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(15 points) A BOD test is run using 50 mL of treated wastewater mixed with 250 mL of pure water. The initial DO of the mix is 9.0 mg/L. After 5 days, the DO is 4.5 mg/L. After a long period of time, the DO is 2.5 mg/L, and it no longer seems to be dropping. Assume there is no nitrogenous BOD and that all BOD is carbonaceous. a. What is the 5-day BOD of the wastewater

Answer :

ajeigbeibraheem

Answer:

27 mg/L

Explanation:

From the given information:

To determine the 5-day BOD using the formula:

[tex]BOD_5 = \dfrac{DO_i -DO_f}{P} \ \ ---(1)[/tex]

where;

BOD = biochemical oxygen demand

[tex]DO_i[/tex] = initial dissolved oxygen

[tex]DO_f[/tex] = final dissolved oxygen

P = dilution factor

The dilution factor [tex]P = \dfrac{V_w}{V_m}[/tex]

where;

[tex]V_w = \text{volume of waste water} \\ \\ V_m = \text{volume of mixture}[/tex]

[tex]P = \dfrac{50}{300} \\ \\ P = 0.167[/tex]

replacing the value of P into equation  (1);

[tex]BOD_5 = \dfrac{(9.0 - 4.5)mg/L}{0.167}[/tex]

[tex]BOD_5 = \dfrac{(4.5)mg/L}{0.167}[/tex]

[tex]\mathtt{BOD_5 = 26.946 \ mg/L}[/tex]

[tex]\mathtt{BOD_5 \simeq 27 \ mg/L}[/tex]

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