The ultimate BOD of a river just below a sewage outfall is 55.0 mg/L, and the...

The ultimate BOD of a river just below a sewage outfall is 55.0 mg/L, and the DO is at a saturation value of 12.0 mg/L. The deoxygenation rate coefficient kd is 0.30/day, and the reaction rate coefficient kr is 0.90/day. The river is flowing at the speed if 52.0 miles per day. The only source of BOD on this river is this single outfall.
1. Find the critical distance downstream at which DO is a minimum.
2. Find the minimum DO.
3. If a wastewater treatment plant is to be built, what fraction of the BOD would have
  
  to be removed from the sewage to assure a minimum of 5.0 mg/L everywhere
  
  downstream?

Solutions

Expert Solution

Ally Wells answered 1 year ago

Next > < Previous

Related Solutions

Determine the Theorethical Oxygen Demand(ThOD) for sewage containing 150 mg/L of CH2(NH2)COOH and 40 mg/L of...

Determine the Theorethical Oxygen Demand(ThOD) for sewage containing 150 mg/L of CH2(NH2)COOH and 40 mg/L of NH3. Find, (1) mg/L of oxygen for both carbon and nitrogen compounds (2) grams of oxygen per gram of CH2(NH2)COOH (oxygen used only for carbon oxidation) (3) grams of oxygen per gram of NH3 (oxygen used only for nitrification)

A 3 diameter biotower has been designed to treat sewage from 22,000 PE.Influent BOD = 250mg/l...

A 3 diameter biotower has been designed to treat sewage from 22,000 PE.Influent BOD = 250mg/l and wastewater flow through the primary clarifier removes 35 percent of the BOD and the flows into the biotower the constant r andom media n= 0.44, recycle ratio = 2 and operating temperature =2500c. a determine the reaction rate constant k25 if the existing BOD =50mg/l b what would be the biotower effluent at 200c if the recycle ratio is increase to 4

An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The...

An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The sludge is aerated for 6 hr. The food-to-microorganism ratio is 0.30. The recirculation ratio is 0.2. The surface loading rate of the secondary clarifier is 800 gal/day-ft2. The final effluent has a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the activated sludge treatment processes (secondary BOD removal), (b) aeration tank volume,...

A river is contaminated with 0.65 mg/L of dichloroethylene (C2H2Cl2). What is the concentration (in ng/L)...

A river is contaminated with 0.65 mg/L of dichloroethylene (C2H2Cl2). What is the concentration (in ng/L) of dichloroethylene at 21 C in the air breathed by a person swimming in the river (kH for C2H2Cl2 in water is 0.033 mol/L * atm)

The average flow velocity is 0.305 m/s. At the source, the BOD concentration is 10.0 mg/L...

The average flow velocity is 0.305 m/s. At the source, the BOD concentration is 10.0 mg/L and the dissolved oxygen deficit is 0.0 mg/L. From field sampling, de-oxygenation (kd) and re-aeration (ka) rate constants are estimated to be 0.6 d-1 and 2.0 d-1, respectively. Find the critical distance, the critical deficit, and the time at which the minimum D.O. occurs

Which wastewater treatment process is responsible for the removal of most of the BOD in sewage?

(7) The following values were obtained for the nitrate concentration (mg/L) in a sample of river...

(7) The following values were obtained for the nitrate concentration (mg/L) in a sample of river water: 0.403, 0.410, 0.401, and 0.380. The last measurement is suspect. Based on this data set only, should it be rejected? Justify your answer. (8) Three further measurements were added to those given in question (7) so that the complete results became: 0.403, 0.410, 0.401, 0.380, 0.400, 0.413, and 0.411. Should “0.380” still be retained/rejected? Justify your answer. (9) Based on questions (7) and...

You have wastewater plant discharging into a clean river. The BOD of the Wastewater flow is...

You have wastewater plant discharging into a clean river. The BOD of the Wastewater flow is 300 mg/L, the wastewater flow is 15 m3/sec, the river flow upstream of the mixing point is 120 m3/sec and BOD upstream of river is 0. The DO of the river at the mixed point is 8 mg/L and the saturation DO in the river would be 10 mg/L, the deoxygenation rate coefficient is 0.30/day and the reaeration constant is 0.90/day. The river is...

River water between Back Falls and Angel Falls contains 0.0769 mg Al3+ per L and has...

River water between Back Falls and Angel Falls contains 0.0769 mg Al3+ per L and has a pH of 8.20. Total carbonates are 0.005 mol L-1 . (a) [5 marks] Calculate the concentrations of HCO3 - (aq) and CO3 2- (aq) (in mol L-1 ). For H2CO3(aq), the hydrolysis constants are Ka1 = 4.45 × 10-7 and Ka1 = 4.69 × 10-11 (b) [5 marks] Calculate the alkalinity

Calculate the molarity of the following: (a) 90 mg/L HNO3, (b) 150 mg/L CaCO3,

Subjects