A 5-day BOD test is performed on a sample of chlorinated wastewater effluent. Fi

Question

A 5-day BOD test is performed on a sample of chlorinated wastewater effluent. Fifty milliliters of wastewater sample and 2 mL of "seed" are added to the 300-mL BOD bottles and mixed with enough dilution water to fill the bottles. "Seed" must be added since the wastewater effluent is chlorinated, which kills pathogens and bacteria. For the blanks or controls, 2 mL of "seed" are added to 300-mL BOD bottles and mixed with enough dilution water to fill the bottles. The average dissolved oxygen concentrations of the diluted wastewater samples and blanks on day 1 of the test were 7.0 and 7.6 mg/L respectively. After incubation at 200 C for 5 days, the average DO concentrations of the diluted wastewater samples andblanks were 2.7 and 7.3 mg/L respectively. Calculate the 5-day BOD of the chlorinated effluent and its ultimate BOD if the rate constant k = 0.34 d-1

Explanation / Answer

BOD, mg/L = [(Initial DO Final DO) x 300]/mL sample

BOD of the blank

Initial DO = 7.6 mg/L

Final DO = 7.3 mg/L

BOD mg/L = [(7.6 7.3) x 300]/300 = (0.3 x 300)/300 = 0.3 mg/L

For the sample

Initial DO = 7.0 mg/L

Final DO = 2.7 mg/L

Sample size = 50 mL

BOD mg/L = [(7.0 2.7) x 300]/50 = (4.3 x 300)/50 = 1290-(0.3 x 250 mL)/50 = 1215/50 = 24.3 mg/L

The 5-day BOD = 24.3 mg/L

The ultimate BOD is given by

BOD5 = BODultimate (1-e-kt)

24.3 = BODultimate(1-e-0.34 day-1 x 5 days)

24.3 = BODultimate(1-e-1.7)

24.3 = BODultimate(0.817)

BODultimate = 24.3/0.817

BODultimate = 29.73 mg/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.