Solution Manual for Water Supply and Pollution Control, 8th Edition

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Solutions ManualWater Supply and Pollution ControlEighth EditionWarren Viessman, Jr., P.E.University of FloridaMark J. Hammer, Emeritus EngineerLincoln, NebraskaElizabeth M. Perez, P.E.Palm Beach Gardens, FloridaPaul A. Chadik, P.E.University of Florida

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage3CHAPTER 1NO SOLUTIONS REQUIREDCHAPTER 2WATER RESOURCES PLANNING AND MANAGEMENT2.1The Internet is an excellent source of information on this topic.The level of integratedwater resources management varies by state.2.2Virtually all of the laws listed in Table 2.1 provide some protection for preventing andcontrolling water pollution. Information on each law may be found on the Internet. It isalso important to note that the EPA only regulates at the Federal level and much of thecleanup and protection is now delegated to states and local governments.2.3Point source pollution =Pollution that originates at one location with discrete dischargepoints.Typicalexamplesincludeindustrialandwastewatertreatmentfacilities.Nonpoint source pollution =Pollution that is usually input into the environment in adispersed manner.Typical examples include stormwater runoff that contains fertilizers,pesticides, herbicides, oils, grease, bacteria, viruses, and salts.2.4Adverse health effects of toxic pollutants are numerous and can include a variety ofconditions.Some pollutant-related conditions include asthma, nausea, and variouscancers—among many others.2.5Agencies that are responsible for water quantity and quality significantly vary by state.2.6This is a subjective question and one that has been and will continue to be debated in thewater resources community.2.7Integrated water resources management is difficult to achieve because it involves both afinancial and resources investment over time.It is also important to obtain concensus onthis approach from all of the involved stakeholders. This difficulty is perhaps why thereare so few examples of true integrated water resources management.2.8This question is subjective but the student should research specific examples to supporttheir argument.

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage4CHAPTER 3THE HYDROLOGIC CYCLE AND NATURAL WATER SOURCES3.1The answer to this question will vary bylocation.3.2reservoir area = 3900/640=6.1 sq. mi.annual runoff = (14/12)(190–6.1)(640) = 137,704 ac-ftannual evaporation = (49/12)(3900) = 15,925 ac-ftdraft = (100 X 365 X 106)/(7.48 X 43,560) = 112,022 ac-ftprecipitation on lake = (40/12)(3900) = 13,000 ac-ftgain in storage = 137,704 + 13,000 = 150,704loss in storage = 112,022 + 15,925 = 127,947net gain in storage = 22,757 ac-ft3.3reservoir area = 1700 hec = 17 X 106sq. metersannual runoff = 0.3(500 X 106–17 X 106) = 144 X 106sq. metersannual evaporation = 1.2 X 17 X 106= 20.4 X 106sq. metersdraft = 4.8 X 24 X 60 X 60 X 365 = 151.37 X 106m3precipitation on lake = 0.97 X 17 X 106= 16.49 X106m3gain in storage = 144 X 106+16.49 X 106= 160.49 106loss in storage = 151.37 X 106+ 20.4 X 106= 171.77 X 106net loss in storage = 11.28 X 106m33.4To complete a water budget, it is first important to understand how the water budget willbe used and what time step will be necessary to successfully model the system. Once thebudget is conceptually designed, a variety of online sources can usually be used to collectthe data. These sourcesinclude—but are not limited to:•state regulatory agencies•special water districts•weather agencies,•local governments•geological surveys•agricultural agenciesHistorical data and previous reports can also yield important information on the system.Verification and calibration data should also be considered as part of the data collectioneffort.3.5The solution for this problem will vary based on location.

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage53.6Event (n)Precip (inches)Tr = n/mFreq. (% years)13310102295203283.33304282.5405272506261.67607221.4708211.25809191.19010181100n= 10, m = rank, Tr = n/m, Freq = (1/Tr) X 100 Then plot precipitation versus frequency.3.7Event (n)Precip (inches)Tr = n/mFreq. (% years)18910102755203723.33304702.5405692506661.67607561.4708541.25809481.19010461100n = 10, m = rank, Tr = n/m, Freq = (1/Tr) X 100 Then plot precipitation versus frequency.

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage63.8Once the data is organized in a table (see below), the solution can be found. Note that thecumulativemax deficiency is 131.5 mg/mi2, which occurs in September. The number ofmonths of draft is 131.5/(448/12) = 3.53. Therefore, enough storage is needed to supplythe region for about 3.5 months.* Only positive values of cumulative deficiency are tabulated.3.9S = 128,000/10*100*640 = 0.203.10S = 0.0002 = volume of water pumped divided by the average decline in piezometrichead times surface area0.0002 = V/(400 X 100)Noting that there are 640 acres per square mileV= 0.0002 X 400 X 100 X 640 = 5120 acre-feetMonthInflowIDraftOCumulativeInflowΣIDeficiencyO-ICumulativeDeficiencyΣ (O–I)*Feb3137.3316.36.3March5437.385-16.70April9037.3175-52.70May1037.318527.327.3June737.319230.357.6July837.320029.386.9Aug237.320235.3122.2Sep2837.32309.3131.5Oct4237.3272-4.7126.8Nov10837.3380-70.756.1Dec9837.3478-60.70Jan2237.350015.315.3Feb5037.3550-12.72.6

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage73.11Draft = (0.726 mgd) X (30 days/mo) = 21.8 mg/month*Maximum storage deficiency is January 85.6 mg/mo/sq. mi.Storage capacity = 85.6 mg/mo/sq.mi.3.12Pn= (1–1/Tr)nlog Pn= n Log (1–1/Tr)n = log Pn/log (1–1/Tr)A straight line can be defined by this equation and the following probability curves willappear.MonthInflowIDraftODeficiencyO-ICumulativeDeficiencyΣ (O–I)*April9721.8-75.20May13621.8-114.20June5921.837.20July1421.87.87.8Aug621.815.823.6Sep521.816.840.43Oct321.818.859.2Nov721.814.874Dec1921.82.876.8Jan1321.88.885.6Feb7421.8-52.233.4*March9621.8-74.20April3721.8-15.20May6321.8-41.20June4921.8-27.20

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage83.1320 month flow equals the sum of 12 + 11 + 10 + 12 + … + 6 + 7 + 9 = 169 cfs3.14

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage93.153.16Reservoir capacity =750 acre-feetReservoir yield is the amount of water which can be supplied during a specified timeperiod. Assume the reservoir is to be operated continuously for 1 year without recharge.Also assume that evaporation, seepage, and other losses are zero.Max continuous yield is 750 acre-ft/yearOr 750 X 43,560 X 0.304 = 917, 846 cubic meters per yearOr 750 X 43,560 X 7.48 X 365 X 24 X 60 = 465 gpm continuously for 1 year3.17Constant annual yield = 1500 gpmReservoir capacity = ? Time of operation without recharge = 1 yrRes. Capacity = 1500 X 365 X 24 X 60 X 0.134 X (1/43,560) = 2,425 ac-ft/yrThis storage will provide a yield of 1,500 gpm for one year without any recharge3.18mean draft = 100 mgd, catchment area = 150 sq. mi., reservoir area = 4000 acresrainfall = 38 inches, runoff = 13 inches, evaporation = 49 inches (mean annual)(a)gain or loss in storage = ?ΔS = rainfall + runoff–evaporation–draftrainfall = 38 X 4000 X (1/12) = 12,667 ac-ftrunoff = [(150 X 640)–4000] X 13 X (1/12) = 99,967 ac-ftevaporation = 49 X 4000 X (1/12) = 16,333 ac-ftdraft = 100,000,000 X 365 X 0.134 X (43,560) = 112,282 ac-ftΔS = 12,667 + 99,667–16,333–112,282 =-16,281 ac-ftThe net loss in storage is 16, 281 ac-ft

