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Lecture No. 4

Fate and Transport of Contaminants

Chapter 4, p.133

This chapter deals with key release mechanisms, contaminate transport and the fate of

contaminants in the environment.

1. Contaminant Release

Three phases: liquid, solid and/or gas.

A. Air Emissions

1.) General

Types:

- Point: Landfill vent or incinerator stack

- Line: Dust from road and/or vehicle

- Area: Lagoon or construction site

- Volume: Building with open windows

- Puff: Volatile emissions from an accidental spill

2.) Volatilization

Transfer of a chemical substance from a liquid to a gaseous phase. Predominant source

of atmospheric emissions at most uncontrolled hw sites.

Function of: temperature, vapor pressure, difference in concentrations.

Can be measured or modeled using Fick's Law.

Usually insignificant at undisturbed, inactive sites. Remediation activities, however,

may substantially increase emissions perhaps several thousand fold.

3.) Particulate Emissions

Originate from materials handling and surface areas.

The Soil Conservation Service has a graphical method for predicting annual averagewind and erosion and resulting soil loss.

Transportable and suspendible amount is that portion of the total soil loss represented

by particles 100 m in diameter or smaller.

Inhalable size 10 m in diameter or smaller.

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Lecture No. 4, Fate and Transport of Contaminants, Page No. 2

The greatest source of fugitive dust is remediation involving soil handling.

E = k(.0032)

U

51.3

M

21.4

eq. 4-2 p.140

The emission factor, E, may be multiplied by the tons of material handled to achieve an

estimate of total emissions.

Example:Given: A site to be remediated has he following soil characteristics:

The average particle size < 15mm

The mean wind speed in the worst season in the Spring is 22 mph

The moisture content is only 60% can be raised to 85% by judicious spraying. Overspraying is

not a good idea as a hw leachate may be formed.

The soil that needs to be moved occupies a vol. of 346' x 123' x 56' of which 1000 tons can be

move per day.

The average soil density is 75 lb/ft3

Find: Estimate the particulate emissions on a daily and total site basis.

1.) Particulate Emission per Day

From p. 140 for particle size < 15 mm, k=.48

E = k(.0032)

U

51.3

M

21.4

= .48(.0032)

22

51.3

85

21.4

= .001536 x 6.86/190.4

E = 5.53 x 10-5 lb of particulate/ton of soil moved

Daily emissions = Daily tons of soil moved x E

Daily emissions = 1000 tons/day x 5.53 x 10-5 lb of particulate/ton of soil moved

Daily emissions = .055 lbs of particulates / day

2.) Emissions for Entire Site

Total tonnage = 346' x 123' x 56' x 75 lb/ft 3 x 1 ton/2000lbs

Total tonnage = 89,372 tons

Total emissions = Total tons of soil moved x E

Total emissions = 89,372 tons/day x 5.53 x 10-5 lb of particulate/ton of soil moved

Total emissions = 4.94 lbs of particulates / total site

B. Water Releases (covered in CE457, Solid Waste)

Controlled releases. Virtually every industrial and commercial facility generates a

wastewater and it not possible to clean the water to 100% standards. Landfill leachate.

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2. Transport of Contaminants in the Subsurface

A. Hydrologic Cycle

The hydrological cycle is the continuous circulation of water in the atmosphere,

surface and subsurface.

Waters reaching the saturated zone in the subsurface will flow from areas of high

hydraulic head to areas of low hydraulic head. If the stratum readily permits this flow it

is know as an aquifer and if it does not, it is an aquitard or confining layer. If the

aquifer is free to move up and down, it is an unconfined aquifer. The stratum is

homogeneous if the properties of the soil do not vary with location. An isotropic

stratum does not vary with orientation or direction. An anisotropic stratum has a

higher hydraulic conductivity in the horizontal direction and less in the vertical. Mostsoils are anisotropic and heterogeneous, but they are usually considered homogeneous.

