Solutions Manual Water Supply and Pollution Control Eighth Edition Warren Viessman, Jr., P.E. University of Florida Mark J. Hammer, Emeritus Engineer Lincoln, Nebraska Elizabeth M. Perez, P.E. Palm Beach Gardens, Florida Paul A. Chadik, P.E. University of Florida Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 3 CHAPTER 1 NO SOLUTIONS REQUIRED CHAPTER 2 WATER RESOURCES PLANNING AND MANAGEMENT 2.1 The Internet is an excellent source of information on this topic. The level of integrated water resources management varies by state. 2.2 Virtually all of the laws listed in Table 2.1 provide some protection for preventing and controlling water pollution. Information on each law may be found on the Internet. It is also important to note that the EPA only regulates at the Federal level and much of the cleanup and protection is now delegated to states and local governments. 2.3 Point source pollution = Pollution that originates at one location with discrete discharge points. Typical examples include industrial and wastewater treatment facilities. Nonpoint source pollution = Pollution that is usually input into the environment in a dispersed manner. Typical examples include stormwater runoff that contains fertilizers, pesticides, herbicides, oils, grease, bacteria, viruses, and salts. 2.4 Adverse health effects of toxic pollutants are numerous and can include a variety of conditions. Some pollutant - related conditions include asthma, nausea, and various cancers — among many others . 2.5 Agencies that are responsible for water quantity and quality significantly vary by state. 2.6 This is a subjective question and one that has been and will continue to be debated in the water resources community. 2.7 Integrated water resources management is difficult to achieve because it involves both a financial and resources investment over time. It is also important to obtain concensus on this approach from all of the involved stakeholders. This difficulty is perhaps why there are so few examples of true integrated water resources management. 2.8 This question is subjective but the student should research specific examples to support their argument. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 4 CHAPTER 3 THE HYDROLOGIC CYCLE AND NATURAL WATER SOURCES 3.1 The answer to this question will vary by location . 3.2 reservoir area = 3900/640 = 6.1 sq. mi. annual runoff = (14/12)(190 – 6.1)(640) = 137,704 ac - ft annual evaporation = (49/12)(3900) = 15,925 ac - ft draft = (100 X 365 X 10 6 )/(7.48 X 43,560) = 112,022 ac - ft precipitation on lake = (40/12)(3900) = 13,000 ac - ft gain in storage = 137,704 + 13,000 = 150,704 loss in storage = 112,022 + 15,925 = 127,947 net gain in storage = 22,757 ac - ft 3.3 reservoir area = 1700 hec = 17 X 10 6 sq. meters annual runoff = 0.3(500 X 10 6 – 17 X 10 6 ) = 144 X 10 6 sq. meters annual evaporation = 1.2 X 17 X 10 6 = 20.4 X 10 6 sq. meters draft = 4.8 X 24 X 60 X 60 X 365 = 151.37 X 10 6 m 3 precipitation on lake = 0.97 X 17 X 10 6 = 16.49 X10 6 m 3 gain in storage = 144 X 10 6 +16.49 X 10 6 = 160.49 10 6 loss in storage = 151.37 X 10 6 + 20.4 X 10 6 = 171.77 X 10 6 net loss in storage = 11.28 X 10 6 m 3 3.4 To complete a water budget, it is first important to understand how the water budget will be used and what time step will be necessary to successfully model the system. Once the budget is conceptually designed, a variety of online sources can usually be used to collect the data. These sources include — but are not limited to: • state regulatory agencies • special water districts • weather agencies, • local governments • geological surveys • agricultu ral agencies Historical 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 collection effort. 