Biochemical Oxygen Demand

Incoming wastewater, with BOD5 equal to about 200 mg/l, is treated in a well-run secondary treatment plant that removes 90 percent of the BOD. You are to run a five-day BOD test with a standard 300-ml bottle, using a mixture of treated sewage and dilution water (no seed). Assume the initial DO is 9.2 mg/l. Thus we have: BOD5 = 200 mg/l; 90% efficient removal; initial DO = 9.2 mg/l The general relation between BOD5, DOi, DOf and P is: BOD5= (DOi-DOf)/P a) Roughly what maximum volume of treated wastewater should you put in the bottle if you want to have at least 2.0 mg/l of DO at the end of the test (filling the rest of the bottle with water)? b) If you make the mixture half water and ... click for more