Percent Removal

One of the basic math equations we use on a regular basis in the wastewater and drinking water industries is percent removal. One way we use it is to help determine how efficient our treatment process is. The following is the formula that we use.

In the wastewater industry, we use this formula to determine our percent removal of biochemical oxygen demand (BOD) and total suspended solids (TSS). These are normally listed as secondary removal standards on your KPDES permit.

Let’s work one for fun.

If the influent BOD was 174 mg/l and the effluent BOD was 12 mg/l, what is the percent removal of BOD?

In the drinking water industry, this equation may be used to determine the effectiveness of our flocculation sedimentation process. By measuring turbidity at the beginning of these processes, you can determine what percentage of turbidity you removed in the process. It can also be used to help determine the amount of water loss from the distribution system.

Now let’s work one for a flocculation sedimentation process.

The turbidity going into the flash mixer was 129.4 NTUs and at the end of the sedimentation basin the turbidity was 0.3 NTUs. What is the percent removal?

Another use of this equation is percent water loss, which is shown in the following equation.

As you can see, this is a very useful multipurpose equation that can be used in both the wastewater and drinking water industries. If you can learn the concept behind percent removal, then these equations will be invaluable to you in your career.

If you would like to practice additional problems similar to the one above, check out the Test Your Knowledge – Percent Removal post.

This entry was posted in Educational Tools, Tim Ricketts and tagged , , , , , . Bookmark the permalink.

One Response to Percent Removal

  1. Pingback: Test Your Knowledge – Percent Removal |

Comments are closed.