Transcript Solutions to the Advection

Solutions to the
Advection-Dispersion
Equation
1
Road Map to Solutions
We will discuss the following solutions
Instantaneous injection in infinite and semi-infinite 1dimensional columns
Continuous injection into semi-infinite 1-D column
Instantaneous point source solution in twodimensions (line source in 3-D)
Instantaneous point source in 3-dimensions
Keep an eye on:
the initial assumptions
symmetry in space, asymmetry in time
2
Recall the Governing Equation
C
  3  u 3 c   3  D'  3c  c
t
What have we assumed thus far?
Dispersion can be expressed as a Fickian process
Diffusion and dispersion can be folded into a single
hydrodynamic dispersion
First order decay
What do we need next?
More Assumptions!
3
Adding Sorption
Thus far we have addressed only the solute behavior
in the liquid state.
We now add sorption using a linear isotherm
Recall the linear isotherm relationship
cs  k d cl
where cl and cs are in mass per volume of water and
mass per mass of solid respectively
The total concentration is then
C  b c s  cl
 b k d cl  cl
4
Retardation factor
We have
C  b k d   c l
Which may be written as
 b k d 
C  1 
c


Rc

l
l
 
where we have defined the retardation
factor R to be
bk d
R = 1

5
Putting this all together
The ADE with 1st order decay & linear isotherm
 ( Rcl )
   (u 3 cl )     ( D 3 ' )cl   Rcl  0
t
What do we need now? More assumptions!
 constant in space (pull from derivatives + cancel)
D’ constant in space (slide it out of derivative)
R constant in time (slide it out of derivative)
Use the chain rule:
  (u3cl )  u3  (cl )  cl   u3
the divergence of
a scalar =gradient;
 u3  (cl )
divergence of a
 u3cl
6
constant is zero
Applying the previous assumptions
The divergence operators turn into
gradient operators since they are applied
to scalar quantities.
What does this give us? The new ADE
cl u3
D3' 2
 cl 
 cl  cl  0
t R
R
7
Looking at 1-D case for a moment
2
d cl u d cl D L d cl


2   cl  0
dt R d x
R dx
To see how this retardation factor works, take
t* = t/R, and  = 0. With a little algebra,
2
c l
 cl
c l
* = DL
2 - u
t
x
x
The punch line:
the spatial distribution of solutes is the same in the
case of non-adsorbed vs. adsorbed compounds!
For a given boundary condition and time t*, the
solution is unique and independent of R
8
1-D infinite column Instantaneous Point Injection
Column goes to +and -
Area A
steady velocity u
mass M injected at x = o and t=0
(boundary condition)
initially uncontaminated column.
i.e. c(x,0) = 0 (initial condition)
linear sorption (retardation R)
first order decay ()
velocity u
x=0
 x  ut/ R2 
expt 
c L (x,t) 
exp
2AR Dt / R
 4Dt / R 
M
9
2

M
x  ut/ R 

expt 
c L (x,t) 
exp 
2AR Dt / R
 4Dt / R 
Features of solution:
Gaussian, symmetric in space, 2 = Dt/R
Exponential decay of pulse
Except for decay, R only shows up as t/R
0.30
Upstream
solutes
0.25
0.0 7
t = 0.1
Peak at 1.23 hr
t = 0.5
t= 2
0.0 6
t= 8
0.20
0.0 5
C onc en tration
C onc en tration
t= 4
0.15
0.10
0.0 4
0.0 3
Center
of Mass
2 hr
0.0 2
0.05
0.0 1
0.00
0.0 0
-5
0
5
10
Po siti on (m)
Spatial Distribution
15
0
2
4
6
8
10
Ti me
Temporal Distribution
10
Putting this in terms of Pore Volumes
•If solute transport is dominated by
advection and dispersion, then the
process is really only dependent on total
water displacement (the two transport
processes scale linerarly with time).
•This would not be true if there were
mass transfer between the mobile
solution and liquid within the particles, or
if diffusion were significant.
11
Pore Volume form of Solution
Want concentration at x=L, in vs. porevolumes passed through the system, P.
Key substitutions:
Coefficient n relates time and volume: nt=P
At P=1, ut=L. This implies that u=Ln.
This gives us:
  x  LP 2 
M

