Determine the underflow volume of sludge wastage An activated sludge process (ASP) is designed for secondary treatment of 7500 m³/dayof municipal wastewater. After primary claritier, the ultimate BOD of the influent, whichenters into ASP reactor is 200 mgt. Treated effluent after secondary clarifier is requiredto have an ultimate BOD of 20 mg. Mix liquor volatile suspended solids (MLVSS)concentration in the reactor and the underflow is maintained as 3000 mg and 12000 mg/L.respectively.y. The hydraulic retention time and mean cell residence time are 0.2 day and10 days, respectively. A representative flow diagram of the ASP is shown below.Inflent afterPrimary carASPReturn foSecondary TreatedclarifierUnderflowShudge wastage

Given: Municipal waste water discharge Hydraulic retention time Mix Liquor Volatile Suspended Solids…

An activated sludge process (ASP) is designed for secondary treatment of 7500 m³/day of municipal wastewater. After primary claritier the ultimate BOD of the influent, which enters into ASP reactor is 200 mgt. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg and 12000 mg/L. respectively. T y. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below. Influent after Primary car ASP Return fo Secondary Treated clarifier Underflow Sludge wastage

An activated sludge process (ASP) is designed for secondary treatment of 7500 m³/day of municipal wastewater. After primary claritier the ultimate BOD of the influent, which enters into ASP reactor is 200 mgt. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg and 12000 mg/L. respectively. T y. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below. Influent after Primary car ASP Return fo Secondary Treated clarifier Underflow Sludge wastage

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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Q/ A wastewater treatment use an activated sludge process for secondary treatment of 10000 m/day for municipal wastewater. After primary clarification, The BOD is 150 mg/L, and it is desired to have not more than 7 mg/L of soluble BOD in the effluent. The mixed liquor has a concentration of 3000 mg/L and underflow solids concentration is 12,000 mg /L from the secondary clarifier. The kinetic constants from a bench scale treatability study are Y=0.5 and K=0.05 day". Assume the cell residence time is 10 days. Determine: a. The volume of the activated sludge tank & the hydraulic residence time. b. The sludge production rate & the sludge wasting flow rate. c. The sludge recycle ratio and recycle flow rate. d. The organic loading rate.
The waste water 1MLD with 500 mg/L 5-day BOD is to be treated by activated sludge process (ASP). The desired effluent BOD is 30 mg/L. (a) Volume of aerator (b) If the ASP is designed as completely mixed reactor without sludge recycle, determine the hydraulic detention time and biological solids retention time of the unit. (c) Draw the flow diagram of this process. Assume BOD loading rate of aeration tank to be 0.5 kg of BOD removed/kg of MLSS/ day. Concentration of mixed suspended solids (MLSS) in aeration tank is 2,000 mg/L.
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
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 complete-mix flow reactor is designed to treat an influent waste stream containing 130mg/L of casein at a flow rate of 80 liters per minute (lpm). Assume that casein removal follows first-order removal kinetics with a rate constant, k, of 0.5h-1 and that the effluent should contain 13 mg/L of casein at steady state. Determine: a. the detention time in hours b. the volume of the reactor in cubic meters
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.
1. A CM reactor receives influent containing 10 mg/L of tracer dye for 2 hours. The tracer addition is terminated but the flowrate of water is maintained. The volume of the reactor is 10 L and the flowrate is 2 L/hr. What is the concentration of tracer in the reactor 1 hr after the dye addition is terminated? The reactor had an initial concentration of tracer of 1 mg/L when the tracer addition commenced.
The volume of a CFSTR is 45 m' and average constant flow in the reactor is 300 m /d. The reactor flow regime was established by feeding continuously 6 g/L stock tracer solution in the influent line at a rate of 125 L per day. Determine the theoretical tracer concentration in the effluent at 3 and 10 h after the tracer injection. Plot the tracer profile as a function of (a) concentration versus t, and (b) C/Co versus t/0.
(D) Sludge in the secondary tank after Activated sludge is the (A)Aerated sludge in the aeration unit. (B) Sludge settled in the humus tank. (C) Sludge in the secondary tank after aeration and rich in microbial masses. (D)Sludge in the secondary tank after aeration and rich in nutrients.
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
4 An activated sludge system is to be used for secondary treatment of 10,000 m³/d of municipal wastewater. After primary clarification, the BOD is 150 mg/L, and it is desired to have not more than 5 mg/L of soluble BOD in the effluent. A completely mixed reactor is to be used, and pilot-plant analysis has established the following kinetic values: Y = 0.5 kg/kg, k, = 0.05 d-¹. Assuming an MLSS concentration of 3000 mg/L and an underflow concentration of 10,000 mg/L from the secondary clarifier (MLVSS=0.8MLSS) determine (1) the volume of the reactor, (2) the mass and volume of solids that must be wasted each day, and (3) the recycle ratio.