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?

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?

Solid Waste Engineering

3rd Edition

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Worrell, William A.

Chapter2: Municipal Solid Waste Characteristics And Quantities

Section: Chapter Questions

Problem 2.14P

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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 2.950
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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.
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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?
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A BOD test is run using 100mL of treated wastewater. The initial DO of the mixture after diluting the waste sample with distilled water is 9.0mg/L. After 5-day incubation, the DO measured was 4.0mg/L. After a long period of time, the DO is 2.0mg/L and no longer seems to be decreasing. Assume that nitrification has been inhibited. Determine the following: a)BOD5 & BODu of the waste sample b) BOD5 of the mixture and c) k THANK YOU
IL Test sample weight= احتمان Sieve size Weight retained (g) Calculations: Sieve size in. (mm) 3/8" 9.5 mm No.4 4.75 mm No.8 2.36 mm No.16 1.18 mm No.30 0.6 mm No.50 0.3 mm No.100 0.15 mm Pan gm 3/8" No.4 No.8 No.16 No.50 No.100 0.15 mm 9.5 mm 4.75 mm 2.36 mm 1.18 mm mm 0.3 mm 0.0 од 1597 21419 від 50д 3589 174д Weight retained (g) Percent retained Beitel 15-11-2021 Fineness modulus отсивно о u 4/8²/00 = 5.3 15 діни з 2141-44871.3 61-448203 Божнизис 1.66 50 78.63 21.3 80.29 19.71 35 84421 23 92.22 7.78 174448518 98.02 1.98 чудд 4 /100 = Cumulative = No.30 Percent cumulative 0 53 94.7 766 23-4 Percent passing loo 222 Sieve deviation = 262 (5) ASTM Limit 100 95 100 80 100 50_85 25 60 10_30 2_10
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