Question

What is the F/M ratio given the following Influent sewage flow = 6500000 gal/day Primary influent...

What is the F/M ratio given the following Influent sewage flow = 6500000 gal/day Primary influent BOD= 170 mg/L MLVSS = 4500 mg/L Aeration tank dimentions 28' x 120' x 15'*

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 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 ?
A 2,250 m3/day extended aeration system is used to treat sewage from Taman Maju Jaya. The...
A 2,250 m3/day extended aeration system is used to treat sewage from Taman Maju Jaya. The average BOD influent of the sewage treatment plant is 250 mg/L and is reduced by 65% after the primary clarifier. It has two rectangular aeration tanks (working in parallel) with a dimension of 4 m width, 6 m length and 3.5 m water depth. The aeration tank maintains a biomass concentration of about 4000 mg/L. (i) Determine the food-to-microorganism ratio of the aeration tank...
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...
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%...
A 1.0 � 106 gallon reactor is used in a sewage-treatment plant. The influent biochemical oxygen...
A 1.0 � 106 gallon reactor is used in a sewage-treatment plant. The influent biochemical oxygen demand (BOD) concentration is 100 mg/L, the effluent BOD concentration is 25 mg/L, and the flow rate through the reactor is 500 gal/min. (a) What is the first-order rate constant for decay of BOD in the reactor? Assume that the reactor can be modeled as a comply mixed flow reactor (CMFR). Report your answer in units of per hour. (b) Assume that the reactor...
F/M ratio is the equilibrium parameter for activated sludge systems . It is the balance between...
F/M ratio is the equilibrium parameter for activated sludge systems . It is the balance between food substrate and microorganisms.If the F/M of a 0.4380 m3/sec activated sludge plant is 0.2 mg/mg-d (inlet BOD 150 mg/l. sludge 2,200 mg/l) what is the required volume of the aeration tank? What can the plant operator do to maintained an acceptable F/M if the inlet volume increases by 20%?
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...
A wastewater treatment plant designed for a community of 60,000 people has a flow of 100...
A wastewater treatment plant designed for a community of 60,000 people has a flow of 100 gal/person/day and a biochemical oxygen demand (BOD) loading of 0.2 lb/person/day. The treatment plant has a permit limit for BOD equal to 15 mg/L. a. (6 pts) What is the concentration (in mg/L) of BOD coming into the plant (i.e., as raw, untreated sewage)? b. (2 pts) When the treatment plant is treating its waste and meeting its permit limit, what is the removal...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by an activated sludge process composed of aeration tanks and a secondary clarifiers. This problem focuses on the activated sludge process. The activated sludge process receives a flow of 15 million gallons per day (MGD). The primary sedimentation process effluent has a BOD5 concentration of 180 mgBOD5/L (this concentration is often referred as S). Before flowing into the activated sludge process, the primary...
Problem #3: Biochemical Oxygen Demand (BOD) and Stream Quality A wastewater treatment plant designed for a...
Problem #3: Biochemical Oxygen Demand (BOD) and Stream Quality A wastewater treatment plant designed for a community of 45,000 people has a flow of 100 gal/person/day and a BOD5 loading of 0.2 lb/person/day. The upstream characteristics of the receiving waters are: stream flow rate = 20 cfs and BOD5 = 2 mg/L The water quality standards require the in-stream BOD5 to be less than 7 mg/L in this stretch of the river. The minimum BOD removal efficiency that the treatment...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT