Lecture series 6 - Civil and Environmental Engineering | SIU

Report
CE 510
Hazardous Waste Engineering
Department of Civil Engineering
Southern Illinois University Carbondale
Instructors: Jemil Yesuf
Dr. L.R. Chevalier
Lecture Series 6:
Volatilization
Course Goals
 Review the history and impact of environmental laws in the
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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
Volatilization
Evaporation of solid or liquid into the gaseous phase
Air emissions from hazardous management facilities
regulated under Clean Air Act.
May be considered Hazardous Air Pollutants (HAP)
Engineered Systems: Air
Stripping
Diagram of an air stripping
tower. Water is piped into
the top, and pours over the
packing. A Counter current
of air is blown into the
bottom of the tower, and
blows by the water,
removing the
contamination.
vent
contaminated
water in
packing material
packing support
clean water out
clean air in
air blower
Packing media: new and after
four years
Source
Engineered Systems: Soil Vapor
Extraction
vent to
atmosphere
air vent or injection well
vapor
treatment
C.F.
LNAPL
LNAPL
water table
residual NAPL
impermeable boundary
saturated
unsaturated
surface soil cap
vacuum
A typical vapor phase GAC
system. The grey cylinder
on the right is an air/water
separator. This unit is
necessary to keep water
from fouling the GAC
vessels. Water removed
from the soil gas is stored
in the black tank on the
right. The two white
cylinders contain the GAC,
which adsorbs organic
contaminants from the soil
gas. The grey unit on the
left contains the blower
that pulls the soil gas from
the wells and through the
vessels.
Source
Approach
Volatilization
from Open
Containers
Volatilization
from Surface Soils
First we need to review the definition of vapor
pressure and Henry’s Law
Volatilization from
Deep Soil
Contamination
Vapor Pressure
Range @ 20 C
10-10 mm Hg
760 mm Hg
T  VP 
The temperature that causes the
vapor pressure to reach 760 mm
Hg is the boiling point of the
compound.
See Appendix J p. 705
Table 6.1 p. 309
Higher driving force
Class Problem
Determine the vapor pressure for anthracene
and carbon tetrachloride. Which is more
volatile?
Henry’s Law
 For a closed system
 Equilibrium between gaseous and
aqueous phase
 Dilute contaminant
P  HX
P= partial pressure (atm)
H = Henry’s Law const. (atm-m3/mole)
X = concentration (mole/m3)
H<10-7 atm-m3/mole less volatile than water, conc. will increase.
H>10-3 atm-m3/mole volatilization is rapid
Henry’s Law
VP
H
S
Henry’s Law constant may also be considered a
partition coefficient between air and water,
analogous to the octanol-water partition
coefficient
Here, S is water solubility
High water solubilities and low vapor pressure
tend to decrease the potential for volatilization
of dilute species
Henry’s Law
B  Correction for temperature

H  exp A   See values Table 6.2 p. 310
T

H
- Dimensionless quantity that may be
H '
used in design
RT
- the ratio of the mass of compound in
the vapor phase to the mass of
compound in the aqueous phase.
Class Problem
Estimate Henry’s Law constant for benzene at
30C.
Governing Equation
B

