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ISSN: 2347-3215 Volume 2 Number 5 (May-2014) pp. xx-xx
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The Mathematical Modelingof the Atmospheric Diffusion Equation
1
Khaled. S.M. Essa*, 2M.M.Abdel El-Wahab, 3H.M.ELsman, 4A.Sh.Soliman,
5
S.M.ELgmmal and 6A.A.Wheida
1
Department of Mathematics and Theoretical Physics, Nuclear Research Centre, Atomic
Energy Authority, Cairo, Egypt
2
Astronmy department, Faculty of science, Cairo university, Egypt
3,5
Physics department, Faculty of science, Monofia university, Egypt
4,6
Theoretical physics department, National Research Center, Cairo, Egypt
*Corresponding author email: [email protected]
KEYWORDS
A B S T R A C T
Advection Diffusion
Equation,
Laplace Transform,
Predicted Normalized
Crosswind Integrated
Concentrations
The advection diffusion equation (ADE) is solved in two directions to obtain
the crosswind integrated concentration. The solution is solved using Laplace
transformation technique and considering the wind speed depends on the
vertical height and eddy diffusivity depends on downwind and vertical
distances. We compared between the two predicted concentrations and
observed concentration data are taken on the Copenhagen in Denmark.
Introduction
The analytical solution of the atmospheric
diffusion equation has been containing
different shaped depending on Gaussian
and non- Gaussian solutions. An analytical
solution with power law for the wind speed
and eddy diffusivity with the realistic
assumption was studied by [1]the solution
has been implemented in the KAPPA-G
model [2],and [3] extended the solution of
[1] under boundary conditions suitable for
dry deposition at the ground. The
mathematics of atmospheric dispersion
modeling is studied by [4]. In the analytical
solutions of the diffusion-advection
equation, assuming constant along the
whole planetary boundary layer (PBL) or
following a power law was studied by [57], [2] and[8].
Estimating of crosswind integrated
Gaussian and non-Gaussian concentration
through different dispersion schemes is
studied by [9].Analytical solution of
diffusion equation in two dimensions using
two forms of eddy diffusivities is studied
by[9]. In this paper the advection diffusion
equation (ADE) is solved in two directions
to obtain crosswind integrated ground level
1
concentration in unstable conditions. We
use Laplace transformation technique and
considering the wind speed and eddy
diffusivity depends on the vertical height
and downwind distance. We compare
between observed data from Copenhagen
(Denmark) and predicted concentration
data using statistical technique.
(3)
Equation (3) is subjected to the following
boundary conditions
i. It is assumed that the pollutants are
reflected at the ground surface
Analytical Method
Time dependent advection
equation is written as [10]
diffusion
i.e.
ii. The flux at the top the mixing layer can
be given by
where:
C is the average concentration of air
pollution ( g/m3).
Where h is the
mixing height
iii. The mass continuity is written in the
form:-
u is the mean wind speed (m/s).
,
and
are the eddy diffusivities
coefficients along x, y and z axes
respectively (m2/s).
uC y ( x , z ) Q
z
hs at x
0 wher Q is
the source strength, is Dirac
function and hs is the stake height.
For steady state, taking
, and the
diffusion in the x-axis direction is assumed
to be zero compared with the advective in
the same directions, hence:
delta
The concentration of the pollutant tends to
zero at large distance of the source, i.e.
C y (x , z ) 0 at z
If L is the operator of the Laplace
transform then;
(2)
C y (s , z )
We must
e
sx
C y (x , z )dx Applying the
0
Laplace transform on equation (3) to have:
2
C y (s , z )
K
suC y (s , z )
uC y (x , z )
2
assumethat
,
integrating the equation (2) with respect to
y,we obtain the normalized crosswind
integrated
concentration
of
contaminant from - to at a point
of
the
atmospheric
advection diffusion
equation is written in the form [11];
z
Fromboundary condition (iii)this
equation becomes
2
C y (s , z )
K
suC y (s , z ) Q
2
z
2
z
hs
Dividing on K to obtain;
2
C y (s , z ) su
C y (s , z )
2
K
z
Q
K
hs (4)
z
Q
z hs
0 then;
K
2
