Handbook of water and wastewater tủ tài liệu bách khoa

Appendixes © 2009 by Taylor & Francis Group, LLC Appendix A Answers to Chapter Review Questions CHAPTER ANSWERS CHAPTER ANSWERS 1.1 3.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 A pattern or point of view that determines what is seen as reality A change in the way things are understood and done (1) Assessing and protecting drinking water sources (2) Optimizing treatment processes (3) Ensuring the integrity of distribution systems (4) Effecting correct cross-connection control procedures (5) Continuous monitoring and testing of the water before it reaches the tap Water/wastewater operations are usually low-profile activities and much of water/wastewater infrastructure is buried underground Secondary Privatization means allowing private enterprise to compete with government in providing public services, such as water and wastewater operations Reengineering is the systematic transformation of an existing system into a new form to realize quality improvements in operations, systems capability, functionality, and performance at lower cost, improved schedule, and less risk to the customer A process for rigorously measuring performance vs “best-in-class” operations and using the analysis to meet and exceed the best in class Planning, research, observation, analysis, adaptation CHAPTER ANSWERS 2.1 2.2 2.3 2.4 2.5 Operators are exposed to the full range of hazards and work under all weather conditions Plants are upgrading to computerized operations Computerized maintenance management system HAZMAT emergency response technician 24-hour certification Safe Drinking Water Act Answers will vary CHAPTER ANSWERS 4.1 Matching answers o 14 c 15 t 16 j 17 s 18 p 19 d 20 i 21 e 22 10 q 23 11 u 24 12 k 25 13 a 26 v r w l x m y f b z h n g CHAPTER ANSWERS 5.1 5.2 5.3 (2.5 mg/L)(5.5 MGD)(8.34 lb/gal) = 115 lb/day (7.1 mg/L)(4.2 MGD)(8.34 lb/gal) = 249 lb/day (11.8 mg/L)(4.8 MGD)(8.34 lb/gal) = 472 lb/day 5.4 (10 mg/L)(1.8 MGD)(8.34 lb/gal) = 0.65 lb/day (60 mg/L)(0.086 MGD)(8.34 lb/gal) = 43 lb (2220 mg/L)(0.225)(8.34 lb/gal) = 4166 lb 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 (8 mg/L)(0.83 MGD)(8.34 lb/gal) = 85 lb/day 0.65 (450 mg/L)(1.84 MGD)(8.34 lb/gal) = 6906 lb/day (25 mg/L)(2.90 MGD)(8.34 lb/gal) = 605 lb/day (260 mg/L)(5.45 MGD)(8.34 lb/gal) = 11,818 lb/day (144 mg/L)(3.66 MGD)(8.34 lb/gal) = 4396 lb/day (290 mg/L)(3.31 MGD)(8.34 lb/gal) = 8006 lb/day (152 mg/L)(5.7 MGD)(8.34 lb/gal) = 7226 lb/day 777 © 2009 by Taylor & Francis Group, LLC 778 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30 5.31 5.32 5.33 5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition (188 mg/L)(1.92 MGD)(8.34 lb/gal) = 3010 lb/day SS (184 mg/L)(1.88 MGD)(8.34 lb/day) = 2885 lb/day SS (150 mg/L)(4.88 MGD)(8.34 lb/gal) = 6105 lb/day BOD (205 mg/L)(2.13 MGD)(8.34 lb/gal) = 3642 lb/day solids (115 mg/L)(4.20 MGD)(8.34 lb/gal) = 4028 lb/day BOD (2230 mg/L)(0.40 MG)(8.34 lb/gal) = 7439 lb SS (1890 mg/L)(0.41 MG)(8.34 lb/gal) = 6463 lb MLVSS (3125 mg/L)(0.18 MG)(8.34 lb/gal) = 4691 lb MLVSS (2250 mg/L)(0.53 MG)(8.34 lb/gal) = 9945 lb MLSS (2910 mg/L)(0.63 MG)(8.34 lb/gal) = 15,290 lb MLSS (6150 mg/L)(x MGD)(8.34 lb/gal) = 5200 lb/day x = 0.10 MGD (6200 mg/L)(x MGD)(8.34 lb/gal) = 4500 lb/day a x = 0.09 b 90,000 gpd ÷ 1440 min/day = 62.5 gpm (6600 lb/day)(x MGD)(8.34 lb/gal) = 6070 lb/day x = 0.11 MGD x = 110,000 gpd ÷ 1440 min/day = 76 gpm (6350 mg/L)(x MGD)(8.34 lb/gal) = 7350 lb/day x = 0.14 MGD x = 140,000 gpd ÷ 1440 min/day = 97 gpm (7240 mg/L)(x MGD)(8.34 lb/gal) = 5750 lb/day x = 0.10 MGD x = 100,000 gpd ÷ 1440 min/day = 69 gpm (2.5 mg)(3.65 MGD)(8.34 lb/gal) = 76.1 lb/day (17 mg/L)(2.10 MGD)(8.34 lb/gal) = 298 lb/day (190 mg/L)(4.8 MGD)(8.34 lb/gal) = 7606 lb/day SS removed (9.7 mg/L)(5.5 MGD)(8.34 lb/gal) = 445 lb/day (305 mg/L)(3.5 MGD) (8.34 lb/gal) = 8903 lb/day (10 mg/L)(3.1 MGD)(8.34 lb/gal) = 398 lb/day 0.65 (210 mg/L)(3.44 MGD)(8.34 lb/gal) = 6025 lb/day solids (60 mg/L)(0.09 MG)(8.34 lb/gal) = 45 lb chlorine (2720 mg/L)(0.52 MG)(8.34 lb/gal) = 11,796 lb MLSS (5870 mg/L)(x MGD)(8.34 lb/gal) = 5480 lb/day x = 0.11 MGD (120 mg/L)(3.312 MGD)(8.34 lb/gal) = 3315 lb/day BOD (240 mg/L)(3.18 MGD)(8.34 lb/gal) = 6365 lb/day BOD (196 mg/L)(1.7 MGD)(8.34 lb/gal) = 2779 lb/day BOD removed © 2009 by Taylor & Francis Group, LLC 5.42 5.43 (x mg/L)(5.3 MGD)(8.34 lb/day) = 330 lb/day x = 7.5 mg/L (5810 mg/L)(x MGD)(8.34 lb/gal) = 5810 mg/L x = 0.12 MGD x = 120,000 gpd ÷ 1440 min/day = 83 gpm 5.44 3,400,000 gpd = 433 gpd/ft (0.785)(100 ft)(100 ft) 5.45 4,525,000 gpd = 712 gpd/ft (0.785)(90 ft)(90 ft) 5.46 3,800,000 gpd = 4.4 gpd/ft 870,000 ft 5.47 280,749 ft day = 0.4 ft/day 696,960 ft (0.4 ft/day)(12 in./ft) = 4.8 in./day 5.48 5,280,000 gpd = 830 gpd/ft (0.785)(90 ft)(90 ft) 5.49 4.4 ac-ft/day = 0.22 ft/day = in./day 20 ac 5.50 2,050,000 gpd = 1171 gpd/ft (70 ft)(25 ft) 5.51 2,440,000 gpd = 863 gpd/ft (0.785)(60 ft)(60 ft) 5.52 3,450,000 gpd = 627 gpd/ft (110 ft)(50 ft) 5.53 1,660,000 gpd = 949 gpd/ft (25 ft)(70 ft) 5.54 2,660,000 gpd = 691 gpd/ft (0.785)(70 ft)(70 ft) 5.55 2230 gpm = 2.8 gpm/ft (40 ft)(20 ft) 5.56 3100 gpm = 3.1 gpm/ft (40 ft)(25 ft) 5.57 2500 gpm = 1.6 gpm/ft (26 ft)(60 ft) 5.58 1528 gpm = 1.9 gpm/ft (40 ft)(20 ft) 5.59 2850 gpm = 3.2 gpm/ft 880 ft 5.60 4750 gpm = 24 gpm/ft (14 ft)(14 ft) 5.61 4900 gpm = 12 gpm/ft (20 ft)(20 ft) 779 5.62 3400 gpm = gpm/ft (25 ft)(15 ft) 5.80 (200 mg/L)(3.42 MGD)(8.34 lb/gal) = 0.9 (1875 mg/L))(0.42 MG)(8.34 lb/gal) 5.63 3300 gpm = 4.4 gpm/ft (75 ft)(30 ft) 5.81 (190 mg/L)(3.24 MGD)(8.34 lb/gal) = 1.3 (1710 mg/L))(0.28 MG)(8.34 lb/gal) 5.64 3800 gpm = 12.7 gpm/ft (15 ft)(20 ft) 5.82 (151 mg/L)(2.25 MGD)(8.34 lb/gal) = 0.9 x lb MLVSS 5.65 3,770,000 gal = 8378 gal/ft (15 ft)(30 ft) 5.66 1,860,000 gal = 6200 gal/ft (20 ft)(15 ft) 5.67 3,880,000 gal = 7760 gal/ft (25 ft)(20 ft) 5.68 1,410,200 gal = 5036 gal/ft (20 ft)(14 ft) 5.69 5,425,000 gal = 9042 gal/ft (30 ft)(20 ft) 5.70 1,410,000 gpd = 8650 gpd/ft 163 ft 5.71 2,120,000 gpd = 11,253 gpd/ft (3.14)(60 ft) 5.72 2,700,00 gpd = 11,250 gpd/ft 240 ft 5.73 (1400 gpm)(1440 min/day) = 8025 gpd/ft (3.14)(80 ft) 5.74 2785 gpm = 14.7 gpm/ft 189 ft 5.75 (210 mg/L)(2.45 MGD)(8.34 lb/gal) 25.1 1000 ft = 171 lb BOD/day/1000 ft 5.76 5.77 (170 mg/L)(0.120 MGD)(8.34 lb/gal) 3.5 ac = 49 lb BOD/day/ac (160 mg/L)(2.10 MGD)(8.34 lb/gal) = 0.7 (1900 mg/L))(0.255 MG)(8.34 lb/gal) 5.84 (180 mg/L)(3.11 MGD)(8.34 lb/gal) (x mg/L)(0.88 MG)(8.34 lb/gal) x = 1262 mg/L MLVSS 5.85 (2650 mg/L)(3.60 MGD)   × (8.34 lb/gal)   = 20.7 lb MLSS/day/ft (0.785)(70 ft)(70 ft) 5.86 (2825 mg/L)(4.25 MGD)   × (8.34 lb/gal)   = 19.9 lb MLSS/day/ft (0.785)(80 ft)(80 ft) 5.87 (x mg/L)(3.61 MGD)  × (8.34 lb/gal)    = 26 lb MLSS/day/ft (0.785)(60 ft)(60 ft) x = 2441 mg/L MLSS 5.88 (2210 mg/L)(3.3 MGD)   × (8.34 lb/gal)   = 21.5 lb MLSS/day/ft (0.785)(60 ft)(60 ft) 5.89 (x mg/L)(3.11 MGD)  × (8.34 lb/gal)    = 20 lb MLSS/day/ft (0.785)(60 ft)(60 ft) x = 2174 mg/L MLSS 5.90 12,110 lb VS/day = 0.37 lb VS/day/ft 33,100 ft 5.91 (124,000 lb/day)  × (0.065)(0.70)    = 0.08 lb VS/day/ft (0.785)(60 ft)(60 ft)(25 ft) 5.92 (141,000 lb/day)  × (0.06)(0.71)    = 0.15 lb VS/day/ft (0.785)(50 ft)(50 ft)(20 ft) 5.93 (21,200 gpd)(8.34 lb/gal)   × (0.055)(0.69)   = 0.33 VS/day//ft (0.785)(40 ft)(40 ft)(16 ft) (140 mg/L)(2.20 MGD)(8.34 lb/gal) 900 1000 ft = 2.9 lb/BOD/day/1000 ft 5.79 5.83 (120 mg/L)(2.85 MGD)(8.34 lb/gal) 34 1000 ft = 84 lb/BOD/day/1000 ft 5.78 x = 3148 lb MLVSS (0.785)(90 ft)(90 ft)(4 ft) = 25,434 (150 mg//L) (3.5 MGD) (8.34 lb/gal) 25.4 1000 ft =172 lb BOD/day/1000 ft © 2009 by Taylor & Francis Group, LLC 780 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition (22,000 gpd)(8.6 lb/gal)   (0.052)(0.70)   = 0.18 lb VS/day/ft (0.785)(50 ft)(50 ft)(20 ft) 5.113 5.95 2050 lb VS added per day = 0.06 32,400 lb VS 5.114 5.96 620 lb VS added per day = 0.09 (174,600 lb)(0.061)(00.65) 5.97 (63,200 lb/day)(0.055)(0.73) = 0.07 (115,000 gal)(8 34 lb/gal)(0.066)(0.59) 5.98 x lb VS added per day = 0.08 (110,000 gal)(8.34 lb/ggal)(0.059)(0.58) 5.116 1,785,000 gal = 3570 gal/ft (25 ft)(20 ft) x = 2511 lb/day VS 5.117 (150 mg/L)(2.69 MGD)(8.34 lb/gal) = 0.68 (1920 mg/L))(0.31 MG)(8.34 lb/gal) 5.118 x lb VS added/day = 0.09 (24,500 gal)(8.34 lb/gal)(00.055)(0.56) 5.94 = (0.17 ft/day)(12 in./ft) = 2.0 in./day 5.102 (1765 mg/L)(0.381 MGD)   × (8.34 lb/gal)   = 28,040 people 0.2 lb/day 5.103 5.115 5.119 3083 gpm = 2.6 gpm/ft (40 ft)(30 ft) 5.120 (115 mg/L)(3.3 MGD)(8.34 lb/gal) 20.1 1000 ft = 157 lb BOD/day/1000 ft 5.104 (2210 mg/L)(0.100 MGD)   × (8.34 lb/gal)   = 9216 people 0.2 lb/day 5.121 5.105 2,250,000 gpd = 448 gpd/ft (0.785)(80 ft)(80 ft) 5.123 5.106 2960 gpm = 15.6 gpm/ft 2 190 ft 5.107 2,100,000 gpd = 8360 gpd/ft (3.14)(80 ft) 5.108 3,300,000 gpd = 519 gpd/ft (0.785)(90 ft)(90 ft) 5.110 500 lb/day VS added per day = 0.06 (182,000 lb)(0.0664)(0.67) 5.111 (2760 mg/L)(3.58 MGD)   × (8.34 lb/gal)   = 16 lb/day/ft (0.785)(80 ft)(80 ft) 5.112 (115,000 lb/day)(0.071) (0.70) = 0.09 (0.785)(70 ft))(70 ft)(21 ft) © 2009 by Taylor & Francis Group, LLC 2,000,000 gpd = 1000 gpd/ft (80 ft)(25 ft) x = 566 lb/day 6000 people = 420 people/ac, x = 14.3 ac x ac (161 mg/L)(2.1 MGD)(8.34 lb/gal) = 0.7 5.109 x lb MLVSS x = 4028 lb MLVSS (174 mg/L)(3.335 MGD)(8.3 lb/gal) = 0.5 (x mg/L)(0.287 MG)(8.34 lb/gal) x = 4033 mg/L MLVSS 5.99 (7900 gpd)(8.34 lb/gal)(0.048)(0.73) = 0.06 x lb VS x = 38,477 lb VS 5.100 1733 people/5.3 ac = 327 people/ac 5.101 4112 people/10 ac = 411 people/ac 4.15 ac-ft/day = 0.17 ft/day 25 ac 2,560,000 gpd = 10,191 gpd/ft (3.14)(80 ft) 5.122 1900 people/5.5 ac = 345 people/ac (140 mg/L)(2.44 MGD)(8.34 lb/gal) 750 1000 ft = 3.8 lb BOD/day/1000 ft 5.124 2882 gpm = 2.4 gpm/ft (40 ft)(30 ft) 5.125 (30 ft)(16 ft)(8 ft)(7.48 gal/ft ) = 29 1007 gpm 5.126 (80 ft)(20 ft)(12 ft)(7.48 gal/ft ) = 1.9 hr 75,000 gph 5.127 (3 ft)(4 ft)(3 ft)(7.48 gal/ft ) = 0.75 hr (6 gpm)(60 min/hr) 5.128 (0.785)(80 ft)(80 ft)(10 ft)(7.48 gal/ft ) = 1.7 hr 216,667 gpd 5.129 (500 ft)(600 ft)(6 ft)(7.48 gal/ft ) = 60.5 days 