Lecture series 5 - Civil and Environmental Engineering | SIU

Report
CE 510
Hazardous Waste Engineering
Department of Civil Engineering
Southern Illinois University Carbondale
Instructor: Jemil Yesuf
Dr. L.R. Chevalier
Lecture Series 5:
Partitioning
Course Goals
 Review the history and impact of environmental laws in the






United States
Understand the terminology, nomenclature, and
significance of properties of hazardous wastes and
hazardous materials
Develop strategies to find information of nomenclature,
transport and behavior, and toxicity for hazardous
compounds
Elucidate procedures for describing, assessing, and
sampling hazardous wastes at industrial facilities and
contaminated sites
Predict the behavior of hazardous chemicals in surface
impoundments, soils, groundwater and treatment systems
Assess the toxicity and risk associated with exposure to
hazardous chemicals
Apply scientific principles and process designs of hazardous
wastes management, remediation and treatment
Partitioning
AIR
WATER
WATER
SOLID
WATER
General phenomenon
that describes
tendency of
contaminants to exist
at equilibrium
between phases
BIOTIC
Partitioning
sorption
WATER
SOLID
Accumulation of contaminants on or in the solid
adsorption
absorption
Partitioning
sorption
WATER
SOLID
Sorbents include:
Soils
Sludges
Hazardous waste containment material (clay liner)
Granular activated carbon
In hazardous waste, the solvent is generally water
Factors of Sorption
 Properties of the sorbent, the sorbate and the liquid
media
 Sorbent


Hydrophobic
Specific surface area
 Sorbate



Hydrophobic
Low water solubility
Octanol-water partition coefficient, Kow
 Solvent


Mostly water
Air in emissions and unsaturated zone
Sorption Mechanisms
 Exchange adsorption (ion exchange)– electrostatic due to
charged sites on the surface. Adsorption goes up as ionic
charge goes up and as hydrated radius goes down.
 Physical adsorption: Van der Waals attraction between
adsorbate and adsorbent. The attraction is not fixed to a
specific site and the adsorbate is relatively free to move on
the surface. This is relatively weak, reversible, adsorption
capable of multilayer adsorption.
 Chemical adsorption: Some degree of chemical bonding
between adsorbate and adsorbent characterized by strong
attractiveness. Adsorbed molecules are not free to move
on the surface. There is a high degree of specificity and
typically a monolayer is formed. The process is seldom
reversible.
 Generally some combination of physical and chemical
adsorption is responsible for activated carbon adsorption.
Aqueous PhaseTransport in Soil
C
C
C Bd C  C 
 DL 2  vx



t
x
x  t  t  rxn
2
dispersion
*
advection
sorption
reaction
Bd = bulk density of the aquifer
Θ = volumetric moisture content (porosity for saturated soils)
C* = amount of solute sorbed per unit weight of solid
rxn = subscript indicating a biological or chemical reaction of the solute,
or radioactive decay (other than sorption)
Aqueous Phase Transport in Soil
C
 2C
C Bd C *  C 
 DL 2  vx



t
x
x  t  t  rxn
C
C
C
R
 DL 2  v x
t
x
x
2
Properties of Soil
 Pre-regulatory hazardous waste
disposal occurred in soils
 Need to review fundamental properties
of soil
 Compared to aqueous and air systems,
soils are heterogeneous composed of
solid, liquid and gaseous phases
Properties of Soil
Organic
5%
Air
20%
Mineral Water
45% 30%
Air
Water
Mineral
Organic
Properties of Soil
 Gravel

2.0-15 mm
 Sand

0.075-2.0 mm
 Silt

0.002-0.075 mm
 Clay

<0.002 mm
USDA Soil
Particle Size
Separates
Relationship between soil class and particle size
distribution
Soil Triangle
Calculator
Class Example
Determine the classification of the following:
22 % Sand , 42 % Silt , 36 % Clay
40 % Sand , 30 % Silt , 30 % Clay
60 % Sand , 30 % Silt , 10 % Clay
64 % Sand , 26 % Silt , 10 % Clay
46 % Sand , 30 % Silt , 24 % Clay
Class Example Solution
Soil Triangle
Calculator
Determine the classification of the following:
22 % Sand , 42 % Silt , 36 % Clay
Clay Loam
40 % Sand , 30 % Silt , 30 % Clay
Clay Loam
60 % Sand , 30 % Silt , 10 % Clay
Sandy Loam
64 % Sand , 26 % Silt , 10 % Clay
Sandy Loam
46 % Sand , 30 % Silt , 24 % Clay
Loam
How to Measure Sorption
 Batch experiments are
conducted to determine the
amount of solute removed


varying the concentration of
solute in solution
varying the amount of soil
 The capacity of a solid to
remove a solute is a function
of the concentration of the
solute
How to Measure Sorption
 If the adsorbent and adsorbate are contacted long
enough, an equilibrium will be established between
the amount of adsorbate adsorbed and the amount
of adsorbate in solution.
 The results of the experiment are plotted on a
graph, called an isotherm
 Equilibrium sorption isotherm

