the suspended solids concentration of the primary clarifier influent is 305 mg/l. if the effluent suspended solids concentration is 137 mg/l, what is the removal efficiency?

55%
35%
45%
65%

Question

the suspended solids concentration of the primary clarifier influent is 305 mg/l. if the effluent suspended solids concentration is 137 mg/l, what is the removal efficiency?

55%
35%
45%
65%

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Solids Removal Calculation" ], "new_concepts": [], "current_concepts": [ "Solids Removal Calculation", "Wastewater Process Control" ] } ## Identify given values and formula $$ \begin{aligned} &\text{Influent Concentration } (C_{\text{in}}) = 305\text{ mg/L} \ &\text{Effluent Concentration } (C_{\text{out}}) = 137\text{ mg/L} \ &\text{Removal Efficiency } (\eta) = \frac{C_{\text{in}} - C_{\text{out}}}{C_{\text{in}}} \times 100\% \end{aligned} $$ ## Calculate removal efficiency $$ \begin{aligned} \eta &= \frac{305 - 137}{305} \times 100\% \ &= \frac{168}{305} \times 100\% \ &\approx 0.5508 \times 100\% \ &\approx 55\% \end{aligned} $$

(A) 55%

(B) 35%

(C) 45%

(D) 65%

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Environmental Science", "Wastewater Treatment Calculations" ] }

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Explanation:

Identify given values and formula

Calculate removal efficiency

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