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Example: A sample of sewage from a town is found to have a BOD after 5 d (BOD5) of 180mg/L. Estimate the Ultimate BOD (the BODL) of the sewage assuming that k = 0.1/d for this waste water.

Solution

BODt = BODL X (1 10-kt)

180 = BODt = BODL X (1 10-kt) , It implies that 180 = BODL X (1-10-0.1X5)

   Therefore 180 = BODL X (1- 0.316) ; 180 = BODL X 0.684

Rearranging terms to solve for BODL gives

BODL = 180/0.684 = 260 mg/L Rounded off.

This is particularly true when the BOD data are used to monitor the efficiency of a water pollution control plant. It has been found that more than two-third of the BODL is usually exerted within the first 5d of decomposition. For instance, in the preceding example, the 5d BOD is 180/260 = 0.69, or 69% of the ultimate BOD. For practical purposes, the 5d BOD, or BOD5, has been chosen as a representation of the organic content of water or waste water. For standardization of results, the test must be conducted at a temperature of 20oC.

In summary, the parameter of BOD5 is the amount of dissolved oxygen used by microbes in the 5 d to decompose organic substances in water at 20oC.

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