H  exp A  
T

Solution
From Table 6.2, A = 5.53 and B = 3190.
B

H  exp A  
T

3190

 exp 5.53 
  0.0068
303 

compare to H = 0.0055 at 25 C
Class Problem
Experimental determination and application of Henry’s
constant
Air is comprised by 21% oxygen on a molar basis. If we bubble a
large amount of air through a liter of water until it is saturated
with the air, the amount of oxygen in the water at this
condition of equilibrium is dictated by Henry’s law. If the
amount of oxygen is measured using a DO probe and is 9.3
mg/L,
a) Calculate Henry’s law constant
b) Determine how much oxygen will be dissolved in water at
20 ºC if pure oxygen (PO2 = 1 atm) is bubbled through the
water until it is saturated.
Density of air at 20 ºC = 1.2 g/L
Density of water at 20 ºC = 998 g/L
Solution
a) Mass ratio of oxygen in the water:
= 9.3 mg/L X 0.001 g/mg X 1L/998 g
= 9.3x10-6 g-O2/g-water
Mass ratio of oxygen in the air:
Remember that 1 mole of any gas occupies 24.05 L of volume
at 1 atm. pressure and 20ºC (Ideal Gas Law) and density of air
is 1.2 g/L,
Number of moles of oxygen in 1 L of air is:
= 0.21 X (1/24.05) moles = 0.0087 moles
Mass ratio of oxygen in the air ( 1 liter basis) is then
calculated as:
= (0.0087 moles X 32 g/mole)/(1.2 g) = 0.233 g-O2/g-air
Solution
Henry’s constant in non-dimensional form (H’) is then:
= (0.233 g-O2/g-air)/ 9.3x10-6 g-O2/g-water
Converting g-air and g-water into volume basis,
multiplying the above expression by,
(1.2x103 g/m3)/(0.998x106 g/m3)
H’ = 30.12 (mol O2/m3-air)/(mol O2/m3-water)
And from H’ = H/RT,
H = H’RT = 30.12 x 8.21x10-5 x 293
= 0.725 atm-m3/mol
Solution
b) From Henry’s law equation:
PO2= H.X,
And for pure oxygen, PO2 = 1 atm; thus
X = PO2 /H
= 1 atm/(0.725 atm-m3/mol)
= 1.38 mol/m3
= 44.1 g/m3 = 44.1 mg/L
......end of example
Estimation of Flux from an Open
Container
Q  VP  P
This term represents the driving force
Q is the evaporation rate (mass/time)
VP is the vapor pressure (atm)
P is the partial pressure of the compound above the
liquid
If open, P = 0
Estimation of Flux from an Open
Container
MKA VP  P 
Q
RT
where
Q = mass flux (evaporation rate)
M = molecular weight
K = mass transfer coefficient per area
Estimation of Flux from an Open
Container
The mass transfer coefficient K can be estimated using
water as a reference
Kw = 0. 83 cm/s
Molecular weight of water = 18 g/m
 M2 

K1  K 2 
 M1 
1
3
 18 

 0.83
 M1 
1
3
Class Problem
A container of benzene has been left open. Estimate
the rate of volatilization across the surface of the
container. The dimensions of the container are 1.25m x
0.75m x 0.3 m deep. The temperature is 20C.
Approach
Governing Equation
MKA VP  P 
Q
RT
 M2 

K1  K 2 
 M1 
1
3
 18 

 0.83
 M1 
1
3
Solution
1. Need to solve for K.
2. Molecular weight of benzene is 6(12)+6 = 78 g/mol
 M2 

K1  K 2 
 M1 
1
 18 
 0.83 
 78 
3
1
3
 18 

 0.83
 M1 
 .509cm s
1
3
Solution
3. Area = 1.25 m x 0.75 m = 0.94 m2
4. Vapor pressure = 76 mm Hg = 0.1 atm
MKAVP  P 
Q
RT

78 g mol 0.00509m s  0.94m3 0.1 atm  0

3

m
atm 

 8.21105

293
K


m ol K 





 1.55 g s
Compare to Ex. 6.1 p. 313
Saturation in an Enclosed Area
Spills in
• waste transfer areas
• drum storage areas
Need to assess the saturated vapor concentration in
order to assess toxicity or explosivity of the vapor
Factors to consider:
volume of enclosed space
ventilation rate
contaminant flow rate
out of enclosed space
contaminant volatilization rate
Saturation in an Enclosed Area
C ppm
Qm RT
6

10
kQv PM
where
Qm = volatilization rate of the compound [M/T] (g/s)
Qv = ventilation rate of the enclosed area [M/T] (m3/s)
k = factor for incomplete mixing (0.1-0.5)
M = molecular weight (g/mol)
Class
A container of benzene has been left open in a
warehouse. The dimensions of the container are 1.25 m
x 0.75 m x 0.3 m deep. The temperature is 20C and
pressure is 0.1 atm. The facility is 220 m3 in volume.
Ventilation is 12 changes of air per hour. Using k=0.2,
determine the steady state benzene concentration in
the warehouse.
Governing Equation
MKA VP  P 
Q
 Qm
RT
C ppm
Qm RT
6

10
kQv PM
Solution
1. From previous problem Qm = 1.55 g/s
2. Calculate the ventilation rate
Qv = (12 changes of air/hr)(220 m3/change)(1 hr/3600 sec)
= 0.73 m3/s
Solution Continued
3. Determine the steady-state benzene concentration
C ppm
Qm RT

