally and
does not provide a useful framework for correlation.
Sutton solved this difficulty by proposing the following definition for
a dispersion coefficient.
1
2
2 n
2
(37)
sx 
C ut 
2
with similar relations given for sy and sz. The dispersion coefficients,
sx, sy, and sz represent the standard deviations of the concentration in the
downwind, crosswind.
page 19
with similar relations given for sy and sz. The dispersion coefficients,
sx, sy, and sz represent the standard deviations of the concentration in the
downwind, crosswind, and vertical (x,y,z) directions, respectively. Values
for the dispersion coefficients are much easier to obtain experimentally
than eddy diffusivities.
The dispersion coefficients are a function of atmospheric conditions
and the distance downwind from the release. The atmospheric conditions
are classified according to 6 different stability classes shown in Table 2.
The stability classes depend on wind speed and quantity of sunlight.
During the day, increased wind speed results in greater atmospheric
stability, while at night the reverse is true. This is due to a change in
vertical temperature profiles from day to night.
page 20
The dispersion coefficients, sy and sz for a continuous source were
developed by Gifford and given in Figures 10 and 11, with the
corresponding correlation given in Table 3. Values for sx are not provided
since it is reasonable to assume sx = sy. The dispersion coefficients sy and
sz for a puff release are given in Figures 12 and 13. The puff dispersion
coefficients are based on limited data (shown in Table 3) and should not be
considered precise.
The equations for Cases 1 through 10 were rederived by Pasquill using
relations of the form of Equation 37. These equations, along with the
correlation for the dispersion coefficients are known as the PasquillGifford model.
page 21
Table 2 Atmospheric Stability Classes for Use with the PasquillGifford Dispersion Model
Day radiation intensity
Night cloud cover
Wind
speed (m/s)
Strong
Medium
Slight
<2
A
A–B
B
2–3
A–B
B
3–5
B
5–6
C
Cloudy
Calm &
clear
C
E
E
B–C
C
D
E
C–D
D
D
D
C
D
D
>6
C
D
Stability class for puff model :
A,B : unstable
C,D : neutral
E,F : stable
page 22
Figure 10 Horizontal dispersion coefficient for Pasquill-Gifford plume
model. The dispersion coefficient is a function of distance downwind and
the atmospheric stability class.
page 23
Figure 11 Vertical dispersion coefficient for Pasquill-Gifford plume
model. The dispersion coefficient is a function of distance downwind and
the atmospheric stability class.
page 24
Figure 12 Horizontal dispersion coefficient for puff model. This data is
based only on the data points shown and should not be considered reliable
at other distances.
page 25
Figure 13 Vertical dispersion coefficient for puff model. This data is
based only on the data points shown and should not be considered reliable
at other distances.
page 26
Table 3 Equations and data for Pasquill-Gifford Dispersion
Coefficients
Equations for continuous plumes
Stability class
sy (m)
A
sy = 0.493x
0.88
B
sy = 0.337x
0.88
C
sy = 0.195x
0.90
D
sy = 0.128x
0.90
E
sy = 0.091x
0.91
F
sy = 0.067x
0.90
page 27
Stability
class
x (m)
A
100 – 300
300 – 3000
sZ = 0.087x
log10sz = -1.67 + 0.902 log10x + 0.181(log10x)²
B
100 – 500
500 – 2 × 104
sZ = 0.135x
log10sz = -1.25 + 1.09 log10x + 0.0018(log10x)²
C
100 – 105
sZ = 0.112x
D
100 – 500
500 – 105
sZ = 0.093x
log10sz = -1.22 + 1.08 log10x - 0.061(log10x)²
E
100 – 500
500 – 105
sZ = 0.082x
log10sz = -1.19 + 1.04 log10x - 0.070(log10x)²
F
100 – 500
500 – 105
sZ = 0.057x
log10sz = -1.91 + 1.37 log10x - 0.119(log10x)²
sz (m)
0.88
0.95
0.91
0.85
0.82
0.80
page 28
Data for puff releases
Stability
condition
x = 100 m
x = 4000 m
sy (m)
sz (m)
sy (m)
sz (m)
Unstable
10
15
300
220
Neutral
4
3.8
120
50
Very stable
1.3
0.75
35
7
page 29
This case is identical to Case 7. The solution has a form similar to
Equation 33.
C  x, y , z , t  
2




x  ut
y2
z 2 
 1





exp




2
2 (38)


3 2
2  s x 
s y s z  
2 s xs ys z




Qm*
The ground level concentration is given at z = 0.
C  x, y ,0, t  
2


Q
y 2 
 1  x  ut 





exp



2 
 s

2
s
2 3 2s xs y s z

x

y 




*
m
(39)
page 30
The ground level concentration along the x-axis is given at y = z= 0.
C x,0,0, t  
 1  x  ut  2 
 
exp 
3 2
2 s xs ys z
 2  s x  
Qm*
(40)
The centre of the cloud is found at coordinates (ut,0,0). The
concentration at the centre of this moving cloud is given by
Qm*
C ut ,0,0, t  
(41)
3 2
2 s xs y s z
The total integrated dose, Dtid received by an individual standing at
fixed coordinates (x,y,z) is the time integral of the concentration.
Dtid x, y, z  


0
C x, y, z, t dt
(42)
page 31
The total integrated dose at ground level is found by integrating
Equation 39 according to Equation 42. The result is  1 y2 

Dtid x, y,0 
exp 
2
 2s 
s ys z u
y 

Qm*
(43)
The total integrated dose along the x-axis on the ground is
Dtid x,0,0 
Qm*
s ys z u
(44)
Frequently the cloud boundary defined by a fixed concentration is
required. The line connecting points of equal concentration around the
cloud boundary is called an isopleth.
page 32
For a specified concentration, <C>*, the isopleths at ground level are
determined by dividing the equation for the centreline concentration,
Equation 40, by the equation for the general ground level concentration,
Equation 39. This equation is solved directly for y.
y sy
 C x,0,0, t  

2 ln



C
x
,
y
,
0
,
t


(45)
The procedure is
1. Specify <C>*, u, and t.
2. Determine the concentrations, <C> (x,0,0,t), along the x-axis using
Equation40. Define the boundary of the cloud along the x-axis.
3. Set <C> (x,y,0,t) = <C>* in Equation 45 and determine the values
of y at each centreline point determined in step 2.
The procedure is repeated for each value of t required.
page 33
This case is identical to Case 9. The solution has a form similar to
Equation 35.
2 
 1  y2
z
C x, y, z  
exp  2  2 
s ys z u
 2  s y s z 
Q
(46)
The ground level concentration is given at z = 0.
2

 y  
Q
1
 
C  x, y ,0  
exp 
s y s z u
 2  s y  


(47)
page 34
The concentration along the centreline of the plume directly
downwind is given at y = z= 0.
C x,0,0 
Q
s ys z u
(48)
The isopleths are found using a procedure identical to the isopleth
procedure used for Case 1.
For continuous ground level releases the maximum concentration
occurs at the release point.
page 35