CE 510Hazardous Waste Engineering

Department of Civil EngineeringSouthern 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 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)

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.

contaminated water in

clean water out

packing support

air blower

clean air in

packing material

ventEngineered Systems: Air Stripping

Source

Packing media: new and after four years

water table

C.F

.

un

satu

rate

dsa

tura

ted

impermeable boundary

LNAPLLNAPL

residual NAPL

Engineered Systems: Soil Vapor Extraction

air vent or injection well

vacuum vapor treatment

vent to atmosphere

surface soil cap

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

First we need to review the definition of vapor pressure and Henry’s Law

Volatilization from Deep Soil

ContaminationVolatilization

from Surface Soils

Volatilization from Open Containers

Vapor Pressure

10-10 mm Hg

760 mm Hg

Ran

ge @

20

C

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. 705Table 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

HXP 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

S

VPH

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

RT

HH

T

BAH

'

expCorrection for temperatureSee values Table 6.2 p. 310

- Dimensionless quantity that may be used in design

- 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 30C.

Governing Equation

T

BAH exp

Solution

From Table 6.2, A = 5.53 and B = 3190.

0068.0303

319053.5exp

exp

T

BAH

compare to H = 0.0055 at 25 C

Class ProblemExperimental determination and application of Henry’s constantAir 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 constantb) 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/LDensity of water at 20 ºC = 998 g/L

Solutiona) 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 molesMass 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

SolutionHenry’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

Solutionb) 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

PVPQ

This term represents the driving forceQ is the evaporation rate (mass/time)VP is the vapor pressure (atm)P is the partial pressure of the compound above the liquidIf open, P = 0

Estimation of Flux from an Open Container

RT

PVPMKAQ

where Q = mass flux (evaporation rate)M = molecular weightK = 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/sMolecular weight of water = 18 g/m

31

1

31

1

221

1883.0

MM

MKK

Class ProblemA 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 20C.

Approach

RT

PVPMKAQ

Governing Equation

31

1

31

1

221

1883.0

MM

MKK

Solution

1. Need to solve for K. 2. Molecular weight of benzene is 6(12)+6 = 78 g/mol

scm

MM

MKK

509.78

1883.0

1883.0

31

31

1

31

1

221

Solution

3. Area = 1.25 m x 0.75 m = 0.94 m2

4. Vapor pressure = 76 mm Hg = 0.1 atm

sg

sm

molg

KKmol

atmm

atmm

RT

PVPMKAQ

55.1

2931021.8

01.094.000509.0783

5

3

Compare to Ex. 6.1 p. 313

Saturation in an Enclosed AreaSpills in • waste transfer areas• drum storage areasNeed to assess the saturated vapor concentration in order to assess toxicity or explosivity of the vaporFactors to consider:

ventilation rate

volume of enclosed space

contaminant flow rate out of enclosed space

contaminant volatilization rate

Saturation in an Enclosed Area

610PMkQ

RTQC

v

mppm

whereQm = 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 20C 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

mQ

RT

PVPMKAQ

610PMkQ

RTQC

v

mppm

Solution1. From previous problem Qm = 1.55 g/s2. Calculate the ventilation rate

Qv = (12 changes of air/hr)(220 m3/change)(1 hr/3600 sec) = 0.73 m3/s

Solution Continued

ppm

atm

K

PMkQ

RTQC

molgm

Kmolatmm

sg

v

mppm

3274

10781.073.02.0

2931021.855.1

10

6

sec

5

6

3

3

3. Determine the steady-state benzene concentration

Volatilization from SoilsSoil particles

air

waterSorption/desorption

diffu

sion

diffusion

Volatilization from SoilsSoil particles

air

watersorptiondi

ffusi

on

diffusion

ktot

tC

C

eCC

dtkC

dC

kCdt

dC

Cdt

dC

t

o

0

Volatilization from SoilsSoil particles

air

watersorptiondi

ffusi

on

diffusion

DOW researchers determined an empirical relationship for a first-order decay rate for contaminant loss from a surface spill, kv

whereVP = vapor pressure (mm Hg)Koc = soil adsorption coeff. (mL/g)S = solubility (mg/L)

SK

VPk

ocv

7104.4

Example

A carrier has spilled lindane on soil. Estimate the time required for 70% volatilization.

Governing Equations

ktot

ocv

eCC

SK

VPk

7104.4

Solution

1

67 028.0

3.71995

104.9104.4

daykv

From reference tables in the appendix

S = 7.3 mg/LKoc = 1995 mL/gVP = 9.4 x 10-6 mm Hg

SK

VPk

ocv

7104.4

Solution

dayst

eC

C

eCC

t

o

t

ktot

43

3.0 028.0

Volatilization in Deep Soils

More complexModels are mostly specific to matrix

Hamaker Equationone 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

1

2

2

1

2

M

M

D

D

DtCQ ot

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 EquationsMW Benzene 78 g/molMW Ethanol 46 g/molMW Ethylene dibromide 188 g/molD for ethanol 0.042 cm2/s D for ethylene dibromide 0.01 cm2/s

1

2

2

1

2

M

M

D

D

DtCQ ot

Solution

016.078

188

01.0

032.078

46

042.0

1

1

D

D

Daverage = 0.024 cm2/s

1. Determine D using both equations and take the average

Solution

2. Convert 500 ppm benzene to g/cm3

oCcm

g

cm

g

g

g

g

g

kg

mg

kg

mgppm

3300075.05.10005.0

0005.0500

500500

Solution

3

2

3

038.0

86400024.000075.02

2

cmg

scm

cmg

ot

s

DtCQ

3. Determine Q

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