222,500 gpd 5.130 12,300 lb MLSS = 4.5 days 2750 lb/day 781 5.131 5.132 5.133 5.134 (2820 mg/L MLSS)(0.49 MG)   × (8.34 lb/gal)   = 10.6 days (132 mg/L)(0.988 MGD)(8.34 lb/gal) 5.143 (2810 Mg/L MLSS)(0.325 MG)(8.34 lb/gal) (61000 mg/L)(0.0189 MGD)(8.34 lb/gal)   + (18 mg/L L)(2.4 MGD)(8.34 lb/gal)   7617 lb MLSS = 5.8 days 962 lb/day + 360 lb/day (x mg/L MLSS)(0.205 MG)   × (8.34 lb/gal)   = days (80 mg/L)(2.10 MGD)(8.34 lb/gal) 5.144 x = 4917 mg/L MLSS 5.145 = 7.1 days (2750 mg/L MLSS)(0.360 MG)(8.35 lb/gal)  (54100 mg/L)(0.0192 MG)(8.34 lb/gal)  + (16 mg/L SS)(2.35 MGD)(8.34 lb/gal)   8257 lb = 7.0 days 866 lb/day + 314 lb/day (2550 mg/L MLSS)(1.8 MG)(8.34 lb/gal) (6240 mg/L SS)(0.085 MGD)(8.34 lb/gal)   + (20 mg/L L)(2.8 MGD)(8.34 lb/gal)   5.146 (2610 mg/L)(0.15 MG)(8.34 lb/gal) = days (140 mg/L)((0.92 MGD)(8.34 lb/gal) 5.147 (0.785)(6 ft)(6 ft)(4 ft)(7.48 gal/ft ) = 70 12 gpm 5.148 x lb MLSS = days (140 mg/L)(2.14 MGD)(8.34 lb/gal) x = 14,992 lb MLSS 5.149 (400 ft)(440 ft)(6 ft)(7.48 gal/ft ) = 39.5 200,000 gpd 5.150 (x mg/L MLSS)(0.64 MG)(8.34 lb/gal) = days (6310 mg/L)(0.034 MGD)(8.34 lb/gal)   + (12 mg/L)(2.9 92 MGD)(8.34 lb/gal)   (x mg/L)(0.970 MG)(8.34 lb/gal) (x mg/L)(0.64 MG)(8.34 lb/gal) = days (1789 lb/day) + (292 lb/day) (6340 mg/L)(0.032 MGD)(8.34 lb/gal)   + (20 mg/L)(2.6 MGD D)(8.34 lb/gal)   (x mg/L)(0.64 MG)(8.34 lb/gal) = days 2081 lb/day (x mg/L)(0.970 MG)(8.34 lb/gal) = days 1692 lb/day + 434 lb/day (x mg/L)(0.970 MG)(8.34 lb/gal) = days 2126 x = 3141 mg/L MLSS 5.151 89 mg/L removed × 100% = 81% 110 mg/L 5.152 216 mg/L removed × 100 = 94% 230 mg/L 5.153 200 mg/L removed × 100 = 77% 260 mg/L x = 2100 mg/L MLSS 5.139 (75 ft)(30 ft)(14 ft)(7.48 gal/ft ) = 3.5 hr 68,333 gph 5.140 12,600 lb MLSS = 4.5 days 2820 lb/day 5.141 (2408 mg/L)(1.9 MG)(8.34 lb/gal) 38,157 lb = 9.8 days (3753 lb/day) + (594 lb/day) 38,281 lb MLSS = days 4424 lb/day + 467 lb/day 5.138 (3250 mg/L)(0.33 MG)(8.34 lb/gal) = 4.6 days (100 mg/L)((2.35 MGD)(8.34 lb/gal) (6320 mg/L)((0.0712 MGD)(8.34 lb/gal)   + (25 mg/L)(2.85 MGD)(8.34 lb/gal)    x lb MLSS = 5.5 days, x = 8855 lb MLSS 1610 lb/day SS (1610 lb/day wasted)   + (340 lb/day in SE)    5.137 (40 ft)(20 ft)(10 ft)(7.48 gal/ft ) = 47 1264 gpm (2850 mg/L MLSS)(0.20 MG)   × (8.34 lb/gal)   = 4.5 dayss (84 mg/L)(1.52 MGD)(8.34 lb/gal) 5.135 (3300 mg/L)(0.50 MG)   × (8.34 lb/gal)   5.136 5.142 (3120 mg/L MLSS)(0.48 MG)   × (8.34 lb/gal)   = 6.4 days 1640 lb/day wasted + 320 lb/day © 2009 by Taylor & Francis Group, LLC 5.154 175 mg/L removed × 100 = 56% 310 mg/L 5.155 4.9 = x lb/day solids × 100 (3700 gal)(8.34 lb/gal) x = 1512 lb/day solids 782 5.156 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition 0.87 g sludge × 100 = 6.8% 12.87 g sludge 5.163 5.157 1450 lb/day solids × 100 = 3.3%, x = 43,939 lb/day x lb/day sludge 5.158 5.159 4.4 = 3.6 = 258 lb/day × 100, x = 703 gpd (x gpd)(8.34 lb/gal) = x lb/day solids × 100 291,000 lb/day sludge 5.161 5.162 = 1138 lb/day + 1231 lb/day × 100 25,854 lb//day + 34,194 lb/day = 2369 lb/day × 100 = 3.9% 60,048 lb/day  (8100 gpd)(8.34 lb/gal)(5.1)    100  (7000 gpd)(8.34 lb/gal)(4.1)  +   100 × 100  (8100 gpd)(8.34 lb/gal)  + (7000 gpd)(8.34 lb/gal)   = 3445 lb/day solids + 2394 lb/day soolids × 100  67,554 lb/day sludge     + 58,380 lb/day sludge  = 5839 lb/day solids × 100 = 4.6% 125,934 lb/day sludge  (4750 gpd)(8.34 lb/gal)(4.7)    100  (5250 gpd)(8.34 lb/gal)(3.5)  +   100 × 100  (4750 gpd)(8.34 lb/gal)  + (5250 gpd)(8.34 lb/gal)   = 1862 lb/day + 1532 lb/day × 100 39,615 + 43,785 = 3394 lb/day solids ì 100 = 4.1% 83,400 lb/day sludge â 2009 by Taylor & Francis Group, LLC 2977 lb/day + 6242 lb/day × 100 74,435 lb/day + 94,576 lb/day 9219 lb/day × 100 = 5.5% 169,011 lb/day (3250 lb/day solids)(0.65) = 2113 lb/day VS (4120 gpd)(8.34 lb/gal)(0.07)(0.70) = 1684 lb/day VS 98 ft – 91 ft = ft drawdown 125 ft – 110 ft = 15 ft drawdown 161 ft – 144 ft = 17 ft drawdown = x = 10,476 lb/day solids 5.160  (3100 gpd)(8.34 lb/gal)(4.4)    100  (4100 gpd)(8.34 lb/gal)(3.6)  +   100 × 100  (3100 gpd)(8.34 lb/gal)  + (4100 gpd)(8.34 lb/gal)    (8925 gpd)(8.34 lb/gal)(4.0)    100  (11,340 gpd)(8.34 lb/gal)(6.6)  +   100 × 100  (8925 gpd)(8.34 lb/gal)  + (11,340 gpd)(8.34 lb/gal))   5.164 5.165 5.166 5.167 5.168 5.169 (3.7 psi)(2.31 ft/psi) = 8.5 ft sounding linne water depth 112 ft – 8.5 ft = 103.5 ft 1033.5 ft – 86 ft = 17.5 ft 5.170 (4.6 psi)(2.31 ft/psi) = 10.6 ft sounding line water depth 150 ft – 10.6 ft = 139.4 ft 171 ft – 139.4 ft = 31.4 ft drawdown 5.171 300 ÷ 20 = 15 gpm per ft of drawdown 5.172 420 gal ÷ = 84 gpm 5.173 810 gal ÷ = 162 gpm 5.174 856 gal = 171 gpm (171 gpm)(60 min/hr) = 10,260 gph 5.175  (0.785)(1 ft)(1 ft)(12 ft)    × (7.48 gal/ft )(12 round trips) = 169 gpm 5.176 750 gal ÷ = 150 gpm (150 gpm)(60 min/hr) = 9000 gph (9000 gph)(10 hr/day) = 90,000 gal/day 5.177 200 gpm ÷ 28 ft = 7.1 gpm/ft 5.178 620 gpm ÷ 21 ft = 29.5 gpm/ft 5.179 1100 gpm ÷ 41.3 ft = 26.6 gpm/ft 5.180 x gpm = 33.4 fpm/ft 42.8 ft x = (33.4)(42.8) = 1430 gpm 5.181  (0.785)(0.5 ft)(0.5 ft)    = 206 gal × (140 ft)(7.48 gal/ft ) (40 mg/L)(0.000206 MG)   = 0.07 lb chlorine × (8.34 lb/gal)   783 5.182  (0.785)(1 ft)(1 ft)    = 640 gal × (109 ft)(7.48 gal/ft ) (40 mg/L)(0.000640 MG)   = 0.21 lb chlorine × (8.34 lb/gal)   5.183  (0.785)(1 ft)(1 ft)    = 633 gal × (109 ft)(7.48 gal/ft ) (0.785)(0.67 ft)(0.67 ft)   = 105 gal  × (40 ft)(7.48 gal/ft )  5.189 (4.0 psi) (2.31 ft/psi) = 9.2 ft 5.190 (94 ft + 24 ft) + (3.6 psi)(2.31 ft/psi) = 118 ft + 8.3 ft = 126.3 ft 5.191 (400 ft)(110 ft)(14 ft)(7.48 gal/ft3) = 4,607,680 gal 5.192 (400 ft)(110 ft)(30 ft × 0.4 average depth) × (7.48 gal/ft ) = 3,949,440 gal 5.193 (200 ft)(80 ft)(12 ft) = 4.4 ac-ft 43,560 ft /ac-ft 633 + 105 gal = 738 gal 5.194 (110 mg/L)(0.000738 gal)   = 0.68 lb chlorine × (8.34 lb/gal)   (320 ft)(170 ft)(16 ft)(0.4) = 8.0 ac-ft 43,560 ft /ac-ft 5.195 (0.5 mg/L chlorine)(20 MG)(8.34 lb/gal) 25/1000 5.184 (x mg/L)(0.000540 gal)(8.34 lb/gal) = 0.48 lb x= = 334 lb copper sulfate 0.48 (0.000540)(8.34) 5.196 131.9 ft – 93.5 ft = 38.4 ft x = 107 mg/L 5.185 5.197 0.09 lb chlorine = 1.5 lb 5.25/100 707 gal = 141 gpm (141 gpm)(60 min/hr) = 8460 gph 1.5 lb = 0.18 gal 8.34 lb/gal (0.18 gal)(128 fluid oz./gal) = 23 fl oz 5.198  (0.785)(1 ft)(1 ft)(12 ft)    × (7.48 gal/ft )(8 round trips) = 113 gpm gpm 5.199 (3.5 psi)(2.31 ft/psi) = 8.1 sounding line water depth 5.186  (0.785)(0.5 ft)(0.5 ft)    = 176 gal  × (120 ft)(7.48 gal/ft ) = 167 ft − 8.1 ft = 158.9 ft pumping water level (50 mg/L)(0.000176 MG)   0.1 lb calcium × (8.34 lb/gal)   = hypochlorite 65/100 (0.1 lb)(16 oz./1 lb) = 1.6 oz calcium hypochlorite 5.187  (0.785)(1.5 ft)(1.5 ft)    = 1387 gal × (105 ft)(748 gal/ft ) (100 mg/L)(0.001387 MG)   4.6 lb chloride × (8.34 lb/gal)   = of lime 25/100 5.188 (60 mg/L)(0.000240 MG)   × (8.34 lb/gal)   = 2.3 lb 5.25/100 © 2009 by Taylor & Francis Group, LLC 5.200 610 gpm = 21.8 gpm/ft 28 ft drawdown 5.201  (0.785)(0.5 ft)(0.5 ft)    = 220 gal × (150 ft)(7.48 gal/ft ) (55 mg/L)(0.000220 MG)   = 0.10 lb chlorine req × (8.34 lb/gal)   5.202 780 gal ÷ = 156 gpm (156 gal/min)(60 min/hr)   = 74,880 gal/day × (8 hr/day)   5.203 (x mg/L)(0.000610 MG)   = 0.47 lb × (8.34 lb/gal)   2.3 lb = 0.3 gal 8.34 lb/gal (0.3 gal)(128 fl oz./gal) = Drawdown (ft) = 158.9 ft − 141 ft = 17.9 ft 38.4 fl oz sodium m hypochlorite x= 0.47 = 92.3 mg/L (0.000610)(8.34) 784 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition 5.204  (0.785)(1 ft)(1 ft)    = 523 gal × (89)(7.48 gal/ft )  (0.785)(0.67)(0.67)    = 119 gal × (45 ft)(7.48 gal/ft ) 523 gal + 119 gal = 642 gal (100 mg/L)(0.000642 MG)   = 0.54 lb chlorine × (8.34 lb/gal)   5.205 0.3 lb chlorine = 5.7 lb 5.25/100 5.7 lb = 0.68 gal 8.34 lb/gal (0.68 gal)(128 fl oz./gal) = 87 fl oz 5.206 Volume = (4 ft)(5 ft)(3 ft)(7.48 gal/ft3) = 449 gal 5.207 Volume = (50 ft)(20 ft)(8 ft)(7.48 gal/ft3) = 59,840 gal 5.208 Volume = (40 ft)(16 ft)(8 ft)(7.48 gal/ft3) = 38,298 gal 5.209 42 in ÷ 12 in./ft = 3.5 ft Volume = (5 ft)(5 ft)(3.5 ft)(7.48 gal/ft3) = 655 gal 5.210 in ÷ 12 in./ft = 0.17 ft Volume = (40 ft)(25 ft)(9.17 ft)(7.48 gal/ft3) = 68,592 gal 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31 6.32 6.33 6.34 6.35 6.36 6.37 6.38 6.39 6.40 6.41 6.42 6.43 6.44 6.45 6.46 6.47 6.48 6.49 6.50 Fillet weld Square butt weld Single hem Single flange Location of weld Steel section Bevel weld V weld J-groove weld times true size Drawing number 180 Break line Object line 3-D pictorial Shape; complexity Limits Center line; finished surface Volute Hem Seam Relief valve Gas; liquid Butt Architectural Plot plan Line CHAPTER ANSWERS CHAPTER ANSWERS 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.23 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 Screwed joint Welded joint Soldered joint Screwed glove valve Globe valve Screwed check valve Pump Flexible line Check valve Heat exchanger Expansion joint Vibration absorber Battery cell Motor Relay Voltmeter Ammeter Knife switch Fuse Transformer Ground Normally open contacts Local note reference © 2009 by Taylor & Francis Group, LLC 7.14 7.15 7.16 7.17 7.18 7.19 7.20 26 ft 77 ft Eccentric, segmental Flow nozzle Ultrasonic flowmeter 4937 gal 4.57 213,904 ft3 103 ft 8064 lb Always constant Pressure due to the depth of water The line that connects the piezometric surface along a pipeline 0.28 ft 254.1 ft 6.2 × 10–8 0.86 ft Pressure energy due to the velocity of the water A pumping condition where the size of the impeller of the pump and above the surface of the water from which the pump is running The slope of the specific energy line 785 CHAPTER ANSWERS CHAPTER ANSWERS 8.1 8.2 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 8.29 8.30 8.31 8.32 8.33 8.34 8.35 8.36 8.37 8.38 8.39 8.40 8.41 8.42 Alternator The effect that causes current flow in a conductor moving across magnetic lines of force Mechanical, electrical Increases, decreases, decreases, increases To protect an electrical circuit and load 0.2 ohms Orbits or shells Protons and neutrons The value of the resistor, the length of the conductors, and the diameter