Introduction to environmental Engineering. C David Cooper, Chapter 6 Homework 6.14 For municipal wastewater that receives...
Introduction to environmental Engineering. C David Cooper, Chapter 6 Homework
6.14 For municipal wastewater that receives treatment with the activated sludge process, plot the effluent BOD5 as a function of solids residence time using values of SRT that range from 1 to 20 days. The influent BOD5 concentration is 220 mg/L. Assume the following wastewater characteristics: Variable order kinetics k = 5 mg BOD5/mg VSS-day, Ks = 60 mg BOD5/L, Ymax = 0.6 mg VSS/mg BOD5, Ke = 0.06 day^-1.
Find the SRT
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Introduction to environmental Engineering. C David Cooper,
Chapter 6 Homework
6.14 For municipal wastewater that receives...
6.8 a. Using the variable-order kinetic information from Example 6.8 for a municipal wastewater, ...
6.8 a. Using the variable-order kinetic information from Example 6.8 for a municipal wastewater, determine the effluent BODs concentration for a complete-mix activated sludge facility operated at a five-day solids residence time. b. For a wastewater flow rate of 2 million gallons per day with character- istics of a typical wastewater from Table 6.1, determine the discharge rate of biomass (Ibs/day). in the lagooll lall x biomass Rearranging Eq.(6.37) V X HRT EXAMPLE 6.8 Determine the required volume of an...
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....
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 municipal wastewater flow of 20,000 m3/day has an influent bsCOD = 152 g/m3 and an...
A municipal wastewater flow of 20,000 m3/day has an influent bsCOD = 152 g/m3 and an influent nbVSS = 30 g/m3. The WWTP effluent has 6 g bsCOD/m3 when the MLVSS is maintained at 2500 g/m3. Referring to Table 7-8 in Metcalf and Eddy, you may assume: k = 5 g bsCOD/(g VSS•day), KS = 40 g bsCOD/m3 Y = 0.4 g VSS/g bsCOD, b = 0.1 g VSS/(g VSS•day) Also, fd = 0.1 g VSS/g biomass VSS depleted by...
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...
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...
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...
(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...
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...
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,...
6.8 a. Using the variable-order kinetic information from Example 6.8 for a municipal wastewater, determine the effluent BODs concentration for a complete-mix activated sludge facility operated at a five-day solids residence time. b. For a wastewater flow rate of 2 million gallons per day with character- istics of a typical wastewater from Table 6.1, determine the discharge rate of biomass (Ibs/day). in the lagooll lall x biomass Rearranging Eq.(6.37) V X HRT EXAMPLE 6.8 Determine the required volume of an...
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....
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 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...
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...
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...
(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...
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...
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,...
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