A sample of sewage from a town is found to have a BOD5 of 210 mg/L...
A sample of sewage from a town is found to have a BOD5 of 210 mg/L and a k value of 0.11/day. Estimate the ultimate biological demand for this sewage (mg/L).
Homework Answers
ANSWER :-
BOD (at a particular time) 5 = L0 ( 1 - e-kt )
where,
1. k is the speed of BOD reaction
2. L0 is the ultimate biological demand
3. ' t ' is the time period of 5 days
BOD5 = L0 × [ 1 - e(-0.11)×5 ]
210 mg/L = L0 × ( 1 - 0.5769 )
210 = L0 × (0.4231)
So, L0 = 210 / 0.4231 = 496.33 mg/L
CONCLUSION :- The ultimate Biological Oxygen Demand for the sewage is 496.33 mg/L.
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