106
kQv PM

1.55 s 8.2110

293 K 
6


10
0.20.73m sec 0.1 atm78 g mol 
5
g
3
 3274ppm
m3
atmmol  K
Volatilization from Soils
Soil particles
air
diffusion
water
Sorption/desorption
Volatilization from Soils
Soil particles
diffusion
air
sorption
water
dC

C
dt
dC

 kC
dt
Ct
t
dC

 k  dt
0
C
Co
Ct  C o e
 kt
Volatilization from Soils
Soil particles
diffusion
air
water
sorption
VP
kv  4.4 10
K oc S
7
DOW researchers
determined an empirical
relationship for a first-order
decay rate for contaminant
loss from a surface spill, kv
where
VP = vapor pressure (mm
Hg)
Koc = soil adsorption coeff.
(mL/g)
S = solubility (mg/L)
Example
A carrier has spilled lindane on soil. Estimate the
time required for 70% volatilization.
Governing Equations
VP
kv  4.4 10
K oc S
7
Ct  Co e
 kt
Solution
From reference tables in the appendix
S = 7.3 mg/L
Koc = 1995 mL/g
VP = 9.4 x 10-6 mm Hg
VP
kv  4.4 10
K oc S
7
6
9
.
4

10
kv  4.4 107
 0.028 day1
19957.3
Solution
Ct  Co e  kt
Ct
 0.028 t
 0 .3  e
Co
t  43 days
Volatilization in Deep Soils
More complex
Models are mostly specific to matrix
Hamaker Equation
one of the better generalized models and assumes that
the contaminant zone is semi-infinite
i.e. the contaminant zone extend into the aquifer and
the source is large
Volatilization in deep soils
Qt  2Co
D1

D2
M2
M1
Dt

Qt = volatilization of compound per unit
surface area (g/cm2)
Co = initial concentration (g/cm3)
D = diffusion coefficient of vapor
through coils (cm2/s)
t = time (sec)
Very few data are available for diffusion
coefficients. This equation allows a
prediction of D based on the known value
of 0.01 cm2/s for ethylene dibromide and
0.042 cm2/s for ethanol.
Problem
Estimate the flux of benzene from a deeply contaminated
soil over 1 day with a concentration of 500 ppm and a bulk
density of 1.50 g/cm3.
Data and Governing Equations
MW Benzene 78 g/mol
MW Ethanol 46 g/mol
MW Ethylene dibromide 188 g/mol
D for ethanol 0.042 cm2/s
D for ethylene dibromide 0.01 cm2/s
Qt  2Co
D1

D2
M2
M1
Dt

Solution
1. Determine D using both equations and take the average
D1
46

 0.032
0.042
78
D1
188

 0.016
0.01
78
Daverage = 0.024 cm2/s
Solution
2. Convert 500 ppm benzene to g/cm3
mg
kg
mg
g
500
 0.0005
kg
g
500ppm  500

g 
g 
g
 0.0005 1.5 3   0.00075 3  Co
g  cm 
cm

Solution
3. Determine Q
Qt  2Co
Dt

 20.00075g cm3
 0.038 g cm3

0.024 86400s 


cm2
s
Summary of Important Points
and Concepts
 The two most important parameters for
assessing volatilization are vapor pressure
and Henry’s Law constant
 Vapor pressure of hazardous waste range
from essentially nonvolatile to those that
rapidly evaporate
 Henry’s Law constant is a useful predictor of
volatilization from water
 Henry’s Law constant is analogous to the
octanol-water partition
Summary of Important Points
and Concepts
 Vapor pressure and the mass transfer
coefficient K are needed to determine the
flux across an open container
 For enclosed areas, an equation based on
mass balance can be used to determine the
concentration of contaminant in the air
 Equations for determining the volatilization
for soils are more complex because of
variability in characteristics (e.g. sorption,
water content, diffusion)
Summary of Important Points
and Concepts
 Researchers at Dow developed an empirical
equation for estimating the first order decay
rate for surface soils
 Hamaker’s equation works reasonable well
for a range of problems involving the
volatilization of contaminants in deep
aquifers

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