Mathematics 108 Professor Alan H. Stein March 5, 2008 SOLUTIONS Solutions are given for questions involving calculations. Symbols which have standard meanings when used in class are used without explanation. Other symbols, of course, are defined. 1. Consider the Barkhamsted-New London Landfill, which is on the EPA s National Priorities List. Locate the site on a map and obtain driving directions to it. Briefly describe the site and the reasons it is on the NPL. Describe the remediation steps taken and the current status of the site. Page 1 of 6
Page 2 of 6 2. (Note: This is an artificial problem and the results obtained will not necessarily seem reasonable.) Consider an underground sand aquifer 50 feet thick and 75 feet wide. Suppose the hydraulic head at one point is 110 feet while the hydraulic head 100 feet downstream is 90 feet. (a) Determine the hydraulic gradient. Solution: i = H l = 110 90100 = 0.2. (b) Determine the minimum and maximum possible values for the flux. q = Ki. Since 0.01 K 1000, 0.01 0.2 K 1000 0.2, 0.002 K 200. The flux is between 0.002 and 200 feet per day. For the rest of this question, assume that the flux is the average of the two values you obtained in the first part. The average is 0.002+200 = 100.001, so we will use q = 100.001 feet per day in the 2 rest of the question. (c) Determine the volumetric flow rate. Solution: A = 50 75 = 3, 750, so Q = qa = 100.001 3750 = 375, 003.75 cubic feet per day. (d) Based on the range of possible porosities for gravel, determine the maximum possible interstitial velocity. Solution: Since the porosity is between 0.25 and 0.5, the maximum possible interstitial velocity is obtained taking a porosity of 0.25. Hence, v = q = 100.001 = η 0.25 400.004. Thus the maximum possible interstitial velocity is 400.004 feet per day. For the rest of this question, assume the value you just obtained is the actual interstitial velocity. (e) Approximately how long will it take for water to travel from the place where the hydraulic head is 110 feet to where the head is 90 feet? Solution: t = l = 100 0.2499975. It thus takes approximately 0.2499975 v 400.004 days (about a quarter of a day, or about six hours) to get to the point where the hydraulic head is 90 feet. (f) Suppose it is discovered that approximately 20 pounds of a toxic chemical has been leeching into the aquifer per week. Find the long term average concentration of the chemical, measured in both pounds per cubic foot and parts per million. Solution: Since the volumetric flow rate of 375, 003.75 cubic feet per day translates into 7 375, 003.75 = 2, 625, 026.25 cubic feet per day, so the 20 pounds of the 20 chemical leeching into the aquifer per week corresponds to 7.61897143 2,625,026.25 10 6 pounds per cubic foot. Since a cubic foot of water weights pounds, this corresponds to 7.61897143 10 6 1.22098901 10 7 pounds per pound of water, 0.122098901 pounds in a million pounds of water. The concentration is thus a miniscule 0.122098901 parts per million.
Page 3 of 6 3. Consider an aquifer where the water moves from point A to point B along a path that may be divided into four segments. The hydraulic conductivity is 40 feet per day, the porosity is 0.2, the hydraulic heads go from 100 feet to 95 feet to 90 feet to 85 feet to 80 feet and the segment lengths are 200 feet, 400 feet, 300 feet and 100 feet. (a) Determine how long, measured in days, it takes for water to travel the length of each segment along with the time it takes to travel the entire path. Solution: The following spreadsheet printout shows the calculations. Groundwater Travel Time Conductivity = 40 feet/day Porosity = 0.2 Segment Length h1 h2 i v(ft/day) t(days) 1 200 100 95 0.025 5 40 2 400 95 90 0.0125 2.5 160 3 300 90 85 0.016666667 3.333333333 90 4 100 85 80 0.05 10 10 Total 300 So it takes 300 days for water to travel the entire length. (b) What can you say about the variation in the cross-sectional area of the aquifer? 4. Consider the following one-dimensional diffusion problem. Fifteen (15) grams of a contaminant is released into a liquid which fills a long tube. The diffusion constant is D = 0.14 cm 2 /sec. Calculate the concentration of the contaminant 0.25, 0.5, 1, 1.5 and 2 cm from the point of release of the contaminant 1, 5 and 10 seconds after the release of the contaminant. M Solution: We use the one-dimensional diffusion formula C = e x2 4Dt. Using 4πDt M = 15, D = 0.14, we get the following spreadsheet. Each cell has the concentration in grams per centimeter x (distance) \ t (time) 1 5 10 0.25 10.11467524 4.945876114 3.536513112 0.5 7.236704581 4.625524802 3.420063445 1 1.896253733 3.538605392 2.991367171 1.5 0.203464724 2.264387088 2.392924469 2 0.008939616 1.212039067 1.75070094 5. Research the February 12 tractor-trailer accident involving the release of compressed hydrogen gas. Find the time and location of the accident. Describe the weather conditions, including the temperature, and precipitation and wind conditions. Find the amount of hydrogen gas released and the number of homes evacuated. Critique the response of the emergency services.
