Determination of BOD dilution
Consider the BOD data below (no dilution,no seed).
a. What is the ultimate carbonaceous BOD?
b. What might have caused the lag at the beginning of the test?
c. Calculate k’, (the reaction rate constant).
Day DO(mg/L) ~~
0 8
1 8
3 7
5 6.5
7 6
15 4
20 4
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7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200...
7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200 mL of dilution water (containing no BOD). The initial DO of the mix is 9.0 mg/L After 5 days, the DO is 4.0 mg/L. After a long period of time, the DO is 2.0 mg/L, and it no longer seems to be dropping. Assume nitrification has been inhibited so that only carbonaceous BOD is measured (a) What is the 5-day BOD of the wastewater?...
Q2. A BOD3 test was run and the results were: Dilution = 20 Initial DO =...
Q2. A BOD3 test was run and the results were: Dilution = 20 Initial DO = 10.6 mg/L Final DO = 2.9 mg/L The reaction rate constant k has been found to be 0.25 day?. (a) What is the BOD3? (2 marks) (b) What would be the ultimate carbonaceous BOD? (3 marks) (c) What is the BOD5? (3 marks) (d) What uld be the remaining oxygen demand after five days (2 marks)
Determination of BOD dilution
5.15 A sample of wastewater is estimated to have a BOD of 200 mg/L. a. What dilution is necessary for this BOD to be measured by the usual technique? b. If the initial and final dissolved oxygen of the test thus conducted is 9.0 and 4.0 mg/L, and the dilution water has a BOD of 1.0 mgL, what is the dilution?
2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted;...
2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted; data are summarized in the table below. Ten mL of wastewater was added per 300 mL bottle to determine the dissolved oxygen demand of the aged, settled wastewater seed (test A). The seeded test bottles (test B) contained 2.5 mL of industrial wastewater and 1.2 mL of seed wastewater a) What is the five-day BOD in this industrial wastewater? What is the k-value using...
Determination of BOD dilution
A sample of wastewater is estimated to have a BOD of 200 mg/L. What dilution is necessary for this BOD to be measured by the usual technique?
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...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L....
Question 1 (a) What is water quality? Why is water quality testing important? (b) Write down...
Question 1 (a) What is water quality? Why is water quality testing important? (b) Write down the name of two organizations responsible for setting water quality standard? (c) The dilution factor D for an unseeded mixture of wastewater is 0.04. The DO of the mixture is initially 8.0 mg/L, and after 5 days it has dropped to 1.5 mg/L. The reaction rate constant K has been found to be 0.21 day!. (1) What is the 5-day BOD of the waste?...
(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...
7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200 mL of dilution water (containing no BOD). The initial DO of the mix is 9.0 mg/L After 5 days, the DO is 4.0 mg/L. After a long period of time, the DO is 2.0 mg/L, and it no longer seems to be dropping. Assume nitrification has been inhibited so that only carbonaceous BOD is measured (a) What is the 5-day BOD of the wastewater?...
Q2. A BOD3 test was run and the results were: Dilution = 20 Initial DO = 10.6 mg/L Final DO = 2.9 mg/L The reaction rate constant k has been found to be 0.25 day?. (a) What is the BOD3? (2 marks) (b) What would be the ultimate carbonaceous BOD? (3 marks) (c) What is the BOD5? (3 marks) (d) What uld be the remaining oxygen demand after five days (2 marks)
2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted; data are summarized in the table below. Ten mL of wastewater was added per 300 mL bottle to determine the dissolved oxygen demand of the aged, settled wastewater seed (test A). The seeded test bottles (test B) contained 2.5 mL of industrial wastewater and 1.2 mL of seed wastewater a) What is the five-day BOD in this industrial wastewater? What is the k-value using...
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...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L....
Question 1 (a) What is water quality? Why is water quality testing important? (b) Write down the name of two organizations responsible for setting water quality standard? (c) The dilution factor D for an unseeded mixture of wastewater is 0.04. The DO of the mixture is initially 8.0 mg/L, and after 5 days it has dropped to 1.5 mg/L. The reaction rate constant K has been found to be 0.21 day!. (1) What is the 5-day BOD of the waste?...
(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...
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