3.5 The solution for this problem will vary based on location. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 5 3.6 Event (n) Precip (inches) Tr = n/m Freq. (% years) 1 33 10 10 2 29 5 20 3 28 3.33 30 4 28 2.5 40 5 27 2 50 6 26 1.67 60 7 22 1.4 70 8 21 1.25 80 9 19 1.1 90 10 18 1 100 n = 1 0, m = rank, Tr = n/m, Freq = (1/Tr) X 100 Then plot precipitation versus frequency. 3.7 Event (n) Precip (inches) Tr = n/m Freq. (% years) 1 89 10 10 2 75 5 20 3 72 3.33 30 4 70 2.5 40 5 69 2 50 6 66 1.67 60 7 56 1.4 70 8 54 1.25 80 9 48 1.1 90 10 46 1 100 n = 10, m = rank, Tr = n/m, Freq = (1/Tr) X 100 Then plot precipitation versus frequency. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 6 3.8 Once the data is organized in a table (see below), the solution can be found. Note that the cumulative max deficiency is 131.5 mg/mi 2 , which occurs in September. The number of months of draft is 131.5/(448/12) = 3.53. Therefore, enough storage is needed to supply the region for about 3.5 months. * Only positive values of cumulative deficiency are tabulated. 3.9 S = 128,000/10*100*640 = 0.20 3.10 S = 0.0002 = volume of water pumped divided by the average decline in piezometric head times surface area 0.0002 = V/(400 X 100) Noting that there are 640 acres per square mile V = 0.0002 X 400 X 100 X 640 = 5120 acre - feet Month Inflow I Draft O Cumulative Inflow Σ I Deficiency O - I Cumulative Deficiency Σ ( O – I )* Feb 31 37.3 31 6.3 6.3 March 54 37.3 85 - 16.7 0 April 90 37.3 175 - 52.7 0 May 10 37.3 185 27.3 27.3 June 7 37.3 192 30.3 57.6 July 8 37.3 200 29.3 86.9 Aug 2 37.3 202 35.3 122.2 Sep 28 37.3 230 9.3 131.5 Oct 42 37.3 272 - 4.7 126.8 Nov 108 37.3 380 - 70.7 56.1 Dec 98 37.3 478 - 60.7 0 Jan 22 37.3 500 15.3 15.3 Feb 50 37.3 550 - 12.7 2.6 Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 7 3.11 Draft = (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. 3.12 P n = (1 – 1/Tr) n log P n = n Log (1 – 1/Tr) n = log P n /log (1 – 1/Tr) A straight line can be defined by this equation and the following probability curves will appear. Month Inflow I Draft O Deficiency O - I Cumulative Deficiency Σ ( O – I )* April 97 21.8 - 75.2 0 May 136 21.8 - 114.2 0 June 59 21.8 37.2 0 July 14 21.8 7.8 7.8 Aug 6 21.8 15.8 23.6 Sep 5 21.8 16.8 40.43 Oct 3 21.8 18.8 59.2 Nov 7 21.8 14.8 74 Dec 19 21.8 2.8 76.8 Jan 13 21.8 8.8 85.6 Feb 74 21.8 - 52.2 33.4* March 96 21.8 - 74.2 0 April 37 21.8 - 15.2 0 May 63 21.8 - 41.2 0 June 49 21.8 - 27.2 0 Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 8 3.13 2 0 month flow equals the sum of 12 + 11 + 10 + 12 + … + 6 + 7 + 9 = 169 cfs 3.14 Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 9 3.15 3.16 Re servoir capacity = 750 acre - feet Reservoir yield is the amount of water which can be supplied during a specified time period. 