Hazardous Waste Management, 2nd ed. Instructors’ Manual Chapter 13: Land Disposal CHAPTER 13 LAND DISPOSAL Supplemental Questions: Which of Shakespeare's plays is the source of the opening quote? The Tempest [1611-1612],Act: I, Scene: i, Line: 70. 13-1. Cite four reasons landfills remain a viable hazardous waste management alternative. Landfills remain a viable hazardous waste management alternative because other hazardous waste management techniques such as source reduction and recycling cannot totally eliminate the waste generated. Also, other methods such as incineration and biological treatment produce residues that must be eliminated. Landfills are also alternatives as a component of remediation of uncontrolled hazardous waste sites. Finally this waste needs a final disposal site and from our choices of land, water or air, land seems to be the best choice for final disposal. 13-2. Redo Example 13-1 for three additional hazardous waste management techniques. (a) 1) 2) 3) 4) (b) 4) Stabilization/Solidification Improve the physical nature/engineering properties of the waste Reduces contaminant toxicity Economical treatment process where large quantities of soil containing contaminants are encountered Reduces contaminant migration 1) 2) 3) 4) Biodegradation Can break contaminants down to less toxic forms Uses "natural" phenomena Can be used in situ Can be used to remediate sites with non-aqueous phased liquids (NAPL) 1) 2) 3) (c) 13-3. Incineration Can significantly reduce waste volume Can significantly reduce waste toxicity Can handle wastes not amenable to other treatments; i.e. liquids Can break down organic contaminants into a less toxic forms Explain the relationship between aquifers and the siting of landfills. The natural environment is very important when selecting a landfill site. It should provide a measure of redundancy to the landfill to protect the environment and public health. Locating a landfill in an area providing an upward gradient from deeper aquifers, for example, would carry contaminants away from the aquifer rather than into the aquifer in case of leakage or a spill. Also, siting a landfill where there is a downward gradient from the land to the aquifer would assure any contaminant leaving the land through the water system would reach the aquifer. Thus aquifers provide a level of redundancy and natural protection. 2001Environmental Resources Managment Page 1 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-4. Chapter 13: Land Disposal Considering the results of Example 13-3, discuss the advantages and disadvantages of employing 0.2 m of daily cover. Use calculations to support your answer. Total landfill space is: 1.88 x 106 cubic meters = V 1 Each daily lift has a volume of: 300 cubic meters = V2 Daily cover volume: 15 x 10 x 0.2 = 30 cubic meters = V3 Life of landfill using daily cover: t1 = V1/(V2 + V3) = 5697 days Life without cover: 6267 days = t2 Landfill life is extended by: t3 = t2 – tl = 570 days Based on these results it would be more beneficial to use a 0.2 m daily cover than a 0.3 m cover. The life of the landfill is increased by roughly 1.6 years with 0.2 m (compared to over 2 years with 0.3 in). The 0.2 m daily cover is less expensive than the 0.3 m cover and provide an extra 248 days to the life of the landfill. 13-5. Assume 17% of the infiltration is absorbed by the landfill and daily cover in Example 13-4. What will be the new rate of leachate generation? Given: Example 13-4 : Assume 17% of infiltration absorbed by landfill and daily cover Solution: 17% absorbed by daily cover 83% infiltrates landfill L = 1000 x 1000 x .83 = 8.3 x 105 cubic meters/year = (8.3 x 105 cubic meters/year) / (365 days/year) = 2274 cubic meters/year = (2274 cubic meters/year) x 1 day/24 hrs x 1 hr/60 min = 1. 