Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- An activated sludge tank is 30 m long and 10 m wide and has an SWD of 4 m. The wastewater flow is 4.0 ML/day and Compute: a) the BOD concentration which will be applied to the aeration tank b) the food-to-microorganism ratio for the system. O a) 105 mg/L b) F/M=0.35 O a) 195 mg/L b) F/M=0.65 O a) 105 mg/L b) F/M=0.22 O a) 195 mg/L b) F/M=0.35 the raw 5-day BOD is 300 mg/L. Assuming that 35 percent of the raw BOD is removed in the primary clarifier, and the MLSS concentration is 1000 mg/L.arrow_forwardNeed complete solution with explanationarrow_forwardThe 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 = 4/1arrow_forward
- The influent suspended solids concentration to a primary settling (clarifier) tank is 300 mg/L. The average flow rate is 0.037 m3/s. If the suspended solids removal efficiency is 65 percent, how many kilograms of suspended solids are removed in the primary settling tank each day?arrow_forwardA waste sludge flow of 40 m³/day is gravity thickened in a circular tank with a diameter of 6.8 m. The solids concentration is increased from 4.5% to 7.5% with 85% suspended- solids capture. Calculate the solids loading in kg/m² d and the quantity of thickened sludge in m³/d.arrow_forwardThe 750-bed Associated Center of Medical Excellence (ACME) Hospital has a small activated sludge plant to treat its wastewater. The average daily Medical Center discharge is 1,250 L per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide by 25.0 m long by 3.5 m deep. The plant operating parameters are as follows: MLVSS = 2,750 mg/L MLSS = 1.25 (MLVSS) Settled sludge volume after 30 min = 250 mL/LDetermine the following: aeration period, F/M ratio, SVI, solids concentration in return sludge.arrow_forward
- Coagulation aided sedimentation produces 135m³ of sludge per day. The solid content of sludge is 1% by weight and specific gravity of sludge solid is 2.7. If 20 my of suspended solids are removed from the raw water by the Al(OH)3 precipitate formed, and flow through the treatment plant is 40 MLD, then the amount of alum required per day in kg isarrow_forwardOnly find the BOD of the mixture.arrow_forwardA complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3)arrow_forward
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