(a)The 5-day BOD of a sewage sample is 180 mg/L at 15°C. What is the BODu...
(a)The 5-day BOD of a sewage sample is 180 mg/L at 15°C. What is the BODu when k is 0.52 d-1 at 25°C? (use the Arrhenius equation, kt2=kt1?t2-t1 where ?=1.135 for 4-20°C and 1.056 for 20 -30°C
(b)Graph the BOD as a function of time for the previous problem for both 15°C and 25°C. What do you observe?
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(a)The 5-day BOD of a sewage sample is 180 mg/L at 15°C. What is
the BODu...
2. The 5-day, 20°C BOD of a wastewater (K 0.23/day, 0 1.047) is 185 mg/L. What...
2. The 5-day, 20°C BOD of a wastewater (K 0.23/day, 0 1.047) is 185 mg/L. What is the corresponding 10-day demand and ultimate BOD (in mg/L)? If the bottle had been incubated at 33°C, what would the 5-day BOD value have been?
5. The following data is for a BOD determination run on a raw sewage sample incubated...
5. The following data is for a BOD determination run on a raw sewage sample incubated for 5 days at 20°C in the dark; calculate the BOD5,20. Raw sewage contains a sufficient population of bacteria to conduct a BOD test. Therefore, this is an unseeded sample. Data: BOD bottle volume mL raw sewage added to BOD bottles DO, in bottle containing waste and dilution water DO in bottle containing waste and dilution water 300 mL 10.00 mL 7.70 mg/L 2.35...
A wastewater sample has a temperature of 10 °C. The 5-day BOD (BOD) of the wastewater...
A wastewater sample has a temperature of 10 °C. The 5-day BOD (BOD) of the wastewater sample is determined as 120 mg/L at 10 °C, and the BOD decay constant (k) is known as 0.20 day 1 at 20°C. Calculate the BOD decay constant (k) at 10 °C.
2. A 5-day BOD of a sample was determined to be 250.0 mg/L and the rate...
2. A 5-day BOD of a sample was determined to be 250.0 mg/L and the rate constant is 0.35 d. What is the ultimate BOD of the sample?
Calculate the BOD Ultimate value given the following information BOD5= 316 mg/L k = 0.23 day-1...
Calculate the BOD Ultimate value given the following information BOD5= 316 mg/L k = 0.23 day-1 Temperature is 20 C for both BOD5 and BOD Ultimate (L)
QUESTION 5 5.1 Determine the 4-day BOD and the Ultimate BOD (1st stage) for a wastewater...
QUESTION 5 5.1 Determine the 4-day BOD and the Ultimate BOD (1st stage) for a wastewater whose BODs at 20 °C is 250 mg/l. The reaction constant is k=0.23d" (base e). Determine the BODs if the test had been done at 25 °C. (10) 5.2. The chlorination unit in a WWTP uses 10 containers of chlorine on a monthly basis. A chlorine dosage is 4 mg/l is used to treat an average monthly wastewater effluent of 500 000 mDetermine the...
(20 pts) The ultimate BOD of a domestic wastewater is 275 mg/L and the e based...
(20 pts) The ultimate BOD of a domestic wastewater is 275 mg/L and the e based rate constant k=0.24 d a. (4 pts) What will be the 5-day BOD (Sample-1)? b. (4 pts) What will be the 7-day BOD? (Sample-1)? c. (4 pts) A different sample has the same ultimate BOD given in the question, but the 20 °C k value (e based) is equal to 0.36 dl Determine the 5-day BOD of this sample (Sample-2). d. (8 pts) Please...
\The flow rate to treat is 5.6 MGD and the BOD concentration is 150 mg/L. The...
\The flow rate to treat is 5.6 MGD and the BOD concentration is 150 mg/L. The design solids concentration (X) at steady-state is 2,000 mg/L. The design MCRT is 7 days. The kinetic coefficients are as follows: k = 2 g BOD/g cells*day; Ks = 25 mg BOD/L; kd = 0.06 1/day; Y = 0.5 g cells/g BOD. The influent ammonia concentration is 40 mg/L and nitrification is needed. It takes 1400 ft3 of air per pound of BOD. Use...
QUESTION 1 6 point City of Hope's Wastewater has a ultimate carbonaceous BOD of 300 mg/L...
