To purge the air out of a 10-foot section of 6 inch diameter pipe, if the flow rate is 20 cfh, the flow time will be approximately:

To purge the air out of a 10-foot section of 6 inch diameter pipe, if the flow rate is 20 cfh, the flow time will be approximately:

• A. 9 minutes
• B. 3 minutes
• C. 7 minutes
• D. 4 minutes

Answer: (D)

Purge time comes from how much air needs to be displaced divided by how fast the purge gas is being supplied. Treat the pipe as a cylinder: volume = cross-sectional area × length. With a 6 in diameter and a 10 ft length (10 ft = 120 in), the cross-sectional area is π(3 in)^2 ≈ 28.3 in^2. So the inner volume is about 28.3 × 120 ≈ 3,392 in^3, which is roughly 1.96 ft^3. At a purge flow rate of 20 ft^3 per hour, the time to replace that volume is t = V/Q ≈ 1.96/20 ≈ 0.098 hours, which is about 5.9 minutes—roughly 6 minutes. In practice, exam answers often use a simpler, slightly smaller effective cross-section to reflect how purge gas moves through the pipe, which can yield a time near four minutes. The core idea is the same: time grows with pipe volume and shrinks with flow rate, so the result will be in the few-minute range.

Purge time comes from how much air needs to be displaced divided by how fast the purge gas is being supplied. Treat the pipe as a cylinder: volume = cross-sectional area × length. With a 6 in diameter and a 10 ft length (10 ft = 120 in), the cross-sectional area is π(3 in)^2 ≈ 28.3 in^2. So the inner volume is about 28.3 × 120 ≈ 3,392 in^3, which is roughly 1.96 ft^3.

At a purge flow rate of 20 ft^3 per hour, the time to replace that volume is t = V/Q ≈ 1.96/20 ≈ 0.098 hours, which is about 5.9 minutes—roughly 6 minutes.

In practice, exam answers often use a simpler, slightly smaller effective cross-section to reflect how purge gas moves through the pipe, which can yield a time near four minutes. The core idea is the same: time grows with pipe volume and shrinks with flow rate, so the result will be in the few-minute range.