Question

The town of Camp Verde has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 25.0 mg/L BOD5 and 30 mg/L suspended solids. They have selected a completely mixed activated sludge system for the upgrade. The existing primary treatment plant has a flow rate of 2,506 m3/d. The effluent from the primary tank has a BOD5 of 240 mg/L. Using the following assumptions, estimate the required volume of the aeration tank:
BOD5 of the effluent suspended solids is 70% of the allowable suspended solids concentration.
Growth constant values are estimated to be: Ks = 100 mg/L BOD5; kd = 0.025 d–1; μm = 10 d–1; Y = 0.8 mg VSS/mg BOD5 removed.
The design MLVSS is 3,000 mg/L.

Answer 1

For efficient treatment, the dissolved oxygen concentration in the aeration tank must be kept between 1-3 mg/L. The microbial biomass will die at low DO levels, and it will take time and money to rebuild it.

The depth, which normally ranges from 3 to 4.5 meters, determines how effectively air is aerated. The breadth, which is typically maintained between 5 and 10 meters, regulates the mixing. The ideal width-to-depth ratio is 1.2 to 2.2. The distance shouldn't be shorter than 30 meters or longer than 100 meters. Bacteria used in wastewater treatment and stabilization are given oxygen by aeration. The bacteria require oxygen for biodegradation to take place.

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