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage10(b)volume of water evaporated = 16,333 ac-ftgiven a community of 100,000 people, assume a consumption of 150 gpcdwaterdemand = 100,000 X 150 X 365 = 5,475mg/yearvolume evaporated = 16,333 X 43,560 X 7.48 = 5,304 mg/yearevaporated water could supply the community with their water needs for5304/5475 = 0.97 or for about one year3.19Use equation 3.29K = 0.000287h = 43m = 8n = 15q = 0.000287*8*43/15 = 0.006582Total Q is therefore 50*0.006582 = 0.325 cfs3.20q= 000084*8*22/15 = 0.000986Q = 0.0007872*35 = 0.0345 m3/s3.21u = (1.87r2Sc)/Tt= (1.87 * 1 * 6.4 * 10-4)/(6200 * 7.5 * 24 * 60) = 8.58 x 10-10Interpolating, W(u) = 20.3S = (114.6 * 60,000* 7.5 * 20.3)/(6,200 * 7.5 * 24 * 60) = 15.63.22)()/log(5281212hhmrrQKf−=2554)110(*90)10log(*850*528ftgpdKf=−=3.23Equation 3.20is applicable)/log(1055)(1/22122rrhhKQf−=37107.0)100235log()log(12==rrfthfth8.772.22100792110012=−==−=

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage11gpmQ44.63437107.0*10558.7779(1320)22=−=3.24Using Equation 3.35, u can be computed544210*23.612*10*310*3*200*87.1−−==uReferring to Table 3.5and interpolating, we estimate W(u) to be 9.1. Then usingEquation 3.34, the drawdown is found to be:fts41.1010*3300*1.9*6.1144==3.25(a) Using Equation 3.35, u can be computed as follows:71.00028.0*1000*400098.0*90*90==uThen from Table 3.5, W(u) is found to be 0.36. Applying Equation 3.33, the drawdowncan be determinedms039.00028.0**436.0*0038.0=(c)Follow the procedure used in (a)0098.00028.0*72000*400098.0*90*90==uThen from Table 3.5, W(u) is found to be 4.06. Applying Eq. 3.33, the drawdown can bedeterminedms44.0000028.0**406.4*0038.0==3.26(a) Using Equation 3.31, u can be computed as follows:25.00028.0*3600*4001.0*100*100==uThen from Table 3.5, the drawdown can be determined,

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage12ms12.00028.0**407.1*004.0==(b) Follow the procedure used in (a)01.00028.0*60*60*24*4001.0*100*100==uThen from Table 3.5,W(u) is found to be 4.04Applying Equation3.33, the drawdown can be determinedms46.00028.0**404.4*004.0==3.27(a) Using Equation 3.31, u can be computed as follows:46.00028.0*60*60*12*4001.0*150*150==uThen from Table 3.5, the drawdown can be determined,ms05.00028.0**436.0*003.0==(b) Follow the procedure used in (a)023.00028.0*60*60*12*4001.0*500*500==uThen from Table 3.5, W(u) is found to be 3.24Applying Equation 3.33, the drawdown can be determinedms28.00028.0**424.3*003.0==3.28392,13)45/120(log*5288*100*1416.3*2*600)45/120(log*528)(**2*101012==−=hhKQfgal/min3.2962.40728.1*100)75/500(log*1200*528)()/(log**52810121210==−=hhmrrQKfgpd/ft23.30hQT=*264From a plot of drawdown versus t, drawdown per log cycle is 28.2–10.5 = 17.1

