Question

The following information is available for a seeded 5-day BOD test conducted on a wastewater sample. 15 mL of the waste sample was added directly into a 300-mL BOD incubation bottle and then 285 mL of dilution water was added to fill the bottle. The initial DO of the diluted sample was 8.8 mg/L and the final DO after 5 days was 1.9 mg/L. 1. Assuming that the BOD5 of 285 mL dilution water is 0 mg/L. What is the 5-day BOD (BOD5) of the wastewater sample 2. In reality, the dilution water usually has its own BOD. In a control experiment of determining the BOD5 of dilution water with seed microorganisms, the initial DO of dilution water was 9.1 mg/L and the final DO after 5 days was 7.9 mg/L. What is the 5-day BOD (BOD5) of the wastewater sample

Answer 1

(a) BOD₅ = 138 mg/L

(b) BOD₅ = 115.2 mg/L

Step-by-step explanation:

Given Data;

volume of sample = 15 mL

volume of sample bottle

Volume of dilution water = 285 mL

initial DO = 8.8 mg/L

Final DO = 1.9 mg/L

Calculating the volumetric fraction of the sample using the fromula;

p = volume of sample/volume of sample bottle

= 15/300

= 0.05

(a) Calculating the 5-day BOD of the waste water sample using the formula;

BOD₅ = D₀ - D₅/p

= 8.8 -1.9/0.05

= 138 mg/L

(b) Calculating the dilution percentage of sample, we have

Dilution percentage of sample = 285/300 *100

= 95%

Calculating the value of ratio f using the formula;

f = %of dilution water in diluted sample/%of dilution water in dilution water

= 95/100

= 0.95

Calculating the 5 day BOD of the waste water sample using the formula;

BOD₅ = [(D₀ - D₅) - f(B₀ -B₅)]/p

= [(8.8 - 1.9) - 0.95*(9.1 -7.9)]/0.05

= 6.9-1.14/0.05

= 5.76/0.05

= 115.2 mg/L