The BOD5 of a waste water is determined to be 240 mg/l at 20℃. The k value at 20℃ is known to be 0.23 day-1 . Calculate the BOD7 if the test were run at 10℃.ans.196.89 mg/l224.05 mg/l240 mg/l258.12 mg/l Previous Marked for Review Next

The BOD (Biochemical Oxygen Demand) is a measure of the amount of oxygen required by bacteria to decompose the organic matter in the water under aerobic conditions. The BOD5 is the BOD measured after 5 days, and the BOD7 is the BOD measured after 7 days.

The BOD at any time t can be calculated using the formula:

BODt = BOD0 * (1 - e^(-kt))

where:

• BOD0 is the initial BOD (in this case, the BOD5),
• k is the deoxygenation constant (day^-1), and
• t is the time (in days).

First, we need to calculate the initial BOD (BOD0) using the given BOD5 and k values:

240 mg/l = BOD0 * (1 - e^(-0.23*5)) Solving this equation for BOD0 gives BOD0 = 311.69 mg/l.

Next, we can calculate the BOD7 at 20℃ using the BOD0 and k values:

BOD7 = 311.69 mg/l * (1 - e^(-0.23*7)) = 290.68 mg/l.

However, the question asks for the BOD7 at 10℃, not 20℃. The deoxygenation constant k is temperature-dependent and typically decreases with decreasing temperature. Without knowing the exact relationship between k and temperature for this particular wastewater, we cannot calculate the BOD7 at 10℃.

Therefore, the answer cannot be determined from the information given.