### Lecture series 8 - Civil and Environmental Engineering | SIU

```CE 510
Hazardous Waste Engineering
Department of Civil Engineering
Southern Illinois University Carbondale
Instructor: Jemil Yesuf
Dr. L.R. Chevalier
Lecture Series 8:
Contaminant Release and Transport from Source
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
Pathways of Contaminants
Dispersion Of Air Pollutants
 Focus on a basic point source Gaussian
dispersion model
 Assumptions:



Atmospheric stability is uniform
Turbulent diffusion is a random activity
Dilution in both the horizontal and vertical
direction can be described by Gaussian or
normal equations
Dispersion Of Air Pollutants
 Assumptions (continued)



Contaminated gas stream is released into the
atmosphere at a distance above the ground equal to
the physical stack height plus the plume rise
Degree of dilution of the effluent plume is inversely
proportional to the wind speed
Pollutant reaching the ground level is totally
reflected back into the atmosphere like a beam of
light striking a mirror at an angle ( image source)
Instantaneous
Plume
Boundary
H h
Time Averaged
Plume Envelope
z
x
H
y
(x,-y, z)
(x,0,0)
h
(x,-y,0)
Gaussian Plume Model

C  x , y ,0 , H   
 


Q
  exp
s y s z u  

2




1 y
    exp
 
 2  s y    


 1  H 2 
 
  

 2  s z   

Eqn. 8.22 (Textbook)
where:
C (x,y,0,H) = downwind conc. at ground level(z=0), g/m3
Q = emission rate of pollutants, g/s
sy, sz = plume standard deviation, m
u = wind speed, m/s
x,y,z and H = distance, m
Effective Stack Height
The value for the effective stack height, H, is
the sum of the physical stack height, h, and
the plume rise DH.
DH may be computed from Holland’s formula
(J.Z. Holland, 1953, A Meteorological Survey
of the Oak Ridge Area, U.S. Atomic Energy
Commission Report No. ORO-99, Washington
D.C., U.S. Government Printing Office)
Plume Rise
vs d 
DH 
1 . 5 
u 

 Ts  Ta   
2
 2 . 68  10  P 
 d  
 Ts   

where
vs = stack velocity, m/s
d = stack diameter, m
u = wind speed, m/s
P = pressure, kPa
Ts = stack temperature, K
Ta = air temperature, K
Gaussian Plume Model
K E Y T O S T A B IL IT Y C A T E G O R IE S
S u rface w in d
D ay
In co m in g so lar rad iatio n
sp eed
(at 1 0 m ) (m /s) S tro n g M o d . S lig h t
<
2
3
5
>
2
-3
-5
-6
6
A
A -B
B
C
C
A -B
B
B -C
C -D
D
B
C
C
D
D
N ig h t
T h in ly o v ercast o r
< 3 /8 clo u d
> 4 /8 lo w clo u d
E
D
D
D
(see Table 8.1) p. 416
F
E
D
D
Gaussian Plume Model

C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
   exp
 
 2  s y    


Need to evaluate these terms,
which are the standard
deviations of the plume
 1  H 2 
 
  

2
s

 z   
Gaussian Plume Model
Martin, D.O., 1976. Comment on “The Change of
Concentration Standard Deviation with Distance”, J. Air
Pollut. Control Assoc., 26:145-147.
sy = ax0.894
sz = cxd + f
where x is the distance downwind, expressed in km, s
is in m, and a,c,d, and f are constants found in the
following table:
NOTE: Alternatives are graphs 8.10 and 8.11
Gaussian Plume Model
S tability
C ategory
A
B
C
D
E
F
a
213
156
104
68
50.05
34
c
440.8
106.6
61
33.2
22.8
14.35
x  1 km
d
1.941
1.149
0.911
0.725
0.678
0.74
f
9.27
3.3
0
-1.7
-1.3
-0.35
c
459.7
108.2
61
44.5
55.4
62.6
x  1 km
d
2.094
1.098
0.911
0.516
0.305
0.18
f
-9.6
2
0
-13
-34
-48.6
Example
Consider the emission of SO2 from a coal fired power
plant, at a rate of 1,500 g/s. The wind speed is 4.0 m/s
on a sunny afternoon. What is the centerline
concentration of SO2 3 km downwind (Note: centerline
implies y=0).
