Question

A city has a population of 50,000 people, an average household generated wastewater flow of 430 L/day-person, and the average BOD5 of the untreated wastewater in population equivalents is 0.1 kg BOD5/day-person. If the BOD reaction rate constant for the waste stream is 0.4/day. Determine the ultimate BOD of the wastewater.

Answer 1

Answer:

The ultimate BOD of waste water is 265 milligram per Liter.

Explanation:

To start with and properly determine the ultimate BOD of the waste water, we need to determine the volume of water used by the population in the city within five days.

Therefore,   50000 people × 430 Liters ÷ day-person × 5 days

1.08 × 10∧8 Liters

The mass BOD5 by the city population in 5 days is therefore,

0.1 kilogram ×  BOD5 ÷ day-person × 50000 people × 5 days

2.5 × 10 ∧ 4 BOD5

The 5 days BOD of waste water is therefore,

5d = 2.5 × 10 ∧ 4 / 1.08 × 10 ∧ 8 Liters × 1000grams  ÷ 1 kilogram × 1000mg ÷ 1gram

233 milligram / liter

The ultimate BOD, is therefore,

ULo = 5d ÷ (1 - e ∧ -k5)

Solving this further, we derive

265 milligram per Liter.