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. Thedesign 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 takesstants from a bench-scale treatability study are Yinto account the anticipated variations in wastewater flows and strengths. The kinetic con-0.60 mg VSS/mg soluble BOD andk₁ = 0.06 per day. Calculate the volume of the aeration tank, aeration period, foodganism ratio, and excess biomassth

The objective of the question is to calculate the volume of the aeration tank, aeration period, food…

Answered: 12.60 The aeration tank for a…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

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Problem 1P

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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
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
A 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).
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 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.
The results of a jar test on a surface water supply sample revealed that the optimum dose of alum (Al2(SO4)3.14H2O) as coagulant is 17 mg/L. Furthermore, the water has total alkalinity = 50 mg/L as CaCO3; pH = 7.8; and residual turbidity at optimum alum dose = 4 NTU. The water treatment plant has a mean flow of 10m3/hour. Determine: The kg of alum per year needed in the operation of the plant. The residual alkalinity in the sample, following the addition of alum at optimum dose. Solve parts (a) and (b) if instead of alum, ferric chloride (FeCl3) were used as coagulant. Assume the numerical jar test results are the same as those with alum.
The town of Gatesville has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30.0 mg/L BOD5 and 30.0 mg/L suspended solids (SS). They have selected a completely mixed activated sludge system. Assuming that BOD5 of the SS may be estimated as equal to 57 percent of the SS concentration, estimate the required volume of the aeration tank. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow =0.150 m3/s BOD5 =82.0 mg/L Assume the following values for the growth constants: Ks= 100 mg/L BOD5; µm =2.5 d-1; kd =0.050 d-1; Y = 0.50 mg VSS/ mg BOD5 removed. Assume the concentration of MLVSS is 2000 mg/L.
A secondary treatment using a mixed water activated sludge system, having 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.By choosing two optimum sludge age, find the volume of the reactor.
A secondary clarifier handles a total flow of 9600 m³/d from the aeration tank of a conventional activated-sludge treatment system. The concentration of solids in the flow from the aeration tank is 3000 mg/l. The clarifier is required to thicken the solids to 12000 mg/l, and hence it is to be designed for a solid flux of 3.2 kg/m²h. The surface area of the designed clarifier for thickening (in m², in integer)
The aeration tank for a completely mixed aeration process is being sized for a designwastewater flow of 4500 m3/d. The influent COD is 150 mg/L. The design effluent COD is7 mg/L. Recommended design parameters are a sludge age of 10 days and MLVSS of 1400 mg/L. The expected Sludge Volume Index is 100 ml/g. The selection of these values takes into account the anticipated variations in wastewater flows and strengths. The kinetic constants from a bench-scale treatability study are Y = 0.60 mg VSS/mg COD and ke = 0.06 per day. Calculate:a. the hydraulic residence time and volume of the aeration tankb. the food/microorganism ratioc. the sludge production rated. the sludge recycle ratioe. the oxygen requirement
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?
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 F/M ratio is nearly?

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