Percent removal of BOD in a clarifier is 35%, if 70 mg/L are removed, what is the influent BOD?

To determine the influent BOD, we can use the information provided about the percent removal and the amount removed. The question states that 35% of the BOD is removed in the clarifier, resulting in a removal of 70 mg/L.

To find the influent BOD, we can set up the following relationship:

1. Let the influent BOD be denoted as "X" mg/L.

2. Since 35% of the influent BOD is removed, we can express the amount removed as:

( 0.35X = 70 , \text{mg/L} ).

3. To find the influent BOD (X), we can rearrange the equation:

( X = \frac{70}{0.35} ).

Calculating this gives:

( X = 200 , \text{mg/L} ).

Thus, the influent BOD is 200 mg/L. This value is derived directly from the percentage removal and the amount removed, clearly demonstrating the relationship between influent concentration and removal efficiency in wastewater treatment processes.