Presentation on theme: "CE 510 Hazardous Waste Engineering"— Presentation transcript:
1 CE 510 Hazardous Waste Engineering Department of Civil EngineeringSouthern Illinois University CarbondaleInstructors: Jemil YesufDr. L.R. ChevalierLecture Series 6:Volatilization
2 Course GoalsReview the history and impact of environmental laws in the United StatesUnderstand the terminology, nomenclature, and significance of properties of hazardous wastes and hazardous materialsDevelop strategies to find information of nomenclature, transport and behavior, and toxicity for hazardous compoundsElucidate procedures for describing, assessing, and sampling hazardous wastes at industrial facilities and contaminated sitesPredict the behavior of hazardous chemicals in surface impoundments, soils, groundwater and treatment systemsAssess the toxicity and risk associated with exposure to hazardous chemicalsApply scientific principles and process designs of hazardous wastes management, remediation and treatment
3 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)
4 Engineered Systems: Air Stripping contaminated water inclean water outpacking supportair blowerclean air inpacking materialventDiagram 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.
6 Engineered Systems: Soil Vapor Extraction vent to atmosphereair vent or injection wellvacuumvapor treatmentsurface soil capunsaturatedLNAPLLNAPLC.F.saturatedwater tableresidual NAPLimpermeable boundary
7 A typical vapor phase GAC system 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
8 ApproachVolatilizationfrom OpenContainersVolatilization fromDeep SoilContaminationVolatilizationfrom Surface SoilsFirst we need to review the definition of vapor pressure and Henry’s Law
9 Vapor Pressure 10-10 mm Hg Range @ 20 C 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. 705Table 6.1 p. 30920 C760 mm HgHigher driving force
10 Class ProblemDetermine the vapor pressure for anthracene and carbon tetrachloride. Which is more volatile?
11 Henry’s Law For a closed system Equilibrium between gaseous and aqueous phaseDilute contaminantP= 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
12 Henry’s LawHenry’s Law constant may also be considered a partition coefficient between air and water, analogous to the octanol-water partition coefficientHere, S is water solubilityHigh water solubilities and low vapor pressure tend to decrease the potential for volatilization of dilute species
13 Henry’s Law Correction for temperature See values Table 6.2 p. 310 Dimensionless quantity that may be used in designthe ratio of the mass of compound in the vapor phase to the mass of compound in the aqueous phase.
14 Class ProblemEstimate Henry’s Law constant for benzene at 30C.
16 Solution From Table 6.2, A = 5.53 and B = 3190. compare to H = at 25 C
17 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,Calculate Henry’s law constantDetermine 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
18 Solution a) Mass ratio of oxygen in the water: = 9.3 mg/L X g/mg X 1L/998 g= 9.3x10-6 g-O2/g-waterMass ratio of oxygen in the air:Remember that 1 mole of any gas occupies 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 = molesMass ratio of oxygen in the air ( 1 liter basis) is then calculated as:= ( moles X 32 g/mole)/(1.2 g) = g-O2/g-air
19 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’ = (mol O2/m3-air)/(mol O2/m3-water) And from H’ = H/RT, H = H’RT = x 8.21x10-5 x 293 = atm-m3/mol
20 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
21 Estimation of Flux from an Open Container 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
22 Estimation of Flux from an Open Container whereQ = mass flux (evaporation rate)M = molecular weightK = mass transfer coefficient per area
23 Estimation of Flux from an Open Container The mass transfer coefficient K can be estimated using water as a referenceKw = cm/sMolecular weight of water = 18 g/m
24 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.
26 Solution Need to solve for K. Molecular weight of benzene is 6(12)+6 = 78 g/mol
27 Solution3. Area = 1.25 m x 0.75 m = 0.94 m24. Vapor pressure = 76 mm Hg = 0.1 atmCompare to Ex. 6.1 p. 313
28 Saturation in an Enclosed Area Spills inwaste transfer areasdrum storage areasNeed to assess the saturated vapor concentration in order to assess toxicity or explosivity of the vaporFactors to consider:volume of enclosed spacecontaminant flow rate out of enclosed spaceventilation ratecontaminant volatilization rate
29 Saturation in an Enclosed Area 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 ( )M = molecular weight (g/mol)
30 ClassA 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.
32 Solution From previous problem Qm = 1.55 g/s Calculate the ventilation rateQv = (12 changes of air/hr)(220 m3/change)(1 hr/3600 sec)= 0.73 m3/s
33 Solution Continued3. Determine the steady-state benzene concentration
34 Volatilization from Soils Soil particlesairdiffusiondiffusionwaterSorption/desorption
35 Volatilization from Soils Soil particlesdiffusionairdiffusionwatersorption
36 Volatilization from Soils Soil particlesDOW researchers determined an empirical relationship for a first-order decay rate for contaminant loss from a surface spill, kvwhereVP = vapor pressure (mm Hg)Koc = soil adsorption coeff. (mL/g)S = solubility (mg/L)diffusionairdiffusionwatersorption
37 ExampleA carrier has spilled lindane on soil. Estimate the time required for 70% volatilization.
41 Volatilization in Deep Soils More complexModels are mostly specific to matrixHamaker Equationone of the better generalized models and assumes that the contaminant zone is semi-infinitei.e. the contaminant zone extend into the aquifer and the source is large
42 Volatilization in deep soils 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 cm2/s for ethanol.
43 ProblemEstimate 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.
44 Data and Governing Equations MW Benzene 78 g/molMW Ethanol 46 g/molMW Ethylene dibromide 188 g/molD for ethanol cm2/sD for ethylene dibromide 0.01 cm2/s
45 Solution Daverage = 0.024 cm2/s 1. Determine D using both equations and take the averageDaverage = cm2/s
48 Summary of Important Points and Concepts The two most important parameters for assessing volatilization are vapor pressure and Henry’s Law constantVapor pressure of hazardous waste range from essentially nonvolatile to those that rapidly evaporateHenry’s Law constant is a useful predictor of volatilization from waterHenry’s Law constant is analogous to the octanol-water partition
49 Summary of Important Points and Concepts Vapor pressure and the mass transfer coefficient K are needed to determine the flux across an open containerFor enclosed areas, an equation based on mass balance can be used to determine the concentration of contaminant in the airEquations for determining the volatilization for soils are more complex because of variability in characteristics (e.g. sorption, water content, diffusion)
50 Summary of Important Points and Concepts Researchers at Dow developed an empirical equation for estimating the first order decay rate for surface soilsHamaker’s equation works reasonable well for a range of problems involving the volatilization of contaminants in deep aquifers
Your consent to our cookies if you continue to use this website.