I need the answer as soon as possible An activated sludge treatment plant treats a sewage at a flow of 3*10^6I/day with a BOD of 350 mg/I, The dimension of the aeration tank is 25 mlong, 10 m width and 4 m depth. The concentration of the mixed liquorsuspended solids (MLSS) is 2500 mg/l. Calculate the F/M ratio.O 0.55О 0.43O 0.29О 0.15

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Answered: An activated sludge 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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The shrimp processing plant generates 0.012 m3/s of wastewater each day. The wastewater is treated in an activated sludge plant. The average BOD5 of the raw wastewater before primary settling is 1400 mg/L. The aeration tank has effective liquid dimensions of 8.0 m wide by 8.0 m long by 5.0 m deep. The plant operating parameters are as follows: Soluble BOD5 after primary settling = 966 mg/L MLVSS = 2000 mg/L MLSS = 1.25 (MLVS) Settled sludge volume after 30 min = 225 mL/L Aeration tank liquid temperature = 15°C Determine the following: Return Activated Sludge (RAS) flow rate and Wasting Activated Sludge (WAS) flow rate.
A coagulation treatment plant with a flow of 0.5 m3/s is dosing alum at 20 mg/l. Concentration of suspended solid is 25 mg/l. The effluent suspended solid concentration is 12 mg/l. The sludge content is 1 % and specific gravity of sludge is 3 %. What volume of sludge must be disposed of each day?
Determine the Aeration Basin Dimensions for a town of 200,000 population. Mixed Liquor Suspended Solids is 3500 mg/L. The BOD Loading rate is 0.48 kg/ (?3 –day), Sludge Volume Index is 120 mL/g; Return sludge is 35%; the concentration of BOD is 0.093 Kg/(Capita –day). Primary sedimentation tank removes 35% of BOD.
As an operator of a wastewater treatment facility, you have learned that your plant runs well at an SRT (based on MLVSS (X) in the aeration basin) of 12 days. If today’s MLVSS (X) concentration is 2,700 mg/L and the RAS (Xr) concentration is 5,500 mg/L, how many L of sludge should you waste from the secondary clarifier today? Other design conditions: Average influent flow = 12,000 m3/d Influent soluble BOD = 150 mg/L Required effluent TSS = 30 mg/L Aeration tank Volume = 1,000 m3
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 wastewater treatment plant will receive a flow of 30,000 m3/day. The plant has a 3.6 m deep circular primary clarifier with a surface overflow rate of 60.0 m3/m2-day.a. Calculate the surface area (m2), diameter (m), volume (m3), and hydraulic retention time of the primary clarifier.b. How long will it take a 1.0 mm particle with a density of 1500 kg/m3to settle in the primary clarifier? show all work
How to determine the needed surface area of the secondary clarifier tank given that the flow is 40,000 m3 per day and the recycling ratio is 0.5 . The necessary solids concentration in the underflow is 13,000 mg/L, while the Mixed liquor suspended solids concentration in the activated sludge is 4,500 mg/L. The table below shows the settling properties of sludge: Mixed liquor suspended solids Mg/l Settling velocity m/hr 2000 6 4000 3 6000 1 8000 0.2 10000 0.1 13000 0.1
A sedimentation tank of dimensions 33m x 12m x 6m is filtering wastewater at a flow rate of 0.3 m3/s. The particles present in the wastewater have a density of 2.4 g/cm3 and the dynamic viscosity of water is 0.01 g/(cm.s). Calculate the removal efficiency of the tank for particle size of 25mm.
The density of water at 4°C is 1000 kg/m3 at atmosphericpressure. The gravity acceleration constant (g) is 9.81 m/s2. Thespecific gravity of an industrial wastewater (SGww) is 1.32. Whatis the specific weight of this wastewater?
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)
It is desired to design an activated sludge process to treat 20,000 m3/d of settled wastewater with a BOD of 250 mg/L and nbVSS concentration of 10 mg/L. If the effluent BOD is not to exceed 20 mg/L, calculate the volume of the aeration tank required, sludge wastage rate, sludge return rate, oxygen requirements, F/M ratio, specific utilization rate, specific growth rate, and maximum specific growth rate. The following data and coefficients are applicable. i. Mean cell residence time = 5 daysii. Y = 0.5 gVSS/g BODiii. Kd = 0.10 /div. K = 5 /dv. Return sludge concentration = 8,000 mg/L VSSvi. Mixed liquor volatile suspended solids concentration = 2500 mg/Lvii. Effluent contains 20 mg/L suspended solids of which 65% are biodegradable.viii. Fraction of biomass remaining as cell debris = 0.15ix. BOD5/BODu = 0.68
Downvote for incorrect solution..A domestic wastewater contains 350 mg/L of suspended solids. Primary sedimentation facilities remove 65%. Approximately how many cubic meters of sludge containing 5% solids will be produced per million cubic meters of wastewater settled?

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