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage13==1.17*264TQConverting T to gal/day/ftT=51003301.17*2645100==Qgpm3.31From plot of data, t0=1.25 minutes = 20.87 x 10-3ft/day, and from plot,Dh14 feet565714300*264==Tgpd/ft00041.06010*87.0*5657*3.0**3.02320===−rtTSc3.3200011.0**87.12==TtSrucW(u)=-0.577216-ln(u)Substituting and solving, using loge(u)W(u)=8.53784.810*1.3537.8*280*6.114)(**6.1144===TuWQSfeet3.33Use Equation 3.22477.0)ln(12=rr468,13477.0*5289*100**2*600==Qgpm3.34Use Equation 3.2322.4338.10*130)65500ln(*1300*528ftgpdKf==3.35Use Equation 3.37and refer to figure which follows

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage14T=700*7.5 = 5250 gpd/ftFrom Fig change in head is 9.53 feet5.18926453.9*5250==Qgpm

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage153.36Use Equation 3.1941683.0)(log1210=rrQ=88141683.0*1055)5.77*5.774.79*4.79(*1300=−gpm3.37Use Equations 3.34and 3.35refer to the following figure determine s and r2/t from thefigure = 1.36 and 20,000Determine u and W(u) from the figure = 0.09 and 1.9ftgpdT050,80365.19.1*500*6.114==1926.020000*87.180050*09.0==cS

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage163.38Use Equation 3.1921222100)60100(6650y−−=156021=y, y1=39.5Drawdown is 100-39.5=60.5 feet3.39Use Equation 3.23Log of the ratio = 0.185628.428)9597(*801856.0*700*528ftgpdKf=−=

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage17CHAPTER 4ALTERNATIVE SOURCES OF WATER SUPPLY4.1Varies by location—students should be encouraged to research this on the Internet and inother local publications. It is useful for the student to learn what technologies are used intheir community since they will be explained further in later chapters of the book and thiswillprovide them with a vested interest.4.2The Internet is a good source of baseline information on this topic, in addition to themany water publications that are available to students in the library that will provide amore in-depth coverage.The social, political, and financial constraints of anywaterconflicts should also be discussed—in addition to the technical challenges.4.3The Internet is a good source of baseline information on this topic, in addition to themany water publications that are available to students in the library that will provide amore in-depth coverage.The social, political, and financial constraints of any conflictsshould also be discussed—in addition to the technical challenges.This problem wasincluded in the text to encourage students to contrast their local conditions with the manypressing water issues throughout the world.4.4Water conservation can include the following measures, among many others:•shutting off the water while brushing your teeth•low-flow shower heads•the many water conservation toilets•loading water-based appliances to full capacity before running•using intelligent irrigation measuresand generally reducing irrigation•rain barrels and cisterns•rain gardens•washing your car in a facility that recycles the water•conserving, recycling, and preserving in general (it is important to note the waterthat goes into creation of our consumer goods and food)

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage18CHAPTER 5WATER USE TRENDS AND FORECASTING5.12.75 x 106ac-ft = 0.90 x 1012gallonsTime = (0.90 x 1012)/(180 x 100,000) = 49,913 days= 136.7 years5.2Q = 200 x 0.1 = 20 cfs = 12.93 mgdPopulation = 12.39 x 106/ 175 = 73,886 people5.3Answers will vary.5.4Answers will vary.5.5Answers will vary.5.6Irrigation water use = 3 ac-ft/acreAssume irrigation season = 180 days (about 5 months)Irrigation water use only on a daily basis during the irrigation season is(1,500 x 3)/180 = 25 acre ft/dayConverting to mgd43,560 x 1 x 7.7 = 326,700 gallons per acre ft25 x 326,700 = 8.2 mgdFor the city, the water use is130,000 x 180 = 23.4 mgdThus during the irrigation season, the municipal water use per day is23.4/8.2 = about 2.9 times greater than the irrigation water use5.7Answers will vary.5.8Answers will vary.5.9(a) Assume lot sizes = 10,000 sq ftUsing Figure 5.2, find a max day value of 800 gpd/dwelling unitFor 4400 dwelling units, 4000 * 800 = 3,520,000 gpdOr 3,520,000/(24*60)=2,444 gpmThe combined draft is therefore 1,000+2,444=3,444 gpm(b) From Fig5.3, for 4400 dwelling units, and a density of 4, find peak hour = 5,000 gpmTherefore, the peak hour controls.5.10Answers will vary.5.11number of dwelling units = 1000 x 4 = 4000

