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Solution Manual For Introduction to Environmental Engineering 6th Edition By Mackenzie L. Davis, David A. Cornwell Chapter 1-14 $17.49
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Solution Manual For Introduction to Environmental Engineering 6th Edition By Mackenzie L. Davis, David A. Cornwell Chapter 1-14

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Solution Manual For Introduction to Environmental Engineering 6th Edition By Mackenzie L. Davis, David A. Cornwell Chapter 1-14

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  • June 28, 2024
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1-1 Solution Manual For Introduction to Environmental Engineering 6th Edition By Mackenzie L. Davis, David A. Cornwell Chapter 1 -14 CHAPTER 1 1-1 Total daily withdrawal Given: Population in 2000 = 281,421,906 Solution: a. Using the total daily withdrawal of 5,400 Lpcd: (281,421,906 people)(5,400 Lpcd) = 1.52 x 1012 L/d b. Converting to m3 dm1052.1mL 1000dL1052.13 9
312
 1-2 Estimate per capita withdrawal for public supply Given: Population data from 1950 to 2000 and corresponding public supply withdrawal Solution: a. Use a spreadsheet to estimate the withdrawal Problem 1 -2 Intro. To ENE 5th Edition Population Withdrawal, m3/d Year Withdrawal, Lpcd 151325798 5.30E+07 1950 350.24 179323175 7.95E+07 1960 443.33 203302031 1.02E+08 1970 501.72 226542203 1.29E+08 1980 569.43 248709873 1.46E+08 1990 587.03 281421906 1.64E+08 2000 582.75 1-2 Figure S -1-2: Per capita daily water withdrawal 1-3 Additional average daily water production required Given: 280 houses and, from text: 1,320 L/d - house Solution: (280 houses)(1,320 L/d - house) = 3.7 x 105 L/d 1-4 Additional average water production required with low -flush toilets 0.0050.00100.00150.00200.00250.00300.00350.00400.00450.00500.00550.00600.00650.00
1940 1950 1960 1970 1980 1990 2000 2010Liters per capita per day Year Problem 1 -2:Per Capita Daily Water Withdrawal for Public Supply Estimated = 575 Lpcd 1-3 Given: 320 houses that have low flush valves that reduce water consumption by 14% and, from text, 1,320 L/d - house Solution: Additional demand = (320 houses)(1,320 L/d - house)(1 – 0.14) = 3.18 x 105 L/d 1-5 Repeat Prob. 1 -3 for peak demand Given: 280 metered houses, AWWA average Solution: a. From text: peak hour = 5.3(avg. day) (5.3)(280 houses)(1,320 L/d - house) = 1.96 x 106 or 2 x 106 L/d at the peak hour 1-6 Water lost (in liters) in one year Given: One drop per second, 0.150 mL per drop Solution: (0.150 mL/s)(86,400 s/d)(365 d/y)(1 x 10-3 L/mL) = 4,730 L/y 1-7 Monthly cost of not repairing valve Given: Valves deliver 130.0 L/min, Water cost = $0.45 per cubic meter Solution: a. Assuming 30 d/mo (130.0 L/min)(1440 min/d)(30 d/mo)(1 x 10-3 m3/L) = 5,616 m3/mo (5,616 m3/mo)($0.45/m3) = $2,527.20 or $2,530/mo 1-8 Value of water lost Given: Year 2000 data from Prob. 1 -2, 15% water loss, cost of water = $0.45/m3 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. 4-4 Solution: a. Amount of water lost (1.6 x 108 m3/d)(0.15) = 2.4 x 107 m3/d b. Value (2.4 x 107 m3/d)($0.45/m3) = $1.08 x 107 or $1.1 x 107/d 1-9 Cost of bottled water Given: 0.5 L bottle of water costs $1.00 Solution: a. Convert L to m3 0005.0mL 1000L5.0
3 or 5.0 x 10-4 m3 b. Cost 3
3 4102$m100.500.1$ or $2000/m3 1-10 Daily per capita withdrawal for South Carolina Given: USGS circular 1268 at /usgs.gov Solution: a. From the web site Domestic withdrawal for SC = 63.5 x 106 gal/d Population = 4,010 x 103 b. Per capita   94.59Lgal 2642.0 people10 4010dgal105.63
36
 or Lpcd

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