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/year Or 750 X 43,560 X 0.304 = 917, 846 cubic meters per year Or 750 X 43,560 X 7.48 X 365 X 24 X 60 = 465 gpm continuously for 1 year 3.17 Constant annual yield = 1500 gpm Reservoir capacity = ? Time of operation without recharge = 1 yr Res. Capacity = 1500 X 365 X 24 X 60 X 0.134 X (1/43,560) = 2,425 ac - ft/yr This storage will provide a yield of 1,500 gpm for one year without any recharge 3.18 mean draft = 100 mgd, catchment area = 150 sq. mi., reservoir area = 4000 acres rainfall = 38 inches, runoff = 13 inches, evaporation = 49 inches (mean annual) (a) gain or loss in storage = ? ΔS = rainfall + runoff – evaporation – draft rainfall = 38 X 4000 X (1/12) = 12,667 ac - ft runoff = [(150 X 640) – 4000] X 13 X (1/12) = 99,967 ac - ft evaporation = 49 X 4000 X (1/12) = 16,333 ac - ft draft = 100,000,000 X 365 X 0.134 X (43,560) = 112,282 ac - f t ΔS = 12,667 + 99,667 – 16,333 – 112,282 = - 16,281 ac - ft The net loss in storage is 16, 281 ac - ft Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 10 (b) volume of water evaporated = 16,333 ac - ft given a community of 100,000 people, assume a consumption of 150 gpcd water demand = 100,000 X 150 X 365 = 5,475 mg/year volume evaporated = 16,333 X 43,560 X 7.48 = 5,304 mg/year evaporated water could supply the community with their water needs for 5304/5475 = 0.97 or for about one year 3.19 Use equation 3.29 K = 0.000287 h = 43 m = 8 n = 15 q = 0.000287*8*43/15 = 0.006582 Total Q is therefore 50*0.006582 = 0.325 cfs 3.20 q= 000084*8*22/15 = 0.000986 Q = 0.0007872*35 = 0.0345 m 3 /s 3.21 u = (1.87r 2 S c )/Tt = (1.87 * 1 * 6.4 * 10 - 4 )/(6200 * 7.5 * 24 * 60) = 8.58 x 10 - 10 Interpolating, W(u) = 20.3 S = (114.6 * 60,000 * 7.5 * 20.3)/(6,200 * 7.5 * 24 * 60) = 15.6 3.22 ) ( ) / log( 528 1 2 1 2 h h m r r Q K f − = 2 554 ) 1 10 ( * 90 ) 10 log( * 850 * 528 ft gpd K f = − = 3.23 Equation 3. 2 0 i s applicable ) / log( 1055 ) ( 1 / 2 2 1 2 2 r r h h K Q f − = 37107 . 0 ) 100 235 log( ) log( 1 2 = = r r ft h ft h 8 . 77 2 . 22 100 79 21 100 1 2 = − = = − = Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 11 gpm Q 44 . 634 37107 . 0 * 1055 8 . 77 79 ( 1320 ) 2 2 = − = 3.24 Using Equation 3. 35 , u can be computed 5 4 4 2 10 * 23 . 6 12 * 10 * 3 10 * 3 * 200 * 87 . 1 − − = = u Referring to Table 3. 5 and interpolating, we estimate W(u) to be 9.1. Then using Equation 3. 34 , the drawdown is found to be: ft s 41 . 10 10 * 3 300 * 1 . 9 * 6 . 114 4 = = 3.25 (a) Using Equation 3. 35 , u can be computed as follows: 71 . 0 0028 . 0 * 1000 * 4 00098 . 0 * 90 * 90 = = u Then from Table 3. 5 , W(u) is found to be 0.36. Applying Equation 3. 33, the drawdo wn can be determined m s 039 . 0 0028 . 0 * * 4 36 . 0 * 0038 . 0  = (c) Follow the procedure used in (a) 0098 . 0 0028 . 0 * 72000 * 4 00098 . 0 * 90 * 90 = = u Then from Table 3. 5 , W(u) is found to be 4.06. Applying Eq. 3.33, the drawdown can be determined m s 44 . 0 000028 . 0 * * 4 06 . 4 * 0038 . 0 = =  3.26 (a) Using Equation 3. 31 , u can be computed as follows: 25 . 0 0028 . 0 * 3600 * 4 001 . 0 * 100 * 100 = = u Then from Table 3. 5 , the drawdown can be determined, Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 12 m s 12 . 0 0028 . 0 * * 4 07 . 1 * 004 . 0 = =  (b) Follow the procedure used in (a) 01 . 0 0028 . 0 * 60 * 60 * 24 * 4 001 . 0 * 100 * 100 = = u Then from Table 3. 5 , W(u) is found to be 4.04 Applying Equation 3.33 , the drawdown can be determined m s 46 . 0 0028 . 0 * * 4 04 . 4 * 004 . 0 = =  3.27 (a) Using Equation 3. 31 , u can be computed as follows: 46 . 0 0028 . 0 * 60 * 60 * 12 * 4 001 . 0 * 150 * 150 = = u Then from Table 3. 5 , the drawdown can be determined, m s 05 . 0 0028 . 0 * * 4 36 . 0 * 003 . 0 = =  (b) Follow the procedure used in (a) 023 . 0 0028 . 0 * 60 * 60 * 12 * 4 001 . 0 * 500 * 500 = = u Then from Table 3. 5 , W(u) is found to be 3.24 Applying Equation 3. 33 , the drawdown can be determined m s 28 . 0 0028 . 0 * * 4 24 . 3 * 003 . 0 = =  3.28 392 , 13 ) 45 / 120 ( log * 528 8 * 100 * 1416 . 