an oxidation ditch holds 200,000 gallons with an mlss of 2880 mg/l. find the sludge age when it receives an influent of 500,000 gpd and a tss concentration of 125 mg/l.

4.5 days
6.7 days
9.2 days
8.4 days

Question

an oxidation ditch holds 200,000 gallons with an mlss of 2880 mg/l. find the sludge age when it receives an influent of 500,000 gpd and a tss concentration of 125 mg/l.

4.5 days
6.7 days
9.2 days
8.4 days

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Sludge Age Calculation" ], "new_concepts": [], "current_concepts": [ "Sludge Age Calculation", "Wastewater Process Control" ] } ## Calculate the mass of suspended solids in the oxidation ditch $$ \begin{aligned} \text{Mass in ditch (lbs)} &= \text{Volume (MG)} \times \text{MLSS (mg/L)} \times 8.34 \ &= 0.200\text{ MG} \times 2880\text{ mg/L} \times 8.34 \ &= 4803.84\text{ lbs} \end{aligned} $$ ## Calculate the mass of suspended solids entering daily $$ \begin{aligned} \text{Mass entering (lbs/day)} &= \text{Flow (MG/day)} \times \text{TSS (mg/L)} \times 8.34 \ &= 0.500\text{ MGD} \times 125\text{ mg/L} \times 8.34 \ &= 521.25\text{ lbs/day} \end{aligned} $$ ## Calculate the sludge age $$ \begin{aligned} \text{Sludge Age (days)} &= \frac{\text{Mass in ditch (lbs)}}{\text{Mass entering (lbs/day)}} \ &= \frac{4803.84}{521.25} \ &\approx 9.216\text{ days} \end{aligned} $$

(A) 4.5 days

(B) 6.7 days

(C) 9.2 days

(D) 8.4 days

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

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Question

Explanation:

Calculate the mass of suspended solids in the oxidation ditch

Calculate the mass of suspended solids entering daily

Calculate the sludge age

Answer: