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) Half-velocity constant – K(s) 0.1 15 Max. Specific Substrate Utilization Rate – k Biomass Yield - Y 6 0.45 gVSS/gVSS-day mg/L bsCOD g bsCOD/gVSS-day gVSS/g bsCOD used The laboratory has provided the information below for your influent from your primary clarifier. Unit mg/L Item Influent nbVSS - Xo Quantity 30 Fraction of the biomass which remains as cell debris (fd) Temperature 0.18 20 °C Quantity of Oxygen in air 0.017 Lb O₂/ft³ air Influent microorganism & substrate concentration as (So) Influent Inert Inorganics (TSS. – VSS.) 215 10 Mg/L bsCOD mg/L Aeration Basin (Reactor) - Assume 6 basins are required 1. Using Eq. #4, calculate the effluent substrate concentration (S) (mg/L).[ Ans: 2.23 mg/L] 2. Using Eq. #7, calculate the hydraulic detention time (t) required in each aeration basin to maintain the design MLVSS concentration. [Ans: 4.77 hours ] 3. Using the aeration basin hydraulic detention time calculated above and the design flow rate of the plant, calculate: A. Volume of each aeration basin (ft³). [Ans: 177,092 ft³] B. Width of each basin (ft) assuming a basin depth of 20-ft and a W:d ratio of 2.2:1. [ Ans: 44 ft ] C. Length of each basin (ft). [ Ans: 201 ft] 4. Using Eq. #6, calculate the volatile sludge produced (wasted) each day (lb/day). [Ans. 5528.4 lb/day] 5. Using Eq. #8, calculate the total sludge produced (wasted) each day (lb/day). [Ans: 7024 lb/day] 6. Using Eq. #10, calculate the F/M (foot to microorganism ratio in aeration tank). [ Ans: F/M = 0.541] 7. Using Eq. #9, calculate the volume of air needed by the reactor (ft³/day). [Ans: 355,614 ft³/day] Secondary Clarifier - Assume 6 Clarifiers are required In designing the secondary clarifier, the following parameters have been agreed: Item Recycle Suspended Solids (✗R) Quantity 7000 Units Mg/L Clarifier Surface Overflow Rate (SOR) Depth of each clarifier 400 Gal/ft2-day 17 ft 8. Using Eq. #11, calculate: A. Recirculation ratio.[ Ans: R = 0.4] B. Average recycle flow rate (mgd). [ Ans: 2.67 mgd] 9. Calculate the clarifier A. Surface area (ft²). [Ans: 16,667 ft²] B. Diameter (ft). [Ans: 146 ft ] C. Circumference (ft). [Ans: 459 ft] D. Volume (ft³). [Ans: 283,333 ft³]
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) Half-velocity constant – K(s) 0.1 15 Max. Specific Substrate Utilization Rate – k Biomass Yield - Y 6 0.45 gVSS/gVSS-day mg/L bsCOD g bsCOD/gVSS-day gVSS/g bsCOD used The laboratory has provided the information below for your influent from your primary clarifier. Unit mg/L Item Influent nbVSS - Xo Quantity 30 Fraction of the biomass which remains as cell debris (fd) Temperature 0.18 20 °C Quantity of Oxygen in air 0.017 Lb O₂/ft³ air Influent microorganism & substrate concentration as (So) Influent Inert Inorganics (TSS. – VSS.) 215 10 Mg/L bsCOD mg/L Aeration Basin (Reactor) - Assume 6 basins are required 1. Using Eq. #4, calculate the effluent substrate concentration (S) (mg/L).[ Ans: 2.23 mg/L] 2. Using Eq. #7, calculate the hydraulic detention time (t) required in each aeration basin to maintain the design MLVSS concentration. [Ans: 4.77 hours ] 3. Using the aeration basin hydraulic detention time calculated above and the design flow rate of the plant, calculate: A. Volume of each aeration basin (ft³). [Ans: 177,092 ft³] B. Width of each basin (ft) assuming a basin depth of 20-ft and a W:d ratio of 2.2:1. [ Ans: 44 ft ] C. Length of each basin (ft). [ Ans: 201 ft] 4. Using Eq. #6, calculate the volatile sludge produced (wasted) each day (lb/day). [Ans. 5528.4 lb/day] 5. Using Eq. #8, calculate the total sludge produced (wasted) each day (lb/day). [Ans: 7024 lb/day] 6. Using Eq. #10, calculate the F/M (foot to microorganism ratio in aeration tank). [ Ans: F/M = 0.541] 7. Using Eq. #9, calculate the volume of air needed by the reactor (ft³/day). [Ans: 355,614 ft³/day] Secondary Clarifier - Assume 6 Clarifiers are required In designing the secondary clarifier, the following parameters have been agreed: Item Recycle Suspended Solids (✗R) Quantity 7000 Units Mg/L Clarifier Surface Overflow Rate (SOR) Depth of each clarifier 400 Gal/ft2-day 17 ft 8. Using Eq. #11, calculate: A. Recirculation ratio.[ Ans: R = 0.4] B. Average recycle flow rate (mgd). [ Ans: 2.67 mgd] 9. Calculate the clarifier A. Surface area (ft²). [Ans: 16,667 ft²] B. Diameter (ft). [Ans: 146 ft ] C. Circumference (ft). [Ans: 459 ft] D. Volume (ft³). [Ans: 283,333 ft³]
Sustainable Energy
2nd Edition
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:DUNLAP, Richard A.
