A wastewater is to be disinfected with 35 mg/Lof chlorine to obtain 99% kill of micro-organisms. The number of micro- organisms remaining alive (N₂) at time t, is modelled by N₁ = Noekt, where No is number of micro-organisms at t= 0, and k is the rate of kill. The wastewater flow rate is 36 m³/hr, and k= 0.23 min¹. If the depth and width of the chlorination tank are 1.5 m and 1.0 m, respectively, the length of the tank (in m, round off to 2 decimal places) is

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A conventional activated-sludge system treats 11,000 m3/day of wastewaterin an aeration tank with a volume of 3400 m3. The BOD after primary clarification is 180mg/l. Given the operation conditions below, calculate the  sludge residence timeMLVSS in the rector = 2,100 mg/lMLSS = (1.25)*MLVSSQw = 160 m3/daySVI = 125 mL/g (Ans.  SRT = 7.0 days)

A municipal wastewater treatment plant processes an average of 14,000 m3 / day. The peak flow is 1.75 times the average. The wastewater contains 190mg/L BOD and 210 mg/L suspended solids at an average flow and 225 mg/L BOD and 365 mg/L suspended solids at peak flow. Determine the following for a primary clarifier with a 20-m diameter.  Assume eff. BOD of 20 mg/L and effluent TSS of 30mg/L for average and peak flows.    1. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ average flow.  2. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ peak flow.  3. Mass of solids (kg/day) that is removed as sludge from average and peak flow conditions.

Given the following information for wastewater plant, the plant is an activated sludge system with an aeration tank that has a volume of 7500m3. efficiency of aeration tank 85% ,SVI is 88ml/gm., volume of settled solids after 30min, is 220ml, Q=60000 m3/d, effluent BOD from aeration tank = 20mg/l. Calculate the values of BOD load, F/M, and efficiency of plant. %3D

Determine the size (diameter and depth) of the secondary clarifier (s) in a municipal activated sludge treatment facility: The wastewater characteristics are: Population 250,000, Average per-capita flow rate = 380 1/capita-day and Peak factor = 2.2. The suspended solids concentration is assumed to be 2500 mg/L. The over flow rate is 110 and 60 m3/d-m2 for peak flow and average flow respectively and solids loading at average flow is 110 kg/d-m2. Estimate detention time if the depth is 4.5 m. What would be the weir loading? Also answer the following questions. (hint : you should use all 3 loading criteria) a. How many kilograms of solids are loaded onto each clarifier per day? b. If the removal efficiency is 65% then how many kilograms of solids are removed from the total flow per day? C. What is the concentration of solids in the wastewater leaving the clarifier? Determine the size (diameter and depth) of the secondary clarifier (s) at a municipal activated sludge treatment facility:…

QUESTION: An aerated lagoon wastewater treatment is to be designed for a municipal wastewater. The design population is 5000 persons, and the aerated lagoon is to provide secondary treatment. Pertinent data are: flow = 380 L/cap-d, influent BOD5 = 255 mg/L; effluent BOD5 = 20 mg/L, K = 2.18 d-1 oxygen required 1.6 kg O2/kg BOD5 applied to the lagoons, and the primary clarifier removes 33 % of the influent BOD5. Calculate the total lagoon volume (m3)

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An activated sludge system is being used to treat 2000 m³/d of wastewater containing 250 mg/L BOD5. The measured parameters for the system are as follows: K; = 40 mg BOD5/L; ka = 0.01 d1; Y = 0.6 mg VSS/mg BOD5; µmax = 2.0 dª. MLVSS = 2000 mg/L. Calculate the aeration basin volume (m³) to achieve 99 % BOD5 removal.

Hi Please try to solve Last time you solved this problem but result is incorrect.