Transcript Are Florida Landfill liner Systems Clogging and How Would

Subsurface Fate and
Transport of Contaminants
Contaminant Transport
Describes mechanisms to move
contaminants from source to receptor
Important to calculate dose in risk
analysis process
Solutes in saturated media can be
transported by three mechanisms
Diffusion
Advection
Dispersion
Diffusion
 Diffusion - spreading of solute due to
concentration gradient, minor at most GW
velocity
 Diffusion controlled mass transport occurs if
the concentration of a species is greater in
one location than an adjacent location
(concentration gradient)
 Fick’s second law used to calculate flux
Advection
Advection - transport with bulk flow of
groundwater at average velocity of GW.
Mass added to stream tubes remains in
stream tubes, other processes move
mass between stream tubes (diffusion,
dispersion,
Dominant transport mechanism
Dispersion
 Mechanical Dispersion - mechanical mixing
due to the velocity variations as groundwater
moves through tortuous pathways
 Solute transport by advection and dispersion
requires flow of groundwater to carry solutes
along with liquid flow
 At most fluid flows advection and dispersion
dominates over diffusion
Transformation Processes
Sorption
Radioactive decade
Chemical transformation
Volatilization
Colloid transport
Biotic
Contaminant Transport
Advection
Conc.
Sorption, Dispersion
Sorption, Dispersion
And Degradation
Time
Mathematics
Change in mass storage with time =
Mass inflow rate - mass outflow rate + mass
production rate
Mass Balance
qx
dz
dy
dx
Mass Balance - (1-Dimensional)
Change in mass storage with time = 
C
xyz
t
Mass in/out due to advection, dispersion, diffusion
and sources and sinks
Mass Balance (1-Dimensional)
Fadv / disp yz@ x  Fadv / disp yz@ xx  rxyz 
C

xyz
t
Divide by xyz and let x -------> 0
F = mass flux rate of contaminant due to
advection, dispersion, and diffusion, mass area-1 time-1
r = source or sink term
Advective Groundwater Flow
v = Darcy velocity, specific discharge (L/t)
= -K dh
dl
K = Hydraulic Conductivity
K = k

Q = vA
Groundwater Flow
Where:
k = intrinsic or physical
permeability of the porous medium, L2
 = dynamic viscosity of fluid, m/Lt
2 2
=
unit
weight
of
fluid,
m/L
t

 = g
3
 = mass density of fluid, m/L
g = gravitational acceleration, L/t2
vs = Seepage velocity, (L/t)
vs = v/ne
Source or sink terms
R (mass/vol-time):
r   C
r   C
q
r  b
t
Biotic
Radioactive Decay
Sorption
Solution Requires
Definition of parameters
Suitable numerical or analytical solution
Boundary and initial conditions
Analytical solution possible if 1-D,
source or sink terms linear, boundary
and initial conditions known
Solution to Mass Balance Including
Advection and Dispersion Only

 L  vxt 




v
L
L

v
t
C 
x 
  exp x erfc
C
erfc







2 
 DL 
 2 DL t 
 2 DL t 

C
Co
L
t
erfc
DL
vx
=
=
=
=
=
=
=
the concentration at time t and distance x
original concentration
distance
time
complimentary error function
dispersion coefficient
linear velocity
Solution to Mass Balance Including
Diffusion Only
c
 C
 Dd
2
t
x
2
Solution:
C ( x, t )  C erfc
C
Co
x
t
erfc
D*
=
=
=
=
=
=
x
2 D *t
the concentration at time t and distance x
original concentration
distance
time
complimentary error function
effective diffusion coefficient
Example – Diffusion Process
Assume Landfill A contains Na+1 = 10,000 mg/l
and Ca+2 = 5,000 mg/l and assume Landfill B
contains Fe+2 = 750 mg/l and Cr+3 = 600
mg/l. Landfill A has a 6 m clay liner under
the waste and Landfill B has 3 m of clay
under the waste. Assuming diffusion is the
only process affecting solute transport, which
of the four species will break through the clay
layer in either of the landfills first? How long
will that take?
Given
Waste
Clay
Waste
Clay
6m
Land fill B
Land fill A
Given: Effective diffusion coefficients D*:
Species
D* (m2/sec)
Na+1
1.33E-09
Ca+2
7.05E-10
Fe2+
7.19E-10
Cr3+
5.94E-10
3m
Solution
The erfc(z)* function has non-zero values
only at z values less than 3. To solve
this problem assume times and
calculate at edge of clay layer for each
case and keep changing the time until z
has a value of 3. WHY?
*Z=x/2((D*t)^0.5)
Assume Time is 5 years
C ( x, t )  C erfc
Species
Clay
layer (m)
Na+1
x
2 D *t
Conc (mg/l)
D*
(m2/sec)
Z=x/2((D*t)^0.5)
6
10000
1.33E-09
6.550992898
Ca+2
6
5000
7.05E-10
8.997842576
Fe2+
3
750
7.19E-10
4.454905626
Cr3+
3
600
5.94E-10
4.901282453
Z>3
Increase Time to 10 Years
Species
Clay layer
(m)
Concentration
(mg/l)
D* (m2/sec)
Z=x/2((D*t)^0.5)
Na+1
6
10000
1.33E-09
4.632251502
Ca+2
6
5000
7.05E-10
6.362435502
Fe2+
3
750
7.19E-10
3.150093978
Cr3+
3
600
5.94E-10
3.465730059
Z>3
Increase Time to 12 Years
Species
Clay layer
(m)
Concentration
(mg/l)
D* (m2/sec)
Z=x/2((D*t)^0.5)
Na+1
6
10000
1.33E-09
4.228647732
Ca+2
6
5000
7.05E-10
5.808082408
Fe2+
3
750
7.19E-10
2.875629216
Cr3+
3
600
5.94E-10
3.163764219
Fe+2 will break through first
Decrease Time to 11 Years
Species
Clay layer
(m)
Concentration
(mg/l)
D* (m2/sec)
Z=x/2((D*t)^0.5)
Na+1
6
10000
1.33E-09
4.416678511
Ca+2
6
5000
7.05E-10
6.066344227
Fe2+
3
750
7.19E-10
3.00349676
Cr3+
3
600
5.94E-10
3.304443956
Fe+2 will break through ~ 11 years
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Last updated July 16, 2015 by Dr. Reinhart