Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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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?
Transcribed Image Text:### 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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- Question #5. The fluoride feeder at the Lemieux Island WTP breaks down. As shown below, the well mixed tank flows into a 5 km long distribution pipe. The volume of the tank is 2.5 m³ and the treatment system serves 190 000 people who each use 200 L/d. The rest of the necessary data is provided in the figure. If the fluoride concentration in the tank is 1.0 mg/L when the feed stops, how long until the concentration of fluoride is reduced to 0.01 mg/L at the end of the distribution pipe? Assume that fluoride acts as a conservative substance. Fluoride C = 1.0 mg/L 5 km u = 0.17 m/sarrow_forwardFollowing observations were made on a 4% dilution of wastewater. DO of aerated water used for dilution = 4 mg/1 DO of diluted sample after 5 days of incubation = 0.5 mg/1 DO of original wastewater sample = 0.2 mg/1 What is the value of BOD5 (in mg/1)?arrow_forwardAll information given... Don't say incomplete questionarrow_forward
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