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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 vS...
6. The aeration tank for a completely mixed aeration process is being sized for a design...
6. The aeration tank for a completely mixed aeration process is being sized for a design wastewater flow of 7500 m²/day. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/1. The design effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design parameters are a sludge age of 10 days and volatile MLSS of 1400 mg/l. The kinetic constants from a bench-scale treatability study are Y=0.60 mg VSS/mg soluble BOD and Ka...
just part e please Activated sludge design A completely mixed activated sludge process is designed to...
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,...
Q Search The following information is given for an activated sludge system design: Flowrate Influent BOD...
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 wastew...
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...
Environmental Engineering 3&4 So 3. An activated sludge plant treats 5 MGD and the average influent...
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...
A complete-mix activated sludge system is used to treat municipal wastewater after primary sedime...
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....
Secondary Treatment Based on a plant flow rate of 40 mgd, design an activated sludge, secondary t...
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...
(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. Th...
(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...
A completely mixed activated sludge plant is designed to treat 10,000 m3/d of an industrial wastewater....
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...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
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...
6. The aeration tank for a completely mixed aeration process is being sized for a design wastewater flow of 7500 m²/day. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/1. The design effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design parameters are a sludge age of 10 days and volatile MLSS of 1400 mg/l. The kinetic constants from a bench-scale treatability study are Y=0.60 mg VSS/mg soluble BOD and Ka...
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,...
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...
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...
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....
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...
(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...
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...
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...
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