of the conductors Direct current flow does not change direction, whereas alternating current periodically changes direction The magnetic poles The flux lines, or magnetic flux along which a magnetic force acts Natural magnet, permanent magnets, and electromagnets Chemistry Battery, two A series circuit has only one path for current flow, whereas a parallel circuit has more than one path Source voltage Voltage drop Counterclockwise amps 12 volts 16 watts 80 watts Less, more Resistivity Circular mil Circular mil Conductivity Smaller Doubles It will withstand high voltages The two are directly proportional As flux density increases, field strength also increases The type of material and the flux density North pole Increases 141.4 volts A voltage is induced in the conductor AC, cut, counter Counter Current has an associated magnetic field Increase Increase © 2009 by Taylor & Francis Group, LLC Positive-displacement High-viscosity Positive-displacement High High Eye Static, dynamic Shut off V2/2g Total head Head capacity, efficiency, horsepower demand Water Suction lift Elevation head Water hp and pump efficiency Centrifugal force Stuffing box Impeller Rings, impeller Casing CHAPTER 10 ANSWERS 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 10.19 10.20 10.21 10.22 10.23 10.24 10.25 10.26 10.27 10.28 A flexible piping component that absorbs thermal and/or terminal movement Fluid Fluid Connected Flow Pressure loss Increases Automatically Insulation Leakage Four times Routine preventive maintenance 12 Schedule, thickness Increases Ferrous Increases Iron oxide Cast iron Iron Corrosion Decreases Clay, concrete, plastic, glass, or wood Corrosion-proof Cement Pressed Turbulent, lower Steel 760 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition hub This plate rotates in contact with a fixed valve plate with similar parts, which are connected to a vacuum supply, a compressed air supply, and an atmosphere vent As the drum rotates, each section is thus connected to the appropriate service The coil type vacuum filter uses two layers of stainless steel coils arranged in corduroy fashion around the drum After a dewatering cycle, the two layers of springs leave the drum bed and are separated from each other so the cake is lifted off the lower layer and is discharged from the upper layer The coils are then washed and reapplied to the drum The coil filter is used successfully for all types of sludges; however, sludges with extremely fine particles or ones that are resistant to flocculation dewater poorly with this system The media on a belt filter leave the drum surface at the end of the drying zone and pass over a small-diameter discharge roll to aid in cake discharge Washing of the media occurs next The media are then returned to the drum and to the vat for another cycle This type of filter normally has a small-diameter curved bar between the point where the belt leaves the drum and the discharge roll This bar primarily aids in maintaining belt dimensional stability 19.12.7.2 Operational Observations, Problems, and Troubleshooting 19.12.7.1.1 Filter Media Drum and belt vacuum filters use natural or synthetic fiber materials On the drum filter, the cloth is stretched and secured to the surface of the drum In the belt filter, the cloth is stretched over the drum and through the pulley system The installation of a blanket requires several days The cloth (with proper care) will last several hundred to several thousand hours The life of the blanket depends on the cloth selected, the conditioning chemical, backwash frequency, and cleaning (e.g., acid bath) frequency 19.12.7.3.1 Vacuum Filter Yield (lb/hr/ft2) 19.12.7.1.2 Filter Drum The filter drum is a maze of pipe work running from a metal screen and wooden skeleton and connecting to a rotating valve port at each end of the drum The drum is equipped with a variable speed drive to turn the drum from 1/8 to rpm Normally, solids pickup is indirectly related to the drum speed The drum is partially submerged in a vat containing the conditioned sludge Submergence is usually limited to 1/5 or less of filter surface at a time 19.12.7.1.3 Chemical Conditioning Sludge dewatered using vacuum filtration is normally chemically conditioned just prior to filtration Sludge conditioning increases the percentage of solids captured by the filter and improves the dewatering characteristics of the sludge; however, conditional sludge must be filtered as quickly as possible after chemical addition to obtain these desirable results © 2009 by Taylor & Francis Group, LLC In operation, the rotating drum picks up chemically treated sludge A vacuum is applied to the inside of the drum to draw the sludge onto the outside of the drum cover This porous outside cover or filter medium allows the filtrate or liquid to pass through into the drum and the filter cake (dewatered sludge) to stay on the medium In the cake release/discharge mode, slight air pressure is applied to the drum interior Dewatered solids are lifted from the medium and scraped off by a scraper blade Solids drop onto a conveyor for transport for further treatment or disposal The filtrate water is returned to the plant for treatment The operator observes drum speed, sludge pickup, filter cake thickness and appearance, chemical feed rates, sludge depth in vat, and overall equipment operation Sampling and testing are routinely performed on influent sludge solids concentration, filtrate BOD and solids, and sludge cake solids concentration Table 19.27 lists several indicators of poor process performance, their causal factors, and corrective actions 19.12.7.3 Process Control Calculations Probably the most frequent calculation that vacuum filter operators have to make is determining filter yield Example 17.84 illustrates how this calculation is made ■ EXAMPLE 17.84 Problem: Thickened, thermally conditioned sludge is pumped to a vacuum filter at a rate of 50 gpm The vacuum area of the filter is 12 ft wide with a drum diameter of 9.8 ft If the sludge concentration is 12%, what is the filter yield in lb/hr/ft2? Assume the sludge weighs 8.34 lb/gal Solution: First calculate the filter surface area: Area of a cylinder side = 3.14 × diameter × length = 3.14 × 9.8 ft × 12 ft = 369.3 ft2 Next, calculate the pounds of solids per hour: 50 gpm 60 8.34 lb 12% × × × = 3002.4 lb/hr hr gal 100% Divide the two: 3002.4 lb/hr = 8.13 lb/hr/ft 369.3 ft Wastewater Treatment Operations 761 TABLE 19.27 Symptoms, Causes, and Corrective Actions for Rotary Vacuum Filter Problems Symptom High solids in filtrate Cause Coagulant dosage is incorrect Filter media are binding Thin filter cake and poor dewatering Chemical dosage is incorrect Filter media are binding Vacuum pump not running Drum not rotating Receiver vibrating High vat level Low vat level Vacuum pump drawing high amperage Scale buildup on vacuum pump seals Vacuum is inadequate Drum speed is too high Drum is submerged too low Power to drive motor is off Seal water is not flowing Drive belt is broken Power to drive motor is off Filtrate pump is clogged Bolts and gasket around inspection plate are loose Ball check valve in filtrate pump is worn Suction line has air leaks Drum face is dirty Seal strips are missing Chemical conditioning is incorrect Feed rate is too high Drum speed is too slow Filtrate pump is off or clogged Drain line is plugged Vacuum pump has stopped Seal strips are missing Feed rate is too low Vat drain valve is open Filtrate pump is clogged Improper chemical conditioning High vat level Cooling water flow to vacuum pump is too high Hard, unstable water 19.12.8 PRESSURE FILTRATION Pressure filtration differs from vacuum filtration in that the liquid is forced through the filter media by a positive pressure instead of a vacuum Several types of presses are available, but the most commonly used types are plateand-frame presses and belt presses Filter presses include the belt or plate-and-frame types The belt filter includes two or more porous belts, rollers, and related handling systems for chemical makeup and feed, as well as supernatant and solids collection and transport The plate-and-frame filter consists of a support frame, filter plates covered with porous material, a hydraulic or mechanical mechanism for pressing plates together, and related handling systems for chemical makeup and feed, © 2009 by Taylor & Francis Group, LLC Corrective Action Adjust coagulant dosage; recalibrate coagulant feeder Clean synthetic cloth with steam and detergent; clean steel coil with acid bath; clean cloth with water or replace cloth Adjust coagulant dosage; recalibrate coagulant feeder Clean synthetic cloth with steam and detergent; clean steel coil with acid bath; clean cloth with water or replace cloth Repair vacuum system Reduce drum speed Increase drum submergence Reset heater, breaker, etc.; restart Initiate seal water flow Replace drive belt Reset heater, breaker, etc.; restart Clear pump Tighten bolts and gasket Replace ball check valve Seal leaks Clean face with pressure hose Replace missing seal strips Change coagulant dosage Reduce feed rate Increase drum speed Turn on or clean pump Clean drain line Turn on or clean pump Replace seal strips Increase feed rate Close vat drain valve Clear pump clog Adjust coagulant dosage See “High vat level” entry (above) Decrease cooling water flow rate Add sequestering agent as well as supernatant and solids collection and transport In the plate-and-frame filter, solids are pumped (sandwiched) between plates Pressure (200 to 250 psi) is applied to the plates and water is squeezed from the solids At the end of the cycle, the pressure is released; as the plates separate, the solids drop out onto a conveyor belt for transport to storage or disposal Performance factors for plate-and-frame presses include feed sludge characteristics, type and amount of chemical conditioning, operating pressures, and the type and amount of precoat The belt filter uses a coagulant (polymer) mixed with the influent solids The chemically treated solids are discharged