Environmental technology | Education homework help

  1. Explain noise pollution. Reading Assignment In order to access the following resource, click the link below: Ashley, K. (Ed.), Waits, A., Hartzell, E., & Harten, T. (Reps.). (2005). Analytical performance criteria. The U.S.

Environmental Protection Agency Environmental Technology Verification Program--An overview. Journal of Occupational & Environmental Hygiene, 2(11), D87–D90. https://libraryresources.columbiasouthern.edu/login?url=https://search.ebscohost.com/login.aspx?dire ct=true&db=a9h&AN=20504615&site=ehost-live&scope=site Unit Lesson Introduction We are now in Unit VIII. We have learned a lot about environmental technologies. We began with key principles of technologies. Then, we learned about water treatment, piping, solid waste, hazardous waste, air pollution, and noise pollution. This lesson will describe a case study scenario that brings together the course learning objectives. Scenario A fictitious company called GoodProcesses Inc. has (fictitiously) had conversations with state and local government officials. GoodProcesses Inc. is proposing using an existing building for its manufacturing process. It needs to make sure the existing piping and water treatment tanks are sufficient. It also needs to determine its classification under hazardous waste laws, treat gaseous emissions, and address noise concerns. In addition, GoodProcesses Inc. is considering building and operating a solid waste landfill and needs to determine if there is enough land at its selected location. Piping system: GoodProcesses Inc. plans to purchase an old industrial facility for a manufacturing process. The old facility has an elevated tank that is open to atmospheric pressure that delivers water through a pipe to a process at atmospheric pressure but at a lower elevation than the tank. The existing pipe is an 8-inch diameter iron water pipe that has an elevation drop of 1.25 feet over its 250 foot length. GoodProcesses Inc. UNIT VIII STUDY GUIDE Putting It All Together ENV 4303, Environmental Technology 2 UNIT x STUDY GUIDE Title needs to determine if the pipe will carry the 300 gpm of water that they need for its manufacturing process. The flow rate is shown below (Nathanson & Schneider, 2015, p. 33). 𝑄𝑄 = 0.28 𝐢𝐢 𝐷𝐷2.63𝑆𝑆0.54 Q = Flow rate, gpm (gallons per minute) C= Hazen-Williams Coefficient D = Pipe inside diameter, inch S = Slope of hydraulic grade line, dimensionless For the plant, the following equation is relevant. 𝑆𝑆 = Elevation Drop Pipe Length = 1.25 ft 250 ft = 0.005 ft ft Since the pipe is old, C is approximately 100. 𝑄𝑄 = 0.28 𝐢𝐢 𝐷𝐷2.63𝑆𝑆0.54 = (0.28)(100)(8 in. )2.63 οΏ½0.005 ft ft οΏ½ 0.54 = 380 gpm Since the flow rate required is 300 gpm and the pipe can deliver 380 gpm, the existing pipe is adequate. Water treatment: GoodProcesses Inc. needs to filter the suspended sediment out of the water so that the water is clean enough to use for its manufacturing process. There is an existing rapid sand filter in the facility that has an area (looking down on it since water flows through it vertically) of 180 ft2. Is this area sufficient to process the desired flow rate of 300 gpm? From Nathanson and Schneider (2015), a rapid sand filter effectively operates at 2 gpm/ft2 where the ft2 represents the cross-sectional area (looking down) on the tank holding the sand. A 180 ft2 tank can process the flow rate of dirty water shown below. 𝑄𝑄 = 𝑅𝑅𝑅𝑅 Q = Flow rate, gpm R = Design filtration rate = 2 gpm/ft2 A = Tank (filter) area = 180 ft2 Therefore, the equation is as shown below. 𝑄𝑄 = οΏ½2 gpm ft2 οΏ½ (180 ft2) = 360 gpm The rapid sand filter can process 360 gpm, which shows that it is sufficiently large and will easily process the required 300 gpm. Hazardous waste: GoodProcesses Inc. realizes that it will generate hazardous waste from its manufacturing process. However, the company is hopeful that it will be classified as a conditionally exempt small-quantity generator. A conditionally exempt small-quantity generator generates less than 100 kg of hazardous waste per month (Nathanson & Schneider, 2015, p. 322). Though GoodProcesses Inc. knows its processes, it is not well versed on what exactly constitutes hazardous waste. Thus, the company has hired an environmental firm to compute its hazardous waste generation. The environmental firm computed a hazardous waste generation rate of 0.2 kg for each hour that GoodProcesses Inc.'s manufacturing process is in operation. GoodProcesses Inc. plans to operate the process 16 hours per 24-hour period on weekdays but not at all on weekends. ENV 4303, Environmental Technology 3 UNIT x STUDY GUIDE Title The number of hours that the manufacturing process runs per week TW is shown below. π‘‡π‘‡π‘Šπ‘Š = οΏ½16 hr day οΏ½ οΏ½5 day wk οΏ½ = 80 hr wk Since months have variable numbers of weeks, to get a monthly number of hours, it is best to use the value 52 weeks/year. Hours of operation per month TM: 𝑇𝑇𝑀𝑀 = οΏ½ 80 hr wk οΏ½ οΏ½ 52 wk yr οΏ½ οΏ½ yr 12 months οΏ½ = 346.7 hr month The amount of hazardous waste generated per month is shown below. 𝐻𝐻 = 𝑅𝑅 𝑇𝑇𝑀𝑀 H = Mass rate of hazardous waste generation R = Hazardous waste generation rate = 0.2 kg/hr of operation TM = Hours of operation per month Therefore, the equation is as shown below. 𝐻𝐻 = 𝑅𝑅 𝑇𝑇𝑀𝑀 = οΏ½

