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.

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.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

What aerobic reactor volume and operating SRT (8) are required to treat a 20 MGD domestic wastewater flow from an initial ultimate BOD - 300 mg/L to an effluent BODu 5 mg/L assuming a complete mix activated sludge system? Use the design information below assuming 1st order kinetics. MLVSS = 3,000 mg/L - X Y-05 K=0.1 L/mg'day (1st order degradation rate constant) K₂ -0.1 day 1 Answers are given as Aeration Tank Volume, Operational O 0.1 MG, 2.5 days O 19.65 MG, 20 days O 3.93 MG, 6.7 days 5.2 MG, 5.1 days
- An aerobic digester with a total capacity of 50,000 ft³ stabilize waste-activated sludge from an extended-aeration treatment plant without primary clarification. The average daily sludge flow pumped to the digestion tanks is 32,000 gpd with 1.5% solids, of which 65% are volatile. Calculate a) sludge detention time in the digesters b) the amount of sludge produced daily and c) the anaerobic digester dimensions. The population being served is 75,000. SS and BOD = 350 mg/l. %3D
An anaerobic digester is operated with a hydraulic retention time (HRT) of 25 d and a flow rate of 0.85 m³/d. The influent substrate concentration is 18 g/L and the digestion process is assumed to follow first order kinetics with a Ko value of 0.09 d'¹. Calculate: (a) the liquid volume of the anaerobic digester, V (m³) (b) the effluent substrate concentration, So (g/L)
The concentration of the mixed liquor suspended solids, MLSS, in the basin of a step aeration system is 2300 mg/L, and the volume of settled sludge after 30 minutes in a 1 L graduated cylinder is 200 mL. Calculate the SVI in mL/g.
Q1: The design flow for a water treatment plant (WTP) is 1 MGD (3.8x10 m³ /d). The rapid mixing tank will have a mechanical mixer and the average alum dosage will be 30 mg/L. The theoretical mean hydraulic detention time of the tank will be 1 minute. Determine the following: a) the quantity of alum needed on a daily basis in kg/d, b) the dimensions of the tank in meters for a tank with equal length, width, and depth, c) the power input required for a G of 900 sec' for a water temperature of 10 °C. Express the answer in kW, given that u = 1.307 × 10³Kg/m.s.
An activated sludge wastewater treatment system was designed and operated in a steadystate under the following conditions:• Reactor volume of 10,000 m3in total;• Sludge concentration (MLSS) of 3200 mg/L in the reactor;• Return sludge concentration at 8000 mg/L (the underflow of the secondary clarifier)• Each day, 400 m3 of underflow sludge (return sludge) was wasted as excesssludge;What is the average Sludge Residence Time (SRT)?
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
The 750-bed Associated Center of Medical Excellence (ACME) Hospital has a small activated sludge plant to treat its wastewater. The average daily Medical Center discharge is 1,250 L per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide by 25.0 m long by 3.5 m deep. The plant operating parameters are as follows: MLVSS = 2,750 mg/L MLSS = 1.25 (MLVSS) Settled sludge volume after 30 min = 250 mL/LDetermine the following: aeration period, F/M ratio, SVI, solids concentration in return sludge.
For an activated sludge plant of capacity discharge (0.4380 m³/s). The F/M ratio is 0.200 mg/mg -d, the influent BOD5 after primary settling is 150 mg/L and MLSS is 2200 mg/L, volume of aeration tank will be Bidddida
Problem 2: Engineers sometimes use pipes as reactors in treatment plants. Suppose that the plant flow is 3.3ft3/s (about 2.1 million gallons per day, mgd) and a cylindrical pipe is to be used for disinfection.A. Draw a sketch of the system. Indicate the control volume by a dashed line. What kind of hydrodynamicmodel should you use for a pipe? What mixing conditions should you assume?B. Write the integrated mass balance equation for the pollutant in symbols at steady state. (“Integrated” meansthat your mass balance looks like: C = …) Define your symbols.C. For a pipe 3.5 feet in diameter and 50 feet long, what is the hydraulic residence time in minutes?D. For the pipe in Part C, what value of the first-order rate constant (in min–1) is needed to achieve a 99% removal of bacteria?
A complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3)
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