The Effect of Nontreated Sewage

The Effect of Nontreated Sewage Free Online Research Papers Title: To study the effect of disposal of nontreated sewage containing chemicals and fecal matter on COD, BOD and DO in pond water. Miss Vineeta Girdoniya,p.g. college Narsinghpur,M.P., India,E.mail Address:ranugirdoniya@gmail.com Introduction: All the ponds selected for this study are from a village named as Baraheta situated near district Narsinghpur.These all are highly polluted due to disposal of sewage containing fecal matters and some detergents also.Most of the aquatic systems of varying characters worldwide recieves regular inputs of nontreated sewage containing various chemical, detergents and organic matter, which results in increase of total nutrient concentration. This causes the development of microbial, algal and higher aquatic plant species which further results in eutrophication (Hinesly and Jones,1990). Eutrophication is a natural process, but if it occurs uninterruptedly, it results in excessive deficiency of oxygen in water.Thus anaerobic organisms favoured more and more at the expense of aerobic organisms (Mengel and Kirkby,1996). Our present investigation is to evaluate the effect of disposal of non treated sewage on chemical properties like COD, BOD, DO, and Alkalinity of pond water. All the experiments are established two times , first in the month of september(2009) and second time in the month of february(2010). Material and methods COD Test Determination of Chemical Oxygen Demand(COD) Iin selected saple of water: This is determined by refluxing the sample with an excess of potassium dichromate in a highly acid condition and estimating by titration the amount of dichromate used. With a reducing agent like ferrous ammonium sulphate. Organic matter(CHO)+Cr2O7+H+2Cr+3+CO2+H2O 6Fe+Cr2O7+14H+6Fe+ +2Cr+3+7 2H2O Interference- Chlorides -1 mg/l Cl- exerts 0.23 mg/l of COD. Therefore correction as mg/litre Cl-*0.23 should be applied by substracting the COD of Cl- from the total COD. Nitrites exert COD of 1.1 mg/mg N. Limitations- (1) Amino nitrogen gets converted to ammonia nitrogen. (2) All organic compounds with few exceptions (e.g.aromatic hydrocarbons,straight aliphatic compounds and pyridine) are oxidized by this procedure. Requirements:Chemical reagents- (1)Standard potassium dichromate 0.25 N (2)Concentrated H2SO4 (3)Ferroin indicator. (4)Catalysts-Silver sulphate(for 8 straight chain aliphatic compounds) and mercuric sulphate(for Cl-). (5)Sulphamic acid-Required only if interference of NO2 is to be eliminated.Add 10 mg sulphamic acid /mg NO N,if present,in the refluxing flask. Also add in blank in this case. Procedure- (1)Place 50 ml sample in round bottom refluxing flask of 300 ml capacity with ground glass joint . (2)Add 50 ml of distilled water to dilute it. (3)Add gently 25 ml of K2Cr2O7 solution and 75 ml of concentrated H2SO4 and shake. (4)Attach refluxing condenser and reflux the mixture for 2 hours. (5)After refluxing wash the condenser with distilled water. (6)Cool the mixture and dilute it with distilled water. (7)Titrate with ferrous ammonium sulphate(0.25 N) with ferroin indicator till the red colour appears after the intermediary green colour. Reactions: CnHaOb+cCr2/O7 2- +8H+ a+8C n CO2+ -2H2O+2cCr3+ Where,c=2/3n+a/6-b/3. Calculations: COD mg/l=(A-B)N*8000 V Where, A=Vol. in ml ferrous ammonium sulphate for blank B=Vol. in ml ferrous ammonium sulphate for blank. V=Volume of sample. N=Normality of ferrous ammonium sulphate. Result: (1) In first sample of water the value of COD determined is 11.4 mg/l., and 13.5 mg/l. (2)In second sample of pond water COD determined is 13.5 mg/l, and 14.9 mg/l. (3)In third sample of water COD determined is 13.00 mg/l., and 15.5 mg/l. (4)The value of COD determined in fourth sample of water is 16.8 mg/l., and 19.00 mg/l. BOD Test Test to determine Biochemical Oxygen Demand(BOD): This is determind by measuring the loss in DO in the sample of water after incubating it for 5 days at 20 degree centigrade It is expressed as mg/l 5BOD 20degree centigrade. Requirements: (1)Specially prepaired BOD glass bottles provided with exactly fitting ground glass stoppers and surrounding well to accomodate 5 ml of water so as to exclude exchange of gases. (2)BOD incubator working at 20 degree centigrade. (3)Chemical Reagents: (a)Distilled water of highest purity and thoroughly