A 2-tank Step Aeration Activated Sludge (SAAS) system has the followingcharacteristics:V1V2QWasteRecycleOxygendemand(L)(L)(L/min)RatioRatioof Cells500,000450,0001000.450.51.2Xv1Xv2Xd1Xd2SOS1S2XrvXrd(mg/L) (mg/L) (mg/L) (mg/L) (mg/L) (mg/L) (mg/L)(mg/L)(mg/L)1500 ☐ 141400 200 150150102200010001. What is the mean cell residence time and hydraulic residence time in days?2. Write the mass balance equation for required oxygen in tank #1.3. What is the oxygen requirement for tank #1 in kg/day? (equation will be in units ofmg/min)

Step 1:Mean Cell Residence Time and Hydraulic Residence Time: Given…

Answered: A 2-tank Step Aeration Activated Sludge…

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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N 5)( rat ala SIL O 23-A) An activated sludge system is to be used for treatment of 10000 m/day of municipal wastewater. A rimary clarification, the value of BOD, is 150 mg/l, and it is desired to have not more than 5 mg/l. in. ffluent. Assume that Y- 0.5; K, 0.05/day; and ML.VSS-3000 mg/l; Return sludge concentration 10 ng/L. 0 10 days. Determine: Treatment efficiency and the volumetric organic loading. 3) Microorganism ratio 5) The quantity of sludge that must be wasted each day i 2) Reactor volume and detention time. 4) the recycle ratin
design an aeration tank of CASP based on the following data influent flowrate is 100000m3/d, influent CODcr is 200mg/l influent TKN is 40mg/l effluent is CODcr Is 20mg/l. required by Local regulation. Please answer very soon will give rating surely
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 th
2. A scrubber separator is used to remove sulfur dioxide (SO2) from a gas stream. A 0.5 m'/sec inlet stream (350 K, 1 atm) contains 1.2 ppmv SO2. Find the required removal efficiency if the cleaned outlet stream must contain no more than 240 µug/scm.
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
Question 30 In activated sludge system, SRT is governed by O HRT sludge wastage aeration
Q- ASP is used to treat a waste water flow of 2MLD having BOD5 of 250 mg/lit. The biomass concentration in the aeration tank is 2500 mg/lit and the concentration of biomass leaving the system is 80 mg/lit, the aeration tank has a volume of 200m³. Find the hydraulic retention time HRT.
The sludge age 1. Is a measure of the time taken for sewage to stay in the bioreactor 2. Is one of the major parameters that controls the quality of the effluent 3. Determines the amount of solids to be wasted from the system 4. Is unaffected by the amount of solids in the effluent O 1, 2 and 4 O 1, 2 and 3 O 2 and 3 O 1 only Clear my choice
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.
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
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A 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 efficiency

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