Turbulence and Dispersion of Contaminants in the Planetary Boundary Layer 57
By substituting Equation 3.82 in Equation 3.81 one obtains:
−
−
∂
∂
+ α ε
+ α ε
− ν
=
∂
∂
1 1 3 5/3
1 2 3 2/3
2
( , )
( , ) 5
( , ) 2
( , ) 0
3
E k t
E k t
k
k E k t
k E k t
t
k
(3.83)
Considering the following dimensionless parameters, in which w * is the convective
velocity scale and z i is the CBL height:
∗
∗
∗
ε
∗
ε
=
=
ψ =
ν
3
,
,
i
i
e
i
w t
w z
z
t
R
z
w
(3.84)
Equation 3.83 becomes
∗
∗
−
−
ε
ε
∗
∗
∗
∂
∂
′
′
+ α ψ
+ α ψ
−
=
′
′
′
′
′
∂
∂ ′
1 1/3
5/3
1 1/3
2 /3
2
( , )
( , ) 5
2
( )
( )
( , )
( ) ( , ) 0
3
e
E k t
E k t
k
k
E k t
k E k t
t
k
R
(3.85)
where k′ = kz i .
To solve Equation 3.85 the following changes of variables are considered: t → s
and k′ → m. We have in t = 0 (m,0,E(m)):
( ) 1
s t =
(3.86)
−
ε
= α ψ
′
′
1 13 5 3
( )
k s
k
(3.87)
−
ε
⎛
⎞
= − α ψ
+
′
′
⎜
⎟
⎝
⎠
1 13 2 3
2
5
2
( )
( )
3
e
Z s
k
k Z s
R
(3.88)
(0) 0
s =
(3.89)
(0)
k
m
′ =
(3.90)
( ,0)
( ,0)
Z m
E m
=
(3.91)
From Equation 3.86
d
1 or
d
s
s t
t
∗
∗
=
=
(3.92)
and from Equation 3.87
−
ε
′ = α ψ ′
1 13 5 3
d
d
k
k
s
(3.93)
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