Stack parameters:
Height = 130 m
Diameter = 1.5 m
Exit velocity = 12 m/s
Temperature = 320°C (593° K)
Atmospheric conditions: P=100 kPa T=25° C (298° K)
Strategy
Strategy
 Determine the effective stack height


Dh
H = Dh + h
 Determine stability class
 Estimate sy and sz
 Apply governing equation
Equations

C  x , y ,0 , H   
 

Q
  exp
s y s z u  

vs d 
DH 
1 . 5 
u 
sy = ax0.894
sz = cxd + f
2


1 y  
   exp
 
 2  s y    


 1  H 2 
 
  

2
s

 z   

 Ts  Ta   
2
 2 . 68  10  P 
 d  
 Ts   

Solution
12 m s 1 . 50 m  


 593  298 
2
DH 
1 . 5   2 . 68  10
100 
1 . 5  

 4 .0 m s  
593




  4 . 5 m  3 . 43   15 . 44 m
H = effective stack height
= h + DH
= 130 m + 15.4 m = 145.4 m
Solution
Atmospheric stability class: Class B
sy = ax0.894 = 156(3)0.894 = 416.6 m
sz = cxd + f = 108.2(3)1.098 + 2 = 363.5 m
Solution
1

 1  145 . 44
1500
C  x, y, z, H  
exp  0  exp   
  416 . 6  363 . 5  4 . 0 
 2  363 . 5
4

exp   0 .08 
4
0 .923 
 7 . 88  10
 7 . 88  10
 7 . 28  10
4
3
g m of SO 2
 728 . 6  g m
3
... end of problem
2

 
 
Example
Simplify the Gaussian dispersion model to
describe a ground level source with no
thermal or momentum flux, which is the
typical release that occurs at a hazardous
waste sites. In this situation, the effective
plume rise, H, is essentially 0.
Solution

C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
   exp
 
 2  s y    



C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
  exp 0 
 
 2  s y   



C  x , y ,0 , H   
 
2



Q
1 y 
 
 exp   
s y s z u 
 2  s y  


 1  H 2 
 
  

2
s

 z   
Puff Models
 Instantaneous one-time release of
material
 Ground level
C  x , y , z, t  
 1   x  ut
Q 'm

exp
  

3
2  s x
2 2s xs ysz



where
C is the concentration at (x,y,z,t) in mg/m3
Q’m is the mass of contaminant released (mg)
2
2

y
z 

  2  2 

sy
sz 


2
Puff Models
C  x , y , z, t  
 1   x  ut
Q 'm

exp
   
3
2
2  s x
2 s xs ysz



2
2

y
z 

  2  2 

sy
sz 


2
For the plume dimensions (i.e. s) we will need to
refer to Figures 8.8 and 8.9.
Example
A hazardous waste spill has occurred, releasing 10 kg of
TCE into the air. If the spill was at night under mostly
overcast skies, and the wind velocity was 7 m/s in the
x-direction, estimate the TCE concentration 0.5 km
Similar to Example 8.2.
Instead of evaluating the spill at 0.5 km, evaluate the spill at 1
km. Compare s values as well as final value.
Strategy
Strategy
 Determine the class
 Estimate sx, sy, and sz
 Estimate C from the governing equation
C  x , y , z, t  
 1   x  ut
Q 'm

exp    
3
2  s x
2 2s xs ysz



2
2

y
z 

  2  2 

sy
sz 


2
Solution
1. Determine the class given u = 7 m/s, night overcast
Class D – use unstable
2. Estimate sx, sy, and sz for x=1000 m
Assume sx = sy = 70 m(see p. 417)
Also, sz =70 m
Solution
3. Calculate t = s/u = 1000 m/ 7 m/s = 142 s
4. Estimate C (1000m, 142 s)
Compare values and discuss
Subsurface Transport of
Contaminants
 Darcy’s Law
 Pulse model Erfc
 Plume model
 Retardation
 Decay
DARCY’S LAW
The first experimental study of groundwater flow was
performed by Henry Darcy in 1856.
