Answered: Disinfection, sedimentation,…

Disinfection, sedimentation, flocculation, coagulation, and filtration are processes for a surface water treatment facility. Place each process in a sequential order.

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

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').…
A continuous stirred tank reactor (CSTR) Activated Sludge Process is used to treat municipal wastewater after primary sedimentation. The following data is available: Primary Effluent Flow Rate (Q) Primary Effluent BOD Concentration (So) Secondary Effluent Biomass Concentration (Xe) Mixed Liquor Biomass Concentration (X) Solids Retention Time (SRT) Waste Sludge Flow Rate (Qw) Value 450 200 O 0 3,000 5 8 Biomass Concentration in RAS (XR) 8,000 O 250 million gallons, 0.28 gBOD/gVSS.d, 13.3 hours 107 million gallons, 0.28 gBOD/gVSS.d, 5.7 hours O 107 million gallons, 0.42 gBOD/gVSS.d, 5.7 hours 107 million gallons, 0.28 gBOD/gVSS.d, 12.2 hour Unit mgd mg/l mg/1 mg/l days mgd The bioreactor volume (V), F/M ratio, and HRT are respectively: mg/l
This question about Sanitary Engineering ........................ involve a physical change such as screening, grit removal, Separation, Crystallization, Evaporation, Filtration, & other reactions. • A. Unit operations • B. Unit processes • C. Both A and B • D. None
A- An industrial wastewater stream flows at a rate of 13500 m³/d containing 100 mg/L. Suspended solids, ferric sulfate [Fe(SO4)] is used as a coagulant' at a dose of 50 mg/L. a) Assuming that there is very little in water, what is the daily lime dose? b) If the sedimentation tank removes 90% of the inlet solids, what is the daily solids production from the sedimentation tank?
Water treatment plant : SOR = 0.35 mm/sec; The analysis of this water found that there are 6 different particle sizes: 0.1 mm with settling velocity of 0.2 mm/sec; and percentage of particles = 10 %; 0.2 mm with settling velocity of 0.25 mm/sec; and percentage of particles = 20 %; 0.3 mm with settling velocity of 0.3 mm/sec; and percentage of particles = 15 %; 0.4 mm with settling velocity of 0.35 mm/sec; and percentage of particles = 5 %; 0.5 mm with settling velocity of 0.4 mm/sec; and percentage of particles = 30 %; 0.6 mm with settling velocity of 0.5 mm/sec; and percentage of particles = 20 %; Calculate the efficiency of this tank ?
6. After your flash mix the water was sent to a clarification basin that was 24 feet in diameter and 20 feet tall. What was the detention time in hours of 0.43 MGD? 5.4 hours 4.3 hours 3.8 hours 2.3 hours After the water leaves your sedimentation basin, it is ready for filtration. Your two filters measure 10 feet in diameter. What is the filtration rate is gpm for your flow of 0.43 MGD? 19 gpm/sq/ft 29 gpm/sq/ft 39 gpm/sq/ft 49 gpm/sq/ft You have had a filtration breakthrough (failure) and have brought your old roughing filtration system on line. The old filter is 8 feet square and 4 feet deep. What is the filtration rate in gpm/sq/ft for your flow of 0.43 MGD? 10.33 gpm/sq/ft 8.55 gpm/sq/ft 5.99 gpm/sq/ft 4.66 gpm/sq/ft
The concentration of the mixed liquor suspended solids, MLSS, in the basin of a step aeration system is 2300 mg/L, and the volume of settled sludge after 30 minutes in a 1 L graduated cylinder is 200 mL. Calculate the SVI in mL/g.
An activated sludge wastewater treatment system was designed and operated in a steadystate under the following conditions:• Reactor volume of 10,000 m3in total;• Sludge concentration (MLSS) of 3200 mg/L in the reactor;• Return sludge concentration at 8000 mg/L (the underflow of the secondary clarifier)• Each day, 400 m3 of underflow sludge (return sludge) was wasted as excesssludge;What is the average Sludge Residence Time (SRT)?
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
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
COURSE: ENV 2
A jar test reveals that the optimum alum coagulant dose is 50 mg/L at a pH of 8.1 and a temperature of 73 F. How many pounds per day of 49% alum are required for the whole plant? Total flow is 36 MGD with four parallel treatment plants, each designed for 12 MGD. Could use some guidance on setting up the equations.