4. Atmospheric chemical transport models
4.1 Introduction
4.2 Box model
4.3 Three dimensional atmospheric chemical
transport model
4.1 Introduction
Questions
•
What is the contribution of source A to the concentration of
pollutants at site B?
•
What is the most cost-effective strategy for reducing pollutant
concentrations below an air quality standard?
•
What will be the effect on air quality of the addition of the reduction
of a specific air pollutant emission flux?
•
What should one place a future source to minimize its environmental
impacts?
•
What will be the air quality tomorrow or the day after?
The atmosphere is an extremely reactive system in which numerous
physical and chemical processes occur simultaneously.
Mathematical models provide the necessary framework for
integration of our understanding of individual atmospheric processes
and study of their interaction.
Three basic components of an atmospheric model are species
emission, transport and physiochemical transformations
① Eulerian model: describes the concentrations in an array of fixed
computational cells
② Lagrangian model: simulates concentration change of air parcel as
it is advected in the atmosphere.
출처: http://www.romair.eu/model-description.php?l
ang=en
출처: http://www.shodor.org/os411/courses/411f/module03/unit05/
page01.html
Classification based on dimension
① box model(상자 모델): zero-dimensional
Concentrations are functions of time only. C(t)
② column model: one-dimensional
Horizontally homogeneous layers
Concentrations are functions of height and time. C(z, t)
③ two dimensional model: often used in description of global
atmospheric chemistry
④ three dimensional model: c(x,y,z,t)
4.2 Box model (상자모델)
4.1.1 Eulerian box model
Assume that the height (H) of the box equal the mixing layer height.
d
0
(ci xyH )  Qi  Ri xyH  S i uHy (ci  ci )
dt
ci 0
Qi
Ri
Si
:
:
:
:
background concentration
mass emission rate kg/h-1
chemical production rate (kg m-3 h-1)
removal rate(dry deposition, wet deposition)
d
(c H ) = qi + Ri H
dt i
qi = Qi /( ΔxΔy )
si = S i′
/( ΔxΔy )
uH 0
(c
Δx i
g / m2 s
si +
ci )
1) For constant mixing height
dci qi
v
c  ci

 Ri  di ci  i
dt
H
H
r
0
r 
x
u
residence time
Problem 1
Ex1) An inert species has as initial concentration and is emitted at a
rate . Assuming that its background concentration is , calculate its
steady-state concentration over a city characterized by an average
wind speed of 3m/s. Assume that the city has dimensions and a
constant mixing height of 1000m.
2) For changing mixing height with time
dH
① dt  0 For decreasing mixing height
No direct change of the concentration inside the mixed layer
Because the box will be smaller, surface sources and sinks will
have a more significant effect.
dH
② dt  0 For increasing mixing height
Entrainment and subsequent dilution will change the
concentration.
c : the concentration above the box.
a
i
• Mass balance
(ci  ci )( H  H )  ci H  ci H
a
ci H  ci H  ci H  ci H  ci H  ci H
a
Neglecting
ci H
ci H  ci H  ci H
a
Hci  (ci  ci )H
a
ci H (ci  ci )

t
t
H
a
t  0
ci dH (ci  ci )

t
dt
H
a
dci
q
v
c  ci
 i  Ri  di ci  i
dt
H (t )
H (t )
r
0
for
dH
0
dt
dci
q
v
c  ci ci  ci dH
 i  Ri  di ci  i

dt
H (t )
H (t )
r
H (t ) dt
0
a
for
dH
0
dt
4.1.2 Lagrangian Box model
• No advection term
dci
q
v
 i  Ri  di ci
dt
H (t )
H (t )
for
dH
0
dt
dci
q
v
c  ci dH
 i  Ri  di ci  i
dt
H (t )
H (t )
H (t ) dt
a
ci (0)  ci
0
for
dH
0
dt
Problem 2
Ex 2) SO2 is emitted in an urban area with a flux of 2000 g m-3 .
The mixing height over the area is 1000m, the atmospheric
residence time 20h, and SO2 reacts with an average rate of 3 % h-1.
Rural areas around the city are characterized by a SO2 concentration
equal to 2 g m-3 . What is the average SO2 concentration in the
urban airshed for the above conditions? Assume an SO2 dry
deposition velocity of 1cms-1 and a cloud/ fog-free atmosphere
4.3 Three dimensional atmospheric chemical model
ci
c
c
c
c
c
c



