Question

Mass of solids Generated from the activated sludge tanks. 13. Assume: The total design flow is...

  1. Mass of solids Generated from the activated sludge tanks.

13. Assume: The total design flow is 3.75 MGD (15,000 m3/day). The NPDES limit is 25/30. Assume that the waste strength is 170 mg/L BOD after primary clarification. Y = 0.55 kg/kg.

What is the mass of solids generated each day (Kg/day or lbs/day) in the activated sludge tankage?

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
14. Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the...
14. Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the following information from the previous problem. The total design flow is 15,000 m3/day. Theoretical hydraulic detention time (Theta) = 8 hours. The NPDES limit is 25 mg/L BOD/30 mg/L TSS. Assume that the waste strength is 170 mg/L BOD after primary clarification. XA=MLSS = 2200 mg/L, Xw = Xu = XR = 6,600 mg/L, qc = 8 days. Make sure you account for the...
Two identical sedimentation tanks are to be designed for an activated sludge plant with a flow...
Two identical sedimentation tanks are to be designed for an activated sludge plant with a flow of 2 MGD (flow is split evenly between the two clarifiers). The MLSS will be 1950 mg/L. The underflow (return sludge) concentration will be 12,000 mg/L. A settling curve was developed (using a column with Ho = 2.13 ft) and is shown below. For clarification area use a safety factor of 2.0 and for thickening area use a safety factor of 5. Tanks are...
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%...
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.
Design a completely mixed activated sludge process using three different design approaches (i.e. calculate the volume...
Design a completely mixed activated sludge process using three different design approaches (i.e. calculate the volume of the aeration basin). Base your design for the first approach on detention time, for the second one on BOD loading, and the third one on kinetics. The influent design flow rate to the activated sludge process is 10,000 m3 /day with a BOD5 concentration of 200mg/L. Completely mixed activated sludge processes typically have a detention time ranging from 3 to 6 hours and...
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...
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...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT