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While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's...
3. (6 pts) Raw wastewater with initial BOD5 = 265 mg/L is treated in Wastewater Treatment...
3. (6 pts) Raw wastewater with initial BOD5 = 265 mg/L is treated in Wastewater Treatment Plant (WWTP), which can remove 65% of initial BOD. Students were asked to run BODs tests on the treated wastewater AFTER treatment. They used standard 300 ml bottle a. Find the volume of treated wastewater that should be put in the bottle to make sure that DO after 5 days will be at least 2mg/L (assume initial DO = 9.2 mg/L; and the bottle...
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...
You have collected water samples of the influent (raw sewage) and effluent (treated sewage) at Houston Wastewater Treatment Plant. You plan to conduct the BOD5 test to determine the BOD rate constant,...
You have collected water samples of the influent (raw sewage) and effluent (treated sewage) at Houston Wastewater Treatment Plant. You plan to conduct the BOD5 test to determine the BOD rate constant, k. A. Would you expect the rate constants to be the same or different in these two samples? If different, which would be higher and why? B. If the BOD5 for the raw sewage sample is 600 mg/L, and the ultimate BOD is 650 mg/L at 20oC, find...
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
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....
Water with 1.0 mg/L methylene chloride + 0.01 mg/L tetrachloroethylene), 70 kg adult drinks 2L/day for...
Water with 1.0 mg/L methylene chloride + 0.01 mg/L tetrachloroethylene), 70 kg adult drinks 2L/day for 30 years. What is hazard index? RfD = 0.060 for methylene chloride, RfD = 0.010 for tetrachloride 1 0.512 2 0.029 3 0.483 4 0.00029 q2) A test bottle (Blank) containing just seeded dilution water has its DO level drop by 1.1 mg/L in a five-day test. A 300-mL BOD bottle filled with 30 mL of wastewater and the rest seeded dilution water (this...
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...
6) (10 points) The BODs of a wastewater sample is expected to range from 25 to...
6) (10 points) The BODs of a wastewater sample is expected to range from 25 to 200 mg/L. Three different dilutions are prepared to cover the range. The same procedure was used for each analysis and 300 ml BOD bottles are used for the BODs test. Please determine average BODs value with standard deviation Sample Volume Initial DO, mg/L Final DO, mg/L 5 mL 9.2 6.7 10 mL20 mL 9.0 1.5 8.9 4.5
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...
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)).
3. (6 pts) Raw wastewater with initial BOD5 = 265 mg/L is treated in Wastewater Treatment Plant (WWTP), which can remove 65% of initial BOD. Students were asked to run BODs tests on the treated wastewater AFTER treatment. They used standard 300 ml bottle a. Find the volume of treated wastewater that should be put in the bottle to make sure that DO after 5 days will be at least 2mg/L (assume initial DO = 9.2 mg/L; and the bottle...
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
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....
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...
6) (10 points) The BODs of a wastewater sample is expected to range from 25 to 200 mg/L. Three different dilutions are prepared to cover the range. The same procedure was used for each analysis and 300 ml BOD bottles are used for the BODs test. Please determine average BODs value with standard deviation Sample Volume Initial DO, mg/L Final DO, mg/L 5 mL 9.2 6.7 10 mL20 mL 9.0 1.5 8.9 4.5
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...
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)).
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