Sludge volume index ( SVI ) is calculated by dividing the settleability by the MLSS concentration. The SVI is always expressed in mL/g.
right one is 214
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 effective volume of aeration tank of this plant, based on the following information: - Influent BOD5 concentration=200 mg L-1 - MLSS: 2000 mg L-? - HRT=5 hours Volume (m3): 208 308 408 500 1,080 A wastewater treatment plant treats 1000 m
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...
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.
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...
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...
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...
Just tell me which one is right
What are the most anxious emerging pollutants in terms of health problems of human beings? Pollutants: antifebrile pills antibiotics detergents BOD
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 wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume of 24 million gallons per day (MGD) at an average influent 5-day biochemical oxygen demand concentration of 200 mgBODs/L and an average influent total suspended solids concentration of 220 mgTSS/L. The plant operates a primary sedimentation process that remove 65% of the incoming TSS and 35% of the incoming BOD %. This process is followed by a secondary treatment process before discharge to the...
10. Write a one-page summary of the attached paper? INTRODUCTION Many problems can develop in activated sludge operation that adversely affect effluent quality with origins in the engineering, hydraulic and microbiological components of the process. The real "heart" of the activated sludge system is the development and maintenance of a mixed microbial culture (activated sludge) that treats wastewater and which can be managed. One definition of a wastewater treatment plant operator is a "bug farmer", one who controls the aeration...