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Introduction to Environmental
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Order PROPRIETARY MATERIAL . © The McGraw -Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyon d the limited distribution to teachers and educators permitted by McGraw -Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1-1 CHAPTER 1 SOLUTIONS 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 dm1052.1
mL 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 PROPRIETARY MATERIAL . © The McGraw -Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyon d the limited distribution to teachers and educators permitted by McGraw -Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 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 0.00 50.00 100.00 150.00 200.00 250.00 300.00 350.00 400.00 450.00 500.00 550.00 600.00 650.00 1940 1950 1960 1970 1980 1990 2000 2010 Liters per capita per day Year Problem 1 -2:Per Capita Daily Water Withdrawal for Public Supply Estimated = 575 Lpcd PROPRIETARY MATERIAL . © The McGraw -Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyon d the limited distribution to teachers and educators permitted by McGraw -Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1-3 1-4 Additional average water production required with low -flush toilets 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 l oss, cost of water = $0.45/m3 Solution:
