Standard Methods for Examination of Water & Wastewater 5 ppt
Standard Methods for Examination of Water & Wastewater_1 ppt
... 55 .3890 g, what is the suspended solids (SS) content of the sample? Solution: FIGURE 2.4 Imhoff cones. SS 55 .3890 55 . 352 0– 260 g mL 55 .3890 55 . 352 0– 260 1000()1000()== 142.3 mg/L Ans= TX249_Frame_C02.fm ... case of taste, there should be no noticeable odor at the point of use of any drinking water. The secondary standard for odor is 3. Fresh wastewater odor...
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Standard Methods for Examination of Water & Wastewater_2 ppt
... t i V remi−1,i C i,i++ ++ 1 8:00 52 27 76 26 33 23 2 3.44 101 10:00 100 39 137 .5 76 45. 5 33 2 5. 96 ⇒ 0 137 .5 12:00 1 75 52 2 05 137 .5 57 45. 5 2 15. 6 a 196.88 b 14:00 2 35 62 2 05 2 05 56 .5 57 2 54 .2 202.37 16:00 1 75 51 163 2 05 ... 48 56 .5 2 91.8 182.24 18:00 151 45 166 163 48 48 2 112.24 174. 75 20:00 181 51 158 166 45. 5 48 2 127.84 167.78 22:00 1 35 40 1 05 1...
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Standard Methods for Examination of Water & Wastewater_3 potx
... pressure of water) = 2.34 kN/m 2 = 0.239 m of water Assume standard atmosphere of 1 atm = 10.34 m of water. Therefore, theoretical limit of pump cavitation = −(10.34 − 0.239) = −10. 05 m −3.087 Cavitation ... h absi h vi + 3.087– 10.34+() 1. 059 2 2 9.81() + 7.31 m of water Ans== = odh h abso h vo + 57 .31 1.3 2 2 9.81() + 57 .39 m of water Ans== = TDH odh=...
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Standard Methods for Examination of Water & Wastewater_4 potx
... Time (min) Z p /Z o 5 10 15 20 25 30 35 40 0.1 95 a 68 55 30 23 0.2 129 121 73 67 58 48 43 0.3 154 113 92 78 69 60 53 45 a Values in the table are the results of the test for the suspended solids ... be obtained: (5. 32) From Equation (5. 27), again, (5. 33) Substituting in Equation (5. 28), (5. 34) (5. 35) Thus, (5. 36) Example 5. 6 Design the cross section of a grit...
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Standard Methods for Examination of Water & Wastewater_5 ppt
... 3 0.16 D 3 5. 36 D 2 += y 0.22– 0. 15 0.22– 220. 05 25. 77– 2 85. 63 1 25. 77– = 0.22 1 25. 77 y 0.22– 0. 15 0.22– 220. 05 25. 77– 2 85. 63 1 25. 77– = y 220. 05 y 0.18 D m Ans== 0. 15 2 85. 63 TX249_frame_C06.fm ... 30.86()= 1 5. 6 30.86( )5. 51 m Ans== G P µ V P µ V G 2 == µ 10 10 4– ()kg/m s⋅ V 1 ;5( )5( )5. 51()137. 75 m 3 === Therefore, P i for comp...
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Standard Methods for Examination of Water & Wastewater_6 pot
... 0.178 1 35 — 0.76 0.180 136 0.44 0 .56 0.020 0.0 056 20–28 0.70 0. 05 1. 05 0.13 69 0.47 0 .53 0.09 0.02 65 28–32 0 .54 0. 15 1.39 0.10 41 0 .50 0 .50 0.30 0.0 75 32– 35 0.46 0.18 1.72 0.08 29 0 .52 0.48 0.38 ... 0.0864 35 42 0.38 0.18 2.21 0.07 20 0 .54 0.46 0.39 0.0828 42–48 0.32 0.20 3.02 0. 05 13 0 .58 0.42 0.48 0.084 48–60 0.27 0. 15 4.94 0.03 7 0. 65 0. 35 0.43 0. 052 5 60–...
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Standard Methods for Examination of Water & Wastewater_7 doc
... 0.00274 kg/s. Therefore, Test No. PAC Added M (g) [C ] (mg/L) X/M [C ]/X/M 1 0. 25 6.0 0.076 78. 95 2 0.32 1.0 0.0 75 13.33 3 0 .5 0. 25 0.04 95 5. 05 4 1.0 0.09 0.0249 3.61 5 1 .5 0.06 0.0166 3.61 a 3 ... 1.0 0. 25+ +()– 3 3.61 3.61+()2 78. 95 13.33 5. 05+ +()– 0. 45 14 .5 21.66 194.66– 0.081=== b l a∑ 1 l C[] X /M ∑ C[] l 1 – 3 0.081 78. 95 13.33 5. 05+ +()6.0 1.0 0...
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Standard Methods for Examination of Water & Wastewater_8 docx
... experiment for the pur- pose of determining the value of β of a particular wastewater. The [C os,w ] of the wastewater after shaking the jar thoroughly is 7 .5 mg/L. The temperature of the wastewater ... corrected to 20°C. Tap Water at 5. 5°C Time (min) ,(mg/L) 3 0 .5 6 1.7 9 3.1 12 4.4 15 5 .5 18 6.1 21 7.1 Wastewater at 25 C Time (min) [ ](mg/L) 3 0.9 6 1.8 9 2....
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Standard Methods for Examination of Water & Wastewater_9 potx
... 50 .0 M T Ca 1 f Ca –() V += f CaHCO 3 1 15 152 .5 – 0.90 f Ca 1 M T CaHCO 3 50 65. 63 kg/d=== = M T Ca 7 45. 75 kg/d V 25, 000 m 3 /d== Ca 2+ [] meq 12.3 50 65. 63 1 0.90–() 25, 000 50 .0 7 45. 75 ... Ca 2. 255 1 1000 25, 000( )56 .3 75= kgeq/d= 56 .3 75 20. 05( )1130.32 kg/d== f Ca 0. 755 2. 255 0.33== Ca 2+ [] meq 12.3 2076.91 1 0.71–() 25, 000 50 .0 1130.3...
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Standard Methods for Examination of Water & Wastewater_10 docx
... 2003 by A. P. Sincero and G. A. Sincero Water Stabilization 52 5 Therefore, Therefore, Therefore, Therefore, Therefore, B 6.80 10 15 () 0.94 7.24 10 15 ()== C K sp,CaCO 3 γ HCO 3 K 2 Ca 2+ {} ... 0.0004 geq/L= B K w γ OH = µ 2 .5( 10 5 )TDS γ 10 0.5z i 2 µ () 1 1.14 µ ()+ – == µ 2 .5 10 5 ()140()3 .5 10 3– () γ OH 10 0 .5 1() 2 3 .5 10 3– (){} 1+1.14 3 .5 10 3– (){} –...
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