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Secondary Treatment Based on a plant flow rate of 40 mgd, design an activated sludge, secondary t...
Question 2 A complete mix activated sludge process is to be used for biological treatment. Assume the following for the activated sludge process: • Plant influent BODs of 156 mg/L • Sludge age is 8 days Biomass yield (Y) of 0.54 kg biomass / kg BOD • Endogenous decay rate (kd) = 0.05 day! • Ks=10 g BOD/m3 Biomass concentration (X) of 2700 mg/L • recycle biomass concentration (Xr) of 12,000 mg/L The fresh flow rate treated is 1000 m²/day...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by conventional activated sludge process composed of covered aeration tanks and secondary clarifiers. You are asked to calculate the different operation parameters to make sure that the system is operated within normal process ranges listed in the reference tables below. Table 5-20 Typical design information for primary sedimentation tanks U.S. customary units Item Unit Range Typical Primary sedimentation tanks followed by secondary treatment...
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...
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...
2. (40 pts) A wastewater treatment plant has an activated sludge system. The aeration tank is 252 ft long, 25 ft wide, and 18 ft deep. An engineer will model the steady-state performance as first order (k= 8.3 d ?) as a series of 9 equal-volume CFSTRs. The flow rate is 3.1 MGD and the influent BODS is 165 mg/L. a) If the engineer's model is correct, what is the outlet BOD5 concentration (mg/L)? b) If the model is correct,...
4. (40 pts) You need to design a set of activated sludge aeration tanks. The flow rate to treat is 5.6 MGD and the BOD concentration is 150 mg/L. The design solids concentration (X) at steady-state is 2,000 mg/L. The design MCRT is 7 days. The kinetic coefficients are as follows: k = 2 g BOD!g cells*day; K = 25 mg BOD/L; k, = 0.06 1/day; Y = 0.5 g cells/G BOD. The influent ammonia concentration is 40 mg/L and...
(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...
1. A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated sludge treatment plant serving a municipality. The state regulatory agency criteria for final clarifiers used for activated sludge are: Peak SOR = 1,200 gal/day-ft2 Average SOR = 500 gal/day-ft2 Peak solids loading 50 lb/day-ft Peak weir loading-30,000 gal/day.ft The average flow to the aeration basin prior to junction with the recycle line is 3.5 MGD ·The recycled sludge flow is 30 percent of the...
Problem 2. An activated sludge unit is being designed to treat 375,000 gpd with an influent BOD of 250 mg/1. Parameters for the wastewater are: Y 0.6, m -2.0 days-1, kd0.15 days-1 and K, 40 mg/l. The recycle flow is 3750 gpd with a biomass concentration of 50 X, where X is the reactor biomass concentration. The waste sludge flow is 500 gpd. Calculate the mean cell residence time, e, and CSTR volume required for a 95% removal of influent...
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...