Let’s take a look at how efficient a wastewater treatment plant is at removing BOD. This is a percent removal problem, so let’s go to the board and see how to work this problem.
[iframe width=”560″ height=”315″ src=”https://www.youtube.com/embed/S0v2ycyjvWA?list=PL309F532284302ACE”]
The problem reads “What is the percent of BOD removed in a plant when the influent BOD is 245 mg per liter and the effluent BOD is 22 mg per liter?” As always we want to pull the numbers out of the problem and write them down. First, let’s write down our known and our unknown information.
They told us that we are looking for the percent BOD removed. So, I write that down as percent removal. Then, they tell us that our influent BOD is 245 mg per liter. So we write that down. Our effluent BOD is 22 mg per liter. Write that down.
So, coming into the plant is 245 mg per liter of BOD and leaving there’s 22 mg per liter. We have a formula that says efficiency or removal efficiency equals what’s coming in minus what’s coming out then you divide that by what’s coming in and then we multiply by a 100 to convert the decimal to a percent.
Efficiency = [(IN – Out)/IN] x 100
Plugging in our numbers and doing the math we see that we take our 245 mg per liter and subtract from that the 22. Then we divide by 245 which is what’s coming in. When we multiply that by 100 we see that this wastewater treatment plant is removing 91% of the incoming BOD.
And that my friend is how this problem is solved.