Q1: A conventional ASP is to be used for treatment of 10,000 m³/d of municipal wastewater. The influent BOD to the ASP is 160 mg/l. As per requirements, the effluent from the ASP should have a soluble BOD of not more than 10 mg/l. Assuming: MLSS=2500 mg/l Mean cell residence time = 10 d Return sludge from SST = 8000 mg/l Yield coefficient = 0.8 kg/kg Endogenous decay rate constant = 0.06 d-¹ Determine: Question i. The volume of aeration tank ii. The mass and volume of sludge to be wasted per day iii. The recycle ratio iv. Food to microorganism ratio

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A bio-tower composed of a modular plastic medium of 6 m is used as secondary treatment component in a municipal wastewater treatment plant. Flow from the primary clarifier is 25000 m³/day with 200 mg/1 BOD. The treatability constant is 0.06 per min. To produce an effluent with a soluble BOD of 20 mg/1, the surface area as per Eckenfelder (coefficient related to medium is 0.5) is approximately:

2a

Please include Equations. This is for a water and wastewater review, it is not for a grade. I just need to understand the problem.

A complete-mixed activated sludge process is to be designed to treat 5.0 MGD of primary effluent having a BOD5 of 180 mg/L. The requirement is that the effluent BOD5 and TSS concentration 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=2.85d-¹, Ks = 70 mg/L BOD5, and kå = 0.041 d¹¹. 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)

Biomass and substrate mass balance equations can be merged giving QwXR/VX=Y/X (Qo/V) (So-S)-kd (i) Prove that X= SRT/HRT Note that HRT = t and SRT = tc. (ii) If you are an engineer working in a wastewater consultancy company, and obtained the following data: - SRT = 20 days HRT 20 hours = (Y(So-S)/1+kd(SRT)). So = 350 mg/L S = 100 mg/L Y = 0.15 kd = 0.22 per day Calculate the increment of MLSS (%) required in the aerobic tank if the discharge BOD5 in the effluent needs to achieve 10 mg/L.

After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the limiting value of the minimum SRT (d). Parameter Hmax Ks ka Y Value 3 d¹ 60 mg/L BOD, 0.10 d ¹¹ 0.6 mg VSS/mg BODS

The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The F/M ratio is nearly?

1. It is estimated that the BOD of raw sewage received at a treatment plant serving a population of 20,000 will be 300 mg/L. It is estimated that the per capita BOD loading is 0.17 lbm/day. 30% of influent BOD is removed by settling. One single-stage high-rate trickling filter is to be used to reduce the plant effluent to 50 mg/L. Recirculation is from the filter effluent to the primary settling influent. Ten States' Standards is in effect, which specifies 100 gpcd in the absence of other information. (1) What is the design flow rate? (A) 0.9 MGD (B) 1.1 MGD (C) 1.4 MGD (D) 2.0 MGD (2) Assuming a design flow rate of 1.35 MGD, what is the total organic load on the filter? (A) 1700 lbm/day (B) 2100 lbm/day (C) 2400 lbm/day (D) 3400 lbm/day (3) Assuming an incoming volume of 1.35 MGD and an organic loading of 60 lbm/1000ft³.d, what flow should be recirculated? (A) 0.6 MGD (B) 1.1 MGD (C) 1.2 MGD (D) 2.2 MGD (4) What is the overall plant efficiency? (A) 45% (B) 76% (C) 83% (D) 91%