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Just tell me which one is right?
A wastewater treatment plant treats 1000 m3 day- of...
Just tell me which one is right A wastewater treatment plant treats 1000 (m3 day) of...
Just tell me which one is right
A wastewater treatment plant treats 1000 (m3 day) of wastewater. Calculate the amount of excess sludge (kg day l) under steady-state condition, based on the following information: -Influent and effluent BOD concentrations: 200 mg Lland 10 mg L-? -Influent P concentration: 10 mg L-? -Yx/s=0.6; ka=0 (ignore the decay rate) -Content percentage of P in excess sludge: 7% The amount of excess sludge (kg day '); 64 84 114 214 the most anxious...
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. (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 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...
Problem 3 (20 pts) Your firm is doing a wastewater treatment plant upgrade for a plant...
Problem 3 (20 pts) Your firm is doing a wastewater treatment plant upgrade for a plant using activated sludge with an influent BODs concentration of 300 mg/L. Recently, the steady state inflow increased to 50.0 mᵒ/d. To maintain proper performance, a VSS concentration of 150 mg/L in the bioreactor and 2500 mg/L in the WAS line needs to be maintained. a) Draw a diagram of this scenario. b) Design the aeration tank for a cell retention time of 7 days....
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%...
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...
Just tell me which one is right Global amount of water evaporation (T ton year -...
Just tell me which one is right
Global amount of water evaporation (T ton year - 500 200 200 100 40 20 Awa treatment plant treats 1000 m 40 davilnot wastewater Calculate the effective
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....
Just tell me which one is right SVI was 200 (ml/g) and MLSS in aeration tank...
Just tell me which one is right
SVI was 200 (ml/g) and MLSS in aeration tank was set at 2,000 (mg/L). Estimate an adequate recycle ration (R/Q) from the following equation. MLSS(Q+R)=(106)*R/SVI R/Q (-) 0.23 0.42 0.68 1.52
Just tell me which one is right
A wastewater treatment plant treats 1000 (m3 day) of wastewater. Calculate the amount of excess sludge (kg day l) under steady-state condition, based on the following information: -Influent and effluent BOD concentrations: 200 mg Lland 10 mg L-? -Influent P concentration: 10 mg L-? -Yx/s=0.6; ka=0 (ignore the decay rate) -Content percentage of P in excess sludge: 7% The amount of excess sludge (kg day '); 64 84 114 214 the most anxious...
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. (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 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...
Problem 3 (20 pts) Your firm is doing a wastewater treatment plant upgrade for a plant using activated sludge with an influent BODs concentration of 300 mg/L. Recently, the steady state inflow increased to 50.0 mᵒ/d. To maintain proper performance, a VSS concentration of 150 mg/L in the bioreactor and 2500 mg/L in the WAS line needs to be maintained. a) Draw a diagram of this scenario. b) Design the aeration tank for a cell retention time of 7 days....
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%...
Just tell me which one is right
Global amount of water evaporation (T ton year - 500 200 200 100 40 20 Awa treatment plant treats 1000 m 40 davilnot wastewater Calculate the effective
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....
Just tell me which one is right
SVI was 200 (ml/g) and MLSS in aeration tank was set at 2,000 (mg/L). Estimate an adequate recycle ration (R/Q) from the following equation. MLSS(Q+R)=(106)*R/SVI R/Q (-) 0.23 0.42 0.68 1.52
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