Q
Q   KA
dh
dl
where
Q = volumetric discharge (L3/T)
K = hydraulic conductivity (L/T)
A = cross-sectional area (L2)
Q
Specific Discharge
q  K
dh
dl
The specific discharge, also called the darcian velocity, is
slower than the average linear velocity. Darcian velocity
assumes that the total cross sectional area is available for
flow. In reality only a percentage of that area is available
(effective porosity). In reality, water has to move faster to
maintain the same discharge.
Q
A
Controlling Processes
 Three basic mechanisms controlling
contaminant transport in
environmental systems:
 Diffusion
 Dispersion

 Movement of the solute with the bulk
water in a macroscopic sense
 Is the main mechanism driving the
movement of solute
 Advective flux ignores the microscopic
processes, but simply follows the bulk
Darcian flow vectors
Class example
A chemical spill occurs above a sloping, shallow
unconfined aquifer consisting of medium sand with K=1
m/d and a Φ of 0.3. Several monitoring wells are
drilled in order to determine the regional hydraulic
the spill location yielded a value of 5m. At a distance
of 200m down the slope another well yielded a
hydraulic head of 1m. How long it will take for the
contaminants to travel 200m.
Diffusion
 Mass transfer by random molecular motion caused by
 Governed by Fick’s law (second law)
C
t
 C
2
 Dd
with solution
x
2
x
C ( x , t )  C o erfc (

)
2 D t
Where C = concentration at distance x and time t
Co = initial concentration
D* = effective diffusion coefficient
erfc = the complimentary error function
erfc ( y )  1 
2

y
e
0
r
2
dt
Class Problem
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
liner 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? The
effective diffusion coefficients D* for the solutes are:
Species
D* (m2/sec)
Na+1
1.33E-09
Ca+2
7.05E-10
Fe2+
7.19E-10
Cr3+
5.94E-10
Solution
Waste
Clay
Waste
Clay
3m
6m
Land fill B
Land fill A
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
Dispersion
 Mixing which occurs due to differences in
velocity field
 Movement away from the solute mass due to
deflection caused by particles that obstruct
flow
 Dispersion increases with increasing scale as
each new dispersive process is added to
those which occur at all of the lower scales
Dispersion
 Define D, Hydrodynamic dispersion to include
mechanical dispersion and molecular diffusion as;
D = Dmech + D0
 The importance of diffusion and dispersion is
assessed based on Peclet number, Pe, the ratio of
dispersion effects to diffusion effects
Pe =
ud
D
where d is mean grain size.
Position of input water at time t
C/Co
1
0.5
0
Distance, x
This “plug flow” is due to advection alone
C/Co
1
0.5
0
Distance, x
Tracer front due to diffusion
C/Co
1
0.5
0
Distance, x
2-D Flow in Homogeneous, Isotropic
Porous Media
 C
2
DL
x
2
 C
2
 DT
y
2
 vx
C
x

C
t
 Assuming a uniform velocity field
 Direction of flow parallel to x-axis
Hydrodynamic Dispersion
C/Co
1
to
Direction of transport, x
Hydrodynamic Dispersion
t1
C/Co
1
to
x1
Direction of transport, x
Hydrodynamic Dispersion
t1
t2
C/Co
1
to
x1
Direction of transport, x
x2
Hydrodynamic Dispersion
t1
t2
C/Co
1
to
x1
x2
Plan view
Hydrodynamic Dispersion
t1
t2
C/Co
1
to
x1
x2
Plan view
Hydrodynamic Dispersion
t1
t2
C/Co
1
to
x1
x2
Plan view
Hydrodynamic Dispersion
sL
C
The concentration
in this Gaussian
curve can be
modeled to have
an average and a
standard deviation
Hydrodynamic Dispersion
sL
As a result, we
have an alternate
formula for DL and
DT
DL 
C
DT 
s
2
L
2t
s
2
T
2t
2-D Flow in Homogeneous, Isotropic
Porous Media
 C
2
DL
x
2
 C
2
 DT
y
2
 vx
C
x

C
t
 Assuming a uniform velocity field
 Direction of flow parallel to x-axis
Methods of Solution - Analytical
 Need to consider initial and boundary
conditions
 Boundary conditions (in this case our
dependent variable is C as opposed to
h)



Dirichlet: fixed concentration
Variable flux
Applied Problem
Give a physical interpretation of the following
initial and boundary conditions
C 0, t   C o t  0
C  x ,0   0
x0
C  , t   0
t0
Solution
 For all time t  0, at x=0 the
concentration is maintained at Co
(Dirichlet boundary)
 The background concentration is zero
at t =0 for all x  0
 As soon as flow starts, the solute
concentration Co will cross the x=0
boundary
Solution
 The third condition states that the
flow domain is infinitely long, and that
the concentration will be zero at the
end of the system
 This is important to the derivation of
the analytical solution
Slug Injection into a Uniform 2-D
Flow Field
initial point injection
C x, y, t  
Co A
4 t  D L D T

1
2

exp  

 x 
2


xo  v xt
4DLt

y 
2

yo 
4 DT t


Slug Injection into a Uniform 2-D
Flow Field
(0,0)
C x, y, t  
Co A
4 t  D L D T

1
2

exp  

 x 
2


xo  v xt
4DLt

y 
2

yo 
4 DT t


Problem
Given the following equation for a slug
injection into a uniform 2-D flow field, what is
the equation for the maximum concentration
at any given time?
