12.60 The aeration tank for a completely mixed aeration process is being sized for a design waste-water flow of 7500 m³/d. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/l. Thedesign effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design para-meters are a sludge age of 10 d and volatile MLSS of 1400 mg/l. Selection of these values takesstants from a bench-scale treatability study are Yinto account the anticipated variations in wastewater flows and strengths. The kinetic con-0.60 mg VSS/mg soluble BOD andk₁ = 0.06 per day. Calculate the volume of the aeration tank, aeration period, foodganism ratio, and excess biomassth

The objective of the question is to calculate the volume of the aeration tank, aeration period, food…

Answered: 12.60 The aeration tank for a…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A CMAS treatment system must be designed for a municipal wastewater and the following operating conditions: • Effluent flow of 0.39 m³/s leaving the primary clarifier Effluent soluble BOD; of 310 mg/L leaving the primary clarifier 20 mg/L of soluble BOD5 allowed in the effluent from the secondary clarifier Negligible suspended solids in the effluent from the secondary clarifier • Volatile suspended solids concentration of 4300 mg/L in the aeration tank • In the laboratory under log-growth conditions, the microorganisms used in the aeration tank can convert 100 g soluble BODS into 55 g VSS • Microorganisms in the CMAS system have a death rate constant of 0.05 d'. Sludge age in the CMAS treatment system will be 4 days (a) Determine the % treatment efficiency for soluble BOD; required in the CMAS system. (b) Determine the required volume of the aeration tank (in m²) and HRT (in h). (c) Determine the food-to-microorganism ratio (in d') and the specific substrate utilization ratio (in d').…
12.50 Determine the activated-sludge aeration volume required to treat 2.64 mgd with a BOD of 120 mg/l based on the criteria of a maximum BOD loading of 40 lb/1000 ft³/day and a minimum aeration period of 5.0 hr. Assuming an operating F/M of 0.20 lb BOD/day per lb of MLSS, calculate the MLSS to be maintained in the aeration tank. Estimate the operat- ing sludge age (mean cell residence time), assuming an effluent suspended solids of 30 mg/l and the daily amount of waste-sludge solids from Figure 13.1. Determine the diame- ter and side-water depth of two identical final clarifiers of the type shown in Figure 10.12 for this activated-sludge system.
8 Determine the solid loading rate on a secondary circular clarifier (diameter = 20 m) if the plant influent load is 2 ML/d. The recycle flow from secondary clarifier to the activated sludge system (aeration basin) is set to 10% of the influent load. It is also known that the aeration basin produces an MLSS (mixed liquor suspended solids) concentration of 3000 mg/L.
What aerobic reactor volume and operating SRT (8) are required to treat a 20 MGD domestic wastewater flow from an initial ultimate BOD - 300 mg/L to an effluent BODu 5 mg/L assuming a complete mix activated sludge system? Use the design information below assuming 1st order kinetics. MLVSS = 3,000 mg/L - X Y-05 K=0.1 L/mg'day (1st order degradation rate constant) K₂ -0.1 day 1 Answers are given as Aeration Tank Volume, Operational O 0.1 MG, 2.5 days O 19.65 MG, 20 days O 3.93 MG, 6.7 days 5.2 MG, 5.1 days
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(D) Sludge in the secondary tank after Activated sludge is the (A)Aerated sludge in the aeration unit. (B) Sludge settled in the humus tank. (C) Sludge in the secondary tank after aeration and rich in microbial masses. (D)Sludge in the secondary tank after aeration and rich in nutrients.
3. A 100 mL of wastewater sample was drying on a 0.5 g empty dish weight at 104 °C to weight 0.5625 g. Another 100 mL of the same wastewater was first filtered and the filtrate was dried on a 0.5 g dish at 104 °C to weight 0.5325 g. Determine in mg/L the:a) TSb) TSSc) TDSd) VSS assuming it represents 70% of TSS
For an activated sludge plant of capacity discharge (0.4380 m³/s). The F/M ratio is 0.200 mg/mg -d, the influent BOD5 after primary settling is 150 mg/L and MLSS is 2200 mg/L, volume of aeration tank will be Bidddida
The following data for activated sludge process, discharge%=D10 m/min, recycle and wasted solid concentration=15000mg/L , sludge waste from the aeration tank =81.1m³/d, influent BOD concentration=160mg/L, effluent BOD concentration=4 mg/L , Oxygen transfer efficiency=8%, Qactual=........?
Q2. A settling tank sludge sample is tested for solids. The sample and the dish weight (74.69) g, the dish weight (21.2) g. after drying, the dish with dry solids weight (22.3) g. what is the percent of total solids (%TS) in the water sample?
Assume that all solutions are prepared at 25 degrees celsius and the solutes arr completely dissolved. In the lead content analysis of a wastewater sample, 75.00 mL of the original sample was taken and diluted with 125.00 mL deionized water. The diluted sample was found to have a lead content of 19.2 ppb. What is the lead content (in pbb) of the original wastewater sample?

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