B. Ground Water Flow

1.) Darcy's Law

Darcy's Law is the basis of our understanding of subsurface ground water flow:

Q = kiA eq. 4-3 units p.150

k = hydraulic conductivity

k= 1x10-2 cm/sec for medium to fine sandThe hydraulic gradient, i, describes the rate of change in which the head is lost as

water flows through the porous materials:

i = (h1 - h2)/l (hydraulic gradient) eq. 4-4 units p.150

Darcy's Law is empirical

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Rearranging the first equation:

Q/A = ki,

Q/A = specific discharge, Darcy flux, or Darcy velocity, v

v = ki

Flow is in direction of declining hydraulic head requiring a negative sign. In differential

form:v = -k dh/dl (dh/dl = hydraulic conductivity)

ExampleGiven: Water flows from elev. 8.3' to 7.7' over a length of 3'. The hydraulic conductivity is

.3x10-4 ft/sec.

Find:

1.)v, Darcy velocity

2.)time of travel

1.) v

v = ki eq 4-6, p.151

i = (h1 - h2)/l (hydraulic gradient) = (8.3-7.7)/3i = .2 (dimensionless)

v = .3x10-4 ft/sec x .2

v = 6.0 x 10-6 ft/sec

2.) travel time

t = l/v = 3ft/6.0 x 10-6 ft/sec

t = 500,000 sec = 8333 min = 139 hours = 5.79 days

Note: This is NOT the travel time, v, the Darcy velocity is assumes that the water is moving

across the entire cross-sectional area. Since the volume and area is occupied by the soil, the water

cannot be moving through the soil and the actual flow area is less than the total area. The actual

velocity, vs, know as the seepage velocity based on the actual flow area, will yield the correct

travel time. A salient point is that the Darcy velocity, v, also know as the specific discharge and

Darcy flux can NOT be used to calculate travel time.

v is not seepage velocity, vs, which is always higher than v and v can NOT be used to

compute travel time.

vs = vA/Av eq. 4-9 units p.152

The water migrates only through the voids, the water can not travel through a soil

particle. Since the actual flow area is reduced, the actual velocity, vs, is increased for

a given flow.

Porosity (volume of the voids divided by total volume)

n = Vv/Vt x 100% p.153

Example

Given: The volume of solids in a soil sample is 63%Find: n, porosity

n = Vv/Vt x 100% p.153

Vv = 100 - 63

Vv = 37%

n = 37/100 x 100

n = 37% = .37

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Void ratio (volume of the voids divided by the volume of the solids)

e = Vv/Vs eq. 4-12 units p.153

ExampleGiven: Data from the previous problem.

Find: e, void ratio

e = Vv/Vs = 37/63

e = .59

The seepage velocity equals the Darcy velocity divided by the porosity:

vs = v/n

Example:Given: The above data

Find: The true travel time

t = l/vs

vs = v/n = 6.0 x 10-6 ft/sec / .37

v

s

= 1.62 x 10-5 ft/sec

t = 3' / 1.62 x 10-5 ft/sec

t = 185,000 sec = 3083 min = 51.38 hr = 2.14 days vs. 5.79 days when using the incorrect

Darcy velocity.

2.) Hydraulic Head

Hydraulic head is the chief driving force for ground water flow.

Standard hydraulic presentation, review CE332, Hydraulic Engineering notes.

Two dimensional flow. F4-14, p. 159. Method to describe flow patterns. Flow nets are

two-dimensional graphical representations of hydraulic head conditions in the

subsurface. Lines of equal total hydraulic head are know as equipotential lines and the

potential energy at any point on one of these lines is the same. Flow lines areperpendicular to the equipotentials and represent the average path a particle of water

takes as it flows in the subsurface.

C. Hydraulic Conductivity of Geologic Materials (T4-9, p.160)

Impermeable soil such as a tight clay may have a hydraulic conductivity of 1x10 -9

cm/sec; the most permeable material such as a clean gravel may be 1x105 cm/sec.