QUESTION 1 6 point City of Hope's Wastewater has a ultimate carbonaceous BOD of 300 mg/L at 20 oC. The reaction rate k at that temperature has been determined to be 0.2/day. a) Find the BOD5. b)Find the reaction rate coefficient at 15 oC. c) Find BOD5 at 15 oC. O 132 mg/L, 0.18/day, 265 mg/L O 190 mg/L, 0.16/day, 165 mg/L O90 mg/L, 0.16/day, 245 mg/L O 232 mg/L, 0.18/day, 365 mg/L
5. A series of BOD (ATU) determinations was made in a sample to enable calculation of...
5. A series of BOD (ATU) determinations was made in a sample to enable calculation of the ultimate BOD and rate constant. Incubation was carried out on a 5 per cent dilution of the sample at 20°C when the initial saturation DO for samples and blanks was 9.10 mg/. Day Final Do in sample (mg/l) Final Do in dilution water (mg/l) 1 2 3 4 5 6 7 7.10 6.10 5.10 4.20 3.90 3.50 3.00 9.00 9.00 8.90 8.90 8.80...
2. The 5-day, 20°C BOD of a wastewater (K 0.23/day, 0 1.047) is 185 mg/L. What is the corresponding 10-day demand and ultimate BOD (in mg/L)? If the bottle had been incubated at 33°C, what would the 5-day BOD value have been?
5. The following data is for a BOD determination run on a raw sewage sample incubated for 5 days at 20°C in the dark; calculate the BOD5,20. Raw sewage contains a sufficient population of bacteria to conduct a BOD test. Therefore, this is an unseeded sample. Data: BOD bottle volume mL raw sewage added to BOD bottles DO, in bottle containing waste and dilution water DO in bottle containing waste and dilution water 300 mL 10.00 mL 7.70 mg/L 2.35...
A wastewater sample has a temperature of 10 °C. The 5-day BOD (BOD) of the wastewater sample is determined as 120 mg/L at 10 °C, and the BOD decay constant (k) is known as 0.20 day 1 at 20°C. Calculate the BOD decay constant (k) at 10 °C.
2. A 5-day BOD of a sample was determined to be 250.0 mg/L and the rate constant is 0.35 d. What is the ultimate BOD of the sample?
QUESTION 5 5.1 Determine the 4-day BOD and the Ultimate BOD (1st stage) for a wastewater whose BODs at 20 °C is 250 mg/l. The reaction constant is k=0.23d" (base e). Determine the BODs if the test had been done at 25 °C. (10) 5.2. The chlorination unit in a WWTP uses 10 containers of chlorine on a monthly basis. A chlorine dosage is 4 mg/l is used to treat an average monthly wastewater effluent of 500 000 mDetermine the...
(20 pts) The ultimate BOD of a domestic wastewater is 275 mg/L and the e based rate constant k=0.24 d a. (4 pts) What will be the 5-day BOD (Sample-1)? b. (4 pts) What will be the 7-day BOD? (Sample-1)? c. (4 pts) A different sample has the same ultimate BOD given in the question, but the 20 °C k value (e based) is equal to 0.36 dl Determine the 5-day BOD of this sample (Sample-2). d. (8 pts) Please...
QUESTION 1 6 point City of Hope's Wastewater has a ultimate carbonaceous BOD of 300 mg/L at 20 oC. The reaction rate k at that temperature has been determined to be 0.2/day. a) Find the BOD5. b)Find the reaction rate coefficient at 15 oC. c) Find BOD5 at 15 oC. O 132 mg/L, 0.18/day, 265 mg/L O 190 mg/L, 0.16/day, 165 mg/L O90 mg/L, 0.16/day, 245 mg/L O 232 mg/L, 0.18/day, 365 mg/L
5. A series of BOD (ATU) determinations was made in a sample to enable calculation of the ultimate BOD and rate constant. Incubation was carried out on a 5 per cent dilution of the sample at 20°C when the initial saturation DO for samples and blanks was 9.10 mg/. Day Final Do in sample (mg/l) Final Do in dilution water (mg/l) 1 2 3 4 5 6 7 7.10 6.10 5.10 4.20 3.90 3.50 3.00 9.00 9.00 8.90 8.90 8.80...
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