Assignment 5 sol

Assignment 5 sol - 2 ~1.3 mg/L as Cl 2 c) The combined...

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CIVL242: Water and Wastewater Engineering Assignment 5 Solution Q1. Compound Mol wt. Mol of Equivalent Chlorine * Cl 2 71 1 HOCl 52.5 1 Na(OCl) 74.5 1 Ca(OCl) 2 143 2 NH 2 Cl 51.5 1 NHCl 2 86 2 NCl 3 120.5 3 *Number of mol of chlorine having an oxidizing capacity equivalent to one mol of the compound. 2 2 2 2 / 248 . 0 2 71 2 5 . 35 1 14 05 . 0 71 5 . 35 2 1 14 12 . 0 ] [NHCl Cl] [NH Cl as Residual Combined a) Cl as L mg = × × × + + + × + × + = + = 2 - 2 / 037 . 1 71 5 . 35 16 05 . 0 71 5 . 35 16 1 41 . 0 ] [OCl [HOCl] Cl as Residual Free b) Cl as L mg = × + + × + + = + = 2 2 / 285 . 1 037 . 1 248 . 0 Residual Free Residual Combined Cl as Residual Total c) Cl as L mg = + = + =
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Q2. From the Graph a) The chlorine dosage required attaining a combined residual of 0.5 mg/L as Cl 2 is ~0.59 mg/L as Cl 2 or ~0.7 mg/L as Cl 2 b) The chlorine dosage required attaining a free residual of 0.5 mg/L as Cl
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Unformatted text preview: 2 ~1.3 mg/L as Cl 2 c) The combined residual present when a free residual of 0.5 mg/L as Cl 2 exists is ~0.18 mg/L as Cl 2 d) Ca(OCl) 2 is a salt that dissociates to Ca(OCl) 2 Ca 2+ + 2OCl-As such one mole of Ca(OCl) 2 (MW = 143 g) is equal to 2 mole of Cl 2 (MW = 71 g) The amount of Ca(OCl) 2 additioin each day = Q x C d kg mg kg m L OCl Ca mg Cl mg L mg d m QC / 63 . 32 ) 10 1 )( 1000 )( ) ( 143 )( 71 35 . 1 )( 4000 , 2 ( 6 3 2 2 3 = = Combined Residual 0.5 mg/L as Cl 2 Free Residual...
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This note was uploaded on 09/13/2011 for the course CIVL 242 taught by Professor Gh during the Fall '09 term at HKUST.

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Assignment 5 sol - 2 ~1.3 mg/L as Cl 2 c) The combined...

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