Chemical oxygen demand

CHEMICAL OXYGEN DEMAND

1. CHEMICAL OXYGEN DEMAND Presented By :- RAJNEESH KUMAR GAUTAM MTECH 1 ST SEM (ENERGY &ENVIRONMENT) BABA SAHEB BHIM RAO AMBEDKAR UNIVERSITY-LUCKNOW 226001

2. INTRODUCTION The organic matter present in sewage can be measured in number of ways. Organic matter is often assessed in terms of oxygen required to completely oxidise the organic matter to CO 2 , H 2 0 and other oxidised species. The oxygen required to oxidise the organic matter present in a given waste water can be theoretically calculated .If the organic compounds and their concentration are known the oxygen demand of the sample can be accurately computed but it is impossible to

3. know the details of organic compounds Present in raw water or waste water. The COD is therefore determined by performing a laboratory test On given water sample using a strong oxidant like dichromate solution. It is expressed in milligrams per liter (mg/L) also referred to as ppm (parts per million), which indicates the mass of oxygen consumed per liter of solution.

4. DEFINITION COD is the total amount of oxygen required to chemically oxidize the bio degradable and non-biodegradable organic matter.

5. HISTORY E arlier potassium permanganate was used as an oxidizing agent COD values were very much lower than that of 5 th day BOD It indicated that potassium permanganate was not very effective in oxidizing all the organic matter present After that other oxidizing agents like ceric sulphate, potassium iodate and potassium di-chromate were also used

6. HISTORY (cont.) Potassium di chromate was found to be the most effective due to - Completely oxidize all organic matter - Relatively cheap - Easy to purify

7. COD TEST BY DICHROMATE PRINCIPLE Water sample is refluxed in strong acidic solution with a known excess amount of potassium dichromate. After digestion, the remaining unreduced K 2 Cr 2 O 7 is titrated with Ferrous Ammonium Sulfate (FAS)to determine K 2 Cr 2 O 7 consumed. This gives us the oxidizable organic matter in terms of oxygen equivalent.

8. TEST PROCEDURE Wash 300 ml round bottom refluxing flask . In refluxing flask put one spatula of HgSO 4 + 10 ml sample + 5ml K 2 Cr 2 O 7 + 15 ml concentrated H2SO4. Add small amount of silver sulphate Shake well and reflux for 2 hr. Cool and add little amount of distilled water to the flask through the condenser

9. TEST PROCEDURE (CONT.) Titrate the solution in the flask against FAS using Ferroin indicator End point green color to reddish brown NOTE: For blank, add 10 ml distilled water instead of sample. Rest of the procedure is the same.

10. CALCULATIONS The COD in mg/l is determined by the formula, COD mg/l = (A-B)xNx8000 ml sample taken A = ml of FAS required for blank. B = ml of FAS requires for sample .

11. ADVANTAGES OF COD TEST COD result are available much sooner than BOD test results. The COD test requires fewer manipulations of the sample. The COD test oxidizes a wide range of chemical compounds. It can be standardize more easily.

12. DISADVANTAGES OF COD TEST The major disadvantage is that the results are not directly applicable to 5-day BOD results without correlation studies over a long period of time. One more limitation of COD is it’s inability to differentiate between biologically oxidizable and biologically inert organic matter.

13. References Journal on industrial waste water treatment by k.l moed Environmental engineering vol 1&2 by S.K Garg Industrial waste water treatment by metcalf & eddy

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