Soil & GW Remediation – From Site Assessment to Final Closure (I)

Soil & GW Remediation –
From Site Assessment to
Final Closure (I)
ff Kuo, Ph.D., P.E.
uo@fullerton.edu
ttp://instructor.fullerton.edu/jkuo
1

Content
UST removal procedure
Remedial investigation
Groundwater and soil sampling
Characterization of vadoze zone
Characterization of groundwater
movement
Mass partition in different phases
Fate & transport of contaminants
Aquifer restoration
2

3
Content…
Soil remediation
technologies
GW remediation
technologies
Treatment of VOC-laden
air
Selection of remedial
technologies
Final closure
Modeling

UST Removal Procedure
Historical record
Pre-drilling
Permitting
• Fire department
• Local environmental agency
• Air quality management districts
Dig-alert
Sampling locations
• EPA methods (EPA 8015, EPA 8020, EPA 418.1,….)
• Certified labs
Additional excavation
Separation of contaminated soils
4

Remedial
Investigation
What chemicals are present?
Where, on the site, are these chemicals located?
Historical records
• What industrial activities occurred at the Site?
• What chemicals were used at the Site?
• Where were they used?
• Were chemicals stored in tanks? Where? Are
they still in the tanks?
How much of these chemicals were released?
• Records?
• Recollection of employees?
5

Common Remedial
Investigation (RI)
Activities
Removal of contamination source(s) such as
leaky USTs
Installation of soil borings (Boring?)
Installation of groundwater monitoring wells
Soil sample collection and analysis
Groundwater sample collection and analysis
Aquifer testing
6

Data Collected from RI
Types of contaminants present in soil and gw
Concentrations of contaminants in the collected
samples
Vertical and areal extents of contaminant plumes in
soil and gw
Vertical and areal extents of free-floating product or
the DNAPLs
Soil characteristics including the types of soil,
density, moisture content, etc.
Groundwater elevations
Drawdown data collected from aquifer tests
7

Groundwater and Soil Sampling
8

Soil Gas Survey
9

Soil Gas Survey
Good remote sensing tool for locating VOCs
Also used to defined gw plume.
For sandy formation, it may over-estimate the
problem.
Inexpensive and fast.
10

General Considerations
Types of Monitoring
Sampling Protocol
Types of Samples
Sampling Location
Sampling Frequency
Sampling Size, Bottles, Preservation
11

Types of Monitoring/Sampling
Detection Monitoring
• presence of contamination condition
• existing drinking water wells?
Assessment Monitoring
• extent and magnitude of contamination
Evaluation Monitoring
• data for remediation system design
Performance Monitoring
• evaluation of remediation effort
12

Sampling Protocol
Objectives
• Property transfer
• Environmental liability (establish baseline for
insurance/mortgage, clean-up)
Locations, frequencies, types, methods, H&S
Information needed for design a sampling plan
• site history (spill, operation, disposal practices)
• site geology and hydrogeology
Good documentation (always)
Leave room for evolutionary development
• collect more, analyze less
13

Types of Samples
QA/QC are important concerns!
Field Blank (p. 90 EPA manual)
• DI water from lab → field → sample vial→ lab
• Used to determine the limit of detection
Rinse/Cleaning Blank
• contaminated from previous sample?
Duplicate Samples
• collected as back-up
• soil samples are commonly duplicated.
Replicate Samples
• sub-samples of the same sample
• labeled separately to estimate lab precision
14

Types of Samples
Split Samples
QA/QC are important concerns!
• split in half in the field to two labs or
regulatory agency
Spiked Samples
• spiked w/a reference standard in the lab
• estimate recovery and matrix interference
Laboratory Blank
• DI water analyzed to determine lab
contamination
Standard Reference Samples
• detect instrument calibration error or analytical
15

16
Types of Samples ….
Travel Blank
Grab Sample vs. Composite Sample
• cost saving?
• field or lab? depth or area?
• time-weighted or volume/mass-weighted?
• be careful about VOCs
QA/QC are important concerns

Locations & Number
of Samples
Statistically based sampling plans
• Number of samples determined statistically.
• Location of samples determined by selecting
n sample points from m potential points at
random.
Judgmentally based sampling plans
• need some knowledge (soil gas survey,
surface geophysical technique, Hydropunch)
17

Locations & Number
of Samples…
Use US EPA’s Data Quality Objective Process
• Used by USEPA to ensure consistent national
approach to environmental sampling.
• Systematic, rational approach to determine
when, where, and how to collect samples or
measurements.
18

Locations & Number of Samples ...
Kuo’s suggestions/approach
Start from the center of source (if known)
Go outward for a specific distance
Another distance, if contaminated
Cut back half-way, if not contaminated
Vadose zone: every 5 to 10 ft
At the property line?
19

Sampling Frequency
Quarterly (most of time is OK) for GW sampling
• survey all wells quarterly, part of them
monthly
Intermittent or slug source
• need more frequent sampling
20

Sampling Size, Containers, and
Preservation Method
Large enough to be representative & for analysis
Bottles (EPA Table 9-14, p. 147)
• plastics (Teflon?)
• glass (transparent or amber)
• head-space free (40-ml VOA vial)
• metal
Preservation Method
Chain of Custody
Holding Time
On-site Measurement
21

23
adose Zone Sampling –
nalyte selection
Target Compounds
Subsurface
Chemistry
• CEC
Microbial
Geology
• grain size
distribution
• porosity, K

Vadose Zone Sampling –
Surface sampling
Hand Sampling (surface) –
scoops
Hand Sampling (subsurface)
• Hand Augers
• Limited sample depth
Direct Push Samplers
• From greater depths
• Allow stratified sampling.
• Hydraulic hammer
24

Vadose Zone Sampling –
On-site analysis
25
No direct reading instruments yet for contaminants
at needed low level detection limits.
On-site methods may be used in some cases.
Few methods approved by USEPA.
Limited to specific contaminants.
On-site mobile lab.
Use PID (photo-ionization detector), FID (flameionization
detector), OVA (organic vapor analyzer),
LEL (lower explosive limit) for screening.

