Standard Methods for Examination of Water & Wastewater 1 ppt

...   += R xave R xave R x1 D 1 ()R x2 D 2 ()R x3 D 3 ()++ D 1 D 2 D 3 ++ 210 () 21( ) 1 0 .1( )++ 10 1 0 .1+ + 1. 99== = = R 1ave R 11 D 1 ()R 12 D 2 ()R 13 D 3 ()++ D 1 D 2 D 3 ++ 210 ( )11 () 1 0 .1( )++ 10 1 0 .1+ + 1. 90=== R 2ave R 21 D 1 ()R 22 D 2 ()R 23 D 3 ()++ D 1 D 2 D 3 ++ ... 1. 90=== R 2ave R 21 D 1 ()R 22 D 2 ()R 23 D 3 ()++ D 1...

Ngày tải lên: 19/06/2014, 16:20

... 13 7.5 12 :00 17 5 52 205 13 7.5 57 45.5 2 15 .6 a 19 6.88 b 14 :00 235 62 205 205 56.5 57 2 54.2 202.37 16 :00 17 5 51 163 205 48 56.5 2 91. 8 18 2.24 18 :00 15 1 45 16 6 16 3 48 48 2 11 2.24 17 4.75 20:00 18 1 51 ... 3020 8080547 9 011 3020 7 16 4 5 16 3 1 2 3 3 8 10 5 8 11 1 8 10 5 8 4 7 8 9 1 4 7 7 8 4 3 8 10 7 8 11 1 2 4 3 4 1...

Ngày tải lên: 19/06/2014, 16:20

... (vapor 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 ... 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= idh–...

Ngày tải lên: 19/06/2014, 16:20

... v o = 0. 017 4 m/min t(min) 0 60 80 10 0 13 0 200 240 420 c(mg/L) 299 19 0 17 9 16 9 15 7 11 0 79 28 t (min) 0 60 80 10 0 13 0 200 240 420 c (mg/L) 299 19 0 17 9 16 9 15 7 11 0 79 28 x = [c]/[c o ] 1. 0 0.64 ... all C D v(m/s) R e 1. 0 0 .12 4 95. 41 0.90 0 .13 1 10 0 0.88 0 .13 2 10 1 FIGURE 5.8 Settling column analysis of discrete settling. C D 24 Re 3 Re 0.34 for...

Ngày tải lên: 19/06/2014, 16:20

... 1 122.6++ 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( )13 7.75 m 3 === Therefore, P i for compartment no. 1 10 10 −4 ( )13 7.75()40 2 () 0.75 ... 0.008 b D 1. 3+   nbD() 210 .3() 10 5 ν () 0.75D    3 = b 5. 51 2 0.3()– 4. 91 m== νµ / ρ 10 10 4– ()/998 10 6– == = 326.52 0.75()0.9()=220.05 = 0.008...

Ngày tải lên: 19/06/2014, 16:20

... = s 1 n 2 ∑ 1 n 2 α ln() i 1 n 1 ∑ 1 n 1 α ln() i – 1 n 2 ∑ 1 n 2 ∆P–()ln() i 1 n 1 ∑ 1 n 1 ∆P–()ln() i – 1 1 1. 11 10 11 ()[]ln 1 1 1. 63 10 11 ()[]ln– 1 1 46 ,18 0[]ln 1 1 11 1,670[]ln– ... 11 1,670[]ln– == 25.43 25.82– 10 .74 11 .62– 0.443 Ans== α o e 1 n 1 ∑ 1 n 1 α ln() i −s 1 n 1 ∑ 1 n 1 ∆P–()ln[] i = e 1. 11...

Ngày tải lên: 19/06/2014, 16:20

... 0.00000063 18 1. 58 0.00 01 0. 016 5 0.0000 017 20 1. 72 0.00 014 0. 019 0.0000027 23 1. 83 0.00 011 0.0 215 0.0000024 25 2.00 0.00 017 0.024 0.00000 41 ∑ = 0.000 012 8 V V n +1 −− −− V n )// // 10 00 (m 3 ) (( V n +1 V n )// // 10 00)– A s π 0.0254[] 2 4 ... ln( µµ µµ F) 17 23.68 1, 460,780 a 14 .19 0 .12 3 9.56 2.26 413 6.84 3,873,940 15 .17 0 .18 7 14 .54 2.68...

Ngày tải lên: 19/06/2014, 16:20

... = y f [] Gy f 1 [ ]1 x f 1 []–( )1 x f []–()Lx f [ ]1 y f 1 []–( )1 x f 1 []–()x f 1 [ ]1 y f 1 []–( )1 x f []–()–[]+ 1 y f 1 []–( )1 x f 1 []–( )1 x f []–() G Gy f 1 [ ]1 x f 1 []–( )1 x f []–()Lx f [ ]1 y f 1 []–( )1 ... y f 1 []–( )1 x f 1 []–()x f 1 [ ]1 y f 1 []–( )1 x f []–()–[]+ 1 y f 1 []–( )1 x f 1 []–( )1 x f []–() + = G...

Ngày tải lên: 19/06/2014, 16:20

... Mg := f MgCa 1 0.32– 1 0.68; M T Mg 1. 0 1 1000   14 ,750( )14 .75 kgeq/d== = = 14 .75 12 .15 ()= 17 9. 21= kg/d M CaOMgCa 2.30 0.68( )17 9. 21( )280.28= kg/d= M CaOCO 2 1. 27 M CO 2 ( )1. 27 22.0() 1 1000 ... any softening Ions No. of Equiv. (meq/L) Ca 2+ 1. 80 Mg 2+ 0.60 Na + 2.758 ∑ = 5 .15 0.62 4.49 ∑ = 5 .11 K sp,HCO 3 10 10 .33 = A 1 10 10 .33 + 10 10...

Ngày tải lên: 19/06/2014, 16:20

... K sp BaCO 3 8 .1( 10 −9 ) CdCO 3 2.5 (10 14 ) CaCO 3 4.8 (10 −9 ) CoCO 3 1. 0 (10 12 ) CuCO 3 1. 37 (10 10 ) FeCO 3 2 .11 (10 11 ) PbCO 3 1. 5 (10 13 ) MgCO 3 1. 0 (10 −5 ) MnCO 3 8.8 (10 11 ) Hg 2 CO 3 9 (10 17 ) NiCO 3 1. 36 (10 −7 ) Ag 2 CO 3 8.2 (10 12 ) SrCO 3 9.42 (10 10 ) ZnCO 3 6 (10 11 ) HCO 3 aq() − ... K CaOHc 10 1. 49– γ SO 4 0.77=== SO 4 2...

Ngày tải lên: 19/06/2014, 16:20