Date:Pag 396How Astrong municipal ww with soluble BODof 200 mill 2 20°C is treated in a bio-tower with an 5.4mdepth of plastic packing. The hydraulic loading of the rawww. is 0.86 m³/m² h, and the recirculation ratio is 3The constants based on apilot-plant study arek = 0.35/hrand n=0.39. Calculate the effluent Soluble BOOD..

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Answered: Date: Pag 396 How Astrong municipal ww…

Solid Waste Engineering

3rd Edition

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Worrell, William A.

Chapter6: Biological Processes

Section: Chapter Questions

Problem 6.1P

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The test Carried aut en an follow. detaits sludge Somple are Weight of Dish =9.52gm Baw Sludge Sempla and duch =23 27ga Aşter Drying Clnstent) After Burning Clnstant) = 6.38gm 9.46gm %3D %3D Calculkt th percentage of total Salidl ad lloldi Solid in t cboue of cowego.
NaOCI concentration in slip stream, Css = () lb/ft³? (*No unit) Qin = 8000 GPM Cin = 2.0 mg/L Slip Stream Qss = 2.3 L/s Css = ? Ib/ft3 Main service line Qout = 8000 GPM Cout = 4.0 mg/L %3D
6. A tertiary treatment plant treats a municipal secondary effluent and is to employ fixed-bed carbon adsorption columns. Parallel treatment using two rows of two columns in series (in other words, a total of four columns) will be used. From batch-type slurry tests it has been found that 0.42 g COD/g carbon is adsorbed when Ce = Co. Pertinent data are as follows: flow = 5700 m³/d, COD in feed to carbon columns = 20 mg/L, contact time function based on an empty carbon bed = 30 min, and unit liquid flow rate= = 4.4 L/s-m². Each column is kept on-line until the entire mass of carbon in the column is completely exhausted. Determine: The volume of each column, m³. a. b. Page 2 of 3 C. d. e. The diameter and height of each column, m. The mass of carbon in each column if the packed density is 400 kg/m³. The kilograms of carbon exhausted per day if the COD removed is assumed to be 98%. The on-line time for each of the columns.
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1. A wastewater is treated in an aerobic CSTR with no cell recycle. The following hold: Ks= 50mg/l; q^=5mg/mg VSSa-d; Kd= .06d-1; Y= 0.6; S°=220mg/l; fd= 1
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Calculate the velocity gradient in a rapid mixer (coagulation), where the power input to the propeller is 0.5 kW, the plant provides water for 16000 people with average consumption of 160 L/person/day, retention time in rapid mixer is 26 s and water temperature is 20 oC. The dynamic viscosity at 20oC is 1x10-3 Pa/s. Debate the G value and if out of the range what steps to take bring G in the correct range
1. A 770,000 gal conventional activated sludge pro- cess has a recycle ratio of 0.25, an influent flow rate of 1 MGD, a mixed liquor volatile suspended solids (MLVSS) concentration of 700 mg/L, and an influent BOD; of 160 mg/L. What is most nearly the F:M ratio of the process? (A) 0.1 lbm/day-lbm (B) 0.2 lbm/day-lbm (C) 0.3 lbm/day-lbm (D) 0.4 lbm/day-lbm В. D.
A plastic media biotower has vertical flow packing with n = 0.5 andk20 of 0.045 min–1. The tower is cylindrical with a diameter of 10 m.Depth of packing medium is 6 m. Primary effluent flow rate is 2500m3/d with a soluble BOD5 of 250 mg/L. The average wastewatertemperature is 16°C. Direct recirculation is practiced at 1:1 ratio.Calculate the following:a. Organic loading rate of the biotowerb. Hydraulic loading rate of the biotowerc. Recirculation ratiod. Effluent soluble BOD5 concentration
Environment Engg. Civil A surface water treatment plant operates round the clock with a flaw rate of 35 m³/min. The water temperature is 15°C and Jar testing indicated and alum dosage of 25 mg/l with flocculation at a Gt value of 4 x 104 producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant is
Example (Activated Sludge Process) - Completely Mixed with Recycle A completely mixed activated sludge process, treating a municipal wastewater, has a primary clarifier effluent BOD, of 135 mg/L. The design MLSS is 3590 mg/L, and the MLVSS is 80% of the MLSS. The plant permit is for an effluent BOD, of 20 mg/L and 20 mg/L suspended solids. The effluent suspended solids have a BOD, of 0.65 mg BOD, / mg suspended solids. The underflow solid concentration (MLVSS) is 10,000 mg/L. If umax = 3.0 day, Ks 60 mg/L, Y= 0.50 mg MLVSS / mg BOD, removed, ka= 0.1 day, and the influent flow is 0.25 m/sec, determine the following: 1. The required reactor detention time, 0. 2. The reactor basin volume, V, in m³. 3. The net waste activated sludge produced each day (kg/day). 4. The mass of solids (both volatile and inert) to be wasted each day (kg/day). The return sludge flow rate the recycle ratio. 6. The F/M ratio. 7 June 2021 Dr. Dawood Eisa Sachit Biological Wastewater Treatment 47

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