A complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3 )

A complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3 )

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

In an activated sludge treatment unit with tank volume of 700 m³, the imfluent BOD of the sludge at 4000 m³/day of discharge is 250 mg/lt which reduced to 30 mg/lt in the effluent sludge. The MLVSS concentration in the aeration tank is 3000 mg/lt. The F/M ratio is kg/kg/day.
A completely mixed activated sludge process operating at equilibrium is designed to treat 300 MLD of waste water. The influent and effluent BOD are 200 mg/1 and 18 mg/1 respectively If the F M then the hydraulic retention time is A 0.125 day B 4.21 hours C 8.42 hours D 0.225 day = 0.3 and MLSS is 3800 mg/1,
In a municipal wastewater treatment plant, the influent flow rate into a complete mixing activated sludge system is 50 L/sec, its solid concentration MLSS is 30 mg/L, and the BODs in the influent is 245 mg/L. The MLSS concentration at the bottom of the secondary clarifier is 15,000 mg/L and 10 mg/L at the top of the |clarifier. The flow rate of returned activated sludge is 20 L/sec. a. What is the concentration of the MLSS in the activated sludge reactor (assuming that it is fully mixed and has constant concentration)? b. What is the activated sludge basin volume if the hydraulic retention time in the basin is 10 hours and the system is operated at the R/Q ratio given in the problem? |c. If the cross section of the basin is 20 m² (perpendicular to the flow), how long the basin will be? d. The plant has 96% BOD5 removal. What is the effluent BOD5? e. The flowrate of the waste activated sludge (WAS) is 0.2 L/sec. How many solids (in kg) are wasted every day through the secondary…
A completely mixed activated sludge plant treats wastewater at a rate of 2,565 m³/d. The influent BOD is 280 mg/L, and the MLVSS in the aeration tank is 2,500 mg/L. The F/M ratio for the aeration tank is 0.35 kg BOD/kg MLVSS.d. Assuming 30% of BOD is removed in the primary tank, determine the organic loading rate in kg BOD/m³.d and the volume of the aeration tank in m³.
A mixed liquor sample form an activated sludge plant was settled in a one-litre measuring cylinder for 30 minutes after which time, a volume of 350 m/ of sludge was recorded. When 50 m/ of the mixed liquor were filtered through a GFC paper for the determination of MLSS the gain in weight of the paper was 0.1375 g. Determine the SVI.
HRT?.
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?
A complete mixed activated sludge process (CMAS) is to be designed to treat 5MGD of primary effluent having a BOD5 of 180 mg/L. The NPDES permit requires that the effluent BOD5 and TSS concentration be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15C: Y = 0.6 mg VSS/mg BOD5, k = 2.85d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basisn is maintained at 2500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the NPDES permit requirement (d) c. The volume of the aeration basin (ft3)
A 14 mgd wastewater treatment plant has a primary clarifier that treats an influent with 800 mg/L of solids and a removal efficiency of 55%. Sludge at the bottom of the tank is pumped at a rate of 0.1 mgd. 1. What is the effluent solids concentration from the clarifier (in mg/L)? What is the solids concentration in the sludge at the bottom of the tank (in mg/L and %)? The primary sludge is further processed in a digester following treatment. What weight of solids (in lbs) need to be handled each week? а. b. с.
A conventional activated sludge plant is designed to treat 28000 kiloliters per day at a settled sewage of BOD equal to 220 mg/l. The effluent BOD is 15 mg/l. F/M ratio is 0.22, MLSS is 3000 mg/1. The hydraulic retention time of the sludge will be
1. A 770,000 gal conventional activated sludge pro- cess has a recycle ratio of 0.25, an influent flow rate of 1 MGD, a mixed liquor volatile suspended solids (MLVSS) concentration of 700 mg/L, and an influent BOD; of 160 mg/L. What is most nearly the F:M ratio of the process? (A) 0.1 lbm/day-lbm (B) 0.2 lbm/day-lbm (C) 0.3 lbm/day-lbm (D) 0.4 lbm/day-lbm В. D.
A WWTP with activated sludge secondary treatment receives 0.290 m / s of flow from the primary clarifier. There is a concentration of microbes in the aeration basin of 2500 mg VSS/L. The return sludge has a concentration of 11,000 mg VSS/L. The concentration soluble BOD from the primary clarifier is 200 mg BODs/L. The yield is 0.49 mg VSS/mg BOD 5. The solids residence time is 10 days. The effluent soluble BOD concentration is 5 mg BODs/L. The decay coefficient is 0.1 1/d. What is the volume of the activated sludge tank? First draw a picture of the system and label all variables.