In a horizontal pipe, cross-section A1 = 0.01 m^2 with speed v1 = 2 m/s flows into A2 = 0.04 m^2. What is v2 and the pressure difference P1 - P2 if ρ = 1000 kg/m^3?

• A. v2 = 0.50 m/s; P1 - P2 = -1875 Pa
• B. v2 = 0.25 m/s; P1 - P2 = -3750 Pa
• C. v2 = 0.80 m/s; P1 - P2 = -1200 Pa
• D. v2 = 1.0 m/s; P1 - P2 = -500 Pa

Answer: (A)

In incompressible, steady flow through a pipe with changing cross-section, the volumetric flow rate is constant. That means A1 v1 = A2 v2, so v2 = (A1/A2) v1 = (0.01/0.04) × 2 = 0.5 m/s.

With the pipe horizontal and losses neglected, Bernoulli’s equation along a streamline gives P1 + 1/2 ρ v1^2 = P2 + 1/2 ρ v2^2. Rearranging to find the pressure difference, P1 − P2 = 1/2 ρ (v2^2 − v1^2) = 0.5 × 1000 × (0.25 − 4) = −1875 Pa.

So the flow speeds up into the smaller velocity region given the area ratio, and the pressure drops by 1875 Pa from P1 to P2.