Monday, March 30, 2020

Analysis of water treatment problemAssignment

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Initial problem



When the river has an Suspendable Solids (SS) of 5mg/l and alkalinity of 1.5mg/l (expressed as CaCO), 0mg/l of alum was added during the treatment process to make sure that the water passed water standards for the World Health Organisation WHO).



But subsequently the amount of SS was found to have increased to 50mg/l. Logically, when this happens, by doubling the amount of alum added initially, we would be able to generate enough precipitate to filter off the 50mg/l of SS such that the amount of SS left in the solution would be able to pass WHO drinking water standards. This did not happen thus this report is to look into the probable reasons to why the move did not work and the possible solutions to the problem.



Allowable Limits According to the standards of Drinking water from World Health Organization (WHO), the acceptable range of alum dosage is from 5 50mg/l



School papers on Analysis of water treatment problemAssignment



Thus, the 60mg/l of Alum added by the Plant manager has exceeded the acceptable limit.



Analysis



There may be various reasons to why the initial process did not work.



Physical reasons



(Note that we are adding in double the dosage thus the reactors/tanks might not be equipped or designed to take that kind of load)



•The process of mixing alum might not be thorough enough thus the Al(OH) was not precipitated properly



•Not enough time was allowed for precipitation to take place



•The coagulant did not have enough time for it to be saturated with the SS before the coagulant was filtered off.



•As the rate of chemical reaction decreases with decreasing temperature, assume that the temperature near the Hu river does not enhance the rate of reaction thus accounting for the decrease in precipitate formation



Chemical reasons



•There might not be enough alkalinity in the water thus the alum would not be able to precipitate to remove the SS.



•For the water pH between 5 to 8 which we assume to be the optimal pH for which Al(OH) would be precipitated best. When the pH is out of this range the Al(OH) become more soluble(due to the solubility constant) thus less precipitate would be produced and less SS will be removed.



Calculations



Main Equation



 



= = 0.7



0.7mg/l of alkalinity is needed mg/l of Alum.



Thus for an alum dosage of 0mg/l the alkalinity expressed as mg/l HCO- = (1/16) x 0.7 x 0 = 16.5mg/l



Thus when 50mg/l of Alum is added,



Amount of HCO- needed = 16.5x50/0



= 7.5mg/l



Total amount of CaCO to react with 50mdg/l of Alum= (16/1)x7.5 = 6.5mg/l



There is a deficiency of alkalinity thus explaining why the alum will not react to precipitate thus not removing the level SS to meet WHO standards.



Possible Solutions



1)Add in the extra alkalinity need for 50mg/l of alum to react then remove the hardness and acidity generated



)Add in lime to remove the extra SS needed to attain WHO standards (more popular method as it is less expensive but increases the hardness)



)Add in Soda Ash to remove the extra SS needed to attain WHO standards (does not increase hardness but increases the acidity of the water)



Solution



Prior to the addition of alum, add in excess CaCO into the water to allow the alum to react fully when added.



 



Total HCO needed to react 60mg/l of alum = 7.5mg/l



Total HCO in the water = 1/100 X 1.5 = 16.47



Note the fact that alum when reacted with Ca(HCO) produces CO gas which becomes HCO- which in turn reacts with the alum.



CO +…HO. HCO- + H+



Initially the HCO- is the limiting agent thus



Amount of CO produced with 1.5m/l of alkalinity in water = (6x44)/(x16)x1.5 = 7.



Amount of alum reacted with 1.5 of alkalinity



= 1.5/0.7 = 18.4 mg/l



Remaining



Alum = 50-18.4 = 1.51mg/l



HCO- generated = (61/44) x 7. = 10.16mg/l



HCO- needed= 7.5-16.47-10.16 =0.87mg/l



In this case as the reactions are reversible, it would be more complicated to add in the solubility constant for the calculations. Thus we estimate that the amount of HCO- needed would be twice to thrice of that needed to fully react with the alum.



Thus HCO- needed = x0.87 = .61mg/l



Amount of CaCO needed to e added in



= (100/1)x.61 = .14mg/l



In the case of hardness, total Ca in the water



= (40/100)x(1.5+.14) = 15.64mg/l.



Conclusion



What is presented is the possible solution to the problems of Hu river. There are other ways of treating the water such as adding in lime or soda ash but in this case alum was used. The important reason to using alum is because it does not increase the hardness of the water and they remove pathogens and particles better than other coagulants. They decrease the alkalinity with the acidic CO gas they produce. Thus they must be added in specific amounts so as to maintain the pH needed to fulfill WHO standards.



 



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