Florida Wastewater Practice Test

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If your plant has an influent of 220 BOD and 163 TSS, and an effluent of 10 BOD and 6 TSS, what is the BOD removal efficiency?

The correct answer is: 95%

To determine the BOD removal efficiency, we first need to understand the formula used to calculate this efficiency. BOD removal efficiency is calculated using the formula: \[ \text{BOD Removal Efficiency} = \frac{(BOD_{in} - BOD_{out})}{BOD_{in}} \times 100 \] In this scenario: - The influent BOD (BOD in) is 220 mg/L. - The effluent BOD (BOD out) is 10 mg/L. Plugging these values into the formula: 1. Calculate the BOD removed: \[ BOD_{removed} = BOD_{in} - BOD_{out} = 220 - 10 = 210 \, mg/L \] 2. Calculate the BOD removal efficiency: \[ BOD Removal Efficiency = \frac{210}{220} \times 100 \] 3. Simplifying this gives: \[ BOD Removal Efficiency = 0.9545 \times 100 = 95.45 \% \] This means that approximately 95% of the BOD has been removed from the influent as it passes through the treatment process. When rounded, this gives a BOD removal efficiency