Suppose you are to measure the BOD (biochemical oxygen demand) removal rate for a primary wastewater treatment plant. You take two samples of raw sewage on its way into the plant and two samples of the effluent leaving the plant. Standard five day BOD tests are run on the four samples with no seeding, producing the following data (See Figure attached) Find BOD5 for the raw and treated sewage, and the percent removal of BOD in the treatment plant. Find the DO (dissolved oxygen) that would be expected in sample two at the end of the test.What would be the maximum volume of treated sewage for Sample 4 that could be put into the 300-mL BOD bottle and still have the DO after five days remain above 2 mg/L? ### Biological Oxygen Demand (BOD) Analysis**Questions:**(a) What is the five-day BOD of the wastewater?(b) Assuming no nitrification effects, estimate the ultimate carbonaceous BOD.(c) What would be the remaining BOD after five days have elapsed?(d) Estimate the reaction rate constant \(k\) (day\(^{-1}\)).**Experimental Setup:**Suppose you are to measure the BOD removal rate for a primary wastewater treatment plant. You take two samples of raw sewage on its way into the plant and two samples of the effluent leaving the plant. Standard five-day BOD tests are run on the four samples, with no seeding, producing the following data:| Sample | Source | Dilution | DO\(_i\) (mg/L) | DO\(_f\) (mg/L) ||--------|---------|----------|----------------|----------------|| 1 | Raw | 1:30 | 9.2 | 2.2 || 2 | Raw | 1:15 | 9.2 | ? || 3 | Treated | 1:20 | 9.0 | 2.0 || 4 | Treated | ? | 9.0 | >0 |- **Sample Information:** - **Sample 1**: Raw sewage with a dilution of 1:30. Initial DO is 9.2 mg/L, final DO is 2.2 mg/L. - **Sample 2**: Raw sewage with a dilution of 1:15. Initial DO is 9.2 mg/L, final DO is unknown. - **Sample 3**: Treated sewage with a dilution of 1:20. Initial DO is 9.0 mg/L, final DO is 2.0 mg/L. - **Sample 4**: Treated sewage with undetermined dilution. Initial DO is 9.0 mg/L, final DO is greater than 0 mg/L.This analysis helps in understanding how effectively the wastewater treatment process reduces the BOD, indicating the level of organic pollutants in the water.

From the given data,now calculating for the treated sewage is,Substitute all the values, we get…

Answered: Suppose you are to measure the BOD…

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Suppose you are to measure the BOD (biochemical oxygen demand) removal rate for a primary wastewater treatment plant. You take two samples of raw sewage on its way into the plant and two samples of the effluent leaving the plant. Standard five day BOD tests are run on the four samples with no seeding, producing the following data (See Figure attached)

Find BOD₅ for the raw and treated sewage, and the percent removal of BOD in the treatment plant.
Find the DO (dissolved oxygen) that would be expected in sample two at the end of the test.
What would be the maximum volume of treated sewage for Sample 4 that could be put into the 300-mL BOD bottle and still have the DO after five days remain above 2 mg/L?

### Biological Oxygen Demand (BOD) Analysis

**Questions:**
(a) What is the five-day BOD of the wastewater?<br>
(b) Assuming no nitrification effects, estimate the ultimate carbonaceous BOD.<br>
(c) What would be the remaining BOD after five days have elapsed?<br>
(d) Estimate the reaction rate constant \(k\) (day\(^{-1}\)).

**Experimental Setup:**

Suppose you are to measure the BOD removal rate for a primary wastewater treatment plant. You take two samples of raw sewage on its way into the plant and two samples of the effluent leaving the plant. Standard five-day BOD tests are run on the four samples, with no seeding, producing the following data:

| Sample | Source | Dilution | DO\(_i\) (mg/L) | DO\(_f\) (mg/L) |
|--------|---------|----------|----------------|----------------|
| 1 | Raw | 1:30 | 9.2 | 2.2 |
| 2 | Raw | 1:15 | 9.2 | ? |
| 3 | Treated | 1:20 | 9.0 | 2.0 |
| 4 | Treated | ? | 9.0 | >0 |

- **Sample Information:**
- **Sample 1**: Raw sewage with a dilution of 1:30. Initial DO is 9.2 mg/L, final DO is 2.2 mg/L.
- **Sample 2**: Raw sewage with a dilution of 1:15. Initial DO is 9.2 mg/L, final DO is unknown.
- **Sample 3**: Treated sewage with a dilution of 1:20. Initial DO is 9.0 mg/L, final DO is 2.0 mg/L.
- **Sample 4**: Treated sewage with undetermined dilution. Initial DO is 9.0 mg/L, final DO is greater than 0 mg/L.

This analysis helps in understanding how effectively the wastewater treatment process reduces the BOD, indicating the level of organic pollutants in the water.

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