can be used when the sorption process is rapid compared to
the flow velocity
 Kinetic sorption isotherm

needed when the sorptive process is slow compared with
the rate of fluid flow
Equilibrium Sorption Isotherms
 Linear Sorption Isotherm
 Freundlich Sorption Isotherm
 Langmuir sorption Isotherm
Equilibrium Sorption Isotherms
Linear Sorption Isotherm
C  Kd C
C*
*
Kd
1
C
C* = mass of solute sorbed
per dry unit weight of solid
(mg/kg)
C = concentration of solute
in solution at equilibrium
with the solid (mg/L)
Kd = distribution coefficient
Linear Sorption Isotherm
Retardation Factor
Bd
1 Kd  R

The average velocity of the solute front where the
concentration is 1/2 of Co is designated as vc.
If the average linear velocity of groundwater is vx, vc can
be estimated as:
vx
vc 
R
Freundlich Sorption Isotherm
C  KC
*
N
If the sorption characteristics can be described by
the Freundlich sorption isotherm, then C*=f(C) will
be curvilinear, and
log C* = log K + N log C
will be linear. We can easily determine the
coefficients K and N through linear regression.
Freundlich Sorption Isotherm
log C* = log K + N log C
N>1
log C*
N=1
N<1
N
1
log K
log C
Freundlich Sorption Isotherm
Show that the retardation factor, R for
Freundlich Sorption Isotherm is given as:
BdKNC
R  1

N1
Langmuir Sorption Isotherm
Developed with the concept that a solid surface
possesses a finite number of sorption sites.
Once filled, the surface will no longer sorb solute
from solution

 C
C 
1  C
*
C*
C
β = adsorption constant related to the binding energy (L/mg)
α = maximum amount of solute that can be sorbed by the solid
(mg/kg)
Problem
If the sorption characteristics can be described by the
Langmuir sorption isotherm, then C/C*=f(C) is linear.
In addition, we can estimate a and b from this linear
plot. Prove this.
 C
C 
1  C
*
Solution
 C
C 
1  C
1  C
1
 *
 C C
*
1 1 1
1
  *
 C  C
1
1
C
 C *
 
C
Solution
1
1
C
 C *
 
C
C/C*
1
1


C
......end of example
Langmuir Sorption Isotherm
Linear transformation
 C
C 
1  C
*
C
1 C


*
C
 
1
1
1
 
*
C
  C

C
C  
C

Langmuir Sorption Isotherm
Bd   


R  1
2 
  1  C  
Text example
Five beakers are filled with one liter of 400 mg/L 2,4-D. 2 g of
soil are added to the first beaker, 3 g to the second, 4 g to the
third, and 5 g to the last beaker.. The fifth beaker is used as
control and no addition of soil. After letting the beakers reach
equilibrium, the following aqueous concentrations are observed:
Beaker
C of 2,4-D (mg/L)
1
289
2
234
3
179
4
124
Fit these data to the Langmuir and Freundlich isotherms and
determine the empirical constants for the best fit isotherm.
Spreadsheet
Text Problem 5.9
A soil evaluated for malathion sorption has been
found to follow the Langmuir isotherm with a=0.01
and b=4.5. A remediation team has proposed adding
200 kg (441 lb) of the soil to a 500,000 L (132,100 gal)
pond as an emergency response measure against a
spill with malathion concentration of 6 mg/L. If the
aqueous concentration needs to be less than 0.01
mg/L, will the process work?
Solution
Langmuir isotherm:
x
 C
C  
m 1  C
*
a = 0.01,
b = 4.5
Mass of contaminant sorbed from mass balance is:
x = (Co – C) (V) = (6-C)(mg/L)(500000)(L)(g/1000mg)
= (3000-500C) g malathion
The mass of the soil (m) = 200 kg = 200,000 g
Solution
From the Langmuir isotherm:
x
 C
*
C  
m 1  C
3000  500C (0.01)(4.5)(C )

200,000
(1  4.5C )
solving for C ,
C  2.33 mg / L  0.01 mg / L
Not effective!
......end of example
What is the minimum mass of soil (kg) that
would make the response effective?
Additional Comments: Kd
Bd
1 Kd  R

