A complete-mix activated sludge system is used to treat municipal wastewater after primary sedime...
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...
(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...
2. Estimate the weight of net solids produced per day in an activated sludge aeration system in which the influent BOD, is reduced from 250 to 30 m l. Calculate SRT and F/M. The following information is given 0 - 4000 m3/day: Aeration tank volume 700 m3 MLVSS - 3000 mg/L. Assume: Y = 0.5, kd=0.09/day.
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...
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 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...
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...
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...
Question 1 The town of Lawrence has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 11 mg/L BODS. They have selected a completely mixed activated sludge system with an aeration tank (in the secondary plant) that can produce an effluent with 2,000 mg MLSS/L. a) Estimate the required volume of the aeration tank needed in the secondary plant The following data are available from the existing primary plant. Existing primary...
A completely mixed activated sludge plant is designed to treat 10,000 m3/d of an industrial wastewater. The wastewater has a BOD5 of 1200 mg/L. Pilot plant data indicates that a reactor volume of 6090 m3 with an MLSS concentration of 5000 mg/L should produce 83% BOD5 removal. The value for Y is determined to be 0.7 kg/kg and the value of kd is found to be 0.03 d–1. Under ow solids concentra- tion is 12,000 mg/L. The ow diagram is...