A 10-inch water main 2,640 ft long is disinfected by applying 60 mg/L of chlorine and allowing it to stand 24 hours. Approximately how many pounds of dry hypochlorite 65% chlorine are needed?

• A. 6.5 pounds
• B. 12.1 pounds
• C. 15.0 pounds
• D. 8.3 pounds

Answer: (D)

To disinfect at a concentration of 60 mg/L, you start by figuring how much water is in the pipe, since the dose depends on volume. The main is a cylinder with a 10-inch diameter and 2,640 feet long. Convert the length to inches (2,640 ft × 12 = 31,680 in). The cross-sectional area is π(5 in)^2 ≈ 78.54 in^2. So the water volume in cubic inches is 78.54 × 31,680 ≈ 2,488,141 in^3. Converting to gallons (1 gal = 231 in^3) gives about 10,776 gallons, and to liters (1 gal ≈ 3.785 L) about 40,800 liters.

The total chlorine needed is 60 mg per liter times 40,800 L, which equals 2,448,000 mg or 2.448 kg of available chlorine.

Dry hypochlorite is 65% chlorine by weight, so the mass of product required is 2.448 kg / 0.65 ≈ 3.76 kg. Converting to pounds (1 kg ≈ 2.2046 lb) gives about 8.29 pounds, or roughly 8.3 pounds.

The 24-hour standing time is the contact period and doesn’t change the amount of disinfectant needed in this calculation.