0.2 kg hr

οΏ½ οΏ½ 346.7 hr month οΏ½ = 69.3 kg month Therefore, GoodProcesses Inc. is a conditionally exempt small-quantity generator since 69.3 kg/month is less than 100 kg/month. Air pollution control: As a part of its pollution prevention plan, GoodProcesses Inc.'s management has a goal of removing 90% of its particulate matter from its gaseous stream that flows at 140 m3/s during process operation. GoodProcesses Inc. has hired an engineering firm to determine how best to achieve the 90% goal. The engineering firm determined that an electrostatic precipitator would be the most cost-effective option for the company’s process. To size the electrostatic precipitator's plate, the following equation is used (Nathanson & Schneider, 2015, p. 380). 𝐸𝐸 = 100 οΏ½1 βˆ’ 𝑒𝑒 βˆ’π‘€π‘€π‘€π‘€ 𝑄𝑄 οΏ½ E = Percent removal efficiency = 90% w = Effective drift velocity A = Total collector plate area Q = Gas flow rate of gas through the electrostatic precipitator = 140 m3/s The effective drift velocity is a design value set by the manufacturer of the electrostatic precipitator. The manufacturer indicates a value of w=0.1 m/s. The above equation is algebraically rearranged to solve for A. 𝐸𝐸 100 = 1 βˆ’ 𝑒𝑒 βˆ’π‘€π‘€π‘€π‘€ 𝑄𝑄 Then, the equation would be represented as shown below. 𝑒𝑒 βˆ’π‘€π‘€π‘€π‘€ 𝑄𝑄 = 1 βˆ’ 𝐸𝐸 100 ENV 4303, Environmental Technology 4 UNIT x STUDY GUIDE Title Take the natural logarithm of both sides. ln�𝑒𝑒 βˆ’π‘€π‘€π‘€π‘€ 𝑄𝑄 οΏ½ = ln οΏ½1 βˆ’ 𝐸𝐸 100 οΏ½ Thus, the following equation must be resolved. βˆ’ 𝑀𝑀𝑅𝑅 𝑄𝑄 = ln οΏ½1 βˆ’ 𝐸𝐸 100 οΏ½ So, the following equation would be the result. 𝑅𝑅 = βˆ’ 𝑄𝑄 𝑀𝑀 ln οΏ½1 βˆ’ 𝐸𝐸 100 οΏ½ Compute area A. 𝑅𝑅 = βˆ’ 𝑄𝑄 𝑀𝑀 ln οΏ½1 βˆ’ 𝐸𝐸 100 οΏ½ = βˆ’ οΏ½140 m3 s οΏ½ οΏ½0.1 m s οΏ½ ln οΏ½1 βˆ’ 90 100 οΏ½ = 3223 m2 The engineering firm suggests that GoodProcesses Inc. should purchase an electrostatic precipitator that has an area of at least 3223 m2 in order to achieve 90% removal of particulate matter. Production of noise: GoodProcesses, Inc. realizes that it must also address potential noise issues. The company wants to make sure that it does not exceed 80 dB (decibels) for its loudest process after using cushioning, mufflers, and enclosures to reduce noise. Considering all of the equipment in the building and the noise protection measures, the loudest process is rated to put out 1.5 ΞΌbar of sound pressure. GoodProcesses Inc. has staff that are able to convert the sound pressure rating to dB. The safety department uses the equation below (Nathanson & Schneider, 2015, p. 391). 𝑆𝑆𝑆𝑆𝑆𝑆 = 20 log οΏ½ 𝑆𝑆 π‘†π‘†π‘œπ‘œ οΏ½ log = Common (base 10) logarithm SPL = Sound pressure level, dB P = Sound pressure = 1.5 ΞΌbar Po = Reference sound pressure = 0.0002 ΞΌbar Therefore, the SPL from the 1.5 ΞΌbar process is as shown below. 