3 * 2 * 600 ) 45 / 120 ( log * 528 ) ( * * 2 * 10 10 1 2 = = − = h h K Q f  gal/min 3.29 62 . 407 28 . 1 * 100 ) 75 / 500 ( log * 1200 * 528 ) ( ) / ( log * * 528 10 1 2 1 2 10 = = − = h h m r r Q K f gpd/ft 2 3.30 h Q T  = * 264 From a plot of drawdown versus t, drawdown per log cycle is 28.2 – 10.5 = 17.1 Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 13 = = 1 . 17 * 264 T Q Converting T to gal/day/ft T=5100 330 1 . 17 * 264 5100 = = Q gpm 3.31 From plot of data, t 0 =1.25 minutes = 20.87 x 10 - 3 ft/day, and from plot, D h 14 feet 5657 14 300 * 264 = = T gpd/ft 00041 . 0 60 10 * 87 . 0 * 5657 * 3 . 0 * * 3 . 0 2 3 2 0 = = = − r t T S c 3.32 00011 . 0 * * 87 . 1 2 = = Tt S r u c W(u)= - 0.577216 - ln(u) Substituting and solving, using log e (u) W(u)=8.537 84 . 8 10 * 1 . 3 537 . 8 * 280 * 6 . 114 ) ( * * 6 . 114 4 = = = T u W Q S feet 3.33 Use Equation 3. 22 477 . 0 ) ln( 1 2 = r r 468 , 13 477 . 0 * 528 9 * 100 * * 2 * 600 = =  Q gpm 3.34 Use Equation 3. 23 2 2 . 433 8 . 10 * 130 ) 65 500 ln( * 1300 * 528 ft gpd K f = = 3.35 Use Equation 3. 37 and refer to figure which follows Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 14 T=700*7.5 = 5250 gpd/ft From Fig change in head is 9.53 feet 5 . 189 264 53 . 9 * 5250 = = Q gpm Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 15 3.36 Use Equation 3. 19 41683 . 0 ) ( log 1 2 10 = r r Q= 881 41683 . 0 * 1055 ) 5 . 77 * 5 . 77 4 . 79 * 4 . 79 ( * 1300 = − gpm 3.37 Use Equations 3. 34 and 3. 35 refer to the following figure determine s and r 2 /t from the figure = 1.36 and 20,000 Determine u and W(u) from the figure = 0.09 and 1.9 ft gpd T 050 , 80 365 . 1 9 . 1 * 500 * 6 . 114 = = 1926 . 0 20000 * 87 . 1 80050 * 09 . 0 = = c S Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 16 3.38 Use Equation 3. 19 2 1 2 2 2 100 ) 60 100 ( 66 50 y − − = 1560 2 1 = y , y 1 =39.5 Drawdown is 100 - 39.5=60.5 feet 3.39 Use Equation 3. 23 Log of the ratio = 0.1856 2 8 . 428 ) 95 97 ( * 80 1856 . 0 * 700 * 528 ft gpd K f = − = Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 17 CHAPTER 4 ALTERNATIVE SOURCES OF WATER SUPPLY 4.1 Varies by location — students should be encouraged to research this on the Internet and in other local publications. It is useful for the student to learn what technologies are used in their community since they will be explained further in later chapters of the book and this will provide them with a vested interest . 4.2 The Internet is a good source of baseline information on this topic, in addition to the many water publications that are available to students in the library that will provide a more in - depth coverage. The social, political, and financial constraints of any water conflicts should also be discussed — in addition to the technical challenges. 4.3 The Internet is a good source of baseline information on this topic, in addition to the many water publications that are available to students in the library that will provide a more in - depth coverage. The social, political, and financial constraints of any conflicts should also be discussed — in addition to the technical challenges. This problem was included in the text to encourage students to contrast their local conditions with the many pressing water issues throughout the world. 