2. Estimate the weight of net solids produced per day in an activated sludge aeration system...
A complete-mix activated sludge system is used to treat municipal wastewater after primary sedimentation at 20 °C with MLVSS concentration of 2000 g/m3. The discharge standard for soluble BOD, concentration in the effluent is 0.5 g/m. The characteristics of the primary influent are: flow Q -2000 m'/d, soluble BOD -200 g/m3, Using these data and the kinetic coefficients in following, determine 1. Theoretical solids retention time (SRT) in unit of days 2. Pk vss in unit of kg VSS/d 3....
3. Determine the aeration tank volume (mgd), excess biomass in the waste sludge (kg/d), and the BOD f a completely mixed activated-sludge process by assuming the following: k 4.0 day r-0.55 lb vSS/1b BOD, K, 80 mg/l, ks-0.05 day', wastewater flow rate 2.5 mgd, soluble effluent BOD- 15 mgl, soluble influent BOD - 170 mgl, and MLVSS-2200 mg/l, and mean cell residence time -6 days. (15 pts) 3. Determine the aeration tank volume (mgd), excess biomass in the waste sludge...
A sewage treatment plant treats sewage at average flowrate of 10000m3/day. The plant applies activated sludge process for biological treatment. Find; 1. Volume of aeration tank. 2. Design air requirement. 3. Flowrate of waste sludge. Use the following data; BOD of raw sewage=250mg/l. Percent of BOD removal in primary sedimentation tank =32%. MLVSS=2400 mg/l. MLVSS in return sludge=10000mg/l. Y=0.6 and kd=0.06/day. Effluent BOD=20mg/l. Effluent SS=30mg/l.
A wastewater flow of 18,000 m3/day is to be treated in a completely mixed activated sludge system at a concentration of 3000 mg/L MLSS (mixed liquor suspended solids). The secondary clarifier is designed to thicken the sludge to 12,000 mg/L. Assuming a growth yield coefficient of 0.5 Kg/Kg, an influent BOD of 100 mg/L with a residence time of 8 days, determine the following: a) The volume of the reactor b) The mass of the solids and the wet volume...
just part e please Activated sludge design A completely mixed activated sludge process is designed to treat 20,000 m/day of domestic wastewater having a BODs concentration of 250 mg/L following primary treatment. The NPDES permit requires that the effluent BOD, and TSS concentra- tions not exceed 20 mg/L on an annual basis. The following biokinetic coefficients are to be used in the design of the process: Y = 0.6 mg VSS/mg BODs, k = 5d"! Ks = 60 mg/L BODs,...
(30 points) A completely mixed activated sludge system comprised of an aerated reactor and a secondary clarifier with sludge return is going to treat 15,000 m/d of industrial wastewater. The primary effluent going to the reactor has a BODs of 1,100 mg/L that must be reduced to 150 mg/L prior to discharge to a municipal sewer. The pilot study was conducted and generated the following results: 3. Mean cell residence time- 5 day, MLSS concentration in the reactor 7.200 mg/L...
Environmental Engineering 3&4 So 3. An activated sludge plant treats 5 MGD and the average influent BOD is 200 mg/L. The WWTP has 2 aeration basins, each 100 ft x 25 ft x 15 ft deep and 2 final clarifiers with diameters of 20 ft. About 35% of BOD is removed in primary clarifiers. Determine the BOD in the primary effluent, hydraulic retention time, F/M ratio, BOD exiting the aeration basin and mass of solids wasted (1b/d). The operation parameters...
2. A completely mixed activated sludge process is designed to treat 20,000 m/d of domestic wastewater having a BODs concentration of 250 mg/L following primary treatment. The NPDES permit requirement for effluent BODs and TSS is 20 mg/L for each. The biokinetic parameters have beern established as: yield 0.6, k 5/d, K-60 mg/L, and kd 0.06/d. The MLVSS concentration in the aeration tank is to be maintained at 3000 mg/L and the ratio of VSS/TSS is 0.75. Assume that the...
Q Search The following information is given for an activated sludge system design: Flowrate Influent BOD Effluent BOD Unit m'/d mg/L mg/L Value 10,000 150 SRT Synthesis yield, Y, g VSS/g bCOD Wastewater 1 Wastewater 2 Wastewater 3 Cell debris yield, f, 0.40 0.50 0.30 0.15 0.08 g VSS/g VSS Endogenous decay, b, gVSS/g VsSd mg/L nbVSS Temperature Note: Wastewater 1, 2, or 3 to be selected by instructor 10 Determine (a) the aeration tank oxygen requirements in kg/d, (b)...
2. A complete mix activated sludge process with recycle is being operated in the following conditions Unit Parameter Flow rate Influent BOD Value 12,500 mg/L 210 Efluent BOD HRT, λ SRT im hours days 0.50 Synthesis yieldVSS/gbCOD Cell debris yield, fa g VSS/g VSS Endogenous decay, ki nbVSS 0.13] g VSS/ g VSS.d 0.08 im 50 Temperature 10 Assume no nitrification occurs due to the SRT selected and low temperature. Find a) Oxygen requirement for the aeration tank, kg d...