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Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To...
Two samples are brought to the laboratory for the BOD test. Please calculate BOD5 for both,...
Two samples are brought to the laboratory for the BOD test. Please calculate BOD5 for both, using laboratory DO data. a. For sample A, seeded BOD test was performed. The initial DO concentration in the BOD bottle, which was filled with 2 mL of sample and 298 mL of seeded dilution water, was 8.9 mg/L just after preparation. After an incubation period of five days, the final DO concentration was reported as 1.8 mg/L. Meanwhile, DO concentration in the seed...
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...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a standard BOD bottle. The 5- day BOD test provided an initial dissolved oxygen concentration (DO intial) = 9 mg/L and a final dissolved oxygen concentration (DO final) = 3.5 mg/L. The BOD experiment was carried out at 20°C and the BOD rate constant at 20°C is 0.22/day. a. Find the dilution factor (P) and BODs value at 20°CY b. Find the ultimate BOD of...
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...
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?...
Given the following BOD data: Time (days) Diluted sample DO (mg/l). 7.95 3.75 3.45 2.75 2.15...
Given the following BOD data: Time (days) Diluted sample DO (mg/l). 7.95 3.75 3.45 2.75 2.15 1.80 Blank seeded sample Do (mg/l) 8.15 8.1 8.05 8.0 7.95 7.9 Find the BOD at 5 days knowing that the diluted sample was diluted 100 times, i.e., P=1/100-volume wastewater/[volume of wastewater+dilution water (300 ml)).
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's...
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's effluent. For the experimental bottle, you insert 30 mL of test sample into the 300-mL ottle and fill the rest with seeded dilution water. You fill a blank bottle with seeded dilution ater. You measure the following dissolved oxygen (DO) concentrations at the start and end f the 5-day test: Bottle Initial DO (mg/L) Final DO (mg/L) Experimental 8.1 4.6 Blank 7.8 6.7 hat...
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....
A BOD test is run with 150 mL of wastewater mixed with 150 mL of deionized...
A BOD test is run with 150 mL of wastewater mixed with 150 mL of deionized water. The initial DO = 8.5 mg/L, the DO after 5 days = 4.2 mg/L, and the DO ultimately levels off at 1.7 mg/L. Assume all nitrification has been inhibited and the temperature remains constant. a) Calculate the CBOD5 (y5, mg/L) b) Calculate the ultimate CBOD (Lo, mg/L) c) Calculate the CBOD remaining after 5 days (L5, mg/L) d) Calculate the deoxygenation rate constant...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L)...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
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...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a standard BOD bottle. The 5- day BOD test provided an initial dissolved oxygen concentration (DO intial) = 9 mg/L and a final dissolved oxygen concentration (DO final) = 3.5 mg/L. The BOD experiment was carried out at 20°C and the BOD rate constant at 20°C is 0.22/day. a. Find the dilution factor (P) and BODs value at 20°CY b. Find the ultimate BOD of...
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...
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?...
Given the following BOD data: Time (days) Diluted sample DO (mg/l). 7.95 3.75 3.45 2.75 2.15 1.80 Blank seeded sample Do (mg/l) 8.15 8.1 8.05 8.0 7.95 7.9 Find the BOD at 5 days knowing that the diluted sample was diluted 100 times, i.e., P=1/100-volume wastewater/[volume of wastewater+dilution water (300 ml)).
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's effluent. For the experimental bottle, you insert 30 mL of test sample into the 300-mL ottle and fill the rest with seeded dilution water. You fill a blank bottle with seeded dilution ater. You measure the following dissolved oxygen (DO) concentrations at the start and end f the 5-day test: Bottle Initial DO (mg/L) Final DO (mg/L) Experimental 8.1 4.6 Blank 7.8 6.7 hat...
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....
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
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