a) 1st part
BOD or “Bio Chemical Oxygen Demand” is an important factor which describes us how pure the given water sample is. It represents the amount of “Dissolved Oxygen“ consumed by microbes to convert the organics present to carbon dioxide and water. Higher the BOD value higher is the water polluted. A sample of water with lesser BOD value is less polluted than a one with much higher value.
Generally as per global standards BOD value of sample at 5 days which incubated at 20 degrees Celsius is represented as the standard BOD of the sample. The reason for taking 5 days BOD as standard is due to the following fact:
The “Dissolved Oxygen” present in water gets used for the decomposition of organic matter as well as nitrogenous matter.
From the combined BOD curve, we have two oxygen demands: 1. Carbonaceous BOD (initial) , 2. Nitrogenous BOD (final).
The first 5 days demand is the “Carbonaceous Bio Chemical Oxygen Demand” which accounts for 68% of the total BOD value. Hence taking the value of 5 days BOD as a standard one represents the oxygen demand for the decomposition of organic matter only.
2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted;...
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?...
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...
QUESTION 5 5.1 Determine the 4-day BOD and the Ultimate BOD (1st stage) for a wastewater whose BODs at 20 °C is 250 mg/l. The reaction constant is k=0.23d" (base e). Determine the BODs if the test had been done at 25 °C. (10) 5.2. The chlorination unit in a WWTP uses 10 containers of chlorine on a monthly basis. A chlorine dosage is 4 mg/l is used to treat an average monthly wastewater effluent of 500 000 mDetermine the...
(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...
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...
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...
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 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...
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 ...
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