If velocity is 2.85 ft/s in an 8.0 inch pipe, what will the velocity be in a 10.0 inch pipe when the same flow is maintained?

• A. 2.85 ft/s
• B. 3.50 ft/s
• C. 0.92 ft/s
• D. 1.83 ft/s

Answer: (D)

When the flow rate remains the same, the product of velocity and cross-sectional area is constant (Q = vA). For a circular pipe, area is proportional to the square of the diameter, so v scales with the inverse of d^2: v2 = v1 × (d1/d2)^2.

Here, d1 = 8 inches and d2 = 10 inches. (8/10)^2 = (0.8)^2 = 0.64. Multiply by the original velocity: 2.85 ft/s × 0.64 = 1.824 ft/s, which is about 1.83 ft/s.

So the larger pipe slows the flow to roughly 1.83 ft/s when the same amount of water passes through.