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14 July, 03:37

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

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  1. 14 July, 05:04
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    (a) BOD₅ = 138 mg/L

    (b) BOD₅ = 115.2 mg/L

    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
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