Question

(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 1

27 mg/L

Step-by-step explanation:

From the given information:

To determine the 5-day BOD using the formula:

`BOD_5 = (DO_i -DO_f)/(P) \ \ ---(1)`

where;

BOD = biochemical oxygen demand

`DO_i` = initial dissolved oxygen

`DO_f` = final dissolved oxygen

P = dilution factor

The dilution factor
`P = (V_w)/(V_m)`

where;

`V_w = \text{volume of waste water} \\ \\ V_m = \text{volume of mixture}`

`P = (50)/(300) \\ \\ P = 0.167`

replacing the value of P into equation (1);

`BOD_5 = ((9.0 - 4.5)mg/L)/(0.167)`

`BOD_5 = ((4.5)mg/L)/(0.167)`

`\mathtt{BOD_5 = 26.946 \ mg/L}`

`\mathtt{BOD_5 \simeq 27 \ mg/L}`