A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial DO of 8.55 mg/L; after five days, it has a final DO of 2.40 mg/L. Another bottle containing just the seeded dilution water has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L. Find the five-day BOD of the waste.
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A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial DO of 8.55 mg/L; after five days, it has a final DO of 2.40 mg/L. Another bottle containing just the seeded dilution water has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L. Find the five-day BOD of the waste.
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- 2. The BOD5 of a waste sample was found to be 60 mg/L The initial oxygen concentration of the BOD dilution water was equal to 8.6 mg/L, the DO concentration measured after incubation was equal to 8.6 mg/L and the size of the sample used was equal to 10 mL. If the volume of the BOD bottle used was equal to 300 mL, then the initial DO concentration (in mg/L) of the waste sample is: O 2.025 2.050 O 2.250 2.400 2.600 2.850 2.950A standard BOD test is run using seeded dilution water. The blank bottle containing the seeded dilution water has an initial DO of 9.2 mg/L and a final DO of 8.2 mg/L. The waste sample is mixed with seeded dilution water at a 1:25 ratio. It has an initial DO of 9.0 mg/L and a final DO of 2.0 mg/L. Calculate the BOD5.1. A mixture consisting of 30 mL of waste and 270 mL of seeded dilution water has an initial Do of 8.55 mg/L; after 5 days, it has a final DO of 2.40 mg/L. Another bottle containing just the seeded dilution water has an initial DO of 8.75 mg/L and a final DO of 8.53 mg/L. Find the 5-days BOD of the waste.
- A test bottle containing just seeded dilution water has its DO level drop by 1.0 mg/L in a five-day test. A 300-ml BOD bottle just filled with 15 ml of wastewater and the rest seeded dilution water (sometimes expressed as a dilution of 1:20) experiences a drop of 7.2 mg/L in the same period. What would be the five day BOD of the waste?(a) Calculate the ultimate BOD of a waste that has a measured 5-day BOD of 20 mg/L, assuming a BOD rate coefficient of 0.15/day measured at 20 degrees C. (b) Estimate the rate coefficient if the temperature of the waste is raised to 30 degrees C. (c) What would be the BOD5 at 30 deg. C?A BOD test is run using 100 ml of treated waste H20 mixed with 200ml of pure water. The initial DO of the mixture is 9.0 mg/ml. After 5 days, Do id 4.0 mg/ml. What is the 5-day BOD of the waste water?
- A BOD analysis was begun on Monday. Fifteen ml of waste with a DO of zero was mixed with 270 mL of dilution water with a DO of 9 mg/L. The sample was then put in the incubator. Since the 5th day fell on a long weekend (Friday) and lab personnel do not work on Saturday, the final DO did not get measured until Monday, the eighth day. The final DO was measured at 2.8 mg/L. However, it was discovered that the incubator was mistakenly set at 32 ºC. Assume k is 0.15 day-1 at 20 ºC and θ= 1.05. Determine the 5-day 20 ºC BOD of the sample. Also, what is the ultimate CBOD at 32 ºC and 20 ºC; Will it vary?An air classifier is applied to separate a mixture of four solid waste components shown in Figure 5-20 above. All four components have equal mass.a) Determine the mass composition in percent of the four components in theoverflow (extract) if the air classifier is operated at an air velocity of 1,000 ft/min.b) Determine the mass composition in percent of the four components in theoverflow (extract) if the air classifier is operated at an air velocity of 2,000 ft/minc) Complete the following Table Q2 (c) and calculate the moisture content of solid waste. Table Q2 (c) Compositions Wet percent (%) Moisture Dry Percent (%) 51 80 9 12 31 60 9 6 Food Plastics Rubber Miscellaneous
- Font (1) If the BOD5 of a waste is 220.0 mg/L and the ultimate BOD is 320.0 mg/L, what is the rate constant?3. A hard drive manufacturer uses Nickel to plate its disks and then polishes them, resulting in a concentration of 50 mg/l in their 50,000 gallons per day wastewater stream. Their industrial permit standard for Ni discharge into the sewer system and treatment plant is limited to 0.2 Ibs./day. What percentage of the mass of Nickel must the onsite treatment system remove to be in compliance with the discharge standard? Use the following formula to determine the daily Nickel loading of the manufacturing system: Loading (Ibs/day) = Conversion Factor (8. Flow(MGD)For the sample of solid wastes with the typical comp osition given in the following Table component Food Раper Card Plastic Textile Rubber Leath Garden Wood Glass Tin Non Ferro- Drift waste board trimmin er cans us ferrous -gs metals ashes metals bric k.etc Moisture 15 6 2 2 10 60 20 3 2 3 content % Dry 4.5 37.6 3.8 2.9 1.8 0.5 0.4 4.8 1.6 7.8 5.8 1.0 1.9 3.7 weight, Ib Determine : Moisture content of Textile (X) waste if the Overall moisture content ( 21.9 %).