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A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume...
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...
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to...
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BODs and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow = 0.2000 m 3/s soluble BODs-80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%...
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....
:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter...
:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter received flow 44L/Sec , the BOD concentration in raw sewage =440 mg/L , 30% BOD removed in the primary sedimentation tank, volume of filter (first stage )=826000L, volume of filter (second stage)=2563000L, recirculation rate =0.5 in second tank while no recirculation in first tank. Determine the following: 1-The effluent concentration of BOD in each unit. 2-Treatment efficiency of each unit. 3-Treatment efficiency of plant....
A sewage treatment plant treats sewage at average flowrate of 10000m3/day. The plant applies activated sludge...
A sewage treatment plant treats sewage at average flowrate of 10000m3/day. The plant applies activated sludge process for biological treatment. Find; 1. Volume of aeration tank. 2. Design air requirement. 3. Flowrate of waste sludge. Use the following data; BOD of raw sewage=250mg/l. Percent of BOD removal in primary sedimentation tank =32%. MLVSS=2400 mg/l. MLVSS in return sludge=10000mg/l. Y=0.6 and kd=0.06/day. Effluent BOD=20mg/l. Effluent SS=30mg/l.
CIEG 328 HOMEWORK7 Homework 7 (Suspended growth) Due: 4/112019 Design an activated sludge biologi...
CIEG 328 HOMEWORK7 Homework 7 (Suspended growth) Due: 4/112019 Design an activated sludge biological treatment system for a typical medium-strength domestic wastewater with a design daily flow of 2.9 million gallons per day. Assume the plant effluent criteria are 10/10/1" This means that the plant must produce an effuent with average BODS and TSS concentrations of 10 mg/Leach year round, and an effluent ammonia concentration of 1 mg/L as N during the winter months. The layout of the plant will...
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...
A Tapered aeration-system is used to treat 12,500 m3/d of municipal wastewater. The wastewater has received...
A Tapered aeration-system is used to treat 12,500 m3/d of municipal wastewater. The wastewater has received primary treatment and has a BOD 5 of 140 mg/l and a suspended solids of 125 mg/l. The system is to be operated in the following way: (a) Soluble BOD 5 in effluent 5 mg/l (b) Average solids concentration in the reactor= 2000 mg/l (c) Mean cell- retention time= 10 d. The biological constants have been determined by pilot-plant analysis and are: Y= 0.55...
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...
urgent A wastewater treatment bioreactor is designed to conduct phosphorus removal by using organisms A and...
urgent
A wastewater treatment bioreactor is designed to conduct phosphorus removal by using organisms A and B. Given that with the following parameters: Heterotrophic synthesis yield, Y = 0,40 g VSSIG COD Endogenous decay coefficient, Kd = 0.08 g VSS/g VSS day SRT = 5 days Biomass produced = [Y/(1+(K.) SRT)] x (bCOD or bsCOD) Phosphorus content of organism A: 0.30 g Plg VSS Phosphorus content of organism B: 0.02 g Plg VSS Clarifier effluent VSS concentration = 8 g/m3...
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...
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BODs and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow = 0.2000 m 3/s soluble BODs-80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%...
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....
CIEG 328 HOMEWORK7 Homework 7 (Suspended growth) Due: 4/112019 Design an activated sludge biological treatment system for a typical medium-strength domestic wastewater with a design daily flow of 2.9 million gallons per day. Assume the plant effluent criteria are 10/10/1" This means that the plant must produce an effuent with average BODS and TSS concentrations of 10 mg/Leach year round, and an effluent ammonia concentration of 1 mg/L as N during the winter months. The layout of the plant will...
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...
urgent
A wastewater treatment bioreactor is designed to conduct phosphorus removal by using organisms A and B. Given that with the following parameters: Heterotrophic synthesis yield, Y = 0,40 g VSSIG COD Endogenous decay coefficient, Kd = 0.08 g VSS/g VSS day SRT = 5 days Biomass produced = [Y/(1+(K.) SRT)] x (bCOD or bsCOD) Phosphorus content of organism A: 0.30 g Plg VSS Phosphorus content of organism B: 0.02 g Plg VSS Clarifier effluent VSS concentration = 8 g/m3...
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