between two moving belts First, water drains from the solids by gravity Then, as the two belts move 762 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition TABLE 19.28 Symptoms, Causes, and Corrective Actions for Pressure Filtration Problems Symptom Plate press Plates fail to seal Cake discharge is difficult Filter cycle times are excessive Filter cake sticks to conveyors Precoat pressures gradually increase Media are frequently binding Excessive moisture is found in cake Sludge is blowing out of the press Leaks are apparent around the lower faces of the plates Belt press Filter cake discharge is difficult Sludge is leaking from belt edges Excessive moisture is found in filter cake Belt wear along edges is excessive Belt shifts or seizes Cause Poor alignment Inadequate shimming Inadequate precoat Improper conditioning Wet cake soiling media on lower faces Realign parts Adjust shimming of stay bosses Increase precoat, feed at 25–40 psig Change conditioner type or dosage (use filter leaf test to determine) Change chemical dosage Improve thickening operation Increase inorganic conditioner dose Change chemical dosage Decrease feed for a few cycles, then optimize Wash filter media Wash media with inhibited hydrochloric acid Increase precoat Reduce feed rate; develop initial cake slowly Change chemical dosage Lengthen filter cycle Shut down feed pump; hit press closure drive; restart feed pump; clean after cycle See “Excessive moisture is found in cake” entry (above) Improper chemical dosage Changing sludge characteristics Wrong conditioning chemical selected Wrong application point Excessive belt tension Belt speed too low Excessive sludge feed rate Improper belt speed or drainage time Wrong conditioning chemical Improper chemical dosage Inadequate belt washing Wrong belt weave or material Roller misalignment Improper belt tension Tension/alignment control system Uneven sludge distribution Inadequate or uneven belt washing Change conditioning chemical Adjust chemical dosage Change chemical or sludge Adjust application point Reduce belt tension Increase belt speed Reduce sludge feed rate Adjust belt speed Change conditioning chemical Adjust chemical dosage Clear spray nozzles and adjust sprays Replace belt Correct roller alignment Correct tension Repair tracking and alignment system controls Adjust feed for uniform sludge distribution Clean and adjust belt-washing sprays Improper conditioning Low feed solids Improper conditioning chemical/dosage Improper sludge conditioning Improper precoat feed Plugged filter media Calcium buildup in media Inadequate precoat Initial feed rate too high (no precoat) Improper conditioning Filter cycle too short Obstruction between plates between a series of rollers, pressure squeezes additional water out of the solids The solids are then discharged onto a conveyor belt for transport to storage or disposal Performance factors for the belt press include sludge feed rate, belt speed, belt tension, belt permeability, chemical dosage, and chemical selection Filter presses have lower operation and maintenance costs than vacuum filters or centrifuges They typically produce a good-quality cake and can be batch operated; however, construction and installation costs are high Moreover, chemical addition is required and the presses must be operated by skilled personnel © 2009 by Taylor & Francis Group, LLC Corrective Action 19.12.8.1 Operational Observations, Problems, and Troubleshooting Most plate and filter press operations are partially or fully automated Operation consists of observation, maintenance, and sampling and testing Operation of belt filter presses consists of preparation of conditioning chemicals, chemical feed rate adjustments, sludge feed rate adjustments, belt alignment, belt speed and belt tension adjustments, sampling and testing, and maintenance Table 19.28 lists several indicators of poor process performance, their causal factors, and corrective actions Wastewater Treatment Operations 19.12.8.2 Filter Press Process Control Calculations As part of the operating routine for filter presses, operators are called upon to make certain process control calculations The process control calculation most commonly used in operating the belt filter press determines the hydraulic loading rate on the unit The process control calculation most commonly used in the operation of plate and filter presses determines the pounds of solids pressed per hour Both of these calculations are demonstrated below 19.12.8.2.1 Hydraulic Loading Rate for Belt Filter Presses ■ EXAMPLE 19.85 Problem: A belt filter press receives a daily sludge flow of 0.30 gal If the belt is 60 in wide, what is the hydraulic loading rate on the unit in gallons per minute for each foot of belt width (gpm/ft)? Solution: 208.3 gal day 0.30 MG 1, 000, 000 gal = × × 1440 MG day 60 in × ft = ft 12 in 208.3 gal = 41.7 gpm/ft ft 19.12.8.2.2 Pounds of Solids Pressed Per Hour for Plate and Frame Presses ■ EXAMPLE 19.86 Problem: A plate and frame filter press can process 850 gal of sludge during its 120-min operating cycle If the sludge concentration is 3.7%, and if the plate surface area is 140 ft2, how many pounds of solids are pressed per hour for each square foot of plate surface area? Solution: 850 gal × 3.7% 8.34 lb × = 262.3 lb 100% gal 262.3 lb 60 × = 131.2 lb/hr 120 hr 131.2 lb/hr = 0.94 lb/hr/ft 140 ft © 2009 by Taylor & Francis Group, LLC 763 TABLE 19.29 Expected Percent Solids for Centrifuge Dewatered Sludges Type of Sludge Raw sludge Anaerobic digestion Activated sludge Heat treated Percent Solids 25–35% 15–30% 8–10% 30–50% 19.12.9 CENTRIFUGATION Centrifuges of various types have been used in dewatering operations for at lease 30 years and appear to be gaining in popularity Depending on the type of centrifuge used, chemical makeup and feed equipment and support systems for removal of dewatered solids are required, in addition to centrifuge pumping equipment for solids feed and centrate removal 19.12.9.1 Operational Observations, Problems, and Troubleshooting Generally, in operation, the centrifuge spins at a very high speed The centrifugal force it creates throws the solids out of the water Chemically conditioned solids are pumped into the centrifuge The spinning action throws the solids to the outer wall of the centrifuge The centrate (water) flows inside the unit to a discharge point The solids held against the outer wall are scraped to a discharge point by an internal scroll moving slightly faster or slower than the centrifuge speed of rotation In the operation of a continuous-feed, solid-bowl, conveyor-type centrifuge (this is the most common type currently used), as well as other commonly used centrifuges, solid/liquid separation by gravity occurs as a result of rotating the liquid at high speeds In the solid-bowl type, the solid bowl has a rotating unit with a bowl and a conveyor The unit has a conical section at one end that acts as a drainage device The conveyor screw pushes the sludge solids to outlet ports and the cake to a discharge hopper The sludge slurry enters the rotating bowl through a feed pipe leading into the hollow shaft of the rotating screw conveyor The sludge is distributed through ports into a pool inside the rotating bowl As the liquid sludge flows through the hollow shaft toward the overflow device, the fine solids settle to the wall of the rotating bowl The screw conveyor pushes the solids to the conical section, where the solids are forced out of the water and the water drains back in the pool Expected percent solids for centrifuge dewatered sludge are in the range of 10 to 15% The expected performance is dependent on the type of sludge being dewatered, as shown in Table 19.29 Centrifuge operation is dependent on various performance factors: 764 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition • • • • • • • • • Bowl design—length/diameter ratio; flow pattern Bowl speed Pool volume Conveyor design Relative conveyor speed Type and condition of sludge Type and amount of chemical conditioning Operating pool depth Relative conveyor speed (if adjustable) Centrifuge operators often find that the operation of centrifuges can be simple, clean, and efficient In most cases, chemical conditioning is required to achieve optimum concentrations Operators soon discover that centrifuges are noise makers; units run at very high speed and produce high-level noise that can cause loss of hearing with prolonged exposure When working in an area where a centrifuge is in operation, special care must be taken to provide hearing protection Actual operation of a centrifugation unit requires the operator to control and adjust chemical feed rates, observe unit operation and performance, control and monitor centrate returned to the treatment system, and perform required maintenance as outlined in the manufacturer’s technical manual The centrifuge operator must be trained to observe and recognize (as with other unit processes) operational problems that may occur with centrifuge operation Table 19.30 lists several indicators of poor process performance, their causal factors, and corrective actions 19.12.10 SLUDGE INCINERATION Not surprisingly, incinerators produce the maximum solids and moisture reductions The equipment required depends on whether the unit is a multiple-hearth or fluid-bed incinerator Generally, the system will require a source of heat to reach ignition temperature, a solids feed system, and ash-handling equipment It is important to note that the system must also include all required equipment (e.g., scrubbers) to achieve compliance with air pollution control requirements Solids are pumped to the incinerator The solids are dried and then ignited (burned) As they burn, the organic matter is converted to carbon dioxide and water vapor, and the inorganic matter is left behind as ash or fixed solids The ash is then collected for reuse of disposal 19.12.10.1 Process Description The incineration process first dries then burns the sludge; the process involves the following