![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 b) Oxygen uptake, mg/1h c) Aeration tank biomass concentration, mgL](http://img.homeworklib.com/questions/a024fee0-1bfb-11ec-9bf0-7ffdb62bc712.png?x-oss-process=image/resize,w_560)
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2. A complete mix activated sludge process with recycle is being operated in the following conditions...
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)...
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...
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 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....
Question 2 A complete mix activated sludge process is to be used for biological treatment. Assume...
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...
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...
For the following activated sludge process unit, find the correct SRT(hr) and F/M ratio, respectively, as:...
For the following activated sludge process unit, find the correct SRT(hr) and F/M ratio, respectively, as: HRT: 1.5 hr Biological reactor Influent Secondary clarifier Effluent Q: 12000 m3/d X= 3500 mg/L BOD: 10 mg/L Setting BOD: 100 mg/L Derby matter XR=8000 mg/L Retum activated sludge Qw/Q=5% Surplus sludge
Problem 2. An activated sludge unit is being designed to treat 375,000 gpd with an influent...
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...
2. Estimate the weight of net solids produced per day in an activated sludge aeration system...
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.
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,...
4. (40 pts) You need to design a set of activated sludge aeration tanks. The flow...
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...
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)...
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 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....
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...
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...
For the following activated sludge process unit, find the correct SRT(hr) and F/M ratio, respectively, as: HRT: 1.5 hr Biological reactor Influent Secondary clarifier Effluent Q: 12000 m3/d X= 3500 mg/L BOD: 10 mg/L Setting BOD: 100 mg/L Derby matter XR=8000 mg/L Retum activated sludge Qw/Q=5% Surplus sludge
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...
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.
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,...
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...
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