𝑆𝑆𝑆𝑆𝑆𝑆 = 20 log οΏ½ 1.5 ΞΌbar 0.0002 ΞΌbar οΏ½ = 77.5 dB Since 77.5 dB is less than 80 dB, GoodProcesses Inc. has achieved its goal of having 80 dB or less noise for its loudest process. Solid waste: In addition to operating its own manufacturing facility, the management of GoodProcesses Inc. has been in conversations with the state’s environmental protection agency about constructing a new landfill. The site that the company would like to use has an area of 150 acres for the waste, plus additional land area for buildings and buffer zones. If the landfill is to serve an average population of 60,000 people for 30 years and the depth of the landfill is 25 ft, will the available land area be sufficient? ENV 4303, Environmental Technology 5 UNIT x STUDY GUIDE Title From Nathanson and Schneider (2015), the following design values are helpful (p. 310). R = Rate of waste generation per person = 5 pounds per person per day = 5 lb/person-day U = Unit weight of refuse = 1000 pounds per cubic yard = 1000 lb/yd3 C = Percent of volume used by cover material = 25% by volume = 0.25 ft3 cover volume/ft3 refuse volume From GoodProcesses Inc.'s proposal, the following information is relevant. P = Population = 60,000 people T = Landfill lifetime = 30 years Y = Landfill depth = 25 ft A = Landfill area = To be computed Compute total weight of refuse generated per year W. π‘Šπ‘Š = 𝑆𝑆𝑅𝑅 = (60,000 people) οΏ½ 5 lb person βˆ’ day οΏ½ οΏ½ 365 day yr οΏ½ = 1.095x108 lb yr Total yearly volume of refuse generated VR is as shown below. 𝑉𝑉𝑅𝑅 = π‘Šπ‘Š π‘ˆπ‘ˆ = 1.095x108 lb yr 1000 lb yd3 = 109,500 yd3 refuse yr Additional annual volume for cover material VC is as shown below. 𝑉𝑉𝐢𝐢 = 𝐢𝐢𝑉𝑉𝑅𝑅 = οΏ½0.25 yd3 cover yd3 refuse οΏ½οΏ½109,500 yd3 refuse yr οΏ½ = 27,375 yd3 cover yr Total annual required landfill volume VT is as shown below. 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑅𝑅 + 𝑉𝑉𝐢𝐢 = 109,500 yd3 refuse yr + 27,375 yd3 cover yr = 136,875 yd3 yr Total lifetime required landfill volume V is as shown below. 𝑉𝑉 = 𝑇𝑇 𝑉𝑉𝑇𝑇 = (30 yr)οΏ½ 136,875 yd3 yr οΏ½ = 4.106x106 yd3 Landfill area A required is as shown below. 𝑅𝑅 = 𝑉𝑉 π‘Œπ‘Œ = (4.106x106 yd3) (25 ft) οΏ½ 27 ft3 yd3 οΏ½ οΏ½ acre 43,560 ft2 οΏ½ = 102 acres Therefore, the available land area of 150 acres is more than enough space because only 102 acres are required. ENV 4303, Environmental Technology 6 UNIT x STUDY GUIDE Title Conclusion This lesson has discussed a fictitious company's considerations for a manufacturing process facility and a solid waste landfill. The lesson has brought together calculations for pipe sizing, water treatment, hazardous waste generation, air pollution control, noise, and landfill sizing. Now, you are more versed in environmental technology as a result of this course. Reference Nathanson, J. A., & Schneider, R. A. (2015). Basic environmental technology: Water supply, waste management, and pollution control (6th ed.). Pearson. Suggested Reading The following chapters were required in previous units. Review the following chapters to refresh your memory to help you with the final assignment: Chapter 1: Basic Concepts Chapter 2: Hydraulics Chapter 4: Water Quality Chapter 5: Water Pollution Chapter 6: Drinking Water Purification Chapter 7: Water Distribution Systems Chapter 8: Sanitary Sewer Systems Chapter 11: Municipal Solid Waste Chapter 12: Hazardous Waste Management Chapter 13: Air Pollution and Control Chapter 14: Noise Pollution and Control

  • Course Learning Outcomes for Unit VIII
  • Reading Assignment
  • Unit Lesson
  • Suggested Reading
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