58 cubic meters/minute = 1580 L/min = 601, 045 gal/day 13-6. Assume the landfill waste in Example 13-4 consolidates and releases liquids as leachate from the pore spaces equal to 1% of the landfill volume. No infiltration is absorbed by the landfill and daily cover. What will be the new rate of leachate generation? Assume average landfill height of 10 m VT=1000 x 1000 x 10=1 x 107 m3 Leachate = 1 % VT = 1 x 105 m3 Assume 1 % per year ∴Leachate =1 x 105 m3/year = 70,000 gaI/year 2001Environmental Resources Managment Page 2 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-7. Chapter 13: Land Disposal Divide the class into two groups: environmental extremists and industrial party liners. Discuss the following questions employing factual information. (a) We are just making our 55-gallon drums bigger. (b) All landfills leak, so they should be banned. (c) When single liners were found inadequate, double liners were used beneath landfills. Now (1992) hazardous waste landfills have three liners, two of geosynthetic materials and one of natural materials. Are landfills now "safe" methods of hazardous waste disposal or will the future prove these systems inadequate? Class Discussion. 13-8. Comment on the implications of an action leakage rate of 5 gal/acre/day versus the readily obtainable leakage rate of 22 gal/acre/day. Given: Action leakage rate of 5 gal/acre/day. Readily obtainable leakage rate of 22 gal/acre/day Solution: All landfills leak. A leakage rate of 22 gal/acre/day through the primary liner can be expected. A secondary leachate collection zone, previously known as the leak detection system, underlies the primary barrier layer to handle the leachate that has leaked through. The action leakage rate of 5 gal/acre/day requires no corrective action. This is much less than the leakage rate of 22 gal/acre/day that can be readily achieved. 13-9. What are the factors which affect the rate of contaminant migration. The factors that affect the rate of contaminant migration include advection, dispersion, hydrolysis, sorption, cosolvation, ionization, precipitation and biotransformation. 13 -10. Explain the difference between the two types of molecular diffusion. Molecular diffusion occurs in two ways. Liquid phase diffusion through a saturated barrier or gas phase diffusion after partitioning from the liquid phase can occur. 13-11. You are considering the selection of a geomembrane as secondary containment beneath a buried underground fuel (gasoline) storage tank. PVC, HDPE, and neoprene are being considered. Which would you recommend and why? If cost was not a factor, I would recommend neoprene because it exhibits excellent resistance to oils, weathering , puncture, abrasion, and mechanical damage which would be necessary for it to perform well as a secondary containment beneath a buried underground fuel storage tank. HDPE is a less expensive alternative but, it would not provide the same level of resistance to oils as the neoprene. PVC is good if it's buried but it isn't good to use when oils are involved. 2001Environmental Resources Managment Page 3 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-12. Chapter 13: Land Disposal Identify and describe the three steps involved in the transport of a contaminant through a geomembrane. There are three steps involved in the transport of a contaminant through a geomembrane. The solution or adsorption of the permeant into the upstream surface of the geomembrane (partitioning of molecular from leachate to gases) occurs first. Next, diffusion of the contaminant throughout the geomembrane occurs. Finally, volatilization or desorption of the contaminant at the downstream surface of the geomembrane occurs. 13-13. Final cover over a hazardous waste landfill is typically installed to minimize the environmental risk by addressing potential contaminant pathways. Identify four of the contaminant pathways mitigated through the use of a cover system. 1) 2) 3) 4) 13-14. Direct contact with waste. Transport in contaminated precipitation runoff. Generation of leachate. Airborne transport. Explain the importance of the vegetative support layer in a landfill cover system. The vegetative support layer provides several important functions in the performance of the landfill cover. It reduces erosion and precipitation infiltration. It enhances evapotranspiration, the process of returning moisture absorbed by the topsoil layer back to the atmosphere. Evapotraspiration also reduces infiltration. 13-15. Identify the three major components of a landfill cover system and describe their functions. The three major components of a landfill cover system are the vegetative support layer, the filter and drainage layer, and the barrier layer. The vegetative support layer is a topsoil material used to support vegetation necessary for proper erosion control, minimization of precipitation infiltration, and evapotranspiration. The drainage layer is used to enhance lateral drainage of any precipitation that infiltrates through the vegetative soil zone. The filter acts as a separation layer and minimizes the migration of materials. The barrier layer are used as the final layer to minimize infiltration through the cover system. 