Advanced Geotechnical Engineering
ES4D8
Contaminated Land (Lecture 6)
Mohaddeseh Moudsavi-Nezhad
Room: D211
Email:m.mousavi-nezhad@warwick.ac.uk
31/05/2016
The University of Warwick
1
Risk Assessment
Whichever method is used, the user usually has the choice as to whether to
undertake a “forward” risk assessment, or a “backward” risk assessment.
Forward
SOURCE
RECEPTORS
• In a forward risk assessment, the user inputs concentrations of compounds
measured on the site (for example, in soils), and uses the model to predict the
(lower) concentration at a receptor.
• That concentration is then compared against an acceptable concentration.
Risk Assessment
Backward
SOURCE
RECEPTORS
• In a backward risk assessment, the user sets the acceptable concentration at
the receptor, and the model back-calculates what (higher) concentration can be
present at the source on the site (for example, in soil).
• So for the backward risk assessment, the modelling is completely independent
of the actual concentration in the soil at the site. The output of a backward risk
assessment is the acceptable concentration of the compound in the ground
Analytical Methods
Solution is achieved for simple problems
generally for contaminated land with;
• Single source
• Single receptor
• Specified boundary conditions,
• Solution is dependent to geometry of domain and
boundary condition (i.e., location of the sources and
receptors)
Numerical methods are used for complicated problems.
Finite Difference
Finite difference method is a well-established numerical method that has been
used extensively for approximating solutions to both fluid flow and transport
partial differential equations.
Transport equation for 1D problem
x
C(0, t) = C0
t>0
Initial condition
C(x, t) = 0
Finite Difference
j-1/2
j-1
j+1/2
j
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
Finite Difference
j-1/2
j-1
j+1/2
j
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
Finite Difference
j-1/2
j-1
j+1/2
j
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
u ¶C u C j+1/2 - C j-1/2
»
Rd ¶x Rd
Dx
Finite Difference
j-1/2
j-1
j+1/2
j
j+1
Dx
General formula for interface concentration
C j+1/2 = (1- a ) C j + aC j+1
a Is a spatial weighting factor. Most obvious choice for a is 0.5.
Finite Difference
j-1/2
j-1
j+1/2
j
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
Finite Difference
j-1/2
j+1/2
j
j-1
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
DH ¶ C DH
»
2
Rd ¶x
Rd
2
(¶C ¶x)
j+1/2
(
- ¶C
Dx
¶x
)
j-1/2
DH C j+1 - 2C j + C j-1
=
2
Rd
( Dx )
Finite Difference
j-1/2
j+1/2
j
j-1
j+1
Dx
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
n+1
n
C
C
¶C
j
» j
¶t
Dt
Finite Difference
n+1
n
¶C C j - C j
»
¶t
Dt
u ¶C u C j+1/2 - C j-1/2
»
Rd ¶x Rd
Dx
C j+1/2 = (1- a ) C j + aC j+1
DH ¶ C DH
»
2
Rd ¶x
Rd
2
(¶C ¶x)
j+1/2
(
- ¶C
Dx
¶x
)
j-1/2
DH C j+1 - 2C j + C j-1
=
2
Rd
( Dx )
Finite Difference
n+1
n
¶C C j - C j
»
¶t
Dt
¶C
u ¶C DH ¶2C
=¶t
Rd ¶x Rd ¶x 2
u ¶C u C j+1/2 - C j-1/2
»
Rd ¶x Rd
Dx
C j+1/2 = (1- a ) C j + aC j+1
DH ¶ C DH
»
2
Rd ¶x
Rd
2
(¶C ¶x)
j+1/2
(
- ¶C
Dx
¶x
)
j-1/2
DH C j+1 - 2C j + C j-1
=
2
Rd
( Dx )
Finite Difference
n
C n+1
C
u é
j
j
n
n
n
nù
=1a
C
+
a
C
1a
C
a
C
) j
) j-1
j+1 (
jû
ë(
Dt
Rd Dx
+
DH
Rd ( Dx )
n
n
n
C
2C
+
C
j
j-1 )
2 ( j+1
uDt é
n
n
n
nù
C =C 1a
C
+
a
C
1a
C
a
C
) j
) j-1
j+1 (
jû
ë(
Rd Dx
DH Dt
n
n
n
+
C
2C
+
C
j
j-1 )
2 ( j+1
Rd ( Dx )
n+1
j
n
j
Risk Assessment Software, ConSim
• Multiple contaminants in 1 model.
