Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- Note:hand written solution should be avoided.arrow_forwardAn aeration basin for an activated sludge facility has the following characteristics Length = 90 ft Width = 30 ft Liquid depth = 12 ft MLSS=4,000 mg/L The raw wastewater has the following characteristics: Flow = 12 MGD BOD, = 220 mg/L Suspended solids (SS) = 250 mg/L A primary clarifier is ovided that removes 25% of the BOD, an 60% of the suspended solids The return activated-sludge flow rate is 0.8 MGD If the primary sludge solids content is 4% solids and the specific gravity is 1.0, the primary sludge volume (ft³/day) is most nearly OA. 1,500 OB. 60,000 O C.600 O D.4,500arrow_forwardExample (Activated Sludge Process) - Completely Mixed with Recycle A completely mixed activated sludge process, treating a municipal wastewater, has a primary clarifier effluent BOD, of 135 mg/L. The design MLSS is 3590 mg/L, and the MLVSS is 80% of the MLSS. The plant permit is for an effluent BOD, of 20 mg/L and 20 mg/L suspended solids. The effluent suspended solids have a BOD, of 0.65 mg BOD, / mg suspended solids. The underflow solid concentration (MLVSS) is 10,000 mg/L. If umax = 3.0 day, Ks 60 mg/L, Y= 0.50 mg MLVSS / mg BOD, removed, ka= 0.1 day, and the influent flow is 0.25 m/sec, determine the following: 1. The required reactor detention time, 0. 2. The reactor basin volume, V, in m³. 3. The net waste activated sludge produced each day (kg/day). 4. The mass of solids (both volatile and inert) to be wasted each day (kg/day). The return sludge flow rate the recycle ratio. 6. The F/M ratio. 7 June 2021 Dr. Dawood Eisa Sachit Biological Wastewater Treatment 47arrow_forward
- 12.60 The aeration tank for a completely mixed aeration process is being sized for a design waste- water flow of 7500 m³/d. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/l. The design effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design para- meters are a sludge age of 10 d and volatile MLSS of 1400 mg/l. Selection of these values takes stants from a bench-scale treatability study are Y into account the anticipated variations in wastewater flows and strengths. The kinetic con- 0.60 mg VSS/mg soluble BOD and k₁ = 0.06 per day. Calculate the volume of the aeration tank, aeration period, food ganism ratio, and excess biomass tharrow_forwardExercise 4 A treatment plant has the following process diagram. Determine the flow rate expression for each unit. The inflow is Qi. The sludge waste flowrate is Qw. The return flowrate rate is Qr. The mixed liquor return flowrate is Qm. Mixed-liquor return Secondary clarifier Effluent Influent- Anoxic Aerobic Anoxic Aerobic Return activated sludge (e) Bardenpho™ (4-stage) Sludge Exercise 5 Derive the ODE system for above process. There are two species involved in the reaction, as shown below. The process rate for each tank can be simply represented by r1, 12,...15. The gravitational sedimentation is only applied to species X in the clarifier and can be represented by a reaction rate of r5 = 3.5X5. The influent are Xi and S₁. S=X.arrow_forwardwet sludge contains 3% solid by weight is pumped to a filter belt press at the rate of 170,000lb/day. the dewatered sludge is 18% by dry weight and the filtrate has total nitrogen concentration of 100 mg/l as nitrogen. assume the belt press has 100% solid capture and specific gravity of the sludge in all forms is 1. Calculate the amount of nitrogen (lb/day) in the filtrate returned? a) 4 b) 9 c) 14 d) 19arrow_forward
- A 14 mgd wastewater treatment plant has a primary clarifier that treats an influent with 800 mg/L of solids and a removal efficiency of 55%. Sludge at the bottom of the tank is pumped at a rate of 0.1 mgd. 1. What is the effluent solids concentration from the clarifier (in mg/L)? What is the solids concentration in the sludge at the bottom of the tank (in mg/L and %)? The primary sludge is further processed in a digester following treatment. What weight of solids (in lbs) need to be handled each week? а. b. с.arrow_forwardChemistry A sewage treatment plant discharges 4 ML/d of effluent with 20.17 mg/L BOD5 into a stream. The stream discharge is 12 ML/d, and its initial BOD5 is 6.10 mg/L. The initial DO of the wastewater is 2.10 mg/L and in the stream, it is 7.91 mg/L. Assume the stream velocity is 0.31 m/s and the dissolved oxygen saturation concentration is 8.76 mg/L. The kd = 0.15 d-1 and kr = 0.31 d-1 values have been adjusted for temperature. Calculate the minimum DO in the stream and determine the distance (km) downstream that this minimum occurs. Is this a safe level for fish?arrow_forwardThe 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 F/M ratio is nearly?arrow_forward
- 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?arrow_forwardwet sludge contains 3% solid by weight is pumped to a filter belt press at the rate of 170,000lb/day. the dewatered sludge is 18% by dry weight and the filtrate has total nitrogen concentration of 100 mg/l as nitrogen. assume the belt press has 100% solid capture and specific gravity of the sludge in all forms is 1. Calculate the amount of nitrogen (lb/day) in the filtrate returned? Please explain with details a) 4 b) 9 c) 14 d)19 The answer is one of these optionsarrow_forwardA completely mixed activated sludge process is being designed for a wastewater flow of 3 MGD using kinetic equations. The influent BOD of 180 mg/L is essentially all soluble and the design effluent soluble BOD is 10 mg/L. The mean cell residence time is selected to be 8.0 days for sizing the aeration volume tank, and the MLVSS is 2500 mg/L. The kinetic constants from a bench-scale treatability study are Y = 0.60 Ib VSS/Ib BOD and Kd = 0.06 day-1. ---Determine the efficiency of soluble BOD removal. For this, you will have to recognize substrate concentration in the coming flow (So) and the concentration of substrate in the exiting flow (Se).arrow_forward
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