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?

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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?
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?...
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...
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...
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 75 6.5 7 6 15 ...
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%
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...
QUESTION1 A 15-mL sample is added to à 300-ml 8OD bottle. Dilution water is added to the sample bottle and the initial dissoved axygen concentration is measured as 8.5 mg/L. After the sample is sealed, the laboratory incorrectly takes a measurement of dissolved oxygen reading on day 6 of 3 mg/L. If the BOD reaction rate coefticient for the sample is 0.30/day, estimate what the 5-day BOD should have been QUESTION 2 A 15-mL sample is added to a 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....