Two things should be remembered regarding hydraulic conductivity:

- Order of magnitude refinement is appropriate

- The value given for a specific material type may be very different than the hydraulicconductivity for the entire formation.

Transmissivity is the hydraulic conductivity times the thickness of the formation.

T = kt eq.4-19 units p.160

The specific yield is the volume of water that drains from the saturated soil pores as

the water table drops. For unconfined conditions, storativity is synonymous with

specific yield.

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D. Flow in the Unsaturated Zone

Definitions:

- Water table. A surface on which the fluid pressure is atmospheric.

- Phreatic zone. The zone below the water table where the fluid is in compression,

positive pressure.- Vadose zone. The zone above the water table where the fluid is in tension,

negative pressure. This zone is divided into a saturated capillary zone and a partially

saturated zone.

E. Contaminant Transport Mechanisms

Contaminants that are dissolved in water are solutes and the water is the solvent and

the combination is the solution. As the water flows, the contaminants are transported

with the water a process known as advection.

As the water flows around the soil particles, it is mixed, a process known as

mechanical dispersion. The result is dilution or reduction in the contaminant

concentration.

A one time introduction of pollutants is termed a slug and opposed to a continuous

source. If the pollutant is introduced at a discrete location, it is known as a point

source as opposed to a non-point source.

The distribution and extent of contaminants migrating in the subsurface is known as

the plume. F4-18, p.165.

Contaminates, particularly the ionic and molecular constituents, will move from areas

of high concentration to areas of low concentration and this process is termed

diffusion.

In general, advective transport and associated mechanical dispersion dominatethe contaminant transport in formations of medium to high hydraulic

conductivity. In formations of low hydraulic conductivity, including clay liners,

diffusive transport is frequently the controlling mechanism.

F. Real-World Contaminant Transport (T4-10, p.167)

Mechanisms that influence and change idealized flow predictions.

Fractured Media Flow. Flow will occur in the fractures which behave like pipes as

opposed to the bulk media.

Heterogeneity. If the formation is not homogenous, the contaminant may transport a

great deal more quickly through a highly conductive lens.

Nonaqueous Phase Liquid (NAPL). Immiscible organic liquid compounds exist as a

separate liquid phase in the subsurface.

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G. Flow and Transport Equations

1.) Laplace equation for isotropic flow in two dimensions:

2h/x2 + 2h/y2 = 0

May be solved by finite elements and finite difference mathematical methods.

2.) Flow nets.

Flow nets. Flow nets are two-dimensional graphical representations of hydraulic head

conditions in the subsurface. Lines of equal total hydraulic head are know as

equipotential lines and the potential energy at any point on one of these lines is the

same. Flow lines are perpendicular to the equipotentials and represent the average path

a particle of water takes as it flows in the subsurface. Flow nets give a feel for the flow

patterns and are alternate methodology, independent verification, to other methods.

Fundamental properties of flow nets:

- The head difference between any pair of adjacent equipotentials is the same as

any other pair.- Flow lines intersect equipotentials at right angles.

- Figures enclosed by adjacent pairs of equipotentials and flow lines are essentially

square.

- The spacing of equipotentials is inversely proportional to the hydraulic gradient

and to the Darcy velocity.

- Every flow channel transmits the same quantity of seepage.

- The impervious boundary is a flow line; the free water boundary is an

equipotential.

The flow net provides data on the hydraulic gradient and quantity of seepage. Forhomogeneous, isotropic conditions:

The seepage quantity is:

q = khnf/nd eq.4-33 units p.174

The hydraulic gradient is:

i = (h/nd)/l eq.4-34 units p. 174

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Example:Given: The following flow net for a water table aquifer. The hydraulic conductivity, k, is based

on sand + gravel mixtures.

Find: Calculate the subsurface flow for 1000m of stream.

