(h) If F/M is high, theni) Longer SRTiii) Better effluent qualityii) High yieldiv) Microbes are starving(i) During biological P removal, PAOSi) release S in anaerobic conditioniii) release P in aerobic conditionii) accumulate P in anaerobic conditioniv) accumulate P in aerobic condition(j) To remove N in wastewater by air-stripping, pH of wastewater should bei) Neutral (pH 7)iii) Alkali (> pH 10)ii) Acidic (

Answered: (h) If F/M is high, then i) Longer SRT…

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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Which of the following is/are correct about BOD in municipal sewage? 1. It must be removed because it is always pathogenic and thus poses a risk to public health 2. It is utilized by the microorganisms which thus reduces adverse environmental effects 3. It is typically wholly particulate and can be removed by settling 4. It is partly solid and can only be partially removed by settling 5. It is a carbon and energy source for microorganisms in the sludge O 2, 4 and 5 O 1, 3 and 5 O 3, 4 and 5 O 1, 2 and 3 Clear mu choico
wet sludge contains 3% solid by weight is pumped to a filter belt press at the rate of 170,000lb/day. the dewatered sludge is 18% by dry weight and the filtrate has total nitrogen concentration of 100 mg/l as nitrogen. assume the belt press has 100% solid capture and specific gravity of the sludge in all forms is 1. Calculate the amount of nitrogen (lb/day) in the filtrate returned? a) 4 b) 9 c) 14 d) 19
An activated sludge plant treating a flow of 45 litres/sec has an aeration basin volume of 500 m3 and operates with a MLVSS concentration of 3000 mg/1. Concentration of solids in the returned sludge or in the wasted sludge is 15000 mg/1 and volume of wasted sludge per day is 35 m³. Sludge age of the system is -- (Round off to 2 decimal places)
7
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Please answer soon I will give u many likes for correct answer
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A complete mix activated sludge process treats 5 MGD of wastewater containing a total BOD5 concentration of 225 mg/L BOD5 to 20 mg/L BOD5, which is 75% soluble in both the influent and the effluent. The volume of the aeration basin is 111,400 ft3 and the MLSS concentration is 2500 mg/L. Determine the following: a) Detention time in the aeration basin (hr) b) F/M ratio c) Specific substrate utilization rate d) Substrate removal efficiency
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Coagulation aided sedimentation produces 135m³ of sludge per day. The solid content of sludge is 1% by weight and specific gravity of sludge solid is 2.7. If 20 my of suspended solids are removed from the raw water by the Al(OH)3 precipitate formed, and flow through the treatment plant is 40 MLD, then the amount of alum required per day in kg is
Q2-) The waste flow 18000 m'/day is treated by activated sludge process to reduce 97.5% of influent BOD, concentration of 250 mg/L. The ratio of formed cell mass and substrate utilized mass are 0.4 and 0.8, respectively, MLVSS = 3500 mg/L, return sludge (Xr) = 8000 mg/L, decay growth rate= 0.05, and 0 = 10 days. Find the following: 1. Reactor volume and hydraulic retention time. 3. The daily amount of wasted sludge. 5. Food to microorganisms' ratio. 2. The volumetric organic loading 4. The recycle ratio, and treatment efficiency.

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