2001Environmental Resources Managment Page 4 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-16. Chapter 13: Land Disposal Sketch your conceptual design of a cover system for four regions of the United States: Minnesota, Arizona, Washington, and South Carolina. Minnesota • compacted clay cold -> freeze/thaw protection • geomembrane needed Arizona Special considerations • wind erosion • burrowing animals Washington Special considerations • significant rainfall South Carolina Special considerations • significant rainfall 13-17. Water treatment systems often use sand and gravel as filters just as landfills use sand and gravel in leachate collection and filter zones. How are sand filters cleaned in water treatment plants? What are the implications for cleaning sand in leachate collection zones? What are implications for attempting to clean geotextiles? Sand filter in a wastewater treatment plant are cleaned by a backwash process where water is sent in a reverse direction to remove trapped particles. Cleaning sand in leachate zone would be difficult sand could potentially be cleaned in a leachate collection zone by pumping water back through the collection pipes. Geotextiles are presently not cleaned. 2001Environmental Resources Managment Page 5 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-18. Chapter 13: Land Disposal A 12-inch layer of soil-bentonite is planned for placement immediately beneath the secondary geomembrane liner. The layer is to be constructed in two six inch lifts with a bentonite content of 9% by dry weight at a compacted unit weight of 95 pcf. By weight, how much bentonite is needed per square foot of liner per lift? Each lift is 0.5 ft thick yielding: Wt = 0.5 ft x 95 lb = 47.5 per square foot of lift ft Of this 9% is bentonite and the remainder is the base soil. Wt = 0.09 x 47.5 = 4.275 lbs bentonite per square foot per lift 13-19. Using the data from Table 13-7: (a) Calculate the total contaminant migration of xylene through a 10-acre site lined with a single 0.8 mm HDPE geomembrane. (b) Calculate the number of gallons of groundwater that could become contaminated with 1.0 µg/L for this contaminant in one year. Given: xylene, From Table 13.7: 10 acre site, 0.8 mm HDPE, 1 year - contaminant migration SVT = 21.6 g (solvent vapor transmissibility) m2d 10 acres x 43560 ft2 = 0.0929 m2 = 40,469m2 acre ft2 (a) Flux = Rate x Area = 21.6g x 40469m2 x 365 days/yr m2/ day = 3.2 x 108 g/year Flux = 1 µg/l VT VT = 3.2 x 108 g/yr 1µg/1yr / 106 µg/g (b) = 3. 2 x 1014 x 0.2642 g/1 = 8. 5 x 1015 gallons/year 2001Environmental Resources Managment Page 6 of 7 Hazardous Waste Management, 2nd ed. Instructors’ Manual 13-20. Chapter 13: Land Disposal Using the data from Figs. 13-15 and 13-16 and assuming a liner beneath a site of 10 acres, calculate the number of gallons of ground water that could become contaminated per year to the drinking water limit for the contaminant during the first year, at 10 years, and at 100 years. 10 acre liner without flyash: Gallons of groundwater that could become contaminated after1 year: Figure 13-16 total solute flux after 1 year is about 0. 001 g / l x 103 cm2 year : For 10 acres we have: Total = rate x area = 2 ft 2 1kg 0.001g 6 cm × × × × × acres 10 43560 1 . 44 10 3 2 2 acre 1000 g ft 1 × 10 cm = 627.26kg contaminant 0.005 mg = concentration kg 627.26kg × 10 6 Total kg of water = mg 0.005 kg mg kg = 1.2545 × 1011 kg water - necessary 1. 2545 x 1011 kg / 1000 kg = 1.2545 x 108 m3 m3 1. 2545 x 108 m3 x 10-3 L x 0. 2642 gal = 33,144 ga1 m3 L 10 years: 100 years: 13-21. 33,144 x 10 = 331, 440 ga1 33,144 x 100 = 3, 314, 400 ga1 Describe the various ways by which liquids can migrate through a geomembrane liner. Liquids can migrate direct through the liner via defaults such as: 1) 2) 3) improperly completed seams manufacturing defects (pin holes) construction and handling defects (tears, rips, etc.) Liquids phase contaminants can also migrate as a result of phase changes as discussed in Problem 13-12 and described on p. 814. 2001Environmental Resources Managment Page 7 of 7