4.4 Water 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 measures and 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 water that goes into creation of our consumer goods and food) Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 18 CHAPTER 5 WATER USE TRENDS AND FORECASTING 5.1 2.75 x 10 6 ac - ft = 0.90 x 10 12 gallons Time = (0.90 x 10 12 )/(180 x 100,000) = 49,913 days = 136.7 years 5.2 Q = 200 x 0.1 = 20 cfs = 12.93 mgd Population = 12.39 x 10 6 / 175 = 73,886 people 5.3 Answers will vary. 5.4 Answers will vary. 5.5 Answers will vary. 5.6 Irrigation water use = 3 ac - ft/acre Assume 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/day Converting to mgd 43,560 x 1 x 7.7 = 326,700 gallons per acre ft 25 x 326,700 = 8.2 mgd For the city, the water use is 130,000 x 180 = 23.4 mgd Thus during the irrigation season, the municipal water use per day is 23.4/8.2 = about 2.9 times greater than the irrigation water use 5.7 Answers will vary. 5.8 Answers will vary. 5.9 (a) Assume lot sizes = 10,000 sq ft Using Figure 5.2, find a max day value of 800 gpd/dwelling unit For 4400 dwelling units, 4000 * 800 = 3,520,000 gpd Or 3,520,000/(24*60)=2,444 gpm The combined draft is therefore 1,000+2,444=3,444 gpm (b) From Fig 5.3, for 4400 dwelling units, and a density of 4, find peak hour = 5,000 gpm Therefore, the peak hour controls. 5.10 Answers will vary. 5.11 number of dwelling units = 1000 x 4 = 4000 Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 19 Use Fig. 5.3 and read up from a = 4000 Find 5000 gpm as peak hourly flow 5.12 Using 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 is Q=100*4*250*365=36,500,000 gal/year Irrigation water use is Q=100*2.5*43,560*7.47=81,457,200 gal/year This makes for a difference = 44,957,200 gal/year (decrease) 5.13 7/2.5 = 9/X X = 9*2.5/7=3.2 bgd Assume 15% reduction in per capita use Then in 2010, the requirement would be 3.2*0.85 = 2.72 bgd 5.14 1995 = 200,000 = 43 mgd 2010 = 260,000 180 gpcd given (a) No change in use rate 260,000*180 = 46.8 mgd Expansion is needed (b) 160 gpcd 260,000*160 = 41.6 mgd Less capacity required. 5.15 Treatment capacity in 1995 = 35 mgd Year 2010 use rate = 140 gpcd (a) 260,000*140 = 36.4 mgd capacity required Thus capacity in 2010 would not be adequate (b) P opulation increase in 15 years = 60,000 Assume annual increase =4000 4000*140x(X) = 1.4 mgd X = 2.5 years New treatment facilities would already be needed – in 2007 (2010 – 2.5) (c) 260,000(X) = 35 mgd Rate would have to be reduced to about 134 gpcd, a fairly achievable reduction Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 20 5.16 Industrial 1.5*0.1+1.5 = 1.65 bgd in 2010 Municipal Assume 180 gpcd 2,000,000*180 = 360 mgd increase Thus in 2000, use = 1.0 + 0.36 = 1.36 bgd Steam electric Plant factor = 0.6 3,000,000 KW * 50 gal/KWh*24h/d*0.6 = 2.16 bgd Steam electric in 2000 = 2.0 +2.16 = 4.16 bgd Total Year 2010 Withdrawal 1.65+1.36+4.16 = 7.17 bgd 5.17 Answers will vary. 5.18 450 x 4 = 1800 dwellings Use fig 5.3 and find the peak hour water use = 2300 gpm Peak hourly sewage flow Peak hr = 3x avg Assume avg water use = 105 gpcd and 75% return Thus avg sewage flow = 78.8 gpcd Peak hr = 78.8 x 3 = 236.4 gpcd Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 21 CHAPTER 6 CONVEYING AND DISTRIBUTING WATER 6.1 This is one potential solution: Assume the channel is lined with concrete, n = 0.015 Using the geometry given in the problem, the side channel length is found to be 8.49 feet feet P A R 12 . 2 49 . 8 * 2 6 * 12 * 5 . 0 = = = fps V 18 . 5 ) 001 . 0 ( * ) 12 . 2 ( * 015 . 0 49 . 