Question

A BOD test is run using 100ml of treated wastewater mixed with 200 mL of pure water. The initial DO of the mix is 9 mg/L. After 5 days, the DO is 4 mg/L. After a long period of time, the DO is 2 mg/L and it no longer seems to be dropping. Assume nitrification has been inhibited so the only BOD being measured is carbonanous.

-What is the five day BOD of the wastewater?

Answer 1

The five day BOD of the wastewater is calculated as the oxygen depletion after five days, adjusted for the dilution factor (3:1), resulting in a BOD of 15 mg/L.

Step-by-step explanation:

The five day Biochemical Oxygen Demand (BOD) of the wastewater can be calculated by measuring the amount of oxygen consumed by bacteria while they decompose organic matter under aerobic conditions over five days. In this case, you start with a dilution mixture that has an initial dissolved oxygen (DO) concentration of 9 mg/L. After five days, the DO concentration drops to 4 mg/L.

The BOD is the difference in dissolved oxygen concentrations, which is 9 mg/L - 4 mg/L = 5 mg/L. However, since the sample was diluted with pure water (1 part wastewater to 2 parts pure water), the BOD for the undiluted wastewater would be three times higher, as only a third of the volume is wastewater. Therefore, the BOD for the undiluted wastewater is 5 mg/L * 3 = 15 mg/L.