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- C. What is the concentration UI SUINUS 5. Determine the size (diameter and depth) of the secondary clarifier (s) at a municipal activated sludge treatment facility: The wastewater flow is 25,000 m³/d, the recycle ratio (recycle sludge flow/wastewater flow) is 0.25. The suspended solids concentration is assumed to be 2500 mg/L in the activated sludge tank. The over flow rate is 50 m3/d- m2 and solids loading is 125 kg/d-m2. Estimate detention time if the depth is 4.5 m. What would be the weir loading? Department of Civil & Environmental Engineering P.O. Box 519 Mail Stop 2510, Prairie View, Texas 77446 w.pvamu.edu Phone (936) 261-1660/1665(dept) Fax (936) 261 1662 Ensuring o Using Zoom o12.50 Determine the activated-sludge aeration volume required to treat 2.64 mgd with a BOD of 120 mg/l based on the criteria of a maximum BOD loading of 40 lb/1000 ft³/day and a minimum aeration period of 5.0 hr. Assuming an operating F/M of 0.20 lb BOD/day per lb of MLSS, calculate the MLSS to be maintained in the aeration tank. Estimate the operat- ing sludge age (mean cell residence time), assuming an effluent suspended solids of 30 mg/l and the daily amount of waste-sludge solids from Figure 13.1. Determine the diame- ter and side-water depth of two identical final clarifiers of the type shown in Figure 10.12 for this activated-sludge system.The waste water is flowing at a rate of 0.2 m2/s in a settling tank. The specific gravity of particles in the water is 2.7 and the kinematic viscosity of water is 1.02x10-2 cm?/s. Calculate the required area of tank for removing particles of size 0.07 mm.
- 8 Determine the solid loading rate on a secondary circular clarifier (diameter = 20 m) if the plant influent load is 2 ML/d. The recycle flow from secondary clarifier to the activated sludge system (aeration basin) is set to 10% of the influent load. It is also known that the aeration basin produces an MLSS (mixed liquor suspended solids) concentration of 3000 mg/L.Sludge stabilization is an important step prior its disposal. Explain thestabilization process. Based on your understanding, justify if the stabilizationmethod provided sustainable.The City of Marietta has been directed by the City Board to upgrade its primary Wastewater Treatment Plant (WWTP) to include a secondary treatment process that can meet an effluent standard of 20 mg/L of BODs and 20 mg/L of suspended solids (SS). They want to select completely mixed activated sludge system. The BOD5 of the SS is estimated to be 50% of the SS concentration. Assume L:W within 4:1 and maximum depth of 4.5 m and maximum length of 50 m for the tank. Assume the reasonable values of missing parameters from FE Exam Handbook. The effluent characteristics from the primary treatment process are: Flow = 0.150 m's, BOD5 =75 mg/L. Assume the values for the growth constants as K, = 100 mg/L BOD5, Mm = 2.5 d-1, kg = 0.05 d'1. Y = 0.50 mg VSS/mg BODs removed, MLVSS = 2,800 mg/L, MISS - 1.20 (MLVSS), wastewater temperature = 25.0°C. Assume 8% 02 transfer rate and BOD5 to BODu ratio as f = 68%. If SRT (qc) = 14.5 days, the volume of the aeration tank is nearly?
- Discuss thoroughly with the aid of illustration how the coagulation and flocculationprocess occur in the water treatment plant to destabilize particles/colloids before settling down in the sedimentation tank.The concentration of the mixed liquor suspended solids, MLSS, in the basin of a step aeration system is 2300 mg/L, and the volume of settled sludge after 30 minutes in a 1 L graduated cylinder is 200 mL. Calculate the SVI in mL/g.An activated sludge wastewater treatment system was designed and operated in a steadystate under the following conditions:• Reactor volume of 10,000 m3in total;• Sludge concentration (MLSS) of 3200 mg/L in the reactor;• Return sludge concentration at 8000 mg/L (the underflow of the secondary clarifier)• Each day, 400 m3 of underflow sludge (return sludge) was wasted as excesssludge;What is the average Sludge Residence Time (SRT)?
- Given a settling tank operating at 80% TSS removal efficiency, an inflow of 0.5 m3/s, and an influent concentration of 100 mg/L. Assuming steady state operations, at what flow would operators have to remove sludge in order to achieve a TSS concentration of 1500 mg/L in the sludge? Answer: 0.027 m3/sA complete mix activated sludge process treats 5 MGD of wastewater containing a total BOD5 concentration of 225 mg/L BOD5 to 20 mg/L BOD5, which is 75% soluble in both the influent and the effluent. The volume of the aeration basin is 111,400 ft3 and the MLSS concentration is 2500 mg/L. Determine the following: a) Detention time in the aeration basin (hr) b) F/M ratio c) Specific substrate utilization rate d) Substrate removal efficiencyA complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3)