Question

A development is expected to house 500 people, each who will produce on average) 1801/day of sewage and 77g/day of BOD. Find the BOD of the sewage (in mg/L), and therefore suggest how much oxygen will need to be added to the water (kg/day). {428mg/L,38.5kg/day}

Answer 1

The BOD of the sewage is calculated as follows:

BOD (mg/L) = 77 g/day * 1000 mg/g / 180 L/day = 428 mg/L

This means that the sewage contains 428 mg of BOD per liter.

The amount of oxygen that needs to be added to the water is calculated as follows:

Oxygen needed (kg/day) = BOD (mg/L) * Total volume of water (L) / 1000 kg/L = 428 mg/L * 500,000 L / 1000 kg/L = 235 kg/day

This means that 235 kg of oxygen needs to be added to the water every day to keep the BOD at a safe level.

Therefore, the BOD of the sewage is 428 mg/L, and the amount of oxygen that needs to be added to the water is 235 kg/day.