23.23 The following dissolved-oxygen values were found after 5 days of incubation in 305-mL BOD bottles: 8.2, 8.0, and 8.1 in three blank samples and 7.4, 6.5, 4.9, 1.5, and 0.0 mg/L in bottles containing 0.5, 1, 2, 4, and 8 mL of sample, respectively. The 0-day dissolved oxygen of the sample was 6.5 mg/L. What is the 5-day BOD of the sample? Why in the solution he did do only the two samples, 2 milliliters, and 4 milliliters? And if you have another solution for the same question that would be great. ### Calculation of 5-Day Biological Oxygen Demand (BOD₅)1. **Initial BOD Reading (B₁):** \[ B_1 = \frac{8.2 + 8.0 + 8.1}{3} = 8.1 \, \text{mg/L} \] Due to depletion in 0.5 ml and 1 ml samples being sufficient (mg/L), the depletion in the 8 ml sample is too large. Therefore, only two samples are used:2. **BOD Calculations for Samples:** - **BOD₅ for 2 ml Sample:** \[ \text{BOD}_5(2\, \text{ml}) = \frac{8.1 - 4.9}{\left(\frac{2}{305}\right)} = 488 \, \text{mg/L} \] - **BOD₅ for 4 ml Sample:** \[ \text{BOD}_5(4\, \text{ml}) = \frac{8.1 - 1.5}{\left(\frac{4}{305}\right)} = 503 \, \text{mg/L} \]3. **Average BOD Calculation:** Using the average of the two acceptable values: \[ \text{BOD}_5 = \frac{488 + 503}{2} = 496 \, \text{mg/L} \]4. **Final Result:** The 5-day BOD of the sample is **496 mg/L**.This calculation provides the BOD level, indicating the amount of dissolved oxygen needed by aerobic organisms to break down organic material in water over a period of five days.

The dissolved-oxygen values you provided, 8.2, 8.0, and 8.1, were found in three blank samples after…

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

See similar textbooks

Related questions

Concept explainers

Question

23.23 The following dissolved-oxygen values were found after 5 days of incubation in 305-mL BOD bottles: 8.2, 8.0, and 8.1 in three blank samples and 7.4, 6.5, 4.9, 1.5, and 0.0 mg/L in bottles containing 0.5, 1, 2, 4, and 8 mL of sample, respectively. The 0-day dissolved oxygen of the sample was 6.5 mg/L. What is the 5-day BOD of the sample?

Why in the solution he did do only the two samples, 2 milliliters, and 4 milliliters?

And if you have another solution for the same question that would be great.

### Calculation of 5-Day Biological Oxygen Demand (BOD₅)

1. **Initial BOD Reading (B₁):**

\[
B_1 = \frac{8.2 + 8.0 + 8.1}{3} = 8.1 \, \text{mg/L}
\]

Due to depletion in 0.5 ml and 1 ml samples being sufficient (mg/L), the depletion in the 8 ml sample is too large. Therefore, only two samples are used:

2. **BOD Calculations for Samples:**

- **BOD₅ for 2 ml Sample:**

\[
\text{BOD}_5(2\, \text{ml}) = \frac{8.1 - 4.9}{\left(\frac{2}{305}\right)} = 488 \, \text{mg/L}
\]

- **BOD₅ for 4 ml Sample:**

\[
\text{BOD}_5(4\, \text{ml}) = \frac{8.1 - 1.5}{\left(\frac{4}{305}\right)} = 503 \, \text{mg/L}
\]

3. **Average BOD Calculation:**

Using the average of the two acceptable values:

\[
\text{BOD}_5 = \frac{488 + 503}{2} = 496 \, \text{mg/L}
\]

4. **Final Result:**

The 5-day BOD of the sample is **496 mg/L**.

This calculation provides the BOD level, indicating the amount of dissolved oxygen needed by aerobic organisms to break down organic material in water over a period of five days.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 5 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions