Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Disinfection, sedimentation, flocculation, coagulation, and filtration are processes for a surface water treatment facility. Place each process in a sequential order.
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- A CMAS treatment system must be designed for a municipal wastewater and the following operating conditions: • Effluent flow of 0.39 m³/s leaving the primary clarifier Effluent soluble BOD; of 310 mg/L leaving the primary clarifier 20 mg/L of soluble BOD5 allowed in the effluent from the secondary clarifier Negligible suspended solids in the effluent from the secondary clarifier • Volatile suspended solids concentration of 4300 mg/L in the aeration tank • In the laboratory under log-growth conditions, the microorganisms used in the aeration tank can convert 100 g soluble BODS into 55 g VSS • Microorganisms in the CMAS system have a death rate constant of 0.05 d'. Sludge age in the CMAS treatment system will be 4 days (a) Determine the % treatment efficiency for soluble BOD; required in the CMAS system. (b) Determine the required volume of the aeration tank (in m²) and HRT (in h). (c) Determine the food-to-microorganism ratio (in d') and the specific substrate utilization ratio (in d').…arrow_forwardA continuous stirred tank reactor (CSTR) Activated Sludge Process is used to treat municipal wastewater after primary sedimentation. The following data is available: Primary Effluent Flow Rate (Q) Primary Effluent BOD Concentration (So) Secondary Effluent Biomass Concentration (Xe) Mixed Liquor Biomass Concentration (X) Solids Retention Time (SRT) Waste Sludge Flow Rate (Qw) Value 450 200 O 0 3,000 5 8 Biomass Concentration in RAS (XR) 8,000 O 250 million gallons, 0.28 gBOD/gVSS.d, 13.3 hours 107 million gallons, 0.28 gBOD/gVSS.d, 5.7 hours O 107 million gallons, 0.42 gBOD/gVSS.d, 5.7 hours 107 million gallons, 0.28 gBOD/gVSS.d, 12.2 hour Unit mgd mg/l mg/1 mg/l days mgd The bioreactor volume (V), F/M ratio, and HRT are respectively: mg/larrow_forwardThis question about Sanitary Engineering ........................ involve a physical change such as screening, grit removal, Separation, Crystallization, Evaporation, Filtration, & other reactions. • A. Unit operations • B. Unit processes • C. Both A and B • D. Nonearrow_forward
- A- An industrial wastewater stream flows at a rate of 13500 m³/d containing 100 mg/L. Suspended solids, ferric sulfate [Fe(SO4)] is used as a coagulant' at a dose of 50 mg/L. a) Assuming that there is very little in water, what is the daily lime dose? b) If the sedimentation tank removes 90% of the inlet solids, what is the daily solids production from the sedimentation tank?arrow_forwardAn activated sludge wastewater treatment system was designed and operated in a steadystate under the following conditions:• Reactor volume of 10,000 m3in total;• Sludge concentration (MLSS) of 3200 mg/L in the reactor;• Return sludge concentration at 8000 mg/L (the underflow of the secondary clarifier)• Each day, 400 m3 of underflow sludge (return sludge) was wasted as excesssludge;What is the average Sludge Residence Time (SRT)?arrow_forwardA complete mix activated sludge process treats 5 MGD of wastewater containing a total BOD5 concentration of 225 mg/L BOD5 to 20 mg/L BOD5, which is 75% soluble in both the influent and the effluent. The volume of the aeration basin is 111,400 ft3 and the MLSS concentration is 2500 mg/L. Determine the following: a) Detention time in the aeration basin (hr) b) F/M ratio c) Specific substrate utilization rate d) Substrate removal efficiencyarrow_forward
- A jar test reveals that the optimum alum coagulant dose is 50 mg/L at a pH of 8.1 and a temperature of 73 F. How many pounds per day of 49% alum are required for the whole plant? Total flow is 36 MGD with four parallel treatment plants, each designed for 12 MGD. Could use some guidance on setting up the equations.arrow_forwardIL Test sample weight= احتمان Sieve size Weight retained (g) Calculations: Sieve size in. (mm) 3/8" 9.5 mm No.4 4.75 mm No.8 2.36 mm No.16 1.18 mm No.30 0.6 mm No.50 0.3 mm No.100 0.15 mm Pan gm 3/8" No.4 No.8 No.16 No.50 No.100 0.15 mm 9.5 mm 4.75 mm 2.36 mm 1.18 mm mm 0.3 mm 0.0 од 1597 21419 від 50д 3589 174д Weight retained (g) Percent retained Beitel 15-11-2021 Fineness modulus отсивно о u 4/8²/00 = 5.3 15 діни з 2141-44871.3 61-448203 Божнизис 1.66 50 78.63 21.3 80.29 19.71 35 84421 23 92.22 7.78 174448518 98.02 1.98 чудд 4 /100 = Cumulative = No.30 Percent cumulative 0 53 94.7 766 23-4 Percent passing loo 222 Sieve deviation = 262 (5) ASTM Limit 100 95 100 80 100 50_85 25 60 10_30 2_10arrow_forward
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