aerated so as to saturate with DO at a lowered temperature of 20 degree centigrade. (b)Phosphate buffer solution. (c)Magnesium sulphate solution. (d)Calcium chloride solution. (e)Ferric chloride solution. (f)Dilution water Preparation of dilution water: (1)Place required 1 litre of distilled water at 20 degree centigrade. (2)Add 1 ml of phosphate buffer solution. (3)Add 1 ml of magnesium sulphate solution. (4)Add 1 ml of calcium chloride solution in it. (5)Add 1 ml of ferric chloride solution in this solution of distilled water. (6)Add 10 ml of settled sewage in this solution of distilled water to seed the dilution water. Procedure: (1)Add 300 ml of sample water in BOD bottle. (2)Fill this bottle completely withdilution water. (3)All concentration should be in duplicate. (4)Keep one bottle of each concentrate in the incubator for 5 days at 20 degree centigrade. (5)Subject the duplicate of that concentration to DO determination on the same day.That will be 0-day DO. (6)After five days,subject the duplicate bottles of all concentrations to the DOdeterminations.It will be 5-day DO. (7)Similarly put one or two bottles for finding out the depletion of DO in seeded dilution water only. (8)Find out the difference in between 0-day DO and 5-day DO values. (9)Chemical reactions taking place: CnHaobNc +(n+a/4-b/2-3/4C)O2-CO2+(a/2-3/2C)H2O+NH3. Calculations:- 5-BOD mg/litre=Initial DO(mg/l) 5-day DO(mg/l) Result: (1)In first sample of water the BOD determined is 13 mg/litre and 16.5 mg/l. (2)The BOD in second sample of water is 12.5 mg/litre and 16 mg/l. (3)The BOD in third sample of water is 11 mg/l. and 14.5 mg/l. (4)The Bod in fourth sample of water detected is 13.5 mg/l. and 17 mg/l. DO test:- To detect the amount of dissolved oxygen in selected four samplesof pond water and two samples of potable water. 1-Fill the given DO bottle with sample of water. 2-Add three drops of magnese ii sulphate solution in DO bottle filled with water. 3-Add three drops of alkaline potassium iodide. 4-Add eight drops of phosphoric acid in the solution. 5-Take 10 ml of solution in a test tube and add 2 drops of starch solution. 6-Solution turned black. 7-Titrate with sodium thio sulphate solution. 8-Calculate the amount of dissolved oxygen by using obtained readings of titration, with the help of following formula: DO=Volume of hypo(sodium thio sulphate solution) consumed*20 Result: The amount of dissolved oxygen in pond water samples and potable water samples is: 1:Amount of DO in first water sample is 15.00 mg/l. and 9.9 mg/l. 2:Amount of DO in second water sample is 14.00 mg/l. and 9.0 mg/l. 3:DO amount in this sample of water is 12.00 mg/l. and 8.2 mg/l. 4:DO amount in fourth sample of water is 10.00 mg/l. and 7.9 mg/l. Result and discussion: As shown in the table, results obtained as a result of our investigation, it is clear that the disposal of non treated sewage in water bodies, increases the chemical oxygen demand in water due to presence of various chemicals in them. It also increases first DO of water bodies due to development of aquatic plants in high nutrient containing water.But when more and more organic matter diposited in water bodies development and death of plants occurs in them then for the decomposition of organic matters various bacterial species develops in the water.But as a result of death and decay of organisms various gases evolved from water and high amount of oxygen is required for this purpose which further results in increased biological oxygen demand and an considerable decrease in Dissolved oxygen of water as a result of eutrophication. S.NO. COD BOD DO September February September February September February 01 11.4 mg/Liter 13.5 mg/Liter 13.00mg/Liter 16.50 mg/Liter 15.0 mg/Liter 9.9 mg/Liter 02 13.5 mg/Liter 14.9 mg/Liter 12.5 mg/Liter 16.00 mg/Liter 14.0 mg/Liter 9.0 mg/Liter 03 13.0 mg/Liter 15.5 mg/Liter 11.5 mg/Liter 14.50 mg/Liter 12.0 mg/Liter 8.2 mg/Liter 04 16.8 mg/Liter 19.0 mg/Liter 13.5 mg/Liter 17.00 mg/Liter 10.0 mg/Liter 7.9 mg/Liter Table 1: Presenting values of COD , BOD and DO obtained Graph showing increasing values of COD Graph showing increasing values of BOD Graph showing decreasing values of DO References: (1) Khan, Fareed A.; Ansari, Abid Ali; December 1, 2005; HighBeam Research Article: Eutrophication: an ecological vision. (2) Sandra E. 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