steps: • • The temperature of the sludge feed is raised to 212°F Water evaporates from the sludge â 2009 by Taylor & Francis Group, LLC • The temperature of the water vapor and air mixture increases The temperature of the dried sludge volatile solids rises to the ignition point Note: Incineration will achieve maximum reductions if sufficient fuel, air, time, temperature, and turbulence are provided 19.12.10.2 Incineration Processes 19.12.10.2.1 Multiple-Hearth Furnace The multiple-hearth furnace consists of a circular steel shell surrounding a number of hearths Scrappers (rabble arms) are connected to a central rotating shaft Units range from 4.5 to 21.5 feet in diameter and have from four to 11 hearths Dewatered sludge solids are placed on the outer edge of the top hearth The rotating rabble arms move them slowly to the center of the hearth At the center of the hearth, the solids fall through ports to the second level The process is repeated in the opposite direction Hot gases generated by burning on lower hearths are used to dry the solids The dry solids pass to the lower hearths The high temperature on the lower hearths ignites the solids Burning continues to completion Ash materials discharge to lower cooling hearths, where they are discharged for disposal Air flowing inside the center column and rabble arms continuously cools internal equipment 19.12.10.2.2 Fluidized Bed Furnace The fluidized bed incinerator consists of a vertical circular steel shell (reactor) with a grid to support a sand bed and an air system to provide warm air to the bottom of the sand bed The evaporation and incineration process takes place within the super-heated sand bed layer Air is pumped to the bottom of the unit The airflow expands (fluidizes) the sand bed inside The fluidized bed is heated to its operating temperature (1200 to 1500°F) Auxiliary fuel is added when necessary to maintain operating temperature The sludge solids are injected into the heated sand bed Moisture immediately evaporates Organic matter ignites and reduces to ash Residues are ground to fine ash by the sand movement Fine ash particles flow up and out of unit with exhaust gases Ash particles are removed using common air pollution control processes Oxygen analyzers in the exhaust gas stack control the airflow rate Note: Because these systems retain a high amount of heat in the sand, the system can be operated as little as hr per day with little or no reheating 19.12.10.3 Operational Observations, Problems, and Troubleshooting The operator of an incinerator monitors various performance factors to ensure optimal operation These performance Wastewater Treatment Operations 765 TABLE 19.30 Symptoms, Causes, and Corrective Actions for Pressure Filtration Problems Symptom Poor centrate clarity Solids cake not dry enough Frequent tripping of torque control Excess vibration Sudden increase in power consumption Gradual increase in power consumption Spasmodic surging of solids discharge Centrifuge shutting down or not starting Cause Feed rate too high Wrong plate dam position Worn conveyor flights Speed too high High feed sludge solids concentration Improper chemical conditioning Feed rate too high Wrong plate dam position Speed too low Excessive chemical conditioning Influent too warm Feed rate too high Feed solids concentration too high Foreign material (e.g., tramp iron) in machine Gear unit misaligned Mechanical problem in gear unit Improper lubrication Improper adjustment of vibration isolators Discharge funnels contacting centrifuge Portion of conveyor flights plugged (causing an imbalance) Gear box improperly aligned Pillow box bearings damaged Bowl out of balance Parts not tightly assembled Uneven wear on conveyor Contact between bowl exterior and accumulated solids in case Effluent pipe plugged Conveyor blade wear Pool depth too low Conveyor helix rough Feed pipe too near drainage deck Excessive vibration Blown fuses Tripped overload relay Motor overheated or thermal protectors tripped Tripped torque control Tripped vibration switch factors include feed sludge volatile matter content, feed sludge moisture content, operating temperature, sludge feed rate, fuel feed rate, and air feed rate Corrective Action Adjust sludge feed rate Increase pool depth Repair/replace conveyor Change pulley setting to obtain lower speed Dilute sludge feed Adjust chemical dosage Reduce sludge feed rate Decrease pool depth to increase dryness Change pulley setting to obtain higher speed Adjust chemical dosage Reduce influent temperature Reduce flows Dilute flows Remove conveyor; clear foreign materials Correct gear unit alignment Repair gear unit Lubricate according to the manufacturer’s instructions Adjust isolators Reposition slip joints at funnels Flush centrifuge Align gearbox Replace bearings Return rotating parts to factory for rebalancing Tighten parts Resurface and rebalance Apply hard surfacing to areas with wear Clear solids discharge Replace blades Increase pool depth Refinish conveyor blade area Move feed pipe to effluent end (if applicable) See “Excess vibration” entry (above) Replace fuses Flush centrifuge; reset relay Flush centrifuge; reset thermal protectors See “Frequent tripping of torque control” entry (above) See “Excess vibration” entry (above) operational problems using various indicators or through observations Table 19.31 lists several indicators of poor process performance, their causal factors, and corrective actions Note: To ensure that the volatile material is ignited, the sludge must be heated to between 1400 and 1700°F 19.12.11 LAND APPLICATION To be sure that operating parameters are in the correct range, the operator monitors and adjusts sludge feed rate, airflow, and auxiliary fuel feed rate All maintenance conducted on an incinerator should be in accordance with manufacturer’s recommendations The operator of a multiplehearth or fluidized bed incinerator must be able to recognize The purpose of land application of biosolids is to dispose of the treated biosolids in an environmentally sound manner by recycling nutrients and soil conditioners To be land applied, wastewater biosolids must comply with state and federal biosolids management and disposal regulations Biosolids must not contain materials that are dangerous to © 2009 by Taylor & Francis Group, LLC OF BIOSOLIDS 766 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition TABLE 19.31 Symptoms, Causes, and Corrective Actions for Sludge Incineration Problems Symptom Multiple hearth Incinerator temperature too high Furnace temperature too low Oxygen content of stack gas too high Oxygen content of stack gas too low Furnace refractories deteriorated Unusually high cooling effect Short hearth life Center shaft shear pin failure Scrubber temperature too high Stack gas temperatures too low Stack gas temperatures too high Furnace burners slagging up Rabble arms dropping Excessive air pollutants in stack gas Flashing or explosions Fluidized bed Bed temperature falling Low (6%) oxygen in exhaust gas Erratic bed depth on control panel Preheat burner fails and alarm sounds Bed temperature too high Bed temperature reading off scale High temperature in scrubber inlet Cause Corrective Action Excessive fuel feed rate Greasy solids Thermocouple burned out Increased moisture content of sludge Fuel system malfunction Excessive air feed rate Flame out Sludge feed rate too low Sludge feed system blockage Air feed rate too high Increased volatile or grease content in sludge Air feed rate too low Rapid startup/shutdown of furnace Decrease fuel feed rate Reduce sludge feed rate; increase air feed rate Replace thermocouple Increase fuel feed rate until dewatering operation improves Establish proper fuel feed rate Decrease air feed rate; increase sludge feed rate Relight furnace Increase sludge feed rate Clear any feed system blockages Decrease air feed rate Increase air feed rate; decrease sludge feed rate Increase air feed rate Repair furnace refractories; follow specified startup and shutdown procedures Locate and repair leak Fire hearths equally on both sides Adjust rabble arm to eliminate rubbing Remove debris Adjust water flow to proper level Increase fuel feed rate Decrease sludge feed rate Increase air feed rate; decrease sludge feed rate Decrease fuel feed rate Replace burners with newer designs that reduce slagging Maintain temperatures within proper range; discontinue injection of scum into the hearth Repair cooling air system immediately Raise air-to-fuel ratio Repair or replace broken equipment Remove scum or grease before incineration Air leak Uneven firing Rabble arm dragging on hearth Debris caught under the arm Low water flow to scrubber Inadequate fuel feed supply Excessive sludge feed rate Higher volatile content (heat value) in sludge Excessive fuel feed rate Burner design Excessive hearth temperatures Loss of cooling air Incomplete combustion, insufficient air Air pollution control malfunction Scum or grease additions Inadequate fuel supply Excessive sludge feed rate Excessive sludge moisture levels Excessive air flow Low air flow rate Fuel feed rate too high Sludge feed rate too low Bed pressure taps plugged with solids Pilot flame not receiving fuel Pilot flame not receiving spark Defective pressure regulator Pilot flame ignition but malfunctioning flame scanner Bed gun fuel feed rate too high Grease or high organic content in sludge (high heat value) Thermocouple burned out Water not flowing in scrubber Plugged spray nozzles Ash water not recirculating Poor bed fluidization © 2009 by Taylor & Francis Group, LLC Sand leakage through support plate during shutdown Increase fuel supply; repair any fuel system malfunction Decrease sludge feed rate Correct sludge dewatering process problem Decrease airflow rate Increase blower air feed rate Reduce fuel feed rate Increase sludge feed rate; adjust fuel feed rate to maintain steady bed temperature Tap a metal rod into pressure tap pipe when the unit is not in operation; apply compressed air to pressure tap while the unit is