Simplest to use
can be estimated from empirical
equations based on



foc – fraction of organic carbon
Koc – soil adsorption coefficient
Kow – octanol-water coefficient
Let’s
consider
these in
more detail
 Linear Isotherm
 Linear portion of Freundlich
 For soils with a high organic content,
Octanol-Water Partition Coefficient, Kow
 A key parameter for estimating toxicity,
bioaccumulation, and sorption to soils and sediments
 Used for estimating contaminant pore water
concentrations in sediments, global transport of
persistent organic pollutants
octanol
contaminant
water
K ow
conc.in octanol

conc.in water
Range between 0.001 to over 100,000,000 (Table 5.7 p. 273)
Minimum Soil Organic Matter for
Sorption, f*oc
Sa
f 
0.84
200K ow 
*
oc
Kow = octanol water coefficient
Sa = surface area
Sample Range of values for Sa
Sandy loam soil 10-40 m2/g
Clay 120-250 m2/g
Soil Adsorption Coefficient,Koc
mass of contaminan t sorbed to soil organic carbon
K oc 
mass of contaminan t in aqueous phase
 L3 
mL
  typically
g
M 
remember that K d , soil distribution coefficient , is given as
mass of contaminan t sorbed to soil
Kd 
mass of contaminan t in aqueous phase
It follows that
K d  K OC . f OC
Range of KOC values p. 277 Table 5.8
Empirical equations based on Kow Table 5.9, p. 278
Estimate of Kd
If foc > f*oc
Kd = Kocfoc
Bd
1 Kd  R

Class Example
Determine R for the herbicide atrazine given the
following soil properties:
Bulk density: 1.4 g/cm3
Porosity: 0.3
Surface area: 8.2 m2/g
Soil organic carbon: 0.3%
Solution
1.
2.
3.
4.
Step one, estimate Kow = 2.68 from Table 5.7
Calculate f*oc = 0.018%
f*oc < 0.3%
From Table 5.8 , Koc = 175
K d  K oc f oc  (175)(0.003)  0.525
 1.4 
R  1
Kd  1 
0.525

 0.3 
 1  2.45  3.45
Bd
Text Problem 5.20
The following data have been collected for batch isotherm analysis of
hexachlorocyclopentadiene in well cuttings from a contaminated
ground water system.
Volume of aqueous solution used: 500 mL
Mass of well cuttings: 50 g
Organic carbon content of well cuttings: 0.2%
Co (mg/L)
0.4
1.8
3.7
5.1
7.6
9.2
11
12.4
C (mg/L)
0.02
0.04
0.05
0.1
1.2
3.8
4.9
5.3
Based on these data,
estimate Kd, and Koc for
the groundwater system.
Spreadsheet
Metals
 Rate of metals migration is also slower
than water
 Factors include




pH
Cation exchange capacity
Iron oxide content
Redox potential
Metals
 Kd from Table 5.11 p. 283
 Empirical estimate of velocity
VC
Co
V 
   A1 X1  A2 X 2  ......An X n  K 
 25 
Where
VC
= velocity at relative concentration evaluated
V = pore water velocity
Ai = regression coefficients
Xi = concentration of parameters
K = constant
Co
Example
A landfill leachate contains 10 mg/L Cd and
migrates with a pore water velocity of 4 cm/day.
The leachate contains 0.08% total suspended solids
(TSS) and 0.2% total organic carbon (TOC). The soil
composition is:
2% free iron oxide
8% clay
22% sand
70% silt
Determine the time required for a concentration of
5 mg/L Cd to migrate 50 meters.
Solution
Based on Table 5.13
Look down the center column for C/Co = 5/10 = 0.5
From here you get the values for the regression coefficients
Column 1 Power
-1
% Clay
% Sand
% Sand2
% FeO2
-1
% FeO
TSS
TSS2
TOC
Constant
-1
1
2
2
-1
1
2
1
Values in Coeff. For
Problem C/Co = .5
AiXi
8
22
22
2
2
0.08
0.08
0.2
29.91
-0.217
0.00108
0.0101
8.532
84.13
-205.1
0.442
-1.76
3.739
-4.774
0.523
0.040
4.266
6.730
-1.313
0.088
-1.760
SUM
7.540
spreadsheet
Solution
V0.5 = (4/25)(7.54) = 1.21 cm/d
The time to migrate 50 meters is
T= (5000cm)/(1.21 cm/d) = 4132 days = 11.3 years
Summary of Important Points
and Concepts
 Sorption is the accumulation of chemicals
onto surfaces from the surrounding solution
 Two most common models for sorption onto
soils and solids is the Langmuir and
Freundlich isotherms
 The octanol-water partition coefficient, Koc,
is the most commonly used contaminant
physiochemical property that correlates with
potential for sorption. Sorptivity is also
inversely related to water solubility
Summary of Important Points
and Concepts
 The soil distribution coefficient, Kd, is
defined as the ratio of the mass of sorbed
contaminant to the mass in the soil water.
 In soil-water systems containing significant
concentrations of organic carbon (i.e. > f*oc),
the organic carbon is the primary sorbent.
Koc may be used to describe the partitioning
 Empirical equations are available to
correlate Kow and Koc
Summary of Important Points
and Concepts
 Kd can be predicted from Koc and foc
 Retardation is predicted from



Kd
Bulk density
Porosity
 Metal sorption can also be predicted
through empirical equations

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