23.22 The following dissolved-oxygen values were found after 5 days of incubation in 310-mL BOD bottles: 7.7, 7.9, and 7.9 in three blank samples; 6.5, 4.0, and 0.5 mg/L in bottles containing 2, 5, and 10 mL of sample, respectively. The 0-day dissolved oxygen of the sample was 0.0 mg/L. What is the 5-day BOD of the sample? The 5 days BOD of the given wastewater sample is,BOD5=(DO initial - DO final)x Dilution factorvolume of BOD bottlevolume of waste waterDO initialInitial dissolved oxygenDilution factor =DO final Final dissolved oxygenNow calculating the Data of 5 days BOD from above equation,BOD5=(DO initialDO final)x Dilution factor(1)BOD5=(7.7-6.5)× 155BOD5 = 186BOD5=(DO initial - DO final)x Dilution factor(2) BOD5=(7.9-4)x62BOD5 = 241.8BOD5 = (DO initial - DO final)x Dilution factor(3) BOD5=(7.7-0.5)x31BOD5 = 229.4

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Answered: The following dissolved-oxygen values…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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23.22 The following dissolved-oxygen values were found after 5 days of incubation in 310-mL BOD bottles: 7.7, 7.9, and 7.9 in three blank samples; 6.5, 4.0, and 0.5 mg/L in bottles containing 2, 5, and 10 mL of sample, respectively. The 0-day dissolved oxygen of the sample was 0.0 mg/L. What is the 5-day BOD of the sample?

The 5 days BOD of the given wastewater sample is,
BOD5=(DO initial - DO final)x Dilution factor
volume of BOD bottle
volume of waste water
DO initialInitial dissolved oxygen
Dilution factor =
DO final Final dissolved oxygen
Now calculating the Data of 5 days BOD from above equation,
BOD5=(DO initialDO final)x Dilution factor
(1)BOD5=(7.7-6.5)× 155
BOD5 = 186
BOD5=(DO initial - DO final)x Dilution factor
(2) BOD5=(7.9-4)x62
BOD5 = 241.8
BOD5 = (DO initial - DO final)x Dilution factor
(3) BOD5=(7.7-0.5)x31
BOD5 = 229.4

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