Answer: Δh = 3ρv²/(2(ρ_m - ρ)g).
- A Δh = 3ρv²/(2(ρ_m - ρ)g)
- B Δh = ρv²/2ρ_m g overall
- C Δh = v²/2g in most cases
- D Δh = 4ρv²/ρ_m g under typical conditions
Correct answer: A. Δh = 3ρv²/(2(ρ_m - ρ)g)
Explanation: By continuity: v<sub>t</sub> = 2v. ΔP = ½ρ(4v² - v²) = 3ρv²/2. Manometer: ΔP = (ρ_m - ρ)gΔh. So Δh = 3ρv²/(2(ρ_m-ρ)g).
Since the same volume of fluid must pass every cross-section per second (continuity), the fluid speeds up where the pipe narrows; Bernoulli's equation then says this faster-moving fluid has lower pressure - the principle behind a venturi meter and an aircraft wing's lift.
Concept context
Pressure, buoyancy, Bernoulli equation, viscosity, and surface tension.