26
adose Zone Sampling –
rilling
Auger
• Minimal damages to aquifer
• No drilling fluid required
• Flight acts as temporary
casing
• Good for unconsolidated
deposits
• Limited to 150 ft in depth
Rotary
Cable Tool

Vadose Zone Sampling –
Standard penetration test
A 2" dia, 18" length split core sampler is used in the
Standard Penetration Test (ASTM-D-1586-84), to
measure soil resistance to core penetration.
This test consists of driving the core sampler into
the soil by dropping a 140-lb hammer 30” in
repetitive blows. Using an 18" sampler, the number
of blows for three successive 6" penetration
increments are counted.
The number of blows corresponding to the first 6"
increment is discarded, while the blows for the
remaining two increments are combined and
reported as the number of blows per foot. 27

Standard penetration test
28

29
Amount of Cuttings from Soil Boring
1: Determine the diameter of the boring, d b.
2: Determine the depth of the boring, h.
3: Calculate the volume of the cutting using the
following formula:
Volume of cuttings
π 2
= ∑ ( db
)( h)( fluffy factor)
4

Groundwater Monitoring - Objectives
Groundwater level
Groundwater quality
• Target Compounds
• DO, pH, conductance
• Dissolved gas
• Fe, Mn, carbonates
• If TCE, then DCE, vinyl chloride
Presence of free-floating product
Aquifer testing
Extraction of groundwater
30

Groundwater
Wells - Design
31
Location
No. of wells
Diameter, Depth
Perforation
• interval, size, method
Packing materials
Permit
Drilling method
H & S

Groundwater Wells -
Construction Materials
Main Considerations
• Compatibility
• Cost
• Handling
• Strength
Choices (EPA Table 4-3, p. 44)
• PVC, PTFE, PP, PE
• Steel, SS
Flexible Joints
32

Groundwater Wells - Sampling
Wells need to be developed after installation
Well inspection
Purging
• Remove stagnant water
• 3-5 pore volumes (well casing + filter pack)
• May want to minimize purge water
• Monitor pH, conductivity, and temperature - until
stabilized.
• EPA Figure 9-9 and Table 9-8.
Sample collection
Field blanks/rinse blanks
Storage and transport
33

34
GW Sampling - Containers
Head space free
Minimize aeration and air contact
Dedicated for each well
No or little adsorption
Teflon, PP, PE: good
PVC, Tygon, silicon rubber - NG

GW Sampling - Devices
Bailers
• art, in-line measurement of
pH …. is impossible
• Not good for purging
(homogenize well volume
only)
Bladder Pumps
• cylinder w/ an internal bladder
that compress and expanded
by a gas.
• precise and pulseless flow.
Submersible Pumps
• deep well with high volume,
poor accuracy
35

GW Sampling - Hydropunch
The Hydropunch is pushed to the
desired depth and the push pipes
are retracted, exposing the
Hydropunch screen to the
groundwater.
The groundwater enters and is
allowed to come to equilibrium,
which generally takes less than 15
to 20 minutes.
One-time deal – no permanent well
installation.
36

Well Volumes for
Groundwater
Sampling
Well volume = volume of the groundwater
enclosed inside the well casing + volume of the
groundwater in the pore space of the packing.
π 2 π 2 2
Well volume = [ dc ] h+ [ ( d −d ) h]
φ
4 4
b c
37

38
Well Volumes for GW
Sampling - Example
The water depth inside one of the four
monitoring wells was measured to be 14.5’.
Three well volumes need to be purged out
before sampling. Calculate the amount of
purge water and also the number of 55-gallon
drums needed to store the water. Assume
the porosity of the well packing to be 0.40.

Well Volumes for GW Sampling - Example
π 2 π 2 2
Well volume = [ dc ] h+ [ ( d −d ) h]
φ
4 4
b c
) Well volume = (π/4)(4/12) 2
(14.5) + (π/4)[(10/12) 2 -
(4/12) 2 ](14.5)(0.4) = 13.92 ft 3
) 3 well volumes =
(3)(13.92)=41.8 ft 3 = 313
gallons
) Number of 55-gallon drums
needed= (41.8 ft 3 )(7.48
gal/ft 3 ) ÷ (55 gallon/drum) =
5.7 drums
39

Chemical Compounds of Concern
40

41
Chemical Compounds of Concern
Common examples of chlorinated solvents

Chemical Compounds of Concern
42
Common examples of aromatics
trachlorodibenzo-p-dioxin

43
Chemical Compounds of Concern
Others

Vadose Zone ….
Contaminants may be present in
• Vadose zone
• vapors in the void
• free product in the void
• dissolved in soil moisture
• adsorbed onto the soil matrix
• floating on top of the capillary fringe (for LNAPLs)
• Groundwater
• dissolved in the groundwater
• adsorbed onto the aquifer material
• sitting on top of the bedrock (for DNAPLs)
44

