for A W.W.T.P of daily flow 30,000 m³ with raw BOD = 300 ppm and return sludge = 25% of sewage flow Design Aeration tanks if organic load dosent exceed 540 gm/m³ /d %3D
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- After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the aeration volume (m³). Parameter Value 3d¹ 60 mg/L BODs 0.10 d¹ 0.6 mg VSS/mg BOD5 Hmax Ks ka YProblem 2.Assume that DLSUD lake is 100m² and its average height is 5ft. Wastewater that flows into the lake are the following: a. Food square kitchen waste 5m³/hr; Total Dissolved Solids (TDS) is 50g/L_which operates 8hrs per day b. CIH 5m³/hr kitchen waste; TDS is 20mg/L which operates 4hrs per day c. PCH 20m/day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volume of wastewater which DLSUD lake can still be contained without rainfall. (2)During operational hours a heavy rain fall, the rain gauge reading is 1.5m. Will the DLSUD Lake overflow? What is the volume of water that will over flow? (3). Compute the TDS concentration of the overflow.Problem 2.Assume that DLSUD lake is 100m? and its average height is 5ft. Wastewater that flows into the lake are the following: Food square kitchen waste 5m³/hr; Total Dissolved Solids (TDS) is 50g/L which operates 8hrs per day b. CIH 5m³/hr kitchen waste; TDS is 20mg/L which operates 4hrs per day c. PCH 20m³/day diluted laboratory waste ; TDS = 40mg/L а. (1)Determine the volume of wastewater which DLSUD lake can still be contained without rainfall. (2)During operational hours a heavy rain fall, the rain gauge reading is 1.5m. Will the DLSUD Lake overflow? What is the volume of water that will over flow? (3). Compute the TDS concentration of the overflow.
- X 1. One million gallons per day of municipal wastewater with a BOD5 of 260 mg/L and suspended solids of 220 mg/L are processed in a primary plus secondary activated- sludge plant. If the primary removal efficiencies for the BOD5 and TSS are 25% and 70%, respectively, then, the total sludge (in kg/day) which is removed by the primary clarifier every day isa. The wastewater from a manufacturing treatment plant is discharged to a municipal `treatment plant for processing with domestic wastewater. The characteristics of the wastewater from the manufacturing treatment plant are 3500 mg/l BOD, 4500 mg/l SS, 70 mg/l nitrogen and 2 mg/l phosphorous. The characteristics of the domestic wastewater are 350 mg/l BOD, 285 mg/l SS, 50 mg/l N and 11 mg/l P. If the required BOD/N/P weight ratio is 100/5/1, what is the minimum quantity of domestic wastewater required per 1000 gallons of manufacturing wastewater to provide adequate nutrients for biological treatment and the concentration of SS in the combined wastewater? b. What additional of NH4OH and H3PO4 would be required in pounds per million gallons of wastewater to provide the necessary nutrient levels for the treatment of the manufacturing wastewater described above (a) if the domestic wastewater had not been combined with it.The characteristics of the synthetic textile wastewater are 1500 mg/L BOD, 2000 mg/L SS, 30 mg/L nitrogen, and no phosphorus. The characteristics of the domestic wastewater are 200 mg/L BOD, 240 mg/L SS, 35 mg/L nitrogen, and 7 mg/L phosphorus. Find the nitrogen concentrations in the mixture. The ratio of flow rate is domestic/textile = 4/1
- An industrial wastewater discharged at a flowrate of 1000 m³/d having 220 mg/l initial concentration of total suspended solids (TSS). If this wastewater is to be treated individually by one of the following processes: 1. Primary settling tank which should remove 60% of TSS. 2. Coagulation unit using 40 g ferric chloride (FeCl3) as a coagulant /m²³ of wastewater. The coagulation unit will remove 85% of TSS. Estimate the mass and volume of the sludge generated from each suggested unit. Assume the following sludge specification: Source of sludge Primary settling tank Coagulation tank Specific gravity 1.03 1.05 Moisture content 94% 92%A municipal wastewater treatment plant processes an average of 14,000 m3 / day. The peak flow is 1.75 times the average. The wastewater contains 190mg/L BOD and 210 mg/L suspended solids at an average flow and 225 mg/L BOD and 365 mg/L suspended solids at peak flow. Determine the following for a primary clarifier with a 20-m diameter. Assume eff. BOD of 20 mg/L and effluent TSS of 30mg/L for average and peak flows. 1. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ average flow. 2. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ peak flow. 3. Mass of solids (kg/day) that is removed as sludge from average and peak flow conditions.The wastewater from a manufacturing treatment plant is discharged to a municipal 'treatment plant for processing with domestic wastewater. The characteristics of the wastewater from the manufacturing treatment plant are 3500 mg/I BOD, 4500 mg/l SS, 70 mg/l nitrogen and 2 mg/l phosphorous. The characteristics of the domestic wastewater are 350 mg/l BOD, 285 mg/l SS, 50 mg/I N and 11 mg/l P. If the required BOD/N/P weight ratio is 100/5/1, what is the minimum quantity of domestic wastewater required per 1000 gallons of manufacturing wastewater to provide adequate nutrients for biological treatment and the concentration of SS in the combined wastewater?
- i). A town produces 8 million litre of sewage per day. BOD of sewage is 260 mg/L, it is to be discharged into a river flowing at a rate of 0.2 m³/s and it's BOD is 5 mg/L. Find out the BOD of river after mixing? [Assume mixing is instantaneous]Design an activated sludge system for the wastewater flowing at the average rate of (10650)GPM having (10650/500)mg-BOD/L. Also determine the amount of air (m3/min), power (kW) and major nutrients (kg/day) required by the process for 80% BOD removal. Assume any missing data. Subject: environmental engineering 2A 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 BOD5 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 m3 /s soluble BOD5 = 80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50% of the suspended solids concentration, the microorganism concentration in the aeration tank (MLVSS) is 2000 mg/L, and the net activated sludge produced is 400.0 kg/d; also assume the following values for the growth constants: bacterial decay rate = 0.10 day-1 , yield coefficient = 0.60 mgVSS ∙ mg-1 BOD5 removed, maximum specific growth rate constant = 2.50 day-1 , and half saturation constant = 100 mg/L. (1) Estimate the volume of the aeration tank; (2) Compute the F/M ratio; (3) If the BOD5 is 60% of the ultimate…