Standard Methods for Examination of Water & Wastewater 2 ppt

... H 2 BO 3 − , HPO 4 2 H 2 PO 4 − RC – OH || || || || O CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O || C – OH R O || || || || C – OR′ TX249_Frame_C 02. fm Page 144 ... (2. 43) MPN 1 R 1ave MPN x R xave MPN 2 R 2ave R xave R x1 D 1 ()R x2 D 2 ()R x3 D 3 ()++ D 1 D 2 D 3 ++ = R 1ave R 11 D 1 ()R 12 D 2 ()...

Ngày tải lên: 19/06/2014, 16:20

... 2( ) 18 22 + 2 2( )+    == 19 18+ 2 2( ) 27 19+ 2 2( ) 39 27 + 2 2( ) 52 39+ 2 2( ) 62 52+ 2 2( )+++++ 51 62+ 2 2( ) 45 51+ 2 2( ) 51 45+ 2 2( ) 40 51+ 2 2( ) 40 51+ 2 2( )    +++++ 37.7 ... Ans= V basin pos of Q i Q i−1 + 2 Q mean –   t i t i 2 –() i =2 i= ξ ∑ pos of 22 26 + 2 37.7–   2( )    == 18 22 + 2 37.7–  ...

Ngày tải lên: 19/06/2014, 16:20

... ∆P ρω 2 ⇒ L 2 t 2 1 t 2 () 2 L 2 == ∆P ρω 2 D 2 ∆P ρω 2 D 2   ⇒ L 2 L 2 = Therefore ∏ 1 ∆P ρω 2 D 2 = ∏ 2 Q ω D 3   = ∆P ρω 2 D 2 Ψ Q ω D 3   = Hg ω 2 D 2 Ψ Q ω D 3 ... TDH P 2 γ P 1 γ – V 2 2 2g V 1 2 2g – h fs ++== h fd h st ++ P 2 γ P 1 γ –0= V 1 2 2g 0 h fs h fd + 20 == m h st 30 m= V 2 2 2g 1.3...

Ngày tải lên: 19/06/2014, 16:20

... microstrainers. V 1 2 V 2 2 2gh 2 h 1 –()+= QA 2 V 2 A 2 2gh 1 h 2 –() 1 A 2 A 1 – A 2 2g∆h 1 A 2 A 1 – == = QC d A 2 2g∆h 1 A 2 A 1 – = ∆h Q 2 1 A 2 A 1 –   2gC d 2 A 2 2 = TX249_Frame_C05 ... area of tank, π D 2 4 0.69 28 24 60()60() 21 29.14==D 52. 07 m, too large= Use two tanks, therefore π D 2 4 21 29.14 2 = D 36. 8...

Ngày tải lên: 19/06/2014, 16:20

... 0 .22 – 22 0.05 25 .77– 28 5.63 125 .77– = 0 .22 125 .77 y 0 .22 – 0.15 0 .22 – 22 0.05 25 .77– 28 5.63 125 .77– = y 22 0.05 y 0.18 D m Ans== 0.15 28 5.63 TX249_frame_C06.fm Page 317 Friday, June 14, 20 02 ... the P Q γ h f = V 1 2 2g y 1 h f –+ V 2 2 2g y 2 += h f y 2 y 1 –()q 2 y 2 y 1 +()2gy 1 2 y 2 2 –[] 2gy 1 2 y 2 2 = F ∑ ∂ ∂ t v CV ∫ ρ V dv A...

Ngày tải lên: 19/06/2014, 16:20

... 2. 26 ∑h i = 4.64 h d 3 0.793– 2. 21 m== 900 0.414 x 120 0– 120 0 900– 2. 21 0.7 62 0.7 62 0.414– = 120 0 0.7 62 x 24 48 min= x 2. 21 x 2. 21 x 1500– 1500 21 00– 2. 21 2. 26– 2. 26 4.64– = 1500 2. 26 ... the values of b averaged are obtained. h d = 0.051(q) 2. 22 for uniform sand, 0.5 mm diameter h d = 0. 022 (q) 2. 22 for uniform sand, 0.7 mm diameter h d...

Ngày tải lên: 19/06/2014, 16:20

... d⋅ π f 23 5.80 17.17 9.93++ 26 2.9 ==kN/m 2 2 62, 900= N/m 2 P p 101, 325 N/m 2 = ; π p 0, assumed.= 26 2,900 0–[]–1= ,460,780 N/m 2 s 1 2. 68 2. 26– 15.17 14.19– – 1 0.43– 0.57=== α mo 1 0.57–()14.19 2. 26– 1 ... membrane according to the following reaction: C=O (CH 2 CH 2 NH) n CH 2 CH 2 NH + 2 (CH 2 CH 2 N) n CH 2 CH 2 N – C (CH 2 CH 2 N) n CH 2 CH...

Ngày tải lên: 19/06/2014, 16:20

... [C]// // ββ ββ [C os ] 3 0.9 8. 12 −0.1049 6 1.8 7 .22 −0 .22 24 9 2. 4 6. 62 −0.3091 12 3.3 5. 72 −0.4553 15 4.0 5. 02 −0.5858 18 4.7 4. 32 −0.7360 21 5.3 3. 72 −0.8855 ∑ = 84 ∑ = 40.74 ∑ = −3 .29 9 C K L a ∑ m=1 m=n C os []C[]– C s [] ... X 2 [] X 1 []–() 0.0585T k 16.31–()X 2 [] Y 1 []– 1L of water 1000 g 1,000,000 mg=== X 2 [] Ӎ 25 /N 1,000,000 25 –()/H 2 O 2...

Ngày tải lên: 19/06/2014, 16:20

... = = 14.75 12. 15()= 179 .21 = kg/d M CaOMgCa 2. 30 0.68()179 .21 ( )28 0 .28 = kg/d= M CaOCO 2 1 .27 M CO 2 ()1 .27 22 .0() 1 1000   14,750()4 12. 12 kg/d == = M CaOExcess 0. 028 V kg 0. 028 14,750()413= ... to the second stage = (0. 32( 14,750) + 1.0(10 ,25 0)) /25 ,000 = 0.60 meq/L. M CaO 920 .65 0 28 0 .28 4 12. 12 413+ + + + 20 26.05= kg/d of pure lime= 20 26.05 0.90...

Ngày tải lên: 19/06/2014, 16:20

... 0.001995 2 2 ()0.0004938 2 2 ()0.0 02 1() 0.0 025 1() 0.0 022 2 2 ()++++[]= 0. 023 Ans= TX249_frame_C11.fm Page 515 Friday, June 14, 20 02 2 :27 PM © 20 03 by A. P. Sincero and G. A. Sincero Water Stabilization ... HOH l() = ∆H 29 8 o ∆H 29 8 o HOH l() H 2 1 2 O 2 +→ HOH l()  H (aq) + OH (aq) − + H 2 CO 3 ∗ H 2 C6O 2 ∆H 29 8 o ++→+167.0 kcal/gmmol of H 2 CO...

Ngày tải lên: 19/06/2014, 16:20