S t r e a m

e q u i p o t e n t i a l

l i n e

f l ow l in e

1 4 0 m 1 2 0 m 1 0 0 m 1 2 0 m 1 4 0 m

From T4-9, p.160 for sand+gravel mixtures, k varies from 1.0 to 1x10-3 cm/sec. Choose mid-

range = .01 cm/sec

q = khnf/nd units p.174

k = .01 cm/sec

h = 140-100

h = 40m

nf = 2 (the flow channels are the paths between the flow lines)

nd = 2 (number of equipotential drops)q = .01 cm/sec x 40m x 2/2 x 1m/100cm

q = 4x10-3 m3/sec.m

Q = qL = 4x10-3 m3/sec.m x 1000m

Q = 4 m3/sec

3.) Finite Difference Method

A two-dimensional grid is constructed in which each nodal point is modeled to

represent the average head for the area enclosed by the square. Also it is assumed that

the head at a particular node is the average head at the four surrounding nodes.

Boundaries conditions must be established either as constant head or no-flowboundaries.

Numerical head values are assigned.

The heads are recalculated using the finite difference equation using the trial and error

relaxation technique until a specified tolerance is achieved.

Example: use ex. 4-8,p176

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H. Ground Water Modeling

Once the model is created, "what-if" scenarios may be easily evaluated.

Most models are mathematical, although analog, electrical etc. are possible. Once the

numerical model is selected, it must be digitized so that the data can be fed into a

computer. Calibration or adjusting the model by comparing the model data to real data

comes next.

Once calibrated, the model can be used to:

- Guide the placement of monitoring wells.

- Predict contaminant concentrations.

- Assess remedial alternatives.

- Predict residual contaminants

3. Fate of Contaminants in the Subsurface (T4-11, p.180)

A. General

Retardation refers to processes that impede the transport of contaminants by

removing or immobilizing them from a free state. The contaminants are NOT

transformed and the process is reversible.

Attenuation refers to 2 processes:

- irreversible removal

- transformation

Soil. Soil is a mixture of different inorganic and organic materials. The predominant

inorganic elements are silicon, aluminum and iron. Classification by mineral grain

include clay to gravel, T4-12, p.182. Hazardous material is naturally occurring such asarsenic and cadmium, p. 182. Naturally occurring hazardous materials are generally

insoluble. Organic matter consists of decomposed plant matter known as humus.

Typical organic content is .2-3%. The organic matter acts as a stabilizer to bind

inorganic particles together as aggregates. Much of the retardation and attenuation

takes place in the aggregate microscale.

B. Retardation Processes

1.) Sorption

Sorption. The accumulation of organic chemicals at soil surfaces for example the

adherence of organic molecules to naturally occurring humic matter in soil. Desorption

is also possible which may limit the use of "pump and treat" cleanup technology. The

potential retention capability can be estimated by saturating undisturbed soil with a

liquid contaminant and allowing the sample to drain. The retained contamination is

termed residual saturation or retention capacity.

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Linear Sorption Model.

S = KdC eq.4-36 units p.187

For saturated conditions and non-polar organics. Kd can be related to organic content:

Kd = Kocfoc eq. 4-37 units p.187A reasonable estimate is forthcoming if the following conditions are met:

Minimum: Greater than .1%; few tenths of a percent; 1% (different sources)

Maximum: 20%

Example:

Given: A phenol underground storage tank is leaking such that the immediate

surrounding ground water contains 7.3 mg/l of phenol. The soil contains 2.5%

organic matter, a fact ascertained by analyzing the soil.

Find: Assuming a linear sorption model, what is the concentration of phenol

sorbed in to the soil? Calculate the sorbed concentration, S.

1.) KdFrom App. A, p.1048, Koc for phenol = 14.2 ml/g Note units.

Kd = Kocfoc units p.187 = 14.2 ml/g x (.025)

Kd = .355 ml/g

2.) S

S = KdC units p.187 = .355 ml/g x 7.3 mg/l

S= 2.59 mg/kg

2.) Ion Exchange

Ion exchange involves the sorption of ions in solution onto oppositely charged,

discrete sites on the surface of a soil particle. Ion exchange applies to metals whilesorption applies to organics, in general.

