The 500-bed Lotta Hart Hospital has a small activated sludge plant to treat its wastewater. The...
The 500-bed Lotta Hart Hospital has a small activated sludge plant to treat its wastewater. The average daily hospital discharge is 1200 L/d per bed, and the average soluble BOD5 after primary settling is 500 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide x 10.0 m ling x 4.5 m deep. The allowable BOD5 and the suspended solid concentration in the effluent of the secondary treatment unit are 30.0 mg/L and 30.0 mg/L, respectively. Assume that the BOD5 of the suspended solids may be estimated as equal to 33.3% of the suspended solid concentration. Given: the plant operating parameters are as follows: MLVSS = 2000 mg/L; Microorganism growth constants: bacterial decay rate = 0.060 day-1 , maximum specific growth rate constant = 2.50 day-1 , and half saturation constant = 100 mg/L.
c) Calculate the solid retention time.
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The 500-bed Lotta Hart Hospital has a small activated sludge
plant to treat its wastewater. The...
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 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...
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...
For the completely mixed activated sludge process shown below, the system boundary is shown by the...
For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Qis the wastewater flow rate into the aeration tank; Q, is the flow rate of liquid containing microorganisms to be wasted: X is the microorganism concentration (VSS) entering the aeration tank: X is the microorganism concentration (MLVSS) in the aeration tank; X is the microorganism concentration (VSS) in the effluent from secondary settling tank X, is the microorganism (VSS) in sludge being...
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.
2. (40 pts) A wastewater treatment plant has an activated sludge system. The aeration tank is...
2. (40 pts) A wastewater treatment plant has an activated sludge system. The aeration tank is 252 ft long, 25 ft wide, and 18 ft deep. An engineer will model the steady-state performance as first order (k= 8.3 d ?) as a series of 9 equal-volume CFSTRs. The flow rate is 3.1 MGD and the influent BODS is 165 mg/L. a) If the engineer's model is correct, what is the outlet BOD5 concentration (mg/L)? b) If the model is correct,...
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....
(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...
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...
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. [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 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...
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...
For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Qis the wastewater flow rate into the aeration tank; Q, is the flow rate of liquid containing microorganisms to be wasted: X is the microorganism concentration (VSS) entering the aeration tank: X is the microorganism concentration (MLVSS) in the aeration tank; X is the microorganism concentration (VSS) in the effluent from secondary settling tank X, is the microorganism (VSS) in sludge being...
2. (40 pts) A wastewater treatment plant has an activated sludge system. The aeration tank is 252 ft long, 25 ft wide, and 18 ft deep. An engineer will model the steady-state performance as first order (k= 8.3 d ?) as a series of 9 equal-volume CFSTRs. The flow rate is 3.1 MGD and the influent BODS is 165 mg/L. a) If the engineer's model is correct, what is the outlet BOD5 concentration (mg/L)? b) If the model is correct,...
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....
(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...
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...
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...
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