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Question 2: (Marks 10) Wastewater with flow rate of 10,000 mº/d containing 280 mg/L of ethanolamine...
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...
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%...
Situation (note: Problem 2 is not linked with Problem 1): Assuming the design flow rate of...
Situation (note: Problem 2 is not linked with Problem 1): Assuming the design flow rate of raw wastewater to be treated is 20 MGD in the design year, the dissolved BODs is 380 mg/L, and SS is 360 mg/L. The wastewater generated by the sludge treatment process is about 4 MGD with the dissolved BODs being 1800 mg/L, and SS being 1200 mg/L, and will be returned to the wet well in the pumping station at the beginning of the...
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....
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...
(20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and...
(20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and a total Kjeldahl nitrogen content (TKN) of 35 mg/L. (a) (b) Find the ultimate carbonaceous oxygen demand (CBOD). Find the ultimate nitrogenous oxygen demand (NBOD). Find the total ultimate (or theoretical) oxygen demand (TBOD). Find the remaining BOD (nitrogenous plus carbonaceous) after five days have elapsed.
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...
1. Some pond water contains 20.O mg/L of some algae, which can be represented by the chemical for...
1. Some pond water contains 20.O mg/L of some algae, which can be represented by the chemical formula CsHisO&N. Using the following reactions: CoHisO&N606CO2 6H2ONHa NHs 202 NOs H2O (a) Find the theoretical carbonaceous oxygen demand CBOD (b) Find the nitrogenous oxygen demand NBOD (c) Find the total theoretical biochemical oxygen demand BOD 2. A sandy soil has a hydraulic conductivity of 6.1 x104 m/d, a hydraulic gradient of 0.00141, and porosity of 20%. (a)what is the Darcy velocity (b)what...
2. A tannery wastewater treatment plant has 2250m/d of effluent having de-oxygenation rate of 0.23/d at...
2. A tannery wastewater treatment plant has 2250m/d of effluent having de-oxygenation rate of 0.23/d at 20°C. Determine the ultimate Biochemical Oxygen Demand (BOD) (L.) and the remaining O, after 5 days (L.). Then, compute de-oxygenation rate constant (kd) and determine the rate of de-oxygenation in mg/L/d at 25°C. Note that the stream has 5 days BOD of the waste (BOD) OF 450 mg/L. The avg speed, u, of the stream flow is 0.03 m/d. The depth.h, is 1.2m and...
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...
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%...
Situation (note: Problem 2 is not linked with Problem 1): Assuming the design flow rate of raw wastewater to be treated is 20 MGD in the design year, the dissolved BODs is 380 mg/L, and SS is 360 mg/L. The wastewater generated by the sludge treatment process is about 4 MGD with the dissolved BODs being 1800 mg/L, and SS being 1200 mg/L, and will be returned to the wet well in the pumping station at the beginning of the...
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....
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...
(20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and a total Kjeldahl nitrogen content (TKN) of 35 mg/L. (a) (b) Find the ultimate carbonaceous oxygen demand (CBOD). Find the ultimate nitrogenous oxygen demand (NBOD). Find the total ultimate (or theoretical) oxygen demand (TBOD). Find the remaining BOD (nitrogenous plus carbonaceous) after five days have elapsed.
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...
1. Some pond water contains 20.O mg/L of some algae, which can be represented by the chemical formula CsHisO&N. Using the following reactions: CoHisO&N606CO2 6H2ONHa NHs 202 NOs H2O (a) Find the theoretical carbonaceous oxygen demand CBOD (b) Find the nitrogenous oxygen demand NBOD (c) Find the total theoretical biochemical oxygen demand BOD 2. A sandy soil has a hydraulic conductivity of 6.1 x104 m/d, a hydraulic gradient of 0.00141, and porosity of 20%. (a)what is the Darcy velocity (b)what...
2. A tannery wastewater treatment plant has 2250m/d of effluent having de-oxygenation rate of 0.23/d at 20°C. Determine the ultimate Biochemical Oxygen Demand (BOD) (L.) and the remaining O, after 5 days (L.). Then, compute de-oxygenation rate constant (kd) and determine the rate of de-oxygenation in mg/L/d at 25°C. Note that the stream has 5 days BOD of the waste (BOD) OF 450 mg/L. The avg speed, u, of the stream flow is 0.03 m/d. The depth.h, is 1.2m and...
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