A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the total n...
A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the total nitrogen concentration in the wastewater effluent if the wastewater contains 2 mg/L NH4+ and 10 mg/L NO3−. Report your answer in mg/L as N.
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A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the total n...
A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the...
A wastewater treatment plant does secondary treatment consisting of BOD oxidation and nitrification. What is the total nitrogen concentration in the wastewater effluent if the wastewater contains 2 mg/L NH4+ and 10 mg/L NO3−. Answer in mg/L as N.
A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume...
A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume of 24 million gallons per day (MGD) at an average influent 5-day biochemical oxygen demand concentration of 200 mgBODs/L and an average influent total suspended solids concentration of 220 mgTSS/L. The plant operates a primary sedimentation process that remove 65% of the incoming TSS and 35% of the incoming BOD %. This process is followed by a secondary treatment process before discharge to the...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by conventional activated sludge process composed of covered aeration tanks and secondary clarifiers. You are asked to calculate the different operation parameters to make sure that the system is operated within normal process ranges listed in the reference tables below. Table 5-20 Typical design information for primary sedimentation tanks U.S. customary units Item Unit Range Typical Primary sedimentation tanks followed by secondary treatment...
A treatment plant has the following species in the influent wastewater: NH3-N = 24 mg-N/L, 0.13...
A treatment plant has the following species in the influent wastewater: NH3-N = 24 mg-N/L, 0.13 mg NO2-N/L, 0.41 mg NO3-N/L, and a pH of 7. Using MINEQL+5.0 available on the COE computers generate a single pC-pH diagram showing the species, NH4+, NH3(aq), NO2-, NO3-, H+ and OH- as a function of pH (from 0.5 to 13.5 in 26 steps). Answer the following question, which nitrogen species is predominating at pH 6 and which is predominating at pH 11?
A Tapered aeration-system is used to treat 12,500 m3/d of municipal wastewater. The wastewater has received...
A Tapered aeration-system is used to treat 12,500 m3/d of municipal wastewater. The wastewater has received primary treatment and has a BOD 5 of 140 mg/l and a suspended solids of 125 mg/l. The system is to be operated in the following way: (a) Soluble BOD 5 in effluent 5 mg/l (b) Average solids concentration in the reactor= 2000 mg/l (c) Mean cell- retention time= 10 d. The biological constants have been determined by pilot-plant analysis and are: Y= 0.55...
:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter...
:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter received flow 44L/Sec , the BOD concentration in raw sewage =440 mg/L , 30% BOD removed in the primary sedimentation tank, volume of filter (first stage )=826000L, volume of filter (second stage)=2563000L, recirculation rate =0.5 in second tank while no recirculation in first tank. Determine the following: 1-The effluent concentration of BOD in each unit. 2-Treatment efficiency of each unit. 3-Treatment efficiency of plant....
7. You work for the Ministry of Environment to evaluate the maximum ultimate BOD concentration in...
7. You work for the Ministry of Environment to evaluate the maximum ultimate BOD concentration in the effluent of a new wastewater treatment plant such that the oxygen deficit never exceeds 1.5 mg/L in the receiving river. During dry weather (i.e., lowest river flow), the wastewater will be diluted 10 times in the river water and the river water contains 3.5 mg/L of ultimate BOD. Also during dry weather, you know that the river velocity after the wastewater discharge would...
You are invited to prepare a report of expertise on the treatment of a wastewater produced...
You are invited to prepare a report of expertise on the treatment of a wastewater produced by an oil company. The wastewater was analyzed in advance. You are provided with the following values: Soluble COD = 15.300 kg/ m3 Soluble BOD = 13.83 kg/ m3 The Total COD of the effluent was 0.2588 kg/ m3 The treatment is performed in a reactor having a total volume of 500 m3. The first step of developing the secondary treatment is a batch...
Nitrogen in a wastewater treatment plant is in the form of ammonia and ammonium ion and...
Nitrogen in a wastewater treatment plant is in the form of ammonia and ammonium ion and has a total ammonia concentration of 7.1 x 10^-4 M at pH = 7. Total ammonia concentration = [NH3] + [NH4^+]. NH3 partial pressure = 5 x 10^-10 atm. KH at 25-degree Celcius is 57 mol/(L x atm). Ka at 25-degree Celcius = 10^-9.26. A. Calculate NH3 and [NH4+] in M at the present condition where pH = 7 at 25 degrees Celcius. B....
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to...
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BODs and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow = 0.2000 m 3/s soluble BODs-80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%...
A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume of 24 million gallons per day (MGD) at an average influent 5-day biochemical oxygen demand concentration of 200 mgBODs/L and an average influent total suspended solids concentration of 220 mgTSS/L. The plant operates a primary sedimentation process that remove 65% of the incoming TSS and 35% of the incoming BOD %. This process is followed by a secondary treatment process before discharge to the...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by conventional activated sludge process composed of covered aeration tanks and secondary clarifiers. You are asked to calculate the different operation parameters to make sure that the system is operated within normal process ranges listed in the reference tables below. Table 5-20 Typical design information for primary sedimentation tanks U.S. customary units Item Unit Range Typical Primary sedimentation tanks followed by secondary treatment...
7. You work for the Ministry of Environment to evaluate the maximum ultimate BOD concentration in the effluent of a new wastewater treatment plant such that the oxygen deficit never exceeds 1.5 mg/L in the receiving river. During dry weather (i.e., lowest river flow), the wastewater will be diluted 10 times in the river water and the river water contains 3.5 mg/L of ultimate BOD. Also during dry weather, you know that the river velocity after the wastewater discharge would...
2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BODs and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow = 0.2000 m 3/s soluble BODs-80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%...
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