Question

2. Incoming wastewater, with BOD5 equal to 200 mg/L, is treated in a well-run secondary treatment plant that removes 90 percent of the BOD. You are to run a five-day BOD test with a standard 300-mL bottle, using a mixture of treated sewage and dilution water (no seed). Assume the initial DO is 9.2 mg/L. a. Roughly what maximum volume of treated wastewater should you put in the bottle if you want to have at least 2.0 mg/L of DO at the end of the test (filling the rest of the bottle with water). (answer in mL)

Answer 1

Answer:

10.8 ml

Explanation:

The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.

See attached file

Answer 2

The maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

The given parameters;

• wastewater density = 200 mg/L
• standard volume = 300 mL
• initial DO = 9.2 mg/L

The dilution factor (P) is calculated as follows;

$200 \ mg/L= \frac{9.2 \ mg/L \ - \ 2\ mg/L}{P} \\\\P = \frac{7.2 \ mg/L}{200 \ mg/L} \\\\P = 0.036$

The maximum  volume of treated wastewater that will be in the bottle to have at least 2.0 mg/L DO;

$0.036 = \frac{V_w}{300 \ mL} \\\\V_w = 0.036 \times 300 \ mL\\\\V_w = 10.8 \ mL$

Thus, the maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

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