C y (s , z ) su
C y (s , z )
2
K
z
To find particular solution put
0 (5)
The nonhomogeneous partial differential equation has a solution in the form of particular
solution plus complementary solution, particular solution has the following form:
C y (s , z ) C 1e
z
su
K
z
su
K
C 2e
z
su
K
(6)
And complementary solution;
C y (s , z ) C 1 z e
C2 z e
su
K
z
(7)
Using properties of partial differential equations[12]to obtain:
C1 z e
And
z
su
K
C2 z e
su
C1 z e
K
z
su
K
0 (8)
su
K
z
z
su
C2 z e
K
su
K
Q
z
hs (9)
su
to obtain
K
Multiplying equation (8) by
su
su
z
z
su
su
C1 z e K
C 2 z e K 0 (10)
K
K
Summing equations (9) and (10) to obtain
Q
e
su
2K
K
C2 z
hs
0
C 2 z dz
Then
Q
su
2K
K
C2 z
su
K
z
hs
0
e
Q
e
2 suK
hs by integrating
z
z
su
K
hs
z
hs dz
su
K
2
(11)
From equation (8)we can deduce that C 1 z e
Substituting by C 2 z then C 1 z
z
su
K
su
K
z
Q
e
2 suK
3
C2 z e
z
z
su
K
hs by integrating then
hs
Q
e
2 suK
C1 z
su
K
1 (12)
usingequations (11), (12) and equation (7) to obtain general solution of equation (4)
C y (s , z )
C y (s , z )
C y (s , z )
C y (s , z )
su
K
hs
Q
e
2 suK
1
Q
e
2 suK
0 when z
su
z hs
K
Q
e
2 suK
su
z hs
K
su
K
z
e
then
su
K
z
1e
Q
e
2 suK
Q
e
2 suK
hs
su
K
2
su
z hs
K
e
su
K
z
2e
z
su
K
(13)
0
2
su
K
z
e
1
(14)
Since L 1 is the operator of Laplace inverse transform [12]
L
1
L
1
e
a s
a2 4x
e
s
e
x
a
a s
2
x
3
e
a2 4x
L 1{C y (s , z )} C y x , z
Taking
C y (x , z )
Laplace
inverse
Q
e
2 uxK
u z hs
2
of
4 Kx
1
z
2x
equation
u
e
xK
z 2u 4 Kx
(14)
we
get
(15)
Differentiation equation (15) with respect to z then
C y (x , z )
z
C y (x , z )
z
z
Q
e
2 uxK
Qu z
4Kx
hs
uxK
e
u z hs
2
4 Kx
1
z
u z hs
2
4 Kx
uz 2
4Kx 2
1
z
2x
u
e
xK
u
e
xK
z 2u 4 Kx
z 2u 4 Kx
1
2x
u
e
xK
iplying both sides of equation (16) by K and using boundary condition (i)
C y (x , z )
at z=0then
K
0
z
2
Quhs
uK
1
0
e uhs 4 Kx
2x
x
4 x uxK
4
z 2u 4 Kx
(16)Mult
Q u h
4 x
s
u x K
2 x Q u h
1
1
4 x
Quhs
e
2uK
u h
e
2
s
4 K x
u K
x
1
2 x
x
e
u K
s
u x K
u h
2
s
4 K x
uhs 2 4 Kx
Substituting with in equation (14) to obtain
C y (x , z )
Q
e
2 uxK
C y (x , z )
Q
e
2 uxK
2
u z hs
u z hs
2
Quhs
e
2uK
4 Kx
Quzhs
4 Kx
2xK
Q
u z
e
2 uxK
1
C y (x , z ) Q
e
2 uxK
hs
C y (x , z )
2
4 Kx
u z hs
uhs 2 4 Kx
2
uxK
e
uzhs u z 2 hs 2
e
2xK
uzhs u z 2
4 Kx
e
2xK
z
2x
u
e
xK
u z 2 hs 2
z 2u 4 Kx
4 Kx
4 Kx
hs 2
4 Kx
(17)
In unstable case we take the value of the vertical eddy diffusivity in the
z
form K z
(18)
k vw z 1
h
Substituting from equation (18) in equation (3), we get that:
kvw z 1
Cy
z
h
u (z )
x
2
Cy
2
z
kvw
1
2z
h
u (z )
Cy
z
(19)
Applying the Laplace transform on equation (19) respect to x and considering that:
C y (s , z ) L p {C y x , z ; x
s}
C y (s , z )
Lp
C
x
0
e
sx
C y x , z dx
sC y s , z
C y 0, z (20)
5
WhereLp is the operator of the Laplace transform Substituting from (20) in equation (19), we
obtain that:
2
Cy s , z
z
2
2z
h
z2
z
h
1
Cy s , z
us
2
z
z
kvw* z
h
Cy s , z
u
Cy 0, z (21)
z2
kvw* z
h
Substituting from (ii) in equation (21) we get:2
Cy s, z
z
1
2z
h
2
z
z
Cy s, z
2
us
z
kvw* z
h
z
Q z hs
Cy s, z
2
kvw* z
h
z
2
(22)
h
After integrated equation (22) with respect to z, we obtain that:
Cy s,z
z h
z
Cy s,z
kv w *
u s ln
z
Q
(23)
hs
h
kvw *hs 1
Equation (23) is non homogeneous differential equation then, above equation has got two
solutions, one is homogeneous and other is special solution, in order to solve the homogeneous,
Q
we put,
=0 in equation (23),the solution becomes:
hs
k v w * hs 1
h
s u ln
C
k
s , z
y
c
Q
2
e
v
z
h
z
w
z
*
(24)
After taking Laplace transform in equation (24) and substitute from (ii), we obtain that: 1
(25)
c2
z hs
us
Substituting from equation (25) in equation (24) we get that:hs h
hs
kvw *
s u ln
C
y
s,z
Q
1
e
u s
z
(26)
6
The special solution of equation (23) becomes:
s u ln
C y s,z
Q
h
z
kvw *
1
k v w hs
z
hs
h
z
e
(27)
1
Then, the general solution of equation (23) is as follows:hs h
hs
z
kvw *
z h
z
z
kvw *
s u ln
Cy s,z
Q
1
e
us
s u ln
1
h
kvw hs s 1
h
Taking Laplace inverse of equation (28), we get that:C y x ,z
1
Q
h h
u ln s
h
hs
k v w hs s
u x
h
kv w
e
(28)
1
u ln
1
x
A L
1
z
(29)
h
z
kv w
Since:
L
1
, L
AB
1
exp
L
1
as
B
1
,L
and L
x a
L-1is the operator of the Laplace inverse transform by[13] (Shamus, 1980).
