Calculate Overflow Rate w/ Interpolation: BOD Removal 80%

In summary, the conversation is about finding the dimensions of a circular sedimentation tank for a treatment plant with a primary sedimentation tank, trickling filter, and constructed marsh system. The aim is to achieve an 80% removal of influent BOD with a detention time of 1.5 hours. Interpolation is mentioned as a possible method for finding the overflow rate at 80% BOD removal. Stacy is unsure of how to use interpolation and is seeking advice.
  • #1
Stacyg
25
0
Interpolation ?

A treatment plant consists of a primary sedimentation tank followed by a trickling filter that removes 80% of the influent BOD. The flow is then directed to a constructed marsh system that has an influent target BOD concentration of 30 mg/l before being discharged to a small stream.

Using the following table of % BOD removal to overflow rate and calculate the dimensions of a circular sedimentation tank suitable for the imposed flow with a detention time of 1.5 hours. Assume BOD and flow of wastewater entering the plant to be 214.245 mg/l and 848 kL/day.
BOD Removal % 22.0 31.5 33.5 35.5 36.5 37.5
Overflow rate m3/m2/day 100 50 40 30 25 20

How do I find the value of overflow rate when the BOD removal percentage is 80% from this table?
Do I use interpolation ? I'm not sure on how to use interpolation.

Thanks Stacy.
 
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  • #2


hi stac
2 questions
- first, do you have any ideas how to do it?
- 2nd are you sure its 80% BOD removal? this is a very large extrapolation of the table... could it be calculate BOD rate by interpolating from a given overflow rate?
 
  • #3


I know how to answer the question, just not how to find the overflow rate at 80% BOD.
The question is exactly what I have written so i have to find the overflow rate at 80% BOD.
Thanks Stacy
 

Related to Calculate Overflow Rate w/ Interpolation: BOD Removal 80%

1. What is the purpose of calculating overflow rate with interpolation for BOD removal at 80%?

The purpose of this calculation is to determine the maximum flow rate that a wastewater treatment plant can handle while still achieving a BOD removal rate of 80%. This is important for designing and operating treatment plants to ensure efficient and effective treatment of wastewater.

2. How is the overflow rate with interpolation for BOD removal 80% calculated?

The overflow rate with interpolation for BOD removal 80% is calculated by first determining the BOD removal rate at different flow rates, then using interpolation to estimate the flow rate at which the BOD removal rate reaches 80%. This is typically done by collecting data on BOD removal rates at different flow rates and using mathematical equations to estimate the flow rate at 80% BOD removal.

3. Why is 80% BOD removal rate important for wastewater treatment?

80% BOD removal rate is important for wastewater treatment because it indicates that a significant amount of organic matter has been removed from the wastewater, reducing its pollution potential. This is a common standard for wastewater treatment plants to meet in order to comply with environmental regulations and protect public health.

4. What factors can affect the overflow rate with interpolation for BOD removal 80%?

There are several factors that can affect the overflow rate with interpolation for BOD removal 80%, including the design and size of the treatment plant, the efficiency of treatment processes, and the characteristics of the influent wastewater. Changes in any of these factors can impact the BOD removal rate and ultimately the overflow rate with interpolation.

5. How can the overflow rate with interpolation for BOD removal 80% be used in wastewater treatment plant operations?

The overflow rate with interpolation for BOD removal 80% can be used to optimize the operation of a wastewater treatment plant, ensuring that it is operating at the most efficient and effective flow rate. This information can also be used for plant design and expansion to accommodate future increases in wastewater flow while maintaining the desired BOD removal rate.

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