Mass & volume of soil excavated
during tank removal
Mass & volume of contaminated soil
left in thevadosezone
Mass of contaminants in the vadose
zone
Mass & volume of the free-floating
product
Volume of contaminated groundwater
Mass of contaminants in the aquifer
GW flow gradient and direction
Hydraulic conductivity of the aquifer
45
Common Engineering
Calculation Related to RI

Mass-concentration
Relationship (ppm)
Liquid: 1 ppm = 1/1,000,000 ~ 1 mg/L
1% by wt. = 10,000 ppm
Solid: 1 ppm = 1 mg/kg =1,000 ppb
Air: ppmV
1ppmV =
MW 224 .
[ mg/ m ] at 0 C
MW
=
2405 . mg m
3
[ / ] at
o
20 C
MW
=
245 . mg m
3
[ / ] at
o
25 C
3
o
6
1ppmV −
= × 10
359
3
[ lb/ ft ]
o
at 32 F
MW
−6 = × 10
385
3
[ lb/ ft ]
o
at 68 F
MW
−6 = × 10
392
3
[ lb/ ft ]
o
at 77 F
46

47
Mass-concentration
Relationship ….
Mass of contaminant in liquid =
(liquid volume)(liquid concentration)
= (V l
)(C)
Mass of contaminant in soil = (X)(M s
)
= (X)[(V s
)(ρ b
)]
Mass of contaminant in air =
(air volume)(concentration in mass/vol)
= (V a
)(G)

Mass-concentration Relationship (Example)
Which of the following media contains the largest
amount of xylene (show your calculations)?
(a) 1 million gallons of water containing 10 ppm of
xylene
(b) 100 cubic yards of soil (bulk density = 1.8
g/cm 3 ) with 10 ppm of xylene
(c) An empty warehouse (200' x 50' x 20') with 10
ppmV xylene in air.
48

Mass-concentration Relationship (Example)
(a) Mass of contaminant in liquid = (V)(C)
= (1,000,000 gallon)(3.785 L/gallon)(10 mg/L) = 3.79 x 10 7 mg
(b) Mass of contaminant in soil = (X)[(V s )(ρ b )]
= [(100 yd 3 )(27 ft 3 /yd 3 )(30.48cm/ft) 3 ][(1.8g/cm 3 )(kg/1000g)](10
mg/kg) = 1.37 x 10 6 mg
(c) Molecular weight of xylene [C 6 H 4 (CH 3 ) 2 ] = 106 g/mole
10 ppmV = (10)(MW of xylene/24.05) mg/m 3 = (10)(106/24.05)
mg/m 3 = 44.07 mg/m 3
Mass of contaminant in air = (V)(G)
= [(200 x 50 x 20 ft 3 )(0.3048m/ft) 3 ](44.07 mg/m 3 ) = 2.5x10 5 mg
49

ass & Volume of Soil
om a Tank Pit
xample)
Step 1: Measure the dimensions of the
tank pit.
Step 2: Calculate the volume of the tank
pit from the measured dimensions.
Step 3: Determine the number and
volumes of the USTs removed.
Step 4: Subtract the total volume of the
USTs from volume of the tank pit.
Step 5: Multiply the value from Step #4
50

Mass & Volume of
Soil from a Tank Pit
(Example)
(Example) Two 5,000-gallon USTs and one
4,000-gallon UST were removed. The
excavation resulted in a tank pit of 50’ x 24’ x
18’. The excavated soil was stockpiled on-site.
The bulk density of soil in-situ (before
excavation) = 1.8 g/cm 3 and bulk density of soil
in the stockpiles is 1.64 g/cm 3 .
Estimate the mass and volume of the excavated
soil.
51

Mass & Volume of Soil from
a Tank Pit (Example)
Volume of the tank pit = (50’)(24’)(18’) = 21,600 ft 3 .
Total volume of the USTs = (2)(5,000) + (1)(4,000)
= 14,000 gallons
= (14,000 gallon)(ft 3 /7.48 gallon) = 1,872 ft 3 .
Volume of soil in the tank pit before removal =
(volume of tank pit) - (volume of USTs)
= 21,600 - 1,872 = 18,728 ft 3 .
Volume of soil excavated (in the stockpile)
= (volume of soil in the tank pit) x (fluffy factor)
= (18,728)(1.10) = 20,600 ft 3
= (20,600 ft 3 )[yd 3 /27 ft 3 ]= 763 yd 3 .
52

Mass & Volume of Soil
from a Tank Pit
(Example)
Mass of soil excavated = (volume of the soil in the
tank pit)(bulk density of soil in-situ)
= (volume of the soil in the stockpile)(bulk density
of soil in the stockpile)
Soil density in situ =
(1.8 g/cm 3 )[(62.4lb/ft 3 )/(1g/cm 3 )]= 112 lb/ft 3 .
Soil density in stockpiles =
(1.64)(62.4) = 102 lb/ft 3 .
Mass of soil excavated = (18,728 ft 3 )(112 lb/ft 3 ) =
2,098,000 lb = 1,049 tons
= (20,600 ft 3 )(102 lb/ft 3 ) = 2,101,000 lb = 1,051
53

