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 control decreased from 9.1 to 8.9 mg/L within five days. What is the 5-day BOD of the sample A?
b. For sample B, an unseeded BOD test was performed. What would be the 5- day BOD of the sample if the same dilution is performed as that of Sample A, and found out that the DO changed from 8.9 to 7.2 mg/L?
c. Discuss the possible origins of samples A and B: i. Are they taken from (i) the tap in someone’s home, (ii) a river, (iii) the inlet of a wastewater treatment plant (raw wastewater), (iv) outlet of a wastewater treatment plant (treated effluent). ii. What does conducting a seeded vs. unseeded BOD test mean about the possible origins of a sample?
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Two samples are brought to the laboratory for the BOD test.
Please calculate BOD5 for both,...
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...
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...
Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To...
Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To run this test, you mix 25 mL of the wastewater sample with clean (unseeded) dilution water to bring the total volume in the bottle to 300 ml, After measuring the initial dissolved oxygen concentration, you cap the BOD bottle and monitor the dissolved oxygen concentration of the diluted sample in the BOD bottle as a function of time (see Figure below). Previous tests have...
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?...
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...
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....
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...
The following data have been obtained in a bOD test that is made to determine how...
The following data have been obtained in a bOD test that is made to determine how well a wastewater treatment plant is operating: Initial Do Final DO Vol. of Wastewater Vol. of Dilution Water (mg/L) (mg/L) (mL) (mL) Untreated Sewage 5.4 10 290 Treated Sewage 8.7 3 280 3 20 What percentage of the BOD is being removed by this treatment plant? 72% 86% 16% 32%
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
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 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...
Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To run this test, you mix 25 mL of the wastewater sample with clean (unseeded) dilution water to bring the total volume in the bottle to 300 ml, After measuring the initial dissolved oxygen concentration, you cap the BOD bottle and monitor the dissolved oxygen concentration of the diluted sample in the BOD bottle as a function of time (see Figure below). Previous tests have...
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?...
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...
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....
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...
The following data have been obtained in a bOD test that is made to determine how well a wastewater treatment plant is operating: Initial Do Final DO Vol. of Wastewater Vol. of Dilution Water (mg/L) (mg/L) (mL) (mL) Untreated Sewage 5.4 10 290 Treated Sewage 8.7 3 280 3 20 What percentage of the BOD is being removed by this treatment plant? 72% 86% 16% 32%
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
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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