1. A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated...
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...
(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...
Given: There is a wastewater treatment plant that has a Primary Clarifier, Activated Sludge, Secondary clarifier and the average daily flow was 1 mgd. Calculate the volume needed if the activated sludge is 24 hours 10 Mgal 100 Mgal 1 Mgal 1000 Mgal
A wastewater flow of 18,000 m3/day is to be treated in a completely mixed activated sludge system at a concentration of 3000 mg/L MLSS (mixed liquor suspended solids). The secondary clarifier is designed to thicken the sludge to 12,000 mg/L. Assuming a growth yield coefficient of 0.5 Kg/Kg, an influent BOD of 100 mg/L with a residence time of 8 days, determine the following: a) The volume of the reactor b) The mass of the solids and the wet volume...
primary clarifier is to be designed for a municipal wastewater treatment plant with a flow of 4160 m/d and peak flow during the day of 3.10 times the average hourly flow. Pertinent data are: peak overflow rate 73.3 m/d.-m2, maximum peak weir loading 373 m3/d-m, and minimum tank depth 2.13 m. Determine: A. The clarifier diameter B. The side water deptlh C. The peak weir loading. Is it acceptable? D. The percent suspended solids removal on the basis of the...
Problem 7) Determine the recommended size of a new circular secondary clarifier for an activated-sludge system with a design flow of 26,000 m3 /d with a peak hourly flow of 32,000 m3 /d. Use maximum overflow rates of 33 m3 /m2 /day at design monthly flow and 66 m3 /m2 /day at peak hourly flow. The minimum side water depth is 10 ft (3.1 m), 11 ft for a 50-ft diameter tank, and 12 ft for 100-ft diameter tank or...
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....
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...
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...