Standard Methods for Examination of Water & Wastewater 3 potx

... 60-mL portions: 30 , 32 , 33 , 31 , 25; and 500-mL portions: 35 0, 34 0, 36 0, 37 0, 34 0. What is the number of coliforms per 100 mL of the sample? Solution: Using the criterion of 20 to 80 as the most ... 1 and 3, 1, 1. Therefore, For reading 2, 1, 1, MPN = 9 = MPN 1 ; for reading 3, 1, 1, MPN = 14 = MPN 2 MPN 90 fecal coliforms/100 mL Ans= Average count for the 60-m...

Ngày tải lên: 19/06/2014, 16:20

... 30 20 164 932 4 26 3 3020 205 034 4 30 3 3020 30 5 638 5 40 5 30 20 4060405 50 6 30 20 5066446 60 7 30 20 6070486 70 8 930 20 7076507 80 9 30 20 8080547 9011 30 20 7 16 4 5 16 3 1 2 3 3 8 10 5 8 11 1 8 ... Q mean –   t i t i−2 –() i=2 i= ξ ∑ pos of 22 26+ 2 37 .7–   2()    == 18 22+ 2 37 .7–   2() 19 18+ 2 37 .7–   + 2() 27 19+ 2 37 .7– ...

Ngày tải lên: 19/06/2014, 16:20

... 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 ... 7 .31 –+= h abso 57 .31 m= idh 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...

Ngày tải lên: 19/06/2014, 16:20

... A ave 0.48 m 2 := Z o 3 2 A mpk 1 /3 w mpk A 2 /3 3 2 1. 03 1 /3 1.5 0.48 2 /3 ()0.62 m== = For A min 0.19 m 2 := Z o 3 2 A mpk 1 /3 w mpk A 2 /3 3 2 1. 03 1 /3 1.5 0.19 2 /3 ()0 .33 m== = w constant v h k ... 2129.14 3 = D 30 .06 m, say 30 m= Depth of tank, H 6210 /3 2129.14 /3 2.92 m. Add freeboard of 0 .3 m== Actual depth of tank 2.92 0 .3+...

Ngày tải lên: 19/06/2014, 16:20

... 0.9() = 32 6.52 N⋅m/s 0.44 hp Ans== V 2 25 11.02()275.50 m 3 == Therefore, P i for compartment no. 2 10 10 −4 ()275.50( )30 2 () 0.75 0.90() 0.44 hp Ans== V 3 25 14 .33 ( )35 8.25 m 3 == TX249_frame_C06.fm ... at a rate of 1.12 m 3 per m 3 of water treated and that the detention time of the tank is 2.2 min. The power dissipation of the mixer is 3. 24 hp. The pres...

Ngày tải lên: 19/06/2014, 16:20

... 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 0.0525 60–65 0. 23 0.07 6.48 0. 03 5 ... (mg/cm 3 ) h di (m 3 ) q i (mg/cm 3 ) h di (m 3 ) 0——————— 8 −8.2 8 −1. 03 37.08 1 .36 51.91 2.76 16 −4.0 8 −0.5 18 0 .3 25.2 0.61 24 −1.8 8 −0. 23 8.28 0.06 11....

Ngày tải lên: 19/06/2014, 16:20

... 100() 104.9 3, 000()0.61 102.18()90()– 104.9 3, 000() 100()== 98.2%= Ans b()R 104.9 3, 000 30 0 400++()0.61 102.18()9068++()– 104.9 3, 000 30 0 400++() 100()= 38 8, 130 6,482 .30 – 38 8, 130 = 100()98 .3% = Ans TX249_frame_C08.fm ... 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...

Ngày tải lên: 19/06/2014, 16:20

... 0.0 03 1 .38 0.0110 0.00026 0.0078 30 9.9 1.0111 0.0 03 0.94 0.0 130 0.00 031 0.0092 261 1.0 132 0.002 0. 53 0.0160 0.00 038 0.011 210.7 1.01 63 0.0 03 0.64 0.0190 0.00044 0.0 13 172 1.0194 0.0 03 0. 53 0.0210 ... 1 0. 03, = y i 30 x i 0. 03; 30x f 1 x f 1 ; 0.001=== G 1 1 0.0 031 – 1 1 0. 03 –   L 1 10– 1 1 0.001– –   = L/G 28= G 10 2 73 30 2 73+   10 3...

Ngày tải lên: 19/06/2014, 16:20

... CaHCO 3 M CaHCO 3 V M T MgHCO 3 M MgHCO 3 V [HCO 3 ] CO 3 meq HCO 3 [] OHCO 3 me q [HCO 3 ] CO 3 meq [HCO 3 − ] meqtent HCO 3 − [] meqtent HCO 3 [] OHCO 3 meq 12 .3 M T CaHCO 3 1 f CaHCO 3 –() V += 13. 7 M T ... Ans= HCO 3 − [] meq HCO 3 [] OHCO 3 meq 12 .3+ M T Ca HCO 3 1 f CaHCO 3 –() V    = 13. 7 M T Mg HCO 3 1 f MgHCO 3 –() V  ...

Ngày tải lên: 19/06/2014, 16:20

... Ca 2 + SO 4 2 − + Ca aq() 2+ [] Ca aq() 2+ []Ca T []CaCO 3 o []CaHCO 3 + []CaOH + []CaSO 4 o []+++()–= ∆H 298 o CaCO 3 o CaHCO 3 + CaSO 4 o CaCO 3 o CaCO 3s  Ca 2+ CO 3 2− K sp,CaCO 3 + Ca 2+ CO 3 2−  + CaCO 3 o 1 K CaCO 3 c CaCO 3s  CaCO 3 o K sp,CaCO 3 1 K CaCO 3 c   TX249_frame_C11.fm ... saturation pH of a solution, thus (11.26) CO 3 2− []CO 3...

Ngày tải lên: 19/06/2014, 16:20