Chapter14: Ocean Thermal Energy Conversion And Ocean Salinity Gradient Energy
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Problem 20P
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![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)
Half-velocity constant – K(s)
0.1
15
Max. Specific Substrate Utilization Rate – k
Biomass Yield - Y
6
0.45
gVSS/gVSS-day
mg/L bsCOD
g bsCOD/gVSS-day
gVSS/g bsCOD used
The laboratory has provided the information below for your influent from your primary clarifier.
Unit
mg/L
Item
Influent nbVSS - Xo
Quantity
30
Fraction of the biomass which remains as cell debris (fd)
Temperature
0.18
20
°C
Quantity of Oxygen in air
0.017
Lb O₂/ft³ air
Influent microorganism & substrate concentration as (So)
Influent Inert Inorganics (TSS. – VSS.)
215
10
Mg/L bsCOD
mg/L
Aeration Basin (Reactor) - Assume 6 basins are required
1. Using Eq. #4, calculate the effluent substrate concentration (S) (mg/L).[ Ans: 2.23 mg/L]
2. Using Eq. #7, calculate the hydraulic detention time (t) required in each aeration basin to maintain the design MLVSS concentration. [Ans: 4.77 hours ]
3. Using the aeration basin hydraulic detention time calculated above and the design flow rate of the plant, calculate:
A. Volume of each aeration basin (ft³). [Ans: 177,092 ft³]
B. Width of each basin (ft) assuming a basin depth of 20-ft and a W:d ratio of 2.2:1. [ Ans: 44 ft ]
C. Length of each basin (ft). [ Ans: 201 ft]
4. Using Eq. #6, calculate the volatile sludge produced (wasted) each day (lb/day). [Ans. 5528.4 lb/day]
5. Using Eq. #8, calculate the total sludge produced (wasted) each day (lb/day). [Ans: 7024 lb/day]
6. Using Eq. #10, calculate the F/M (foot to microorganism ratio in aeration tank). [ Ans: F/M = 0.541]
7. Using Eq. #9, calculate the volume of air needed by the reactor (ft³/day). [Ans: 355,614 ft³/day]
Secondary Clarifier - Assume 6 Clarifiers are required
In designing the secondary clarifier, the following parameters have been agreed:
Item
Recycle Suspended Solids (✗R)
Quantity
7000
Units
Mg/L
Clarifier Surface Overflow Rate (SOR)
Depth of each clarifier
400
Gal/ft2-day
17
ft
8. Using Eq. #11, calculate:
A. Recirculation ratio.[ Ans: R = 0.4]
B. Average recycle flow rate (mgd). [ Ans: 2.67 mgd]
9. Calculate the clarifier
A. Surface area (ft²). [Ans: 16,667 ft²]
B. Diameter (ft). [Ans: 146 ft ]
C. Circumference (ft). [Ans: 459 ft]
D. Volume (ft³). [Ans: 283,333 ft³]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F379782e3-36e2-4b06-9a33-e2601af30423%2F7cdcb10f-ef69-46ff-81c6-7c4562819c7a%2Ft4psidr_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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)
Half-velocity constant – K(s)
0.1
15
Max. Specific Substrate Utilization Rate – k
Biomass Yield - Y
6
0.45
gVSS/gVSS-day
mg/L bsCOD
g bsCOD/gVSS-day
gVSS/g bsCOD used
The laboratory has provided the information below for your influent from your primary clarifier.
Unit
mg/L
Item
Influent nbVSS - Xo
Quantity
30
Fraction of the biomass which remains as cell debris (fd)
Temperature
0.18
20
°C
Quantity of Oxygen in air
0.017
Lb O₂/ft³ air
Influent microorganism & substrate concentration as (So)
Influent Inert Inorganics (TSS. – VSS.)
215
10
Mg/L bsCOD
mg/L
Aeration Basin (Reactor) - Assume 6 basins are required
1. Using Eq. #4, calculate the effluent substrate concentration (S) (mg/L).[ Ans: 2.23 mg/L]
2. Using Eq. #7, calculate the hydraulic detention time (t) required in each aeration basin to maintain the design MLVSS concentration. [Ans: 4.77 hours ]
3. Using the aeration basin hydraulic detention time calculated above and the design flow rate of the plant, calculate:
A. Volume of each aeration basin (ft³). [Ans: 177,092 ft³]
B. Width of each basin (ft) assuming a basin depth of 20-ft and a W:d ratio of 2.2:1. [ Ans: 44 ft ]
C. Length of each basin (ft). [ Ans: 201 ft]
4. Using Eq. #6, calculate the volatile sludge produced (wasted) each day (lb/day). [Ans. 5528.4 lb/day]
5. Using Eq. #8, calculate the total sludge produced (wasted) each day (lb/day). [Ans: 7024 lb/day]
6. Using Eq. #10, calculate the F/M (foot to microorganism ratio in aeration tank). [ Ans: F/M = 0.541]
7. Using Eq. #9, calculate the volume of air needed by the reactor (ft³/day). [Ans: 355,614 ft³/day]
Secondary Clarifier - Assume 6 Clarifiers are required
In designing the secondary clarifier, the following parameters have been agreed:
Item
Recycle Suspended Solids (✗R)
Quantity
7000
Units
Mg/L
Clarifier Surface Overflow Rate (SOR)
Depth of each clarifier
400
Gal/ft2-day
17
ft
8. Using Eq. #11, calculate:
A. Recirculation ratio.[ Ans: R = 0.4]
B. Average recycle flow rate (mgd). [ Ans: 2.67 mgd]
9. Calculate the clarifier
A. Surface area (ft²). [Ans: 16,667 ft²]
B. Diameter (ft). [Ans: 146 ft ]
C. Circumference (ft). [Ans: 459 ft]
D. Volume (ft³). [Ans: 283,333 ft³]
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