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage19Use Fig. 5.3 and read up from a = 4000Find 5000 gpm as peak hourly flow5.12Using Figure 5.2, find an average annual water use of 250 gpd/dwelling unit,Lot size = 10,000 sq ft (4 lots per acre)Thus annual urban water use isQ=100*4*250*365=36,500,000 gal/yearIrrigation water use isQ=100*2.5*43,560*7.47=81,457,200 gal/yearThis makes for a difference = 44,957,200 gal/year (decrease)5.137/2.5 = 9/XX = 9*2.5/7=3.2 bgdAssume 15% reduction in per capita useThen in 2010, the requirement would be 3.2*0.85 = 2.72 bgd5.141995 = 200,000 = 43 mgd2010 = 260,000180 gpcd given(a) No change in use rate260,000*180 = 46.8 mgdExpansion is needed(b) 160 gpcd260,000*160 = 41.6 mgdLess capacity required.5.15Treatment capacity in 1995 = 35 mgdYear 2010 use rate = 140 gpcd(a) 260,000*140 = 36.4 mgd capacity requiredThus capacity in 2010 would not be adequate(b) Population increase in 15 years = 60,000Assume annual increase =40004000*140x(X) = 1.4 mgdX = 2.5 yearsNew treatment facilities would already be needed–in 2007 (2010–2.5)(c) 260,000(X) = 35 mgdRate would have to be reduced to about 134 gpcd, a fairly achievable reduction

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage205.16Industrial1.5*0.1+1.5 = 1.65 bgd in 2010MunicipalAssume 180 gpcd2,000,000*180 = 360 mgd increaseThus in 2000, use = 1.0 + 0.36 = 1.36 bgdSteam electricPlant factor = 0.63,000,000 KW * 50 gal/KWh*24h/d*0.6 = 2.16 bgdSteam electric in 2000 = 2.0 +2.16 = 4.16 bgdTotal Year 2010 Withdrawal1.65+1.36+4.16 = 7.17 bgd5.17Answers will vary.5.18450 x 4 = 1800 dwellingsUse fig 5.3 and find the peak hour water use = 2300 gpmPeak hourly sewage flowPeak hr = 3x avgAssume avg water use = 105 gpcd and 75% returnThus avg sewage flow = 78.8 gpcdPeak hr = 78.8 x 3 = 236.4 gpcd

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage21CHAPTER 6CONVEYING AND DISTRIBUTING WATER6.1This is one potential solution:Assume the channel is lined with concrete, n = 0.015Using the geometry given in the problem, the side channel length is found to be8.49 feetfeetPAR12.249.8*26*12*5.0===fpsV18.5)001.0(*)12.2(*015.049.12132==Q = VA =5.18 * 36 = 187 fps or 187*0.0283 = 5.28 cubic meters per second6.2Using the given channel geometry, the side channel length is found to be 1.8 mR = A/P = [(2.5 x 0.5 x 1 x 1.5) + (1 x 1.5)] / [(2 x 1.8) + 1] = 0.652V = (1/n)(R2/3S1/2)= (1/0.012)(0.6522/3)(0.0051/2)= 4.43 m/secQ = AV= 4.43 x 3 = 13.3 cubic meters/sec6.3Want capacity of 45 cfs, use Figure 6.1 to find a diameter of 30 inches.6.4V = (1/n)(R2/3S1/2)R = (Vn/S1/2)3/2= [(10.2 x 0.013)/(0.0151/2)]3/2= 1.13 mA = Q/V = 150/10.2 = 14.7 mR = A/P; P = A/R = 14.7/1.13 = 13.0 mLet Y = depth of channel, X = width of channelSet up two simultaneous equations for wetted perimeter and area2Y +X = 13XY = 14.7X = 14.7/Y)12/(]))35.714()5.6(()5.6([035.75.607.1413213/7.1422/1222−−−==+−=+−=+YYYYYYYY = 5.04, 1.46X = 14.7/Y = 2.9, 10.1There are two possible solutions:depth = 5m, width = 2.9 m

