21. Determine the biomass concentration of the activated sludge system based on the following information: Flow rate 18 300 m³/day 160 mg/L Influent BOD Effluent BOD 5 mg/L Solids Retention Time 9 days 0.55 Y kd 0.04 per day 6100 m³ Volume of reactor

21. Determine the biomass concentration of the activated sludge system based on the following information: Flow rate 18 300 m³/day 160 mg/L Influent BOD Effluent BOD 5 mg/L Solids Retention Time 9 days 0.55 Y kd 0.04 per day 6100 m³ Volume of reactor

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

Assume that DLSUD lake is 100m2 and its average height is 5ft. Wastewater that flows into the lake are the following: Food square kitchen waste 7 m3/sec; Total Dissolved Solids (TDS) is 25,000 mg/L which operates 8hrs per day CIH 5m3 /hr kitchen waste; TDS is 20mg/L which operates 4hrs per day PCH 20m3 /day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volumetric flow rate of the lake water can still accommodate prior to rainfall.
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)
The shrimp processing plant generates 0.012 m3/s of wastewater each day. The wastewater is treated in an activated sludge plant. The average BOD5 of the raw wastewater before primary settling is 1400 mg/L. The aeration tank has effective liquid dimensions of 8.0 m wide by 8.0 m long by 5.0 m deep. The plant operating parameters are as follows: Soluble BOD5 after primary settling = 966 mg/L MLVSS = 2000 mg/L MLSS = 1.25 (MLVS) Settled sludge volume after 30 min = 225 mL/L Aeration tank liquid temperature = 15°C Determine the following: Return Activated Sludge (RAS) flow rate and Wasting Activated Sludge (WAS) flow rate.
1) Design a UASB reactor to treat wastewater at 30°C from a food processing plant. The wastewater fow rate is 500 m3/d with a solubleCOD concentration of 6000 mg/L. The design parameters are asfollows:Reactor effectiveness factor (E) = 0.85Upfow velocity = 1.2 m/hOrganic loading rate = 12 kg sCOD/m3 ·dY = 0.08 kg VSS/kg CODkd = 0.03 d–1μmax = 0.25 d–1Ks = 360 mg/LHeight for gas collection = 3 mUsing the given information, determine the following:a. The reactor area and diameterb. The reactor liquid volumec. Liquid depth and total height of the reactord. The average SRT (assuming 97% degradation of sCOD)
1) Design a UASB reactor to treat wastewater at 30°C from a food processing plant. The wastewater fow rate is 500 m3/d with a solubleCOD concentration of 6000 mg/L. The design parameters are asfollows:Reactor effectiveness factor (E) = 0.85Upfow velocity = 1.2 m/hOrganic loading rate = 12 kg sCOD/m3 ·dY = 0.08 kg VSS/kg CODkd = 0.03 d–1μmax = 0.25 d–1Ks = 360 mg/LHeight for gas collection = 3 mUsing the given information, determine the following:a. The reactor area and diameterb. The reactor liquid volumec. Liquid depth and total height of the reactord. The average SRT (assuming 97% degradation of sCOD) 2) Determine the methane gas production rate (m3/d) for the reactorfrom problem 1) Assume COD equivalent of VSS equals 1.42 kgCOD/kg VSS.
The quantity of sludge produced from an aeration tank is 2,453 kg/day. The volume of the aeration tank is 8,500 m3 and the yield is 0.55 mg VSS/mg BOD5. The water retention time is 1 day and the inlet BOD5 is 540 mg/L. Calculate the outlet BOD5 in the tank?
As an operator of a wastewater treatment facility, you have learned that your plant runs well at an SRT (based on MLVSS (X) in the aeration basin) of 12 days. If today’s MLVSS (X) concentration is 2,700 mg/L and the RAS (Xr) concentration is 5,500 mg/L, how many L of sludge should you waste from the secondary clarifier today? Other design conditions: Average influent flow = 12,000 m3/d Influent soluble BOD = 150 mg/L Required effluent TSS = 30 mg/L Aeration tank Volume = 1,000 m3
Assume that taal lake is 100m2 and its average height is 5ft. Wastewater thatflows into the lake are the following: a. Food square kitchen waste 5m3/hr; Total Dissolved Solids (TDS) is 50g/L whichoperates 8hrs per dayb. CIH 5m3/hr kitchen waste; TDS is 20mg/L which operates 4hrs per dayc. PCH 20m3/day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volume of wastewater which taal lake can still be contained withoutrainfall.(2)During operational hours a heavy rain fall, the rain gauge reading is 1.5m. Will the taalLake overflow? What is the volume of water that will over flow?(3). Compute the TDS concentration of the overflow.
How to determine the needed surface area of the secondary clarifier tank given that the flow is 40,000 m3 per day and the recycling ratio is 0.5 . The necessary solids concentration in the underflow is 13,000 mg/L, while the Mixed liquor suspended solids concentration in the activated sludge is 4,500 mg/L. The table below shows the settling properties of sludge: Mixed liquor suspended solids Mg/l Settling velocity m/hr 2000 6 4000 3 6000 1 8000 0.2 10000 0.1 13000 0.1
Dear I need solution for this problem also The system of two basins shown in the picture, full flow stream 3785x1000 l/day in which the concentration of non-conservative pollutant is 20mg/l. The rate of decomposition of a non-conservative pollutant is 0.3/day. The volume of the first pool is 18925x1000 l and the second one is 11355x1000 l. Assuming complete mixing in each basin, determine the concentration of the non-conservative pollutant that leaves the individual basins. Please se attachemnt . Thank You very much
The influent suspended solids concentration to a primary settling (clarifier) tank is 300 mg/L. The average flow rate is 0.037 m3/s. If the suspended solids removal efficiency is 65 percent, how many kilograms of suspended solids are removed in the primary settling tank each day?
A wastewater treatment plant discharges 2.16x107 L/hr of treated effluent with an ultimate BOD of 22.0 mg/L and dissolved oxygen concentration of 5.0 mg/L to a river. Upstream of the waste discharge, the river has a flow of 9.0 m3/s, an ultimate BOD of 10.0 mg/L, and dissolved oxygen concentration of 7.0 mg/L. Both the river and effluent are at 20°C. After the river and effluent mix, the river flows at a velocity of 0.13 m/s and has a depth of 1.8 m. The deoxygenation rate constant for the waste is 0.45 day-1. A fish hatchery is located 9.6 km downstream of the waste discharge. The hatchery requires 5.0 mg/L DO for healthy fish growth and reproduction. Is the hatchery requirement being met? show all work