Question

A sewage treatment plant influent is estimated to have a typical BOD5 value of 320 mg/L. What efficiency must the treatment plant have in order to meet the NPDES BOD5 requirement of less than 30 mg/L? Note: Your calculated efficiency must be entered with units of %

Answer 1

Answer:

The required efficiency of BOD5 removal of the plant needs to be at least 91%

Explanation:

Even though a WWTP (Waste water treatment plant) has several unit operations, it needs to have an overall treatment efficiency. In this case, it is only required to know how much BOD5 do we need to lower by removing organic matter from the water. It goes as following:

$Efficiency(%) =\frac{actual concentration-goal concentration}{actual concentration}*100%$

This will work for any efficiency, for any parameter you need to define. They need to be in the same concentration unit, of course.

Replacing known values, we have:

$Efficiency (%)=\frac{320 mg/L-30 mg/L}{320 mg/L}*100%= 90,625%$

Which we will approximate to 91%, because decimal percentages do not say anything and are not currently used.