Page 4 of 6 6. What do you consider the single most important environmental issue today? Briefly describe that issue and justify it as your choice. 7. Define each of the following terms. If a symbol has been associated with a given term, indicate what that symbol is. (a) aquifer (b) aquitard (c) interstitial velocity (d) volumetric flow rate (e) hydraulic gradient (f) flux (g) porosity 8. It has been discovered that one of the storage tanks at a nearby gasoline station has been leaking for the last six months. The community relies on drinking water drawn from wells a half mile from the gasoline station. Discuss measures that may be taken to determine whether residents have been drinking contaminated water and, if so, for how long and at what concentration. Explain how it may be determined whether there is a future risk.
Page 5 of 6 9. Consider a portion of an aquifer where the water flows from an industrial site to a reservoir four thousand feet downstream. (Obviously, not the best situation.) Assume the hydraulic head at the industrial site is 225 feet and the hydraulic head at the reservoir is 185 feet. The aquifer flows through sand with a hydraulic conductivity 50 feet per day and porosity 0.35. The aquifer is 100 feet wide and 20 feet deep. Suppose also that a volatile organic compound (VOC) is leaching from the industrial site at a rate of 2 pounds per day. Calculate each of the following. (a) The flux q. Solution: K = 50 and i = h l flux is thus 0.5 feet per day. = 225 185 400 = 0.01, so q = KI = 50 0.01 = 0.5. The (b) The volumetric flow rate Q. Solution: A = 100 20 = 2000, so Q = KiA = 50 0.01 2000 = 1000. The volumetric flow rate is thus 1000 cubic feet per day. (c) The concentration C of the VOC, expressed in pounds per cubic foot. Solution: Since 2 pounds of the VOC leak into 1000 cubic feet of water each day, C = 2 = 0.002. The concentration is thus 0.002 pounds per cubic foot. 1000 (d) The interstitial flow rate v. Solution: v = q = 0.5 = 10 1.42857. The interstitial velocity is approximately η 0.35 7 1.42857 feet per day. (e) The amount t of time it takes for water to flow from the landfill to the well. Solution: t = l = 4000 = 2800. It takes approximately 2, 800 days for the water v 10/7 to flow from the landfill to the well. 10. Suppose a VOC with the same density of water is dissolved in water and has a concentration of 0.05 pounds per cubic foot. Assuming that a cubic foot of water weighs pounds, express the concentration in parts per million (ppm). Solution: 0.05 pounds per cubic foot per pound of water, which corresponds to 0.05 pounds for every pounds of water or 0.05 pounds per pound of water, which corresponds to 0.05 1, 000, 000 pounds for every million pounds of water. 1, 000, 000 801.282051282, the concentration is 801 parts per million. Since 0.05 11. Consider the formula given for Two-Dimensional Diffusion. Describe each of the variables involved.
Page 6 of 6 12. Assume 10 grams of a pollutant is released into a liquid that fills a long tube and diffuses from the center of the tube. Assuming a diffusion constant D = 0.2cm 2 /sec, find the concentration 0.5 cm from the center after 1 second and after 2 seconds. Solution: Using the one-dimensional diffusion formula C = M e x2 4Dt, 4πDt when t = 1 we get C = 10 4π0.2 1 e 0.52 4 0.2 1 4.61490796754, so the concentration is approximately 4.61490796754 grams per centimeter. 10 and when t = 2 we get C = e 0.52 4 0.2 2 3.81510556522, so the concentration is 4π0.2 2 approximately 3.81510556522 grams per centimeter.