Use the following data to size (volume in gallons) an activated sludge reactor basin (CFSTR with...
Use the following data to size (volume in gallons) an activated sludge reactor basin (CFSTR with recycle and Monod)
MLSS = 3500 mg/l
MLVSS = 80%
Yield Coeffiecient = 0.6 MLVSS
BOD5 influent to the activated sludge basin = 230 mg/l
BOD5 effluent to the activated sludge basin = 50 mg/l
MCRT is 6 days
Endogenous decay coefficient is 0.06d^-1
flow rate = 7 MGD
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Use the following data to size (volume in gallons) an activated
sludge reactor basin (CFSTR with...
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...
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 wastewater flow of 18,000 m3/day is to be treated in a completely mixed activated sludge...
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...
(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...
- of waste? PROBLEM 3 (20) An activated sludge wastewater treatment system uses a 8-million-L aeration basin. The mean cell retention time is 12 days. The MLSS is kept at 3100 mg/L and the recyc...
- of waste? PROBLEM 3 (20) An activated sludge wastewater treatment system uses a 8-million-L aeration basin. The mean cell retention time is 12 days. The MLSS is kept at 3100 mg/L and the recycled activated sludge (RAS) is 11000 mg/L. What is the wasted activated sludge (WAS) rate if the sludge is wasted from a) the aeration basin and b) the recycle line? Read the question. I already posted it before and 'Rashid' copied it from an earlier wrong...
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...
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...
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
1. A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated...
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...
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...
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...
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 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...
(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...
- of waste? PROBLEM 3 (20) An activated sludge wastewater treatment system uses a 8-million-L aeration basin. The mean cell retention time is 12 days. The MLSS is kept at 3100 mg/L and the recycled activated sludge (RAS) is 11000 mg/L. What is the wasted activated sludge (WAS) rate if the sludge is wasted from a) the aeration basin and b) the recycle line? Read the question. I already posted it before and 'Rashid' copied it from an earlier wrong...
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...
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...
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
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...
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...
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