C x, y, t  
Co A
4 t  D L D T

1
2

exp  

 x 
xo   v xt 
4DLt
2

y 
2
yo  

4 D T t 
Solution
 Assume the initial injection is at x=y=0
 Assume the flow is in the x-direction
 The center of the mass would be at


y=0
x=vxt
Solution
(0,0)
x=vxt1
x=vxt2
Solution
C max 



Co A
4 t  D L D T

1
2
Co A
4 t  D L D T

1
2
Co A
4 t  D L D T


 x 

exp  

v x t  0   v x t  2
exp 0 
1
2
Co A
4 t  D L D T

exp  

1
2
xo   v xt 
2

4DLt
4DLt

y 
2
yo  

4 D T t 
0  0 


4 D T t 
2
C max 
Co A
4 t  D L D T
Recall,
s
x


1
2
2DLt
s
y

2 DT t
By definition then, 99.7% of the mass of contamination
will be contained within an area represented by 3s away
from the center of mass. Thus the plume can be defined
by the location of the center of the mass, 3sx and 3sy
(0,0)
x=vxt
3 2 DT t
3 2 D Lt
Applied Problem
Given the following data at 140 days, determine the
extent of the plume.
Cmax = 450 ppm
DL = 0.984 m2/d
DT = 0.1DL
Solution
s
s
x
y

2 0 . 984

275 . 52 m

2 0 . 0984

27 . 552 m
m
140 d 
2
d
2
m
 16 . 6 m
2
d
2
140 d 
 5 .2 m
The leading edge of the plume is 3sx = 3(16.6) =50m
ahead of the center of mass (located at vxt). The
plume has spread out in the y-direction 3sy = 3(5.2)
=15.6 m.
Laboratory Experiment
t=0
Inject solute
Soil column
t=to
Inject solute
Field Scenario
aquifer
Analytical Solution
C
C C
DL 2  vx

x
x t
2
For the boundary conditions
C 0, t   Co t  0
C  x ,0   0 x  0
C  , t   0 t  0
Co 
C 
 erfc
2 
 L  v xt 
 vx L 

  exp 
 erfc
 DL 
 2 D Lt 
 L  v xt  

 
 2 D Lt 
Analytical Solution
Co 
C 
 erfc
2 
 L  v xt 
 vx L 

  exp 
 erfc
 DL 
 2 D Lt 
 L  v xt  

 
 2 D Lt 
Under certain conditions, we can drop this term.
Erfc
 Mathematical function related to the
normal, or Gaussian distribution
 This means that the solute
concentration is normally distributed
 erfc(B) = 1 - erf(B)
 erfc(-B)=1 + erf(B)
 erfc(0) =1
 erfc(B) = 0 for B>3
Erfc
erf  B  
2

B
e
t
2
dt
0
This equation cannot be solved analytically. Can
find tables of values. It can also be approximated
by the analytical expression:
erf  B  
  4B 
1  exp 

  
2
Erfc
1
0.8
0.6
erf
erfc
0.4
0.2
0
0
1
2
3
Problem
Compute concentration at L=200m for the
following conditions:
Co = 1757 g/m3
vx = 7.67x10-6 m/s
DL = 3.8 x 10-5 m2/s
t = 6 months (1.58 x 107 s)
Make two calculations, one with the short and one
with the long form of the equation. What is the %
difference?