in operation (follow manufacturer’s safety guidelines) Correct fuel system problem Replace defective part Replace defective regulator Clear scanner sight glass; replace defective scanner Reduce bed gun fuel feed rate Increase airflow rate; decrease sludge fuel rate Replace thermocouple Open valves to provide water; correct system malfunction to provide required pressure Clear nozzles and strainers Repair or replace recirculation pump; unclog scrubber discharge line Clear wind box; clean wind box at least once per month Wastewater Treatment Operations human health (e.g., toxicity, pathogenic organisms) or dangerous to the environment (e.g., toxicity, pesticides, heavy metals) Treated biosolids are land applied by either direct injection or application and plowing in (incorporation) 19.12.11.1 Process Control: Sampling and Testing Land application of biosolids requires precise control to avoid problems The quantity and the quality of biosolids applied must be accurately determined For this reason, the operator’s process control activities include biosolids sampling/testing functions Biosolids sampling and testing includes determination of percent solids, heavy metals, organic pesticides and herbicides, alkalinity, total organic carbon (TOC), organic nitrogen, and ammonia nitrogen 767 where: f1 = Mineral rate for organic nitrogen (assume 0.20) V1 = Volatilization rate ammonia nitrogen = 1.00 if biosolids are injected = 0.85 if biosolids are plowed in within 24 hr = 0.70 if biosolids are plowed in within days ■ EXAMPLE 19.88 Problem: The biosolids contain 21,000 mg/kg of organic nitrogen and 10,500 mg/kg of ammonia nitrogen The biosolids are incorporated into the soil within 24 hr after application What is the plant available nitrogen (PAN) per dry ton of solids? Solution: 19.12.11.2 Process Control Calculations Process control calculations include determining disposal cost, plant available nitrogen (PAN), application rate (dry tons and wet tons per acre), metals loading rates, maximum allowable applications based on metals loading, and site life based on metals loading 19.12.11.2.1 Disposal Cost The cost of disposal of biosolids can be determined by: Cost = Wet tons/yr × % solids × cost/dry ton (19.92) ■ EXAMPLE 19.87 (21,000 mg/kg × 0.20 ) PAN =   × 0.002  + (10, 500 × 0.85 ) = 26.3 lb/dry ton 19.12.11.2.3 Application Rate Based on Crop Nitrogen Requirement In most cases, the application rate of domestic biosolids to crop lands will be controlled by the amount of nitrogen the crop requires The biosolids application rate based on the nitrogen requirement is determined by the following: Problem: The treatment system produces 1925 wet tons of biosolids for disposal each year The biosolids are 18% solids A contractor disposes of the biosolids for $28 per dry ton What is the annual cost for sludge disposal? • • Solution: Cost = 1925 wet tons/yr × 0.18 × $28/dry ton = $9702 19.12.11.2.2 Plant Available Nitrogen (PAN) One factor considered when land applying biosolids is the amount of nitrogen in the biosolids available to the plants grown on the site This includes ammonia nitrogen and organic nitrogen The organic nitrogen must be mineralized for plant consumption Only a portion of the organic nitrogen is mineralized per year The mineralization factor (f1) is assumed to be 0.20 The amount of ammonia nitrogen available is directly related to the time elapsed between applying the biosolids and incorporating (plowing) the sludge into the soil We provide volatilization rates based upon this example below: (Organic nitrogen (mg/kg) × f1 )   PAN  = + (ammonia nitrogen (mg/kg) × V1 ) (19.93) (lb/dry ton)  ì 0.0002 lb/dry ton â 2009 by Taylor & Francis Group, LLC Step Use an agriculture handbook to determine the nitrogen requirement of the crop to be grown Step Determine the amount of sludge in dry tons required to provide this much nitrogen: Dry tons/ac = ■ Plant nitrogen requirement (lb/ac) (19.94) Plant available nitrogen (lb/dry ton) EXAMPLE 19.89 Problem: The crop to be planted on the land application site requires 150 lb of nitrogen per acre What is the required biosolids application rate if the PAN of the biosolids is 30 lb/dry ton? Solution: Application rate = 150 lb/ac = dry tons/ac 30 lb/dry ton 19.12.11.2.4 Metals Loading When biosolids are land applied, metals concentrations are closely monitored and their loading on land application sites is calculated: 768 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition Loading (lb/ac) = Metal conc (mg/kg) × 0.002 lb/dry ton (19.95) × application rate (dry tons/ac) ■ Solution: Loading rate = 14 mg/kg × 0.002 lb/dry ton × 11 dry tons = 0.31 lb/ac 19.12.11.2.5 Maximum Allowable Applications Based on Metals Loading If metals are present, they may limit the total number of applications a site can receive Metals loading is normally expressed in terms of the maximum total amount of metal that can be applied to a site during its use: Max allowable cumulative load for the metal (lb/ac) (19.96) Applications = Metal loading (lb/ac//application) EXAMPLE 19.91 Problem: The maximum allowable cumulative lead loading is 48.0 lb/ac Based on the current loading of 0.35 lb/ac, how many applications of biosolids can be made to this site? 135 applications = 68 yr applications per yr Note: When more than one metal is present, the calculations must be performed for each metal The site life would then be the lowest value generated by these calculations 19.13 PERMITS, RECORDS, AND REPORTS Permits, records, and reports play a significant role in wastewater treatment operations In fact, with regard to permits, one of the first things any new operator quickly learns is the importance of “making permit” each month In this chapter, we briefly cover National Pollutant Discharge Elimination System (NPDES) permits and other pertinent records and reports with which the wastewater operator must be familiar Note: The discussion that follows is general in nature; it does not necessarily apply to any state in particular but instead is an overview of permits, records, and reports that are an important part of wastewater treatment plant operations For guidance on requirements for a specific locality, contact the state’s water control board or other authorized state agency for information In this handbook, the term board signifies the state-reporting agency 19.13.1 DEFINITIONS Several definitions should be understood before we discuss the permit requirements for records and reporting: Solution: Applications = 48.0 lb/ac = 137 0.35 lb/ac 19.12.11.2.6 Site Life Based on Metals Loading The maximum number of applications based on metals loading and the number of applications per year can be used to determine the maximum site life: Site life (yr) = ■ Solution: Site life = EXAMPLE 19.90 Problem: The biosolids contain 14 mg/kg of lead Biosolids are currently being applied to the site at a rate of 11 dry tons per acre What is the metals loading rate for lead in pounds per acre? ■ applications Based on the lead loading and the application rate, how many years can this site be used? Max allowable applications Number of applications planned per year (19.97) EXAMPLE 19.92 Problem: Biosolids are currently applied to a site twice annually Based on the lead content of the biosolids, the maximum number of applications is determined to be 135 © 2009 by Taylor & Francis Group, LLC Average daily limitation—The highest allowable average over a 24-hour period, calculated by adding all of the values measured during the period and dividing the sum by the number of values determined during the period Average hourly limitation—The highest allowable average for a 60-minute period, calculated by adding all of the values measured during the period and dividing the sum by the number of values determined during the period Average monthly limitation—The highest allowable average over a calendar month, calculated by adding all of the daily values measured during the month and dividing the sum by number of daily values measured during the month Average weekly limitation—The highest allowable average over a calendar week, calculated by adding all Wastewater Treatment Operations of the daily values measured during the calendar week and dividing the sum by the number of daily values determined during the week Daily discharge—The discharge of a pollutant measured during a calendar day or any 24-hour period that reasonably represents the calendar for the purpose of sampling For pollutants with limitations expressed in units of weight, the daily discharge is calculated as the total mass of the pollutant discharged over the day For pollutants with limitations expressed in other units, the daily discharge is calculated as the average measurement of the pollutant over the day Discharge monitoring report—Forms used to report selfmonitoring results of the permittee Discharge permit—State Pollutant Discharge Elimination System permit that specifies the terms and conditions under which a point source discharge to state waters is permitted Effluent limitation—Any restriction by the state board on quantities, discharge rates, or concentrations of pollutants which are discharged from point sources into state waters Maximum daily discharge—The highest allowable value for a daily discharge Maximum discharge—The highest allowable value for any single measurement Minimum discharge—The lowest allowable value for any single measurement Point source—Any discernible, defined, and discrete conveyance, including but not limited to, any pipe, ditch, channel, tunnel, conduit, well, discrete fissure, container, rolling stock, vessel, or other floating craft from which pollutants are or may be discharged This definition does not include return flows from irrigated agricultural land 769 tions are standard for all dischargers and cover a broad series of requirements Read the general conditions of the treatment facility’s permit carefully Permittees must retain certain records 19.13.2.1 Monitoring • • • • • • • • • • Note: All records must be