Problem 1
Question
A water treatment plant processes \(10.5\) MGD. What is the required filter bed area needed to maintain a filtration rate of \(1.75 \mathrm{gpm} / \mathrm{ft}^{2}\) ?
Step-by-Step Solution
Verified Answer
The required filter bed area is 4,166.67 ft².
1Step 1 - Convert flow rate to gallons per minute (gpm)
Given the plant processes 10.5 million gallons per day (MGD). Convert this to gallons per minute (gpm). There are 1,440 minutes in a day, so: $$10.5 \text{ MGD} \times \frac{1,000,000 \text{ gallons}}{1 \text{ MG}} \times \frac{1}{1,440 \text{ minutes}}$$ Simplify the calculation: $$10,500,000 \text{ gallons/day} \times \frac{1}{1,440 \text{ minutes}} = 7,291.67 \text{ gpm}$$
2Step 2 - Using the filtration rate
The filtration rate is given as 1.75 gpm/ft². We need to determine how many square feet of filter bed area are required to handle 7,291.67 gpm. Use the formula: $$\text{Filter Bed Area} = \frac{\text{Flow Rate}}{\text{Filtration Rate}}$$ Substituting the values: $$\text{Filter Bed Area} = \frac{7,291.67 \text{ gpm}}{1.75 \text{ gpm/ft²}}$$
3Step 3 - Calculate the required filter bed area
Complete the calculation from Step 2 to find the filter bed area. $$\text{Filter Bed Area} = \frac{7,291.67}{1.75} = 4,166.67 \text{ ft²}$$
Key Concepts
flow rate conversionfiltration ratefilter bed area calculation
flow rate conversion
Understanding flow rate conversion is crucial when working with water treatment plants. A common unit used is Million Gallons per Day (MGD), which needs to be converted to Gallons Per Minute (GPM) to calculate other parameters like filtration rate and filter bed area. To convert MGD to GPM:
\[10.5 \text{ MGD} \times \frac{1,000,000 \text{ gallons}}{1 \text{ MG}} \times \frac{1}{1,440 \text{ minutes}}\] This equation simplifies to:
\[10,500,000 \text{ gallons/day} \times \frac{1}{1,440 \text{ minutes}} = 7,291.67 \text{ gpm}\] By converting MGD to GPM, you can now proceed to other calculations.
This step is foundational as it standardizes the flow rate unit.
- Remember that 1 MG equals 1,000,000 gallons.
- Also, note that there are 1,440 minutes in a day (24 hours x 60 minutes).
\[10.5 \text{ MGD} \times \frac{1,000,000 \text{ gallons}}{1 \text{ MG}} \times \frac{1}{1,440 \text{ minutes}}\] This equation simplifies to:
\[10,500,000 \text{ gallons/day} \times \frac{1}{1,440 \text{ minutes}} = 7,291.67 \text{ gpm}\] By converting MGD to GPM, you can now proceed to other calculations.
This step is foundational as it standardizes the flow rate unit.
filtration rate
The filtration rate is vital in determining the effectiveness of a filter bed in a water treatment plant. It usually expresses how quickly water can pass through a given area of the filter bed and is often measured in gallons per minute per square foot (gpm/ft²). In our exercise:
\[\text{Filter Bed Area} = \frac{7,291.67 \text{ gpm}}{1.75 \text{ gpm/ft}^2}\] This tells us how much area is needed to maintain the specified filtration rate. Properly understanding the filtration rate helps in ensuring the efficiency and capacity planning of the water treatment facility.
- The filtration rate given is \[1.75 \text{ gpm/ft}^2\]
\[\text{Filter Bed Area} = \frac{7,291.67 \text{ gpm}}{1.75 \text{ gpm/ft}^2}\] This tells us how much area is needed to maintain the specified filtration rate. Properly understanding the filtration rate helps in ensuring the efficiency and capacity planning of the water treatment facility.
filter bed area calculation
The filter bed area calculation is an essential step to ensure that the filter can handle the flow rate specified by the water treatment plant's design. Using the prior values from flow rate conversion and filtration rate, we can now complete the calculation to determine the filter bed area. Using the formula:
\[\text{Filter Bed Area} = \frac{7,291.67 \text{ gpm}}{1.75 \text{ gpm/ft}^2} = 4,166.67 \text{ ft}^2\] This tells us that an area of 4,166.67 square feet is needed for the filter bed to handle a flow rate of 7,291.67 gpm while maintaining a filtration rate of 1.75 gpm/ft². Understanding this calculation ensures the correct design and operational efficiency of the water treatment plant, preventing overloading and ensuring proper water treatment.
- \(\text{Filter Bed Area} = \frac{\text{Flow Rate}}{\text{Filtration Rate}}\)
\[\text{Filter Bed Area} = \frac{7,291.67 \text{ gpm}}{1.75 \text{ gpm/ft}^2} = 4,166.67 \text{ ft}^2\] This tells us that an area of 4,166.67 square feet is needed for the filter bed to handle a flow rate of 7,291.67 gpm while maintaining a filtration rate of 1.75 gpm/ft². Understanding this calculation ensures the correct design and operational efficiency of the water treatment plant, preventing overloading and ensuring proper water treatment.
Other exercises in this chapter
Problem 1
What is the detention time in hours of a \(100 \mathrm{ft}\) by \(20 \mathrm{ft}\) by \(10 \mathrm{ft}\) sedimentation basin with a flow of \(5 \mathrm{MGD}\) ?
View solution Problem 2
A \(15 \mathrm{ft}\) by \(17 \mathrm{ft}\) filter needs to be back washed at a rate of \(17 \mathrm{gpm} / \mathrm{ft}^{2}\) for a minimum of 20 minutes. How ma
View solution Problem 2
What is the detention time in a circular clarifier with a depth of \(30 \mathrm{ft}\) and a \(70 \mathrm{ft}\) diameter if the daily flow is \(4.5\) MG. (Expres
View solution