A wastewater plant with the activated sludge process received 5 × 106 L/day of wastewater with...

A wastewater plant with the activated sludge process received 5 × 10⁶ L/day of wastewater with a BOD₅ of 500 mg/L. The primary clarifier removes 30% of the BOD and primary clarify effluent has no biomass (Xi=0 mg/L). Recycle (Q_r) and waste sludge (Q_w) flows are 25% and 2% of Qin, respectively. Effluent BOD₅ from secondary clarifier and waste sludge is 20 mg/L. Microorganism concentration in the waste sludge (X_w) is 6,000 mg/L. Soluble BOD₅ is biologically converted into CO₂ and Y (yield coefficient) is 0.5 mgVSS /mgBOD
1. Calculate the value of X and the volume of the tank

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Ally Wells answered 4 months ago

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