Math Problem Statement

In a conventional activated sludge treatment plant, the mixed liquor suspended solids were 2,800 mg/L. If the aeration basin is 140 feet long by 50 feet wide by 12 feet deep, how many pounds of solids are under aeration?

Solution

To calculate the pounds of solids under aeration in the aeration basin, we need to follow these steps:

1. Calculate the volume of the aeration basin:

The dimensions of the aeration basin are given as:

  • Length = 140 feet
  • Width = 50 feet
  • Depth = 12 feet

The volume of the aeration basin can be calculated by multiplying the length, width, and depth:

Volume=140ft×50ft×12ft=84,000cubic feet\text{Volume} = 140 \, \text{ft} \times 50 \, \text{ft} \times 12 \, \text{ft} = 84,000 \, \text{cubic feet}

2. Convert volume from cubic feet to gallons:

Since 1 cubic foot = 7.48 gallons, we can convert the volume:

Volume in gallons=84,000ft3×7.48gallonsft3=628,320gallons\text{Volume in gallons} = 84,000 \, \text{ft}^3 \times 7.48 \, \frac{\text{gallons}}{\text{ft}^3} = 628,320 \, \text{gallons}

3. Calculate the total pounds of solids:

We know the concentration of the mixed liquor suspended solids (MLSS) is 2,800 mg/L. To convert this into pounds per gallon:

  • 1 mg/L = 1 part per million (ppm) = 1 milligram per liter (mg/L)
  • 1 mg = 1×1061 \times 10^{-6} kg
  • 1 liter = 1 kg of water = 1 gallon of water weighs approximately 8.34 pounds

Thus, 2,800 mg/L can be converted to pounds per gallon as:

Solids concentration (pounds per gallon)=2,800mg/L×1lb1,000,000mg×8.34lb/gal\text{Solids concentration (pounds per gallon)} = 2,800 \, \text{mg/L} \times \frac{1 \, \text{lb}}{1,000,000 \, \text{mg}} \times 8.34 \, \text{lb/gal} Solids concentration=2,800×8.34×106=0.0234lb/gal\text{Solids concentration} = 2,800 \times 8.34 \times 10^{-6} = 0.0234 \, \text{lb/gal}

4. Calculate total solids in pounds:

Finally, multiply the volume in gallons by the concentration in pounds per gallon:

Total solids=628,320gallons×0.0234lb/gal=14,711.57lbs\text{Total solids} = 628,320 \, \text{gallons} \times 0.0234 \, \text{lb/gal} = 14,711.57 \, \text{lbs}

Final Answer:

The total pounds of solids under aeration is approximately 14,712 pounds.


Would you like further details on any of these steps or have additional questions?

Related questions:

  1. How would the solids concentration change if the depth of the aeration basin were increased?
  2. How does the volume of the aeration basin affect the efficiency of the treatment process?
  3. What is the significance of mixed liquor suspended solids (MLSS) in wastewater treatment?
  4. How can you adjust the aeration basin size to meet a specific MLSS requirement?
  5. What are common methods for measuring and controlling MLSS in activated sludge systems?

Tip:

Ensure to regularly monitor the MLSS concentration to avoid overloading the system and maintain effective biological treatment.

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Math Problem Analysis

Mathematical Concepts

Volume Calculation
Unit Conversion
Solids Concentration

Formulas

Volume = Length × Width × Depth
1 cubic foot = 7.48 gallons
Solids Concentration (lb/gal) = mg/L × (1 lb / 1,000,000 mg) × 8.34 lb/gal
Total Solids = Volume in gallons × Solids Concentration (lb/gal)

Theorems

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Suitable Grade Level

Grades 11-12