Question

A complete mixed activated sludge process aeration tank treats 5 MLD sewage having the influent and...

A complete mixed activated sludge process aeration tank treats 5 MLD sewage having the influent and effluent soluble BOD concentrations of 220 and 20 mg/L respectively. The volume of the aeration tank is 5060 m3 and the mean cell residence time is 10 days. MLVSS in aeration tank is 3,400 mg/L, MLVSS/MLSS = 0.8 and sludge wastage rate is 1 m3/hour. Calculate the concentration of return sludge suspended solids ?

Engineering   Civil-Engineering   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1


0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A completely mixed activated-sludge process is being designed for a wastewater flow of (2.64 mgd) using...
A completely mixed activated-sludge process is being designed for a wastewater flow of (2.64 mgd) using the kinetics equations. The influent BOD of 120 mg/Lis essentially all soluble and the design effluent soluble BOD is 7 mg/L, which is based on a total effluent BOD of 20 mg/L. For sizing the aeration tank, the mean cell residence time is selected to be10 days and the MLVSS 2000 mg/L. The kinetic constants from a bench-scale treatability study are as follows: Y...
A completely mixed activated sludge process is designed to treat 20,000 m3 /d of domestic wastewater...
A completely mixed activated sludge process is designed to treat 20,000 m3 /d of domestic wastewater having a BOD5 concentration of 250 mg/L following primary treatment. The NPDES permit requirement for effluent BOD5 and TSS is 20 mg/L for each. The biokinetic parameters have been established as: yield = 0.6 , k = 5/d, Ks = 60 mg/L, and kd = 0.06/d. The MLVSS concentration in the aeration tank is to be maintained at 3000 mg/L and the ratio of...
Sooner City Reclamation Plant is a completely mixed activated sludge process designed to treat 20,000 m3/d...
Sooner City Reclamation Plant is a completely mixed activated sludge process designed to treat 20,000 m3/d of domestic wastewater having an influent BOD5 concentration of 350 mg/L (assume a 28.6% reduction in during primary treatment). Use typical activated sludge kinetic constants except for kd = 0.06 1/d. Assume that the MLVSS concentration in the aeration basin is maintained at 3,000 mg/L and the ratio of VSS:TSS is 0.75. Permit requires that the effluent BOD5 and TSS concentrations do not exceed...
If the F/M of a 0.44 m3/s completely mixed activated sludge plant is 0.20 mg/mg.d, the...
If the F/M of a 0.44 m3/s completely mixed activated sludge plant is 0.20 mg/mg.d, the influent BOD5 after primary settling is 180 mg/L, and the MLVSS is 3,200 mg/L, Calculate the following:(you can assume any missing data) The volume of the aeration tank? The HRT The cell residence time The recirculated sludge The waste activated sludge produced each day from the plant in kg/d The mass of oxygen to be supplied (kg/d).
A conventional activated sludge plant treating a domestic flow of 150 ML/d is operated at a...
A conventional activated sludge plant treating a domestic flow of 150 ML/d is operated at a SRT of 10 d with a MLVSS of 3500 ppm. The activated sludge plant is required to reduce the influent BOD from 200 ppm to less than 10 ppm prior to discharge. The discharge from the aeration basin is clarified, such that the clarifier overflow contains no particulate material and the MLVSS of the recycle stream is 8000 ppm. You are required to determine...
ANSWER as much as possible: Design an activated sludge process consisting of a complete mix flow...
ANSWER as much as possible: Design an activated sludge process consisting of a complete mix flow through aeration tank followed by final settling tanks. The sizing of the reactor and all appurtenances should be based on a solids retention time, (SRT), of 4 days and the particulates given below. Biological characteristics: Ks = 20 mg/L COD, kd = 0.1 per day, mumax = 3 gVSS/gVSS-day,    Y = 0.4 gVSS/gCOD Primary settling Tank Effluent Characteristics: Q= 20MGD, biodegradable COD, bCOD...
You have been retained to design a wastewater treatment plant using a conventional activated sludge process...
You have been retained to design a wastewater treatment plant using a conventional activated sludge process for a town of 40,000 people. The industrial flow contribution to the sewer system is 300,000 gallons per day. Effluent discharge standards require effluent BOD5 and total suspended solids concentration of <10 mg/l. Assume the following design criteria are applicable: Domestic sewage flow = 100 gal/cap/day (including I/I) BOD loading = 0.18 lbs/cap/day TSS loading = 0.20 lbs/cap/day BOD removal in primary = 33%...
An activated sludge tank has the following characteristics: Q = 4.18 m3/sec; Influent BOD5 = 225...
An activated sludge tank has the following characteristics: Q = 4.18 m3/sec; Influent BOD5 = 225 mg/L; Effluent BOD5 = 4.5 mg/L; Ks = 100 mg/L; μm = 2.5 day-1; kd = 0.05 day-1. The mean cell residence time (MCRT) in days is most nearly: a.12.4 b.7.0 c.2.5 d.17.4 QUESTION 38 The hydraulic detention time, θc, of the activated sludge tank above (in Question #37) in hours is most nearly: (Note: Cells in the activated sludge tank, X = 3,500...
Estimate the biomass concentration in a CSTR aeration tank with the following operating conditions: hydraulic residence...
Estimate the biomass concentration in a CSTR aeration tank with the following operating conditions: hydraulic residence time = 3 h; mean cell residence time = 6 d; yield coefficient = 0.6 mg VSS/mg BOD; decay rate = 0.1 d-1; influent soluble BOD = 200 mg/L; effluent BOD = 2 mg/L.
Estimate the volume of air to be supplied (m3/d) for an activated sludge system with a...
Estimate the volume of air to be supplied (m3/d) for an activated sludge system with a flow of 0.25 m3/sec, net waste activated sludge production of 340 kg/day VSS, and an influent soluble BOD5 of 74 mg/L. Assume that soluble BOD5 is 58% of BODL, BOD5 in the aeration tank is 14 mg/L, and that the oxygen transfer efficiency is 9%. Density of air is 1.185 kg/m3, and air has 23.2% oxygen.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT