problem 4:A wastewater treatment plant with an average daily flow of 0.25 m³/s must meet 30 mg/L of totalBODs and 25 mg/L of suspended solids standards. The suspended solids is 65% BODs. Thewastewater is treated in an activated sludge plant. Assume that the soluble BODs after theprimary clarifier is 115 mg/L and the concentration of the microorganism in the aeration tank is2500 mg/L. Use K-60 mg/L; ka=0.06 d¹; m = 5 and Y = 0.6. Determinea) The concentration of the soluble BODs in the effluent of the aeration tankb) The mean cell residence timec) The hydraulic retention timed) The required volume of the aeration basin

Part e) is not in your question but if you want it then only write otherwise leave that.

Answered: problem 4: A wastewater treatment plant…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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An 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.4,500
You have designed a treatment system for contaminant Z. The treatment system consists of two parts: (1) a pipe with a volume of 15 m3 and (2) a CSTR has a volume of 60 m3. The pipe feeds into the CSTR. The inflow to the pipe is 10 m3/d and the influent Z concentration at the start of the pipe is 2,500 mg/L. The effluent from the pipe has a Z concentration of 1500 mg/L. Assume the reaction is first order. What is the reaction rate constant in units of (1/day)?
Chemistry 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?
A completely mixed activated sludge plant treats wastewater at a rate of 2,565 m³/d. The influent BOD is 280 mg/L, and the MLVSS in the aeration tank is 2,500 mg/L. The F/M ratio for the aeration tank is 0.35 kg BOD/kg MLVSS.d. Assuming 30% of BOD is removed in the primary tank, determine the organic loading rate in kg BOD/m³.d and the volume of the aeration tank in m³.
In a municipal wastewater treatment plant, the influent flow rate into a complete mixing activated sludge system is 50 L/sec, its solid concentration MLSS is 30 mg/L, and the BODs in the influent is 245 mg/L. The MLSS concentration at the bottom of the secondary clarifier is 15,000 mg/L and 10 mg/L at the top of the |clarifier. The flow rate of returned activated sludge is 20 L/sec. a. What is the concentration of the MLSS in the activated sludge reactor (assuming that it is fully mixed and has constant concentration)? b. What is the activated sludge basin volume if the hydraulic retention time in the basin is 10 hours and the system is operated at the R/Q ratio given in the problem? |c. If the cross section of the basin is 20 m² (perpendicular to the flow), how long the basin will be? d. The plant has 96% BOD5 removal. What is the effluent BOD5? e. The flowrate of the waste activated sludge (WAS) is 0.2 L/sec. How many solids (in kg) are wasted every day through the secondary…
A wastewater treatment plant uses an activated sludge process to treat 0.2 m3/s of primary effluent. The concentration of the mixed liquor is 2,000 mg/L MLVSS (Mixed liquor volatile suspended solids), while the RAS (recycled activated sludge) concentration is 10,000 mg/L MLVSS. The substrate concentration in the primary effluent is 210 mg/L BOD5. The F/M ratio is 0.5 mg BOD5/(mg MLVSS-d) and the solids retention time is 8.5 d. What is the volume of the activated sludge tank? 2.What is the WAS (waste activated sludge) flowrate? 3.What is the flowrate of the secondary treated effluent? 4.What is the RAS (recycled activated sludge) flowrate? 5.What is the hydraulic residence time for the activated sludge tank?
A coagulation treatment plant with a flow of 0.5 m/s is dosing alum at 23 mg/litre. No other chemicals are being added. The raw water suspended solid concentration is 37 mg/l. The effluent suspended solid concentration is 12 mg/l. Sludge content is 1% and specific gravity of sludge is 3.01. What volume of sludge must be disposed of every day.
The PEF, Inc. firm desires to utilize a complete-mix activated-sludge system with recycle to treat their municipal wastewater after primary sedimentation. The design parameters are: Flow rate=0.92m³/s Influent Soluble BOD5 = 292mg/L MLVSS=2,500mg/L Mean cell residence time = 7.24 d For any missing parameters, use typical values of growth constants for domestic wastewater (provided in your text, page 494). Assume that the solids lost in the secondary clarifier effluent are minor and can be neglected. Assuming a BODs (S) concentration of 24 mg/L, the detention time (residence time or theta) in the activated sludge system is most nearly: O 28 hours O 6.5 hours O 168 hours O 53 hours
Example (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 47
The BOD just downstream from a wastewater pipe is 12.9 mg/L and the DO (dissolved oxygen) is 8.6 mg/L. The saturated value of dissolved oxygen is 9.6 mg/L, the deoxygenation constant (kd) is 0.21/day, the reaeration constant (kr) is 0.42. Find the time where oxygen deficit is at a maximum and find the minimum value of DO.
A secondary treatment using a mixed water activated sludge system, have the flow rate of 10000m^3/day of municipal wastewater. BOD after primary clarification is 150mg/L.It is desired to not reach 5mg/L of soluble BOD in effluent. The kinetic values of the system having X=0.5kg/kg and Kd=0.05d^-1.MLVSS concentration in the secondary treatment is 3000 mg/L, the concentration in return sludge is 10000mg/L. Calculate the volume of solid that must be wasted each day for each of the selected sludge.
A secondary treatment using a mixed water activated sludge system, have the flow rate of 10000m^3/day of municipal wastewater. BOD after primary clarification is 150mg/L.It is desired to not reach 5mg/L of soluble BOD in effluent. The kinetic values of the system having X=0.5kg/kg and Kd=0.05d^-1.MLVSS concentration in the secondary treatment is 3000 mg/L, the concentration in return sludge is 10000mg/L. Calculate the mass of solid that must be wasted each day for each of the selected sludge.

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