1
1
1
s
1
exp a s
1
x
a
To get the crosswind integrated ground level concentration, we put z=0 in equation (23), we
get that:
C y x ,0
1
1
(30)
hs
Q
hs h
k v w hs
1 x
u ln
h
hs
u x
kv w
Validation
The used data was observed from the atmospheric diffusion experiments conducted at the
northern part of Copenhagen, Denmark, under neutral and unstable conditions [13-14] table (1)
7
shows that the comparison between observed, predicted model 1 and predicted model
integrated crosswind ground level concentrations under unstable condition and downwind
distance.
Run
no.
Stability
Down
distance
(m)
Cy/Q *10-4 (s/m3)
1
1
2
2
Very unstable (A)
Very unstable (A)
Slightly unstable (C)
Slightly unstable (C)
1900
3700
2100
4200
6.48
2.31
5.38
2.95
Predicted model 1
K(x) = 0.16 ( w2/u) x.
4.469878
2.295811
3.135609
1.568679
3
3
3
5
5
5
6
6
6
7
7
7
8
8
8
9
9
9
Moderately unstable (B)
Moderately unstable (B)
Moderately unstable (B)
Slightly unstable (C)
Slightly unstable (C)
Slightly unstable (C)
Slightly unstable (C)
Slightly unstable (C)
Slightly unstable (C)
Moderately unstable (B)
Moderately unstable (B)
Moderately unstable (B)
Neutral (D)
Neutral (D)
Neutral (D)
Slightly unstable (C)
Slightly unstable (C)
Slightly unstable (C)
1900
3700
5400
2100
4200
6100
2000
4200
5900
2000
4100
5300
1900
3600
5300
2100
4200
6000
8.2
6.22
4.3
6.72
5.84
4.97
3.96
2.22
1.83
6.7
3.25
2.23
4.16
2.02
1.52
4.58
3.11
2.59
5.454451
2.802046
1.920065
7.136663
2.364105
1.627889
4.961762
1.261926
0.898436
2.647701
1.976309
1.52897
4.26145
2.719159
1.847384
4.657708
1.712648
1.199
Observed
Predicted model 2
K(z)=kv w* z (1-z /h)
5.01
2.62
4.36
2.26
5.01
2.61
1.80
4.50
2.27
1.57
4.35
2.21
1.60
4.57
2.32
1.81
4.89
2.68
1.85
4.34
2.26
1.60
Fig.1 the variation of the tow predicted models ad observed via downwind distance
8
Fig. (1) Shows that the predicted normalized crosswind integrated concentrations values of the
model 2are good to the observed data than the predicted of model 1.
Fig.2the variation between the two predicted models and observed concentrations.
Table (2) Comparison between our two models according to standard statistical Performance
measure
Models
Predicated model 1
Predicated model 2
NMSE
0.30
0.26
FB
-0.40
0.32
COR
0.78
0.67
FAC2
1.56
0.80
Fig. (2) Shows that the predicted data of model 2 is nearer to the observed concentrations data
than the predicted data of model 1.
From the above figures, we find that there are agreement between the predicted normalized
crosswind integrated concentrations of model 2 depends on the vertical height with the
observed normalized crosswind integrated concentrations than the predicted model 1 which
depends on the downwind distance.
Statistical method
Now, the statistical method is presented and comparison between predicted and observed
results will be offered by (Hanna, 1989).The following standard statistical performance
measures that characterize the agreement between prediction (Cp=Cpred/Q) and observations
(Co=Cobs/Q):
9
Where p and o are the standard deviations of Cp and Co respectively. Here the over bars
indicate the average over all measurements. A perfect model would have the following
idealized performance: NMSE = FB = 0 and COR =
Where p and o are the standard deviations
of Cp and Co respectively. Here the over
bars indicate the average over all
measurements. A perfect model would have
the following idealized performance:
NMSE = FB = 0 and COR = 1.0.
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diffusion
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Atmos.
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From the statistical method, we find that
the two models are inside a factor of two
with observed data. Regarding to NMSE
and FB, the predicted two models are good
with observed data. The correlation of
predicated model 1 equals (0.78) and model
2 equals (0.67).
Conclusions
We find that the predicted crosswind
integrated concentrations of the two models
are inside a factor of two with observed
concentration data. One finds that there is
agreement
between
the
predicted
normalized
crosswind
integrated
concentrations of model 2 depends on the
vertical height with the observed
normalized
crosswind
integrated
concentrations than the predicted model 1
which depends on the downwind distance.
10
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11

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