• Multiple sources and receptors in 1 model.
• Can model declining source.
• Probabilistic, includes sensitivity testing.
• Can model attenuation in the unsaturated zone.
• Can model dual porosity rocks.
• 2-D model.
• BUT only designed to do forward modelling.
• Not free. Need to buy it. £££.
• Needs time and experience to set up and run.
• Implements UK guidance.
http://www.consim.co.uk
Implements UK guidance
.
Implements UK guidance
.
Risk Assessment Software, ConSim
• Developed by Golder Associates in conjunction with the EA. Very commonly
used in the UK.
• It also implements the guidance within the Remedial Targets Methodology,
but is probabilistic (Monte Carlo simulation) and other added extras.
• It can model single values, or ranges of values which helps where there is
uncertainty with the input parameters.
• It can model attenuation during leaching in the unsaturated zone, so if there is
a large unsaturated zone (say 10-20m of unsaturated chalk or sandstone), then
the model allows the user to see how attenuation occurs over that distance.
• It can model multiple sources with different potential contaminants, migrating
to multiple receptor points.
• It can produce a visual prediction of contaminant plumes in groundwater.
• With the right input parameters, ConSim model can be set up in an hour to
two, depending on the experience of the assessor. ConSim allows the user to
model risks from both a soil and a groundwater source.
Risk Assessment Software,
FLOWPATH II
• Can model heterogeneity laterally.
• Steady state or time variant.
• Can be used to predict changes to
hydrogeological conditions with time.
• Good for more complex site settings.
• BUT not free. Need to buy it. £££.
• Only designed to do forward modelling.
• Considerable data input requirements.
• Longer timescale to setup, experience
needed to use.
http://www.scisoftware.com/environmental_
oftware/detailed_description.php?products_
=195
Risk Assessment Software,
FLOWPATH II
FLOWPATH II - Developed by Waterloo Hydrogeological.
• One of the most sophisticated 2-D modelling tools on the market.
• Rather than using single values or range of values to describe each parameter, the site
and surrounding area are positioned within a grid system.
• Each “cell” within the grid is assigned a different parameter value (if required), so
laterally, the model accounts for heterogeneity.
• However, the model is still a “single layer” model so will only model a single aquifer
type vertically.
• One of the key differences between FLOWPATH and ConSim is that there is a
definite split between groundwater flow and contaminant transport.
• The model must first be set up for the modelled area (which can be regional in scale)
to model groundwater flow, which is calibrated against known conditions (i.e. the
groundwater elevation contours, and water mass balance).
• Once the groundwater flow predicted by the model closely matches reality, the
contaminant transport is added to the modelling.
• So the model can take a week or more to set up. But, the model can be useful where
an understanding of interactions between contamination and receptors is required (e.g.
where is contamination likely to enter a river?) and if conditions are going to be
changed in the future (e.g. installing a remediation system which will disturb the
groundwater flow).
Risk Assessment Software,
Groundwater - MODFLOW
Visual MODFLOW - Developed by Waterloo
Hydrogeological
• Can model heterogeneity laterally and
vertically.
• Steady state or time variant.
• Can be used to predict changes to
hydrogeological conditions with time.
• Good for more complex site settings.
• BUT not free. Need to buy it.
• Only designed to do forward modelling.
• Considerable data input requirements.
• Longer timescale to setup, experience
needed to use.
Risk Assessment Software,
Groundwater - MODFLOW
• One of the most sophisticated 3-D modelling tools on the market.
