In a horizontal pipe with a sudden expansion from D1 to D2, given V1 = 2.0 m/s, D1 = 0.15 m, and D2 = 0.30 m, what is V2?

• A. 0.25 m/s
• B. 0.50 m/s
• C. 1.00 m/s
• D. 2.00 m/s

Answer: (B)

The main idea is the continuity of flow: in an incompressible fluid, the volume flow rate must stay the same as the fluid moves through the pipe. This means A1 times V1 equals A2 times V2. Since area scales with the square of the diameter, A ∝ D^2, so V2 = V1 × (D1^2 / D2^2).

Compute the ratio: D1^2 = 0.15^2 = 0.0225 and D2^2 = 0.30^2 = 0.09, so D1^2 / D2^2 = 0.25. Then V2 = 2.0 m/s × 0.25 = 0.5 m/s.

As the pipe expands, the velocity decreases to conserve the flow rate; doubling the diameter reduces velocity by a factor of four in this case. The other values would not satisfy the required V2 that keeps the same volumetric flow rate.