(h) If F/M is high, then i) Longer SRT iii) Better effluent quality ii) High yield iv) Microbes are starving (i) During biological P removal, PAOS i) release S in anaerobic condition iii) release P in aerobic condition ii) accumulate P in anaerobic condition iv) accumulate P in aerobic condition (j) To remove N in wastewater by air-stripping, pH of wastewater should be i) Neutral (pH 7) iii) Alkali (> pH 10) ii) Acidic (

An activated sludge system using CSTR with sludge recycle has an influent flow rate (Q°) of 2.5 MGD that contains BOD of 200 mg/L. The volume of the aeration tank (CSTR) is 0.6 million gallons. The MLVSS (assume this is X,) in the aeration tank is 2000 mg VSS/L and the decay rate (b) is 0.1/day. The effluent from the secondary clarifier meets NPDES standards for suspended solids and BOD. The concentration of the settled sludge exiting the bottom of the secondary clarifier is 15,000 mg VSS/L. The sludge waste flow rate (Q") is 1 % of the influent flow rate, and the effluent flow rate (Qª) is ~99% of the influent flow rate. а) What is the hydraulic retention time, 0 (d)? b) What is the solids retention time, 0x (d)? Activated sludge with recycle с) What is the F/Ma ratio?

Q/ A wastewater treatment use an activated sludge process for secondary treatment of 10000 m/day for municipal wastewater. After primary clarification, The BOD is 150 mg/L, and it is desired to have not more than 7 mg/L of soluble BOD in the effluent. The mixed liquor has a concentration of 3000 mg/L and underflow solids concentration is 12,000 mg /L from the secondary clarifier. The kinetic constants from a bench scale treatability study are Y=0.5 and K=0.05 day". Assume the cell residence time is 10 days. Determine: a. The volume of the activated sludge tank & the hydraulic residence time. b. The sludge production rate & the sludge wasting flow rate. c. The sludge recycle ratio and recycle flow rate. d. The organic loading rate.

Q- ASP is used to treat a waste water flow of 2MLD having BOD5 of 250 mg/lit. The biomass concentration in the aeration tank is 2500 mg/lit and the concentration of biomass leaving the system is 80 mg/lit, the aeration tank has a volume of 200m³. Find the hydraulic retention time HRT.

Q- The sludge from the aeration tank of the activated sludge process (ASP) has solids content (by weight) of 2%. This sludge is put in a sludge thickener, where sludge volume is reduced to half. Assume that the amount of solids in the supernatant from the thickener is negligible, the specific gravity of sludge solids is 2.2 and the density of water is 1000 kg/m³. What is the solids content (by weight) of the thickened sludge?

The results of a jar test on a surface water supply sample revealed that the optimum dose of alum (Al2(SO4)3.14H2O) as coagulant is 17 mg/L. Furthermore, the water has total alkalinity = 50 mg/L as CaCO3; pH = 7.8; and residual turbidity at optimum alum dose = 4 NTU.  The water treatment plant has a mean flow of 10m3/hour.  Determine:   The kg of alum per year needed in the operation of the plant. The residual alkalinity in the sample, following the addition of alum at optimum dose. Solve parts (a) and (b) if instead of alum, ferric chloride (FeCl3) were used as coagulant. Assume the numerical jar test results are the same as those with alum.

Waste activated sludge contains 1% dry solids that are 80% volatile. Determine:a. The specific gravity (Ss) of the dry solids.b. The specific gravity (Ssl) of the sludge.c. The volume (m3) of the thickened sludge if 60,000 kg of dry solids is thickened to6% in a gravity belt thickener.

Air is passed through the sludge 1. To provide the biomass with the oxygen required for removing the organic matter from the sewage 2. To keep the biomass suspended thus bringing them in close contact with the organic matter in the sewage 3. To kill pathogens 4. To reduce the hydraulic retention time and thus reduce the size of the bioreactor O 2 and 3 only O 1, 2 and 4 O 1 and 2 only OAll of the above Clear my choice

A wastewater treatment plant uses an activated sludge process to treat 0.2 m3/s of primary effluent. The concentration of the mixed liquor is 2,000 mg/L MLVSS (Mixed liquor volatile suspended solids), while the RAS (recycled activated sludge) concentration is 10,000 mg/L MLVSS. The substrate concentration in the primary effluent is 210 mg/L BOD5. The F/M ratio is 0.5 mg BOD5/(mg MLVSS-d) and the solids retention time is 8.5 d.  What is the volume of the activated sludge tank?  2.What is the WAS (waste activated sludge) flowrate?  3.What is the flowrate of the secondary treated effluent?  4.What is the RAS (recycled activated sludge) flowrate?  5.What is the hydraulic residence time for the activated sludge tank?

Problem 4In water and wastewater treatment processes a filtration device may be used to remove water from the sludgeformed by a precipitation reaction. The initial concentration of sludge from a softening reaction is 20,000 mg/Land the volume of sludge is 100 m3. After filtration the sludge solids concentration is 350,000 mg/L. Assume thatthe sludge does not change density during filtration, and there is no accumulation of sludge in the filtrationdevice. Using the mass balance method, determine the volume of sludge after filtration.