Volume of Soil in the Vadose
Zone (Example)
1. Determine area of the plume at each sampling depth, A i .
2. Determine the thickness interval for each area, h i.
3. Determine the volume of the contaminated soil, V s , using:
4. Determine the mass of the contaminated soil, M s , by
multiplying V s by the density of soil.
V = ∑ A i
h
(Example) After the USTs were removed, five soil borings were
installed., the area of the plume was determined as follow:
Depth (ft bgs) Area of the plume (ft 2 )
15 0
20 350
25 420
30 560
35 810
40 0
54
i

olume of Soil in the
adose Zone (Example)
V = ∑ A i
h
i
i
Thickness interval for each area is the same at 5 ft.
Volume of the contaminated soil
= (5)(350) + (5)(420) + (5)(560) + (5)(810)
= 10,700 ft 3 = 396 yd 3, or
= (22.5-17.5)(350) + (27.5 - 22.5)(420) + (32.5 -
27.5)(560) + (37.5 -32.5)(810) = 10,700 ft 3
Mass of the contaminated soil
3 3
55

Capillary Fringe
Capillary fringe (or capillary zone) is a
zone immediately above the water
table of unconfined aquifers.
It extends from the top of the water
table due to the capillary rise of water.
The capillary fringe often creates complications in site
remediation projects.
If the water table fluctuates, the capillary fringe will
move upward or downward with the water table.
If free-floating product exists, the fluctuation of the
water table will cause the free product moving away
56

57
Capillary Rise
Height of capillary rise
is a function of diameter
of capillary tube
h c1
h c2
h c3
h c4

Capillary Fringe …..
The height of capillary fringe at a site strongly
depends on its subsurface geology.
For pure water at 20 o C in a clean glass tube, the
height of capillary rise can be approximated by
the following equation:
h
c = 0153 .
r
where h c
is the height of capillary rise in cm, and
r is the radius of the capillary tube in cm. This
formula can be used to estimate the height of
the capillary fringe.
58

Mass/Volume of Free-Floating Product
It is now well-known that the thickness of free
product found in the formation (the actual thickness)
is much smaller than that floating on top of the
water in a monitoring well (the apparent thickness).
t = t( 1 − S ) − h
g g a
Ballestero et al. (1994):
• t g = actual (formation) free product thickness
t = apparent (wellbore) product thickness
S g = specific gravity of free product
h a = distance from bottom of the free product to the
water table. If no further data for h a are available,
average wetting capillary rise can be used as h a .
59

Thickness of
Free-Floating
Product
(Example) A recent survey of a GW monitoring well
showed a 75-in thick layer of gasoline floating on top
of the water. The density of gasoline is 0.8 g/cm 3
and the thickness of the capillary fringe above the
water table is one foot. Estimate the actual thickness
of the free-floating product in the formation.
Actual free product thickness
= (75)(1 - 0.8) - 12 = 3 inches
60

Mass/Volume of Free-
Floating Product
Step 1: Determine the areal extent of the free-floating
product.
Step 2: Determine the true thickness of the freefloating
product.
Step 3: Determine the volume of the free-floating
product by multiplying the area with the true
thickness and the porosity of the formation.
Step 4: Determine the mass of the free-floating
product by multiplying the volume with its density.
61

Mass/Volume of Free-
Floating Product
(Example) Recent groundwater monitoring results
at a contaminated site indicate the areal extent of
the free-floating product is approximately a
rectangular shape of 50 ft by 40 ft.
The true thicknesses of the free-floating product in
the four monitoring wells inside the plume are 2,
2.6, 2.8, and 3 ft, respectively.
The porosity of the subsurface is 0.35.
Estimate the mass and volume of the free-floating
product present at the site. Assume the specific
62

Mass/Volume of Free-Floating Product
(a) The areal extent of the free-floating product = (50')(40') =
2,000 ft 2
(b)The average thickness of the free-floating product
= (2 + 2.6 + 2.8 + 3)/4 = 2.6 ft
(c)The volume of the free-floating product
= (area)(thickness)(porosity of the formation)
= (2,000 ft 2 )(2.6 ft)(0.35) = 1,820 ft 3
= (1,820 ft 3 )(7.48 gal/ft 3 ) = 13,610 gallon
(d) Mass of the free-floating product
= (volume of the free-floating product)(density of the freefloating
product)
= (1,820 ft 3 ){0.8 g/cm 3 )[62.4 lb/ft 3 /(1 g/cm 3 )]}= 90,854 lb
= 41,300 kg.
63

Mass of
Contaminants
Present in
Different Phases
A NAPL enters subsurface, it may end up in 4 phases.
• leave the free product and enter the air void.
• dissolve in the liquid.
• adsorb onto the soil grains.
Concentrations in the void, in the soil moisture, and
on the soil grains are interrelated and affected greatly
by the presence or absence of the free product.
Partition of the contaminants in these four phases has
a great impact on the fate and transport of the
64

Mass of Contaminants
Present in Different
Phases…..
Understanding this partition is needed to implement
cost-effective alternatives for cleanup.
Here, we will discuss:
• vapor concentration resulting from the presence
of free-product in the pores
• relationship between the concentration in the
liquid and that in the air
• relationship between the concentration in the
liquid and that on the soil
• relationship among the liquid, vapor and solid
concentrations
65

Equilibrium between Free
Product and Vapor
For an ideal liquid mixture, the
vapor-liquid equilibrium follows
Raoult's law as:
P = ( P )( x )
A
vap
• P A
= partial pressure of A in the
vapor phase
• P vap = vapor pressure of A as a
pure liquid
• x A
= mole fraction of A in the
liquid phase
A
66

Equilibrium between
Free Product & Vapor ...
P = ( P )( x )
A
vap
A
The partial pressure is the pressure that a compound
would exert if all other gases were not present.
This is equivalent to the mole fraction of the
compound in the gas phase multiplied by the entire
pressure of the gas.
Raoult's law holds only for ideal solutions. In dilute
aqueous solutions commonly found in environmental
applications, Henry's law applies.
67

Equilibrium between
Free Product & Vapor ...
Benzene leaked into a vadose zone. Estimate
the maximum benzene concentration (in ppmV)
in the pore space. T = 25 o C.
• Vapor pressure = 95 mm-Hg at 25 o C=
0.125 atm.
• Partial pressure of benzene in the pore space
is 0.125 atm (125,000 x 10 -6 atm), which is
equivalent to 125,000 ppmV.
68

69
Equilibrium between Free Product &
Vapor ...
Discussion: The 125,000 ppmV is the vapor
concentration in equilibrium with the pure benzene
liquid. The equilibrium can occur in a confined
space or a stagnant phase.
If the medium is not totally confined, the vapor
tends to move away from the source and creates a
concentration gradient (the vapor concentration
decreases with the distance from the free liquid).
However, in the vicinity of the free product the
vapor concentration would be at or near this
equilibrium value.

Liquid-Vapor Equilibrium
Henry's law is used to describe the
equilibrium relationship between the
liquid and vapor concentrations.
At equilibrium, the partial pressure
of a gas above a liquid is
proportional to the concentration of
the chemical in the liquid.
P A
= H A
C A
• P A
= partial pressure of A in the
gas phase
• H A
= Henry's constant of A
• C A
= conc. of A in the liquid
70

Liquid-Vapor
Equilibrium…..
Henry's law can also be expressed as:
G = HC
• C = concentration in the liquid phase
• G = concentration in the gas phase.
The units of the Henry's law constant (or Henry's
constant) reported in the literature vary considerably
The units commonly encountered are: atm/mole
fraction, atm/M, M/atm, atm/(mg/L), and
dimensionless.
71

72
Liquid-Vapor Equilibrium…..
Henry's Constant Conversion Table (from Kuo & Cordery, 1988)
Desired unit for Henry's constant
Conversion equation
atm/M, or atm L/mole
H = H * RT
atm m 3 /mole
H = H * RT/1,000
M/atm
H = 1/(H * RT)
atm/(mole fraction in liquid), or atm
H = (H * RT)[1,000γ/W]
(mole fraction in vapor)/(mole fraction in liquid)
H = (H * RT)[1,000γ/W]/P
Note: H * = Henry's constant in dimensionless form
γ = specific gravity of the solution (1 for dilute solution)
W = equivalent molecular weight of solution (18 for dilute aqueous solution)
R = 0.082 atm/(K)(M)
T = system temperature in Kelvin
P = system pressure in atm (usually =1 atm)
M = solution molarity in (g mol/L)

Liquid-Vapor Equilibrium…..
Henry's constant of any given compound varies with
T. The Henry’s constant is practically the ratio of the
vapor pressure divided by solubility.
H = vapor pressure
solubility
The higher the vapor pressure the larger the Henry’s
constant is. The lower the solubility, the higher
Henry’s constant will be. For most organics, the
vapor pressure increases and the solubility decreases
with temperature.
73

74
Liquid-Vapor Equilibrium…..
(Example) Henry's constant for benzene in water
at 25 o C is 5.55 atm/M. Convert this value to
dimensionless units and also to units of atm.
H = H * RT = 5.55 = H * (0.082)(273 +25)
• H * = 0.227 (dimensionless)
H = (H * RT)[1,000γ/W]
• H=(0.227)(0.082)(273+25)][(1,000)(1)/(18)]
= 308 atm

Liquid-Vapor Equilibrium…..
(Example) The subsurface of a site is contaminated with
tetrachloroethylene (PCE). Recent soil vapor survey
indicates that the vapor contained 1,250 ppmV PCE.
Estimate the PCE concentration in the soil moisture.
Assume the subsurface temperature to be 20 o C.
H = H * RT = 25.9 = H * (0.082)(273 +20)
• H * = 1.08 (dimensionless)
Convert ppmV to mg/m 3
• 1,250 ppmV = (1,250)[(165.8/24.05)] mg/m 3
• = 8,620 mg/m 3 = 8.62 mg/L
G = HC = 8.62 mg/L = (1.08)C
75

Solid-Liquid Equilibrium
Adsorption is the process in which a component
moves from liquid/gas phase to solid phase
across the interfacial boundary.
• adsorbent (e.g., vadose zone soil, aquifer
matrix, and activated carbon)
• adsorbate (e.g., the contaminant)
• solvent (e.g., soil moisture and groundwater)
Adsorption is an important mechanism governing
the contaminant’s fate and transport in an
environmental medium.
76

Solid-Liquid Equilibrium….
For a system where solid phase and liquid phase
coexist, an adsorption isotherm describes the
equilibrium relationship between the liquid and
solid phases. The “isotherm” indicates that the
relationship is for a constant temperature.
Two most popular isotherms are the Langmuir
isotherm and the Freundlich isotherm. Both
were derived in early 1900s. The Langmuir
isotherm has a theoretical basis, while the
Freundlich is a semi-empirical relationship.
77

