BOD5 of Domestic Sewage

BOD5 refers to the amount of dissolved oxygen that is needed by microorganisms to break down organic matter in water. It is an important parameter to measure the pollution level of sewage water. The average BOD5 of domestic sewage is given as:

Option D: 0.08 kg/day/person

Explanation:

- BOD5 is measured in terms of the amount of oxygen consumed by microorganisms per unit volume of water over a period of 5 days.
- Domestic sewage is the wastewater generated from households, which includes human waste, kitchen waste, and other waste material.
- The BOD5 of domestic sewage depends on the quantity and quality of the waste material present in it.
- The average BOD5 of domestic sewage is estimated to be around 60-100 mg/L.
- To calculate the BOD5 per person per day, we need to consider the amount of domestic sewage generated by a person in a day.
- According to the World Health Organization (WHO), the average water consumption per person per day is around 100-200 liters.
- Assuming that around 80% of this water is converted into domestic sewage, the average domestic sewage generated per person per day is around 80-160 liters.
- Based on this estimate, the BOD5 per person per day can be calculated as follows:

BOD5 per person per day = (BOD5 of sewage in mg/L) x (volume of sewage generated per person per day in liters) / 1000

- Using the average BOD5 of domestic sewage as 80 mg/L and assuming a sewage volume of 100 liters per person per day, we get:

BOD5 per person per day = (80 x 100) / 1000 = 8 mg/day/person

- Therefore, the correct answer is option D: 0.08 kg/day/person (since 1 kg = 1000 g and 1 g = 1000 mg)