Using continuity, v2 and p1-p2 across nozzle: A1 = 0.015 m^2, A2 = 0.005 m^2, v1 = 3 m/s; ρ = 1000 kg/m^3, Δp = 0.5 ρ (v2^2 - v1^2). What are v2 and p1 - p2?

Using continuity, v2 and p1-p2 across nozzle: A1 = 0.015 m^2, A2 = 0.005 m^2, v1 = 3 m/s; ρ = 1000 kg/m^3, Δp = 0.5 ρ (v2^2 - v1^2). What are v2 and p1 - p2?

• A. v2 = 9 m/s; p1 - p2 = 36 kPa
• B. v2 = 6 m/s; p1 - p2 = 18 kPa
• C. v2 = 12 m/s; p1 - p2 = 72 kPa
• D. v2 = 9 m/s; p1 - p2 = 72 kPa

Answer: (A)

The main idea is that for steady, incompressible flow, mass must be conserved: A1 v1 = A2 v2. Since the area drops, the velocity must rise. With A1 = 0.015 m^2 and A2 = 0.005 m^2, v2 = v1 × (A1/A2) = 3 × (0.015/0.005) = 9 m/s. The pressure change follows from the energy balance for inviscid flow: p1 − p2 = 0.5 ρ (v2^2 − v1^2). Plugging in ρ = 1000 kg/m^3, v2^2 = 81, v1^2 = 9 gives p1 − p2 = 0.5 × 1000 × (72) = 36,000 Pa = 36 kPa. So v2 is 9 m/s and p1 − p2 is 36 kPa. The velocity increase comes from the smaller nozzle area, and the corresponding pressure drop reflects the conversion of static pressure into kinetic energy.

The main idea is that for steady, incompressible flow, mass must be conserved: A1 v1 = A2 v2. Since the area drops, the velocity must rise. With A1 = 0.015 m^2 and A2 = 0.005 m^2, v2 = v1 × (A1/A2) = 3 × (0.015/0.005) = 9 m/s.

The pressure change follows from the energy balance for inviscid flow: p1 − p2 = 0.5 ρ (v2^2 − v1^2). Plugging in ρ = 1000 kg/m^3, v2^2 = 81, v1^2 = 9 gives p1 − p2 = 0.5 × 1000 × (72) = 36,000 Pa = 36 kPa.

So v2 is 9 m/s and p1 − p2 is 36 kPa. The velocity increase comes from the smaller nozzle area, and the corresponding pressure drop reflects the conversion of static pressure into kinetic energy.