 u x i  u y i  u z i  ( K xx i )  ( K yy i )  ( K zz i )
t
x
y
z x
x
y
y
z
z
 Ri (c1 , c 2 , )  Ei ( x, y, z , t )  S i ( x, y, z , t )
K xx , K yy , K zz
: eddy diffusivity
𝐸𝑖 : emission rate
𝑆𝑖 : removal flux
𝑅𝑖 : Reactionrate
Input data: three dimensional meteorological field, emission data, initial
and boundary condition of pollutants
Example of three dimensional model
RADM(Regional Acid Deposition Model)
UAM(Urban Airshed Model),
CMAQ (Community Multiscale air quality) Chemical transport model
4.2.1 Coordinate system
Terrain following coordinates
𝑧 − ℎ(𝑥, 𝑦)
𝜁=
𝐻𝑡 − ℎ(𝑥, 𝑦)
𝑝 − 𝑝𝑡
𝜎=
𝑃𝑠 − 𝑃𝑡
4.2.2 Initial conditions
Start atmospheric simulations some period of time.
At the end of start up period the model should have established
concentration fields that do not seriously reflect the initial conditions.
4.2.3 Boundary condition
Side boundary condition
a function of time.
𝑐 𝑥, 0, 𝑧, 𝑡
𝑐 𝑥, Δ𝑌, 𝑧, 𝑡
𝑐 0, 𝑦, 𝑧, 𝑡
𝑐 Δ𝑥, 𝑦, 𝑧, 𝑡
= 𝑐𝑥0 𝑥, 𝑧, 𝑡
= 𝑐𝑥1 𝑥, 𝑧, 𝑡
= 𝑐𝑥0 𝑦, 𝑧, 𝑡
= 𝑐𝑥0 𝑦, 𝑧, 𝑡
• Unlike initial condition, boundary conditions, especially at the
upwind boundaries, continue to affect predictions throughout the
simulations.
• Therefore, one should try to place the limits of the modeling
domain in relatively clean areas where boundary conditions are
relatively well know and have a relatively small effect on model
predictions.
• Uncertainty of side boundary conditions in urban air pollution
model prediction may be reduced by use of larger scale models
to provide the boundary condition to the urban scale model.
: nesting technique
Upper boundary condition
• Total reflection condition at the upper boundary of the
computation domain
𝜕𝑐
𝜕𝑧
• 𝐾𝑧𝑧 ( )𝑧=𝐻𝑡 = 0
• Alternative boundary condition (Reynolds et al., 1973)
• |𝑢𝑧 |𝑧=𝐻𝑡 𝑐𝑖 𝑎 − 𝑐𝑖 (𝑧 = 𝐻𝑡
𝜕𝑐
𝜕𝑧
• 𝐾𝑧𝑧 ( )𝑧=𝐻𝑡 = 0
𝜕𝑐
𝜕𝑧
= 𝐾𝑧𝑧 ( )𝑧=𝐻𝑡
for 𝑢𝑧 ≤ 0
for 𝑢𝑧 > 0
• 𝑐𝑖 𝑎 : the concentration above the modeling region.
