What is the maximum organic loading capacity of a treatment plant handling BOD of 280 mg/L?

Prepare for the OSSF Designated Representative Exam. Study with multiple choice questions and explanations to help you succeed. Get ready for the exam and ensure you're well-prepared to become a designated representative!

To determine the maximum organic loading capacity of a treatment plant handling a Biochemical Oxygen Demand (BOD) of 280 mg/L, it is essential to understand how to convert the concentration of BOD into a loading rate, typically expressed in pounds per day.

First, we need to understand that BOD measures how much oxygen microorganisms will consume while decomposing organic matter in water. The loading capacity helps evaluate the treatment plant's potential to handle organic waste effectively.

To calculate the organic loading capacity in pounds per day, you can use the formula:

[ \text{Loading (lbs/day)} = \frac{\text{BOD (mg/L)} \times \text{Flow (MGD)} \times 8.34}{1} ]

Where:

  • BOD is given in mg/L (280 mg/L in this case).

  • Flow is the flow rate of the wastewater through the treatment plant, measured in million gallons per day (MGD).

  • The factor 8.34 is a conversion factor that helps convert mg/L into pounds/day.

For this question, while the flow rate is not specifically provided, if we assume a standard flow rate that is commonly used in practice for calculations or is implied in the context of the question

Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy