If the note growth rate of microorganisms is r'g, write a steady-state biomass balance equation for the complete mixed biological reactor (aeration tank) shown in the figure below.
If the influent Q = 0.150m3/s and BOD5 = 84 mg/L, volume of the reactor V = 970 m3, and MLVSS concentration X = 2000 mg/L, determine the F/M ratio.
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If the note growth rate of microorganisms is r'g, write a
steady-state biomass balance equation for...
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. Q is the wastewater flow rate into the aeration tank; Qu 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; Xe is the microorganism concentration (VSS) in the effluent from secondary settling tank; &c is the microorganism (VSS) in sludge...
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...
This is all the information I was given For the completely mixed activated sludge process shown...
This is all the information I was given
For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Q is the wastewater flow rate into the aeration tank; Qy is the flow rate of liquid containing microorganisms to be wasted; Xo is the microorganism concentration (VSS) entering the aeration tank; X is the microorganism concentration (MLVSS) in the aeration tank; He is the microorganism concentration (VSS) in the effluent from secondary settling...
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...
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...
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%...
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,...
Question 2 A complete mix activated sludge process is to be used for biological treatment. Assume...
Question 2 A complete mix activated sludge process is to be used for biological treatment. Assume the following for the activated sludge process: • Plant influent BODs of 156 mg/L • Sludge age is 8 days Biomass yield (Y) of 0.54 kg biomass / kg BOD • Endogenous decay rate (kd) = 0.05 day! • Ks=10 g BOD/m3 Biomass concentration (X) of 2700 mg/L • recycle biomass concentration (Xr) of 12,000 mg/L The fresh flow rate treated is 1000 m²/day...
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...
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...
For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Q is the wastewater flow rate into the aeration tank; Qu 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; Xe is the microorganism concentration (VSS) in the effluent from secondary settling tank; &c is the microorganism (VSS) in sludge...
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...
This is all the information I was given
For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Q is the wastewater flow rate into the aeration tank; Qy is the flow rate of liquid containing microorganisms to be wasted; Xo is the microorganism concentration (VSS) entering the aeration tank; X is the microorganism concentration (MLVSS) in the aeration tank; He is the microorganism concentration (VSS) in the effluent from secondary settling...
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...
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...
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%...
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,...
Question 2 A complete mix activated sludge process is to be used for biological treatment. Assume the following for the activated sludge process: • Plant influent BODs of 156 mg/L • Sludge age is 8 days Biomass yield (Y) of 0.54 kg biomass / kg BOD • Endogenous decay rate (kd) = 0.05 day! • Ks=10 g BOD/m3 Biomass concentration (X) of 2700 mg/L • recycle biomass concentration (Xr) of 12,000 mg/L The fresh flow rate treated is 1000 m²/day...
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...
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