1) A 15% solution of sodium hypochlorite (NaOCl), a 75%-pure granular calcium hy
ID: 792520 • Letter: 1
Question
1)
A 15% solution of sodium hypochlorite (NaOCl), a 75%-pure granular calcium hypochlorite [Ca(OCl)2] , and pure chlorine gas (Cl2) are each available to be used as the primary disinfectant at a 5,000 m3/day water treatment plant. The projected chlorine dose (OCl-) is expected to be 2.3 mg/L.
(a) Determine how much of each disinfectant would be required to provide the 2.3 mg/L dosage (of OCl-) in kg/month (assuming a 30-day month).
b) Of the three potential disinfectants, which one is the most cost effective? The 15% solution of NaOCl costs $0.90/kg; the 75% pure Ca(OCl)2 costs $1.45/kg; and the chlorine gas (Cl2) costs $0.55/kg? Show your cost calculations for each disinfectant.
Explanation / Answer
a) 2.3 mg/L , since vol = 5000 m3 = 5000 x 1000 liters = 5 x10^6
hence mass of OCl- required per day = 5 x10^6 x 2.3 mg = 1.15 x 10^7 mg = 1.15 x10^4 gm per day
per month OCl- mass = 1.15 x10^4 x 30 = 3.45 x10^5 gm = 345 kg
15 % NaOCl means 15 gm NaOCl per 100 gm soln
hence for 15 gm NaOCl OCl- mass = ( 15/74.4) ( 51.4) = 10.363 gm ( here NaOCl mol wt = 74.4,
OCl- mol wt = 51.4) hence 10.363 gm OCl- per 100 gm soln
i.e 0.010363 kg per 0.1 kg soln
hence total NaOCl dosage required = ( 345 x 0.1 /0.010363) = 3329 Kg/month
75 % pure Ca(OCl)2 means 75 gm Ca(OCl)2 per 100 gm
75 gm Ca(OCl)2 gives ( 75/143) ( 2 x 51.4) = 53.916 gm
hence 0.053916 kg OCl- per 0.1 kg Ca(OCl)2
hence dosage per month = ( 345x0.1/0.053916) = 639.9 kg
Cl2 gives 2OCl- hence
for 3.45 x10^ 5 gm OCl- = 3.45x10^5/51.4 = 6712 moles OCl- we require 6712/2 = 3356 moles Cl2
thus Cl2 dosage = ( 3356 x 71/1000) = 238.28 kg /month
b) cost used if used NaOCl = 0.9 x 3329 = 2996 $
cost used if we se Ca(OCl)2 = 1.45 x 639.9 = 927.855 $
cost used if we use Cl2 = 0.55 x 238.28 = 131 $
hence Cl2 usage is cost effective
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