Lower boundary condition
𝜕𝑐𝑖
(𝑣𝑑,𝑖 𝑐𝑖 − 𝐾𝑧𝑧
)𝑧=0 = 𝐸𝑖
𝜕𝑧
𝑣𝑑,𝑖 : deposition velocity of species
𝐸𝑖 : ground-level emission rate of the species.
4.2.4 Numerical solution of chemical transport models
•
𝜕𝑐𝑖
𝜕𝑡
𝜕𝑐𝑖
𝜕𝑐
𝜕𝑐
𝜕𝑐
𝜕𝑐
)𝑎𝑑𝑣 +( 𝑖 )𝑑𝑖𝑓𝑓 +( 𝑖 )𝑐𝑙𝑜𝑢𝑑 +( 𝑖 )𝑑𝑟𝑦 +( 𝑖 )𝑑𝑟𝑦 +𝑅𝑔𝑖
𝜕𝑡
𝜕𝑡
𝜕𝑡
𝜕𝑡
𝜕𝑡
=(
+ 𝐸𝑖
• 𝑐 𝑥, 𝑦, 𝑧; 𝑡 + ∆𝑡 = 𝑐 𝑥, 𝑦, 𝑧; 𝑡 + [𝐴 ∆𝑡 + 𝐷 ∆𝑡 + 𝐶( ∆𝑡 + 𝐺 ∆𝑡 +
𝑃 ∆𝑡 + S ∆𝑡 𝑐(𝑥, 𝑦, 𝑧; 𝑡)
•
•
•
•
•
•
A: Adevection operator
D: Diffusion operator
C: cloud operator
G: Gas-phase chemistry operator
P : Aerosol operator
S: source/sink operator
Operator splitting
• Instead of solving the full equation at once, basic idea is to solve
independently the pieces of the problem corresponding to the
various processes and then couple the various changes resulting
from the separate partial calculations.
∆𝑐 = 𝑐 𝑡 + ∆𝑡 − 𝑐(𝑡)
∆𝑐 𝐴 = 𝐴 ∆𝑡 𝑐 𝑡
∆𝑐 𝐷 = 𝐷 ∆𝑡 𝑐 𝑡
∆𝑐 𝐶 = 𝐶 ∆𝑡 𝑐 𝑡
∆𝑐 𝐺 = 𝐺 ∆𝑡 𝑐 𝑡
∆𝑐 𝑃 = 𝑃 ∆𝑡 𝑐 𝑡
∆𝑐 𝑆 = 𝑆 ∆𝑡 𝑐 𝑡
∆𝑐 = ∆𝑐 𝐴 + ∆𝑐 𝐷 + ∆𝑐 𝐶 + ∆𝑐 𝐺 + ∆𝑐 𝑃 + ∆𝑐 𝑆
𝑐 𝑡 + ∆𝑡 = 𝑐 𝑡 + ∆𝑐
• The other alternatives
𝑐1 𝑡 + ∆𝑡 = 𝐴 ∆𝑡 𝑐 𝑡
𝑐 2 𝑡 + ∆𝑡 = 𝐷 ∆𝑡 𝑐1 (𝑡 + ∆𝑡)
𝑐 3 𝑡 + ∆𝑡 = 𝐶 ∆𝑡 𝑐 2 (𝑡 + ∆𝑡)
𝑐 4 𝑡 + ∆𝑡 = 𝐺 ∆𝑡 𝑐 3 (𝑡 + ∆𝑡)
𝑐 5 𝑡 + ∆𝑡 = 𝑃 ∆𝑡 𝑐 4 (𝑡 + ∆𝑡)
𝑐 𝑡 + ∆𝑡 = 𝑆 ∆𝑡 𝑐 5 (𝑡 + ∆𝑡)
• Order of operator application is another issue.
• McRae et al. (1982a)
∆𝑡
∆𝑡
∆𝑡
∆𝑡
∆𝑡
∆𝑡
𝑐 𝑡 + ∆𝑡 = 𝑇𝑥
𝑇𝑦
𝑇𝑧
𝐺(∆𝑡)𝑇𝑧
𝑇𝑦
𝑇𝑥
𝑐(𝑡)
2
2
2
2
2
2
T: transport operator : advection and diffusion
Diffusion
∂ci
∂ci
∂ci
∂ci
∂
∂
∂
=
( K xx
) + ( K yy
) + ( K zz
)
∂t ∂x
∂x
∂y
∂y
∂z
∂z
𝜕𝑐
𝜕2𝑐
=𝐾 2
𝜕𝑡
𝜕𝑥
Crank-Nicholson algorithm
𝜕2𝑐
𝐾
=
[ 𝑐𝑖−1 𝑛 + 𝑐𝑖−1 𝑛+1 − 2 𝑐𝑖 𝑛+1 + 𝑐𝑖 𝑛 + 𝑐𝑖+1 𝑛 + 𝑐𝑖 𝑛+1 ]
2
2
𝜕𝑥
2∆𝑥
𝜕𝑐 (𝑐𝑖 𝑛+1 − 𝑐𝑖 𝑛 )
=
𝜕𝑡
∆𝑡
𝐾∆𝑡
𝐾∆𝑡
𝐾∆𝑡
𝑛+1
𝑛+1
−
𝑐𝑖−1
+ 1 + 2 𝑐𝑖
−
𝑐𝑖+1 𝑛+1
2
2
2∆𝑥
∆𝑥
2∆𝑥
𝐾∆𝑡
𝐾∆𝑡
𝐾∆𝑡
𝑛
𝑛
=
𝑐𝑖−1 + 1 − 2 𝑐𝑖 +
𝑐𝑖+1 𝑛
2
2
2∆𝑥
∆𝑥
2∆𝑥
Advection
•
𝜕𝑐
𝜕𝑡
•
𝜕𝑐
𝜕𝑡
•
𝜕𝑐
𝜕𝑥
+𝑢
𝜕𝑐
𝜕𝑥
=0
=
(𝑐𝑖 𝑛+1 −𝑐𝑖 𝑛 )
∆𝑡
=
(𝑐𝑖 𝑛 −𝑐𝑖−1 𝑛 )
∆𝑥
• 𝑐𝑖 𝑛+1 = 𝑐𝑖 𝑛 +
•
𝑢∆𝑡
∆𝑥
𝑢∆𝑡
(𝑐𝑖−1 𝑛
∆𝑥
≤ 1 : stable
− 𝑐𝑖 𝑛 ) : upwind finite difference scheme