Question

What is activated sludge? How it is different from primary sludge?

What is activated sludge? How it is different from primary sludge?

Homework Answers

Answer #1

The term sluge denotes the group of microorganisms like bacteria, protozoa and fungi which are added to the wastewater for treating the water for decomposition of the contaminants. It is the biological process by which waste decomposition is carried out. The activated sluge is different from the primary sluge as the microbes (activated) are added to the incoming wastewater for cleanup. In this process the high quality of effluents are produced and comtaminants are removed out biologically.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
What is activated sludge? How it is different from primary sludge?
What is activated sludge? How it is different from primary sludge?
What is activated sludge? How it is different from primary sludge?
What is activated sludge? How it is different from primary sludge?
Mass of solids Generated from the activated sludge tanks. 13. Assume: The total design flow is...
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?
An activated sludge system with recycle (CMFR) uses to treat primary effluent after sedimentation.Use the appropriate...
An activated sludge system with recycle (CMFR) uses to treat primary effluent after sedimentation.Use the appropriate kinetic coefficients for the influent water characteristics to CMFR. Influent Characteristics: Flow = 1000 m3/day; bsCOD= 192 g/m3; nbVSS= 30 g/m3; inert inorganics= 10 g/m3 Aeration Tank Data: MLVSS = 2500; g/m3; SRT or θc = 6-day Calculate: 1.- Effluent bsCOD 2.- Hydraulic retention time 3.- Daily sludge production in Kg/day as a) VSS b)TSS
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...
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%?
Identify two unit process used to thicken waste activated sludge and describe how each process functions.
Identify two unit process used to thicken waste activated sludge and describe how each process functions.
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...
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.
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...