Question

Secondary Treatment Based on a plant flow rate of 40 mgd, design an activated sludge, secondary treatment system with recycle which maintains a MLVSS = 2,000 mg/L in the aeration reactor and has an average Solids Retention Time (SRT)-4 days. The biota kinetic constants are Kinetic Coefficients for the Activated-sludge Process for the Removal of Organic Matter from Domestic Wastewater Endogenous Decay Coefficient- k(d) 0.1 Half-velocity constant- K(s VSS/qVss-da 15 Mg/L bsCOD Max. Specific Substrate Utilization Rate-k 6 G bsCOD/gVSS-da Biomass Yield-Y 0.45 gVSS/g bsCOD used The laboratory has provided the information below for your influent from your primary clarifier. Quantit 30 0.18 20 0.017 215 10 Item Unit Influent nbVSS-Xo Fraction of the biomass which remains as cell debris (fd Temperature Quantity of Oxvgen in air Influent microorganism & substrate concentration as (So Influent Inert Inorganics (TSSo- VSSo Lb O2 / ft3 air Ma/L bsCOD Aeration Basin (Reactor)-Assume 6 basins are required 1. using Eq. #4, calculate the effluent substrate concentration (S) (mg/L) (Ans. 2.23 mg/L) 2. Using Eq. #7, calculate the hydraulic detention time ( ) required in each aeration basin to 3. Using the aeration basin hydraulic detention time calculated above and the design flow rate of the plant, calculate: (a) Volume of each aeration basin (t)..........(Ans. 177,092 ft3) (b) Width of each basin (ft) assuming a basin depth of 20-ft and a W:d ratio of 2.2:1 (Ans. 44 ft) 4. using Eq. #6, calculate the volatile sludge produced (wasted) each day (lb/day) (Ans. 5528.4 Ib/day)

Answer 1

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