• It is based on MODFLOW, developed by the USEPA. As for FLOWPATH,
rather than using single values or range of values to describe each parameter,
the site and surrounding area are positioned within a grid system. Each “cell”
within the grid is assigned a different parameter value (if required), so laterally,
the model accounts for heterogeneity. Additionally, the model can be set up to
model up to 5 layers vertically. The model must first be set up to model
groundwater flow, which is calibrated against known conditions (i.e. the
groundwater elevation contours, and water mass balance). Once the
groundwater flow predicted by the model closely matches reality, the
contaminant transport is added to the modelling. This model can take weeks to
months to set up for complex scenarios, and does not generate Site Specific
Assessment Criteria (Site Specific Assessment Criteria (SSAC)). But, the
model can be useful where an understanding of interactions between
contamination and receptors is required (e.g. where is contamination likely to
enter a river?) and if conditions are going to be changed in the future (e.g.
installing a remediation system which will disturb the groundwater flow).
Runge-Kutta
This method finds an approximate solution to an
ordinary differential equation.
We are going to use the fourth-order Runge-Kutta
formula.
Runge-Kutta
If dy
=
f(x,y) , with initial condition y(0) = y0
dx
Then
Where:
yn+1 = yn + 1/6 (k1 + 2k2 + 2k3 + k4)h
h = step size
k1 = f(xn, yn)
k2 = f( (xn + ½ h) , (yn + ½ h k1) )
k3 = f( (xn + ½ h) , (yn + ½ h k2) )
k4 = f( (xn + h) , (yn + h k3) )
Runge-Kutta
The k values are estimates of the gradient of the
function
dy /dx = f(x,y)
k1 ≈ gradient at start
k2 ≈ gradient at midpoint
based on gradient at k1
4
k3 ≈ gradient at midpoint
based on gradient at k2
k4 ≈ gradient at end based
on gradient at k3
3
k1 = f(xn, yn)
k2 = f( (xn + ½ h) , (yn + ½ h k1) )
2
k3 = f( (xn + ½ h) , (yn + ½ h k2) )
1
Xn
h
Xn+1
k4 = f( (xn + h) , (yn + h k3) )
Runge-Kutta - Example
A lake has been polluted with initial concentration of
pollutant at 106 parts/m3. Addition of fresh water
lowers the concentration of the pollutant, this is
governed by the equation:
dC/dt + 0.1C = 0
with C(0) = 106, time in days
Using the Runge-Kutta 4th order method find the
concentration after 10 days using a step size of 5 days.
Runge-Kutta - Example
dC/dt + 0.1C = 0
dy /dx
= f(x,y)
dC/dt = -0.1C
f(t,C) = -0.1C
h = step size = 5 days
Cn+1 = Cn + 1/6 (k1 + 2k2 + 2k3 + k4)h
Runge-Kutta - Example
Cn+1 = Cn + 1/6 (k1 + 2k2 + 2k3 + k4)h
f(t,C) = -0.1C
Step 1
n= 0
t0 = 0
C0 = 106
k1 = (f(t0, C0)) = f(0,106) = -0.1(106) = -105
k2 = f( (t0 + ½ h) , (C0 + ½ h k1) )=
k3 = f( (t0 + ½ h) , (C0 + ½ h k2) )=
k4 = f( (t0 + h) , (C0 + h k3) )
C1 =
Runge-Kutta - Example
Step 2
n= 1
t1 = 5
C1 =
k1 = (f(t1, C1)) =
k2 = f( (t1 + ½ h) , (C1 + ½ h k1) )=
k3 = f( (t1 + ½ h) , (C1 + ½ h k2) )=
k4 = f( (t1 + h) , (C1 + h k3) )=
C2 =
The approximate concentration of pollutant after 10 days is
368171 parts/m3
Runge-Kutta - Example
In this example we know the exact solution is given by
C(t) = 106 e(-0.1t). Below is a plot of the exact and
estimated solutions.
1200000
dC/dt + 0.1C = 0
1000000
800000
Exact values
600000
Runge-Kutta approximation
400000
200000
0
0
2
4
6
8
10
Runge-Kutta
To be continued next week.