1 2 1 3 2 = = Q = VA = 5.18 * 36 = 187 fps or 187*0.0283 = 5.28 cubic meters per second 6.2 Using the given channel geometry, the side channel length is found to be 1.8 m R = A/P = [(2.5 x 0.5 x 1 x 1.5) + (1 x 1.5)] / [(2 x 1.8) + 1] = 0.652 V = (1/n)(R 2/3 S 1/2 ) = (1/0.012)(0.652 2/3 )(0.005 1/2 ) = 4.43 m/sec Q = AV = 4.43 x 3 = 13.3 cubic meters/sec 6.3 Want capacity of 45 cfs, use Figure 6.1 to find a diameter of 30 inches. 6.4 V = (1/n)(R 2/3 S 1/2 ) R = (Vn/S 1/2 ) 3/2 = [(10.2 x 0.013)/(0.015 1/2 )] 3/2 = 1.13 m A = Q/V = 150/10.2 = 14.7 m R = A/P; P = A/R = 14.7/1.13 = 13.0 m Let Y = depth of channel, X = width of channel Set up two simultaneous equations for wetted perimeter and area 2Y +X = 13 XY = 14.7 X = 14.7/Y ) 1 2 /( ] )) 35 . 7 1 4 ( ) 5 . 6 (( ) 5 . 6 ( [ 0 35 . 7 5 . 6 0 7 . 14 13 2 13 / 7 . 14 2 2 / 1 2 2 2    −  − − = = + − = + − = + Y Y Y Y Y Y Y Y = 5.04, 1.46 X = 14.7/Y = 2.9, 10.1 There are two possible solutions: depth = 5m, width = 2.9 m Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 22 depth = 1.46 m, width = 10.1 m 6.5 Using Equation 6.1 (rearranged) 2 ) * ( 3 2 R n V S = , choose n = 0.012 023 . 0 236 . 036 . 0 4 46 . 0 012 . 0 * 3 2 2 3 2 =       =                     = S 93 . 0 40 * 023 . 0 = = L H meters 6.6 Use Equation 6.15 in the second form 54 . 0 63 . 2 * * * 432 . 0 S D C Q = , choose C=120 3 . 16 004 . 0 * 12 24 * 120 * 432 . 0 54 . 0 63 . 2 =       = Q cfs 6.7 (1) Initially assume that the elevation of the hydraulic grade line at P is 90 ft and therefore there is no flow in pipe 2. Then using the Darcy - Weisbach equation for head loss, Q is determined for pipes 1 and 3. Since V 2 = Q 2 /A 2 , the Darcy - Weisbach equation can be restated as H L = fLQ 2 /D2gA 2 ; Q = (H L D2gA 2 /fL) 1/2 For Pipe 1 ) 2400 )( 016 . 0 ( ] ) 12 / 4 )( )[( 4 . 64 )( 12 / 8 )( 90 150 ( 2 2 1  − = Q = 2.86 cfs For Pipe 3 ) 5500 )( 016 . 0 ( ] ) 12 / 5 . 10 )( )[( 4 . 64 )( 12 / 21 )( 40 90 ( 2 2 3  − = Q = 19.2 cfs Since Q 3 > Q 1 , assume flow is out of reservoir B. By continuity, assume that Q 1 + Q 2 = Q 3 or Q 3 – (Q 1 + Q 2 ) = 0 The objective of the trial and error solution is to determine P such that Q 3 – (Q 1 + Q 2 ) = 0. In performing a trial and error solution, it is helpful to plot trials of elevation at P versus Q 3 – (Q 1 + Q 2 ). For trial 1, P = 90, Q 2 = 0, and Q 3 – (Q 1 + Q 2 ) = 16.3 cfs. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 23 (2) For trial 2, choose P = 50. ) 2400 )( 016 . 0 ( ] ) 12 / 4 )( )[( 4 . 64 )( 12 / 8 )( 50 150 ( 2 2 1  − = Q = 3.69 cfs ) 1500 )( 016 . 0 ( ] ) 12 / 6 )( )[( 4 . 64 )( 12 / 12 )( 50 90 ( 2 2 2  − = Q = 8.14 cfs ) 5500 )( 016 . 0 ( ] ) 12 / 5 . 10 )( )[( 4 . 64 )( 12 / 21 )( 40 50 ( 2 2 3  − = Q = 8.61 cfs Q 3 – (Q 1 + Q 2 ) = 8.61 – (3.69 + 8.14) = - 3.22 cfs This point is plotted and joined by a straight line with the point from trial 1. This line intersects the P axis at approximately P = 56. (3) For trial 3, try P = 56 ) 2400 )( 016 . 0 ( ] ) 12 / 4 )( )[( 4 . 64 )( 12 / 8 )( 56 150 ( 2 2 1  − = Q = 3.57 cfs ) 1500 )( 016 . 0 ( ] ) 12 / 6 )( )[( 4 . 64 )( 12 / 12 )( 56 90 ( 2 2 2  − = Q = 7.50 cfs ) 5500 )( 016 . 0 ( ] ) 12 / 5 . 10 )( )[( 4 . 64 )( 12 / 21 )( 40 60 ( 2 2 3  − = Q = 10.9 cfs Q 3 – (Q 1 + Q 2 ) = 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 . 56 150 ( 2 2 1  − = Q = 3.57 cfs ) 1500 )( 016 . 0 ( ] ) 12 / 6 )( )[( 4 . 64 )( 12 / 12 )( 5 . 56 90 ( 2 2 2  − = Q = 7.45 cfs ) 5500 )( 016 . 0 ( ] ) 12 / 5 . 10 )( )[( 4 . 64 )( 12 / 21 )( 40 6 . 56 ( 2 2 3  − = Q = 11.06 cfs Q 3 – (Q 1 + Q 2 ) = 11.06 – (3.57 + 7.45) = 0.04 cfs  0. Therefore P = 56.5 ft, Q 1 = 3.57 cfs, Q 2 = 7.45 cfs, and Q 3 = 11.06 cfs. