Answer :
a. The total log reduction of E. coli in the wastewater treatment plant if the primary treatment reduces E. coli by 50%, the secondary treatment reduces it by 90%, and the disinfection process further reduces it by 99.9% is 0.348.
b. The total % reduction is 58.8%
a. To calculate the total log reduction of E. coli in the wastewater treatment plant, we need to convert the reduction percentages to log values and then add them together. First, let's convert the reduction percentages to log values. The formula to convert a percentage reduction to a log reduction is:
log reduction = -log(percentage reduction)
- For the primary treatment: log reduction = -log(50%) = -log(0.5) = 0.301
- For the secondary treatment: log reduction = -log(90%) = -log(0.9) = 0.046
- For the disinfection: log reduction = -log(99.9%) = -log(0.999) = 0.001
Now, let's add the log reductions together:
Total log reduction = 0.301 + 0.046 + 0.001 = 0.348
b. To calculate the total percentage reduction, we can convert the total log reduction back to a percentage reduction using the formula:
percentage reduction = 1 - 10^(-total log reduction)
percentage reduction = 1 - 10^(-0.348)
percentage reduction = 1 - 0.412
percentage reduction ≈ 0.588 or 58.8%
Therefore, the total log reduction of E. coli in the wastewater treatment plant is approximately 0.348, and the total percentage reduction is approximately 58.8%.
Your question is incomplete, but most probably your full question was
In a wastewater treatment plant, Primary treatment yields 50% reduction of E. coli, Secondary treatment yields 90% reduction, and Disinfection yields 99.9% reduction. What is the total log reduction of E. coli in the wastewater treatment plant? What is the total % reduction?
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