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage22depth = 1.46 m, width = 10.1 m6.5UsingEquation 6.1 (rearranged)2)*(32RnVS=, choose n = 0.012023.0236.036.0446.0012.0*32232===S93.040*023.0==LHmeters6.6Use Equation 6.15 in the second form54.063.2***432.0SDCQ=, choose C=1203.16004.0*1224*120*432.054.063.2==Qcfs6.7(1)Initially assume that the elevation of the hydraulic grade line at P is 90 ftand therefore there is no flow in pipe 2. Then using the Darcy-Weisbach equationfor head loss, Q is determined for pipes 1 and 3.Since V2= Q2/A2, the Darcy-Weisbach equation can be restated asHL= fLQ2/D2gA2;Q = (HLD2gA2/fL)1/2For Pipe 1)2400)(016.0(])12/4)()[(4.64)(12/8)(90150(221−=Q= 2.86 cfsFor Pipe 3)5500)(016.0(])12/5.10)()[(4.64)(12/21)(4090(223−=Q= 19.2 cfsSince Q3> Q1, assume flow is out of reservoir B. By continuity, assume thatQ1+ Q2= Q3orQ3–(Q1+ Q2) = 0The objective of the trial and error solution is to determine P such thatQ3–(Q1+ Q2) = 0. In performing a trial and error solution, it is helpful to plottrials of elevation at P versus Q3–(Q1+ Q2). For trial 1, P = 90, Q2= 0, andQ3–(Q1+ Q2) = 16.3 cfs.

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage23(2)For trial 2, choose P = 50.)2400)(016.0(])12/4)()[(4.64)(12/8)(50150(221−=Q= 3.69 cfs)1500)(016.0(])12/6)()[(4.64)(12/12)(5090(222−=Q= 8.14 cfs)5500)(016.0(])12/5.10)()[(4.64)(12/21)(4050(223−=Q= 8.61 cfsQ3–(Q1+ Q2) = 8.61–(3.69 + 8.14) =-3.22 cfsThis point is plotted and joined by a straight line with the point from trial 1. Thisline intersects the P axis at approximately P = 56.(3)For trial 3, try P = 56)2400)(016.0(])12/4)()[(4.64)(12/8)(56150(221−=Q= 3.57 cfs)1500)(016.0(])12/6)()[(4.64)(12/12)(5690(222−=Q= 7.50 cfs)5500)(016.0(])12/5.10)()[(4.64)(12/21)(4060(223−=Q= 10.9 cfsQ3–(Q1+ Q2) = 10.9–(3.57 + 7.50) =-0.17 cfs(4)For trial 4, chose P = 56.5)2400)(016.0(])12/4)()[(4.64)(12/8)(5.56150(221−=Q= 3.57 cfs)1500)(016.0(])12/6)()[(4.64)(12/12)(5.5690(222−=Q= 7.45 cfs)5500)(016.0(])12/5.10)()[(4.64)(12/21)(406.56(223−=Q= 11.06 cfsQ3–(Q1+ Q2) = 11.06–(3.57 + 7.45) = 0.04 cfs0.Therefore P = 56.5 ft, Q1= 3.57 cfs, Q2= 7.45 cfs, and Q3= 11.06 cfs.

