Q4: A wastewater treatment plant as shown:The surface area of each unit 800 m². Depth of each unit 2 m. and %BOD removal inPrimary and Secondary sedimentation tanks = 25%. Find:1- Total % BOD removal in the plant.2- BOD removal in the Trickling filter.3- The flow rate (m³/d) in the plant.r-150%BOD, 180 mg/lPri. sed. tankRiverBOD,-20 mg/1Trickling filterSec. sed. tank(25 marks)

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Answered: Q4: A wastewater treatment plant as…

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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2. The sedimentation portion of the in-line treatment plant is to be designed. The proposed design for the sedimentation tank is 130 ft wide, 180 ft long, and 12ft in depth. The length of the effluent weir along the four channels at the end of the tank is 1160ft. Calculate the major parameters used in sizing this unit based on the design flow of 42 mgd. Determine (a) detention time (b).Over flowrate or velocity of effluent water; (c) Weir Loading; (d) Overflow rate.
Q 1. Calculate the volume and weight of sludge generated per day in circular primary settling tank by chemical coagulation. Find HRT of tank also. Data given as: Wastewater flow: 10000 m/d FeCl; added in PST: 500 mg/L Overall particle removal efficiency: 80% Sludge's specific gravity: 1.05 Assume STANDARD chemical reactions and their product's behavior (if any). TSS in influent: 1100 mg/L Ca(OH)2 added in PST: 400 mg/L PST depth: 3 m Sludge's moisture content: 95% Surface overflow rate: 40 m/d
Sedimentation is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity. A water treatment plant has four clarifiers treating 1.5 MGD of water. Each clarifier is 4.88 m wide, 24.4 m long, and 4.57 m deep. Determine the following: (i) The detention time (ii) The overflow rate marks (iii) The horizontal velocity (iv) The weir loading rate assuming the weir length is 2.5 times the basin width.
Question 1 An anaerobic waste stabilization pond is treating poultry wastewater with following design data; 225 Alder Aerial Organic Loading- 500 kg BOD5/ha.day BODs load - 3,500 kg/day 3.v Depth - 2 m BOD-20,000 mg/L Assume no side slope. Calculate: (i) Volume of tank Dl = Disch & BOD ол prch= (ii) The incoming flow in m³/day sl всь 500 20000
A primary wastewater treatment uses centrifugation for dewatering of primary sedimentation sludge. The raw wastewater contains 300 mg/L TSS and 250 mg/L BOD5. The average daily flow to the treatment plant is 5000 m3/d. In primary sedimentation basin, TSS removal is 55%, and BOD5 removal is 30% of the incoming flow. The primary sludge contains 3.5 % solids. The TSS and BOD5 capture efficiency of centrifugation is 90 % each, and the moisture (water) content of the sludge cake is 70 %. Assume the specific gravities of the primary and sludge cake are 1.01 and 1.06 g/cm3. 1. Calculate the primary sludge flowrate (underflow rate) (m3/d). 2. Calculate sludge cake flowrate (m3/d).
A municipal wastewater treatment plant has an average flow of 0.5 m3/s, and two aerated grit chambers are to be used in parallel. Dual units are to be provided for flexibility in operation, since one could be used, should the other require temporarily maintenance. During normal operation, both chambers would be used. Each chamber is to be a rectangular tank with spiral-roll flow. The peak flow is 2.4 times the average flow. The tank width is 2 times the depth, and the length is 4 times the width. Determine tank width in meters, if the detention time is 3 min. Determine the total air flow in m3/min, if 0.35 m3/min per meter of tank length is to be provided.
2 A groundwater (Q1 = 5.0 L/s) reaches a point of convergence (mixing) in a river (Q = 1.0 m3/s) where an industrial wastewater discharge (Q2 = 10 L/s) also converges. /s). If the groundwater has ferrous iron with concentration [Fe2+] = 10.0 mg/L, the industrial wastewater has organic matter with concentration TOC = 50 mg/L, determine the COD of the river downstream of the mixing point.
Q2: A column analysis for an activated sludge suspension yielded the following: MLSS Conc (mg/l) Settling velocity m/h 1000 2.8 2000 3000 1.2 0.4 4000 0.2 5000 Question 0.1 6000 0.06 The influent MLSS concentration is 2000 mg/l and the flow rate is 4200 m³/d. Determine the size of the clarifier that will thicken the solids to 8000 mg/l.
Please Answer this with clear and step by step solution. I need this ASAP. Thankyou. It is Hydrology Subject.
After initial mixing of a wastewater discharge in a 5-m-deep river, the DO concentration in the river is 7 mg/L and the temperature is 22°C. The average flow velocity in the river is 6 cm/s.(a) If the ultimate BOD of the mixed river water is 15 mg/L, the rate constant for BOD at 20°C is 0.5 d−1, there is negligible removal of BOD by sedimentation, and the reaeration rate constant at 20°C is 0.7 d−1, estimate the minimum DO concentration in the river.(b) How far downstream of the outfall location will the minimum DO concentration occur?
Please solve this question queckly
Q2: A wastewater containing 300 mg/l SS and 250 mg/l BOD passes through a primary sedimentation tank of surface area 600 m². The flow is 15*103 m³/d. Estimate the effluent SS and BOD. (Use Figure 1)

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