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 24 6.8 (1) Initially assume that the hydraulic grade at the intersection of the pipes is at an elevation of 55 ft, and therefore there is no flow in pipe 2. Using Equation 6.17 and f = 0.010, determine Q 1 . Q 1 = (H L D2gA 2 /fL) 1/2 ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 55 125 ( 2 2 1  − = Q = 16.8 ft Since 125 ft > 55 ft, flow is out of reservoir 1. Since Q 1 > Q 3 assume that by continuity Q 1 = Q 2 + Q 3 or Q 1 – (Q 2 + Q 3 ) = 0 For trial 1, 16.8 – (0 + 15) = 1.8. Plot [Q 1 – (Q 2 + Q 3 )] vs. P. (2) By continuity, flow must be into both B and C. Since we know that the elevation of B is 55 ft, then the elevation of the intersection must be greater than 55 ft for water to flow to B. Therefore in choosing another value for P, try 60 ft. ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 60 125 ( 2 2 1  − = Q = 16.2 ft Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 25 ) 2200 )( 010 . 0 ( ] ) 12 / 0 . 5 )( )[( 4 . 64 )( 12 / 10 )( 55 60 ( 2 2 2  − = Q = 1.905 ft Q 1 – (Q 2 + Q 3 ) = - 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 [Q 1 – (Q 2 + Q 3 )] = 0. Try P = 58 ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 58 125 ( 2 2 1  − = Q = 16.4 ft ) 2200 )( 010 . 0 ( ] ) 12 / 0 . 5 )( )[( 4 . 64 )( 12 / 10 )( 55 58 ( 2 2 2  − = Q = 1.48 ft Q 1 – (Q 2 + Q 3 ) = - 0.075 ft. (4) Since the flow into C is known, the head loss can be determined: 2 2 2 2 2 ] ) 12 / 9 ( )[ 4 . 64 )( 12 / 18 ( ) 15 )( 1600 )( 010 . 0 ( 2 /  = = gA D fLQ H L = 12 ft Since the flow is into C, the elevation of C must be lower than P, therefore the elevation of C is 58 – 12 = 46 ft. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 26 6.9 (1) Using the data from problem 6.8 trial 1 for P = 55ft: Q 1 = 16.8 cfs Q 2 = 0 Q 3 = 15 cfs Because 125 ft > 55 ft, flow is out of Q 1 . But we know that flow is out of Q 3 also, so the continuity relationship must be: Q 1 + Q 3 = Q 2 or Q 1 + Q 3 – Q 2 = 0 In this case Q 1 + Q 3 – Q 2 = 31.8 (2) We want Q 2 to be much larger, so choose P >> 55 ft. This is consistent with the flow being toward B. Try P = 130 ft. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 27 ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 125 130 ( 2 2 1  − = Q = 4.5 cfs ) 2200 )( 010 . 0 ( ] ) 12 / 0 . 5 )( )[( 4 . 64 )( 12 / 10 )( 55 130 ( 2 2 2  − = Q = 7.37 cfs But now the flow is into A. This changes the continuity expression to: Q 3 = Q 1 + Q 2 or Q 3 – (Q 1 + Q 2 ) = 0 15 – (4.5 + 7.37) = 3.13 cfs Plot P vs. [Q 3 – (Q 1 + Q 2 )]. (3) Try P = 140 ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 125 140 ( 2 2 1  − = Q = 7.79 cfs ) 2200 )( 010 . 0 ( ] ) 12 / 0 . 5 )( )[( 4 . 64 )( 12 / 10 )( 55 140 ( 2 2 2  − = Q = 7.85 cfs Q 3 – (Q 1 + Q 2 ) = - 0.064 cfs (4) Connect the plotted points of the last two trials with a straight line and estimate where the line intersects. Q 3 – (Q 1 + Q 2 ) = 0. Try P = 138 ft. ) 3000 )( 010 . 0 ( ] ) 12 / 5 . 7 )( )[( 4 . 64 )( 12 / 15 )( 125 138 ( 2 2 1  − = Q = 7.25 cfs ) 2200 )( 010 . 0 ( ] ) 12 / 0 . 5 )( )[( 4 . 64 )( 12 / 10 )( 55 138 ( 2 2 2  − = Q = 7.76 cfs Q 3 – (Q 1 + Q 2 ) = - 0.009 cfs Therefore the elevation of the intersection of the hydraulic grade line is 138 ft. Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 28 (5) To determine the elevation of reservoir C, first determine the head loss in pipe 3 from Equation 6.17. This was determined in the solution to problem 6.8 to be 12 ft. Since water flows from reservoir C, its elevation must be higher than P, therefore the elevation of C is 138 + 12 = 150 ft. 6.10 Choose f = 0.0125. Writing the head loss equation in terms of total flow: 2 2 2 / gA D fLQ H L = Since the total head loss of the system is the sum of the head losses for each section, 3 2 1 L L L L H H H H + + = Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 29 Substitution the head loss equation for each section: 2 3 3 2 3 3 2 21 2 2 2 2 2 1 1 2 1 1 2 2 2 gA D Q fL gA D Q fL gA D Q fL H L + + = And since by continuity Q = Q 1 + Q 2 + Q 3 , the equation for head loss can be rewritten as:         + +         = 2 3 3 3 2 2 2 2 2 1 1 1 2 2 A D L A D L A D L g fQ H L       + +         = 2 2 2 2 2 2 2 ) ) 27 . 0 ( )( 54 . 0 ( ) 750 ( ) ) 20 . 0 ( )( 40 . 0 ( ) 450 ( ) ) 30 . 0 ( )( 60 . 0 ( ) 400 ( 6 . 19 ) 2 . 1 )( 0125 . 0 (    = 97. 3 m 6.11 Using the same procedure as for Problem 6.10         + + =         + + = 2 2 2 2 2 2 2 3 3 3 2 2 2 2 2 1 1 1 ) ) 27 . 0 ( )( 54 . 0 ( ) 750 ( ) ) 20 . 0 ( )( 40 . 0 ( ) 450 ( ) ) 30 . 0 ( )( 60 . 0 ( ) 400 ( ) 0125 . 0 ( ) 6 . 19 )( 60 ( 2    A D L A D L A D L f g H Q L = 0.94 m 3 /s 6.12 (a)(1) Keep in mind that LBD LBC LBCD LBCD LBED L DF BCD BED AB H H H and H H H and Q Q Q Q + = = = = + = (2) From Fig. 6.8 , assuming C = 100, the head loss in a 12 in. pipe for 6 cfs is 25 ft/1000 ft. The total head loss for BC is therefore Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 30 H LBC = (25 ft/1000 ft)(450 ft) = 11.25 ft Similarly the head loss for CD is H LCD = (12 ft/1000 ft)(300 ft) = 3.6 ft The total head loss for BCD is 11.25 ft + 3.6 ft = 14.85 ft (3) Since H LBED = H LBCD , the head loss for BED is 14.85 ft. Over a length of 850 ft, this amounts to 850 ) 1000 )( 85 . 14 ( = 17.5 ft/1000 ft Entering this value in Fig. 6. 8 for an 8 in. pipe, the flow rate in BED is determined 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 Q BED at 14 cfs H L = 30 ft/1000 ft for 16 in. pipe; L 16 = (14.85)(1000)/(30) = 495 ft For Q BCD at 6 cfs, H L = 6 ft/1000 ft for 16 in pipe; L 16 = (14.85)(1000)/(6) = 2475 ft The equivalent pipe length for the parallel pipe system is therefore 495 ft + 2475 ft = 2970 ft 6.13 Writing the head loss equation in terms of Q and solving for Q: fL gA D H Q L 2 2 = And since 3 2 1 Q Q Q Q + + = 3 2 3 3 3 2 2 2 2 2 1 2 1 1 1 2 2 2 fL gA D H fL gA D H fL gA D H Q L L L + + = But 3 2 1 L L L L H H H H = = = in a parallel pipe system. Therefore; Solutions Manual Water Supply and Pollution Control, Eighth Edition Page 31 ( ) 2 3 3 2 2 2 2 1 1 * 2 A D A D A D f g H Q L + + = And in terms of H L this becomes: ( ) 2 2 2 2 2 2 2 2 2 2 3 3 2 2 2 2 1 1 2 25 ) ) 225 . 0 ( )( 45 . 0 ( 40 ) ) 1 . 0 ( )( 2 . 0 ( 30 ) ) 15 . 0 ( )( 3 . 0 ( ) 8 . 9 * 2 ( ) 015 . 0 ( ) 1 . 1 ( ) 2 (         + + = + + =    A D A D A D g f Q H L = 0.99 m = 3 2 1 L L L H H H = = ) 30 )( 15 . 0 ( ) ) 5 . 1 ( )( 8 . 9 )( 2 )( 3 . 0 )( 99 . 0 ( 2 2 1  = Q = 0.26 m 3 /s ) 40 )( 15 . 0 ( ) ) 1 . 0 ( )( 8 . 9 )( 2 )( 3 . 0 )( 99 . 0 ( 2 2 2  = Q = 0.08 m 3 /s ) 25 )( 15 . 0 ( ) ) 225 . 0 ( )( 8 . 9 )( 2 )( 3 . 0 )( 99 . 0 ( 2 2 3  = Q = 0.77 m 3 /s Check: 3 2 1 Q Q Q + + = 1.11 m 3 /s (0.01 m error due to rounding) 6.14 Using 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 ( * 2 2 2 2 2 2 2 2 3 3 2 2 2 2 1 1    A D A D A D f g H Q L = 259 cfs 6.15 In order to use Figure 6.8 , one must convert metric units to FPS units: 600 m = 1970 ft 10 cm = 3.9 in 24 cm = 9.5 in 1200 m = 3940 ft 5 cm = 2.0 in 6 cm = 2.4 in 550 m = 1800 ft Assume flow rate of 0.2 cfs. Then CD BC AB Q Q Q = = from Figure 6. 8 H LAB = (12/1000)(1970) = 23.64 ft