Solution
C 
1757
g
m

 erfc

3
2
 18 . 07
g
m
 200 m  7 . 67 x 10  6 m 1 . 58  10 7 s   
s


 2 3 . 8  10  5 m 2 1 . 58  10 7 s   
s


2
Co 
C 
 erfc
2 
 L  v xt 
 vx L 

  exp 
 erfc
 DL 
 2 D Lt 
 L  v xt  

 
 2 D Lt 
This term equals zero
Solution
x
140
160
180
200
220
C (partial eqn)
462.40
206.86
70.60
18.07
3.43
C (full eqn.) ea
533.10
13.26
243.94
15.2
70.60
0
18.07
0
3.43
0
Bear (1979 p. 268) states that the second term can be
neglected when vxL/D > 500, since the error < 3%.
Note, vxL/D is the Pechlet number.
Solution
x
140
160
180
200
220
C (partial eqn)
462.40
206.86
70.60
18.07
3.43
C (full eqn.) ea
533.10
13.26
243.94
15.2
70.60
0
18.07
0
3.43
0
Retardation
As a result of sorption processes, some solutes will
move more slowly through the aquifer than the
ground water that is transporting them
R
C
t
 C
2
 D
x
2
v
C
x
Example: Analytical Solution
Co 
C 
 erfc
2 
 L  v xt 
v L

  exp  x  erfc
 DL 
 2 D Lt 
 RL  v t
Co 
x
C 
 erfc 
 2 RD t
2 
L


 L  v xt  

 
 2 D Lt 

 RL  v t
 vx L 
x
  exp 
 erfc 
 D 

 2 RD t
L


L





 
Retardation
As a result of sorption processes, some solutes will
move more slowly through the aquifer than the
ground water that is transporting them
C
t
D  C
2

R x
2

v C
R x
Sorption decreases the value of the
transport parameters D and v
Retardation
C/Co
x2=vct
x1=vwt (w/ conservative tracer)
The velocity of the contaminant is less than
the velocity of the groundwater
Retardation
x/L = 1
1.00
0.80
R>1
R=1
C/Co
0.60
0.40
0.20
0.00
0.00
0.50
1.00
t*
1.50
2.00
Problem
Given the following graph, sketch C/Co vs x/L
for t* = 1 and t* =2.
x/L = 1
1.00
0.80
R>1
R=1
C/Co
0.60
0.40
0.20
0.00
0.00
0.50
1.00
t*
1.50
2.00
Solution
t* = 1
1.00
C/Co
0.80
0.60
R>1
0.40
R=1
0.20
0.00
0.00
0.20
0.40
0.60
x/L
0.80
1.00
Solution
t* = 2
1.00
C/Co
0.75
0.50
0.25
R>1
R=1
0.00
0.00
0.50
1.00
x/L
1.50
2.00
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
0.10
0.20
0.30
0.40
R > 1 (@ 0 x/L  1
0.50
0.60
0.70
0.80
0.90
1.00
1.00
2.00
t*
R = 1 (@ 0 x/L  1
C/Co
C/Co
Solution
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.00
t*
2.00
Decay
Modify the approximate solution for 1-D
Lv t
x
C 
erfc 
2 D t
2
L

Co




 C
 Lv t
o
x
C 
erfc 
2 D t
 2
L

    ln 2 t

e

 
Summary of Important Points
and Concepts
 Release of contaminants from a hazardous
waste source is a function of their sorptivity,
volatilization rate and transformation rate.
 Transport is controlled by advective and
dispersive processes in air and water.
 Analytical solutions are based on boundary
conditions that mathematically represent
continuous or instantaneous release of the
contaminant
Summary of Important Points
and Concepts
 The advection-dispersion equation for soil
may be modified for contaminant sorption
and transformation using the retardation
factor and a term for first order