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage246.8(1)Initially assume that the hydraulic grade at the intersection of the pipes isat an elevation of 55 ft, and therefore there is no flow in pipe 2. UsingEquation 6.17 and f = 0.010, determine Q1.Q1= (HLD2gA2/fL)1/2)3000)(010.0(])12/5.7)()[(4.64)(12/15)(55125(221−=Q= 16.8 ftSince 125 ft > 55 ft, flow is out of reservoir 1. Since Q1> Q3assume that bycontinuityQ1= Q2+ Q3orQ1–(Q2+ Q3) = 0For trial 1, 16.8–(0 + 15) = 1.8. Plot [Q1–(Q2+ Q3)] vs. P.(2)By continuity, flow must be into both B and C. Since we know that the elevationof B is 55 ft, then the elevation of the intersection must be greater than 55 ft forwater to flow to B. Therefore in choosing another value for P, try 60 ft.)3000)(010.0(])12/5.7)()[(4.64)(12/15)(60125(221−=Q= 16.2 ft

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 25 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage25)2200)(010.0(])12/0.5)()[(4.64)(12/10)(5560(222−=Q= 1.905 ftQ1–(Q2+ Q3) =-0.705(3)Plot this value and join with a straight line to the plotted point from trial 1.Estimate P where the line crosses [Q1–(Q2+ Q3)] = 0. Try P = 58)3000)(010.0(])12/5.7)()[(4.64)(12/15)(58125(221−=Q= 16.4 ft)2200)(010.0(])12/0.5)()[(4.64)(12/10)(5558(222−=Q= 1.48 ftQ1–(Q2+ Q3) =-0.075 ft.(4)Since the flow into C is known, the head loss can be determined:22222])12/9()[4.64)(12/18()15)(1600)(010.0(2/==gADfLQHL= 12 ftSince the flow is into C, the elevation of C must be lower than P, therefore the elevationof C is 58–12 = 46 ft.

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 26 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage266.9(1)Using the data from problem 6.8 trial 1 for P = 55ft:Q1= 16.8 cfsQ2= 0Q3= 15 cfsBecause 125 ft > 55 ft, flow is out of Q1. But we know that flow is out of Q3also,so the continuity relationship must be:Q1+ Q3= Q2orQ1+ Q3–Q2= 0In this case Q1+ Q3–Q2= 31.8(2)We want Q2to be much larger, so choose P >> 55 ft. This is consistent with theflow being toward B. Try P = 130 ft.

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 27 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage27)3000)(010.0(])12/5.7)()[(4.64)(12/15)(125130(221−=Q= 4.5 cfs)2200)(010.0(])12/0.5)()[(4.64)(12/10)(55130(222−=Q= 7.37 cfsBut now the flow is into A. This changes the continuity expression to:Q3= Q1+ Q2orQ3–(Q1+ Q2) = 015–(4.5 + 7.37) = 3.13 cfsPlot P vs. [Q3–(Q1+ Q2)].(3)Try P = 140)3000)(010.0(])12/5.7)()[(4.64)(12/15)(125140(221−=Q= 7.79 cfs)2200)(010.0(])12/0.5)()[(4.64)(12/10)(55140(222−=Q= 7.85 cfsQ3–(Q1+ Q2) =-0.064 cfs(4)Connect the plotted points of the last two trials with a straight line and estimatewhere the line intersects.Q3–(Q1+ Q2) = 0. Try P = 138 ft.)3000)(010.0(])12/5.7)()[(4.64)(12/15)(125138(221−=Q= 7.25 cfs)2200)(010.0(])12/0.5)()[(4.64)(12/10)(55138(222−=Q= 7.76 cfsQ3–(Q1+ Q2) =-0.009 cfsTherefore the elevation of the intersection of the hydraulic grade line is 138 ft.