Solid-Liquid Equilibrium….
Langmuir isotherm:
X
=
X
max 1+
KC
KC
• X = sorbed concentration, C = liquid
concentration, K = equilibrium
constant, and X max
= the maximum
adsorbed concentration.
Freundlich isotherm:
1/
X = KC n
• Both K and 1/n are empirical
constants.
78

Solid-Liquid Equilibrium….
Both isotherms are non-linear. Incorporating
them into the mass balance equation will make
the computer simulation harder or more timeconsuming.
Fortunately, it was found that in many
environmental applications, the linear form of
the Freundlich isotherm applies. It is called the
linear adsorption isotherm, since 1/n =1, thus
X = KC
79

Solid-Liquid
Equilibrium….
For soil-water systems, the linear adsorption isotherm
is often written as:
X = K p
C; or K p
= X/C
• where K p
is called the partition coefficient that
measures the tendency of a chemical to be
adsorbed by soil or sediment from a liquid phase
and describes how the chemical compound
distributes (partitions) itself between the two media.
Henry’s constant, which was discussed earlier, can
be viewed as the vapor-liquid partition coefficient.
80

Solid-Liquid Equilibrium….
For a given organic chemical compound, the partition
coefficient is not the same for every soil.
It was found that K p
increases as the fraction of
organic carbon, f oc
, increases in soil, thus
K p
=f oc
K oc
The organic carbon partition coefficient, K oc
, can be
considered as the partition coefficient for the organic
compound into a hypothetical pure organic carbon
phase. For soil which is not 100% organics, the
81

82
Solid-Liquid Equilibrium….
K oc
is actually a theoretical parameter. Research
has been conducted to relate them to more
commonly available chemical properties such as
solubility in water (S w
) and the octanol-water
partition coefficient.
K ow
is a dimensionless constant:
Coctanol
K ow =
C
• C octanol
= conc. of an organic compound in octanol
• C water
water
= conc. of the organic compound in water

Solid-Liquid Equilibrium….
K ow
indicates how an organic compound will
partition between an organic phase and water.
Values of K ow
range widely, from 10 -3 to 10 7 .
Organic chemicals with low K ow
values are
hydrophilic and have low soil adsorption. Many
equations exist between K oc
and K ow
(or S w
)
The following is also commonly used:
K oc
= 0.63 K ow
83

Solid-Liquid Equilibrium….
(Example) The aquifer is contaminated with PCE. A
groundwater sample contains 200 ppb of PCE.
Estimate the PCE concentration adsorbed on the
aquifer material, which contains 1% of organic
carbon. Assume the adsorption follows a linear
model.
(a) Log K ow
= 2.6 => K ow
= 398
(b) K oc
=0.63K ow
=0.63(398)=251 mL/g=251 L/kg
(c) K p
=f oc
K oc
=(1%)(251)=2.51 mL/g=2.51 L/kg
(d) X = K p
C = (2.51 L/kg)(0.2 mg/L) = 0.50 mg/kg
84

Solid-Liquid-Vapor Equilibrium
The soil moisture in the vadose zone is in contact with
both soil grains and air in the void, and the
contaminant in each phase can travel to the other
phases. The concentration in the liquid, for example,
is affected by the concentrations in the other phases
(i.e., soil, vapor, and free product).
These concentrations are related by the equilibrium
equations. If the system is at equilibrium and the
concentration of one phase is known, the
concentrations at other phases can be estimated.
Although in real applications, the equilibrium condition
does not always exist, the estimate serves as a good
starting point or as the range of the real values.
85

Partition of Contaminants
in Different Phases
M
V
t
= [( φ ) + ( ρ ) K + ( φ ) H ] C
w b p a
K
w b p
= [ ( φ ) ( ρ )
+ +
H H
( φ )] G
a
w
H
= [ ( φ ) + ρb
+ ( φa
) ]
K
K
p
P
X
where M t
/V can be viewed as the average mass
concentration of the plume.
86

Partition of Contaminants
in Different Phases…..
(Example) A new technician collected a
groundwater sample. He filled only half of the 40-
mL sample vial. The benzene concentration in the
collected groundwater was analyzed to be 5 mg/L.
Determine:
(1) benzene concentration of benzene in the head
space before the vial was opened,
(2) percentage of total benzene mass in the
aqueous phase of the closed vial, and
(3) true benzene concentration, if head-space free
sample is collected.
87

88
Partition of Contaminants
in Different Phases…..
Basis: 1-liter container
(1) Concentration of benzene in the head space
= H x C l
= (0.22)(5) = 1.1 mg/L = 340 ppmV.
(2) Mass of benzene in liquid = (5)(0.5) = 2.5 mg
Mass of benzene in the head space = (1.1)(0.5)
= 0.55 mg
Total mass of benzene = 2.5 + 0.55 = 3.05 mg
% of total benzene in liquid = 2.5/3.05 = 82%.
(3) The actual liquid concentration should be
= (3.05)/(0.5) = 6.1 mg/L

Partition of Contaminants
in Different Phases…..
(Example) The aquifer underneath a site is
contaminated with PCE. The aquifer porosity is 0.4
and the bulk density of the aquifer material is 1.8
g/cm 3 . A groundwater sample contains 200 ppb of PCE.
Assuming a linear adsorption model, estimate:
(1) the PCE concentration adsorbed on the aquifer
material, (f oc
= 0.01), and
(2) the partition of PCE in the two phases, i.e.,
89

