Which of the following is/are correct about BOD inmunicipal sewage?1.It must be removed because it is alwayspathogenic and thus poses a risk to public health2.It is utilized by the microorganisms which thusreduces adverse environmental effectsIt is typically wholly particulate and can beremoved by settling3.It is partly solid and can only be partiallyremoved by settling4.5.It is a carbon and energy source formicroorganisms in the sludgeO 3,4 and 5O 1,2 and 3O 1,3 and 5O 2, 4 and 5 The preliminary stage in a wastewater treatmentplant1.Removes only large items that cannot be takenup in the biological process2.Removes the large items to protect the plantequipmentIncludes flow measurement devices to help inthe control and operation of the biological stage3.4.Normally produces significant BOD removalthat aids in meeting regulatory restrictions2 and 3O 11, 2, 34

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

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

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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The preliminary stage in a wastewater treatment plant 1. Removes only large items that cannot be taken up in the biological process 2. Removes the large items to protect the plant equipment 3. Includes flow measurement devices to help in the control and operation of the biological stage 4. Normally produces significant BOD removal that aids in meeting regulatory restrictions O 2 and 3 O 4 O 1 1, 2, 3 Clear my choice
A system is designed to treat contaminant X with two PFRs in series. The concentration of contaminant X entering the first PFR is 120 mg/L and the inflow is 420 L/s. The first PFR is 5000 m³ in size, and the contaminant reacts via first order kinetics with a k value of 0.50 (1/hr). The second tank is being built so that the treatment facility can meet the permit level for contaminant X of 9.4 mg/L. What size do they need to make the second tank in m³?
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
The wastewater from a synthetic textile manufacturing plant is discharged to a municipal treatment plant for processing with domestic wastewater. The characteristics of the synthetic textile wastewater are 1500 mg/L BOD, 2000 mg/L SS, 30 mg/L nitrogen, and no phosphorus. The characteristics of the domestic wastewater are 200 mg/L BOD, 240 mg/L SS, 35 mg/L nitrogen, and 7 mg/L phosphorus. Find the BOD, SS of the mixture. The ratio of flow rate is domestic/textile = 4/1
You have designed a treatment system for contaminant Z. The treatment system consists of two parts: (1) a pipe with a volume of 15 m3 and (2) a CSTR has a volume of 60 m3. The pipe feeds into the CSTR. The inflow to the pipe is 10 m3/d and the influent Z concentration at the start of the pipe is 2,500 mg/L. The effluent from the pipe has a Z concentration of 1500 mg/L. Assume the reaction is first order. What is the reaction rate constant in units of (1/day)?
The BOD just downstream from a wastewater pipe is 12.9 mg/L and the DO (dissolved oxygen) is 8.6 mg/L. The saturated value of dissolved oxygen is 9.6 mg/L, the deoxygenation constant (kd) is 0.21/day, the reaeration constant (kr) is 0.42. Find the time where oxygen deficit is at a maximum and find the minimum value of DO.
An aeration basin for an activated sludge facility has the following characteristics Length = 90 ft Width = 30 ft Liquid depth = 12 ft MLSS=4,000 mg/L The raw wastewater has the following characteristics: Flow = 12 MGD BOD, = 220 mg/L Suspended solids (SS) = 250 mg/L A primary clarifier is ovided that removes 25% of the BOD, an 60% of the suspended solids The return activated-sludge flow rate is 0.8 MGD If the primary sludge solids content is 4% solids and the specific gravity is 1.0, the primary sludge volume (ft³/day) is most nearly OA. 1,500 OB. 60,000 O C.600 O D.4,500
A Granulated Activated Carbon (GAC) adsorption laboratory column 10cm in diameter and 3m deep, is found to produce good results in COD removal when operated at a flow of 50L/h. The porosity of the GAC is 0.5 Calculate the following: You need to design a GAC system consisting of two columns in series for treating the same water at a rate of 0.5 ML/d. Practical considerations limit the diameter of the columns to 3m. Calculate the minimum required height of the columns. Area=7.854×10−3m3 Volume= 6.3660m/hr t=0.47123 Q=2.122 bed volume /hr
The Rajasthan river water suspended matter and particles is quite low at the Kanowit tributary. The planned water treatment plant source of raw water intake is the kanowit tributary. Proposed a suitable process to be adopted is designing the treatment plant. Justify the selected sequence proposed.prescribe the chemicals required in each step if any.
Which of the following statements is false regarding Primary Treatment? Removes solids by sedimentation Significantly reduces the soluble BOD of the influent O Its removal efficiency is enhanced by using chemical coagulants Significantly reduces the TSS of the influent
A municipal wastewater treatment plant is installing completely mixed activated sludge to meet their NPDES permit of 15 mg/L BOD5. The mixed liquor concentration is 1,000 mg/L MLVSS. The wastewater flowrate going to secondary treatment is 0.170 m3/s and the BOD5 in that stream is 110 mg/L. Laboratory measurements of the growth constants show: Ks = 60 mg/L BOD5, um = 2.8 d-1, kd = 0.05 d-1, Y = 0.73 mg MLVSS/mg BOD5. Determine the required volume of the aeration tank.
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).