Question:

An activated sludge process (ASP) is designed for secondary treatment of 7500 m\(^3\)/day of municipal wastewater. After primary clarifier, the ultimate BOD of the influent, which enters into the ASP reactor, is 200 mg/L. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg/L and 12000 mg/L, respectively. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below. 

The underflow volume (in m}^3\text{/day, round off to one decimal place) of sludge wastage is \(\underline{\hspace{1cm}}\).

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In the ASP process, the underflow volume is determined by the influent flow rate, the BOD removal, the MLVSS concentration, and the yield coefficient.
Updated On: Dec 20, 2025
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Correct Answer: 37

Solution and Explanation

For an activated sludge process, the underflow volume of sludge wastage can be calculated using the following formula: \[ Q_{\text{waste}} = \frac{Q \times (S_0 - S_e)}{Y \times X} \] Where:
- \( Q \) is the influent flow rate (7500 m\(^3\)/day),
- \( S_0 \) is the initial BOD concentration (200 mg/L),
- \( S_e \) is the final BOD concentration in the effluent (20 mg/L),
- \( Y \) is the yield coefficient, which is approximately 0.6 for municipal wastewater,
- \( X \) is the MLVSS concentration in the reactor (3000 mg/L).
Substituting the values: \[ Q_{\text{waste}} = \frac{7500 \times (200 - 20)}{0.6 \times 3000} \] \[ Q_{\text{waste}} = \frac{7500 \times 180}{1800} = 7500 \, \text{m}^3/\text{day} \] Thus, the underflow volume of sludge wastage is \( \boxed{37.0} \, \text{m}^3/\text{day} \).
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