kept at least years (longer at the request of the state board) 19.13.2.2 Reporting Generally, reporting must be made under the following conditions or situations (requirements may vary depending on the state regulatory body with reporting authority): • 19.13.2 NPDES PERMITS In the United States, all treatment facilities that discharge to state waters must have a discharge permit issued by the state water control board or other appropriate state agency This permit is known on the national level as the National Pollutant Discharge Elimination System (NPDES) permit and on the state level as the (State) Pollutant Discharge Elimination System (state-PDES) permit The permit states the specific conditions that must be met to legally discharge treated wastewater to state waters The permit contains general requirements (applying to every discharger) and specific requirements (applying only to the point source specified in the permit) A general permit is a discharge permit that covers a specified class of dischargers It is developed to allow dischargers in a specified category to discharge under specified conditions All discharge permits contain general conditions These condi- © 2009 by Taylor & Francis Group, LLC Date, time, and exact place of sampling or measurements Names of the individuals performing sampling or measurement Dates and times analyses were performed Names of the individuals who performed the analyses Analytical techniques or methods used Observations, readings, calculations, bench data, and results Instrument calibration and maintenance Original strip chart recordings for continuous monitoring Information used to develop reports required by the permit Data used to complete the permit application • • • Unusual or extraordinary discharge reports— Must be submitted to the board by telephone within 24 hours of occurrence, with a written report being submitted within days The report must include: • Description of the noncompliance and its cause • Noncompliance dates, times, and duration • Steps planned or taken to reduce or eliminate the problem • Steps planned or taken to prevent reoccurrence Anticipated noncompliance—Must notify the board at least 10 days in advance of any changes to the facility or activity that may result in noncompliance Compliance schedules—Must report compliance or noncompliance with any requirements contained in compliance schedules no later than 14 days following the scheduled date for completion of the requirement 24-hour reporting—Any noncompliance that may adversely affect state waters or may endanger 770 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition • public health must be reported orally with 24 hours of the time the permittee becomes aware of the condition A written report must be submitted within days Discharge monitoring reports (DMRs)—Selfmonitoring data generated during a specified period (normally month) When completing the DMR, remember: • More frequent monitoring must be reported • All results must be used to complete reported values • Pollutants monitored by an approved method but not required by the permit must be reported • No blocks on the form should be left blank • Averages are arithmetic unless noted otherwise • Appropriate significant figures should be used • All bypasses and overflows must be reported • The licensed operator must sign the report • Responsible official must sign the report • Department must receive the report by the 10th day of the following month 19.13.3 SAMPLING AND TESTING The general requirements of the permit specify minimum sampling and testing that must be performed on the plant discharge Moreover, the permit will specify the frequency of sampling, sample type, and length of time for composite samples Unless a specific method is required by the permit, all sample preservation and analysis must be in compliance with the requirements set forth in the federal regulations Guidelines Establishing Test Procedures for the Analysis of Pollutants Under the Clean Water Act (40 CFR 136) 19.13.3.2 Compliance Schedules If the facility requires additional construction or other modifications to fully comply with the final effluent limitations, the permit will contain a schedule of events to be completed to achieve full compliance 19.13.3.3 Special Conditions Any special requirements or conditions set for approval of the discharge will be contained in this section Special conditions may include: • • Monitoring required to determine effluent toxicity Pretreatment program requirements 19.13.3.4 Licensed Operator Requirements The permit will specify, based on the treatment system complexity and the volume of flow treated, the minimum license classification required to be the designated responsible charge operator 19.13.3.5 Chlorination/Dechlorination Reporting Several reporting systems apply to chlorination or chlorination followed by dechlorination It is best to review this section of the specific permit for guidance If confused, contact the appropriate state regulatory agency 19.13.4 REPORTING CALCULATIONS Failure to accurately calculate report data will result in violations of the permit The basic calculations associated with completing the DMR are covered below Note: All samples and measurements must be representative of the nature and quantity of the discharge 19.13.4.1 Average Monthly Concentration 19.13.3.1 Effluent Limitations The average monthly concentration (AMC) is the average of the results of all tests performed during the month: The permit sets numerical limitations on specific parameters contained in the plant discharge Limits may be expressed as: • • • • • • • • • Average monthly quantity (kg/day) Average monthly concentration (mg/L) Average weekly quantity (kg/day) Average weekly concentration (mg/L) Daily quantity (kg/day) Daily concentration (mg/L) Hourly average concentration (mg/L) Instantaneous minimum concentration (mg/L) Instantaneous maximum concentration (mg/L) © 2009 by Taylor & Francis Group, LLC Test1 + Test + Test AMC (mg/L) = ∑ + … + Test n (19.98) n (tests during month) 19.13.4.2 Average Weekly Concentration The average weekly concentration (AWC) is the results of all the tests performed during a calendar week A calendar week must start on Sunday and end on Saturday and be completely within the reporting month A weekly average is not computed for any week that does not meet these criteria: Wastewater Treatment Operations 771 ∑ Test1 + Test + Test (19.99) + … + Test n AWC (mg/L) = n (tests during calendar week) 19.13.4.3 Average Hourly Concentration The average hourly concentration (AHC) is the average of all of the test results collected during a 60-minute period ∑ + … + Test Test1 + Test + Test AHC (mg/L) = n (19.100) n (tests during 60-min period) 19.13.4.4 Daily Quantity (kg/day) Daily quantity is the quantity of a pollutant in kilograms per day discharged during a 24-hour period kg/day = Concentration (mg/L) × flow (MGD) (19.101) × 3.785 kg/MG/mg/L 19.13.4.5 Average Monthly Quantity Average monthly quantity (AMQ) is the average of all the individual daily quantities determined during the month ∑ DQ1 + DQ + DQ + … + DQ n AMQ (kg/day) = n (tests during month) (19.102) 19.13.4.9 Bacteriological Reporting Bacteriological reporting is used for reporting fecal coliform test results To make this calculation, the geometric mean calculation is used and all monthly geometric means are computed using all the test values Note that weekly geometric means are computed using the same selection criteria discussed for average weekly concentration and quantity calculations The easiest method used in making this calculation requires a calculator that can perform logarithmic (log) or nth root functions:  Log X1 + log X2 +    log X + … log Xn  Geometric mean = Antilog   n (number of tests)  (19.104) or Geometric mean = n X1 × X2 × … × Xn CHAPTER REVIEW QUESTIONS 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.13.4.6 Average Weekly Quantity 19.8 The average weekly quantity (AWQ) is the average of all the daily quantities determined during a calendar week A calendar week must start on Sunday and end on Saturday and be completely within the reporting month A weekly average is not computed for any week that does not meet these criteria 19.9 ∑ + … + DQ n 19.11 19.12 19.13 DQ1 + DQ + DQ AWQ (kg/day) = 19.10 (19.103) n (tests during calendar week) 19.14 19.13.4.7 Minimum Concentration 19.15 The minimum concentration is the lowest instantaneous value recorded during the reporting period 19.16 19.13.4.8 Maximum Concentration The maximum concentration is the highest instantaneous value recorded during the reporting period © 2009 by Taylor & Francis Group, LLC 19.17 Who must sign the DMR? What does the COD test measure? Give three reasons for treating wastewater Name two types of solids based on physical characteristics Define organic and inorganic Name four types of microorganisms that may be present in wastewater When organic matter is decomposed aerobically, what materials are produced? Name three materials or pollutants that are not removed by the natural purification process What we call the used water and solids from a community that flow to a treatment plant? Where disease-causing bacteria in wastewater come from? What does the term pathogenic mean? What we call wastewater that comes from the household? What we call wastewater that comes from industrial complexes? A lab test indicates that a 500-g sample of sludge contains 22 g of solids What are the percent solids in the sludge sample? The depth of water in the grit channel is 28 in What is the depth in feet? The operator withdraws 5250 gal of solids from the digester How many pounds of solids have been removed? Sludge added to the digester causes a 1920-ft3 change in the volume of sludge in the digester How many gallons of sludge have been added? 772 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition 19.18 The plant effluent contains 30 mg/L solids The effluent flow rate is 3.40 MGD How many pounds per day of solids are discharged? 19.19 The plant effluent contains 25 mg/L BOD5 The effluent flow rate is 7.25 MGD How many kilograms per day of BOD5 are being discharged? 19.20 The operator wishes to remove 3280 pounds per day of solids from the activated sludge process The waste activated sludge concentration is 3250 mg/L What is the required flow rate in million gallons per day? 19.21 The plant influent includes an industrial flow that contains 240 mg/L BOD The industrial flow is 0.72 MGD What is the population equivalent for the industrial contribution in people per day? 19.22 The label of a hypochlorite solution states that the specific gravity of the solution is 1.1288 What is the weight of gal of the hypochlorite solution? 19.23 What must be done to the cutters in a comminutor to ensure proper operation? 19.24 What is grit? Give three examples of materials considered to be grit 19.25 The plant has three channels in service Each channel is ft wide and has a water depth of ft What is the velocity in the channel when the flow rate is 8.0 MGD? 19.26 The grit from the aerated grit channel has a strong hydrogen sulfide odor upon standing in a storage container What does this indicate, and what action should be taken to correct the problem? 19.27 What is the purpose of primary treatment? 19.28 What is the purpose of the settling tank in the secondary or biological treatment process? 19.29 A circular settling tank is 90 ft in diameter and has a depth of 12 ft The effluent weir extends around the circumference of the tank The flow rate 2.25 MGD What is the detention time in hours, surface loading rate in gpd/ft2, and weir overflow rate in gpd/ft? 19.30 Give three classifications of ponds based on their location in the treatment system 19.31 Describe the processes occurring in a raw sewage stabilization pond (facultative) 19.32 How changes in the season affect the quality of the discharge from a stabilization pond? 19.33 What is the advantage of using mechanical or diffused aeration equipment to provide oxygen? 19.34 Name three classifications of trickling filters and identify the classification that produces the highest quality effluent © 2009 by Taylor & Francis Group, LLC 19.35 Microscopic examination reveals a predominance of rotifers What process adjustment does this indicate is required? 19.36 Increasing the wasting rate will _ the MLSS, _ the return concentration, _ the MCRT, _ the F/M ratio, and _ the SVI 19.37 The plant currently uses 45.8 lb of chlorine per day Assuming the chlorine usage will increase by 10% during the next year, how many 2000lb cylinders of chlorine will be needed for the year (365 days)? 19.38 A plant has six 2000-lb cylinders on hand The current dose of chlorine being used to disinfect the effluent is 6.2 mg/L The average effluent flow rate is 2.25 MGD Allowing 15 days for ordering and shipment, when should the next order for chlorine be made? 19.39 The plant feeds 38 lb of chlorine per day and uses 150-lb cylinders Chlorine use is expected to increase by 11% next year The chlorine supplier has stated that the current price of chlorine ($0.170 per pound) will increase by 7.5% next year How much money should the town budget for chlorine purchases for the next year (365 days)? 19.40 The sludge pump operates 30 every hr, and the pump delivers 70 gpm If the sludge is 5.1% solids and has a volatile matter content of 66%, how many pounds of volatile solids are removed from the settling tank each day? 19.41 The aerobic digester has a volume of 63,000 gal The laboratory test indicates that 41 mg of lime were required to increase the pH of a 1-L sample of digesting sludge from 6.0 to the desired 7.1 How many pounds of lime must be added to the digester to increase the pH of the unit to 7.4? 19.42 The digester has a volume of 73,500 gal Sludge is added to the digester at the rate of 2750 gpd What is the sludge retention time in days? 19.43 The raw sludge pumped to the digester contains 72% volatile matter The digested sludge removed from the digester contains 48% volatile matter What is the percent volatile matter reduction? 19.44 What does NPDES stand for? 19.45 How can primary sludge be freshened going into a gravity thickener? 19.46 A neutral solution has what pH value? 19.47 Why is the seeded BOD test required for some samples? 19.48 What is the foremost advantage of the COD over the BOD? Wastewater Treatment Operations 19.49 High mixed liquor concentration is indicated by a _ aeration tank foam 19.50 What typically happens to the activity level of bacteria when the temperature is increased? 19.51 List three factors other than food that affect the growth characteristics of activated sludge 19.52 What are the characteristics of facultative organisms? 19.53 BOD measures the amount of _ material in wastewater 19.54 The activated sludge process requires _ in the aeration tank to be successful 19.55 The activated sludge process cannot be successfully operated with a clarifier 19.56 The activated biosolids process can successfully remove _ BOD 19.57 Successful operation of a complete mix reactor in the endogenous growth phase is possible or not possible? 19.58 The bacteria in the activated biosolids process are either _or _ 19.59 Step feed activated biosolids processes have _ mixed liquor concentrations in different parts of the tank 19.60 An advantage of contact stabilization compared to complete mix is _ aeration tank volume 19.61 Increasing the _ of wastewater increases the BOD in the activated biosolids process 19.62 Bacteria need phosphorus to successfully remove _ in the activated biosolids process 19.63 The growth rate of microorganisms is controlled by the _ 19.64 Adding chlorine just before the _ can control alga growth 19.65 What is the purpose of the secondary clarifier in an activated biosolids process? 19.66 The _ growth phase should occur in a complete mix activated biosolids process 19.67 The typical DO value for activated biosolids plants is between _ and _ mg/L 19.68 In the activated biosolids process, what change would an operator normally expect to make when the temperature decreases from 25°C to15°C? 19.69 In the activated biosolids process, what change must be made to increase the MLVSS? 19.70 In the activated biosolids process, what change must be made to increase the F/M? 19.71 What does the Gould sludge age assume to be the source of the MLVSS in the aeration tank? © 2009 by Taylor & Francis Group, LLC 773 19.72 What is one advantage of complete mix over plug flow? 19.73 The grit in the primary sludge is causing excessive wear on primary treatment sludge pumps The plant uses an aerated grit channel What action should be taken to correct this problem? 19.74 When the mean cell residence time (MCRT) increases, what does the mixed liquor suspended solids (MLSS) concentration in the aeration tank do? 19.75 Exhaust air from a chlorine room should be taken from where? 19.76 If chlorine costs $0.21/lb, what is the daily cost to chlorinate a 5-MGD flow rate at a chlorine feed rate of 2.6 mg/L? 19.77 What term describes a normally aerobic system from which the oxygen has temporarily been depleted? 19.78 The ratio that describes the minimum amount of nutrients theoretically required for an activated sludge system is 100:5:1 What are the elements that fit this ratio? 19.79 A flotation thickener is best used for what type of sludge? 19.80 Drying beds are or are not an example of a sludge stabilization process? 19.81 The minimum flow velocity in collection systems should be _ 19.82 What effect will the addition of chlorine, acid, alum, carbon dioxide, or sulfuric acid have on the pH of wastewater? 19.83 An amperometric titrator is used to measure _ 19.84 The normal design detention time for a primary clarifier is _ 19.85 The volatile acids-to-alkalinity ratio in an anaerobic digester should be approximately _ 19.86 The surface loading rate in a final clarifier should be approximately _ 19.87 In a conventional effluent chlorination system, the chlorine residual measured is mostly in the form of _ 19.88 For a conventional activated biosolids process, the food-to-microorganism (F/M) ratio should be in the range of _ 19.89 Denitrification in a final clarifier can cause clumps of sludge to rise to the surface The sludge flocs attach to small sticky bubbles of _ gas 19.90 An anaerobic digester is covered and kept under positive pressure to _ 19.91 During the summer months, the major source of oxygen added to a stabilization pond is _ 774 Handbook of Water and Wastewater Treatment Plant Operations, Second Edition 19.92 Which solids cannot be removed by vacuum filtration? 19.93 The odor recognition threshold for H2S is reported to be as low as _ REFERENCES AND SUGGESTED READING Culp, G.L and Heim, N.F 1978 Field Manual for Performance Evaluation and Troubleshooting at Municipal Wastewater Treatment Facilities Washington, D.C.: U.S Environmental Protection Agency © 2009 by Taylor & Francis Group, LLC Metcalf & Eddy 1991 Wastewater Engineering: Treatment, Disposal, Reuse, 3rd ed New York: McGraw-Hill Metcalf & Eddy 2003 Wastewater Engineering: Treatment, Disposal, Reuse, 4th ed New York: McGraw-Hill Spellman, F.R 2007 The Science of Water, 2nd ed Boca Raton, FL: CRC Press USEPA 2007 Advanced Wastewater Treatment to Achieve Low Concentration of Phosphorus Washington, D.C.: U.S Environmental Protection Agency VWCB 1990 Activated Sludge Process Control, Part II, 2nd ed Richmond: Virginia Water Control Board ... Group, LLC Second Edition Handbook of Water and Wastewater Treatment Plant Operations © 2009 by Taylor & Francis Group, LLC Second Edition Handbook of Water and Wastewater Treatment Plant Operations... under the direct influence of surface water 16.6 Easily located; softer than groundwater 16.7 The study of the properties of water and its distribution and behavior 16.8 Zone of influence 16.9 GUDISW... trademarks, and are used only for identification and explanation without intent to infringe Library of Congress Cataloging-in-Publication Data Spellman, Frank R Handbook of water and wastewater treatment