1. 1 The town of Camp Verde has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 25.0 mg/L BOD, and 30 mg/L suspended solids. They have selected a completely mixed activated sludge system for the upgrade. The existing primary treatment plant has a flow rate of 2,506 m³/d. The effluent from the primary tank has a BODs of 240 mg/L. Using the following assumptions, estimate the required volume of the aeration tank: a. BOD5 of the effluent suspended solids is 70% of the allowable suspended solids concentration. b. Growth constant values are estimated to be: K, = 100 mg/L BODs; kd =0.025 d 1; μm-10 d1; Y= 0.8 mg VSS/mg BOD, removed. c. The design MLVSS is 3,000 mg/L.

8 Determine the solid loading rate on a secondary circular clarifier (diameter = 20 m) if the plant influent load is 2 ML/d. The recycle flow from secondary clarifier to the activated sludge system (aeration basin) is set to 10% of the influent load. It is also known that the aeration basin produces an MLSS (mixed liquor suspended solids) concentration of 3000 mg/L.

The aeration tank for a completely mixed aeration process is being sized for a designwastewater flow of 4500 m3/d. The influent COD is 150 mg/L. The design effluent COD is7 mg/L. Recommended design parameters are a sludge age of 10 days and MLVSS of 1400 mg/L. The expected Sludge Volume Index is 100 ml/g. The selection of these values takes into account the anticipated variations in wastewater flows and strengths. The kinetic constants from a bench-scale treatability study are Y = 0.60 mg VSS/mg COD and ke = 0.06 per day. Calculate:a. the hydraulic residence time and volume of the aeration tankb. the food/microorganism ratioc. the sludge production rated. the sludge recycle ratioe. the oxygen requirement

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 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…

A circular secondary clarifier will be used to treat an average daily flow (ADF) of 6,000 m3 /d from a high-purity oxygen activated sludge process with a MLSS concentration of 8,500 mg/L. The RAS flow is 2,500 m3 /d. Key design criteria include an overflow rate of 15 m3 /(d.m2 ) and a solids loading rate of 100 kg/(d.m2 ). Determine the minimum diameter (m) of the clarifier to achieve these criteria.

The City of Marietta has been directed by the City Board to upgrade its primary Wastewater Treatment Plant (WWTP) to include a secondary treatment process that can meet an effluent standard of 20 mg/L of BODs and 20 mg/L of suspended solids (SS). They want to select completely mixed activated sludge system. The BOD5 of the SS is estimated to be 50% of the SS concentration. Assume L:W within 4:1 and maximum depth of 4.5 m and maximum length of 50 m for the tank. Assume the reasonable values of missing parameters from FE Exam Handbook. The effluent characteristics from the primary treatment process are: Flow = 0.150 m'/s, BOD5 = 75 mg/L. Assume the values for the growth constants as K, = 100 mg/L BOD5, Mm = 2.5 d-1, kg = 0.05 d'1. Y = 0.50 mg VSS/mg BODs removed, MLVSS = 2,800 mg/L, MLSS - 1.20 (MLVSS), wastewater temperature = 25.0°C. Assume 8% 02 transfer rate and BOD5 to BODu ratio as f = 68%. If SRT (qc) = 5.65 days, the volume of the aeration tank is nearly.

The aeration tank for a completely mixed aeration process is being sized for a design wastewater flow of 4500 m³/d. The influent COD is 150 mg/L. The design effluent COD is 7 mg/L. Recommended design parameters are a sludge age of 10 days and MLVSS of 1400 mg/L. The expected Sludge Volume Index is 100 ml/g. The selection of these values takes into account the anticipated variations in wastewater flows and strengths. The kinetic constants from a bench-scale treatability study are Y = 0.60 mg VSS/mg COD and ke = 0.06 per day. Calculate: a. the hydraulic residence time and volume of the aeration tank b. the food/microorganism ratio c. the sludge production rate d. the sludge recycle ratio e. the oxygen requirement

Assume that taal lake is 100m2 and its average height is 5ft. Wastewater thatflows into the lake are the following: a. Food square kitchen waste 5m3/hr; Total Dissolved Solids (TDS) is 50g/L whichoperates 8hrs per dayb. CIH 5m3/hr kitchen waste; TDS is 20mg/L which operates 4hrs per dayc. PCH 20m3/day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volume of wastewater which taal lake can still be contained withoutrainfall.(2)During operational hours a heavy rain fall, the rain gauge reading is 1.5m. Will the taalLake overflow? What is the volume of water that will over flow?(3). Compute the TDS concentration of the overflow.

The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The aeration time is nearly?

For an activated sludge basin with a flow rate of 500 L/s if the influent BOD5 = 100 mg/L, the MLVSS = 3000 mg/L and the volume of the aeration basin = 766100 gallons Find the F/M ratio If the WAS  X= 10000 mg/L, find the WAS flowrate (m3/d)