cL ( x , t ) 
exp 

4
DP
/
n
2A DP / n


Now how do we get the n out?
12
Pore Volume Form of Solution
  x  LP 2 
M

cL ( x , t ) 
exp 

4
DP
/
n
2A DP / n


Note that D/n = DL/u = luL/u = lL
With this the result is entirely written in terms
of the pore volumes and the media
dispersivity, as we desired!
 x  LP 
M
c L ( x, t ) 
exp  
2A l LP
 4 l LP
2




13
What about Retardation?
No Problem, just put it in as before:
2


M
x  LP / R  

c L ( x, t ) 
exp  

4

LP
/
R
2AR l LP / R
l


14
1-D semi-infinite Instantaneous Point Injection
How do we handle a surface application?
Use the linearity of the simplified ADE
Can add any two solutions, and still a solution
By uniqueness, any solution which satisfies initial
and boundary conditions is THE solution
Boundary and initial conditions
c(0,t)
0
x
(no flux at the soil surface)
C(x,0) = 0 (initially uncontaminated)
M(0,0) = M
15
Semi-infinite solution
  x  ut / R 2 


exp

4Dt / R 
M expt   
c L (x, t) 


2
2 AR Dt / R
 x  ut / R  



exp

4Dt / R 

Upward pulse and downward.
Only include region of x > 0 in domain
Solution symmetric about x = 0, therefore slope of
dc(x=0,t)/dx = 0 for all t, as required
Compared to infinite column, c starts twice as high,
but in time goes to same solution
16
Continuous injection, 1-D
Since the simplified ADE is linear, we use
superposition. Basically get a continuous
injection solution by adding infinitely many
infinitely small Gaussian plumes.
Use the complementary error function: erfc
x
2
 y2
e dy
erfc(x) = 1 

 0
erfc(x) = 1 - erf(x)
erf(-x) = -erf (x)
17
Plot of erfc(x)
2.00
1.80
1.60
erfc(x)
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
x
18
Solution for continuous injection, 1-D
  x  x  ut / R 

exp
erfc

•

m
 x   2   2 utR 
cl (x,t) 
exp 

2A u
2
x


x

ut

/
R
 

exp
erfc






2
2 utR 

  1  4R / u
 = lon gitu din al dispersiv ity
•
m  c o Au
19
Plot of solution
1.00
t =1
0.90
t = 10
t=20
0.80
t=30
t=40
Concentration (C/Co)
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0
10
20
30
40
50
Dis tance
• =1
 = 0.1, R = 1, = 0.02, u = 1.0, and m
20
1-D, Cont., simplified
With no sorption or degradation this reduces to
.
m   x  ut 
x   x  ut 
cl (x, t) 
erfc

exp
erfc
  2 ut 
2A u 
 2 ut 

21
2-D and 3-D instantaneous solutions
  x  ut / R 2
M
y2 
 exp t 
cL ( x , y , t ) 
exp 


4
D
t
/
R
4
D
t
/
R
4t Dl Dt
l
t


2
2
 x  ut / R 2
M expt 

y
z


c L (x, y,z,t) 
exp 


3/ 2
Dl Dy Dz
4Dy t / R 4Dz t / R 
4t  
 4Dl t / R
Note:
- Same Gaussian form as 1-D
- Note separation of longitudinal and
transverse dispersion
22
Review of Assumptions
Assumption
Effects if Violated
 constant in space
-R higher where lower
-Velocity varies inversely with 
-Increased overall dispersion due
to heterogeneity
- Plume will grow more slowly at
first, then faster.
- Increase plume spreading and
overall region of contamination
- Increased “tailing” and spreading
- Higher peak C and faster travel
- Stretching & smearing along beds
- Greater scale effects of D and ALL
EFFECTS DISCUSSED ABOVE 23
D constant in space
D independent of scale
Reversible Sorption
Equilibrium Sorption
Linear Sorption
Anisotropic media
Heterogeneous Media