Example (Activated Sludge Process) - Completely Mixed with RecycleA completely mixed activated sludge process, treating a municipal wastewater, has a primary clarifier effluentBOD, of 135 mg/L. The design MLSS is 3590 mg/L, and the MLVSS is 80% of the MLSS. The plant permit isfor an effluent BOD, of 20 mg/L and 20 mg/L suspended solids. The effluent suspended solids have a BOD, of0.65mg BOD, / mg suspended solids. The underflow solid concentration (MLVSS) is 10,000 mg/L. If umax = 3.0day, Ks 60 mg/L, Y= 0.50 mg MLVSS / mg BOD, removed, ka= 0.1 day, and the influent flow is 0.25m/sec, determine the following:1. The required reactor detention time, 0.2. The reactor basin volume, V, in m³.3. The net waste activated sludge produced each day (kg/day).4. The mass of solids (both volatile and inert) to be wasted each day (kg/day).The return sludge flow ratethe recycle ratio.6. The F/M ratio.7 June 2021Dr. Dawood Eisa Sachit Biological Wastewater Treatment47

Answered: Image /qna-images/answer/3179e868-04b8-4783-9344-5204e9b7a1db.jpg

1. The required reactor detention time, 0. 2. The reactor basin volume, V, in m³. 3. The net waste activated sludge produced each day (kg/day). 4. The mass of solids (both volatile and inert) to be wasted each day (kg/day). 5. The return sludge flow rate and the recycle ratio. 6. The F/M ratio.

1. The required reactor detention time, 0. 2. The reactor basin volume, V, in m³. 3. The net waste activated sludge produced each day (kg/day). 4. The mass of solids (both volatile and inert) to be wasted each day (kg/day). 5. The return sludge flow rate and the recycle ratio. 6. The F/M ratio.

Solid Waste Engineering

3rd Edition

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Worrell, William A.

Chapter4: Mechanical Processes

Section: Chapter Questions

Problem 4.2P

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