A sudden contraction from D1 = 0.30 m to D2 = 0.15 m is analyzed in a pipe with Q such that V1 = 0.80 m/s. Compute V2.

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Multiple Choice

A sudden contraction from D1 = 0.30 m to D2 = 0.15 m is analyzed in a pipe with Q such that V1 = 0.80 m/s. Compute V2.

Explanation:
Continuity in steady, incompressible flow means the volume flow rate stays the same along the pipe, so velocity changes inversely with cross-sectional area: Q = A1 V1 = A2 V2. With D1 = 0.30 m and D2 = 0.15 m, areas scale with D^2. The ratio A1/A2 equals (D1^2)/(D2^2) = (0.30^2)/(0.15^2) = 0.09/0.0225 = 4. So V2 = V1 × (A1/A2) = 0.80 m/s × 4 = 3.2 m/s. A quick check: A1 ≈ π(0.30)^2/4 ≈ 0.0707 m^2, Q ≈ A1 V1 ≈ 0.0707 × 0.80 ≈ 0.0566 m^3/s. A2 ≈ π(0.15)^2/4 ≈ 0.0177 m^2, V2 ≈ Q/A2 ≈ 0.0566/0.0177 ≈ 3.20 m/s.

Continuity in steady, incompressible flow means the volume flow rate stays the same along the pipe, so velocity changes inversely with cross-sectional area: Q = A1 V1 = A2 V2.

With D1 = 0.30 m and D2 = 0.15 m, areas scale with D^2. The ratio A1/A2 equals (D1^2)/(D2^2) = (0.30^2)/(0.15^2) = 0.09/0.0225 = 4. So V2 = V1 × (A1/A2) = 0.80 m/s × 4 = 3.2 m/s.

A quick check: A1 ≈ π(0.30)^2/4 ≈ 0.0707 m^2, Q ≈ A1 V1 ≈ 0.0707 × 0.80 ≈ 0.0566 m^3/s. A2 ≈ π(0.15)^2/4 ≈ 0.0177 m^2, V2 ≈ Q/A2 ≈ 0.0566/0.0177 ≈ 3.20 m/s.

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