What were the pipe dimensions used in the disinfection calculation?

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Multiple Choice

What were the pipe dimensions used in the disinfection calculation?

Explanation:
The key idea is that disinfection calculations for a pipe segment depend on the water volume inside that segment, since the amount of disinfectant and the contact time are based on how much water is being treated. For a cylinder, volume is V = π(D/2)^2 × L, and you can convert that volume to gallons (1 ft^3 ≈ 7.4805 gal). Using 3.00 ft in diameter and 1,040 ft long: radius is 1.5 ft, cross-sectional area is π × 1.5^2 ≈ 7.0686 ft^2. Multiply by length: V ≈ 7.0686 × 1,040 ≈ 7,356 ft^3, which is about 7,356 × 7.4805 ≈ 55,000 gallons. The other options yield notably different volumes: the 2.50 ft by 1,200 ft case gives roughly 44,000 gallons; the 3.50 ft by 900 ft case about 65,000 gallons; the 4.00 ft by 1,000 ft case about 94,000 gallons. The calculation in question is based on a volume near 55,000 gallons, so the 3.00 ft diameter and 1,040 ft length dimensions fit the calculation best.

The key idea is that disinfection calculations for a pipe segment depend on the water volume inside that segment, since the amount of disinfectant and the contact time are based on how much water is being treated. For a cylinder, volume is V = π(D/2)^2 × L, and you can convert that volume to gallons (1 ft^3 ≈ 7.4805 gal).

Using 3.00 ft in diameter and 1,040 ft long: radius is 1.5 ft, cross-sectional area is π × 1.5^2 ≈ 7.0686 ft^2. Multiply by length: V ≈ 7.0686 × 1,040 ≈ 7,356 ft^3, which is about 7,356 × 7.4805 ≈ 55,000 gallons. The other options yield notably different volumes: the 2.50 ft by 1,200 ft case gives roughly 44,000 gallons; the 3.50 ft by 900 ft case about 65,000 gallons; the 4.00 ft by 1,000 ft case about 94,000 gallons. The calculation in question is based on a volume near 55,000 gallons, so the 3.00 ft diameter and 1,040 ft length dimensions fit the calculation best.

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