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 28 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage28(5)To determine the elevation of reservoir C, first determine the head loss in pipe 3from Equation 6.17. This was determined in the solution to problem 6.8 to be 12ft.Since water flows from reservoir C, its elevation must be higher than P, thereforethe elevation of C is 138 + 12 = 150 ft.6.10Choose f = 0.0125. Writing the head loss equation in terms of total flow:222/gADfLQHL=Since the total head loss of the system is the sum of the head losses for each section,321LLLLHHHH++=

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 29 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage29Substitution the head loss equation for each section:2332332212222211211222gADQfLgADQfLgADQfLHL++=And since by continuity Q = Q1+ Q2+ Q3, the equation for head loss can be rewritten as:++=23332222211122ADLADLADLgfQHL++=2222222))27.0()(54.0()750())20.0()(40.0()450())30.0()(60.0()400(6.19)2.1)(0125.0(= 97. 3 m6.11Using the same procedure as for Problem 6.10++=++=222222233322222111))27.0()(54.0()750())20.0()(40.0()450())30.0()(60.0()400()0125.0()6.19)(60(2ADLADLADLfgHQL= 0.94 m3/s6.12(a)(1)Keep in mind thatLBDLBCLBCDLBCDLBEDLDFBCDBEDABHHHandHHHandQQQQ+====+=(2)From Fig. 6.8, assuming C = 100, the head loss in a 12 in. pipefor 6 cfs is 25 ft/1000 ft. The total head loss for BC is therefore

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 30 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage30HLBC= (25 ft/1000 ft)(450 ft) = 11.25 ftSimilarly the head loss for CD isHLCD= (12 ft/1000 ft)(300 ft) = 3.6 ftThe total head loss for BCD is 11.25 ft + 3.6 ft = 14.85 ft(3)Since HLBED= HLBCD, the head loss for BED is 14.85 ft. Over a length of 850 ft,this amounts to850)1000)(85.14(= 17.5 ft/1000 ftEntering this value in Fig. 6.8for an 8 in. pipe, the flow rate in BED isdetermined to be 14 cfs.(b)The total flow is therefore 14 + 6 = 20 cfs.(c)To calculate the length of equivalent 16 in. pipe:For QBEDat 14 cfs HL= 30 ft/1000 ft for 16 in. pipe;L16= (14.85)(1000)/(30) = 495 ftFor QBCDat 6 cfs, HL= 6 ft/1000 ft for 16 in pipe;L16= (14.85)(1000)/(6) = 2475 ftThe equivalent pipe length for the parallel pipe system is therefore495 ft + 2475 ft = 2970 ft6.13Writing the head loss equation in terms of Q and solving for Q:fLgADHQL22=And since321QQQQ++=323332222212111222fLgADHfLgADHfLgADHQLLL++=But321LLLLHHHH===in a parallel pipe system. Therefore;

Solution Manual for Water Supply and Pollution Control, 8th Edition - Page 31 preview image

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Solutions ManualWater Supply and Pollution Control, Eighth EditionPage31()233222211*2ADADADfgHQL++=And in terms of HLthis becomes:()222222222233222211225))225.0()(45.0(40))1.0()(2.0(30))15.0()(3.0()8.9*2()015.0()1.1()2(++=++=ADADADgfQHL= 0.99 m =321LLLHHH==)30)(15.0())5.1()(8.9)(2)(3.0)(99.0(221=Q= 0.26 m3/s)40)(15.0())1.0()(8.9)(2)(3.0)(99.0(222=Q= 0.08 m3/s)25)(15.0())225.0()(8.9)(2)(3.0)(99.0(223=Q= 0.77 m3/sCheck:321QQQ++= 1.11 m3/s(0.01 m error due to rounding)6.14Using the formula developed for problem 6.13()++=++=60))12/5.10()(12/21(95))12/4()(12/8(50))12/9()(12/18(*024.0)4.64)(45(*2222222233222211ADADADfgHQL= 259 cfs6.15In order to use Figure 6.8,one mustconvert metric units to FPS units:600 m = 1970 ft10 cm = 3.9 in24 cm = 9.5 in1200 m = 3940 ft5 cm = 2.0 in6 cm = 2.4 in550 m = 1800 ftAssume flow rate of 0.2 cfs. ThenCDBCABQQQ==from Figure 6.8HLAB= (12/1000)(1970) = 23.64 ft

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