Partition of
Contaminants
in Different
Phases…..
(a) PCE concentration adsorbed onto the solid has
been determined as 0.50 mg/kg.
(b) Basis: 1-liter aquifer formation
Mass of PCE in the liquid phase
= (C)[(V)(φ)] = (0.2)[(1)(0.4)] = 0.08 mg
Mass of PCE adsorbed on the solid
90

91
Partition of Contaminants
in Different Phases…..
Total mass of PCE = mass in liquid + mass on
the solid
= 0.08 + 0.9 = 0.98 mg
Percentage of total PCE mass in the aqueous
phase
= 0.08/0.98 = 8.2%.
Discussion: 91.8% of total PCE is adsorbed onto
the aquifer materials. This partially explains
why the clean-up of aquifer takes a long time
using the pump-and-treat method

Groundwater
Movement
Why do we have to know?
Advection is the primary transport mechanisms
of contaminants.
How far does it go? - Site Assessment
How fast can we remove it? - Remediation
Where is it from?
Where is it going?
92

Darcy’s Law
Q = -KA (∆h/dL)
v
Q
= = −
A
K dh
dl
93

Calculating Gradient from Well Data
Graphical method
42 m
45 m
40 m
39 m
40 m
45 m
49 m
Linear interpolation between well points
94

Darcy Velocity vs.
Seepage Velocity
Does this Darcy velocity represent the
groundwater flow velocity?
The Darcy velocity in Darcy’s equation assumes
the flow occurs through the entire cross-section
of the porous medium.
The actual fluid velocity through the porous
medium would be larger than the Darcy velocity.
This flow velocity is often called the seepage
velocity or the interstitial velocity.
Q
vs = =
φ A
v
φ
95

Hydraulic Conductivity vs.
Intrinsic Permeability
The intrinsic permeability of a soil core sample is
1 Darcy. What is the hydraulic conductivity of
this soil for water at 25 o C?
• density of water (25 o C) = 0.99703 g/cm 3
• viscosity of water (25 o C) = 0.00890 g/s•cm
K
=
kρg
µ
=
(9.87 × 10
−9
2
cm )(0.99703g/
cm
0.00890g/
s • cm
3
)(980cm/
s
2
)
= 1.09×
10
−3
cm/
s
=
(1.09×
10
−3
)(2.12×
10
4
)
=
23.0gpd
/
ft
2
96

Methods of Determining K
Laboratory Tests
• Sieve Analysis
• Constant Head Permeameter
• Falling-Head Permeameter
Aquifer Tests
• Pumping Test
• Slug Test
97

Constant Head Permeameter
Used for sands,
noncohesive sediment
Cylindrical chamber to
hold sample
Maintain a constant head
Water moves through
sample at constant rate
Measure Q (volume/ time)
Q = -KA (∆ h/L)
Rearrange
K = QL/A(-∆h)
98

Falling Head Permeameter
Continuity: flow in = flow out
K =
d
d
2
t
2
c
L
t
h
ln(
h
o
)
K = hydraulic conductivity
L = sample length
h 0
= initial head in falling tube
h = final head in falling tube
t = time
d t
= diameter of falling head tube
d c
= diameter of chamber
99

Aquifer Tests
T
S
K
Usually involve pumping a well at a constant rate (stress
the aquifer) for several hours to several days and
measuring water levels in nearby observation wells
located at different distances from the pumping well.
100

Water Wells
GW Monitoring Wells
101

Pumping Test
One pumping well
Several monitoring wells
Expensive and time consuming
Cover a large area
Water disposal?
Interference
Get flow rate and contaminant concentrations
102

Steady State Analysis
for Confined System
T = transmissivity (L
r 2 /T)
Q
T =
2
2π(h 2
–h 1
) ln ( r ) Q = pumping rate (L 3 /T)
1
h 1
= head at distance r 1
(L)
h 2
= head at distance r 2
(L)
rouble is steady-state can take a long time time to reach,
and cannot give us information regarding storage! 103

K = hydraulic conductivity (L/T)
Q = pumping rate (L 3 /T)
b 1
= saturated thickness at distance r 1
(L)
b = saturated thickness at distance r (L)
104
Steady State
Analysis
for Unconfined
System
K =
r 2
r 1
Q
π(b 22
–b 12
) ln ( )

105
Slug Tests
Quick and cheap: “instantaneously” raising or lowering
the water level in a well and by dropping a long object
into the well to displace the water monitoring the
recovery of the water level
Localized
No information on Q and C
Initial condition t = 0 t = t 1 t = t 2 t = t 3

106
Perched Aquifer
Lens is “Perched” Above Unconfined System
Lens of Low Permeability Material
in Unsaturated Zone

Retardation Factor
for Plume Migration in Groundwater
Physical, chemical, and biological processes in
the subsurface that can affect the fate and
transport of contaminants.
The processes include:
• biotic degradation
• abiotic degradation
• dissolution
• ionization
• volatilization
• adsorption.
107

108
Retardation Factor
for Plume Migration in Groundwater…..
For transport of dissolved plume in groundwater,
adsorption of contaminants is probably the most
important and most studied mechanism for
removal of contaminants from the groundwater.
If adsorption is the primary removal mechanism
in the subsurface, the reaction term in advectiondispersion
equation can then be written as
(ρ b
/φ)∂S/∂t, where ρ b
= dry bulk density of soil
(or the aquifer matrix), φ = porosity, t = time,
and S = contaminant concentration on soil.