The capacity of a soil to retain and exchange cations is quantified as the cation

exchange capacity e.g. 150 meg/100grams. Clay has a much higher cation exchange

capacity than other inorganic particles because of its extremely large surface area that

contains many negative sites.

Cations are replaced in the following order:Na+ < Li+ < K+ < Rb+ < Cs+ < Mg++ < Ca++ < Ba++ < Cu++ < Al+++ < Fe+++ < Th++++

3.) Precipitation

Precipitation occurs when the concentration of a solution exceeds the solubility of thatcompound and any excess solute changes to a solid and falls out of the solution.

Reversible.

Particularly applicable to heavy metals. F4-32, p.192.

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C. Attenuation Processes

Chemical Oxidation-Reduction. Redox involves the gain or loss of electrons.

Hydrolysis. Chemical substances reacting with water molecules.

Volatilization. The conversion of volatile chemical constituents in ground water to

vapor; the vapor ending up in the atmosphere.

D. Mobility Enhancement

Opposite of retardation and attenuation i.e. the contaminant is sped along.

Cosolvation. The presence of bulk solvents promotes increased interaction between a

solute and the solvent that would not occur in water and can dramatically increase the

mobility of contaminants.

Ionization. Increased solubility

Dissolution. Dissolving of chemical substances such as a leachate.

Complexation. Also chelation is the formation of a coordinate bond between a metal

ion and an anion know as a ligand.

Atmospheric Transport of Contaminants is covered in CE351, the first

environmental course.

HOMEWORK

Read Chapter 4, Fate and Transport of Contaminants, pp. 133-223

Problems, p.215, 4-2, 4-3, 4-5, 4-6, 4-11, 4-13, 4-17,

ANCILLARY PROBLEMS

4A. Given: The BadKnews Landfill which has been since shut down has the following

characteristics.

The average soil particle at a site is .41mm

The maximum average wind speed is 17.8 mph.

The average moisture content is 52% with a density of 83 lb/ft3

The site is trapezoidal in cross section with the top being 710 ft across, the bottom being 250 ft

across and the depth being 93 ft. The length of the trapezoid is 1/4 mile.

Find:

1.) Estimate the particulate matter attendant to remediation for the entire project.

2.) They can load a 5-ton truck in 15 minutes. Assuming several trucks but only one loaded at a

time, an eight hour day, 5 days per week, how long will it take to finish the job.

3.) How much particulates do they get on a daily basis. Calculate this number and check it by

comparing the number obtained by using the first two parts.

4B. Given: For a particular soil formation, the Darcy velocity, v, is 3.5x10-4 cm/sec , the

hydraulic gradient is .33 and the porosity is 52%. The distance from the bottom of the leaking

landfill to the ground water table is 57.8ft.

Find:

1.) The hydraulic conductivity.

2.) Travel time for the leachate to reach the ground water table.

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4C. Given: A continuous waste source such as leaking landfill.

Find: Draw the plume if the ground water gradient is from southwest to northeast.

4D. Given: Same as 4C. but the waste source is a 5000 gallon spill.

4E. Given: k=5x10-2cm/sec

S t r e a m

e q u i p o t e n t i a l

l i n e

f l o w l i n e

1 4 0 m 1 2 0m 1 0 0 m 1 2 0 m 1 4 0 m Find: Calculate the subsurface flow for a mile of stream.

4F. Given: A mustard gas underground storage tank is leaking such that the immediate

surrounding ground water contains 1.9 mg/l of mustard gas. The soil contains 1.1% organic

matter, a fact ascertained by analyzing the soil.

Find: Assuming a linear sorption model, what is the concentration of mustard gas sorbed in to

the soil? Calculate the sorbed concentration, S.

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