The pipe length used in the 50.1 mg/L dosage problem was which of the following?

• A. 1,248 ft
• B. 4,088 ft
• C. 2,400 ft
• D. 980 ft

Answer: (A)

The essential idea is tying how much chemical is added to how much water is present in a segment of pipe, so you end up with the target mg/L concentration.

When you inject chemical into a flowing pipeline, the concentration you achieve depends on the amount of water in the portion of pipe affected and the rate at which the chemical is added. To relate this to pipe length, you treat a section of pipe of length L as a water volume V = π(D^2/4) × L, where D is the pipe’s diameter. The target concentration (50.1 mg/L) means the mass of chemical in that water volume should equal concentration × volume. So you set up mass_in_segment = 50.1 mg/L × V and solve for L, using the known injection rate and the system’s flow rate to connect how much chemical is being added per unit time to how much water passes through per unit time.

The correct option is the one that makes this equation balance with the target 50.1 mg/L under the given pipe diameter and flow/injection rates. The other candidate lengths would yield too high or too low a concentration because they correspond to different volumes of water behind the injection point, which changes the mg/L result. In short, that length is the one where the injected mass and the water volume together produce the exact target dosage.