109
∂ C
∂ C ∂ C
= D − v ±
∂ x
Retardation Factor
∂ t
2
∂ x
for Plume Migration in Groundwater…...
Assume a linear adsorption isotherm (e.g., S =
K p
C), thus
∂ S
= K p
∂ C
2
RXN
∂ S
∂ t
∂ S ∂ C
= ( )( )
∂ C ∂ t
=
K
p
∂ C
∂ t
∂ C
∂ t
ρ b ∂ C ρ bK
p ∂ C ∂ C
+ ( ) K p = (1 + ) = D −
φ ∂ t φ ∂ t
2
∂ x
2
v
∂ C
∂ x

110
∂ C
Retardation Factor
∂ t
2
∂ x
for Plume Migration in Groundwater…...
2
∂ C ∂ C
= D − v ±
∂ x
RXN
∂ C
∂ t
ρ b ∂ C ρ bK
p ∂ C ∂ C
+ ( ) K p = (1 + ) = D −
φ ∂ t φ ∂ t
2
∂ x
Divide both sides by (1 + ρ b
K p
/φ):
2
v
∂ C
∂ x
∂ C
∂ t
2
D ∂ C
= −
R 2
∂ x
v ∂ C
R ∂ x
where
R
= 1 +
ρ
b
K
φ
p

Retardation Factor
for Plume Migration in Groundwater…...
R
= 1 +
ρ
b
K
φ
p
R = retardation factor (dimensionless) and ≥ 1.
R reduces the impact of dispersion and migration
velocity by a factor of R.
All of the mathematical solutions that are used to
solve the transport of inert tracers can be used for
the contaminants, if the groundwater velocity and
the dispersion coefficient are divided by R.
From the definition of R, one can tell that R is a
function of ρ b
, φ, and K p
.
For a given aquifer, ρ b
and φ would be the same
for different contaminants. Consequently, the
greater the partition coefficient, the greater the R.
111

Effect of the distribution coefficient on contaminant
retardation during transport in a shallow groundwater.
112

Retardation Factor
for Plume Migration in Groundwater…...
The groundwater underneath a landfill is
contaminated with benzene, and pyrene.
Estimate the R: φ = 0.40; ρ b
= 1.8 g/cm 3; f oc
=
0.015; K oc
= 0.63 K ow
(a) Log(K ow
) = 2.13 for benzene K ow
= 135
Log(K ow
) = 4.88 for pyrene K ow
= 75,900
(b) K oc
= (0.63)(135) = 85 (for benzene)
K oc
= (0.63)(75,900) = 47,800 (for pyrene)
(c) K p
= (0.015)(85) = 1.275 (for benzene)
K p
= (0.015)(47,800) = 717 (for pyrene)
113

114
Retardation Factor
for Plume Migration in Groundwater…...
(d) benzene:
R
ρbK
p (.)(. 18 1275)
= 1+ = 1+ =
φ
04 .
674 .
R
pyrene:
ρ bK
p ( 18 . )( 717)
= 1+ = 1+ =
φ
04 .
3227 ,

Migration Speed of the Dissolved Plume
The retardation factor relates the plume
migration velocity to the groundwater seepage
velocity as:
R
V s
Vs
= or Vp
=
V
R
p
Where V s
is the groundwater seepage velocity and
V p
is the velocity of the dissolved plume.
When the value of R is equal to unity (for inert
compounds), the compound will move at the
same speed as the groundwater flow without any
"retardation”.
When R = 2, for example, the contaminant will
115

Migration Speed of the Dissolved Plume…..
(Example) The groundwater underneath a landfill
is contaminated with benzene and pyrene. A
recent groundwater monitoring in September
1997 indicated that benzene has traveled 50m
down-gradient; while pyrene was not detected in
the down-gradient well.
Time when benzene first entered the aquifer?
φ = 0.40; K = 30 m/day;i = 0.01
ρ b
= 1.8 g/cm 3 ;f oc
= 0.015
116

R
V s
Vs
= or Vp
=
V
R
Migration Speed of the Dissolved Plume…..
p
(a) Darcy velocity: v = ki = (30)(0.01) = 0.3 m/d
(b) GW velocity (or the seepage velocity)
v s
= v/φ = (0.3)/(0.4) = 0.75 m/d
(c) Migration speeds of the plumes:
v p
= (0.75)/(6.74) = 0.111 m/d (for benzene)
v p
= (0.75)/(3227) = 0.0002 m/d (for pyrene)
(d) Time for benzene to travel 50 meters:
t = (50 m)/(40.6 m/yr) = 1.23 yr
Benzene entered the gw in June of 1996.
117

Migration
Speed of the
Dissolved
Plume…..
Discussion:
The estimate is the time when the benzene
entered the aquifer. The information given is not
sufficient to estimate the time the leachates
leaked through the landfill liner.
The migration of pyrene is extremely small, 0.08
m/yr, therefore, it was not detected in the
downstream monitoring wells. Most, if not all, of
the pyrene compounds will be adsorbed onto the
118

Migration Speed of
the Dissolved
Plume…..
The estimates are rough, because lots of factors may
affect the accuracy of the estimates.
Factors include uncertainty of the hydraulic conductivity
porosity, i, K ow
,f oc
, etc.
Neighborhood pumping will affect the natural
groundwater gradient and, consequently, the migration
of the plume.
Other subsurface reactions such as oxidation and
biodegradation may also have large impacts on the fate
119