Suppose you are to measure the BOD removal rate for the DC wastewater treatment

Question

Suppose you are to measure the BOD removal rate for the DC wastewater treatment plant. For comparison pur you take two samples the same time of raw sewage on its way into the plant and of BOD_5 tests, without seeding, are below: Find BOD_5 for samples #1, #3 and the percentage removal. Find DO_final. in Sample #2 and dilution ratio of sample #4. What does the BoD of wastewater indicate? Is a high BOD of water good or bad? Why? Why is the BOD_5 test conducted in the darkroom Why it is tested for only 5 days? Why is it often necessary to dilute the wastewater? And why is it often to seed the test sample?

Explanation / Answer

1. BOD for sample 1 =(9.5-2.2)*dilution ratio=(9.5-2.2)*(30/1)= 219 mg/L

for sample 3 BOD=(9-2)*20=140mg/L

b. Dofinal=DOintial -(BOD)*(1/Dil ratio)=9.5-219*(1/25/1)=9.5-219*(1/25)= 0.7400

Dil. ratio of sample 4 =(BOD/(DOin-DOfin)=140/9=15.5556

c.Biological Oxygen Demand is one of the most common measures of pollutant organic material in water..BOD measure the amount of organic compounds in water.It may be considered severely polluted when BOD values exceed certain limit of BOD (mg/L).Fish and aquatic insects may die when oxygen is depleted by microbial metabolism.

Therefore, a small BOD is an indicator of good quality water, while a high BOD indicates polluted water.

d. conducted in dark room to prevent Dissolve Oxygen production via photosynthesis.

DO used for the decomposition of organic matter as well as nitrogenous matter.there are two type of oxygen demands: Carbonaceous BOD initial, and Nitrogenous BOD (final).Since, first 5 days demand is the “Carbonaceous Bio Chemical Oxygen Demand” which accounts for 68% of the total BOD value. Hence, it represents the oxygen demand for the decomposition of organic matter only.

e.Dilution is needed to prevent the depletion of entire oxygen present in in BOD bottle in less than incubation period of 5 days.

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